diff --git "a/stack-exchange/math_overflow/shard_4.txt" "b/stack-exchange/math_overflow/shard_4.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_4.txt" +++ /dev/null @@ -1,33686 +0,0 @@ -TITLE: Complex structure on flag manifolds -QUESTION [13 upvotes]: Let $G$ be a compact Lie group and $T$ a maximal torus of $G$. Then the flag manifold $G/T$ is a complex manifold and a symplectic manifold. One way to see the symplectic structure is to view $G/T$ as the co-adjoint orbit of a generic element $F_0 \in Lie(T)^*$. Then the symplectic structure is given by -$$ -\omega_F(X^+_F, Y^+_F) = F([X,Y]), -$$ -where $X^+,Y^+$ are the fundamental vector fields corresponding to $X,Y \in Lie(G)$ and $F \in Orbit(F_0)$. -All the references I've seen get the complex structure on $G/T$ by showing it is isomorphic to $G^{\mathbb C}/B$ where $G^{\mathbb C}$ is the complexification of $G$ and $B$ is a Borel subalgebra. -My question is if there is a way to get the complex structure explicitly in terms of the Lie algebras of $G$ and $T$, in a similar vein to how I defined the symplectic structure. - -REPLY [6 votes]: Let $G_C$ denote the complexification of the compact connected Lie group $G$ and let $H$ be the centralizer of any toral subgroup (not necessarily maximal) in $G$. Then there is a (complex) parabolic subgroup $Q \subset G_C$ with $G$ (as a subgroup of $G_C$) transitive on the complex manifold $Z := G_C/Q$ and $H = G \cap Q$. So $G/H$ is realized as the complex flag manifold $Z$. Here $H_C$ is the reductive component of $Q$, and the choice of $Q$ with given reductive part $H_C$ gives the complex structure: the antiholomorphic tangent space is the Lie algebra of the unipotent radical of $Q$. These choices are parameterized by the quotient $W_G/W_H$ of the Weyl groups of $G$ and $H$. -Your question is about the special case where $H = T$, a maximal torus in $G$. -The flag manifolds have many beautiful properties. They were first described by Jacques Tits in his thesis published by the Belgian Academy of Sciences in 1954.<|endoftext|> -TITLE: Polytopes with few vertices. -QUESTION [10 upvotes]: Suppose I have a convex polytope in $\mathbb{R}^d$ which I know has few vertices (in the case which prompted this question, I seem to have a polytope in $\mathbb{R}^9$ which has sixteen vertices). Is there some constructive way to enumerate the possibilities? If the polytope has $k$ vertices, is there some not-too-horrible upper bound on the number of possibilities? - -REPLY [4 votes]: I've got some information about a lower bound. -We know how many binary codes there are of length 5 on n words - http://oeis.org/A034190. Each corresponds to a subset of the vertices of a cube and thus a convex polytope. -For length 5 with 16 words we get 169112, and a total of 1,226,525 up to symmetry. -I've computed the flag vectors for all such polytopes in dimension 5 - http://arxiv.org/abs/1011.4269. Including polars I get 688,298 flag vectors. (I did this for a good reason.) -From this calculation I can conclude that in dimension 5 with up to 32 vertices there are at least 344,149 such polytopes that are combinatorially distinct.<|endoftext|> -TITLE: Incompressible surfaces in an open subset of R^3 -QUESTION [13 upvotes]: Let $U$ be a connected open subset of $\mathbb R^3$. Furthermore, we have: - -$\mathbb R^3\setminus U$ has exactly two connected components (thus by Alexander duality, $H_2(U;\mathbb Z)=\mathbb Z$). -$U$ may be "very complicated": we make no assumptions on the regularity of $\partial U$. - -I would like to understand the incompressible surfaces in $U$ representing a generator of $H_2(U;\mathbb Z)$. -Question: If I have a collection of incompressible surfaces $F_1,\ldots,F_p$ (all representing the same generator of $H_2(U)$), is there a canonical "lower envelope" incompressible surface $F$? What I want morally is that $F=\partial\left(\bigcap_{i=1}^p\operatorname{int}(F_i)\right)$. Of course, $\partial\left(\bigcap_{i=1}^p\operatorname{int}(F_i)\right)$ depends on how the $F_i$'s are embedded, and it may not be incompressible, so it's not the right choice. Perhaps though as long as there aren't any intersections $F_i\cap F_j$ which are inessential in either $F_i$ or $F_j$, then this guarantees (the isotopy class of) $\partial\left(\bigcap_{i=1}^p\operatorname{int}(F_i)\right)$ is unchanged by isotoping $F_i$. -Motivation: Secretly there is a group action on $U$, and I can easily construct incompressible surfaces $F_1,\ldots,F_p$ which are cyclically permuted by the group, but what I really need is a single incompressible surface which is fixed by the action. Taking some sort of "lower envelope" of $F_1\ldots,F_p$ is my first try at constructing such a surface. - -REPLY [16 votes]: Yes, I think if I understand your question, what you're asking for is true. Take minimal area representatives for your surfaces $F_1,\ldots, F_p$, and take the boundary of the component of the complement of $F_1\cup\cdots \cup F_p$ which contains infinity (I think this is the same as your surface $F$). If this surface is compressible to the "inside", then you may choose a maximal compression body to the inside. The inner boundary will be unique up to isotopy, and will be incompressible inwardly. One can show that each compression may be made disjointly from $F_i$ for all $i$ so this surface may be made disjoint from isotopic copies of the $F_i$ simultaneously. Repeat this compressing outwardly and inwardly, until you obtain an incompressible surface $F'$, and isotopic surfaces $F_i$ which are disjoint. Now it's a fact that if two incompressible surfaces are disjoint, then their minimal area representatives will also be disjoint. So a minimal area representative of $F'$ will be disjoint from the original $F_i$s. So I think this will give your the desired "lower envelope", modulo checking certain details. -Addendum: One would like to see that the surface $F$ is unique up to isotopy, in particular independent of the Riemannian metric. I think this is true, but it requires some proof. I'll sketch a possible argument. There is a partial ordering on isotopy classes of incompressible surfaces in the homology class, where $[E]\leq [F]$ if there are disjoint representatives $E'\sim E, F'\sim F$ such that $E' \cap F'=\emptyset$, and $E'$ is ``inside" $F'$. Now, one needs to show that if $[E]\leq [F], [F]\leq [E]$, the $[E]=[F], i.e. E\sim F$ are isotopic. This follows from standard facts. -The key property here is that this partial order is a lattice: given isotopy classes of incompressible surfaces $[F_1],\ldots, [F_k]$, there is a unique surface $[F]$ such that $[F]\geq [F_i], i=1,\ldots,k$ (and similarly there is a unique infimum). The argument I gave before shows the existence of some $[F]\geq [F_i]$. If we have two such $[F], [F']\geq [F_i]$, then a similar argument shows that there exists $[F'']$ such that $[F]\geq [F''], [F']\geq [F'']$, and $[F'']\geq [F_i]$. Repeating this argument, and using a compactness argument, I think one can show that there exists a unique minimal such surface with is $\geq [F_i]$. This is independent of the Riemannian metric, since a 3-manifold has a unique smooth structure, and the partial ordering is independent of the differentiable structure.<|endoftext|> -TITLE: Are morphisms from affine schemes to arbitrary schemes affine morphisms? -QUESTION [19 upvotes]: To put this question in precise language, let $X$ be an affine scheme, and $Y$ be an arbitrary scheme, and $f : X \rightarrow Y$ a morphism from $X$ to $Y$. Does it follow that $f$ is an affine morphism of schemes? While all cases are interesting, a counterexample that has both $X$ and $Y$ noetherian would be nice. - -REPLY [46 votes]: Though it is not true in general, it is true whenever $Y$ is separated. The map $f$ from -$X$ to $Y$ factors as a composition -$$ X \stackrel{f'}{\rightarrow} X \times Y \stackrel{f''}{\rightarrow} Y$$ -The map $f''$ is a pullback of the projection map from $X$ to a point -(or Spec $\mathbb{Z}$, or whatever base you are working over), and therefore affine. -The map $f'$ is a pullback of the diagonal $Y \rightarrow Y \times Y$, and therefore a closed immersion if $Y$ is separated (and in particular affine). - -REPLY [35 votes]: No, here is an example of a morphism $f:X\to Y$ which is not affine although $X$ is affine. -Take $X=\mathbb A^2_k$, the affine plane over the field $k$ and for $Y$ the notorious plane with origin doubled: $Y=Y_1\cup Y_2$ with $Y_i\simeq \mathbb A^2_k$ open in $Y$ and $Y\setminus Y_i= \lbrace O_i\rbrace$, a closed rational point of $Y$. -We take for $f:X\to Y$ the map sending $X$ isomorphically to $Y_1$ in the obvious way. -Then, although the scheme $X$ is affine, the morphism $f$ is not affine because the inverse image $f^{-1}(Y_2)$of the affine open subscheme $Y_2\subset Y$ is - $X \setminus \lbrace 0 \rbrace=\mathbb A^2_k \setminus \lbrace 0 \rbrace$, the affine plane with origin deleted, well known not to be affine.<|endoftext|> -TITLE: Is there for every variety X an abelian variety A such that their 1st l-adic cohomologies are isomorphic? -QUESTION [5 upvotes]: This question is somewhat inspired by Kevin Buzzard's answer to What is the interpretation of complex multiplication in terms of Langlands? and somewhat from my own curiosity about such topics. -Let $X$ be a variety over $\mathbb{Q}$. This variety induces a pure motive of weight $1$ (it induces pure motives of other weights, but I will focus on the one with weight $1$). I understand that weight $1$ motives come (conjecturally, of course) from weight $2$ newforms. -Okay, now let's trace it back. Let's start with a weight $2$ newform. Then, as James D. Taylor alluded to (and from what I know from http://staff.science.uva.nl/~bmoonen/MTGps.pdf), the corresponding newform must be the pure motive of weight 1 that is induced by an abelian variety (this is special to the weight $2$ newforms). -If so, then it seems that this proves that any pure motive of weight 1 is equal to the pure motive of weight 1 of a motive coming from some abelian variety. -Put back in words that are not conjectural: -Is it true that for every variety $X$ over $\mathbb{Q}$ there is an abelian variety over $\mathbb{Q}$, $A$, such that $H^1_{et}(X,\mathbb{Q}_l) \cong H^1 _{et} (A,\mathbb{Q}_l)$ as $Gal(\mathbb{Q})$-representations? -Or perhaps is the following weaker statement true (if I somehow managed to get something wrong in the above): For every variety $X$ over $\mathbb{Q}$, $L(X,s)$ (coming from the action on the pure motive of weight 1 -- which is well defined even without motives, since one can create it using $l$-adic cohomology) = $\prod_i L(A_i,s)$ where the $A_i$'s are (finitely many) abelian varieties over $\mathbb{Q}$ and the $L$'s are coming from their pure motives of weight $1$. -I would very much like to know, if the above is wrong, where exactly the fallacy was. But if everything above is right, then is this known without assuming crazy conjectures like the standard conjectures or forms of Langlands? - -REPLY [6 votes]: The answer to the OP's question in his para. 5 is ``yes'', I think. Even if $X/\mathbb Q$ has no $\mathbb Q$-point there is still an Albanese torsor $T$ (Lang, Abelian Varieties, p. 45, para. 3), universal with respect to morphisms from $X$ to abelian torsors, and then $H^1(X)$ is isomorphic to $H^1(T)$. Define $A=Aut_T^0$, so that $T$ is a torsor under $A$. The isomorphism $A\times T\to T\times T: (a,t)\mapsto (a(t),t)$ gives $H^1(A)\times H^1(T)\cong H^1(T)\times H^1(T)$. Then divide both sides by the copy of $H^1(T)$ coming from the second projection to get $H^1(A)\cong H^1(T)$, so $A$ is the abelian variety you want: $H^1(X)$ is isomorphic to $H^1(A)$.<|endoftext|> -TITLE: Is there a standard name for the intersection of all maximal linearly independent subsets of a given set in a vector space? -QUESTION [5 upvotes]: The title more or less says it all.... Let $V$ be a vector space (over your favorite field; $V$ not necessarily finite dimensional), and let $S$ be a subset of $V$. A maximal linearly independent subset of $S$ is exactly that: a subset of $S$ that is linearly independent yet not properly contained in any other linearly independent subset of $S$. (Equivalently, it is a basis for the subspace of $V$ that is spanned by $S$.) -Let $T$ be the intersection of all maximal linearly independent subsets of $S$. This $T$ might be as large as $S$, when $S$ itself is linearly independent. Alternatively, $T$ might be empty: if $\{v_1,v_2,v_3\}$ is a basis for $V$, then both examples $S = {}${$v_1,2v_1$} and $S = {}${$v_1,v_2,v_3,v_1+v_2+v_3$} have corresponding $T=\emptyset$. There are plenty of intermediate cases as well: in the same notation, if $S = {}${$v_1,v_2,v_3,v_2+v_3$} then $T={}${$v_1$}. -Does this object $T$ have a standard name? - -REPLY [10 votes]: For a matroid the elements that are contained in every basis are called coloops, dual to the notion of a loop, which is an element not contained in any basis. Since you are interested in linearly independent sets perhaps adopting the language of matroids is not such a bad idea.<|endoftext|> -TITLE: frechet manifolds book -QUESTION [8 upvotes]: hi, does anyone know a good book or some lecture notes on the theory of frechet manifolds ? - -REPLY [11 votes]: You could try - -Hamilton, Richard S. (1982). "The inverse function theorem of Nash and Moser". Bull. Amer. Math. Soc. (N.S.) 7 (1): 65–222. doi:10.1090/S0273-0979-1982-15004-2 - -And there's a paper by Milnor on infinite dimensional Lie groups which could be useful. - -REPLY [8 votes]: There is the book by Kriegl and Michor called "Convenient setting of global analysis" published by the AMS. It goes much beyond Fréchet and really gives a big panorama. However, it is not easy reading and requires really some work. But I guess that is due to the subject...<|endoftext|> -TITLE: An example of a beautiful proof that would be accessible at the high school level? -QUESTION [72 upvotes]: The background of my question comes from an observation that what we teach in schools does not always reflect what we practice. Beauty is part of what drives mathematicians, but we rarely talk about beauty in the teaching of school mathematics. -I'm trying to collect examples of good, accessible proofs that could be used in middle school or high school. Here are two that I have come across thus far: -(1) Pick's Theorem: The area, A, of a lattice polygon, with boundary points B and interior points I is A = I + B/2 - 1. -I'm actually not so interested in verifying the theorem (sometimes given as a middle school task) but in actually proving it. There are a few nice proofs floating around, like one given in "Proofs from the Book" which uses a clever application of Euler's formula. A very different, but also clever proof, which Bjorn Poonen was kind enough to show to me, uses a double counting of angle measures, around each vertex and also around the boundary. Both of these proofs involve math that doesn't go much beyond the high school level, and they feel like real mathematics. -(2) Menelaus Theorem: If a line meets the sides BC, CA, and AB of a triangle in the points D, E, and F then (AE/EC) (CD/DB) (BF/FA) = 1. (converse also true) See: http://www.cut-the-knot.org/Generalization/Menelaus.shtml, also for the related Ceva's Theorem. -Again, I'm not interested in the proof for verification purposes, but for a beautiful, enlightening proof. I came across such a proof by Grunbaum and Shepard in Mathematics Magazine. They use what they call the Area Principle, which compares the areas of triangles that share the same base (I would like to insert a figure here, but I do not know how. -- given triangles ABC and DBC and the point P that lies at the intersection of AD and BC, AP/PD = Area (ABC)/Area(DBC).) This principle is great-- with it, you can knock out Menelaus, Ceva's, and a similar theorem involving pentagons. And it is not hard-- I think that an average high school student could follow it; and a clever student might be able to discover this principle themselves. -Anyway, I'd be grateful for any more examples like these. I'd also be interested in people's judgements about what makes these proofs beautiful (if indeed they are-- is there a difference between a beautiful proof and a clever one?) but I don't know if that kind of discussion is appropriate for this forum. -Edit: I just want to be clear that in my question I'm really asking about proofs you'd consider to be beautiful, not just ones that are neat or accessible at the high school level. (not that the distinction is always so easy to make...) - -REPLY [3 votes]: Fermat's Little Theorem: If $p$ is prime and does not divide $a$, then $a^{p-1} \equiv 1 (\mbox{mod } p)$. -Proof: List the multiples of $a$ up to $a(p-1)$: -$$ a, a2, a3, \dots , a(p-1).$$ -For any $r$ and $s$ with, $ra \equiv sa (\mbox{mod } p)$, we have $r \equiv s (\mbox{mod } p)$, so that the list above contains $p-1$ many distinct numbers. -Thus, the list above is some ordering of the list $1, 2, 3, \dots p-1$ modulo $p$. This gives us -$$ a\cdot a2 \cdot a3 \cdot \cdots a(p-1) \equiv (p-1)! (\mbox{mod } p) $$ -Finally we see -$$ a^{p-1} \equiv 1 (\mbox{mod } p). $$<|endoftext|> -TITLE: How to Interpret the Euler Characteristic of Complex Algebraic Varieties -QUESTION [6 upvotes]: Let $X$ be a projective (or affine) variety over $\mathbb{C}$ defined by some homogenous ideal $I = (f_1,\ldots,f_n)$. How can we interpret the Euler characteristic of $X$ other than as just an invariant to distinguish non isomorphic objects? What I mean is what could it tell us if anything about either $X$ or the polynomials defining $I$ if $\chi(X) > 0$? How about if $\chi(X) = 0$ or $\chi(X) < 0$? How about in the case of $X$ being a hypersurface cut out by the single (homogeneous) polynomial $f$. What can we deduce about $f$ if $\chi(X) > 0, \chi(X) = 0,$ or $\chi(X) < 0$. I've been trying to relate conditions on the Euler characteristic to anything explicit about the polynomials defining $X$ but haven't had any success so anything that could be said would be interesting and helpful. - -REPLY [2 votes]: Hi Dori! I think that you will find Paolo Aluffi's paper "Computing characteristic classes of projective schemes" useful, if not for the exact formulas, at least for the algorithm. -Among other things it proves the version of the formula in J.C. Ottem's answer for singular hypersurfaces. If $f\in \mathbb C[x_0,\dots,x_n]$ is a non-constant homogeneous polynomial, let $X=\mathcal Z(f)$ and $\Sigma = \mathcal Z_{\mathbb P^n}(\partial_0 f,\dots,\partial_n f)$. If we denote by $g_0,g_1,\dots,g_{n-1}$ the degrees of the gradient morphism -$$\mathbb P^n\backslash \Sigma\to \mathbb P^n, x=(x_0,\dots,x_n)\to (\partial_0f(x),\dots,\partial_nf(x))$$ -then we have -$$\chi(X)=n+\sum g_i (-1)^{n+1-i}.$$ -As far as restricting to graph hypersurfaces, I would say look at the recent papers of Aluffi and Marcolli but you know more than me about these matters so... :) -Added: A toy application of this is in the case of graphs $\Gamma_n$ which have two vertices and $n$ parallel edges, so called banana graphs. the formula above gives -$$\chi(X_{\Gamma_n})=n+(-1)^n$$ -so the Euler characteristic determines the graph. The motivic generalization is done in "Feynman motives of banana graphs". The motivic viewpoint also shows that one shouldn't expect a nice relation to hold for all graphs, since graph hypersurfaces are generators for the Grothendieck ring (details and actual statement are here).<|endoftext|> -TITLE: What does actually being a CW-complex provide in algebraic topology? -QUESTION [50 upvotes]: From time to time, I pretend to be an algebraic topologist. But I'm not really hard-core and some of the deeper mysteries of the subject are still ... mysterious. One that came up recently is the exact role of CW-complexes. I'm very happy with the mantra "CW-complexes Good, really horrible pathological spaces Bad." but there's a range in the middle there where I'm not sure if the classification is "Good" or just "Pretty Good". These are the spaces with the homotopy type of a CW-complex. -In the algebraic topology that I tend to do then I treat CW-complexes in the same way that I treat Riemannian metrics when doing differential topology. I know that there's always a CW-complex close to hand if I really need it, but what I'm actually interested in doesn't seem to depend on the space actually being a CW-complex. But, as I said, I'm only a part-time algebraic topologist and so there may be whole swathes of this subject that I'm completely unaware of where actually having a CW-complex is of extreme importance. -Thus, my question: -In algebraic topology, if I have a space that actually is a CW-complex, what can I do with it that I couldn't do with a space that merely had the homotopy type of a CW-complex? - -REPLY [5 votes]: There is a fundamental reason why CW-complexes $X$ say are useful, namely for the purpose of constructing continuous functions $f: X \to Y$ by induction on the skeletal filtration $X^n, n \geqslant 0$, and for going from $X^{n-1} $ to $X^{n}$ knowing that you have to construct the extension only on each $n$-cell $e^n$. The topology on $X$ is arranged precisely for this purpose. So a complicated space is put together from simple ones. -It is this possibility of constructing continuous functions, and also homotopies, which allows for the proof of the Theorem of Whitehead mentioned above. -It interesting to trace the development of this concept from his earlier papers for example -Whitehead, J H.C. On incidence matrices, nuclei and homotopy types. Ann. of Math. (2) 42 (1941) 1197--1239, -which introduced the term "membrane complexes, which are so to speak more elastic than simplicial complexes". The idea was to avoid the subdivision into simplices by amalgamating them. This paper, and others, were rewritten after the war, in terms of the CW-complexes we know and love. -A crucial point was to be able to construct not only continuous functions but also continuous homotopies of these. For these a result on the product of CW-complexes was needed, which I was told took him a year to prove. He also early on formulated basic results on adjunction spaces, which are now clear in terms of pushouts. -I feel the emphasis on filtrations is crucial, and allows for other methods in algebraic topology. -22 Nov 2013: Grothendieck in his "Esquisse d'un programme" Section 5 (English version available here, see p.258) discussed what he feels is the inadequate nature of topological spaces for certain geometric considerations. -When foundations are under consideration, one should take a "no holds barred" attitude. This agrees with the advice of Einstein: -"What is essential and what is based only on the accidents of development?... Concepts which have proved useful for ordering things easily assume so great an authority over us, that we forget their terrestrial origin and accept them as unalterable facts. They then become labelled as "conceptual necessities", "a priori situations", etc. The road of scientific progress is frequently blocked for long periods by such errors. It is therefore not just an idle game to exercise our ability to analyse familiar concepts, and to demonstrate the conditions on which their justification and usefulness depend, and the way in which these developed, little by little..."<|endoftext|> -TITLE: Subadditive Kingmans theorem for lattices. -QUESTION [5 upvotes]: I am looking for a multidimensional version of Kingman's subadditive theorem. I found this but it is not exactely what I need. -I would rather have something like that: -Let us consider $\mathbb{Z}^2_+$ and a family $(X)$ of random variables indexed by pairs of points in $\mathbb{Z}^2$ i.e. $X_{z_1, z_2}$ is a random variable associated with subgrid of $\mathbb{Z}^2$ "starting from" $z_1$ and ending on $z_2$. Assume that for any rectangular subgrid $\Lambda \subset \mathbb{Z}^2_+$ and point $x\in \Lambda $ we have -$X_{\Lambda}\leq X_{\Lambda_1} + X_{\Lambda_2}+X_{\Lambda_3}+X_{\Lambda_4},$ -where $\Lambda_1,\Lambda_2,\Lambda_3,\Lambda_4$ is $\Lambda$ split in point $x$ into four subgrids. I suppose that this together with some "usual" ergodic theorem conditions should imply that -$X_{(0,0),(n,n)}/n^2$ converges in $L^1$ and a.s. -May be some one you could give me some references. -Being in the topic of subadditivity. I found a multidimensional version of Fekete lemma. It is surprising for me that it was not done before 2007. But may be it was. Again I will be happy to know any. - -REPLY [2 votes]: Have a look at the paper by Nguyen, Xuan-Xanh Ergodic theorems for subadditive spatial processes, Z. Wahrsch. Verw. Gebiete 48 (1979), no. 2, 159–176, MR0534842 (82c:60056). The setup there is somewhat different, but it seems that the main result there should contain your claim.<|endoftext|> -TITLE: exact sequence of logarithmic differential sheaves associated to an effective Cartier divisor on a smooth variety -QUESTION [5 upvotes]: Let $X$ be a smooth variety over $\mathbb{C}$. -Let $D \subset X$ be an effective Cartier divisor. - -Question 1. What is the definition of the logarithmic differential sheaf $\Omega^1_X (\log D)$ ? - -I saw the definition in the book by Esnault-Viehweg (meromorphic form $\alpha$ such that $\alpha, d \alpha$ has pole of order 1 along $D$ ?). -If there is another definition, I would be happy. -And I have another question. If $D$ is normal crossing, there is an sequence $$0\rightarrow \Omega^1_X \rightarrow \Omega^1_X(\log D) \rightarrow \nu_* \mathcal{O}_{\tilde{D}} \rightarrow 0, $$ -where $\nu: \tilde{D} \rightarrow D$ is the normalization. - -Question 2. If $D$ is not normal crossing, is there a similar exact sequence? Is there a suitable way to define a residue map $\Omega^1_X(\log D) \rightarrow \nu_* \mathcal{O}_D$? - -REPLY [7 votes]: Here is a way to define this sheaf algebraically over any field of characteristic zero. Let $\mathrm{T}_{X}$ denote the tangent sheaf on $X$. Choose a local equation $\phi_U$ for $D$ on $U$. Consider the following submodule: -$\mathrm{T}_X(-\log\phi_U)=\ ${$\partial\in\mathrm{Der}(\mathcal{O}_X(U))\mid \partial\phi_U\in (\phi_U)$}$\ \subset \mathrm{T}_X(U)$. -It is an easy exercise to check that this does not depend on the choice of equation and glues into a sheaf $\mathrm{T}_X(-\log D)$. Now take the dual $\Omega^1_X(\log D)=\mathrm{T}_X(-\log D)^*$. -The good thing about having normal crossings is that in this case $\Omega^1_X(\log D)$ becomes a locally free sheaf. -For your second question, look at Definition 2.5 in this paper by Dolgachev, arXiv:math/0508044<|endoftext|> -TITLE: What is the modern understanding of the order of a mock theta function? -QUESTION [5 upvotes]: Ramanujan introduced mock theta functions and described them by an "order" which he did not define. As a result of the work of Zwegers and others we now have a better understanding of mock theta functions. They appear as the holomorphic projection of weight 1/2 harmonic Maass forms and in the theta expansions of meromorphic Jacobi forms. Given this modern understanding one wonders if there is a natural definition of the "order" which agrees with Ramanujan's. On the Wikipedia page on mock modular forms one finds the statement "Ramanujan's notion of order later turned out to correspond to the conductor of the Nebentypus character of the weight 1/2 harmonic Maass forms which admit Ramanujan's mock theta functions as their holomorphic projections." I can check that this is true in a few specific cases (e.g. the order 3 mock theta functions studied by Bringmann and Ono) but have not been able to find this statement in the literature, hence my questions. First, does the definition of the order in the Wikipedia article agree with the orders (2,3,5,6,7,8,10) of the mock theta functions given later in the same article? Second, is there a reference to the literature for this definition? - -REPLY [5 votes]: I'm afraid there isn't very much in the literature about the order. If you really want to know, you'll have to check it yourself... -I think Ramanujan actually only defined the order for the functions of order 3, 5 and 7 (in his last letter to Hardy). Mock theta functions of different order show up in his Lost Notebook, but if I´m not mistaken, he doesn´t label those. By various other people they were later called of order 2,6,8,10,etc. It could very well be that those were not labeled according to Ramanujan's "definition" and/or the modern definition.<|endoftext|> -TITLE: free group of finite rank can contain free groups of infinite rank as a subgroup -QUESTION [7 upvotes]: how a free group of rank greater than or equal to 3 contains every free group of countable ranks as a sugroup? - -REPLY [3 votes]: There is a very nice theorem which states: -Thm: If $F$ is a free group on $n$ generators and $H$ is a normal subgroup of index $j$ then if both $n$ and $j$ are finite $H$ is a free group on $j(n-1)+1$ generators. If $n$ is infinite and $j$ is finite then $H$ is a free group on infinitely many generators. Finally, if $j$ is infinite, then $H$ may be finitely or infinitely generated; however, if $H$ contains a normal subgroup $N$ of $F$, $N\neq 1$, then $H$ is a free group on infinitely many generators. -This is theorem 2.10 in Magnus, Karrass and Solitar "Combinatorial Group Theory" (and has a name...Schreier's formula?) -Once you have this theorem you can prove, for example, that the commutator subgroup is a free group on infinitely many generators (it contains a normal subgroup and is itself characteristic so you can apply the last line of the theorem). But doing it the way Benjamin Steinberg did it is much neater. I just thought it would be useful to mention this formula!<|endoftext|> -TITLE: Intersection of field extensions of torsion points of non-isogenous elliptic curves -QUESTION [5 upvotes]: Let $E$ and $E'$ be non-isogenous elliptic curves over a field $k$ (characteristic 0) such that $Gal(k(E[p^{\infty}])/k)=Gal(k(E'[p^{\infty}])/k) = SL_2(\mathbb{Z}_p)$ with $p \geq 5$ (where $E[p^{\infty}]$ is the set of $p^n$ torsion points of $E$ for all $n$). Then is it true that $k(E[p^{\infty}])\cap k(E'[p^{\infty}]) = k$, or can someone provide a counterexample? - -REPLY [6 votes]: By the way, I think that under your hypotheses, your question is really about group theory, not about algebraic geometry. Namely: the action of Galois on E[p^infty] x E'[p^infty] gives you a homomorphism -G_K -> SL_2(Z_p) x SL_2(Z_p). -Call the image H. By your hypothesis, H projects surjectively onto both copies of SL_2(Z_p). You also know that H is not contained in any conjugate of the diagonal (if it were, E[p^infty] and E'[p^infty] would be isomorphic Galois representations and I'm presuming you're in a situation where Faltings rules that out -- you'd better be, if you want an affirmative answer to your question.) -Now what you have to prove is that a subgroup of SL_2(Z_p) x SL_2(Z_p) which projects surjectively onto each direct summand and which is not conjugate to a subgroup of the diagonal must be finite-index in SL_2(Z_p) x SL_2(Z_p). This is true for SL_2(F_p) by Hall's lemma and I think you can induct from there (but didn't think about it carefully.)<|endoftext|> -TITLE: Generalizing the Catalan number (enumerative combinatorics) -QUESTION [5 upvotes]: Consider any sequence consisting of n A's and n B's so that in any of its initial partial segments, the number of B's never exceed the number of A's. It is well known that the number of such sequences is the Catalan number $\frac{1}{n+1}\binom{2n}{n}$. -Now consider sequences consisting of n A's, n B's, and n C's, so that in any initial segments, the number of B's never exceed the number of A's AND the number of C's never exceed the number of B's. Is the number of such sequences known? - -REPLY [16 votes]: Yes. This is the generalized ballot problem. -The directed walks on $\mathbb Z_{+}^k$ starting from the origin and ending at $(\lambda_1,\dots,\lambda_k)$, that satisfy $0\le x_1\le \cdots \le x_k$ at every point are in bijection with the number of standard Young tableaux of shape $\lambda$, and this can be found by the Hook-length formula. -In your case we need the number of standard Young tableaux of shape $(n,n,n)$ which is -$\frac{2(3n)!}{n!(n+1)!(n+2)!}$. -For even more generality, see the paper "Random walk in a Weyl chamber" by I. M. Gessel and D. Zeilberger.<|endoftext|> -TITLE: Smooth and topological bordism and homology -QUESTION [5 upvotes]: I have some questions about homology, manifolds and bordism. First of all, if X is a smooth manifold, in general an integral homology class in X cannot be represented by a smooth embedded submanifold, as Thom proved. -1) If X is a topological manifold, does the same result hold? Are there in general singular homology classes, which are not representable by a topological embedded submanifold? -Then, let us consider the oriented bordism groups of a topological space X. Its elements are represented by couples (M, f), for M a smooth oriented manifold and f: M -> X continuous. There is a natural map to singular homology, defined as $[(M, f)] \rightarrow f_{*}([M])$, which, in general, is not surjective. -2) If we define the "topological bordism", requiring that M is a topological manifold (not necessarily smooth), is the corresponding map to singular homology surjective? -3) If X is a smooth manifold, and we define bordism requiring that f is a smooth map (not only continuous), do we obtain the same bordism groups (up to isomorphism)? - -REPLY [2 votes]: (3) Yes, this is exactly the differential bordism groups $D_k(Y)$ of Conner (see §1.9 in P.E. Conner, Differentiable Periodic Maps, second edition, Springer Lecture Notes in Mathematics 738, Springer-Verlag, Berlin, 1979.) -He proves in Theorem I.9.1 that the natural projection $D_k(Y) → MSO_k(Y)$ is an isomorphism. Hence the so defined bordism groups are isomorphic to the usual definition of bordism groups $MSO_k(Y)$. (As in Conner, or Atiyah (M.F. Atiyah, Bordism and cobordism, Proc. Camb. Phil. Soc. 57 (1961), 200–208.))<|endoftext|> -TITLE: Deforming ample line bundles vs cohomology group -QUESTION [5 upvotes]: Let X be a smooth projective variety over the complex numbers, of dimension at least two. $D$ is an ample divisor on X. Then we know for $m>>0$, $H^i(mD)=0$. Now suppose $E$ is another divisor that is algebraiclly equivalent to $mD$, i.e. the line bundle $\mathcal{L}(E-mD)\in Pic^0(X)$ lies in the identity component of of Picard variety of X. Then is it true that even if they are not linearly equivalent, we still have $dim H^0(\mathcal{L}(E))=dim H^0(\mathcal{L}(mD))$, for $m>>0$ ? -Since the Euler character is a topological invariant, we know $\chi(\mathcal{L}(E))=\chi(\mathcal{L}(mD))$. Therefore if we know $H^i(\mathcal{L}(E))=0$ for $i>0$, we are done. However it is not obvious to me if that is true or not. -Some of my mumbling which may or may not be related: -For a general line bundle $\mathcal{L}\in Pic^0(X)$, if $\mathcal{L}\neq \mathcal{O}_X$, then $H^0(\mathcal{L})=0$ since for effective divisor in a projective variety we have a notion of degree, see Hartchorne chap II Exer 6.2. But I don't know if $H^i(\mathcal{L})=0$ or not. -In a series of paper by Green and Lazarsfeld, they looked at the case where X is compact Kahler, not necessarily projective, the behavior where $\mathcal{L}\in Pic^0(X)$ but $H^i(\mathcal{L})\neq 0$. see paper. But I don't know how to use that to construct an example where $E\sim_{alg}mD$ but $H^i(\mathcal{L}(E))\neq 0$ for $i>0$, or $H^0(\mathcal{L}(E))\neq H^0(\mathcal{L}(mD))$. - -REPLY [5 votes]: What you want follows easily from the Kodaira vanishing theorem: -If $m$ is sufficiently large then $mD - K$, where $K$ is the canonical divisor, is ample (this is true for $K$ replaced by any divisor). Since ampleness is preserved by algebraic equivalence, for example by Kleiman's criterion, it follows that $E - K$ is also ample if $E$ is algebraically equivalent to $mD$. Kodaira's vanishing theorem then implies that $H^i(\mathcal{L}(E)) = 0$ for $i>0$.<|endoftext|> -TITLE: Sums of four fourth powers -QUESTION [7 upvotes]: Apologies in advance if this is a naive question. -If I understand correctly, it's well-known that the Fermat quartic surface -$X = \lbrace w^4 +x^4+y^4+z^4 =0 \rbrace \subset \mathbf{P}^3$ -has points over every finite field $\mathbf{F}_q$ except $\mathbf{F}_5$. If I still understand correctly, one way to check this is to use the Weil bound relating the number of $\mathbf{F}_q$–points on $X$ to the number on $\mathbf{P}^2$; this shows that $X$ must have points over $\mathbf{F}_q$ for all but a small number of (small) values of $q$, and the remaining cases can be checked by hand. -The point of my question is that this proof uses the Weil bound, and I am curious if there is an elementary proof. In other words, I am looking for the most elementary proof of the following statement: - - Let $p$ be a prime number different from 5. Show that there exists an $n$ such that $np$ is a sum $a^4+b^4+c^4+d^4$ of four fourth powers (where $a$, $b$, $c$, $d$ are not all divisible by $p$). - -Of course, this has something to do with Waring's problem, but a quick search on that topic didn't turn up anything on this kind of variant. -Edit: Thanks to all for the very nice answers. It seems unfair to select just one, so I'll hold on to my checkmark for the time being. - -REPLY [12 votes]: This special case of the problem of estimating the number of ${\bf F}_p$-points on a variety is much easier than the Weil bound: the standard estimates on Gauss sums suffice. I hope that this is elementary enough. I recite at the end of this answer a derivation of a bound that generalizes to arbitrary "diagonal hypersurfaces" and reduces the present problem to an easy finite computation (through $p=73$). -With more precise information about quartic Gauss sums (somewhat less elementary, but still known to Gauss — it's in an appendix to Disquisitiones Arithmeticæ), one can obtain an exact formula for the number of rational points. Assume that $p \equiv 1 \bmod 4$, else the count is $(p+1)^2$ and is entirely elementary (we may as well count on the quadric $w^2+x^2+y^2+z^2=0$ because in this case every number mod $p$ has as many square roots as it has fourth roots). Write $p = a^2 + b^2$ with $a$ odd. Then the number of rational points is $p^2+mp+1+4a^2$, where $m=18$ or $-6$ according as $p$ is congruent to $1$ or $5 \bmod 8$. This shows that $p=5$ is the only case where there are no rational points. The formula also fits in with the fact that the diagonal Fermat quartic is a K3 surface of maximal Picard number (= Néron-Severi rank), so I'll add the "k3-surfaces" tag. -To connect the enumeration problem with Gauss sums we argue as follows. Again it is enough to consider the case $p \equiv 1 \bmod 4$, since if $p \equiv 3 \bmod 4$ the problem is equivalent to the easier task of enumerating rational points on $w^2+x^2+y^2+z^2=0$, for which the same technique works (using only quadratic Gauss sums) but is overkill. -For $n \in \bf Z$ write $e(n) = \exp(2\pi i n /p)$. Then the number of solutions of $w^4+x^4+y^4+z^4 = 0$ in ${\bf F}_p$ is $p^{-1} \sum_{k=0}^{p-1} S_k^4$ where $S_k = \sum_{t=0}^{p-1} e(kt^4)$. [Proof: expand $\sum_{k=0}^{p-1} S_k^4$ as -$$ -\sum_{k=0}^{p-1} \left[ \mathop{\sum\sum\sum\sum}_{w,x,y,z\phantom0\bmod \phantom0p} - \phantom. e(k(w^4+x^4+y^4+z^4))\right] -$$ -and switch the order of summation: the sum over $k$ is $p$ if $w^4+x^4+y^4+z^4 = 0$, -and zero otherwise. NB we're counting affine solutions, including $(w,x,y,z)=(0,0,0,0)$, not projective solutions.] -Now $S_0 = p$. I claim that $|S_k| \leq 3\sqrt{p}$ for $k \neq 0$. It will follow that the number of solutions is within $81(p-1)p^2$ of $p^4$. In particular the number of solutions exceeds $1$ once $p \geq 81$, so the number of projective ${\bf F}_p$-points is positive. Checking the remaining cases $p=13,17,29,37,41,53,61,73$ is routine. -To prove that $|S_k| \leq 3 \sqrt{p}$ for $k \neq 0$, write $S_k = \sum_{n=0}^{p-1} c_n e(kn)$ where $c_n$ is the number of solutions of $t^4 \equiv n \bmod p$. Now $c_n = 1 + \chi(n) + \chi^2(n) + \chi^3(n)$ where $\chi$ is a Dirichlet character mod $p$ of order $4$. Thus -$$ -S_k(n) = \sum_{j=0}^3 \left[ \sum_{n=0}^{p-1} \phantom. \chi^j(n) \phantom. e(kn) \right]. -$$ -For $j=0$, the inner sum vanishes; for $j = 1,2,3$ it's a Gauss sum, which has absolute value $\sqrt p$. Hence ${}$$|S_k| \leq \sum_{j=1}^3 \sqrt p = 3 \sqrt p$ and we're done. -I'll leave the proof of the formula $p^2+mp+1+4a^2$ as an "exercise" because this answer is already rather long...<|endoftext|> -TITLE: So, did Poincaré prove PBW or not? -QUESTION [16 upvotes]: This seems to be a question whose answer depends on whom you ask. Maybe we can come up with a final answer? -It is known that Poincaré, at least, invented something that can be called Poincaré-Birkhoff-Witt theorem (PBW theorem) in 1900. Okay, it was a version of the PBW theorem that required a basis of the Lie algebra, and it was formulated over $\mathbb R$ or $\mathbb C$ only, but this is not unexpected of a 1900 discovery, and generalizing the statement is pretty straightforward from a modern perspective (generalizing the proofs, not so much). -What is unclear is whether Poincaré gave a correct proof. Poincaré's proof appears in two places: his paper -Henri Poincaré, Sur les groupes continus, Transactions of the Cambridge Philosophical Society, 18 (1900), pp. 220–255 = Œuvres de Henri Poincaré, vol., III, Paris: Gauthier-Villars (1934), pp. 173–212, -and the modern paper -T. Ton-That, T.-D. Tran, Poincaré's proof of the so-called Birkhoff-Witt theorem, Rev. Histoire Math., 5 (1999), pp. 249-284, also arXiv:math/9908139. -I have no access to the former source, so all my knowledge of the proof comes from the latter. -Ton-That and Tran claim that Poincaré's proof was long misunderstood as wrong, while in truth it is a correct, if somewhat incomplete proof. The incompleteness manifests itself in the fact that a property of what Poincaré called "symmetric polynomials" (and what we nowadays call "symmetric tensors") was used but not proven. However, in my opinion this is not a flaw: This property (which appears as Theorem 3.3 in the paper by Ton-That and Tran and is proven there in an overly complicated, yet nice way) is simply the fact that the $k$-th symmetric power of a vector space $V$ over a field of characteristic $0$ is generated by $k$-th (symmetric) powers of elements of $V$. This fact is known nowadays and was known in 1900 (I think it lies at the heart of umbral calculus). -What I am not sure about is the actual proof of PBW. Since I don't have the original Poincaré source (nor, probably, the understanding of French required to read it), I am again drawing conclusions from the Ton-That and Tran paper. My troubles lie within this paragraph on pages 277-278: -"The first four chains are of the form -$U_1 = XH_1,\ U'_1 = H'_1Z,\ U_2 = YH_2,\ U'_2 = H'_2T$, -where each chain $H_1$, $H'_1$, $H_2$, $H'_2$ is a closed chain of degree $p - 1$; therefore by induction, each is the head of an identically zero regular sum. It follows that $U_1$, $U'_1$, $U_2$, $U'_2$ are identically zero, and therefore each of them can be considered as the head of an identically zero regular sum of degree $p$." -Question: Why does "It follow[] that $U_1$, $U'_1$, $U_2$, $U'_2$ are identically zero"? This seems to be equivalent to $H_1 = H'_1 = H_2 = H'_2 = 0$, which I don't believe (the head of an identically zero regular sum isn't necessarily zero). The authors, though, are only using the weaker assertion that each of $U_1, U'_1, U_2, U'_2$ is the head of an -identically zero regular sum of degree $p$ - but this isn't obvious to me either. -What am I missing? Is this a mistake in Poincaré 1900? Or have the authors of 1 misrepresented Poincaré's argument? Has anybody else tried to decipher Poincaré's proof? -PS. I have asked more or less asked this question some months ago, but it was hidden in another question and did not receive any answer. Mailing the authors did not help either. So my last hope is a public discussion. - -REPLY [16 votes]: I don't know if Poincaré proved PBW in 1900, but Alfredo Capelli -did it for $\mathfrak{g}\mathfrak{l}_n$ ten years before. -Here is the link to Capelli's paper (in French).<|endoftext|> -TITLE: Foundations: Existence of uncountable ordinals. -QUESTION [8 upvotes]: This isn't really a research question, but at least it's research-level mathematics. I'm talking with some other people about the first uncountable ordinal, and I want some facts to inform this discussion. Specifically, what useful or interesting foundations of mathematics do or don't allow one to prove the existence of an uncountable ordinal? -If you don't have a better interpretation, then for "useful", you can probably take "capable of encoding most if not all rigorous applied mathematics"; for "interesting", you can probably take "popular for study by researchers in foundations". For "existence of an uncountable ordinal", you could take "existence of a well-ordered uncountable set", "existence of a set whose elements are precisely the countable ordinals", etc. -Hopefully there is a body of known results or obvious corollaries of such, since it could be a matter of some work to apply this question to foundational system X, and I don't expect anybody to do that. - -REPLY [18 votes]: For the existence of an uncountable set with a definable well-ordering, it suffices to have $\mathcal P(\mathcal P(\mathbb N))$, where $\mathcal P$ means the power set. Any countable order-type is represented by a well-ordering of $\mathbb N$, and can therefore be coded by a subset of $\mathbb N$. Call two such codes equivalent if the well-orderings they code are isomorphic. Then the equivalence classes are elements of $\mathcal P(\mathcal P(\mathbb N))$, they correspond naturally to the countable ordinals, and so they constitute an uncountable well-ordered set. (If you want the well-ordering itself to be a set, rather than given by a definition, and if you want it to be literally a set of ordered pairs, with ordered pairs defined in the usual Kuratowski fashion, then you'll need some more assumptions to ensure the existence of the relevant ordered pairs, etc. But if you're willing to settle for some other coding of ordered pairs, then $\mathcal P(\mathcal P(\mathbb N))$ seems to suffice.) -In particular, the existence of an uncountable well-ordered set is well within the power of Zermelo set theory (like ZF but without the axiom of replacement). If, on the other hand, you want the set of countable ordinals and if you use von Neumann's (nowadays standard) representation of ordinals by sets, then Zermelo set theory is not enough. One natural model of Zermelo set theory is the collection of sets of rank smaller than $\omega+\omega$; this contains plenty of uncountable well-ordered sets, but its ordinals are just those below $\omega+\omega$. (The moral of this story is that, in Zermelo set theory and related systems, ordinals should not be defined using the von Neumann representation but rather as isomorphism classes of well-orderings.)<|endoftext|> -TITLE: Is there an "undecided" assertion of which a proof that it's not undecidable is known? -QUESTION [5 upvotes]: Just a curiosity: - -Is there an assertion of which a proof (formalizable, say, in ZFC) is not known but a proof that it's not undecidable (in ZFC) is known? - -Edit: after the comments, I think the actual question was - -Is there an ("interesting") assertion of which neither a proof (formalizable, say, in ZFC) of it or its negation is known but a proof that it's not undecidable (in ZFC) is known? - -REPLY [5 votes]: There's tons of assertions like that in finite combinatorics. For example the Ramsey numbers R(5,5) and R(6,6) can be "straightforwardly" (i.e. given impractically large computing resources) found by direct enumeration. It's known that $43\le R(5,5) \le 49$ and $102\le R(6,6)\le 165$. But Wikipedia's article on Ramsey theory quotes Joel Spencer: - -Erdős asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of R(5, 5) or they will destroy our planet. In that case, he claims, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they ask for R(6, 6). In that case, he believes, we should attempt to destroy the aliens. - -The article Graham's number has another interesting example.<|endoftext|> -TITLE: Constructible sheaves and dg-modules -QUESTION [6 upvotes]: Let $M$ be a smooth manifold, $A_M$ the de Rham algebra of $M$, $D_{A_M}$ the derived category of the category of differential graded (dg) $A_M$-modules and $D^+_c(M)$ the bounded below constructible derived category of sheaves of real vector spaces on $M$. The category $D^+_c(M)$ knows a lot about the topology of $M$; for example, it allows one to compute the real cohomology of $M$ together with all Massey products. However, unsurprisingly, this category is also quite complicated. Informally, the question I'd like to ask is: can one describe at least some pieces of $D^+_c(M)$ in terms of dg-modules (which is something much more manageable)? -In "Equivariant sheaves and functors", 12.3, J Bernstein and V. Lunts construct two mutually inverse equivalences $\gamma_M:\langle \mathbb{R}_M\rangle\to \langle A_M\rangle$ and $\mathcal{L}_M: \langle A_M\rangle\to \langle \mathbb{R}_M\rangle$. Here $\langle\cdot\rangle$ stands for the full triangulated subcategory generated by $\cdot$, $\mathbb{R}_M$ is the constant sheaf on $M$ and $\langle \mathbb{R}_M\rangle$ and $\langle A_M\rangle$ are subcategories of $D^+_c(M)$ and $D_{A_M}$ respectively. -The functors are defined as follows: $\gamma_M$ takes a sheaf, multiplies it by the de Rham complex and then takes the global sections; $\mathcal{L}_M$ takes a module, replaces it with a $\mathcal{K}$-projective resolution and multiplies the result by the de Rham complex. (A complex of dg $A_M$-modules is $\mathcal{K}$-projective, if $Hom$ from it to an acyclic complex is acyclic.) -Notice that $\gamma_M$ is in fact defined on the whole of $D^+(M)$. I would like to ask: is there a subcategory $D$ of $D^+_c(M)$ larger than the one generated by the constant sheaf such that $\gamma_M$ restricted to $D$ is fully faithful? In particular, if $i:N\subset M$ is a submanifold and $M,N,M\setminus N$ are all simply connected, what happens if we take $D=\langle\mathbb{R}_M, i_*\mathbb{R}_N\rangle$? - -Here is a related result. Suppose we fix a stratification of $M$ with all strata and their closures simply-connected. Consider the subcategory $D\subset D^+_c(M)$ formed by complexes which are constructible with respect to the chosen stratification and let $I^*$ be an injective resolution of the direct sum of the constant sheaves on the strata. Then, due to a result by B. Keller (Deriving dg-algebras, Ann ENS, 1994, no 1, 63-102) by taking $C^*\to Hom(C^*,I^*)$ we get a fully faithful functor from $D$ to $D_{End(I^*)}$ where $End(I^*)$ is the (global) endomorphism algebra of $I^*$. However, this is not exactly what I'm looking for since the endomorphism algebra is still quite difficult to describe explicitly in the example I'm interested in. - -REPLY [7 votes]: This is not a helpful answer to your main question, but merely a negative answer to your "In particular..what happens.." question. But, the general idea may be helpful in figuring out what more precise things (weaker than Keller's result) would be reasonable to ask for. -$\newcommand{\RR}{\mathbb{R}}\newcommand{\RHom}{\mathrm{RHom}}\newcommand{\pt}{\mathrm{pt}}\DeclareMathOperator{\deg}{deg}\newcommand{\RGamma}{R\Gamma}\renewcommand{\mod}{\text{-mod}}\newcommand{\C}[1]{C^\bullet(#1)}$ -Since I'll lapse into this notation anyway, let me make it explicit: Identity $\gamma_M$ with the functor -$$ \RGamma(M, -)\colon D \to \C{M}\mod $$ where by $\mod$ I'm implicitly working in a dg-setting. -At some point below, I'll assume that $N$ is compact oriented of dimensions $n$. (This is not strictly necessary, but allows me to avoid extra notation.) -Claim: Suppose that $M$ is simply connected, $\dim M \geq 2$, and that $i \colon N \hookrightarrow M$ is not the identity. Then, the functor $\gamma_M \colon D= \langle \RR_M, i_* \RR_N \rangle \to D_{A_M}$ is not fully faithful. -Fuzzy Remark: -Before sketching an argument, here's a "philsophical" remark about why things will go wrong: -The category $D_c(M)$ feels the topology (or maybe even geometry) of $M$. In particular, it has a Proper Base-change Theorem saying something like $q^* p_! = (p')_! (q')^*$ where the maps take part in a fiber-square of (actual) topological spaces. The category $D_{A_M} = \C{M}\mod$ feels only the homotopy theory of $M$. You should expect a Base-Change Theorem in this context, but now with a fiber-square of homotopy types -- more correctly, a homotopy fiber-square. -A simpler sort of 'no go' result that this heuristic implies: Suppose you had wanted to include two sub-manifolds $i_k\colon N_k \hookrightarrow M$, $k=1,2$. The $\C{M}\mod$ images would be unable to tell them apart if the $i_k$ were homotopic -- e.g., the inclusion of any two points. While the constructible theory would certain care whether the two points were the same or not. -Sketch of Claim: -To see this, note that -$$ \RHom_{D_c(M)}(i_* \RR_N, i_* \RR_N) = \RHom_{D_c(N)}(i^* i_* \RR_N, \RR_N) = \RHom_{D_c(N)}(\RR_N, \RR_N) = \C{N} $$ -while -Sub-Claim: Letting $\stackrel{h}\times_M$ denote the homotopy fiber product, -$$ \RHom_{A_M}(\gamma_M(i_* \RR_N), \gamma_M(i_* \RR_N)) =\RHom_{\C{M}}\left(\C{N}, \C{N}\right) \approx C_{\bullet}\left(N \stackrel{h}\times_M N\right)[-n] $$ -Assuming the sub-claim: to conclude it suffices to produce homology classes on $\Omega M$ in arbitrarily positive degrees, whose images under the composite -$$ H(\Omega M) \to H_*(N \stackrel{h}\times_M N) \to H_*(\Omega (M/N)) $$ -are non-zero. I think the following should do this upon filling in the details: Equip $M$ with a base-point in $N$, take some non-zero element of $\pi_i M$, with $i \geq 2$, that remains non-zero in $\pi_i (M/N)$. Use it to produce an $(i-1)$-homology class on $\Omega M$, and then take its Pontrjagin products. -Sketch of sub-claim: -Underlying the Eilenberg-Moore spectral sequence is the statement that, letting $\boxtimes$ denote derived co-tensor of co-modules over a co-algebra, -$$ C_\bullet(N) \boxtimes_{C_\bullet(M)} C_\bullet(N) \approx C_\bullet(N \stackrel{h}\times_M N) $$ -Poincare duality gives an equivalence $\C{N} \approx C_\bullet(N)[-n]$ of $\C{M}$-modules (or $C_\bullet(M)$-comodules). It remains to identify -$$ \RHom_{\C{M}}(\C{N}, \C{N}) \approx C_\bullet(N) \boxtimes_{C_\bullet(M)} \C{N} $$ -by term-wise identifying the co-simplicial cobar constructions on both sides. -Example: Note that if $N = \pt$ the sub-claim is a familiar statement in Koszul duality: That for $M$ simply-connected $\RHom_{\C{M}}(\RR,\RR) \approx C_\bullet(\Omega M)$. In certain cases, e.g. $M = S^{2k+1}$, you can just see it. As an aside: $C_\bullet(\Omega M)\mod$ knows about all locally-constant things, not just local systems finitely-buildable from the trivial one. (But will run into the same issues if you try to include submanifolds without explicitly adding in extra generators for the strata.) -Remark: Though $\gamma_M$ is not fully-faithful here, it does get the maps into/out of $\RR_M$ right. Logic as above shows that -$$ \RHom_D(\RR_M, i_* \RR_N) = \C{N} = \RHom_{\C{M}}(\C{M}, \C{N}) $$ -and then Verdier Duality (for $D_c(M)$) + something like Grothendieck Duality (for $\C{M}$) give the other direction as well.<|endoftext|> -TITLE: A family of polynomials with symmetric galois group -QUESTION [6 upvotes]: Consider the following family of polynomials in $K[x,y]$, where $K$ has characteristic zero: -$f_n(x,y)=(x+y)^n+(x-1)y^n,$ -for $n\geq 3$. I can prove that $f_n(x,y)$ has an irreducible factor of degree $n-1$ in $x$. I also know that the galois group of $f_n$ over $K(y)$ is the symmetric group of degree $n-1$, but am having trouble proving this. -Here is an alternative form for $f_n$: make the substitution $x\rightarrow xy$ and divide by $y^n$. this gives: -$g_n(x,y)=(x+1)^n+yx-1.$ -Substituting $-n$ for $y$ in $g_n(x,y)$ we get: -$g_n(x,-n)=x^2h(x),$ -where $h(x)$ is separable (EDIT: $h(x)$ is the subject of this question). -Alternatively, substituting $x\rightarrow x-1$ into $g_n(x,y)$ gives us a polynomial which factors as: -$(x-1)(x^{n-1}+x^{n-2}+\ldots +x^2+x+y+1)$ -It seems as though it shouldn't be hard to show that some specialisation of $y$ into this gives a polynomial with galois group $S_{n-1}$ over $K$...but I'm well and truly stuck. -Any advice much appreciated! - -REPLY [18 votes]: [Edited mostly to incorporate references etc. from Michael Zieve] -The polynomial $(x^n-1)/(x-1) - y$ has Galois group $S_{n-1}$ over ${\bf C}(y)$ for each $n$, as expected. This answers one of the three problems; the question statements asserts that they are equivalent, which doesn't seem to be the case (see my comment there), so I hope I'm answering the intended one. -The proof combines group theory and polynomial algebra (as might be expect) with algebraic topology, which might be a bit less expected. The connection is as follows. Let $S$ be the Riemann sphere with coordinate $y$, let $P(x,y)$ be any polynomial of degree $d>0$ in $x$, and let $S'$ be the Riemann surface corresponding to $P(x,y)=0$. The rational function $y$ on $S'$ gives a map $S' \rightarrow S$; let $B = \lbrace y_1,\ldots,y_b\rbrace$ be its set of branch points, and fix some $\eta \in S$ not in $B$. Then $\pi_1(S-B,\phantom.\eta)$ is a free group on $b-1$ generators, with generators $g_1,\ldots,g_b$ (where each $g_j$ is a loop from the base point $\eta$ around $y_j$ and back to $\eta$) subject to the single relation $g_1 g_2 \cdots g_b = 1$. These lift to automorphisms $\tilde g_j$ of $S' - y^{-1}(B)$ such that $y = y \cdot \tilde g_j$, and thus to permutations $\pi_j$ of the $d$ sheets of $S'$ above $S$. Then the key fact is that - -The Galois group of $P(x,y)$ as a polynomial over ${\bf C}(y)$ is the subgroup of $S_d$ generated by the permutations $\pi_j$. - -In our case $P\phantom.$ has the form $p(x) - y$. In general I think it is known which polynomials (or even rational functions) $p$ yield $p(x)-y$ with Galois group other than $S_d$, but it's not easy and involves the classification of finite simple groups! -[Actually even less is known; the groups have been determined, but the polynomials, not quite, even in the polynomial case. See below. But this is tangential to the question at hand (albeit a fascinating tangent) because the ensuing argument is enough.] -Fortunately in our case the following sufficient condition is all we need: - -Let $p\in {\bf C}[X]$ be a polynomial of degree $d$. Assume that the derivative $p'$ has distinct roots $x_1,\ldots,x_{d-1}$, and that the numbers $y_j = p(x_j)$ are also pairwise distinct. Then $p(x)-y$ has Galois group $S_d$ over ${\bf C}(y)$. - -[Naturally this result is not new, but I was still surprised to learn from Mike Zieve of its true age and pedigree: Hilbert (1892)! See below.] -Proof: Here $b=d$ and the branch points are $y_1,\ldots,y_{d-1}$ and $y_d = \infty$. Since $P$ is a polynomial, $\pi_d$ is a $d$-cycle. (This already proves that the Galois group is transitive, that is, that the polynomial is irreducible; but we already knew this because it is obviously irreducible over ${\bf C}[y]$.) Each $\pi_j$ for $j3$, even though $T'_d$ has distinct roots $x_j$, because $T_d(x_j) = \pm 1$ for each of them.] -It remains to check that the hypothesis on $p$ is satisfied when $d=n-1$ and $p=(X^n-1)/(X-1)$. For lack of a better idea I did this by computing the discriminant of the polynomial -$$ -q(y) = \frac{(n-1)^{n-1} y^n - n^n (y-1)^{n-1}}{(y-n)^2} \qquad (1) -$$ -of degree $n-2$ whose roots are the $y_j$, and checking that it is nonzero; in fact -$$ -\mathop{\rm disc} q(y) = \pm 2 \Delta_{n-1} \Delta_n -$$ -where $\Delta_m := m^{(m-1)(m-3)}$ (and the sign depends on $n \bmod 4$). The formula for $q$ was obtained from the familiar expression $\pm \bigl( (n-1)^{n-1} A^n - n^n B^{n-1} \bigr)$ for the discriminant of the trinomial $x^n - Ax + B$. It follows that the numerator of (1) is $\pm$ the discriminant of $x^n - xy + (y-1) = (x-1) (p(x)-y)$ as a polynomial in $x$, whence (1) soon follows; substituting $y = nz/(nz-(n-1))$, and using the covariance of the discriminant under ${\rm PGL}_2$ together with the same trinomial formula, soon gives the claimed $\pm 2 \Delta_{n-1} \Delta_n$ (the factor of $2$ arises at the end from a double application of L'Hôpital's rule), QED. - -$\phantom.$ - -Here are some relevant references and additional information. -Polynomials with distinct critical values: This is in section 4.4 of Serre's Topics in Galois Theory (Boston: Jones & Bartlett 1992), as is the reference to Hilbert: "Ueber die irreduzibilität genzen rationalen Funktionen mit ganzzahligen Koeffizienten", J. reine angew. Math. ("Crelle's J.") 110, 104-129 (= Ges. Abh. II, 264-286). Serre gives such polynomials the suggestive name "Morse functions". M. Zieve also notes that a 1959 paper by Birch and Swinnerton-Dyer in Acta Arith. (MR0113844 (22 #4675)) attributes to Davenport the equivalent formulation - -[The Galois group of $f(x) = y$] is the symmetric group if the discriminant of the discriminant of $f(x)-y$ does not vanish. - -They call this a "more euphonious form"; that's a matter of taste, but at any rate it's a memorable formulation, and the one that ended up being used here. -Exceptional Galois groups of $p(x) = y$. The primitive groups that can occur are described by Peter Müller in a paper Primitive monodromy groups of polynomials. They are the symmetric and alternating groups, plus cyclic and dihedral groups (for polynomials equivalent to powers and Čebyšev), and a finite but substantial list of exceptional possibilities. It is hopeless to classify all cases of alternating Galois group. For the rest it can be done, but not easily. Some are exhibited by Cassou-Nogues and Couveignes in an Acta Arith. paper Factorisations explicites de $g(y)-h(z)$; most others were done by Mike Zieve himself; and the final case, polynomials of degree $23$ with Galois group the Mathieu group $M_{23}$, I computed only a few days ago after Mike noted it was still open (they're defined over the quadratic extension of ${\bf Q}(\sqrt{-23})$ of discriminant $3 \cdot 23^3$). -When $p$ is allowed to be a rational function, even the list of groups is not yet known. Mike notes that this "ties in with the 'genus zero program' initiated by Guralnick and Thompson" that was completed in a 2001 paper by Frohardt and Magaard in the Annals of Math. Thanks again to Mike Zieve for all this information.<|endoftext|> -TITLE: binary code with constant hamming distance -QUESTION [5 upvotes]: I want as many 80-bits words as possible with the constraint that the hamming distance between any couple of words is exactly 40. How many can I generate? Is there a generic formula telling me how many n-bits words I can generate with the constraint that any couple of words is at hamming distance exactly n/2? Any general algorithm to generate them? -For 2 bits codewords, "00, 01" are at HD=1. For 4-bits codewords, "0000, 0011, 0101, 1001" are all at HD=2. And then? -Thank You, -JMC - -REPLY [11 votes]: There cannot be more than $n$ such words. To see this, consider the words as elements of $\{-1,1\}^n$. If two such words have Hamming distance $\Delta$, then their scalar product in $\mathbb R^n$ is $n-2\Delta$. In particular, if they all have distance $n/2$, then they are orthogonal, hence linearly independent, hence there are at most $n$ many. -An array of $n$ $\pm 1$ vectors such that every two are orthogonal is called a Hadamard matrix. Such a matrix does not exist unless n=1,2 or n is a multiple of 4. The famous Hadamard conjecture asserts that when $n$ is a multiple of 4, there exists a Hadamard matric of order $n$, This is known to be true in many cases and, in particular, when $n<668$. -An important special case is when $n=2^k$ is a power of two. Then you can construct a set of $n$ such words as follows. We identify bits with elements of the field $\mathbb F_2$, and coordinates $x< n$ with elements of $\mathbb F_2^k$, and thus consider words as functions $f\colon\mathbb F_2^k\to\mathbb F_2$. For any $x,y\in\mathbb F_2^k$, let $f_x(y)=\langle x,y\rangle:=\bigoplus_{i< k}x_iy_i$, where $\oplus$ denotes addition in $\mathbb F_2$ (i.e., modulo $2$). Then $\{f_x:x\in\mathbb F_2^k\}$ is a set of words of size $2^k$, and pairwise Hamming distance $2^{k-1}$, as for any distinct $x,x'$, we have $f_x(y)=f_{x'}(y)$ iff $\langle x\oplus x',y\rangle=0$, which holds for exactly one half of all $y$’s.<|endoftext|> -TITLE: Introductory reading on the Scholz reflection principle? -QUESTION [10 upvotes]: The Scholz reflection principle says, among other things, that if $D < 0$ is a negative fundamental discriminant, not $-3$, then the 3-ranks of the class group of $\mathbb{Q}(\sqrt{D})$ is either equal to that of $\mathbb{Q}(\sqrt{-3D})$, or one larger. -Does anyone know of (and recommend) any introductory reading on this fact? Why it is true, what context to view it in, etc.? Googling reveals some highbrow perspectives on it, some interesting applications, and citations to Scholz's 1932 article (which I'm having difficulty accessing for the moment). All of this is interesting, but there doesn't seem to be any obvious place to begin. -Thank you! - -REPLY [7 votes]: Hi Frank, -There are two places that I remember reading, and enjoying, when learning about the reflection principle: -i) Washington's book on cyclotomic fields section 10.2.(This book is just so great, so in case you don't own a copy this might be a good excuse to buy it.) -ii) Reflection principles and bounds for class group torsion By Ellenberg and Venkatesh--This is a VERY cool paper, but in case you are short on time you just need to read Lemmas 4 and 5. -Also some of the answers to this question Explicit map for Scholz reflection principle might help a bit.<|endoftext|> -TITLE: Is this pleasing polynomial irreducible? -QUESTION [12 upvotes]: Let: -$f(x)=x^n+2x^{n-1}+3x^{n-2}+4x^{n-3}+\ldots + (n-1)x^2+nx+(n+1)$. -Is $f(x)$ irreducible? -In light of the answers to this question, I now know that this is true when $n+1$ is prime. What about when $n+1$ is composite? I have checked a lot of cases and it seems to be true. -(Unnecessary background information: this is linked to this recent question of mine. At the point where I say: -"$g_n(x,−n)=x^2h(x)$," -the $h(x)$ in question has the property that substituting $x-1$ for $x$ puts it in the form of $f(x)$ above. If I can show that $h(x)$ is irreducible, then I will have shown that the galois group of the original polynomial is doubly transitive. Not that I am trying to draw attention back to my original question!) - -REPLY [19 votes]: It is a conjecture that for any $k < n$ one has -$$\frac{d^k}{dx^k}(1+x+\cdots+x^n)$$ -is irreducible in $\mathbb Z[x]$. Your question is the case $k=1$. It is known that the set of integers $n\in [0,t]$ for which the polynomial $nx^{n-1}+(n-1)x^{n-2}+\cdots+1$ is reducible has size at most $O(t^{1/3+\epsilon})$, see the paper - -A. Borisov, M. Filaseta, T. Y. Lam, O. Trifonov, "Classes of polynomials having only one non-cyclotomic irreducible factor", Acta Arith. 90 (1999) 121-153<|endoftext|> -TITLE: I'm looking for a Virasoro-module whose character is 1+ 240q+ 2160q^2+ 6720q^3... -QUESTION [13 upvotes]: Let $E_4(q)=1+ 240q+ 2160q^2+ 6720q^3+\ldots $ be the Eisenstein series of weight 4, -also known as the theta-series of the $E_8$-lattice. -I'm looking for a $\mathbb N$-graded vector space $V$ of graded dimension -$(1, 240, 2160, 6720,\ldots )$. -Moreover, I would like $V$ to be a module for -the Virasoro algebra of central charge $c=4$. -I think that this is indeed possible: -Take $1$ times the Verma module of character $1+q+2q^2+3q^3+5q^4+7q^5+\ldots$ -plus $239$ times the Verma module of character $q(1+q+2q^2+3q^3+5q^4+\ldots)$ -plus $1919$ times the Verma module of character $q^2(1+q+2q^2+3q^3+\ldots)$, etc. -But that's not what I'm looking for. -I'm wondering whether there exists a natural construction that produces the above Virasoro module. -Does anybody know of such a construction? - -Variant: -In the above, I said that I wanted central charge $c=4$ and minimal energy $h=0$. -But I'm flexible: if you know a natural construction of a $Vir_c$-module (pick you $c$) whose character is $q^hE_4(q)$, I'll be happy to hear about it. - -REPLY [5 votes]: Would you be happy with a supercharacter $str \ q^{L(0)-c/24}$ as opposed to the usual graded dimension ${\rm tr} \ q^{L(0)-c/24}$? Also, from the $q$-exapansion of $E_4$, it seems to me that the correct central charge should be zero, not $4$. -You could take $V(E_8)$, the $E_8$ - lattice vertex algebra of central charge $c=8$, and tensor it with the $\mathbb{Z}_2$-graded symplectic fermion vertex -superalgebra $\mathcal{SF}^8$ of central charge $c'=-8$. The tensor product $V(E_8) \otimes \mathcal{SF}^8$ is $\mathbb{Z}_2$-graded by the fermionic part. -Since ${\rm tr}({\mathcal{SF}^8}) q^{L(0)-c'/24}=q^{1/3} \prod_{i=1}^\infty (1+q^i)^8$, -after taking the supertrace of the tensor product, you will clear the denominator in -$\frac{\theta_{E_8}}{\eta^8}$, and end up with the supertrace equal $\theta_{E_8}$. Actually you can think of it as a virtual character (the Virasoro generator is even). -This construction is perhaps silly, but it has some motivation. In rational CFT, all characters are of weight zero. In rational SCFT there are also supercharacters, again of weight zero. Symplectic fermions $\mathcal{SF}^{2d}$, $d \geq 1$ are not rational, but "quasi-rational" or sometimes called "logarithmic", and their supercharacters exhibit "higher weight" phenomena.<|endoftext|> -TITLE: Clique sizes in a unit disk graph -QUESTION [11 upvotes]: This is a spiritual successor to a question that Peter Shor answered here: -Generalized Euclidean TSP -Are there any results known on the asymptotic behavior of clique sizes in a unit disk graph with uniformly sampled points? That is, suppose I sample $N$ points uniformly in a square (or disk, or whatever figure is easiest to deal with) of size $L \times L$, and I draw an edge between two points if the distance between them is less than $1$. As $N$ becomes large, is anything known about the distribution of clique sizes in this graph? - -REPLY [7 votes]: Hi Gunnar, -The behaviour of the clique number = max.clique size will depend on L (you may want to keep L fixed, -or let it grow with $N$ in some way). -Dropping $N$ points in a $L\times L$ box and connecting when the distance is less than 1 is -the same as dropping $N$ points in the unit square and connecting is the distance is -less than $r = 1/L$. -The book by Penrose and an earlier paper by McDiarmid (random channel assignment in the plane", -Random Structures & Algorithms, Volume 22, Issue 2, pages 187–212) describe the -"first order" behaviour of the clique number. -Basically there are three case depending on whether the "average degree" $N \pi r^2$ is $\ll \log N$, $\Theta( \log N )$ or $\gg \log N$. -In each case there is an expression $f(N,r)$ such that clique.no divided by $f(N,r)$ tends to one in -probability. -This of course does not answer all questions about the clique number. One may for instance wonder whether the clique number can somehow be normalized to tend to some known limiting distribution. -(This is not very chique, I know, but) you may also want to have a look at a paper by me (tobias mueller) -"Two point concentration in random geometric graphs", Combinatorica, Volume 28, Number 5, 529-545. -It solves a conjecture on the probability distribution of the max.clique size posed by Penrose -in his book. -Namely that the probability mass of the max.clique size becomes concentrated on two consecutive integers when the parameters are chosen such that $N \pi r^2 \ll \log N$. -I.e. if $N, r$ satisfy this condition then there is a function $k(N,r)$ such that -${\mathbb P}( k \leq \text{clique.no} \leq k+1 ) \to 1$, -as $N\to\infty$. -For other choices of the parameters, if $N\pi r^2 = \Theta( \log N )$ or $\gg \log N$, the (asymptotic) probability distribution of the clique number is an open problem -- see the conclusion of that paper.<|endoftext|> -TITLE: When is a closed differential form harmonic relative to some metric? -QUESTION [35 upvotes]: Let $\omega$ be a closed non-exact differential $k$-form ($k \geq 1$) on a closed orientable manifold $M$. -Question: Is there always a Riemannian metric $g$ on $M$ such that $\omega$ is $g$-harmonic, i.e., $\Delta_g \omega = 0$? -Here $\Delta_g$ is the Laplace-deRham operator, defined as usual by -$\Delta_g = d \delta + \delta d$, where $\delta$ is -the $g$-codifferential. Note that non-exactness is important, since if $\omega$ were to be exact and harmonic, then by the Hodge decomposition theorem $\omega = 0$. -For instance, if $\omega$ is a 1-form on the unit circle, then it is not hard to see that $\omega$ is harmonic with respect to some metric $g$ if and only if it is a volume form (i.e., it doesn't vanish). This observation generalizes to forms of top degree on any $M$. -What can be said in general for forms which are not of top degree? - -REPLY [5 votes]: 1 - Eugenio Calabi affirms that every closed non singular 1-forms in closed manifolds are intrinsecally harmonic ("An instrinsic characterization of harmonic 1-forms", 1969). This is easy proved observing first that such forms are transitive. The dual case is a open problem, but I believe that is true. A closed $(n-1)$-form non singular and non null in cohomology is intrinsecally harmonic iff the voluming-preserving flow induced by this form admit cross section (closed comdimension one submanifold cutting every orbit of the flow) or iff admit complementar foliation (foliation transversal to the flow; at leas $C^2$). Is possible proof this fact in the spirit of Calabi's work. This problem have consequences in flat characterization of circle bundles. -2 - In the case of closed $p$-forms of the rank $p$, the problem have a translate for foliation theory. Is possible to show that if a closed $p$ form $\omega$ of the rank $p$ is transitive, then exists complementar form $\eta$, that is, $\eta$ is a closed form such that $\omega\wedge\eta$ is volume form. This is proved using the theory of foliations cycles (see the Sullivan's paper "Cycles for the dynamical study of foliated manifolds and complex manifolds"). However, in a fiber bundle $\xi=(\pi,F,E,M)$ with symple connected base, compact total space, if $\Omega_M$ is any volume form in $M$, the form $\pi^*\Omega$ is intrinsecally harmonic iff $\xi$ is trivial. We can run away from trivial cases showing that $[F]\neq 0\in H_{\dim F}(E;\mathbb{R})$ and exists exemples those bundles with section. This give us the examples cited by Dan Fox. The problem is the dimension of the kernel of $\eta$. In the cases $p=1$ or $p=n-1$ (without singularities), in the condition of transitivity, the dimension of the kernel of $\eta$ is $n-1$ (case $p=1$) or 1 (case $p=n-1$), and the Calabi argument applies. -3 - Is too a open problem to show that harmonic forms are transitive, as observed by Katz ("Harmonic forms and near-minimal singular foliations"). We can to show that if $\omega$ is a harmonic $p$-form of rank $p$, then the have in $M$ two complementary $SL(*)$-foliations induced by $\ker\omega$ and $\ker *\omega$. -4 - I have studied the problem of decomposable forms. By the Tischler's argument (and others considerations; see "On fibering certain foliated manifolds over $\mathbb{S}^1$") is sufficient considering bundles $\xi=(F,\pi,E,\mathbb{T}^{p})$. If this bundle admit transversal foliation with holonomy group contained in $SL(*)$, then the form $\pi^*(\Omega_{\mathbb{T}^{p}})$ is intrinsecally harmonic. The ideia of study particular examples is know if we can rule out the hypothesis of transitivity. In any bundle $\xi=(\pi,F,E,M)$, with compact total space, $[F]\neq 0$ and $\pi_1(M)$ finite, the form $\pi^*(\Omega)$ is intrinsecally harmonic. -Ps. The above remarks is part of development of my doctoral project and are is under analysis.<|endoftext|> -TITLE: Independence of being an integer -QUESTION [27 upvotes]: In this MO question, the OP asked for an example of a statement which was known not to be independent of ZFC, but for which the truth value was unknown. I immediately thought of a question I asked on math.SE: is $e^{e^{e^{79}}}$ an integer? This is apparently an open question, but I realized after some thought that I don't know how to prove it is decidable in ZFC. -If the number is not an integer, this can be proved in ZFC, because that fact could be expressed by an arithmetical sentence saying there is an integer $n$ such that the sum of a certain definable series is greater than $n$ and less than $n+1$. This sentence can be seen to be $\Sigma^0_1$ by standard techniques, and any true $\Sigma^0_1$ sentence is provable in ZFC. -But if the sum is an integer, it does not seem obvious that this must be provable in ZFC. In general, only $\Sigma^0_1$ statements have to be provable if they are true, and the claim that a certain definable series sums to an integer is $\Sigma^0_2$ rather than $\Sigma^0_1$. -Moreover, it's not hard to see that there are definitions of sequences $(a_n)$ in ZFC such that ZFC proves that $\sum a_n$ converges but ZFC doesn't prove this sum is an integer and ZFC doesn't prove it is not an integer. These sequences can be constructed using the incompleteness theorem in the usual way. In fact, we can make $0 \leq a_n \leq 2^{-n}$ for all $n$, so there is no issue with the rate of convergence. -But there must be something special about $e^{e^{e^{79}}}$ that means either ZFC can prove it's an integer, or can prove it's not an integer - right? - -REPLY [18 votes]: Here is a conditional answer. It was shown by Macintyre and Wilkie that if (a weak variant of) Schanuel's Conjecture is true, then the first-order theory of the real exponential field $(\mathbb{R};0,1,+,\times,\exp)$ is decidable. In particular, the (very unwieldy) first-order sentence -$$\bigvee_{n=A}^B \exp\exp\exp 79 = n,$$ -where $A = 2^{2^{2^{79}}}$ and $B = 3^{3^{3^{79}}}$ is then decidable by finitary means. Of course, it is conceivable that Schanuel's Conjecture is false... - -As pointed out by Dave Marker and George Lowther, Schanuel's Conjecture directly implies that $e^{e^{e^{79}}}$ is not an integer. Indeed, since $e^{79}$ is irrational, $79$ and $e^{79}$ are linearly independent over $\mathbb{Q}$, which means that $e^{79},e^{e^{79}}$ are algebraically independent over $\mathbb{Q}$. Since $79,e^{79},e^{e^{79}}$ are therefore linearly independent over $\mathbb{Q}$, it follows that $e^{79},e^{e^{79}},e^{e^{e^{79}}}$ are algebraically independent over $\mathbb{Q}$. In particular, $e^{e^{e^{79}}}$ is not an integer.<|endoftext|> -TITLE: Does the inverse function theorem hold for everywhere differentiable maps? -QUESTION [124 upvotes]: (This question was posed to me by a colleague; I was unable to answer it, so am posing it here instead.) -Let $f: {\bf R}^n \to {\bf R}^n$ be an everywhere differentiable map, and suppose that at each point $x_0 \in {\bf R}^n$, the derivative $Df(x_0)$ is nonsingular (i.e. has non-zero determinant). Does it follow that $f$ is locally injective, i.e. for every $x_0 \in {\bf R}^n$ is there a neighbourhood $U$ of $x_0$ on which $f$ is injective? -If $f$ is continuously differentiable, then the claim is immediate from the inverse function theorem. But if one relaxes continuous differentiability to everywhere differentiability, the situation seems to be much more subtle: - -In one dimension, the answer is "Yes"; this is the contrapositive of Rolle's theorem, which works in the everywhere differentiable category. (The claim is of course false in weaker categories such as the Lipschitz (and hence almost everywhere differentiable) category, as one can see from a sawtooth function.) -The Brouwer fixed point theorem gives local surjectivity, and degree theory gives local injectivity if $\det Df(x_0)$ never changes sign. (This gives another proof in the case when $f$ is continuously differentiable, since $\det Df$ is then continuous.) -On the other hand, if one could find an everywhere differentiable map $f: B \to B$ on a ball $B$ that was equal to the identity near the boundary of $B$, whose derivative was always non-singular, but for which $f$ was not injective, then one could paste infinitely many rescaled copies of this function $f$ together to produce a counterexample. The degree theory argument shows that such a map does not exist in the orientation-preserving case, but maybe there is some exotic way to avoid the degree obstruction in the everywhere differentiable category? - -It seems to me that a counterexample, if one exists, should look something like a Weierstrass function (i.e. a lacunary trigonometric series), as one needs rather dramatic failure of continuity of the derivative to eliminate the degree obstruction. To try to prove the answer is yes, one thought I had was to try to use Henstock-Kurzweil integration (which is well suited to the everywhere differentiable category) and combine it somehow with degree theory, but this integral seems rather unpleasant to use in higher dimensions. - -REPLY [3 votes]: S. Radulescu and M. Radulescu gave a proof of the inverse function theorem for everywhere differentiable maps in 1989. The title of the paper, published in J. Math. Anal. Appl., is “Local inversion theorems without assuming continuous differentiability”.<|endoftext|> -TITLE: Recognizing a measure whose moments are the motzkin numbers -QUESTION [9 upvotes]: The Catalan numbers are the moments of the Wigner semicircle distribution. -$$ \frac{1}{2\pi} \int_{-2}^2 x^{2n} \sqrt{4 - x^2} dx = \frac{1}{n+1} \binom{2n}{n} $$ -Motzkin numbers enumerate the number of paths from (0,0) to (0,n) in steps of (1,1),(1,-1),(1,0) remaining above the x-axis. -For this particular sequence of numbers 1, 1, 2, 4, 9, 21, 51, 127... satisfying the recursion -$$ M_n = \frac{3(n-1)M_{n-2} +(2n+1)M_{n-1}}{n+2} $$ -Is it possible to reconstruct a measure $\rho(x)$ whose moments $\langle x^n \rangle$ follow that sequence? - -REPLY [10 votes]: The formula for the measure of the Motzkin numbers as stated by Gjergij follows from the formula for the Catalan numbers if we write the formula for the Catalan numbers in the form -$\frac{1}{{2\pi }}\int_{ - 2}^2 {x^{2n} \sqrt {4 - x^2 } } dx = C_n $ and observe that the corresponding orthogonal polynomials are the Fibonacci polynomials $F_{n + 1} (x, - 1)$ defined by $F_n (x, - 1) = xF_{n - 1} (x, - 1) - F_{n - 2} (x, - 1)$ with initial values $F_0 (x, - 1) = 0$ and $F_1 (x, - 1) = 1.$ -For the Motzkin numbers the corresponding orthogonal polynomials are $F_{n + 1} (x - 1, - 1).$ Therefore a simple transformation of the integral gives the result. -Edit: -More generally (as answer to the comments by Gjergji and Brendan): -Let $f(z) = \sum {r(n,c,d)z^n } $ satisfy $f(z) = 1 + czf(z) + d^2 z^2 f(z)^2. $ -Then -$r(n,c,d) = \frac{1}{{2\pi d^2 }}\int_{c – 2d}^{c + 2d} {x^n \sqrt {4d^2 - (x - c)^2 } } dx.$<|endoftext|> -TITLE: I was wondering if the set of singular loops is a (somewhere) submanifold of loop space? -QUESTION [7 upvotes]: The set of all smooth maps $S^1\to M^n$ ($M$ is a smooth manifold) is a generalized manifold(see http://ncatlab.org/nlab/show/smooth+loop+space). -I was wondering if the set of singular loops (maps with selfcrossings or zeros of derivative) is a (Fréchet,Frolicher,diffeological)submanifold of loop space? -EDIT: it is clear that answer is no (see answer below). So, I rewrite question: is it true that set of singular loops is a collection of submanifolds different codimension(set of loops with one selfintersection has codimension $dim M-2$, loops with zero of derivative has codimension $dim M -3$ and so on. Set of loops with infinite number of singularities should have infinite codimension... Let's forget about them.) -So, the question is about local situation: for example, let's consider a loop with one self-intersection (and without other singlarities). Is it true that set of near loops with one self-intersection is a submanifold in sense of (Fréchet|Frolicher|diffeological)? -EDIT 2. I reformulate question. $map(S^1 \to \mathbb R^3)$ is a functional space, so we can apply a technique of singularity theory. Generic map $f$ with one self-intersection has one-parameter versal deformation $V:[0,1]\times S^1\to \mathbb R^3$ - and any deformation are induced from $V$. Does it imply that $D$(set of singular loops) near the $f$ is a submanifold of codimension 1 in sense of Fréchet or Frolicher? -ADDED. Andrew Stacey explains in his answer and in http://ncatlab.org/nlab/show/on+the+manifold+structure+of+singular+loops why a stratum of a loop space is not a submanifold (the reason is tha same as in a standart smooth injection of line to plane where image is not a submanifold). -But locally, as Ryan said, each stratum is a submanifold (in sense of Frechet). And the kast question is: -For a smooth submanifold $X⊂Y$ of codimension 2, for a general point $x∈X$ we always have a map from small neghbourhood $U$ of $x$ to $D^2$ ($x∈U⊂Y,f:U→D^2$) such that $U\cap X= f^{−1}(0)$. I belive that in situation of space of loops there is no such map... Is it true? - -REPLY [5 votes]: My interpretation of your question is that you would like to see a stratification of the space of smooth maps $S^1\to M^n$. It suffices to consider maps $\newcommand{\r}{\mathbb R} \r \to \r^n$, and for simplicity I will take $n=3$. -I think this is a very interesting question. I suspect the answer is no, there is not a nice stratification "all the way down", but that you have to go to fairly high codimension before you run into problems. (See the answers to this question and the references therein.) -In particular, I don't think Ryan's example presents a problem. We just have to recognize that the set of maps with a single double point (and no other singularities) consists of several strata: a codimension 1 stratum where the double point is transverse, a codimension 3 stratum where the tangents at the double point are colinear but the higher derivatives are in general position, and higher codimension strata where the two curves are tangent to higher order. -More specifically, a generic colinear double point is (after change of coordinates in the domain and range) of the form $[t \mapsto (t, 0, 0); t \mapsto (t, t^2, t^3)]$. I claim that singularities of this form are a codimension 3 submanifold of the space of all smooth maps. As in Ryan's answer, a 3-dimensional space of perturbations transverse to this stratum has the form $t\mapsto (t, t^2+a, t^3+bt + c)$. -The link of this stratum is a 2-sphere with an embedded "figure 8". The nonsingular parts of the figure 8 correspond to single transverse double points. The central (singular) point of the figure 8 corresponds to a pair of transverse double points. Thus there is no reasonable way to make the subset of maps with a single not-necessarily-transverse double point into a nice stratified set, much less into a manifold. I think this was the point of Ryan's answer. The point of my answer is that the space of all smooth maps nevertheless has a perfectly nice stratification near this sort of singularity. -The list of indecomposable strata for maps $\r\to \r^3$ starts out like this: - -codimension 1: single transverse double point -codim 2: zero derivative with 2nd, 3rd and 4th derivatives linearly independent, i.e. conjugate to $t\mapsto (t^2, t^3, t^4)$ -codim 3: (a) triple point with tangents linearly independent; (b) double point with tangents colinear and higher derivatives independent (as above) -codim 4: (a) triple point with tangents coplanar; (b) double point with one of the tangents zero (and higher derivatives generic); (c) colinear double point with some higher derivatives non-generic (like $[t\mapsto (t, 0, 0); t\mapsto (t, t^2, t^4)]$; not sure about this one) - -Items from the above list can of course be combined. For example, two generic double points and four generic triple points has codimension $2\cdot 1+ 4\cdot3 = 14$. -It would be nice if one could continue the above list so that the complement of all the strata had infinite codimension. I'm not sure whether that's possible.<|endoftext|> -TITLE: Semi-simplicial versus simplicial sets (and simplicial categories) -QUESTION [14 upvotes]: Hi, -Let's denote by "semi-simplicial set" a simplicial set without degeneracies, i.e. a contravariant functor from the category $\Delta_{inj}$ of finite linearly ordered sets and order preserving injections to sets (this is also known as $\Delta$-set). -We have an inclusion functor $j: \Delta_{inj} \rightarrow \Delta$ giving us an adjunction between semi-simplicial sets and simplicial sets: The right adjoint is $j^*: Set^{\Delta^{op}} \rightarrow Set^{\Delta_{inj}^{op}}$ given by precomposition with $j$ (i.e. forgetting degeneracies), the left adjoint is $j_!:Set^{\Delta_{inj}^{op}} \rightarrow Set^{\Delta^{op}} $ given by left Kan extension (i.e. taking realization in simplicial sets). It is known from the article referred to in this MO-answer that boosting a semi-simplicial set up to a simplicial one and then forgetting degeneracies again results in an equivalent semi-simplicial set (in the sense that their realizations are equivalent). -I would like to know about the other direction: Does anyone know conditions on a simplicial set $K$ which ensure that the map $j_!j^*K \rightarrow K$ is a weak equivalence? -I am especially interested in the case where $K$ is the nerve of a category. - -A related question is the following: Given a Reedy cofibrant semi-cosimplicial object in the category of simplicial categories, which is equivalent to the underlying semi-cosimplicial object of the usual cosimplicial object, one can by the usual yoga set up an adjunction between semi-simplicial sets and simplicial categories. -Starting with a simplicial set $K$ one can view it as a semi-simplicial set, then produce a simplicial category. I would like to know under which conditions on $K$ this simplicial category is equivalent to the simplicial category which one gets by applying the usual functor from simplicial sets to simplicial categories (the left adjoint to the coherent nerve) directly to $K$. Again the case of biggest interest for me is when $K$ is the nerve of a category. - -REPLY [17 votes]: The map $j_! j^{\ast} K \rightarrow K$ is never a Joyal equivalence unless $K$ is empty. -For example, if $K = \Delta^{0}$, then $j_{!} j^{\ast} K$ is the nerve of the category with one object $X$ and a single nonidentity morphism $e: X \rightarrow X$ satisfying $e^2 = e$. -You can think of $j_{!} j^{\ast} K$ as "obtained from $K$ by freely adjoining new identity morphisms".<|endoftext|> -TITLE: Example of a group which has $\text{SL}_{n}(\mathbb{Z})$ as the automorphism group -QUESTION [15 upvotes]: For the past one week, I have been trying to learn more about automorphism groups of different groups. Very recently one of my friend asked this question to me: - -What is the automorphism group of $(\mathbb{Q}^{\ast},\times)$. In short, what is $\text{Aut}(\mathbb{Q}^{\ast})$? - -I emailed couple of friends and got the answer as: - -$\text{Aut}(\mathbb{Q}^{\ast})$ is isomorphic to the automorphism group of a free abelian group of countable rank. In particular, it will contain $\text{GL}(n,\mathbb{Z})$ for all $n$. - -My question would be : - - -Can we realize $\text{SL}_{n}(\mathbb{Z})$ to be the automorphism group of some group? - - -Are there groups which are which are "very difficult" to be realized as the automorphism group of a certain group. - - -So suppose someone comes and asks me: Is $S_{3}$ or $\text{GL}_{2}(\mathbb{Z})$ the automorphism group of some group, then how can I answer the question? I am particularly interested in seeing how to think for a solution. - -REPLY [6 votes]: Another not quite answer: In the beautiful paper -Automorphism groups of polycyclic by finite groups and arithmetic groups, by Oliver Baues and Fritz Grunewald, -Publ. Math. IHES 104 (2006), no. 1 (arXiv version, MathReviews, journal version) -the authors show that in any cases the automorphism groups of polycyclic-by-finite groups are "arithmetic" (basically subgroups of $SL(n, \mathbb{Z})$ (results of this sort have actually been know for quite a while for narrower classes of groups -- check the references of the paper).<|endoftext|> -TITLE: Automorphisms of a matrix in Smith normal form? -QUESTION [9 upvotes]: Added: Amritanshu Prasad's answer makes it clear that I am really asking for a description of the group of integer unimodular matrices $P$ such that $D^{-1}PD$ is also integer. These matrices are characterized by the property that the elements below the diagonal satisfy certain divisibility properties, namely that for $j\lt i$, the element $p_{ij}$ is divisible by $d_i/d_j$. (The latter is integer by assumption on $D$.) My question was whether there is a simple set of generators for this group. -Amritanshu Prasad's answer provides a nice set of generators when the elements of each row, rather than being integers, are taken modulo a certain number. I will have to think about whether this helps with the problem that motivated the question originally. Meanwhile, I am still interested in finding out what is known about this question in the integer case. -Original post: -Let $M$ be a nonsingular integer $n\times n$ matrix with invariant factors $d_1,\ldots,d_n$ satisfying $d_j\mid d_{j+1}$ for $1\le j\lt n$ and $d_j\gt0$ for $1\le j\le n$. Let $D=\mathrm{diag}(d_1,\ldots,d_n)$ be the Smith normal form of $M$. There is a pair of integer unimodular matrices $(P_1,Q_1)$ such that $P_1MQ_1=D$, but $(P_1,Q_1)$ is not uniquely determined. I am trying to understand this nonuniqueness. -Suppose that $P_1MQ_1=P_2MQ_2=D$. Define $P$ and $Q$ to be the integer unimodular matrices that satisfy $P_2=PP_1$ and $Q_2=Q_1Q$. Then $PDQ=D$. We call such a pair $(P,Q)$ an automorphism of $D$, and are interested in characterizing the group consisting of all automorphisms of $D$. -Define the elementary matrices $S_{ij}$, $N_i$, $L_{ij}(a)$ as follows: - -$S_{ij}M$ interchanges rows $i$ and $j$ of $M$; -$N_iM$ multiplies row $i$ of $M$ by $-1$; -$L_{ij}(a)M$ adds $a$ times row $j$ of $M$ to row $i$ of $M$, where $a$ is a nonzero integer. - -With these definitions, some elementary pairs that satisfy $PDQ=D$ are: - -$(P,Q)=(S_{ij},S_{ij})$ for any $1\le i\lt j\le n$ such that $d_i=d_j$, -$(P,Q)=(N_i,N_i)$ for any $1\le i\le n$, -$(P,Q)=(L_{ij}(1),L_{ij}(-d_j/d_i))$ for any $1\le i\lt j\le n$, -$(P,Q)=(L_{ij}(-d_i/d_j),L_{ij}(1))$ for any $1\le j\lt i\le n$. - -My question is: Do these four types of pair generate the entire automorphism group? -I initially thought that this would be a straightforward question to answer, and that the answer would be 'yes', but now I am fairly sure it is not so simple. For example, consider the smallest nontrivial form, $D=\begin{bmatrix}1 & 0\\ 0 & r\end{bmatrix}$ with $r>1$. Writing $P=\begin{bmatrix}a & b\\ c & d\end{bmatrix}$ with $\lvert ad-bc\rvert=1$, the relation $Q=D^{-1}P^{-1}D$ implies that $Q=(ad-bc)^{-1}\begin{bmatrix}d & -br\\ -c/r & a\end{bmatrix}$, which is integer when $r\mid c$. Hence the most general pair is $(P,Q)=\left(\begin{bmatrix}a & b\\ rc' & d\end{bmatrix},(ad-rbc')^{-1}\begin{bmatrix}d & -rb\\ -c' & a\end{bmatrix}\right)$ with $\lvert ad-rbc'\rvert=1$. For the subgroup satisfying $ad-rbc'=1$, we therefore require that $P$ be an element of the congruence subgroup $\Gamma_0(r)$ and that $Q=\rho(Q)$ where $\rho:\Gamma_0(r)\rightarrow\Gamma^0(r)$ is the map $\begin{bmatrix}a & b\\c & d\end{bmatrix}\mapsto\begin{bmatrix}d & -rb\\-c/r & a\end{bmatrix}$. We obtain the full automorphism group by including, in addition to the generators $(\gamma,\rho(\gamma))$ where $\gamma$ is a generator of $\Gamma_0(r)$, the generators $(N_1,N_1)$ and $(N_2,N_2)$. -The problem with this is that the set of generators 1–4 appears not to be adequate for the case $r=5$, for example. Andy Putman's question Generators for congruence subgroups of SL_2 seems relevant in this regard, although it is concerned with generators of $\Gamma(r)$ rather than $\Gamma_0(r)$. The Grosswald and Frasch references in Ignat Soroko's answer to that question provide a set of generators that freely generates $\Gamma(p)$ for $p$ an odd prime; this set contains many generators in addition to 1–4, and the number of generators grows as $p^3$. -It would therefore appear that, if the picture for $\Gamma_0(r)$ is similar to that of $\Gamma(r)$, and if Frasch's requirement of free generation is not the origin of all this complication, then the answer to my question is no, at least in the case where $n=2$ and $r$ is a prime greater than 3. On the other hand, a remark in Andy Putman's question suggests to me that the situation may be considerably simpler for $n>2$, and that there's a chance that the generators 1–4 suffice. I am not, however, sure that congruence subgroups are the relevant concept for $n>2$. Also, for $n=2$, I wonder whether adding the single extra generator $L_{12}(1)$ to Frasch/Grosswald's set would generate all $P$? -This leads to the following additional questions: - -Is the above understanding of $n=2$ correct? If so, what is the smallest set of generators one can write down? -Do 1–4 generate the automorphism group for $n>3$? If so, how and where is this proved? - -REPLY [2 votes]: If you have $P$, I think you can recover $Q$ as $(D^{-1}PD)^{-1}$. -Therefore, you are looking for invertible integer matrices $P$ such that $D^{-1}PD$ is also invertible (i.e., $P\in GL_n(\mathbf Z)\cap D GL_n(\mathbf Z) D^{-1}$). -Going modulo the subgroup group consisting of $I+T$, where $T$ is an endomorphism of $\mathbf Z^n$ such that $T(\mathbf Z^n)\subset D\mathbf Z^n$, you get the automorphism group of the finite abelian group $A=\mathbf Z/d_1\mathbf Z\times\dotsb\mathbf\times Z/d_n\mathbf Z$. This group is a product of the automorphism groups of the primary components of $A$. -A $p$-primary component of $A$ is of the form $\mathbf Z/p^{\lambda_1}\mathbf Z\times\dotsb\times\mathbf Z/p^{\lambda_n}\mathbf Z$. This group is generated by the Birkhoff moves (see Subgroups of Finite Abelian Groups by Garrett Birkhoff in Proceedings of the London Math. Society, 1935): - -Scaling any row/column by a $p$-free integer -Adding $\alpha$ times the $i$th row (column) to the $j$th row (column) so long as $p^{\max\{0,\lambda_i-\lambda_j\}}$ divides $\alpha$. -Interchanging rows or columns with the same invariant factors.<|endoftext|> -TITLE: Rotational symmetry group of QxQ -QUESTION [6 upvotes]: What is the rotational symmetry group of $\mathbb{Q}\times \mathbb{Q}$, the subset of the real plane consisting of rational points? i.e. are there infinite rotations, is this a named group ? - -REPLY [4 votes]: I might as well add another two interpretations of this group from the theory of algebraic tori - this should make the isomorphism between $SO_2(\mathbb{Q})$ and $\mathbb{Q}[i]^\times/\mathbb{Q}^\times$ mentioned in another answer look a bit more natural. The group you seek is the group of rational points of the norm torus of the field extension $\mathbb{Q}[i]/\mathbb{Q}$, and it is also the group of rational points of the quotient of the Weil restriction (along the same extension) of the multiplicative group by the counit subgroup. -Recall that restriction of scalars (also known as Weil restriction) of the multiplicative group $\mathbb{G}_m$ along $\mathbb{Q}[i]/\mathbb{Q}$ produces a two dimensional algebraic torus, that in particular is a subgroup of $GL_{2,\mathbb{Q}}$. The restriction of determinant to this subgroup is called the norm map, and its kernel (i.e., the intersection with $SL_2$) is the norm torus. The norm torus is equal to $SO_2$ as an algebraic subgroup of $SL_2$ - this is special to $\mathbb{Q}[i]$ among quadratic extensions, as other quadratic fields have norms that are not isomorphic to the standard diagonal quadratic form. -Since restriction of scalars is right adjoint to base change, you get a counit homomorphism $\mathbb{G}_{m, \mathbb{Q}} \to \operatorname{Res}_{\mathbb{Q}}^{\mathbb{Q}[i]} \mathbb{G}_{m, \mathbb{Q}[i]}$ of group schemes. The resulting quotient is an anisotropic rank one torus whose rational points are in natural bijection with elements of $\mathbb{Q}[i]^\times/\mathbb{Q}^\times$. -An isomorphism between these two descriptions can be seen by passing to character lattices as Galois modules (this passage is an antiequivalence - see SGA3 Exp. 9). The character lattice of $\mathbb{G}_m$ is a copy of $\mathbb{Z}$ with trivial action, and restriction of scalars corresponds to taking a tensor product with the group ring of the Galois group (i.e., it is the induced representation - there is also a way to think of this as a pushforward of étale sheaves on the corresponding spectra of fields). The norm torus construction corresponds to taking the quotient by the invariant sublattice, and the counit quotient corresponds to taking the Galois anti-invariant sublattice. As abstract Galois modules, both character lattices are free of rank one with isotropy group equal to the absolute Galois group of $\mathbb{Q}[i]$. -You can show that there are elements of infinite order by noting that $m$th roots of unity have degree $\phi(m)$ for all $m \geq 1$, so you only have to find a point that is neither a 4th root of unity nor a 6th root of unity. Any nondegenerate Pythagorean triple will suffice to produce a suitable rotation, and there is a neat characterization of such triples using cohomology given in this MathOverflow answer.<|endoftext|> -TITLE: Global applications of eigenvarieties -QUESTION [28 upvotes]: A Sunday question for specialists of eigenverieties: -In their important paper "the eigencurve", Coleman and Mazur globalized the earlier construction of Coleman's families, constructing a beautiful eponymous rigid analytic space -that parametrizes all systems of Hecke eigenvalues of finite-slope overconvergent modular forms. Since them many generalizations of this construction have been performed (the "eigenvarieties") which each time appear as globalizations of local constructions generalizing Coleman's families. This process of globalization has even been axiomatized by Buzzard ("the eigenvariety machine"). -Yet, I wonder: - -What are the benefits of working with a global object (which is - considerably more difficult to construct and to deal with) rather than just the - local objects with which it is constructed (the families of Coleman and their - generalizations) ? - -Of course, having a global, canonical, object is much more satisfying on esthetic grounds. -As a mathematician formed after Grothendieck's revolution, this reason alone would be for me -a sufficient one to consent the effort of constructing global eigenvarieties. -But my question is meant to be understood a little bit more specifically: - -What are the applications, or expected applications (to our knowledge of the arithmetic of automorphic forms, Galois representations, L-functions, etc.) of the global existence and geometry of eigenverities that are not already consequences of the existence and geometry - of their local pieces? - -Of course, there are already an enormous amount, still growing fast, of arithmetic informations obtained from the local pieces of eigenvarieties. But what for the global structure? Let me mention the only one I know: the global existence of the eigencurve (say) -is necessary to be able to attached to any overconvergent finite slope modular form -a Galois representation. With Coleman's families alone, we could construct those representations only for these forms having a weight sufficiently close p-adically to a non-negative integer (for example the one with negative weights). Yet I find this application -not very convincing, as why do we care about overconvergent form with weight far away from integers except for their being the "flesh" of the eigencurve? -So what other applications do you have in mind? -(edited for one typo) - -REPLY [7 votes]: This is something I've told people privately for a while, and now have enough ingredients written up (jointly with Liang Xiao) to claim on the web somewhere: the validity of the parity conjecture is constant in $p$-adic analytic families. (Fine print: the family must be symplectic self-dual, and be equipped with a sort of Panchishkin triangulation, but that's all.) Now that the triangulation of the entire eigencurve has been constructed, and the conjecture is known in weight two by any number of authors, it follows that the parity conjecture holds for any finite-slope form lying on an irreducible component of the eigencurve that admits a classical weight two point. Thus, the parity conjecture for all finite-slope forms reduces to the claim that Buzzard's observation noted above in $p=2,N=1$ holds generally (apply Coleman's classicality theorem to the low-slope, weight two points near the boundary), which is a question of the global geometry of the eigencurve.<|endoftext|> -TITLE: Sheaf with free stalks -QUESTION [9 upvotes]: Say we are given a complex manifold $X$ and an $\mathcal{O}_X$-module $\mathcal{F}$. Assume that for any point $P\in X$ the stalk $\mathcal{F}_P$ is a free $(\mathcal{O}_X)_P$-module of finite rank. Does it imply that $\mathcal{F}$ is locally free? If not, what do you need to know additionally about $\mathcal{F}$ to make it true? -Note that if we were looking at the case of schemes then it would be wrong in general. Mathoverflow answer to a related question is -here -Remark: As it was pointed out by Francesco Polizzi, this is true if $\mathcal{F}$ is coherent. What if we do not know it apriori? - -REPLY [4 votes]: This is a small modification of Donu's answer. -Let $\mathcal F=(i_x)_\ast\mathcal O_x$ (the skyscraper sheaf of $\mathcal O_x$ over $x$) for some $x\in X$. Then $\mathcal F$ is locally free of rank $1$ at $x$, and is locally free of rank $0$ everywhere else. Clearly $\mathcal F$ is not locally free near $x$, since it doesn't even have locally constant rank.<|endoftext|> -TITLE: Which polynomials are determinants of a symmetric matrix with linear entries? -QUESTION [13 upvotes]: Let $k$ be a field. Can each degree $n$ polynomial $P(t) \in k[t]$ be written as the determinant of the matrix $A + tB$, where $A$ and $B$ are two symmetric $(n \times n)$-matrices with entries in $k$? -Over an algebraically closed field this is pretty obvious, but is this also true for non-closed fields? - -REPLY [10 votes]: Here is a simple argument showing that you can get any polynomial up to a constant factor. As Rob Israel and my comments above show, you might not be able to get rid of that constant. -If $\det(A_1 t + B_1) = f_1(t)$ and $\det(A_2 t + B_2) = f_2(t)$, then $\det \left( \begin{smallmatrix} A_1 t+B_1 & 0 \\ 0 & A_2t + B_2 \end{smallmatrix} \right) = f_1(t) f_2(t)$. Thus, we are immediately reduced to the case that $f$ is irreducible. -We also may ignore the case that $f(t) = t$, as that is easy to achieve. -Let $k$ be our ground field, and let $L = k(t)/f(t)$. Since $f$ is irreducible, this is a field. Choose a nonzero $k$-linear map $\tau: L \to k$. (If our extension is separable, it would be elegant to choose the trace, but this choice doesn't actually matter.) Let $u \in L$ be such that $\tau(u) \neq 0$. -Define two $k$-linear symmetric bilinear forms on $L$: $\langle x,y \rangle_1 = \tau(xy)$ and $\langle x,y \rangle_2 = \tau(xyt)$. These are symmetric because multiplication is commutative. The first is non-degenerate because, for any non-zero $x \in L$, we have $\langle x, u x^{-1} \rangle_1 \neq 0$. A similar argument applies to $\langle \ , \ \rangle_2$, using that $t \neq 0$. (This is why we had to exclude $f(t)=t$.) -Observe that -$$\langle x,y \rangle_2 = \langle x, T y \rangle_1$$ -where $T$ is the linear map $L \to L$ given by multiplication by $t$. -Choose a $k$-basis for $L$. Let $A$ and $B$ be the matrices of $\langle \ , \ \rangle_1$ and $\langle \ , \ \rangle_2$ in this basis. Then the above equation shows that $A = BT$. -Since our bilinear forms are symmetric, $A$ and $B$ are symmetric matrices. Since our bilinear forms are nondegenerate, they are invertible. Since $A = BT$, we have $\det(A \lambda+B) = \det(B) \det(\lambda T + \mathrm{Id})$. The characteristic polynomial of $T$ is $f(t)$, so we are done. -ADDED: It looks like Allison Miller has just about the same argument, and also has some references to where this idea can be found in the literature.<|endoftext|> -TITLE: Analogue of Wick formula for orthogonal polynomials -QUESTION [8 upvotes]: n-point correlations of Gaussian random variables can be simplified with Wick expansion. -$$ \langle x_{i_1} x_{i_2} \dots x_{i_{2n-1}} x_{i_{2n}} \rangle = \int_{\mathbb{R}^n} x_{i_1} \dots x_{i_{2n}} e^{- \frac{1}{2}\sum x_i^2} = \sum_{\sigma \in \text{matchings}} \; \prod_{\{i,j\}\in \sigma} \langle x_i x_j \rangle$$ -A matching is a partition of $\{ 1,2,\dots ,2n\}$ into two-element sets, e.g. $\{ \{1,2\},\dots, \{2n-1,2n\}\}$. -The Chebyshev polynomials are orthogonal with respect to $\mu(dx) = \sqrt{4-x^2} dx$ but there probably isn't a way to simplify n-point correlations, -\[ \int_{S^n} x_{i_1}\dots x_{i_{2n}} \sqrt{1 - (x_1^2+\dots + x_{n}^2)} \; dx \] -Is there an analogue of Wick's formula for other orthogonal polynomials? It seems rather unlikely since the measure can be arbitrary. -Maybe look at Wick expansion as the sum over matchings. Is there a version of the Wick expansion for non-crossing matchings or for Schröder paths? - -REPLY [8 votes]: The integral you want to compute is related to multivariable beta functions, the Dirichlet distribution and Dirichlet integrals. In particular one has -$$\int\int\cdots\int f(x_1+\cdots+x_n) x_1^{k_1}\cdots x_n^{k_n} dx_1\cdots dx_n$$ -$$=\frac{\Gamma(k_1+1)\cdots \Gamma(k_n+1)}{\Gamma(k_1+\cdots+k_n+n)}\int_0^1 f(t)t^{k_1+\cdots+k_n+n-1}dt$$ -where the first integral is over the simplex $\sum x_i\le 1$, and in your case $f(t)=\sqrt{1-t}$. Since you are considering moments the $k_i$'s will end up being half-integers and therefore the result will be some sort of generalized Catalan number. -There is also the "Free probability" point of view, which I think is closer to what you want. In fact the joint distribution of free Gaussian random variables has an expansion similar to the Wick expansion but over noncrossing matchings. Of course in free probability the role of the normal distribution is played by the Wigner semicircle distribution. -From the perspective of orthogonal polynomials I believe the correspondence goes like this: Hermite polynomials correspond to the normal distribution and complete matchings. Chebyshev polynomials of second kind correspond to the semicircle distribution and to noncrossing matchings. Charlier polynomials correspond to set partitions, Laguerre polynomials correspond to permutations etc.<|endoftext|> -TITLE: Second eigenvalue of suspension of a graph -QUESTION [5 upvotes]: Suppose I have some $d$-regular graph $G$. Let $\lambda = \max\{\lambda_2(G), |\lambda_n(G)|\}$ be the second largest eigenvalue of the adjacency matrix of $G$. Now take $\tilde{G}$, the suspension of $G$, obtained by adding a new vertex $v$ and connecting $v$ to all vertices of $G$. Is it true that $\lambda(\tilde{G}) \leq \lambda(G)$? (or more generally, is it true if instead of suspending with one vertex we suspend with a clique of size $n$) -The intuition is that $\tilde{G}$ should exhibit better mixing (hence, smaller $\lambda$), since it's easier for a random walk to get from one vertex to any other. - -REPLY [6 votes]: The eigenvalues of $G$ determine the eigenvalues of $\tilde G$ in the regular case. If I calculated correctly, just replace the eigenvalue $d$ by the two zeros of $x^2-dx-n$, and keep the other eigenvalues the same. This is for the adjacency matrix. I agree with Alon that you might be looking at the wrong matrix, and I would also call this the cone of $G$.<|endoftext|> -TITLE: Exceptional zeros and Liouville's $\lambda$ function -QUESTION [13 upvotes]: This originated from an textbook exercise (recently posted to math.stackexchange -https://math.stackexchange.com/questions/62883/quadratic-characters-and-liouvilles-function -with no success) but I think it has research level implications so please bear with me. -I'm working through the problems in Montgomery and Vaughan's Multiplicative Number Theory. In Section 11.2 'Exceptional Zeros', Exercise 9a says that for a quadratic character $\chi$, show that for all $k\ge 0, x\ge1 $ -$$ -\sum_{n0$ -$$ -L(\sigma,\chi)>0. -$$ -I expect one is meant to use a Mellin transform with Cesaro weighting, S 5.1 in Montgomery and Vaughan. The difficulty is that $\chi(n)/n$ are the Dirichlet series coefficients of $L(s+1,\chi)$, not $L(s,\chi)$. Thus (5.18) gives -$$ -L(\sigma+1,\chi)>0 -$$ for all $\sigma>0$. - -Question: Show, as the exercise states, that if there exists $k$ such that for all $x\ge1$ - $$ -\sum_{n0$, - $$ -L(\sigma,\chi)>0. -$$ - -By a theorem of Landau, the positivity hypothesis for some $k$ and all $x$ would imply the Riemann Hypotheses as well; I don't see how to use this (and I think Landau's theorem is beyond the scope of the book.) -Alternately, the method of part a will show that -$$ -\sum_{n0$ for all quadratic $\chi$. -The reason one cares which version of part a is used, is that the numerics for small $k$ and moderate $x$ indicate that positivity is at least plausible for the original part a. It is not plausible for the revised version. -Moreover, several conjectures in the literature were known to imply the Riemann Hypothesis, but have since been disproved. Polya conjectured that -$$ -\sum_{nd^{-1/2}$, while Summation by Parts and the Polya-Vinograov inequality bounds the tail of the infinite series by -$$ -\sum_{xd^{-1/2}-\frac{5}{3}\cdot \frac{d^{1/2}\log(d)}{x}. -$$ -The above is positive for -$$ -x>\frac53 d\log(d),\quad\text{or}\quad d<\frac{3x}{5W(3x/5)}, -$$ -where $W(x)$ is the Lambert function, that is, the inverse function of $x=w\exp(w)$. Thus assuming one could show that -$$ -\sum_{n0 -$$ -for some very large $x$ (Haselgrove's disproof of Turan's conjecture suggests that $x=\exp(853)$ might be possible), one would be able (if the premise of the problem were correct) to rule out the possibility of an exceptional zero for a very large collection of $d$. For example, with $x=\exp(853)$ one would get all $d<2\cdot 10^{367}.$ -A similar argument with fixed $k>0$ would give still more $d$.<|endoftext|> -TITLE: Counting isomorphism classes via extensions -QUESTION [9 upvotes]: Given a group $Q$ and an abelian group $C$, I want to determine the number $I(Q,C)$ of isomorphism classes of all groups $G$ having a central subgroup $C'$ isomorphic to $C$ such that the quotient $G/C'$ is isomorhic to $Q$. Thus $I(Q,C)$ equals the number of groups (up to isomorphism), that fit into a central extension -$$ 1 \to C \to G \to Q \to 1$$ -The number of such extensions (up to equivalence) is given by $|H^2(Q,C)|$ (trivial coefficients). This gives an upper bound. But often it is much larger than $I(Q,C)$ (cf. the example below). -Question: Is it possible to obtain the correct value for $I(Q,C)$ out of $H^2(Q,C)$, or at least an improved upper bound ? -Example: Let $C_n$ denote the cyclic group of order $n$. Each $p$-group $G$ of order $n$ has a central subgroup $C_p$. Hence $I(G/C_p,C_p)$ is the number of isomorphism classes of $p$-groups of order $n$. -Take $n=2$. Then there are exactly two groups of order $p^2$, i.e $I(C_p,C_p) =2$, while -$|H^2(C_p,C_p)| = |\mathbb{Z}/p\mathbb{Z}| = p$. From these $p$ extensions, $p-1$ belong to the isomorphism class of $C_{p^2}$. - -REPLY [8 votes]: Let $Q$ be a group and $A$ a $Q$-module. Call two extensions $G, G'$ of $Q$ by $A$ weakly equivalent, if there is a commutative diagramm -$$ 0 \to A \to G \to Q \to 1 $$ -$$ \hspace{2pt} \downarrow \hspace{20pt} \downarrow \hspace{20pt} \downarrow $$ -$$ 0 \to A \to G' \to Q \to 1 $$ -with vertical isomorphisms. Denote the corresponding set of equivalence classes by $W(Q,A)$. Since $G,G'$ are isomorphic, if the extensions are weakly equivalent, $W(Q,A)$ is finer than isomorphism classes, but coarser than $H^2(Q;A)$. -In order to describe the relation between $W(Q,A)$ and $H^2(Q;A)$, some notation is needed: Call $(\varphi, \alpha) \in Aut(Q) \times Aut(A)$ compatible, if $\alpha(\varphi(q)\cdot a) = q \cdot \alpha(a)$ for all $q \in Q, a \in A$. Such a pair induces an automorphism $(\varphi,\alpha)^*$ of $H^2(Q;A)$ (see Brown: Cohomology of Groups, III, after Cor. 8.2). -Taking into account that cohomology is contravariant in the first argument, let $T\subseteq Aut(Q) \times Aut(A)$ be the subgroup of all pairs $(\varphi, \alpha)$ such that $(\varphi^{-1}, \alpha)$ is compatible. Then, $T$ operates on $H^2(Q;A)$ through $(\varphi,\alpha) \cdot x = (\varphi^{-1},\alpha)^*(x)$. Now, the central result is: - -There is a bijection between $W(Q,A)$ and the orbits of $H^2(Q;A)$ under - the action of $T$. - -The proof consists in essential of the fact, that for a 2-cocycle $f: Q\times Q \to A$ -and a compatible pair $(\varphi, \alpha)$, the extensions -corresponding to $f$ and $f':= \alpha \circ f \circ (\varphi^{-1} \times \varphi^{-1})$ -are weakly equivalent. -As noted above, weak equivalence is finer than isomorphism. But in some situations, $W(Q,A)$ will directly classify isomorphism classes. - -a) If the center of $Q$ is trivial, then $|W(Q,A)|=I(Q,A)$ (as defined in the question). - -b) Suppose $A$ is finite and let the integer $k$ be coprime to $|A|$. Thus, multiplication by $k$ is an automorphism of $A$ that is compatible with $\operatorname{id}_Q$. Hence, for $x \in H^2(Q;A)$, its orbit contains $kx$ for all $k$ comprime to $|A|$. In particular: - -If $|H^2(Q;A)|$ is a prime, then $|W(Q,A)| = 2$ and $I(Q,A) \le 2$. - -Hence, in your example ahead, there are at most two isomorphism classes of groups of -order $p^2$ and since $C_p \times C_p$, $C_{p^2}$ aren't isomorphic, we are done. -Edit: Including a proof of a) -Let $\mathcal{C}$ be the class of groups $G$, satisfying $Z(G) \cong A$ and $G/Z(G) \cong Q$. By fixing such isomorphisms each $G \in \mathcal{C}$ exhibits a central extension -$$\mathcal{E}_G: \quad\quad A \cong Z(G) \hookrightarrow G \twoheadrightarrow G/Z(G) \cong Q.$$ -The first key observation is: An isomorphism $\phi: G \to H$ maps $Z(G)$ isomorphically onto $Z(H)$ and therefore induces a commutative diagramm with vertical isomorphisms: -$$\mathcal{E}_G: \quad\quad A \cong Z(G) \hookrightarrow G \twoheadrightarrow G/Z(G) \cong Q$$ -$$\hspace{17pt} \phi \downarrow \hspace{17pt} \phi \downarrow \hspace{17pt} \bar{\phi} \downarrow $$ -$$\mathcal{E}_H: \quad\quad A \cong Z(H) \hookrightarrow H \twoheadrightarrow H/Z(H) \cong Q$$ -Hence $\mathcal{E}_G$ and $\mathcal{E}_H$ are weakly equivalent and we obtain a map -$$\mathcal{C}/\cong \to \mathcal{W}(Q,A),\quad [G] \mapsto [\mathcal{E}_G].$$ -Since a weak equivalence between $\mathcal{E}_G$ and $\mathcal{E}_H$ implies $G \cong H$, this map is injective. Surjectivity follows from the second key observation: Let -$$\mathcal{E}: \quad\quad A \hookrightarrow G \overset{\kappa}{\twoheadrightarrow} Q$$ -be a central extension, i.e. $A \le Z(G)$. Since $\kappa$ is epi, $\kappa(Z(G)) \le Z(Q) = 1$, implying $Z(G) = A$. Thus $G \in \mathcal{C}$ and $[\mathcal{E}_G] = [\mathcal{E}]$.<|endoftext|> -TITLE: How important is it for one on the job market to have thought about suitable REU projects? -QUESTION [16 upvotes]: An REU is a "Research Experience for Undergraduates" and entails undergrads (usually after their junior year) coming for 8-10 weeks out of the summer to University X to learn about and work on a problem under the supervision of one of the faculty at X. They are often funded by the NSF and the students get a stipend. REUs seem to be very popular among small colleges because they want to be able to sell themselves as a place undergrads can do research even if there are no grad students around. As a grad student who will be on the job market in the not-too-distant future, I'm starting to wonder about how best to prepare and how to fit REUs into my job application material. I work in stable homotopy theory, and I have never come across an open problem that I think an undergrad could do meaningful work on (ignoring the small set of really amazing undergrads who frequent MO and would probably not end up at my hypothetical REU). If I did come across such a problem, I'd probably solve it myself rather than hang on to it as a potential REU problem. -A bit more background: I will be applying for a large number and variety of jobs because with the job market as it is, it seems prudent to cast a wide net. I'll be applying for post-docs in the US and in Europe, and I'll also be applying for tenure track jobs at schools which are more focused (but not entirely focused) on teaching. Long-term I hope to end up at a good liberal arts college, but in the short term I'll look for post-docs to strengthen my research profile. If I end up in a post-doc in Europe, I assume they will want me to focus on research and not running undergraduate research projects. Thus, this question is more focused on applications to colleges where teaching matters at least as much as research. - -(1) should I spend some time seeking out a suitable REU problem in case a job asks me for project ideas? - -I have no idea how common it is to ask a job applicant this type of question, but it seems like the sort of thing a small teaching college might want to know. - -(2) Supposing that I did find a good problem and stopped myself from solving it, where would be the best place to display that in my job application material? - -I've read up on teaching statements (on MO and elsewhere), but it seems that is not really the place to discuss plans for REUs. Similarly, it shouldn't go in the research statement. Is this just something I bring to the interview and hope they ask about? - -(3) Do REU ideas need to be in the same field as my PhD research or could I just think up a good problem in a different field (e.g. graph theory where the background is more reasonable for an undergrad) - -I'm amazed this hasn't been asked on MO before. I hope it's suitable, since many grad students frequent this site and would I suspect care about questions they may be asked in the job process. If it has been asked, please forgive the duplication. All I could find were questions about the merits of research if you are an undergrad, where to find info on REU type experiences, and the pros/cons for publishing with undergrads if you're looking to get tenure. - -REPLY [19 votes]: 1) If you are genuinely seriously interested in mentoring undergraduates in REU projects, then you would naturally be fantasizing about projects you might do. If you aren't all that interested in it, you can't fake it just by thinking up some projects. Note that there is some middle ground between 'genuinely seriously interested' and 'not all that interested'. Thinking up projects does show that you are interested in it. The ones you think up don't need to be great; they just need to be not delusional. Keep in mind that weaker schools will only have someone who is potential grad student material only once every several years, so those schools, if they are interested in undergraduate research, will want projects that someone who is not grad student material can tackle. (As a rough and not particularly accurate guide, by 'weaker school' I mean median SAT of incoming freshman below 1150 or so for a small liberal arts college; larger universities may have some stronger students even if the average is somewhat lower.) People in applied and computational areas have a huge advantage here because they can propose research that does not involve proving any theorems. -The truth is that you'll probably have a failure or two before you have a success, and it's not a big deal unless it is a total failure where the student gets nothing out of it. So don't worry too much about the specifics of the project; the point of having one is to show you are serious and not completely delusional (given the school and its students) about it. -2) Unless the advertisement does not ask for a research statement or asks for undergraduate research to be addressed elsewhere, it goes in a section of your research statement. -Keep in mind that many schools that care seriously about undergraduate research will care only a little about your research except inasmuch as it generates projects for undergraduates. When applying for positions at such schools, you may need to reshape your research statement to say only generalities about your own research (since no one there will understand most of it anyway) and focus on the accessible parts and the undergraduate research. It may also be the case that such a school will not ask for or read a research statement, in which case you need to be prepared to put the material as a section in your teaching statement. -3) It absolutely does NOT have to be in your primary research area, as long as it is an area where you have enough of an idea what is going on to actually guide a project. (In other words, you should know enough that having the project turn out to be a fairly well-known solved problem is not at all a risk.) In fact, many professors at undergraduate institutions have gradually migrated out of their dissertation areas to research areas that are more friendly to their students.<|endoftext|> -TITLE: Ring with Z as its group of units? -QUESTION [19 upvotes]: Is there a ring with $\mathbb{Z}$ as its group of units? -More generally, does anyone know of a sufficient condition for a group to be the group of units for some ring? - -REPLY [18 votes]: The example provided by Noam answers the first question. The second question is very old and, indeed, too general. See e.g. the notes to Chapter XVIII (page 324) of the book "László Fuchs: Pure and applied mathematics, Volume 2; Volume 36". In particular, rings with cyclic groups of units have been studied by RW Gilmer [Finite rings having a cyclic multiplicative group of units, Amer. J. Math 85 (1963), 447-452], by K. E. Eldridge, I. Fischer [D.C.C. rings with a cyclic group of units, Duke Math. J. 34 (1967), 243-248] and by KR Pearson and JE Schneider [J. Algebra 16 (1970) 243-251].<|endoftext|> -TITLE: Including a Jordan arc into a Jordan loop (Can the Magi go home by another way?) -QUESTION [13 upvotes]: The title refers, of course, to Matthew (2:12) ''And being warned in a dream not to return to Herod, they departed to their own country by another way''. To be honest, it is not that specific particular case I'm more interested in. -I'd like to have a reference, or a hint here, for a simple proof of the following fact (intuitive, but not easy to proof, as usual in these matters). - -Let $\Gamma$ be a simple Jordan arc in - $\mathbb{R}^2$ (a homeomorphic image - of the interval $[0,1]$). Then, $\Gamma$ - can be included in a simple Jordan - loop $ \Sigma $ (a homeomorphic image - of $\mathbb{S}^1$). - -By the (generalized) Jordan's theorem, we know that $\mathbb{R}^2\setminus\Gamma$ is connected; and being open, it is even connected by piece-wise linear paths. The difficulty is that we need a path connecting the end-points of $\Gamma$; in other words, the question is how to show that $(\mathbb{R}^2\setminus\Gamma)\cup \partial\Gamma $ is path-connected (after that, an injective path could always be extracted). -It seems to me everything would follow easily from this lemma: - -Assume that $B(0,2)\setminus \Gamma$ - has at least two connected components - that meet $B(0,1)$. Then, there are - three consecutive points of $\Gamma$, - resp. $y_1$, $y_2$, and $y_3$ such - that $\|y_2\|=1$, and - $\|y_1\|=\|y_3\|=2$ - -I'm also a bit puzzled by the quantitative aspect of this problem: - -Assume that $\Gamma$ is parametrized - by a homeomorphism - $\gamma:[0,1]\to\Gamma$ with a modulus - of continuity $\omega(t)$ (say, a - continuous concave function vanishing - at $t=0$) and let $\omega_1$ be - another modulus of continuity such - that $\omega_1(t) > \omega(t)$ for $t > 0$. - Is there a Jordan loop $\Sigma$ with parametrization $\gamma_1[0,2 -> ]\to\Sigma$ that has modulus of - continuity $\omega_1$ ? - -REPLY [7 votes]: Yes they can! Every Jordan arc can be included in a Jordan loop. I'll give you the simplest proof I could think of, which only uses very basic topology. It also shows that the lemma you propose does indeed hold, although I'll skip the quantitative aspect. [I see that Jim Conant linked to a very similar question in a comment, but the proof I'll give here is very different from the answers given to the other question. It is also, in my opinion, more elementary, and I prove the lemma asked for here.] -The idea is to use the Jordan arc $\Gamma\subseteq\mathbb{R}^2$ to construct an undirected graph $G_\Gamma$ as follows. Write $S^1=\bar B_1(0)\setminus B_1(0)$ for the unit circle. Then, let the nodes $V_\Gamma$ of the graph consist of the connected components of $\mathbb{R}^2\setminus(\Gamma\cup S^1)$, which naturally split into those inside the circle and those outside. Let the edges $E_\Gamma$ of the graph consist of the connected components of $S^1\setminus\Gamma$. Each edge lies in the closure of precisely two elements of $V_\Gamma$ (one inside and one outside the circle), and we consider it to join these two nodes. Then, the main fact used to prove your lemma is the following. - -1) $G_\Gamma$ is a tree. - -I'll give a proof of this statement below. As it seems like a rather simple fact, there should be a reference somewhere - maybe some knows one? -Using this, we can easily dispatch with your lemma, which I'll state as follows. I'm using $\Gamma={\rm Im}(\gamma)$ for a continuous one to one map $\gamma\colon[0,1]\to\mathbb{R}^2$. - -2) If $\epsilon < 1$ and at least two connected components of $B_1(0)\setminus\Gamma$ meet $B_\epsilon(0)$ then there exists $t_1 < t_2 < t_3\in[0,1]$ with $\Vert\gamma(t_1)\Vert,\Vert\gamma(t_3)\Vert\ge1$ and $\Vert\gamma(t_2)\Vert < \epsilon$. - -If the curve $\Gamma$ did not intersect with $S^1$ then $G_\Gamma$ would only have one edge and, hence, only two nodes (the inside and outside of the circle), so only one region could intersect with $B_\epsilon(0)$. On the other hand, suppose that $\Gamma\cap S^1$ is nonempty. Then, let $t_1,t_3\in[0,1]$ be respectively the first and last times at which $\gamma(t)\in S^1$. Let $\gamma^\prime=\gamma\vert\_{[t_1,t_3]}$ and $\Gamma^\prime={\rm Im}(\gamma^\prime)$. -Note that $\Gamma\cap S^1=\Gamma^\prime\cap S^1$, so $E_\Gamma=E_{\Gamma^\prime}$. Also, we can map $V_\Gamma$ to $V_{\Gamma^\prime}$ by taking each connected component $U\in V_\Gamma$ to the connected connected component of $\mathbb{R}^2\setminus(\Gamma^\prime\cup S^1)$ containing it. This is one-to-one on the components outside the circle (as $\Gamma\setminus B_1(0)=\Gamma^\prime\setminus B_1(0)$. By (1), it must be one-to-one on all of $V_\Gamma$. Otherwise, if $U,V\in V_\Gamma$ mapped to the same element of $V_{\Gamma^\prime}$, then a simple path from $U$ to $V$ in $G_\Gamma$ would map to a closed loop in $G_{\Gamma^\prime}$, which must pass through a node outside of the unit circle (as any path alternates between inside and outside the circle), and only passes through this node once (as $V_\Gamma\to V_{\Gamma^\prime}$ is 1-1 on the regions outside the unit circle). This would contradict the statement that $G_{\Gamma^\prime}$ is a tree unless $U=V$. So, $G_{\Gamma^\prime}$ must also have at least two components intersecting with $B_\epsilon(0)$, so $\Gamma^\prime\cap B_\epsilon(0)\not=\emptyset$ and $t_2\in[t_1,t_3]$ exists as required. -As you suggest, this shows that any point $P\in\mathbb{R}^2\setminus\Gamma$ can be connected to $\gamma(0)$ by a by a curve not intersecting with $\Gamma$. Wlog, suppose that $\gamma(0)=0$ and let $t_n\in(0,1]$ be the first time at which $\gamma$ exits from $B_{1/n}(0)$. Then, choose points $P_n\in\mathbb{R}^2\setminus\Gamma$ with $\Vert P_n\Vert\le\min\_{t\ge t_n}\Vert\gamma(t)\Vert$. We can define $f\colon[0,1]\to\mathbb{R}^2$ by letting $f\vert\_{[1/2,1]}$ be a path in $\mathbb{R}^2\setminus\Gamma$ joining $P_2$ to $P$ and, for each $n\ge 2$, let $f\vert\_{[1/{(n+1)},1/n]}$ be a path in $B_{1/n}(0)\setminus\Gamma$ joining $P_{n+1}$ to $P_n$ (which exists, by the lemma). Then, setting $f(0)=0$, this gives the desired curve. - -I'll now give a proof of (1), that $G_\Gamma$ is a tree, based on the following facts. - -$\mathbb{R}^2\setminus\Gamma$ is connected. -If $P,Q,R,S$ are points occuring clockwise (in that order) on $S^1$ and $C_1,C_2\subseteq \bar B_1(0)$ are curves joining $P$ to $R$ and $Q$ to $S$ respectively, then $C_1$ and $C_2$ intersect. - -Neither of these facts requires anything particularly difficult. Both of them follow from the methods in the notes Brouwer's Fixed Point Theorem and The Jordan Curve Theorem from this link (Section 5) -- see the proof of Lemma 5.6 for the first fact and, reducing to the case $P=(0,1)$, $Q=(1,0)$, $R=(0,-1)$, $S=(-1,0)$, see Lemma 5.5 for the second fact. -The first fact immediately implies that $G_\Gamma$ is connected, so we just need to show that it has no simple closed loops. Suppose, on the contrary, that there was a simple loop with edges $E_1,E_2,\ldots,E_n$ (which are just open segments of $S^1$). Then, as any path alternates between inside and outside the unit circle, $n$ is even. So, the connected components of $S^1\setminus(E_1\cup\cdots\cup E_n)$ consists of $n$ closed connected segments $I_1,\ldots,I_{n/2},J_1,\ldots,J_{n/2}$. I'm labeling these alternately by $I_k$ and $J_k$ as you go round the circle. As each of these closed segments intersects with $\gamma$, there must exist $t_1,t_2\in[0,1]$ with $\gamma(t_1)\in\bigcup_k I_k$ and $\gamma(t_2)\in\bigcup_k J_k$. Wlog, suppose that $t_1 < t_2$. Also, choose $t_1$ maximal in the range $[0,t_2]$ and, then, choose $t_2$ minimal in $[t_1,1]$. Then, $\gamma^\prime=\gamma\vert\_{[t_1,t_2]}$ is a curve joining a point in $I_i$ to $J_j$ (say), and intersects $S^1$ only at its end points. Wlog, we can assume that $\Gamma^\prime={\rm Im}(\gamma^\prime)$ lies in $\bar B_1(0)$ (otherwise, circular inversion can be applied to reduce to this case). -Then, $S^1\setminus(I_i\cup J_j)$ consists of two connected open arcs each of which contains an odd number of the intervals $E_k$. However, the edges of $G_\Gamma$ joining a single node $V\in V_\Gamma$ must all lie in the same arc otherwise, by connectedness of $V$, we would contradict the second fact above. So, each pair of edges $E_k,E_{k+1}$ in the path joining to a node inside the circle lies in the same arc, and hence each arc must contain an even number of the edges $E_k$. This gives the desired contradiction.<|endoftext|> -TITLE: Physicist's request for intuition on covariant derivatives and Lie derivatives -QUESTION [48 upvotes]: A friend of mine is studying physics, and asks the following question which, I am sure, others could respond to better: -What is the difference between the covariant derivative of $X$ along the curve $(t)$ and a Lie derivative of $X$ along $y(t)?$ I know the technical stuff about not needing to define a connection with a Lie derivative, needing to define the fields $X$ and $Y$ over a greater neighborhood, etc. -I am looking for a more physical sense. If a Lie derivative gives the sense of the change of a vector field along the direction of another field, how does the covariant derivative differ? - -REPLY [5 votes]: I know I'm really late in the game here, but I'm teaching a differentiable manifolds class for the first time this semester and I have been thinking a bit about this question. At least for the case of a torsion-free connection, there is a certain intuitive picture of the relation between the Lie and covariant derivatives that I find appealing. -Let $X$ be a smooth vector field and let $\nabla$ be a torsion-free affine connection. Then $\nabla X$ defines a $C^\infty(M)$-linear endomorphism of vector fields: $Y \mapsto \nabla_Y X$. Therefore functorially $\nabla X$ induces $C^\infty(M)$-linear endomorphisms of tensor fields, that is, sections of $TM^{\otimes s} \otimes T^*M^{\otimes r}$. -The intuitive picture is that the covariant derivative is pointwise a sort of affine transformation, with the Lie derivative $L_X$ playing the part of the translation and $\nabla X$ playing the part of the linear transformation. For example, on vector fields, -$$ -\nabla_X Y = L_X Y + (\nabla X)(Y) = [X,Y] + \nabla_Y X -$$ -which is the torsion-free condition. This is more general than just vector fields, of course. If $T$ is any tensor, then -$$ -\nabla_X T = L_X T + (\nabla X)(T), -$$ -where $\nabla X$ acts on $T$ in the natural way.<|endoftext|> -TITLE: Longest element of a Weyl group -QUESTION [6 upvotes]: Let $G$ an algebraic (reductive) group. $T$ a maximal torus, $B$ a Borel subgroup containing $T$, and $w_0$ the longest element of the Weyl group. -I'm looking for a reference explaining why when you conjugate $B$ by $w_0$, the result is the opposite Borel subgroup $B^-$. -Is there a proof involving roots of $G$ relative to $T$ ? -I've found a proof in a book of M. Geck, but this proof doesn't involve roots at all, but only the fact that $B^-$ is uniquely defined by the relation $B \cap B^- = T$. - -REPLY [8 votes]: The role of the longest element in $W$ emerges only gradually in the Chevalley structure theory. This is developed similarly but in slightly different styles in the three books with the same title Linear Algebraic Groups (by Borel, Springer, and me). -It's important to understand that the desired statement about opposite Borel subgroups of a reductive group depends on a long series of steps culminating in the Bruhat decomposition. In my book, the later steps occupy Sections 26-28. -There's already a proof in 26.2 of the existence of a (necessarily unique) Borel subgroup intersecting a given Borel subgroup $B$ in a specified maximal torus $T$ of $B$. Theorem 26.3(b) almost gives the answer you want, but before enough details of the structure theory are in place. The underlying strategy is to fix $B$ and the flag variety $G/B$, then see that the fixed points of $T$ on the latter are in natural bijection with both the elements of $W$ and the Borels containing $T$. (I don't think this got articulated explicitly enough, however.) -Only in my Section 27 is the root system explored, followed in Section 28 by the full details of the Bruhat decomposition. There is still some work to be done -on the internal structure of $W$ (including the length function and longest element) in relation to the root system. In my treatment this gets folded into the more general study of $BN$-pairs (Tits systems), which axiomatize the Bruhat decomposition efficiently. Along the way it turns out for instance that $W$ has a natural structure of Coxeter group. -Unfortunately, the precise statement you want to see a proof of is left somewhat implicit in all these textbook treatments. But this is due partly to the fact that so much heavy theory has to be developed systematically before that kind of statement becomes obvious. It's hard to take any real shortcuts, but also unnecessary to get into the theory of buildings and such. -EDIT. The basic outline here looks deceptively simple: Start with the fixed data $B$ and $T$ and Weyl group $W$. Then $W$ has a unique longest element $w_0$, which has order 2 and interchanges positive and negative roots. Here the positive roots are defined by the choice of $B$, so that $B= T U$ with $U$ the product in any order of 1-dimensional root subgroups corresponding to positive roots. Then the Borel subgroup $B^- = w_0 B w_0$ contains $T$ and has the form $B^- =T U^-$ with $U^-$ the product of root subgroups for negative roots. -But to work this out for a reductive group over an algebraically closed field of arbitrary characteristic requires virtually all of the Borel-Chevalley structure theory through the Bruhat decomposition. For instance, it's highly nontrivial to find the root subgroups and define the root system intrinsically for the group, as well as to work out basic facts about the Weyl group $N_G(T)/T$ and its length function.<|endoftext|> -TITLE: To what extent can one prescribe degrees of irreducible representations of a group? -QUESTION [12 upvotes]: Suppose one starts with an (infinite) multiset of positive integers $\mathcal{A} = \{a_i\}_{i\geq 0}$ such that: -$1=a_0\leq a_1\leq a_2\leq\ldots$ -Can one always find a (necessarily infinite) group $G$ such that the set of degrees of its finite dimensional irreducible complex representations is exactly $\mathcal{A}$? -Clearly the answer to the above question is no for arbitrary $\mathcal{A}$; for example if $a_1 = N>1$ (i.e. the trivial representation is the only 1-dimensional representation of the potential group), then a simple argument involving the decomposition of the tensor square of this $N$-dimensional representation shows that $N\leq a_2\leq \frac{N^2+N}{2}$. Using more general forms of such arguments involving tensor powers of decompositions of $GL_n(\mathbb{C})$-representations, one can find other restrictions ruling out certain potential degree sequences. - -$\bf{Question:}$ Given $\mathcal{A}$ such that there are no obstructions arising from tensor power considerations, is it always possible to find/construct a group $G$ whose irreducible degree sequence is exactly $\mathcal{A}$? - -I assume that in general this is still a hard question, so I am also interested in partial results. I would also be interested in negative results along the lines "Even though this sequence $\mathcal{A}$ has no obstructions, it still cannot be the degree sequence of any group for some deeper reason." - -REPLY [2 votes]: There are lots of results like this, but they're pretty piecemeal. For example, the number of 1-dimensional representations divides the size of the group. There are some results about how large the largest dimensions can be (see Durfee & Jensen (J. Alg, 2011) for the most recent results). There are also some results about the small end of the distribution (e.g. if you have a 2-dimensional representation then you can use the classification of subgroups of SU(2) to get info). -In general if you write down a sequence that passes all the obvious tests then no one has any idea whether a group exists which realizes that.<|endoftext|> -TITLE: How to locate an obscure paper? -QUESTION [15 upvotes]: A colleague asked me to locate the following paper on the web: -Kovalenko, I.N.: On the reconstruction of an additive type of distribution based upon a sequence of independent trials. -Memoirs of the All-Union Conference on Probability Theory and Mathematical Statistics, Erevan 1958. -After a few failed attempts, I asked myself, how would I go about locating this paper in the REAL (not virtual) world? What library stocks proceedings of the All-Union Conference... in Erevan, from 1958? -Then it occurred to me that of course, the paper had to be in Russian. Being a Russian speaker, it wasn't too difficult to recover the original title: -Коваленко И.Н., О восстановлении аддитивного типа распределения по последовательной серии независимых испытан��й, Труды Всесоюзн. Совещания по теор. вероятн. и матем. стат., Ереван, 1958, Изд-во АН АрмССР, Ереван,1960. -So if any Russian-based readers out there have access to this archive, I would very much appreciate a PDF of this paper. -But I am more interested in the process. How does a researcher go about locating such things? - -REPLY [2 votes]: University of Yerevan has a copy (http://89.249.207.20/cgi-bin/koha/opac-detail.pl?biblionumber=156826&query_desc=kw%2Cwrdl%3A%20%D0%A1%D0%BE%D0%B2%D0%B5%D1%89%D0%B0%D0%BD%D0%B8%D1%8F%20%D0%BF%D0%BE%20%D1%82%D0%B5%D0%BE%D1%80.%20%D0%B2%D0%B5%D1%80%D0%BE%D1%8F%D1%82%D0%BD.%20%D0%B8%20%D0%BC%D0%B0%D1%82%D0%B5%D0%BC.%20%D1%81%D1%82%D0%B0%D1%82), and you can obtain a copy from them for a fee. -Contact their librarian directly (mirzoyan@ysu.am). I don't know at this stage in history whether English or Russian is the preferred international language of Armenian librarians, but you are clearly fluent in both.<|endoftext|> -TITLE: Geometric interpretation of matrix minors -QUESTION [9 upvotes]: i was recently interested in geometric interpretation of various notions showing up in linear algebra because in most cases linear algebra with geometry courses focus too much on linear algebra not giving rationale or intuitions behind ideas hinting that lecturer doesn't have much idea about mathematics involved staying only on the surface of whole matter (maybe only due to didactic purposes). it's quite funny that answer doesn't show up anywhere on the network or in textbooks: in most cases i find this style very poor showing somewhat that author of the book / script might not have full grasp of it neither. -this can only mean that it hasn't got much of geometric interpretation which i'm eager to reject because i haven't had found in my first handbooks any intuitions regarding determinant or trace (those problems where addressed on mathoverflow.net already: volume of oriented parallelotope and measure of change of edges in respect to edges of unit hypercube, respectively) but what i'm interested most at the moment is: - -what is the geometrical interpretation of matrix minor? - -i can only guess that it gives measure of some object of lower dimension for parallelotope, it sides for minor of highest but one degree; i could only guess that minors are projections on respective axes or something but i'm not sure of my interpretation. probably laplace expansion could give decisive answer but i also fail to interpret it properly; other clue might lie in expanded characteristic polynomial as for then it's coefficients are sums of all principial minors (of degree equal to degree of the term as far as i know). one can also find them in search of invertible coordinates in inverse and implicit function theorems. you're welcome to give interpretation of those using given intuitions! ;) - -REPLY [12 votes]: A minor is a matrix element of the action of a linear transformation $T : V \to W$ on exterior powers $\Lambda^k(T) : \Lambda^k(V) \to \Lambda^k(W)$. The geometric entity is the entire linear transformation $\Lambda^k(T)$ (or its trace, which is what appears in the characteristic polynomial), which describes how $T$ acts on oriented paralleletopes of various intermediate dimensions.<|endoftext|> -TITLE: Computer platforms for combinatorial search problems/mathematical music theory? -QUESTION [5 upvotes]: I'm finding programming various combinatorial searches -(connected to mathematical music theory) in a general purpose computer -language tedious, so I'd like pointers to computer platforms/environment purpose-built to enable such searches. If the answer has the form "Maple can -do that" or "Mathematica can do that," then I'd appreciate pointers specific -enough to get me started. -Within reason I would trade ease of specification against -any optimization of the search algorithm, but I'd feel grateful for speed if I could have that too. -Here's an example of a typical question; I'm really -hoping for an environment where I can specify such a search almost as succinctly -as I now explain it (I use terminology from mathematical music theory -but define everything in purely mathematical terms): -Write $T$ for the set of complex $12$th roots of unity. -Tetrachord means a subset of $T$ of cardinality $4$. Row means a permutation of T. -Fix two distinct rotation classes of tetrachords, $a$ and $b$. By definition row $R\in G_{a,b}$ if applying $R$ to the 24 tetrachords -in $a \cup b$ yields 24 tetrachords pairwise distinct up to isometry. -Find $\bigcup_{a,b} G_{a,b}$. -Thanks in advance for any help!! - -REPLY [3 votes]: A possible solution could be Strasheela, a library of music-related stuff built upon the Mozart programming system which is an interactive environment for the Oz programming language, a multi-paradigm programming language that supports constraint programming. The website states many uses, from Fuxian counterpoint to harmonic analysis but also realtime generation of rhythmic patterns. And due to the integration of many output formats (Lilypond for typeset music, MIDI for ugly beeps, Csound for less ugly beeps, Fomus for other compositional tools, ...) it will be easy to actually use your results. -The problem you describe sounds like it's doable in Strasheela (based on my own tiny bit of experience in it). It will also be able to find all solutions if you ask it to do so, but it might take some time (or be untractable) depending on the problem and the size of the solution space. -The idea of the language is different from your combinatorial approach, but it has great support of music theory and finding stuff, so you can use the terms for music theory without translating everything to a more mathematically oriented lingo. You'll have to phrase your actual problem in another way though: less combinatorial and more artificial intelligence-ish. - -REPLY [2 votes]: A student of mine wrote a progam alg which enumerates models of first-order theories (although it works best for equational theories). Specification is easy, but the program can't beat specialized enumeration techniques. If you can write down your problem without referring to actual complex numbers, it just might do it.<|endoftext|> -TITLE: Degree of maps on the sphere with a property of symmetry -QUESTION [6 upvotes]: Suppose a continuous map $f : S^2 \rightarrow S^2$ verifies : -$$ f(x) - f(-x) = 2 \left(f(x),x\right)x $$ -where $\left(x,y\right)$ is the scalar product in $R^3$. An equivalent way of expressing $f$ could be : -$$ f(x) = X(x) + u(x)x $$ -where $X$ is a tangent vector field of $S^2$ and $u$ a real-valued function with : -$$ X(-x)=X(x) \hspace{15mm} u(-x) = u(x) \hspace{15mm} \left|X\right|^2+u^2 = 1 $$ -For instance, the identity and antipodal maps are solutions of degree 1 and -1. My question is : -Do there exist solutions $f$ of degree $0$ ? -Edit : The answer is no, solutions have odd degree. What about the result for dimension n>2 ? - -REPLY [11 votes]: First, for any $x\in S^2$ we have an endomorphism $A(x)$ of $\mathbb{R}^3$ given by $A(x)(w)=x\times w$. More generally, we have an orthogonal matrix $B(t,x)=\exp(t A(x))$, which is a rotation through angle $t$ around $x$. When $w$ is perpendicular to $x$ we have $B(\pi/2,x)(w)=A(x)(w)$. -Let $F$ be the space of maps $f$ as in the question, and let $G$ be the space of maps $g:S^2\to S^2$ satisfying $g(-x)=-g(x)$ for all $x$. Given a function $f(x)=X(x)+u(x)x$, put $\phi(f)(x)=A(x)(X(x))+u(x)x=x\times X(x)+u(x)x$. This gives a homeomorphism $\phi:F\to G$. Using the maps $x\mapsto B(t,x)(X(x))+u(x)x$ (for $0 \leq t\leq \pi/2$) we see that $\phi(f)$ is homotopic to $f$ and so has the same degree. It is fairly standard that maps in $G$ have odd degree, and it follows that maps in $F$ have odd degree. -UPDATE: Romain asks about a generalisation for $n>2$. This seems harder. The analogous space $F$ is the space of sections of the bundle $W=n+2-L$ over $\mathbb{R}P^n$, where $L$ is the tautological bundle. By thinking about exterior algebras, one can produce an isomorphism $2^nL\simeq 2^n$. In the case $n=2$ this gives $4L\simeq 4$ so $W=4-L=4L-L=3L$, and by working out explicit formulae for this identification one is led to the proof above. If $n>2$ then $2^n>n+2$ so this approach will not work without some additional ideas.<|endoftext|> -TITLE: Thom's Principle: rich structures are more numerous in low dimension -QUESTION [15 upvotes]: Marcel Berger states Thom's Principle as: - -"rich structures are more numerous in low dimension, - and poor structures are more numerous in high dimension." - -This is in -Geometry II -(Springer-Verlag, Berlin, 1987. - -Google books), pp.39-40, after a discussion of regular polytopes -shows that there are only three convex regular polytopes in dimensions -larger than 4, but six in dimension 4, five in dimension 3, and -an infinite number in dimension 2. -He then lists further examples of the principle: -e.g., simple Lie groups illustrate rich structure in low dimensions, -and topological vector spaces illustrate poor structure in -high dimension (all homeomorphic in finite dimensions). -In so far as Thom's principle is true—or at least holds widely—my question is: - -Why should low dimensions exhibit richer structure - than high dimensions? Is there any generic reason to expect this? - -It might also be interesting to track down Thom's own formulation -of his principle, to understand the context in which he proposed it. -Edit. Here is a snapshot of Berger's examples (p.40): - -REPLY [4 votes]: I am reminded of the fact that for geometries of dimension strictly higher than 2, the Desarguesian and Pappian configurations are equivalent, so that (in my view) there is a greater degree of uniformity in higher dimensional geometric spaces. I think Thom's principle (as noted in another post) is a selection principle and tells the reader what Thom thinks of by rich structure. If structures are essentially (in mathematical logic, some notion of conservative) extensions of incidence relations, then I am not too surprised that something like this principle should hold, since a certain amount of uniqueness is lost in a structure that should be invariant under more geometric transformations. I am not a geometer, though, so go ask someone like Hrushovksii what they think (from a mathematical logic perspective) of Thom's principle. -Gerhard "Ask Me About System Design" Paseman, 2011.09.13<|endoftext|> -TITLE: Not quite regular polyhedra -QUESTION [7 upvotes]: Take a naive interpretation of regular polyhedra: -All vertices (including epsilon ball) congruent -All edges congruent -All faces congruent -We can now find interesting families by removing one requirement. For example the uniform polyhedra have all vertices and edges congruent, but not all faces, and their duals have faces and edges congruent, but not vertices. -Are there examples, or interesting families, of polyhedra where every pair of faces is congruent and every pair of vertices, but not every pair of edges? - -REPLY [3 votes]: Allan Edmonds has a few papers studying "equifacetal simplices", i.e. simplices in which any two facets must be congruent. Here are references: -Edmonds, Allan, The center conjecture for equifacetal simplices. Adv. Geom. 9 (2009), no. 4, 563--576. -Edmonds, Allan L., The partition problem for equifacetal simplices. Beitrage Algebra Geom. 50 (2009), no. 1, 195-213. -Edmonds, Allan, The geometry of an equifacetal simplex. Mathematika 52 (2005), no. 1-2, 31-45.<|endoftext|> -TITLE: flatness in complex analytic geometry -QUESTION [18 upvotes]: It is always a pain to move back and forth between definitions in algebraic geometry and complex analytic geometry. Dictionary is much easier when are working with (family of) smooth varieties but the pain grows exponentially when we include singular varieties. -Here is some of that: -Suppose $f: X\rightarrow Y$ is a map of possibly singular complex analytic varieties, then is there a simpler definition of flatness for $f$ ? this one should be hard to answer but how about following, ----Let $f:X\rightarrow Y$ be a family of curves. there might be multiple fibers, non-reduced fibers, nodal curves, cusp curves,... in the family, but fibers are connected. ----When this fibration is flat? What kind of bad fibers mentioned above are allowed in a flat family? ----Same question for higher dimensional family of analytic varieties? -For simplicity you may assume that the base of fibration is smooth. - -REPLY [52 votes]: Instead of trying to say what flatness in analytic geometry means I'll give you some street-fighting tricks for recognizing whether a morphism of analytic spaces ( not necessarily reduced) $f:X\to Y $ is , or has a chance to be, flat. -a) A flat map is always open. . Hence, contraspositely, the embedding -$\lbrace 0\rbrace \hookrightarrow \mathbb C$ is not flat. More generally, the embedding of a closed (and not open !) subspace $X \hookrightarrow Y$ is never flat. -b) An open map need not be flat: think of the open map $ Spec(\mathbb C) \to Spec(\mathbb C[\epsilon ]) $ from the reduced point to the double point, which is open but not flat. - [It is not flat because the $\mathbb C[\epsilon]$- algebra $\mathbb C=\mathbb C[\epsilon]/(\epsilon)$ is not flat : recall that a quotient ring $A/I$ can only be flat over $A$ if $I=I^2$ and here - $I=(\epsilon) \neq I^2=(\epsilon)^2=(0)$] -An example with both spaces reduced is the normalization [see g) below] $f:X^{nor} \to X $ of the cusp $X\subset \mathbb C^2$ given by the equation $y^2=x^3$ .That normalization is a homeomorphism and so certainly open, but it is not flat : this results either from g) or from h) below. -c) Given the morphism $f:X\to Y $ , consider the following property: - $\forall x \in X, \quad dim_x(X)=dim_{f(x)} (Y) +dim_x(f^{-1}(f(x)))$ $\quad (DIM) $ -We then have: $ f \; \text {flat} \Rightarrow f \;\text { satisfies } (DIM)$ -For example a (non-trivial) blowup is not flat. -d) For a morphism $f:X\to Y $ between connected holomorphic manifolds we have: -$$f \text { is flat} \iff f \text { is open} \quad \iff (DIM) \;\text {holds}$$ -For example a submersion is flat, since it is open. -e) Given two complex spaces $X,Y$ the projection $X\times Y\to X$ is flat ( Not trivial: recall that open doesn't imply flat!). -f) flatness is preserved by base change. -g) The normalization $f:X^{nor} \to X $ of a non-normal space is never flat. -For example if $X\subset \mathbb C^2$ is the cusp $y^2=x^3$, the normalization morphism $\mathbb C\to X: t\mapsto (t^2,t^3)$ is not flat. -h) Given a finite morphism $f:X\to Y $, each $y\in Y$ has a fiber $X(y) \subset X$ and for $x\in X$ we can define $\mu (x)= dim_{\mathbb C} (\mathbb C \otimes_{\mathcal O_{Y,y}} \mathcal O_{X,x})$ . Now for $ y\in Y$ we put $\; \nu (y)= \Sigma_{x\in X(y)} \mu(x)$ and we obtain : -$$f \; \text{ flat} \iff \nu :Y\to \mathbb N \text { locally constant}$$ -[Of course for connected $Y$, locally constant = constant] -Bibliography: - A. Douady, Flatness and Privilege, L'Enseignement Mathématique, Vol.14 (1968) - G.Fischer, Complex Analytic Geometry, Springer LNM 538, 1976<|endoftext|> -TITLE: The duel problem -QUESTION [9 upvotes]: The following duel problem is due to Ben Polak (maybe there's earlier origin, which I'll be glad to be informed about). The rule is as follows: -Two players 1 and 2 start a duel $N$ steps away from each other. They take turns to act. When it's somebody's turn, he must make one of the two choices: -Choice A: Shoot at his opponent with probability ${p}_{i}(d)$ of hitting the target, where $i=1,2$, and $d$ is distance (measured in steps) between them. -Choice B: Forsake the opportunity to shoot, and make one step forward toward his opponent. -Now the distribution of ${p}_{i}(d)$ is such that ${p}_{i}(0)=1$, and ${p}_{i}(d)>{p}_{i}(d+1)$, $d=0,1,2,...,N-1$, $i=1,2$. There're no other restrictions. Both players are assume to be rational and intelligent. A player's goal is to maximize his probability of killing the opponent. Player 1 act first. - -My question is: For all possible distributions of ${p}_{i}(d)$, $i=1,2$ described above, is there a simple and uniform decision rule according to which both players can make their choices at each distance? -(For example, the decision rule could be something like: "if ${p}_{1}(d)+{p}_{2}(d-1)>1$, then player 1 should shoot at distance d when it's his turn to move; otherwise step forward") - -Edit: the original statement "a player's goal is to maximizing his surviving probability" is changed to "A player's goal is to maximize his probability of killing the opponent", due to Emil. - -REPLY [6 votes]: Note that the distinction between the objectives of survival and killing the opponent can be important. Suppose some $p_i(n) = 0$ (for both 1 and 2) while $p_i(n-1) > 1/2$. The player -who steps forward first is very likely to be killed, so to maximize your probability of survival your best strategy at distance $n$ is always to shoot. The opponent, also wanting to survive, will also shoot, and the game will go on forever without anybody getting hurt. -But if your objective is to kill the opponent, this strategy is clearly sub-optimal: it would be better to step forward and have a positive probability of killing the opponent. But that's not optimal either: there's no need to step forward right away, you could wait a while in the hope that the opponent steps forward first. Waiting $k+1$ turns before stepping forward dominates waiting $k$ turns, so there is no optimal strategy. -To avoid such problems, let's assume $p_i(d) > 0$ for all $d$. This will ensure that the probability of both players surviving indefinitely is 0. -Then an optimal strategy can be found using dynamic programming. Let $V_i(d)$ be player $i$'s probability of winning under optimal strategies, starting with distance $d$ and $i$'s turn to shoot. -Then $V_i(0) = 1$, otherwise $V_i(d) = \max(1 - V_{3-i}(d-1), W_i(d))$, where $W_i(d)=\min\left(\frac{p_i(d)}{p_1(d) + p_2(d) - p_1(d) p_2(d)}, p_i(d) +(1 - p_i(d)) V_i(d-1))\right)$. It is optimal to shoot if $W_i(d) > 1 - V_{3-i}(d-1)$, to step forward if $\lt$, and both are equally good if $=$. - -REPLY [4 votes]: Let $p_n,q_n$ be the two players chance of hitting at distance $n$, and $P_n,Q_n$ their chance of winning under optimal play if it is their turn. Then $P_0=Q_0=1$ and we have the recursion -$ -P_n = \max(Q_{n-1}, p_n + \overline{p_n} \overline{Q_n}) -$ -and similarly with p,q switched. Plugging the formula for $Q_n$ into that for $P_n$ gives an identity for $P_n$ in terms of $p_n,q_n,P_{n-1},Q_{n-1}$ that is easy to solve, and has a unique solution. The optimal strategy given $P_n,Q_n$ is obvious. -(Above, $\bar{x}= 1-x$.)<|endoftext|> -TITLE: When does the relative differential $df=0$ imply that $f$ comes from the base? -QUESTION [20 upvotes]: Let $A \to B$ be a map of commutative rings, and $d : B \to I/I^2$ be -defined by $df = f\otimes 1 - 1\otimes f$, where $I$ is the kernel of -$B \otimes_A B \to B$, as in [Hartshorne II.8]. - -If $df=0$, I would like to infer that $f \in A$, i.e. "if the - derivative is zero, the function is constant". - -This is certainly $\bf false$, -e.g. $A = {\mathbb F}_p$, $B = A[x]$, and $f=x^p$. -There $B\otimes_A B\to B$ is $A[x_1,x_2] \mapsto A[x]$, with -$I = \langle x_1 -x_2\rangle \ni x_1^p - x_2^p = f\otimes 1 - 1\otimes f$. -(That is, not only is $f\otimes 1 - 1\otimes f$ in $I^2$, it's in $I^p$.) - -What is the right condition, then, on $A\to B\ $? My primary interest is - in $A = {\mathbb Z}[1/d]$. - -REPLY [30 votes]: I will rather regard $I/I^2$ as $\Omega_{B/A}$, the module of differential forms. -First some necessary conditions. If $D$ is a sub-$A$-algebra of $B$ such that $\Omega_{D/A}=0$ (e.g. $D$ is a localization of $A$ or étale over $A$), the canonical map $\Omega_{D/A}\otimes_D B\to \Omega_{B/A}$ shows that $df=0$ for all $f\in D$. So if $A$ is a field (of characteristic $0$), you want it to be algebraically closed in $B$. If $A$ is not a field, to avoid $B$ to contain a localization of $A$, you almost want to suppose $\mathrm{Spec} B\to \mathrm{Spec}A$ be surjective. -These being said, now a sufficient condition. - -Suppose $A$ is noetherian, integrally closed of characteristic $0$, $B$ is an integral domain, $\mathrm{Spec} B\to \mathrm{Spec}A$ is surjective, and its generic fiber is smooth of finite type and geometrically integral. Then the kernel of $B\to \Omega_{B/A}$ is equal to $A$. - -Proof. -(1) Let $C$ be the kernel of $d : B\to \Omega_{B/A}$. This is a sub-$A$-algebra of $B$. The canonical exact sequence -$$ \Omega_{C/A}\otimes_C B\to \Omega_{B/A}\to \Omega_{B/C}\to 0$$ -implies that $\Omega_{B/A}\to \Omega_{B/C}$ is an isomorphism because the first map is identically zero ($dc\otimes b\mapsto bdc=0$ if $c\in C$). -(2) Let $K=\mathrm{Frac}(A)$, $L=\mathrm{Frac}(B)$. Let us first show that $E:=\mathrm{ker}(L\to\Omega_{L/K})$ equals to $K$. Note that $E$ is a field. Applying (1) to the situation $A=K$ and $B=L$, we see that $\Omega_{L/K}\to \Omega_{L/E}$ is an isomorphism. This map is $L$-linear, and $\dim_L\Omega_{L/K}$ is the transcendental degree of $L$ over $K$ (here we use the characteristic zero hypothesis). Similarly for $L/E$. Hence $E$ is algebraic over $K$. But $B_K$ is geometrically integral over $K$, this forces $E=K$ (otherwise $L\otimes_K E$ is not integral). -(3) Let us show $C_K:=C\otimes_A K$ equals to $K$. Tensoring the exact sequence of $A$-modules -$$ 0\to C\to B\to \Omega_{B/A}$$ -by $K$, we get the exact sequence -$$ 0\to C_K \to B_K \to \Omega_{B_K/K}. $$ -First suppose $B_K$ is smooth over $K$. As $\Omega_{L/K}$ is a localization of $\Omega_{B_K/K}$ and the latter is a free (hence torsion free) $B_K$-module, the map $\Omega_{B_K/K}\to \Omega_{L/K}$ is injective. By (2) we then get $C_K=K$. In the general case, some dense open subset $U$ of $\mathrm{Spec}(B_K)$ is smooth. If $f\in B_K$ satisfies $df=0$ in $\Omega_{B_K/K}$, then $d(f_{|U})=0$ in $\Omega_{U/K}$. So $f{|_U}\in K$. As $B_K\to O(U)$ is injective, $f\in K$. -(4) Let us show $C=A$. We have $C\subseteq B\cap K$ (in fact the equality holds). So we have to show $B\cap K=A$. Let $g\in B\cap K$ viewed as a rational function on $\mathrm{Spec}(A)$. Then $A\subseteq A[g]\subseteq B$. For any prime ideal $Q$ of $B$, $g$ is regular at the point $Q\cap A$. So the image of $\pi : \mathrm{Spec}(B)\to \mathrm{Spec}(A)$ is contained in the complementary of the pole divisor of $g$. So the latter is empty by the surjectivity hypothesis on $\pi$. Therefore $g$ is a regular function (here we use the normality of $A$) and $C=A$. -Add Suppose $A, B$ are integral domains of characteristic $0$, $B_K$ is finitely generated over $K=\mathrm{Frac}(A)$, and $\Omega_{B_K/K}$ is torsion free over $B_K$. Then we proved above that the kernel of $d: B\to\Omega_{B/A}$ is contained in $B\cap K^{alg}$. -Add 2 One can remove free or torsion-free condition on $\Omega_{B_K/K}$. The proof is a little modified in the step 3. We just notice that any integral variety in characteristic $0$ has a dense open subset which is smooth !<|endoftext|> -TITLE: Approximating high-dimensional integrals by low-dimensional ones -QUESTION [5 upvotes]: This question is motivated by the following naive one: suppose we have a nice subset $X$ of some Euclidean space, say a polyhedron, and a nice $\mathbb{R}$-valued function $f$ on this subset, say a polynomial. Is it possible to deduce the value of the integral of $f$ along $X$ with respect to the Lebesgue measure from the integrals of $f$ along sets of arbitrarily small Hausdorff dimension? One way to make this precise is as follows. -Let $A_t,0< t\leq 1$ be the subset of $[0,1]$ obtained from $[0,1]$ by first removing the middle $1-t$ part, then removing the middle $1-t$ parts of the resulting two segments, then removing the middle $1-t$ parts of the resulting four segments and so on. (The middle $1-t$ part of a segment $[a,b]$ is the open interval of length $(1-t)(b-a)$ with center $\frac{b-a}{2}$.) Here are some remarks: - -If $t=\frac{2}{3}$, then $A_t$ is the Cantor middle third set. -The Hausdorff dimension of $A_t$ is $\frac{\ln 2}{\ln 2-\ln t}$. - -Now let $f:[0,1]\to\mathbb{R}$ be some ``nice'' function. Set $g(t)$ to be the average of $f$ over $A_t$. Recall that if $A$ is a compact metric space and $f:A\to\mathbb{R}$ is continuous, then the average of $f$ over $A$ is the limit as $i\to \infty$ of the averages of $f$ over $B_i$ where $(B_i)$ is a sequence of finite sets that converges to $A$ in the Hausdorff metric; recall also that the Hausdorff distance between two closed nonempty $A',A''\subset A$ is $$max(min_{a\in A'}dist (a,A''),min_{a\in A''}dist (a,A')).$$ -Is it possible to explicitly compute $g(t)$, say when $f(x)=x^n$ with $n$ a non-negative integer? If not, what can one say about this function? Is it analytic in $t$? If so, does it extend analytically to $\mathbb{C}\setminus (-\infty,0]$? -[upd: I would also be happy with answers to the same questions for $f(x)=e^{nx}$ or $e^{inx}, n\in\mathbb{Z}$.] - -REPLY [5 votes]: Here's how to average $e^{sx}$ over $A_t$. Let $r = t/2$, so at the $N$-th step of the construction of $A_t$ we have the disjoint union $A_t^{(N)}$ of $2^N$ intervals of length $r^N$ whose left endpoints are $\sum_{n=1}^N (r^{n-1}-r^n) \phantom. \epsilon_n$ with each $\epsilon_n \in \lbrace0,1\rbrace$. Thus to choose uniformly at random from $A_t^{(N)}$ we choose for each $n=1,2,\ldots,N$ either $0$ or $r^{n-1}-r^n$ with probability $1/2$, sum these $n$ terms, and add a real number chosen uniformly at random from $[0,r^N]$. This is a convolution of $N$ discrete measures and a single continuous one, so the average of $e^{sx}$ over $A_t^{(N)}$ is the product of $N$ finite sums and one integral: -$$ -\prod_{n=1}^N \frac{1+\exp((r^{n-1}-r^n)s)}{2} \cdot \frac1{r^N} \int_0^{r^N} e^{sx} dx \phantom. . -$$ -letting $N \rightarrow \infty$ we find that the average of $e^{sx}$ over $A_t$ is -$$ -I_s(r) = \prod_{n=1}^\infty \frac{1+\exp((r^{n-1}-r^n)s)}{2} \phantom. . -$$ -(Check: for $r=0$ this is just $(1+e^{s})/2$ because $A_0 = \lbrace 0,1 \rbrace$; and for $r=1/2$ it's $(e^{sx}-1)/s$ because $A_1$ is the interval $[0,1$].) -To average polynomials over $A_t$, it is enough to average $x^k$ for $k=0,1,2,\ldots$. -To compute these averages we expand $I_s(r)$ in a power series about $s=0$. It's easier to do this with the logarithm: -$$ -\log I_s(r) = \sum_{n=1}^\infty \phantom. \log\frac{1+\exp((r^{n-1}-r^n)s)}{2} = \sum_{n=1}^\infty \phantom. \lambda((r^{n-1}-r^n)s), -$$ -where -$$ -\lambda(z) := \log \frac{1+e^z}{2} = \frac{z}{2} + \frac{z^2}{8} - \frac{z^4} {192} + \frac{z^6}{2880} - \frac{17z^8} {645120} + - \cdots. -$$ -Thus the $s$ coefficient of $\log I_s(r)$ is -$\frac12 \sum_{n=1}^\infty (r^{n-1} - r^n) = 1/2$; the $s^2$ coefficient is -$\frac18 \sum_{n=1}^\infty (r^{n-1} - r^n)^2 = \frac18 (1-r)/(1+r)$; the $s^4$ coefficient is $-\frac1{192} (1-r)^3 / (1+r+r^2+r^3)$; and the $s^3$, $s^5$, $s^7$, ... coefficients vanish. So -$$ -I_s(r) = \exp\left( \frac{s}{2} \phantom. + \phantom. \frac{1-r}{1+r} \frac{s^2}{8} \phantom. - \phantom.\frac{(1-r)^3}{1+r+r^2+r^3} \frac{s^4}{192} \phantom.+\phantom. O(s^6) \right). -$$ -From this we may recover the average of $x^k$ over $A_t$ by extracting the $s^k$ coefficient and multiplying by $k!$. If I did this right, the average comes to $1/2$ for $k=1$ (of course), then $1/(2+t)$ for $k=2$, and -$$ -\frac{4-t}{8+4t}, \phantom{\infty} -\frac{8+2t^2-t^3}{(2+t)^2(4+t^2)},\phantom{\infty} -\frac{(4-t)(4+2t^2-t^3)}{(2+t)^2(4+t^2)} -$$ -for $k=3,4,5$. I did at least check that for $t=1$ we recover $1/2,1/3,1/4,1/5,1/6$, and for $t=0$ each average is $1/2$.<|endoftext|> -TITLE: What is the "reason" for modularity results? -QUESTION [14 upvotes]: The question is a little wishy-washy, but I take my cues from other popular questions that relate to the philosophy behind the mathematics as Why do Groups and Abelian Groups feel so different? . -I am aware of the statements of class field theory and the modularity theorem, as well as far-reaching generalizations that have to do with the conjectural Langlands group and motives. But on a basic level, I just don't understand why such statements should be true, other than that there is a lot of evidence that they are. -What is the philosophical impetus behind modularity results? -When I read about number theory, I can very easily understand the intuition behind ramification of primes (because the intuition is geometric), but as soon as we start talking about splitting of primes, and are therefore in the realm of modularity results, I lose all intuition of why things should be true (even though I can read and understand the results as an undergraduate can -- agreeing line by line). -An example of an answer for CFT can be the following thing I've heard, but was somewhat unsatisfied with because I didn't fully understand it: that it grew out of generalizations of Fourier analysis. (if you also think of this as the reason it's true, and can expatiate -- do!) - -REPLY [7 votes]: No one knows. Or at least, no one knows why we know. -I do not mean this flippantly. If you study the proofs of e.g. local and global class field theory (global especially), they use over and over again all kinds of tricks in the yoga of group cohomology to reduce everything down to understanding cases we can do by hand, like cyclotomic extensions of $\mathbf{Q}$ and Kummer extensions more generally. But I think this style of proof is very far from a satisfying "why", and I have heard the same opinion from other people (Tate, Rosen). The most satisfying proof in class field theory for me is the Lubin-Tate construction of totally ramified extensions of local fields, precisely because you can make canonical choices and it's reasonably explicit. -Likewise, the Taylor-Wiles method, while an extremely beautiful and powerful idea, is ultimately unsatisfying (to me) as a reason for why Hecke algebras should match deformation rings. If you read the "context-free version" in Section 2 of Diamond's paper "The Taylor-Wiles method and multiplicity one", you'll notice that a subsequence of (quotients of) the auxiliary modules $H_n$ is chosen using compactness. Compactness! Roughly speaking, this corresponds to controlling the relation between deformation rings and Hecke rings at some fixed level by smooshing together a bunch of modular forms at sporadic higher levels, which are chosen in some gratuitously noncanonical way. -I agree with Emerton that converse theorems provide an extremely persuasive reason for believing modularity results. But I also believe that the ultimately "correct" method of proof has not surfaced yet, and who knows how many decades or centuries until it does? -(Let me stress that these are simply my opinions, and nothing more. But they are not unconsidered.)<|endoftext|> -TITLE: A differential inequality needed to prove a theorem about odd-dimensional souls -QUESTION [9 upvotes]: I need a solution to this problem (which is really a calculus problem) in order to prove a rigidity result for open nonnegatively curved manifolds with odd-dimensional souls: -Suppose that $f,g:\mathbf{R}\rightarrow \mathbf{R}$ are smooth functions. Assume that $f(0)=0$ and that $g(t)$ has a global maximum at $t=0$. Assume for all $t\in\mathbf{R}$ that: -$$f'(t)^2 \leq f(t)^2 + g''(t).$$ -Prove that $f$ and $g$ must both be constant functions. -By the way, this is simple to prove if $g$ is analytic. In this case, the fact that $g$ has a global maximum implies that $g''(t)\leq 0$ on an interval about $t=0$. So you have $f'(t)^2\leq f(t)^2$ which implies that $f(t)=0$ on this interval (establishing the result on an interval is enough). - -REPLY [16 votes]: We may assume $g(0)=g'(0)=0$. -For $t\in(0,1)$, we have -$$ - \int_0^t (f(s)^2+g''(s))\,ds \geq \int_0^t f'(s)^2\,ds - \geq \left(\int_0^t f'(s)\,ds\right)^2 \left(\int_0^t 1^2\,ds\right)^{-1} - =f(t)^2/t\geq f(t)^2, -$$ -hence $g'(t)\geq f(t)^2-\int_0^t f(s)^2\,ds$. Let $h(t)=\int_0^t f(s)^2\,ds$; we have $g'(t)\geq h'(t)-h(t)$. Since $h$ is monotonous we get $g(t)\geq h(t)-\int_0^t h(s)\,ds\geq h(t)-th(t)\geq 0$. So, on our interval we have $g(t)=h(t)=f(t)=0$.<|endoftext|> -TITLE: Coin pusher game -QUESTION [11 upvotes]: While doing laundry at my local laundromat, I saw a coin pusher game. Below is a picture, and here is a video depicting how it works (disregard non-coins). - -Essentially, one has a distribution of coins on a table, and you get to drop one coin at a time at one end, which ends up being pushed into the table, thereby potentially pushing coins off the edge. Note that you can choose where you can drop your coin, width wise. For simplicity, assume coins cannot stack on each other. -My question is, are there known limit laws for this game? That is, if I specify a distribution of coins on the table, and then start dropping coins in randomly, what can be said about how the expected number of dropped coins fluctuates, per turn. Consequently, are there various phase transitions as a function of coin density? As well, if I feed coins at a specific spot, what will the distribution of coin falls look like as a function of the table width? Do the boundary conditions (the side walls and the pusher) create interesting "modes" in the coin falling distribution? -I would think that this has to do with sand stacking cascades and KPZ growth but, do not have much experience in this area. Or perhaps this is just a simple Galton box that produces a normal distribution? - -REPLY [5 votes]: I frequent a restaurant that has one of these machines. It has an upper level that moves toward the player then backwards. You can drop one quarter at a time through a movable slot so that you can drop several coins onto the upper shelf during one cycle. -The guys who work there have figured out pretty effective tactics and strategies for winning. -Strategy: Only play when quarters are built up heavily in the middle of the edge. (The sides don't matter.) The quarters should look like they will fall at any moment. Some may be largely over the edge but stopped from falling by other quarters lying on them. -Tactic: During one cycle, drop four quarters as close together as you can get them at the center of the machine. They act like a bulldozer to push coins in the center of the upper shelf forward. These, in turn, fall to the the lower shelf and push coins in the center toward the edge. This strategy minimizes movement of coins toward the edge slots and maximizes the flow of coins forward. -Sometimes they make quite a killing.<|endoftext|> -TITLE: Characterizing Groupoids via Quotients? -QUESTION [19 upvotes]: A groupoid is a category in which all morphisms are invertible.(*) The groupoids form a very nice subclass of categories. The inclusion of the groupoids into the 2-category of small categories admits both left and right (weak) adjoints. So you can localize (or complete) a category to a groupoid. If E denotes the free walking isomorphism, then the groupoids are precisely the categories which are local with respect to the inclusion of the free walking arrow (let's denote this as [1]) into E. -I have been thinking a lot recently about a certain very nice class of categories, which in many ways is the antithesis of the groupoids. These are the categories X, such that a morphism is an isomorphism in X if and only if it is an identity. Let us call these categories lean.(**) These too form a nice class of categories. The inclusion admits a left adjoint, and the lean categories are precisely those categories which are local for the projection from [1] to the terminal category pt. -What is this left adjoint? (We'll denote it as L). It forms a certain quotient category. You can see it explicitly by taking your given category and forming the quotient where you identify all isomorphisms with identities. -In some cases it is easy to work out what you get. For example if you have a groupoid X and you apply L, you get a discrete category (set) consisting of the components of X. -More generally you have to be more careful. Quotienting out by just the isomorphisms is not compatible with the composition in your category, and so the result is not a category. To get the actual categorical quotient you have to quotient by additional things: compositional consequences of the initial equivalence relation. -For example suppose that f,g,h are morphisms in your category X, where g and fgh are isomorphisms. Then for any lean category Y and any functor $X \to Y$, both g and fgh must map to identities in Y. Hence the composite fh must also map to an identity, even if it is not an isomorphism of X. It is not hard to construct an example where this happens. Nevertheless we can ask the following - -Question: Suppose that X is a category such that the lean quotient L(X) is a set (discrete category). Is X necessarily a groupoid? - -If true, I think this would give a very neat characterization of the groupoids. I have so far been unable to prove this result or come up with a counter example, but I feel I have been staring at the problem too long. It is enough to prove just the case that L(X) is just a single point. The above discussion shows that in general the quotient $X \to L(X)$ identifies things which are not isomorphisms, which suggests the answer is negative. A positive answer would have to somehow make use of the fact that everything is being identified. -(*) For this question when I speak of categories I mean small categories. -(**) If anyone knows of a reference for these categories, I would be interested to know it. - -REPLY [8 votes]: Here's a different (as far as I can see) proof for the same example Todd gave, the monoid of endofunctions of $\mathbb N$. Consider the following elements of this monoid: - -$b$ is the bijection that interchanges $2n$ with $2n+1$ for all $n$. -$e:n\mapsto 2n$ and $f:n:\mapsto 2n+1$. -$p$ maps even numbers by $2n\mapsto n$ and maps all odd numbers to 0$. -$z$ is the constant zero map. - -Then we have - -$f=b\circ e$, -$p\circ e=1$, and $p\circ f=z$. - -In the quotient monoid where all isomorphisms are identified, $b$ becomes 1, so the first of these equations gives $f=e$, and then the remaining equations give $1=p\circ e=p\circ f=z$. But then, for any $x$ in the monoid, we have in the quotient that $x=1\circ x=z\circ x=z$.<|endoftext|> -TITLE: Did any new mathematics arise from Ruffini's work on the quintic equation? -QUESTION [40 upvotes]: The impossibility of solving the general polynomial of degree $\ge 5$ by radicals is surely one of the most celebrated results in algebra. This result is known as the Abel-Ruffini theorem, although it's usually asserted that Ruffini's proof was incomplete. -Still, Paolo Ruffini's contributions to algebra seem to be neither widely known nor well understood. This is perhaps not so surprising since Ruffini's first attempted proof spanned 516 pages and the mathematical argument was difficult to follow. -A modern discussion of Ruffini's proof is found in Ayoub's article 'Paolo Ruffini's contributions to the quintic'. Accoring to Ayoub, Ruffini's argument was not a priori flawed, but it relies on several unproved non-trivial claims. More precisely, Ruffini fails to prove that the splitting field is one of the fields in the tower of radicals which corresponds to a solution expressed in radicals. -The revolutionary proofs of Abel and Galois following Ruffini paved way for group theory and Galois theory. Still, one could wonder what Ruffini actually proved in those 516 pages: - -Did any significant new mathematical concepts, -ideas or theorems arise from Ruffini's -work on the quintic? - -Cauchy seems to be one of the few mathematicians who found inspiration from Ruffini's work. In a letter dated 1821, he writes: -"... your memoir on the general -resolution of equations is a work which has always seemed to me worthy of the -attention of mathematicians and which, in my judgement, proves completely the -insolvability of the general equation of degree $>4$. [...] In another memoir which I read last year to the Academy of Sciences, I cited your work and reminded the audience that your proofs establish the impossibility of solving equations algebraically ..." -In fact, according to Pesic, Cauchy's influential 1815 paper on permutations is clearly based on work of Ruffini. So are there other examples of his influence? - -REPLY [10 votes]: Pierre L. Wantzel acknowledges the works of Niels H. Abel and Paolo Ruffini in the introduction to his complete proof of the insolubility of higher polynomial equations: In meditating on the researches of these two mathematicians, and with the aid of principles we established in an earlier paper, we have arrived at a form of proof which appears so strict as to remove all doubt on this important part of the theory of equations. -Heinrich Burkhard in Die Anfänge der Gruppentheorie holds the opinion that work of the Italien mathematician Pietro Abbati Marescotti could have been inspired by Paolo Ruffini's results. Vice versa it cannot be excluded that Abbati mentioned the first ideas of group theory to Ruffini, who subsequently expanded it. -Whether Ruffini himself or his friend Abbati had the first idea to apply group theory, partially based upon Lagrange's work on permutations, is not clear from their preserved correspondence but since Ruffini determined nearly all subgroups of the symmetric group $S_5$ his result belongs to the foundation of group theory. So his influence extends to class field theory (although modern terminology is connected with the name of Abel). This is certainly an accomplishment important enough to be called "significant new mathematical concepts or ideas".<|endoftext|> -TITLE: the complexity of Lanczos method -QUESTION [6 upvotes]: Hi, all -I am working on an algorithm which uses Lanczos method to compute K smallest eigenvalue(and their eigenvectos) of a sparse matrix, just want some information or links about the complexity of Lanczos method. -Thank you - -REPLY [5 votes]: Complexity analysis of Lanczos seems to be hard to find in the literature. Here are two leads, that might help a bit. -For the largest eigenvalue, you might find the complexity analysis in the following paper to be useful. -Estimating the Largest Eigenvalue by the Power and Lanczos Algorithms with a Random Start by J. Kuczyński and H. Woźniakowski -A very important point raised in the above paper is that even if an eigenvalue is easily estimable, its corresponding eigenvector can be computationally hard to estimate! -Also see Lemma 2 of Fast algorithms for approximate semidefinite programming ... by S. Arora, E. Hazan, and S. Kale, for additional results. - -REPLY [2 votes]: Chapter 9 = Lanczos Methods, of "matrix computations", Golub, van Loan, John Hopkins univ press.<|endoftext|> -TITLE: Exponential sums over finite fields with even characteristic -QUESTION [6 upvotes]: I am looking for an elementary evaluation (if one exists) of the exponential sum -$$ -G_r(a,b) = \sum_{x \in \mathbb{F}_{2^r}} \psi(ax^2 + bx), -$$ -where $a,b \in \mathbb{F}_{2^r}^*$ are both units, $\psi(x) = e(Tr(x)/2)$ and $Tr : \mathbb{F}_{2^r} \to \mathbb{F}_2$ is the usual field Trace map -$$ -Tr(x) = \sum_{i=0}^{r-1} x^{2^i}. -$$ -It should be noted that -$$ -G_r(a,0) = G_r(0,a) = 0, -$$ -since the map $x \mapsto x^2$ permutes the elements of $\mathbb{F}_{2^r}$. -I feel that such a sum must have surely been studied before, but I am having trouble both evaluating the sum and finding references for it. Short of an explicit formula for the sum, any information (or any reference to where this sum might be studied) would be appreciated. I found no information on this sum in the usual suspects: Ireland-Rosen, Iwaniec-Kowalski and "Gauss and Jacobi Sums," by Berndt, Evans, and Williams. - -REPLY [19 votes]: Trace is additive, and ${\rm Tr}(u)={\rm Tr}(u^2)$ for all $u$, so $ax^2+bx$ has the same trace as $(a+b^2)x^2$. Therefore the sum is $2^r$ if $a=b^2$ and zero otherwise. -In general, for a polynomial $P(x)$ over the field of $2^r$ elements, the sum of $\psi(P(x))$ is $2^r$ less than the number of affine points on the "hyperelliptic" curve $y^2+y=P(x)$. Here $P(x) = ax^2+bx$, so (for much the same reason I gave above: polynomials $\eta^2+\eta$ can be absorbed into $y^2+y$) the curve is rational, with $2^r$ points, unless $a=b^2$ when it is the union of two disjoint lines and has $2^{r+1}$ points.<|endoftext|> -TITLE: Calculate the group cohomology classes $H^d[U(1)\rtimes Z_2, Z]$ and $H^d[U(1)\rtimes Z_2, Z_T]$ -QUESTION [7 upvotes]: I would like to know what are the group cohomology classes $H^d[U(1)\rtimes Z_2, Z]$ and $H^d[U(1)\rtimes Z_2, Z_T]$, and/or how to calculate them. -It can be shown that $H^d[U(1), Z]$ is $Z$ for even $d$ and 0 for odd $d$. -Here $\rtimes$ is the semidirect product: $T U T = U^{-1}$ for $U \in U(1)$, where -$T$ is the generator of $Z_2$. -In $H^d[U(1)\rtimes Z_2, Z]$, the module $Z$ is the trivial module. -In $H^d[U(1)\rtimes Z_2, Z_T]$, the module $Z_T$ is still the Abelian group $Z$, - but $U(1)\rtimes Z_2$ has a non-trivial action on $Z_T$: -$(U,1) a = a$ and $(U,T) a = -a$, $a \in Z_T$. -Thanks! - -REPLY [9 votes]: The group $U(1) \rtimes \mathbb{Z}/2$ you describe is nothing but the group $O(2)$ (as $U(1) = SO(2)$). -As such I think one can see the spectral sequence for the extension does collapse, and one obtains -$$H^*(BO(2);\mathbb{Z}) = \mathbb{Z}[x_2, x_3, x_4]/(2x_2, 2x_3, x_3^2-x_2x_4).$$ -Here we can take $x_2 = \beta(w_1)$ and $x_3 = \beta(w_2)$, the Bocksteins on the Stiefel--Whitney classes, and $x_4 = p_1$ the Pontrjagin class. -If you allow me to write $\mathbb{Z}^-$ for your $\mathbb{Z}_T$, there is a short exact sequence of $\mathbb{Z}[\mathbb{Z}/2]$-modules -$$0 \to \mathbb{Z} \overset{1 \mapsto 1+T}\to \mathbb{Z}[\mathbb{Z}/2] \to \mathbb{Z}^- \to 0$$ -and $H^*(BO(2); \mathbb{Z}[\mathbb{Z}/2]) = H^*(BSO(2);\mathbb{Z}) = \mathbb{Z}[e_2]$ is a polynomial algebra on the Euler class. The long exact sequence on cohomology gives the following short exact sequences, once you know that $x_4 \mapsto e_2^2$ under -$H^*(BO(2); \mathbb{Z}) \to H^*(BO(2); \mathbb{Z}[\mathbb{Z}/2])$ (since $p_1 = e^2 \in H^4(BSO(2);\mathbb{Z})$): -\begin{eqnarray*} -0 \to H^{4i}(BO(2); \mathbb{Z}^-) \to H^{4i+1}(BO(2); \mathbb{Z}) \to 0 \\ -0 \to H^{4i+1}(BO(2); \mathbb{Z}^-) \to H^{4i+2}(BO(2); \mathbb{Z}) \to 0 \\ -0 \to \mathbb{Z}\langle e_2^{2i+1} \rangle \to H^{4i+2}(BO(2); \mathbb{Z}^-) \to H^{4i+3}(BO(2); \mathbb{Z}) \to 0\\ -0 \to H^{4i+3}(BO(2); \mathbb{Z}^-) \to H^{4i+4}(BO(2); \mathbb{Z}) \to \mathbb{Z}\langle e_2^{2i+2}\rangle \to 0\\ -\end{eqnarray*} -One can read off the groups $H^*(BO(2);\mathbb{Z}^-)$ from these short exact sequences. -(It is probably more efficient to describe $H^*(BO(2);\mathbb{Z}^-)$ as a module over $R := H^*(BO(2);\mathbb{Z})$.)<|endoftext|> -TITLE: Relative version of Symplectic Thom conjecture. -QUESTION [10 upvotes]: Ozsváth and Szabó proved Symplectic Thom conjecture [Annals of Mathematics, 151(2000), 93-124]. It states: An embedded symplectic surface in a closed, symplectic 4-manifold is genus-minimizing in its homology class. -Does a suitable relative version of above hold true ? More specifically, suppose we start with an embedded symplectic surface $\Sigma $ with boundary in symplectic 4-manifold with contact type boundary then is it true that $\Sigma $ is genus-minimizing in its (relative) homology class? - -REPLY [13 votes]: I think this must be a consequence of the version of the slice-Bennequin inequality proved by Mrowka and Rollin (but I might be wrong). Perhaps the argument also requires the boundary to have the "strong filling" property (Stein near the boundary). -Given a Legendrian knot $L$ in (to start with) the 3-sphere $S^3$, that inequality says -$$ - 2 g_*(L) - 1 \ge tb(L) - r(L). -$$ ` -Every transverse knot $K$ is a push-off of a Legendrian approximation $L$, and the self-linking number of $K$ is related to the invariants of $L$ by -$$ sl(K) = tb(L) - r(L). $$ -So the slice-Bennequin inequality says -$$ - 2 g_*(K) -1 \ge sl(K). -$$ -Unless I'm mistaken, this inequality is an equality for the case of a transverse knot bounding a symplectic surface in the 4-ball. -All this generalizes to the case of a symplectic 4-manifold $X$ that is Stein near its (contact) boundary. Given a Legendrian knot $L$ in the boundary, and a homology class $s$ of surfaces in $X$ with boundary $L$, one has invariants $tb(L,s)$ and $r(L,s)$. Then there is an inequality (as in Mrowka-Rollin), -$$ - 2 g_*(L,s) - 1 \ge tb(L,s) - r(L,s), -$$ -where $g_*(L,s)$ is the smallest possible genus of a surface in the class $s$ with boundary $L$. In terms of a transverse push-off $K$, one again has -$$ - 2 g_*(K,s) - 1 \ge sl(K,s). -$$ -If $K$ is actually the transverse boundary of a symplectic surface, then one has equality, this being (I think) just the adjunction formula in a relative version. -The Mrowka-Rollin version of the result can today be deduced from the existence of concave fillings (caps). We may assume from the outset that $tb(L,s)$ is positive: if it is not, we may sum in a bunch of Legendrian trefoils until it is. Now enlarge $X$ by first adding a 2-handle along $L$ (standard contact surgery) and then closing it up with a concave filling. The inequality one wants is just the adjunction inequality applied to the homology class formed from $s$ and the 2-disk in the core of the handle. -So with less notation, the answer to the original question is supposed to be: take a Legendrian approximation to the transverse knot and alter things so as to make $tb$ postive; then add a 2-handle and a cap to get a closed symplectic manifold. Then apply the adjuntion inequality to the homology class formed from the symplectic surface and the Lagrangian 2-disk.<|endoftext|> -TITLE: Group cohomology of compact Lie group with integer coeffient -QUESTION [5 upvotes]: It is known that group cohomology class $H^d[U(1),Z]$ is Z for even d and 0 for odd d. -Do we know $H^d[G,Z]$ for $G=SO(3)$, $SU(2)$ and other compact Lie group? -Also is the Borel-group-cohomology class $H^d[G,R]$ alway trivial for compact Lie group $G$? - -REPLY [11 votes]: For the group $SU(2)=S^3$ we just have $H^*(BSU(2);\mathbb{Z})=\mathbb{Z}[c_2]$ (where $c_2\in H^4$). More generally, for all $n$ we have -\begin{align*} - H^*(BU(n);\mathbb{Z}) &= \mathbb{Z}[c_1,\dotsc,c_n] \\\\ - H^*(BSU(n);\mathbb{Z}) &= \mathbb{Z}[c_2,\dotsc,c_n] \\\\ - H^*(BSp(n);\mathbb{Z}) &= \mathbb{Z}[p_1,\dotsc,p_n] -\end{align*} -with $c_i\in H^{2i}$ and $p_i\in H^{4i}$. -Now let $V$ be the tautological $3$-plane bundle over the space $X=BSO(3)$. This has Stiefel-Whitney classes $w_2\in H^2(X;\mathbb{Z}/2)$ and $w_3\in H^3(X;\mathbb{Z}/2)$. There is also a Bockstein element $v=\beta(w_2)\in H^3(X;\mathbb{Z})$ (which satisfies $2v=0$) and a Chern class $c=c_2(\mathbb{C}\otimes V)\in H^4(X;\mathbb{Z})$. The mod two reduction map $\rho$ satisfies $\rho(v)=Sq^1(w_2)=w_3$ and $\rho(c)=w_2^2$. If I've got everything straight, one can check using the Bockstein spectral sequence that -$$ H^*(BSO(3);\mathbb{Z}) = \mathbb{Z}[v,c]/(2v). $$ -It is not possible to be similarly explicit about $H^*(BSO(n);\mathbb{Z})$ for general $n$ (although $H^*(BSO(n);\mathbb{Z}/2)$ and $H^*(BSO(n);\mathbb{Q})$ are fairly straightforward).<|endoftext|> -TITLE: Does isomorphic generic fibre imply isomorphic special fibre for smooth morphisms? -QUESTION [14 upvotes]: Let $X$ and $Y$ be regular integral Noetherian schemes. Assume that $X$ and $Y$ are smooth and proper over a base scheme $S=Spec R$, where $R$ is a discrete valuation ring. -If $X$ and $Y$ have isomorphic generic fibres, is it also the case that their special fibres are isomorphic? -Remarks: - -The answer is yes when $X$ and $Y$ are abelian schemes (this follows from the theory of the Néron model). In general though it is not the case that morphisms between the generic fibres extend to the special fibre. -I am also particularly interested in case where $R$ is the localisation of $\mathbb{Z}$ at some prime, and hence the generic fibres are smooth proper varieties over $\mathbb{Q}$. - -REPLY [7 votes]: In a more positive direction, if the special fiber $X_s$ is not uniruled ruled, then $X_s$ is birational to $Y_s$. -This can be found in a paper of Ulf Persson (memoirs of AMS 1977) perhaps over complex numbers. But the idea works over excellent rings (e.g. localization of finitely generated $\mathbb Z$-algebras): consider the graph $\Gamma$ of the birational map from $X$ to $Y$ and normalize it. By a theorem of Abhyankar, any irreducible component of the exceptional locus of $\Gamma\to Y$ is ruled, hence doesn't dominate $X_s$ (because they would be birational). Therefore $\Gamma\to X$ and $\Gamma\to Y$ are isomorphic over the generic points of $X_s$ and $Y_s$. So $X_s$ is birational to $Y_s$.<|endoftext|> -TITLE: Do Denominator Vectors Determine the Cluster Variable -QUESTION [6 upvotes]: Given a cluster algebra $A=A(\mathbf{x},Q)$, the Laurent Phenomenon states that all the cluster variables of $A$ are Laurent polynomials in the elements of $\mathbf{x}$. Thus, any cluster variable $y$ can be written $$y=\frac{p(x_1,\dots,x_n)}{x_1^{d_1}\cdots x_n^{d_n}}$$ where $p$ is a polynomial and $d_i$ are positive integers. We call $d(y):=(d_1,\dots,d_n)$ the denominator vector of $y$. -If $Q$ is mutation equivalent to a simply laced Dynkin diagram, all cluster variables are uniquely determined by their denominator vector. I would like to know to what extent this holds in general. That is: -Is it true that for any cluster algebra, the clusters are determined by their denominator vectors? If not, what classes of cluster algebras have this property? I am particularly interested in surface cluster algebras. - -REPLY [2 votes]: For acyclic initial seeds, this is now Corollary 8 of Rupel and Stella arXiv:1712.08478 .<|endoftext|> -TITLE: Things to keep in mind while looking for a Postdoc overseas -QUESTION [15 upvotes]: Hello, -I would like to receive some suggestions about what you think to be the best important things that should be kept in mind while looking for a postdoc position. I'm not considering (in this question) the opportunities one may have from direct knowledge, e.g. because they were pointed out from your advisor or in any case about positions in some nearby university where you have some contact. -For instance while it is often not written explicitly, for most positions the candidate is expected to have taken personal contacts with one professor that may be interested, and set up a research project together. But this is certainly not the only matter, and I would like to hear some story from who has some experience in this kind of issues, also for what regards things which can vary a lot depending on the country (i did my Master and I am doing Ph.D. in Italy, with a short period in France). -To make the question more focused, i'm mostly interested in the scenario of someone working in pure mathematics (e.g. number theory) making application from Europe to some university in North America, or possibly also in Asia. -(EDITED as suggested by quid, making the question CW, and more focused and about data rather than about advice). - -REPLY [7 votes]: To complement Neil's answers: At Oxford and Cambridge you will find things called "Junior Research Fellowships". These are funded by and administered by the colleges (and not the mathematics department). They could be in any subject, or in, say, "Science and Mathematics", or even just "Mathematics" (but it's very unlikely to be more specific). Probably the college will rotate which subject(s) it wants over the years. -These position receive a very large number applications, and so some sort of filtering system along the lines Ben suggests must happen. Having strong references is perhaps the most important thing. -(I had one of these. I didn't know anyone at the college before, but in the end, a mathematics fellow at the college in question worked in a field rather close to mine, and interviewed me, so presumably understood my work and was sufficiently impressed by it. But the other postdoc in my year at the same college was in a different area, and the college simply invited in an expert from outside to interview him. So my strong advice would be to apply for everything, and get lucky once!) -The best guide to these things remains, I think, Tom Korner's: http://www.dpmms.cam.ac.uk/~twk/fellow.pdf -I think a few other places in the UK (Warwick? Imperial College London?) might offer one postdoc position a year under a similar format (any subject, best candidate wins)??<|endoftext|> -TITLE: Exact Value of a Series -QUESTION [7 upvotes]: It is very easy to show that the series -$$\frac{1-1/2}{1\times2} - \frac{1-1/2+1/3}{2\times3} + \frac{1-1/2+1/3-1/4}{3\times4} - ...$$ -i.e. -$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}[1-\frac{1}{2} + \frac{1}{3} - ...+ \frac {(-1)^{n}}{n+1}]$$ -is convergent. Can one find its exact value? Or is it unreasonable to hope for such a thing? -Thank you for your answers. - -REPLY [15 votes]: The interior sum is equal to $\int_0^1\frac{1-(-x)^{n+1}}{1+x}dx$ and -$$ -\sum _{n=1}^{\infty } \frac{(-1)^{n+1} \left(1-(-x)^{n+1}\right)}{n (n+1) (x+1)}= -\frac{(x-1) \log (1-x)-x+\log (4)-1}{1+x}. -$$ -So the answer is eqaul to -$$ -\int_0^1 \frac{(x-1) \log (1-x)-x+\log (4)-1}{1+x}dx=\frac{\pi ^2}{6}+\log ^2(2)-2. -$$<|endoftext|> -TITLE: Commutative algebra final project -QUESTION [7 upvotes]: I'm looking for a topic for a final project in commutative/homological algebra, for first year master's students (in a decent European university). During the course, they will cover the following topics: commutative rings (as in chapter 1 of Atiyah-McDonald), general module theory and structure of finitely generated modules over a PID, tensor products, basic category theory - including, products, coproducts, Yoneda - complexes and (co)homology, derived functors, flat, injective and projective modules, first properties of Tor and Ext. Not included will be: Artinian modules and length, notions of dimension, completion, localization, valuation rings, regular local rings (though DVR's, Nakayama are fine). -The goal would be to prove some nice result and possibly introduce some new notions along the way - everything in the form of (rather long) series of exercises - without having to develop too much big machinery or new theory. The project should take about 15-20 hours of work. Of course the topic could be (part of) one of the topics which I mentioned above as "not treated in class". Ideally it should be a "synthesis" and use a lot of the techniques learned in the course. -Any suggestions? Classical theorems, things extracted from recent research... -I'm looking forward to your suggestions! - -REPLY [7 votes]: this is similar to some other answers. -when i took basic graduate algebra from Maurice Auslander he handed out 16 pages of very terse notes the first day that he said was our Fall semester final exam. There were four sections and each of us was assigned to read, learn and write up in more detail one section. They were on i) depth, ii) modules of finite projective dimension, iii) regular local rings, iv) unique factorization domains. -To give an idea of the style, the first sentence defined M depth N (modules over any ring) to be (when finite) the smallest degree such that Ext(M,N) is non zero. One page later he proved this integer (if finite) equals the length of a maximal N regular rad(A) sequence where A = ann(M), if R is noetherian and M,N finitely generated. -In the second section he defined projective dimension and related it for fin gen modules over noetherian local rings to the length of a minimal free resolution and the non vanishing of Tor. He then proved the formula relating depth and the dimensions of R,M. He used without proof facts such as tensoring with flat algebras commutes with Ext and (if faithfully flat) leaves projective dimension unchanged. -In section 3 he characterized when noetherian local rings are regular in terms of global dimension, projective dimension of modules, and regular sequences, and equated global dimension with krull dimension for such rings. -In the last section he showed every regular local ring is ufd, and the formal power series ring over any regular ufd is also regular ufd. -I did not yet know what an ideal was when I started the semester. You never forget a class like that.<|endoftext|> -TITLE: Giving a math talk with no blackboard or projector -QUESTION [12 upvotes]: I need to give a math talk to a group of undergraduates. I am asking for advice because this talk will take place at a department picnic and there will be no blackboard or projector. I would like to talk about graph theory/combinatorics and I could conceivably use props or handouts. Does anybody have any suggestions? - -REPLY [2 votes]: Thank you to everybody for the great suggestions. I have decided what to do (see below) and I'll let you know how it goes after the picnic next week. I was inspired by the audience participation aspect of Josephus problem suggestion. -I will ask for a group of say 11 volunteers and ask them to shake hands with each other so that they all shake exactly three hands. Of course this is impossible and I will explain why. I may ask them if they can do this with an even number of people and have them try to come up with the construction. -Next I will describe a triple handshake (a group hug might be awkward...) and ask a group of 7 volunteers if they can do triple handshakes so that every pair of people are involved in exactly one triple handshake. Of course this is challenging without thinking about it for a while, but I will show them how to do it after they try. Then I will ask them if this can happen with 8 people. Again this is impossible and I will explain why. Now that we know there can be no "special triplehandshake system" (STS) on an even number of vertices I will explain why it can't happen with some odd numbers, like 11 for instance.<|endoftext|> -TITLE: Has this "pseudo-quotient" of groups been studied before? -QUESTION [8 upvotes]: Let $G$ be a (finite) group with subgroup $H$. Pick out (left) coset representatives: $x_1=1, x_2, \dots, x_\ell$ (so $x_1H=1H=H$). Now form an $\ell \times \ell$ table with rows and columns labeled by $H, x_2H, \dots, x_\ell H$. -In addition group cosets together according to their double cosets. -For example, if $Hx_2H=x_2H \cup x_3H \cup x_4H$ we group cosets 2, 3, and 4 together, so we would have something like $(H)$ $(x_2H \; x_3H \; x_4H)$ $(x_5H \cdots$ -This (parenthesized) list forms the first row of our table. -The rest of the rows are formed by left multiplying by each representative in turn. So, for example, -if $Hx_2H=x_2H \cup x_3H \cup x_4H$, we would have -$$\begin{array}{rcccccc} -\mbox{Row }x_1H: & \quad & (H) & (x_2H & x_3H & x_4H) & (x_5H & \cdots \\\\ -\mbox{Row }x_2H: & \quad & (x_2H) & (x_2x_2H & x_2x_3H & x_2x_4H) & (x_2x_5H & \cdots \\\\ -\mbox{etc.} \end{array}$$ -Notice that in row $xH$, two cosets $yH$ and $zH$ lie in the same parenthesized group if and only if $x^{-1}yH$ and $x^{-1}zH$ lie in the same double coset. Put another way, cosets $yH$ and $zH$ are connected in row $xH$ if and only if $xHx^{-1}yH = xHx^{-1}zH$. So the parentheses in row $xH$ are recording something about the action of $xHx^{-1}$ on the left cosets of $H$. -Of course if $H$ is a normal subgroup, this is just the Cayley table for the quotient group $G/H$ (with some extra parentheses). -When $H$ is not normal, switching representatives necessarily changes the order in which cosets appear in rows 2+ but the parenthesized groups stay intact. This pseudo-quotient gets as close to a group structure as one can hope for when $H$ fails to be normal. -My question, does this look familiar to anyone? References? -Now for some much needed explanation. About one year ago I got a call out of the blue by someone living close to my university who said he had a short proof of the Feit-Thompson odd order theorem. -I was quite skeptical but told him to email me his proof, I'd look it over and we could talk. It turns out he is a retired mathematician and who has worked off-and-on on his proof for 30+ years as a hobby. While I still have serious doubts his "proof" can be adequately formalized and/or corrected, I did find his fundamental object of study interesting -- these "pseudo-quotients" which he call his "diagrams". -These pseudo-quotients allow one to almost quotient groups even when there are no normal subgroups around. Using these pseudo-quotients, I thought we had a character free proof that the Frobenius Kernel is a normal subgroup, but it fell apart and I haven't been able to repair the proof. -While these quotients haven't helped me prove anything nontrivial (yet), they have caused me to "trip over" a lot of old important theorems. - -REPLY [3 votes]: This is very interesting. I don't really understand what is going on with the double cosets, but this is reminiscent of the standard construction of loops (quasigroups with identity elements) as tranversals in groups. This dates back to Baer in the early 40's. The idea is the following. You have a group $G$, a subgroup $H$, and a transveral $T$, that is a set of coset representatives. Assume further that $T$ is normalized, that is, $1\in T$. For $x,y \in T$, define an operation $\circ$ on $T$ by $(x\circ y)H = xyH$, that is, $x\circ y$ is the unique representative in $T$ of $xyH$. It is easy to see that $1$ is the identity element for $\circ$ and that for all $a,b\in T$, the equation $a\circ x = b$ is uniquely solvable for $x\in T$. If, in addition, one has that $T$ is a transversal for every conjugate of $H$, then it follows that the equation $y\circ a = b$ is also uniquely solvable for $y\in T$. Thus $(T,\circ)$ is a loop. Conversely, it is not hard to show that every loop arises in this way. -The tables you are talking about remind me of this, except that you are keeping track of double coset clusters and you are not worrying about which products of representatives themselves have representatives. It would be interesting to know how this idea relates to loop theory.<|endoftext|> -TITLE: Applications of Rademacher's Theorem -QUESTION [35 upvotes]: Rademacher's Theorem (that every Lipschitz function on $\mathbb{R}^{n}$ is almost everywhere differentiable) is a remarkable result on the structure of the space of Lipschitz functions, but I was wondering whether it has any interesting applications. All of the "useful" results (or maybe "applicable") that I know of about weak versions of differentiability involve estimates (e.g. Sobolev embedding, Lebesgue differentiation theorem). - -REPLY [5 votes]: The following application may not be too far outside the realm of usual applications, as described in the OP's original question statement, but I will let others decide for themselves. -Rademacher's theorem is used to prove a maximum/minimum principle for 2D drift-diffusion equations -$$\partial_t\theta + u\cdot\nabla\theta + \kappa(-\Delta)^{\gamma/2}\theta, \tag{1}$$ -where $\theta:\mathbb{R}^2\rightarrow\mathbb{R}$ is a scalar-valued function, $u:\mathbb{R}^2\rightarrow\mathbb{R}^2$ is a divergence-free vector field possibly depending on $\theta$, $\kappa\geq 0$, and $(-\Delta)^{\gamma/2}$ is the fractional Laplacian of order $\gamma$. In the case where the velocity field $u$ is recovered from $\theta$ through the Biot-Savart law -$$u = (-R_2\theta,R_1\theta),$$ -where $R_1,R_2$ are the usual 2D Riesz transforms, we recover the well-known dissipative Surface Quasi-Geostrophic (SQG) equation. -When $\kappa=0$, it is easy to see from the existence of a flow map that the minimum and maximum of a smooth solution $\theta$ are conserved. For $\kappa>0$, this is no longer necessarily the case; however, the minimum is nondecreasing and the maximum is nonincreasing. To see this, one argue as follows, as for instance shown in this paper by the Cordobas. Denote the supremum $\theta$ at time $t$ respectively by -$$M(t) := \sup_{x\in\mathbb{R}^2} \theta(t,x).$$ -If $\theta(0)\in L^\infty$, then since $\|\theta(t)\|_{L^p} \leq \|\theta(0)\|_{L^p}$, for any $1\leq p\leq \infty$, (see the aforementioned paper), it follows that $M(t)$ is finite. Moreover, if $\theta(t)\in \hat{L}^1$ for each $t$, then it follows from the Riemann-Lebesgue lemma that $\theta(t)\in C_0$. Hence, for each $t$, there exists a point $x(t)\in\mathbb{R}^2$ such that -$$M(t) = \theta(t,x(t)).$$ -It's not hard to show that $M(t)$ is a Lipschitz continuous function. Hence by Rademacher's theorem, $M(t)$ is differentiable for almost every $t$. At such a point $t_0$ of differentiability, we can find by a compactness argument a sequence of positive numbers $h_j\rightarrow 0$ such that $x(t_0+h_j)\rightarrow x_0\in\mathbb{R}^2$ as $j\rightarrow\infty$. Since $M$ is continuous, we then have $M(t_0) = \theta(t_0,x_0)$. Now -\begin{equation*} -\begin{split} -&\frac{M(t_0+h_j)-M(t_0)}{h_j} = \frac{\theta(t_0+h_j,x(t_0+h_j))-\theta(t_0,x_0)}{h_j} \\ -&=\frac{\theta(t_0+h_j,x(t_0+h_j))-\theta(t_0,x(t_0+h_j))}{h_j} + \frac{\theta(t_0,x(t_0+h_j))-\theta(t_0,x_0)}{h_j} \\ -&\leq \frac{\theta(t_0+h_j,x(t_0+h_j))-\theta(t_0,x(t_0+h_j))}{h_j}, -\end{split} -\end{equation*} -since the second term in the penultimate line is nonpositive. Letting $j\rightarrow\infty$ and using that $t_0$ is a point of differentiability by assumption, we obtain that -$$M'(t_0) \leq \partial_t\theta(t_0,x_0) = -u(t_0,x_0) \cdot \nabla\theta(t_0,x_0) - \kappa(-\Delta)^{\gamma/2}\theta(t_0,x_0).$$ -The first term on the RHS of the preceding equality is nonpositive since $\theta(t_0)$ is maximal at $x_0$. The first term on the RHS is also nonpositive for the same reason since -$$(-\Delta)^{\gamma/2}\theta(t_0,x_0) = C_\gamma PV\int_{\mathbb{R}^2}\frac{\theta(t_0,x_0)-\theta(t_0,y)}{|x-y|^{2+\gamma}}dy.$$ -Hence, $M'(t_0)\leq 0$. Note that by considering a sequence $h_j$ of negative numbers, one can show that $M'(t_0) = -\kappa(-\Delta)^{\gamma/2}\theta(t_0,x_0)$. That $m$ is nondecreasing follows by replacing $\theta$ by $-\theta$ in the preceding argument.<|endoftext|> -TITLE: Infinitely many minimal models -QUESTION [12 upvotes]: There are examples of elliptic fiber spaces over a two-dimensional base which have infinitely many relative minimal models (where two abstractly isomorphic models connected by flops are counted separately). The one I know is given by Reid and Kawamata and works by repeatedly flopping two rational curves in a singular fiber. In Matsuki's "Introduction to the Mori Program" he indicates (pg. 366) that this construction can be extended to a non-relative setting to yield a variety $X$ with infinitely many minimal models over Spec k. -I haven't managed to make this extension or find it written down, so a couple questions: where can I find an explicit example of a variety with infinitely many minimal models (over Spec k)? What is the Kodaira dimension in this case? Is it possible to find a Calabi-Yau threefold with infinitely many minimal models? - -REPLY [6 votes]: Dear John, -One example can be found in this paper by Fryers. This example is a Horrocks--Mumford quintic, in particular a Calabi--Yau threefold. He shows there are infinitely many marked minimal models, but they fall into only 8 (unmarked) isomorphism classes. -I think there is another class of examples in higher dimensions in this recent paper of Oguiso. I must admit I haven't looked at the paper too closely, but if I remember correctly a talk I heard him give, there are indeed infinitely many marked models in this case. There is also a nice picture involving the geometry of Coxeter groups, but sadly that didn't seem to make it into the paper.<|endoftext|> -TITLE: Kuenneth-formula for group cohomology with nontrivial action on the coefficient -QUESTION [18 upvotes]: For a trivial action on the coefficient, we have the following Kuenneth formula -for group cohomology: -$$ -H^n(G_1 \times G_2; M) \cong -[\oplus_{i= 0}^n H^i(G_1;M) \otimes_M H^{n-i}(G_2;M)] -\oplus [\oplus_{p =0}^{n+1} \text{Tor}^M(H^p(G_1;M),H^{n+1-p}(G_2;M))] -$$ -where $G_1$ and $G_2$ are finite groups and/or compact Lie groups --Edit-- and $M$ is a PID such as $Z$? -If the group action on the coefficient $M$ is non-trivial, does the above Kuenneth formula remain valid? - -REPLY [16 votes]: I learned this through Ken Brown's textbook "Cohomology of Groups" (and studying under him): I basically use the beginning of his Chapter 5 and solve the two exercises in that section. -Let $M$ (resp. $M'$) be an arbitrary $G$-module (resp. $G'$-module), let $F$ (resp. $F'$) be a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ (resp. $\mathbb{Z}G'$), and consider the map $(F\otimes_GM)\otimes(F'\otimes_{G'}M')\rightarrow (F\otimes F')\otimes_{G\times G'}(M\otimes M')$ given by $(x\otimes m)\otimes(x'\otimes m')\mapsto(x\otimes x')\otimes(m\otimes m')$. -Note that $(F\otimes_GM)$ $\otimes(F'\otimes_{G'}M')$ = $(F\otimes M)_G \otimes(F'\otimes M')_G$ , -which is the quotient of $F\otimes M\otimes F'\otimes M'$ by the subgroup generated by elements of the form $gx\otimes gm\otimes g'x'\otimes m'$. The isomorphism $F\otimes M\stackrel{\cong}{\rightarrow}M\otimes F$ of chain complexes ($M$ in dimension $0$) is given by $x\otimes m\mapsto (-1)^{deg(m)\cdot deg(x)}m\otimes x=m\otimes x$, and so the aforementioned quotient is isomorphic to $F\otimes F'\otimes M\otimes M'$ modulo the subgroup generated by elements of the form $(gx\otimes g'x'\otimes gm\otimes g'm')=(g,g')\cdot(x\otimes x'\otimes m\otimes m')$ where this latter action is the diagonal $(G\times G')$-action. Now this is precisely -$$(F\otimes F'\otimes M\otimes M')_{G\times G'} = (F\otimes F')\otimes_{G\times G'}(M\otimes M')$$ and hence the considered map is an isomorphism. -Assuming now that either $M$ or $M'$ is $\mathbb{Z}$-free, we have a corresponding Künneth formula $\bigoplus_{p=0}^nH_p(G,M)\otimes H_{n-p}(G',M')\rightarrow H_n(G\times G',M\otimes M')$ -$\rightarrow\bigoplus_{p=0}^{n-1}Tor_1^\mathbb{Z}(H_p(G,M),H_{n-p-1}(G',M'))$ by Proposition I.0.8[Brown]. Note that in order to apply the proposition we needed one of the chain complexes (say, $F\otimes_G M$) to be dimension-wise $\mathbb{Z}$-free (and so with a free resolution $F$ this means we needed $M$ to be $\mathbb{Z}$-free). -Actually, the general Künneth theorem has a more relaxed condition and it suffices to choose $M$ (or $M'$) as a $\mathbb{Z}$-torsion-free module}. - -Cohomology Künneth Formula (no proofs, just statements/notes) -Let $M$ (resp. $M'$) be an arbitrary $G$-module (resp. $G'$-module), let $F$ (resp. $F'$) be a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ (resp. $\mathbb{Z}G'$), and consider the cochain cross-product $Hom_G(F,M)\otimes Hom_{G'}(F',M')\rightarrow Hom_{G\times G'}(F\otimes F',M\otimes M')$ which maps the cochains $u$ and $u'$ to $u\times u'$ given by $\langle u\times u',x\otimes x'\rangle=(-1)^{deg(u')\cdot deg(x)}\langle u,x\rangle\otimes\langle u',x'\rangle$. This map is an isomorphism under the hypothesis that either $H_i(G,M)$ or $H_i(G',M')$ is of finite type, that is, the $i$th-homology group is finitely generated for all $i$ (alternatively, we could simply require the projective resolution $F$ or $F'$ to be finitely generated). -For example, if $M=\mathbb{Z}$ then $Hom_G(-,\mathbb{Z})$ commutes with finite direct sums, so we need only consider the case $F=\mathbb{Z}G$. An inverse to the above map is given by $t\mapsto \varepsilon\otimes \phi$, where $\varepsilon$ is the augmentation map and $\phi:F'\rightarrow M'$ is given by $\phi(f')=t(1\otimes f')$. Note that this does not hold for infinitely generated $P=\bigoplus^\infty \mathbb{Z}G$ because $Hom_G(P,\mathbb{Z})\cong\prod^\infty \mathbb{Z}$ is not $\mathbb{Z}$-projective (i.e. free abelian). -Assuming now that either $M$ or $M'$ is $\mathbb{Z}$-free, we have a corresponding Künneth formula $\bigoplus_{p=0}^nH^p(G,M)\otimes H^{n-p}(G',M')\rightarrow H^n(G\times G',M\otimes M')\rightarrow$ -$\bigoplus_{p=0}^{n+1}Tor_1^\mathbb{Z}(H^p(G,M),H^{n-p+1}(G',M'))$ by Proposition I.0.8[Brown].<|endoftext|> -TITLE: Prime Decomposition in Cyclotomic Z_p-extensions -QUESTION [7 upvotes]: In their classic paper "Class fields of abelian extensions of $\mathbf{Q}$", Mazur and Wiles assert that - -"in a cyclotomic - $\mathbf{Z}_p$-extension only finitely - many primes lie above any prime of - $\mathbf{Q}$." - -My only other source in learning this material so far has been Washington's "Introduction to Cyclotomic Fields", and the only result along these lines is that such extensions are unramified outside of $p$. -So apparently, all primes lying above $l \neq p$ stop splitting at some finite level $K_n$, after which they remain inert. I've been unable to make much progress is proving this. - -How can we see that this statement is - true, and what other, more general - results do we have about prime - decomposition in cyclotomic - extensions? - -REPLY [10 votes]: The point is that the Frobenius at $\ell$ is nontrivial in this extension, so it generates an open subgroup, and the fixed field of this subgroup is precisely the field at which $\ell$ stops splitting. -To see that Frob$_\ell$ is nontrivial, recall that in the full $\mathbb{Z}_p^*$ extension $\mathbb{Q}(\mu_{p^\infty})$, Frob$_\ell$ is simply the element $\ell\in \mathbb{Z}_p^*$, which is clearly nontrivial. The cyclotomic $\mathbb{Z}_p$ extension is obtained by quotienting $\mathbb{Z}_p^*$ by the subgroup $\mu_{p-1}$. This subgroup does not contain $\ell$, so Frob$_\ell$ remains nontrivial in the $\mathbb{Z}_p$-extension.<|endoftext|> -TITLE: Donaldson-Thomas Invariants in Physics -QUESTION [16 upvotes]: First of all, I am sorry for there are a bunch of questions (though all related)and may not be well framed. -What are the DT invariants in physics. When one is computing DT invariants for a Calabi-Yau manifold, what is one computing in physics? -What about the generalized (motivic) version? -Also what does the Gromov-Witten/DT correspondence (MNOP) say in terms of physics, are there (strong) physical reasons to believe such a correspondence. -Please suggest some useful references. -Thanks. - -REPLY [35 votes]: Donaldson-Thomas invariants in mathematics are a virtual count of sheaves (or possibly objects in the derived category of sheaves) on a Calabi-Yau threefold. In physics, sheaves (and more generally objects in the derived category) are considered as models for D-branes in the topological B-model and Donaldson-Thomas invariants are counts of the BPS states of various D-branes systems. For example, the "classical" DT invariants that are considered by MNOP count ideal sheaves of subschemes supported on curves and points. You will hear physicists refer to such invariants as "counting the states of a system with D0 and D2 branes bound to a single D6 brane". The single D6 brane here is the structure sheaf $\mathcal{O}_X$ and the D0 and D2 branes form the structure sheaf $\mathcal{O}_C$ of the subscheme $C$ (which is supported on curves and points) and the term "bound to" refers to the map $\mathcal{O}_X \to \mathcal{O}_C$ because they are replacing the ideal sheaf with the above two-term complex (which are equivalent in the derived category. Note that the $k$ in D$k$-brane refers to the (real) dimension of the support. -There is a discussion of the meaning of the motivic DT invariants in physics in the paper "Refined, Motivic, and Quantum" by Dimofte and Gukov (http://arxiv.org/pdf/0904.1420) where the basic claim is that the motivic invariants and the "refined" BPS state counts are the same. "Refined" here refers to the way you count BPS states. BPS states are certain kinds of representations of the super-Poincare algebra and "counting" means just finding the dimension of these representations (I think that little book on super-symmetry by Dan Freed has a good mathematical discussion of this). Sitting inside the super-Poincare algebra is a copy of $\mathfrak{sl}_2\oplus\mathfrak{sl}_2$ and normally one looks at the action of the diagonal $\mathfrak{sl}_2$ on the space of BPS representations and finds the dimensions of the irreducibles, for the "refined" count, you look at both copies of $\mathfrak{sl}_2$. The generating function for the dimensions of these representations thus gets an extra variable which is suppose to correspond to the Lefschetz motive $\mathbb{L}$ in the motivic invariants. -As for the DT/GW correspondence, I'm afraid that I don't really understand the physicist's explanations. There is a few paragraphs in MNOP (presumably written by Nekrasov) about it and I think that physicists regard it as well understood, but I haven't found something that I can understand. Let me know if you do.<|endoftext|> -TITLE: Higher derivatives than Jacobi fields -QUESTION [7 upvotes]: The first and second derivatives of the distance function (either the full $d:M\times M\to \mathbb{R}$ function or the $d(p,\cdot):M\to \mathbb{R}$ function) as well as the derivative of the exponential map (again both of the full $\exp:TM\to M$ map and of the map $\exp_p:T_pM\to M$) may be computed with the aid of Jacobi fields, i.e, solutions to Jacobi's equation. -I have a scenario where I need second derivatives of the `full' exponential map $\exp:TM\to M$. That is, denoting the pushforward of a differentiable map by a '$ _*$', I would like to know a thing or two about $\nabla_X\exp_*\mathcal{V}$ (where $\mathcal{V}$ is a section of $TTM$ and $X$ is an appropriate vector field). In particular, I think I will require some comparison techniques analagous to those for Jacobi fields (e.g., Rauch's comparison theorems). -Can anyone point me in the right direction? - -REPLY [3 votes]: Recently in the context of optimization on manifolds, the covariant derivative of the differential of the exponential map $\nabla \text{d } \text{exp}$ was studied in two papers: -1 ''An accelerated first-order method for non-convex optimization on manifolds'' and -2 ''Curvature-Dependant Global Convergence Rates for Optimization on Manifolds of Bounded Geometry'' -As @Deane Yang describes, both papers show that this object satisfies a second-order inhomogenous linear ODE, see Proposition B.1 in 1, and Proposition 4.1 in 2. The following is from 2: - -Let $(M, g)$ be a Riemannian manifold, and let $\gamma$ be a geodesic -with initial unit vector $v \in T_p M$. For two vectors $w_1, w_2$ -perpendicular to $v$, the vector field along $\gamma$ $$K(t) = (\nabla - \text{d } \text{exp}_p)_{tv}(t w_1, t w_2)$$ satisfies the following -ODE along $\gamma$ $$\ddot{K} + R(K, \dot{\gamma})\dot{\gamma} + Y = - 0, \quad K(0) = 0, \dot{K}(0) = 0$$ where $Y$ is the vector field -along $\gamma$ given by $$Y = 2 R(J_1, \dot{\gamma}) \dot{J}_2 + - 2R(J_2, \dot{\gamma})\dot{J}_1 + (\nabla R_{\dot{\gamma}})(J_2, - \dot{\gamma})J_1 + (\nabla R_{J_2})(J_1, \dot{\gamma})\dot{\gamma}$$ -and $J_1, J_2$ are the Jacobi fields along $\gamma$ with initial -conditions $J_i(0) = 0, \dot{J}_i(0) = w_i$. - -Using this ODE, both papers derive bounds on $||\nabla \text{d } \text{exp}||$ based on bounds for the sectional curvatures and $\nabla R$. Paper 1 uses a bootstrapping technique, while paper 2 uses ODE comparison techniques which yield comparatively tighter bounds albeit requiring a little bit more analysis. For these results, see Theorem B.2 in 1, and Theorems 4.10 and 4.11 in 2. The following is from 2: - -Let $$\text{sn}_{\kappa}(t) = \left\{ \begin{array}{lII} - \frac{\sin(\sqrt{\kappa}t)}{\sqrt{\kappa}} & \quad \kappa > 0 \\ t & \quad - \kappa = 0 \\ \frac{\sinh(\sqrt{-\kappa} t)}{\sqrt{-\kappa}} & \quad - \kappa < 0 \end{array} \right\}$$ and $$\pi_{\kappa} = \left\{ - \begin{array}{lI} \frac{\pi}{\sqrt{\kappa}} & \quad \kappa > 0 \\ - \infty & \quad \kappa \leq 0 \end{array} \right\}.$$ -Let $(M, g)$ be a Riemannian manifold with sectional curvatures -bounded below by $\delta$ and above by $\Delta$. Also assume -$$||(\nabla_u R)(w, u)w + (\nabla_w R)(w, u)u|| \leq 2 \Lambda ||u||^2 - ||w||^2 \quad \forall p \in M \quad u, w \in T_p M.$$ -For a geodesic $\gamma : [0, r] \rightarrow M$ with initial unit -vector $v$, $r < \pi_{(\Delta + \delta)/2}$, and any two vectors $w_1, - w_2 \in T_p M$, we have that $$||(\nabla \text{d } \text{exp}_p)_{r - v}(w_1, w_2)|| \leq \frac{8}{3r^2} \text{sn}_{\delta} - \left(\frac{r}{2} \right)^2\left(\Lambda \cdot - \text{sn}_{\delta}\left(\frac{r}{2}\right)^2 + 2 \max\{|\Delta|, - |\delta|\}\text{sn}_{\delta}(r) \right) ||w_1|| \cdot ||w_2||.$$ - -Finally, these bounds are shown to be tight, in a certain sense -- see the discussion after Theorem B.2 in 1, and subsection 4.6.1 in 2.<|endoftext|> -TITLE: Infinite dimensional central simple algebras -QUESTION [8 upvotes]: When constructing the Brauer group of a field, only the finite-dimensional central simple algebras are considered (because of Artin-Wedderburn's characterization). -But what happens to the infinite-dimensional ones? (I.e., to simple algebras which are infinite-dimensional over their centers). - -Is there an analogue for the Brauer group? -Is there some structure theorem like Artin-Wedderburn's? -What can be said about cross products of these algebras? - -REPLY [7 votes]: Perhaps one approach to thinking about this question would be Dixmier-Douady theory (as described in Brylinski's book "Loop Spaces, Characteristic Classes and Geometric Quantization"). This involves considering certain Banach algebras over the complex numbers as infinite dimensional analogs of central simple algebras, and leads to some useful notions generalizing bundles of central simple algebras over topological spaces. -The point of view here starts with the correspondence between bundles of central simple $\mathbb C$-algebras over a topological space $X$ and torsion elements of the second cohomology $H^2(X, \underline{\mathbb C}^*)$. This comes from the identification of isomorphism classes degree $n$ algebra bundles with $H^1(X, \underline{PGL_n})$ and the nonabelian short exact sequence: -$$ 1 \to \underline{\mathbb C}^* \to \underline{GL_n} \to \underline{PGL_n} \to 1$$ -(here I'm using the underline to denote the corresponding sheaves of continuous functions). -Now, when you do this, you find that those classes which come from $H^1(X, \underline{PGL_n})$ (i.e. from degree $n$ central simple algebra bundles) are necessarily $n$-torsion in $H^2(X, \underline{\mathbb C}^*)$. what Dixmier-Doady theory does is identify all the elements of this $H^2$ as arising from infinite dimensional algebras. -A little more precisely, assuming paracompactness of $X$, on fixes some separable Hilbert space $E$, and defines $\mathcal K$ to be the Banach algebra of compact operators on $E$. It then turns out that one can naturally identify the whole of -$H^2(X, \underline{{\mathbb C}^\ast})$ - with the set of isomorphism classes of locally trivial Banach algebra bundles with fiber isomorphic to $\mathcal K$. What makes this really work is an analog of the above sequence -$$ 1 \to {\mathbb C}^\ast \to {\mathcal L}^\ast \to {\mathcal L}^\ast/{\mathbb C}^\ast \to 1 $$ -where $\mathcal L$ is the Banach algebra of bounded (sup-norm) operators on $E$ (which incidentally happens to have center $\mathbb C$), and one uses an identification $\mathcal L^\ast/\mathbb C^\ast \cong Aut(\mathcal K)$.<|endoftext|> -TITLE: Compact holomorphic symplectic manifolds: what's the state of the art? -QUESTION [18 upvotes]: This came up in a discussion I had yesterday. Since my understanding is limited, -I thought I ask here, because I know there are quite a few experts lurking about. -Recall that a holomorphic symplectic manifold $X$ is a complex manifold which comes -equipped with a nondegenerate holomorphic $2$-form $\omega$, i.e. $\omega^{\dim X/2}$ is nowhere zero. Here I'll be interested exclusively in the compact simply connected Kahler (see Dmitri's answer) examples. -Using Yau's work, the class of these manifolds can be identified with the class hyper-Kähler manifolds subject to the same restrictions (cf. [1]). This means a Riemannian manifold which is Kähler with respect to a triple of complex structures $I,J,K$ which behave like the quaternions, $IJ=K$ etc. Needless to say, such things are exotic. In dimension two, by the classification -of surfaces, the only possible examples are K3 surfaces (or more crudely, things that behave like quartics in $\mathbb{P}^3$). -What little I know in higher dimensions can be summarized in a few sentences. Beauville [1] found a bunch of beautiful simply connected examples as Hilbert schemes of points on a K3 surface and variants for abelian -surfaces: the so called generalized Kummer varieties. More generally, Mukai [2] constructed -additional examples as moduli space of sheaves on the above surfaces. -Huybrechts [3] mentions some further examples which are deformations of these. So now my questions: - -Are there examples which are essentially different, i.e. known to not be deformations of the examples discussed above? If not, then what is the expectation? Is there any sort of -classification in low dimensions? - -I'm aware of some work on hypertoric varieties, which are hyper-Kähler, but I haven't followed this closely. So: - -Are any of these compact? If so, how do they fit into the above picture? - -While I can already anticipate one possible answer "no, none, hell no...", feel free to -elaborate, correct, or discuss anything related that seems relevant even if I didn't explicitly ask for it. -Thanks in advance. - -Refs. -[1] Beauville, Variétés Kähleriennes dont la première classe de Chern est nulle, J. Diff. Geom 1983 -[2] Mukai, Symplectic structure of the moduli space of sheaves on an abelian or K3 surface, -Invent 1984 -[3] Huybrechts, Compact hyper-Kähler manifolds: basic results, Invent 1999 - -REPLY [3 votes]: It's probably bad form to answer one's own question -- even the software has registered its disapproval -- but I thought I'd make a small update. Regarding classification in low dimensions, according to Beauville's nice survey http://arxiv.org/pdf/1002.4321.pdf (which I wasn't aware of last week), it would appear to be open even in dimension 4. Although some -bounds are known in this case. For example, Guan shown that the Betti number $b_2\in [3,8]\cup \lbrace 23\rbrace$, where $7$ and $23$ correspond to the known cases mentioned above.<|endoftext|> -TITLE: How can I write down a point in the Berezinian of a super vector space? -QUESTION [8 upvotes]: A vector space $V$ of dimension $n$ has an associated determinant line $Det(V)$. - An element of $Det(V)$ is represented as a (formal limear combination) of expresstions of the form - $v_1 \wedge v_2 \wedge \ldots \wedge v_n$, subject to the usual multilinearity and antisymmetry relations. - -I'm wondering what is analog of the above fact/construction in the world of super vector spaces. - -Let $V$ be a supervector space of dimension $n|m$. Then there is a line $Ber(V)$ called the Berezinian of $V$, that behaves like a super-determinant. -Here's a naive description of the Berezinian: -for $V=V_0\oplus V_1$, it is given by $$Ber(V)=Det(V_0)\otimes Det(V_1)^*.$$ -That's clearly not a good description of $Ber(V)$, as it relies on the decomposition of $V$ into even and odd parts, which is not a $GL(V)$-invariant thing to do. -I want to make sure that I don't get non-invariant answers. -To ensure that, I'll do things in families (and thus make the question more complicated $-$ sorry for that): - -Let $\Lambda=\Lambda(\theta_1,\ldots,\theta_n)$ be a Grassmann algebra (=exterior algebra) on $n$ variables, and let $V$ be a $Spec(\Lambda)$-parametrized family of super vector spaces, i.e., a super vector bundle $V\to Spec(\Lambda)$. - How can I describe concretely a section of the associated line bundle $Ber(V)\to Spec(\Lambda)$? - -For those who don't like the above language, I can translate into algebra. -Let $\Lambda=\Lambda(\theta_1,\ldots,\theta_n)$, and let $V$ be a free $\Lambda$-module -on $n$ even generators and $m$ odd generators. -How can I describe concretely an even element of the rank one $\Lambda$-module $Ber(V)$? - -REPLY [5 votes]: There is Koszul complex introduced by Yu. Manin in his book "Gauge field theory and complex geometry". The cohomology of this complex yields the Berezinian.<|endoftext|> -TITLE: Finite simple groups and conjugacy classes with 2p elements -QUESTION [11 upvotes]: Let $p$ be an odd prime number. Can a finite simple group have a conjugacy class with $2p$ elements? - -REPLY [13 votes]: I believe the answer is no. A group with a conjugacy class of degree $2p$ has a transitive permutation action of degree $2p$ on that class, and the stabilizer is the centralizer of an element. -If this action is imprimitive, then the blocks have size $p$ or 2. Wtih blocks of size $p$ there would be a subgroup of index 2, contradicting simplicity. With blocks of size $p$, the group would act faithfully on the blocks. Burnside proved that a permutation group of prime degree $p$ is solvable or 2-transitive. The finite 2-transitive permutation groups are now known (using the classification of finite simple groups). They are listed, for example, in -PETER J. CAMERON, FINITE PERMUTATION GROUPS AND FINITE SIMPLE GROUPS, -BULL. LONDON MATH. SOC , 13 (1981),1-22 -and it can be checked that none of the examples of prime degree have point stabilisers that have a subgroup of index 2 with nontrivial centre. -So suppose that the action on the conjugacy class is primitive. In the Cameron paper cited above, a result of Wielandt is mentioned that says that primitive permutation groups of degree $2p$ either have rank 3 or are 2-transitive. Using the classification of finite simple groups, it has been checked that the only rank 3 examples are $A_5$ and $S_5$ in degree 10, and $A_5$ does not have a class of degree $2p$. For 2-transitive groups, we can again check the list and observe that none of the groups has point stabilizer with nontrivial centre.<|endoftext|> -TITLE: Does one need to sheafify when defining the inverse image of a sheaf with respect to an embedding? -QUESTION [8 upvotes]: This seems to be a rather simple (stupid?:)) question; yet I was not able to find an answer quickly. -For a morphism $f:X\to Y$ of schemes (or topological spaces) and an (etale or topological) sheaf $f$ (of abelian groups) on $Y$ one defines the inverse image $f^*(S)$ (or is $f^{-1}$ more standard?) as the sheafification of the inverse image of $S$ considered as a presheaf. -Now suppose that $f$ is a (closed) embedding. Is the sheafification necessary here? It seems that the answer is no, since any covering of an open $U\subset X$ could be presented as a 'limit' of coverings of open $V\subset Y$, $V\supset f(U)$ (in the topogical context); so the 'presheaf inverse image' of a sheaf is a sheaf also. This seems to be easy, as well as carrying this argument over to the etale context; yet I would be deeply grateful for a definite answer and a reference for it. Also, does the situation change when one passes to simplicial schemes and sheaves? Is this fact 'standard' (if it is true)? - -REPLY [2 votes]: Perhaps I'm thinking of sheaves a bit differently from you here, but as long as we think of sheaves as \'{e}tale maps this holds for arbitrary continuous maps $f:X\to Y$ and not just open or closed embeddings. The reason for this is that, \'{e}tale maps are stable under pullback along arbitrary continuous maps. See, e.g., Mac Lane and Moerdijk (Sheaves in Geometry and Logic) Lemma 1 (II.9, page 99). -I guess that you know this though and are asking about the the case where we are not regarding sheaves as \'{e}tale maps. In this case you are right that some sheafification is required (since the construction of the inverse image along $f$ involves taking a colimit in $\text{Sh}(X)$). (I had thought that this was incorrect, but I had misunderstood precisely how you were claiming one should compute the inverse image in this case and so have removed my counterexample since it did not quite address your question.)<|endoftext|> -TITLE: joint papers/collaboration, take two. -QUESTION [12 upvotes]: Some of you may remember I asked this question last night, and it was closed for being too argumentative. I apologize; I didn't mean to be argumentative or offensive in the slightest. Re-reading my question, I can see how I can come across as frivolous, or, "a troll". But I am honestly curious about the attitude the mathematical community takes towards research collaboration, so let me try re-phrasing my question: -I am a graduate student, and I was wondering whether collaboration could be a two-edged sword for me. -Since I am yet unfamiliar with all the workings of research, having an extra body to work with me will be definitely helpful, and I'll work faster and produce more papers. -On the other hand, I am not famous, and people don't know whether I'm a proficient mathematician, or whether I'm capable of conducting research independently. -So if I decided to collaborate, and when the time comes for me to apply for a post-doc, or a tenure-track position, it's quite possible that this joint paper could be my only paper (I think my field publishes quite slowly). And if that were the case, would people wonder about my ability to conduct research on my own? - -REPLY [14 votes]: While I agree with much of JSE's post (recently most of my projects are joint, and I am enjoying them much more than my solo work), I have some reservations about his advice. I think at the beginning of your career, you should make sure that a significant percentage of your best papers are solo. Here are some reasons why: -a) If most of your papers are collaborations, the committees would have to "dig deeper" (to quote Joel David Hamkins). Why make it harder for your supporters to support you? -b) Once people know that you are productive and creative, collaborations would be easier to come by and even more enjoyable. In fact, you would probably get more credits than you deserve! (while this is a nice problem to have, now you should make sure that the graduate students who write with you get the fair share of credits). -c) While working solo is harder and more boring, it does train you in different aspects that would be quite useful for your career: for example you have to be more careful with yourself since nobody will be around to point out your mistakes. -One more thing that is already mentioned by Joel, but I think worth repeating: when you collaborate, make sure you diversify (both in people and topics).<|endoftext|> -TITLE: Convex tilings of the plane -QUESTION [5 upvotes]: For convex polyhedra you have Steinitz's theorem characterizing them as the 3-connected planar graphs. My question is not about spheric tilings, but about periodic tilings of the euclidean plane. Is it here also the case that 3-connectivity corresponds with convexity? -It is easy to construct an example of a 2-connected periodic tiling that is not convex. My guess is that the symmetry group prohibits some tilings from being convex and thus there are also 3-connected periodic tilings that are not convex, but I can't seem to be able to construct a counter example to support this guess. -Is there a known counterexample for this? Or is it actually true? Any reference or hint would help. -edit: I was looking for a periodic tiling which has no equivariantly equivalent (i.e. topologically equivalent and the same symmetry group) tiling that is convex and that is 3-connected when you look at the segments as graph edges. My feeling is that in case of the euclidean plane there might be a tiling in which you can't convexify one face, without needing to make another one nonconvex to keep the same symmetry group. - -REPLY [4 votes]: You are perhaps thinking of Tutte's "spring" theorem that every 3-connected graph has an embedding with convex faces. It indeed a corollary of the (earlier) Steinitz theorem, but Tutte's proof is of independent interest. Now, for periodic networks in the plane, you need to look at the "torus version" of the Tutte theorem. Start with this important paper which gives a version of the result and check some further refs therein.<|endoftext|> -TITLE: Ordinal category theory? -QUESTION [12 upvotes]: Just out of curiosity: Is there a notion of $\alpha$-category for an ordinal number $\alpha$, extending the given notions for $\alpha \leq \omega$? If there is none, which one would you propose? Feel free to draw images. - -REPLY [6 votes]: Let $\Delta_2$ the 2-tronked simplicial category (objects are the finite orders $0=$(0), $1=$(0,1), $2=$(0, 1, 2) and order-preserving functions). Then $Cat$ (category of small categories)is isomorphically to $Fun_c(\Delta_2^{op}, Set)$ (finite limits preserving functors and natural transformations). -More in general the (hyper)category of n-double small categories $n$-$Double$ is isomorphically to $Fun_c((\Delta_2^{op}\times\ldots\times\Delta_2^{op}), Set)$ (n fold products), and $n$-$Cat$ is a subcategory of $n$-$Double$ (elements of $Fun_c((\Delta_2^{op}\times\ldots\times\Delta_2^{op}), Set)$ that send some morphisms to identities). Of course you can generalize this for a infinite cardinal $n$.<|endoftext|> -TITLE: Which Shimura varieties are known to be automorphic? -QUESTION [13 upvotes]: This seems like something that should be well-known, but as an outsider to the field, I'm having trouble locating precise statements. -Hasse-Weil zeta functions of Shimura varieties should be alternating products of automorphic $L$-functions. This seems to be known when the underlying group is $GL_2$ (or a quaternion division algebra) over a totally real field, $GSp_4$ over $\mathbb Q$ (maybe a totally real field), unitary groups in three variables over $\mathbb Q$ (maybe a totally real field), and maybe certain other unitary groups. -In "Where Stands Functoriality Today?", Langlands writes that the "principal factors" of the zeta function of (general) Siegel modular varieties are automorphic $L$-functions attached to spinor representations (which have not been analytically continued), so some more general calculations seem to be known... -My question: For which groups are the Hasse-Weil zeta functions of the associated Shimura varieties known to be alternating products of automorphic $L$-functions (maybe modulo "technical" restrictions and bad prime calculations)? - -REPLY [14 votes]: As far as I understand, what you need to do to prove that the zeta function of a Shimura variety is automorphic is (I'm ignoring the bad primes here, but I think that we now know enough about them too - I can develop later if you want) : -(1) Do some kind of point-counting over finite fields. For PEL Shimura varieties of type A and C, Kottwitz has done it. Actually it's a bit more, you calculate the trace of a Hecke operator times a power of the Frobenius (at a place of good reduction) on the cohomology with compact support. -(2) "Stabilize" the resulting formula so you can compare it with Arthur's stable trace formula. Here there is a choice. As the reason we expect the trace of a Hecke operator to compare well to the trace formula is Matsushima's formula, as Matsushima's formula for noncompact Shimura varieties (due to Borel and Casselman in that case) is a formula for the L^2 cohomology of the Shimura variety, and as the algebraic avatar of this L^2 cohomology is the intersection cohomology of the minimal compactification, you can choose to first extend the result of (1) to this intersection cohomology, and then "stabilize" whatever you get. Or you can ignore this and choose to work with compact support cohomology, at the cost of greater complication on the trace formula side. Laumon followed that approach for Siegel modular threefolds, but as far as I know, in other cases people generally choose the first approach. Anyway, this is not a problem for compact Shimura varieties. For compact PEL Shimura varieties of type A and C, this "stabilization" part is also due to Kottwitz (in the first part of his beautiful Ann Arbor article "Shimura varieties and $\lambda$-adic representations"), we also know some noncompact PEL cases of type A and C by Morel's work, and basically all other type A and C PEL cases should reduce to Kottwitz's calculations, but I don't think this is written anywhere. -Oh, and you need the fundamental lemma for this part. -(3) Now you still need to compare the result of (2) with the stable trace formula, so obviously you need to know the stable trace formula (here you need the weighted fundamental lemma), and to make sense of the results and get a nice formula for your zeta functions you also want to know Arthur's conjectures on the classification of discrete automorphic representations. How you get the zeta function formula if you assume Arthur's conjectures is explained in the second part of Kottwitz's previously-cited beautiful Ann Arbor article. -So the question is, what do we know about Arthur's conjectures ? Well, they are accessible. Arthur is supposed to be writing a proof in the case of symplectic groups, and he is actually making progress on it. Note however that for Siegel modular varieties, you'll need general symplectic groups, so there will be a further reduction step even after Arthur finishes writing his book. (But we are nearing a proof of the automorphy of the zeta function. Yay !) There are a few young and brave ones who are planning to tackle the case of unitary groups (if I remember well, Sug Woo Shin, Tasho Kaletha, Paul-James White and Alberto Minguez). A last word of caution, all this (Arthur's work and the others' future work) depends on the stabilization of the twisted trace formula, which is at the moment not totally written down, I'm afraid, but there's a group of Serious People in Paris (like Clozel, Waldspurger etc) who have vowed to take care of it. -So, to sum up, It's Complicated, but we seem to be close for PEL Shimura varieties of type A and C, especially Siegel modular varieties. Also, if you just want the zeta function to be a product of automorphic L-functions with complex exponents (and not integral exponents), then I think that's known for PEL case A (maybe not written in all cases, though); whether you can make these exponents rational without too much additional work, I am not sure (it seems that this should be an easier thing than proving they're integers).<|endoftext|> -TITLE: Strange canonical cohomology class on topological spaces -QUESTION [9 upvotes]: Pick an integral basis $a_i,\ i=1,\ldots,n=b_1(X)$, for $H^1(X,\mathbb Z)$, then form the element -$$a:=a_1\cup\ldots\cup a_n\in H^{b_1}(X,\mathbb Z).$$ -This is canonical up to sign; any other choice of basis differs by some $g\in GL(H^1(X,\mathbb Z))$, which changes $a$ by $\det g=\pm1$. -There is no sign ambiguity on a Kähler manifold because we can choose the complex orientation on $H^1(X)$. -Has anyone seen this thing before ? Does it even have a name ? -It depends on $X$. For instance examples with $b_1(X)=4$ include $T^4$, where it is the volume form, and $\Sigma_2\times S^2$, where it is zero. ($\Sigma_2:=$ genus $2$ Riemann surface.) -I'm not sure what kind of answer I'm after. I have some curve counting invariants on a projective variety $X$ that depend only on the Chern numbers of $X$, and Chern classes evaluated against this strange canonical class $a$. Is that the best I can say ? - -REPLY [6 votes]: Given a space $X$ you can ask for a universal -$$ X \to (S^1)^n$$ -such that any other map -$$ X \to (S^1)^j$$ -factors (up to homotopy) through a map/homomorphism of groups $(S^1)^n \to (S^1)^j$ -One way to construct this map is to abelianize the fundamental group of $X$ by attaching 2-cells that are commutators. Then kill $\pi_2$ of this space by attaching 3-cells, and so on. So you construct a $K(\pi,1)$ from $X$ by attaching only 2-cells and higher dimensional cells, and the 2-cells are commutators. So this constructed spaces is a $K(AB(\pi_1 X),1)$-space, i.e. its fundamental group is the abelianization $AB(\pi_1 X)$ of $\pi_1 X$. -The torsion part of the abelianization factors off (some lens space product factors) and you're left with just a product of circles. -I'm not sure if that has a name, but that's one way to think of it. -In low-dimensional topology, this is relevant because this is the map whose homotopy-fibre has the homology of the universal (multi-variable) Alexander module. -edit: of course, if $H_1 X$ isn't finitely-generated, this universal map may not be to a product of circles. But there still is a universal map.<|endoftext|> -TITLE: Is there a two-variable prime-representing polynomial (in the sense of Jones-Sato-Wada-Wiens)? -QUESTION [10 upvotes]: In the math.se question Proof of no prime-representing polynomial in 2 variables, Alon Amit asks if Ribenboim's claim that a prime-representing polynomial (a Diophantine polynomial in which the positive values are precisely the primes) must have at least three variables has been proven. Alon suggested that perhaps the number was a typo, that all that is known is that (trivially) no univariate polynomial is prime-representing. -As of Jones 1982 [1, p. 550] the question of the existence of a universal Diophantine equation in two variables was open, so certainly it was not known that the number of variables for the special case of the primes was more than 2 at that time. -[1] James P. Jones, "Universal Diophantine equation", The Journal of Symbolic Logic 47:3 (1982), pp. 549-571. - -REPLY [10 votes]: Davis [1] writes that the two-variable case of universal Diophantine equations is still open as of 2006. (Ribenboim's book was published in 1996.) So the question of a prime-representing polynomial in two variables was (and, presumably, is) still open. -[1] Martin Davis, [FOM] Decidability of Diophantine equations, post to the FOM mailing list, December 14 2006.<|endoftext|> -TITLE: Riemann's $\zeta$ function and the uniform distribution on $[-1,0]$ -QUESTION [18 upvotes]: https://math.stackexchange.com/questions/64566/riemanns-zeta-function-and-the-uniform-distribution-on-1-0 -Stackexchange isn't getting really excited about this, so here it is. -The $n$th cumulant of the uniform probability distribution on the interval $[-1,0]$ is $B_n/n$, where $B_n$ is the $n$th Bernoulli number. -And $-\zeta(1-n)=B_n/n$, where $\zeta$ is Riemann's function. -Those two facts can be derived, but is there some argument that shows, without doing that, that you'd expect the cumulants of the uniform distribution to be those values of the $\zeta$ function? -Might this shed some light either on the $\zeta$ function or the uniform distribution? Or anything else? -Appendix: Cumulants are like moments, but better. The $n$th cumulant of a probability distribution is homogeneous of degree $n$ (like the $n$th moment). If $n=1$ it is shift-equivariant; if $n>1$ it is shift-invariant (like the $n$th central moment). If $X_1,X_2,X_3,\ldots$ are independent random variables, then the $n$th cumulant of the distribution of their sum is just the sum of the $n$th cumulants of their distributions. That last property is shared by central moments only in the cases $n=2,3$ (where the cumulant is just the central moment). Simplest nontrivial example: the 4th cumulant is the fourth central moment minus 3 times the square of the second central moment. -Later addition (should this be a separate question?): The number of independent Bernoulli trials strictly preceding the first success, with probability $1/(1+c)$ of success on each trial, is a geometrically distributed random variable with expected value $c$, taking values in the set $\lbrace 0,1,2,\ldots \rbrace$. The first several cumulants of that distribution are these -$$ -\begin{align} -& c \\ -& c+c^2 \\ -& c + 3c^2 + 2c^3 \\ -& c + 7c^2 + 12c^3 + 6c^4 \\ -& c + 15c^2 + 50c^3 + 60c^4 + 24c^5 \\ -& c + 32c^2 +180c^3+390c^4+360c^5+120c^6 -\end{align} -$$ -The probability distribution of the number of successes in just one trial, with probability $c$ of success on each trial, has cumulants -$$ -\begin{align} -& c \\ -& c-c^2 \\ -& c - 3c^2 + 2c^3 \\ -& c - 7c^2 + 12c^3 - 6c^4 \\ -& c - 15c^2 + 50c^3 - 60c^4 + 24c^5 \\ -& c - 32c^2 +180c^3-390c^4+360c^5-120c^6 -\end{align} -$$ -They're the same except for alternating signs. -If it were possible for the probability to be $-c$, where $c>0$, this sequence would be just $-1$ times the sequence of cumulants of the geometric distribution. Likewise, the sequence of values of $\zeta(1-n)$ is $-1$ times the sequence of cumulants of the probability distribution specified above. Could $\zeta(1-n),\quad n=1,2,3,\ldots$ be the sequence of cumulants of a distribution that would exist if we allowed negative probabilities (but still required the measure of the whole probability space to be $1$)? -If so, do we have two instances of a phenomenon that can be stated in a general way? - -REPLY [8 votes]: $X$ is the uniform distribution on $[-1,0]$, so the moment generating function of $X$ is -$$E(e^{tX})=\lim_{k\to\infty} \frac{1}{k}\left(1+e^{-t/k}+\dots+e^{-(k-1)t/k}\right)=\lim_{k\to\infty} \frac{1}{k}\frac{1-e^{-t}}{1-e^{-t/k}}=\frac{1-e^{-t}}{t}$$ -The cumulant generating function is the logarithm of this, call it $g(t)$. It'll be easier to compute the Taylor series of $tg'(t)+1$: -$$tg'(t)+1=-\frac{t}{1-e^t}=-\sum_{k=0}^\infty te^{kt}=-t\sum_{k=0}^\infty \sum_{n=0}^\infty \frac{k^nt^{n}}{n!}$$ -Rearrange the order of summation: -$$tg'(t)+1=-t\sum_{n=0}^\infty \zeta(-n)\frac{t^n}{n!}$$ -Now compare coefficients on either side: the $n$th cumulant of $X$ is plainly $-\zeta(1-n)$. -What? Why are you looking at me like that?<|endoftext|> -TITLE: When does the set of isometries form a group? -QUESTION [17 upvotes]: Motivation -Its a classic set up. Take a metric space $M$, with distance function $d:M\times M\to \mathbb{R}$. The set of isometries of $M$ is the set of functions $f: M \to M$ which preserve distance. This set has much of the structure of a group without additional assumptions; the composition of two isometries gives an isometry, the identity function takes the place of the identity, etc. -Although every isometry must be injective, however, it is not necessarily a bijection, and so might not have an inverse. For example any injective function from a metric space with the discrete metric ($d:M\times M\to \mathbb{R}, d(a,b) = 1$ if $a \neq b$ and $0$ if $a=b$) to itself is an isometry. -As the group of isometries is quite a useful gadget we can get round this. For example in MathWorld an isometry is assumed to be bijective: http://mathworld.wolfram.com/Isometry.html -Yet in Euclidean space, we do not need any additional assumption: -Lemma -Every isometry of $\mathbb{E}^2$, the Euclidean plane, is a surjection. -Proof -On the plane, for example, assume that a point $a$ does not lie in the image of an isometry $T$. Take three distinct points $T(b_1), T(b_2)$ and $T(b_3)$ that do lie in the image (as $T$ is injective they have a unique preimage). Let $d_i = d(b_i,a)$ be the distance between $d$ and $b_i$. -The three circles radius $d_i$ around $T(b_i)$, intersect together only at $a$, as the distances between the $T(b_i)$ and the $B_i$ are the same, the circles radius $d_i$ around $b_i$ will also intersect at a unique point $a'$. The point $T(a')$ must be $a$ as that is the only point that satisfies all the point to point distances, so $a$ does lie in the image of $T$. -$\square$ -Question -We can extend this argument to higher dimensional Euclidean spaces, yet it uses non-trivial properties, in particular how circles intersect. Is there a simple propetry of a metric space that ensures that the set of isometries forms a group? -Edit To rather strengthen the question, are there simple properties that are necessary and sufficient. - -REPLY [14 votes]: You are asking for conditions on a metric space $X$ for which any distance preserving map $X\to X$ is bijective. (Usually isometry is defined as bijective distance preserving map). - -Well, there are arbitrarily bad spaces $X$ such that the only distance preserving map $X\to X$ is identity. -In this case distance preserving maps (well, the only one) form the trivial group. -So, I do not see a language which could be used to formulate a necessary and sufficient condition. -For sufficient conditions: - -Compactness; -Proper + cocompact isometric group action. (proper=bounded closed sets are compact) -Any complete connected space for which the domain invariance theorem holds; in particular complete connected Riemannian manifolds without boundary.<|endoftext|> -TITLE: Examples of seemingly elementary problems that are hard to solve? -QUESTION [66 upvotes]: I'm looking for a list of problems such that -a) any undergraduate student who took multivariable calculus and linear algebra can understand the statements, (Edit: the definition of understanding here is that they can verify a few small cases by themselves ) -b) but are still open or very hard (say took at least 5 years to solve), -Edit: c) and first proposed in the 20th century or later. -Edit : my motivation is to encourage students to addict to solving mathematical problems. -I know that there are many such problems in number theory and combinatorics, for trivial example, Fermat's last theorem. I'll be more interested in other fields, but less famous problems in number theory or combinatorics would be also welcome. For example, -1) The $n!$ conjecture can be stated in an elementary language, but had been notoriously hard. See section 2.2 in Haiman's paper http://arxiv.org/abs/math.AG/0010246 . -2) Let $r$ be any positive integer, and let $x_1, x_2$ be indeterminates. Consider the sequence $\{x_n\}$ defined by the recursive relation $$x_{n+1} =(x_n^r +1)/x_{n-1}$$ -for any integer n. Prove that $x_n$ is of the form $P/Q$, where $P$ is a polynomial of $x_1$ and $x_2$ with non-negative coefficients, and Q is a monomial. This problem appeared as a special case of a conjecture by Fomin and Zelevinsky in the context of cluster algebras around 2001. They proved that $x_n$ can be written as $\frac{(\text{polynomial})}{(\text{monomial})}$. Proofs of positivity are recently obtained by Nakajima ( http://arxiv.org/abs/0905.0002 ) and Qin ( http://arxiv.org/abs/1004.4171 ). -3) Nagata conjecture : Let $r$ be a positive integer $\geq 10$, but not a square. Consider $r$ random points on the plane $R^2$. Let $m$ be any positive number. Prove that the degrees of plane curves passing through each of the $r$ points at least $m$ times are greater than $m\sqrt{r}$. See "Masayoshi Nagata, On the 14-th problem of Hilbert. Amer. J. Math. 81 (1959) 766–772". This is still wide open. -I'll be grateful for any more examples. - -REPLY [5 votes]: Consider all possible sequences $(x_1,\ldots,x_n)$ in $\mathbb{Z}^2$ such that $x_1=0$, $|x_{i+1}-x_i| = 1$, and $i\neq j \rightarrow x_i \neq x_j$. For any fixed $n$, there are only finitely many of them so we pick one uniformly at random. Question: what is the asymptotic behaviour of $R_n = \mathbb{E}\sup_{i\le n} |x_i|$ as $n \to \infty$? -One trivially has $R_n \le n$ and $R_n \ge \sqrt{n/\pi}$. The conjecture (about 70 years ago by Flory) is that $R_n \approx n^{3/4}$. (The original argument apparently contained conceptual mistakes and reached the "correct" value somewhat by miracle. The modern argument is that if the self-avoiding random walk has a conformally invariant scaling limit, then it has to be one of the SLE($\kappa$), but SLE($8/3$) is the only one with the restriction property, so it has to be that one. The exponent then follows from the known self-similarity properties of SLE.) -To appreciate how far one is from proving this, consider that there is no theorem showing that there exists $\epsilon > 0$ such that either $R_n \lesssim n^{1-\epsilon}$ or $R_n \gtrsim n^{{1\over 2}+\epsilon}$! The best is a relatively recent article by Duminil-Copin and Hammond showing that $\lim_{n \to \infty} R_n/n = 0$... To the best of my knowledge, that argument does not provide any rate at all, so we still don't know that $R_n \lesssim n/\log\log n$ for example...<|endoftext|> -TITLE: What is the non-motivic motivation behind automorphic representations? -QUESTION [11 upvotes]: In one of my last questions: -What is the "reason" for modularity results? -it was pointed out to me that "the notion of automorphic representation developed independently of any concern with modularity or class field theory. As I said, it has its own history, arising ultimately from the theory of elliptic integrals." Since automorphic representations never made intuitive sense to me, and since they are usually presented nowadays in relation to the Langlands program, I wondered what their original motivation was. In particular, can you recommend a readable and introductory source where I can learn about the non-motivic motivation behind automorphic representations? Why did they come onto the scene? What problem did they solve? - -REPLY [23 votes]: These are some comments that originally appeared on the OP's earlier question (linked above), gathered together here as an answer: -The notion of automorphic representation (as an irreducible representation of an adelic group) is a generalization of the notion of Hecke eigenform. This aspect of the theory of automorphic forms (i.e. the theory of Hecke operators and their simultaneous eigenvectors) was initiated by Hecke, as a means of understanding and generalizing Mordell's proof of Ramanujan's conjectured multiplicative relations for the $\tau$ function. The notion of automorphic form itself arose as a generalization of the notion of modular form. The latter arose out of the study of elliptic integrals and elliptic functions. -The generalization to automorphic forms took place over a long period of time, and was placed in a representation theoretic context by Gelfand and his school (as far as I know): they shifted the focus from functions on $G/K$ satisfying an automorphy condition under the action of $\Gamma$ to functions on $\Gamma\backslash G$, which then admit a $G$-action. (Here $G$ is a real semisimple group, say.) From this point of view, the interpretation of Hecke operators in terms of an adelic group action is not so remote. -But there are lots of other traditions feeding into the modern theory of automorphic forms, too. I believe that Maass was motivated to introduce his Maass forms in response to Hecke's theory relating Grossencharacters for imag. quad. fields to CM modular forms; Maass introduced automorphic forms that can play the same role for real quad. fields. I think that Selberg was motivated by Maass's papers to then study the spectrum of the Laplacian on modular curves, which led him to develop his trace formula, and, along the way, to effect the analytic continuation of Eisenstein series. It was generalizing this result to arbitrary groups that then led Langlands to discover general automorphic $L$-functions (see his book Euler products). -From the beginning of the theory of modular forms, theta series (generating functions of quadratic forms) had played a key role, and Siegel's work on more general automorphic forms was aimed at, among other things, generalizing this theory. It was Tamagawa (I think) who saw how to phrase some of Siegel's main results in terms of properties of the adelic quotient $G(\mathbb Q) \backslash G(\mathbb A)$, and he (and then Weil in his book -Adeles and algebraic groups) are thus responsible for introducing adelic groups into the subject (and for a reason not directly related to the theory of Hecke operators). -The realization that algebraic number theory and automorphic forms were related by (what we now call) modularity was something that evolved slowly, over a long period of the twentieth century. Even when it became concretely articulated, in the 60s and 70s, there were several strands of development feeding into it: of course there is the work and ideas of Langlands, -who defined automorphic L-functions in general, and saw directly the relationship between his functoriality conjecture and Artin's conjecture, and more broadly saw that his $L$-functions were candidates to be Hasse--Weil (i.e. motivic) $L$-functions; but there was also the work and ideas of Taniyama and Shimura about modularity of elliptic curves, Shimura's work on (what are now called) Shimura varieties (which gave another, previously unknown, link between arithmetic and automorphic forms), Serre's ideas about 2-dimensional Galois representations attached to modular forms, Weil's converse theorem (which was certainly a -decisive result, showing as it did that if Hasse--Weil $L$-functions had the anticipated properties, they were going to have to be automorphic $L$-functions) --- a result -which generalized earlier results of Hecke, and so on. -To quote Langlands on the subject of automorphic forms: It is a deeper subject than I appreciated and, I begin to suspect, deeper than anyone yet appreciates. To see it whole is certainly a daunting, for the moment even impossible, task.<|endoftext|> -TITLE: Isometry group of a homogeneous space -QUESTION [30 upvotes]: Background -Let $(M,g)$ be a finite-dimensional riemannian (or more generally pseudoriemannian) manifold. Suppose that I know that a certain Lie group $G$ acts transitively and isometrically on $M$ and after a little bit more work I exhibit $M$ as a homogeneous space $G/H$. Let -$$ -\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m}~, -$$ -where $\mathfrak{g}$ and $\mathfrak{h}$ are the Lie algebras of $G$ and $H$ respectively, and suppose that this split is reductive. (This is a nonempty condition in the pseudoriemannian case.) -The group $G$ is a subgroup of the isometry group of $M$, but it need not be the full isometry group. For example, we could have $M$ be a compact Lie group $G$ with the natural bi-invariant metric coming from (minus) the Killing form on the Lie algebra. Clearly $G$ acts transitively on $G$ via left multiplication, but the full isometry group is $G \times G$ (modulo the centre, if we insist that the isometry group acts effectively). -Question -Is there a way to determine the isometry Lie algebra of a homogeneous pseudoriemannian manifold $G/H$, preferably by algebraic means from the data $(\mathfrak{g},\mathfrak{h})$ and the $\mathfrak{h}$-invariant inner product on $\mathfrak{m}$? - -This is not idle curiousity, by the way. In some work I'm doing, I have encountered an explicit 7-dimensional homogeneous lorentzian manifold which I can describe as $G/H$, but it is important that I know whether the isometry Lie algebra is indeed $\mathfrak{g}$ or something larger. I fear, though, that writing down the explicit example I am faced with might be deemed "too localized" (and rightly so), and I'm hoping this more general question is acceptable. -Update -I have now ran the algorithm in Robert's answer for my 7-dimensional homogeneous space in my work and out popped one additional Killing vector field! (Not the result I wished for, but explains something I did not understand.) - -REPLY [36 votes]: Here is an algorithm to compute the Lie algebra of the group of isometries of a homogeneous space $G/H$ endowed with a $G$-invariant (pseudo-)Riemannian metric $g$. It is phrased in terms of essentially algebraic computations using the left-invariant forms on $G$, but it could be reduced completely to computations with the Lie algebra (and the matrix $Q$ that defines the metric) if that is what one wanted to do. -Let $\frak{g}$ and $\frak{h}\subset\frak{g}$ denote the Lie algebras of $G$ and $H\subset G$ respectively. Set $s = \dim\frak{h}$ and let $n>0$ be the dimension of $\frak{m} = \frak{g}/\frak{h}$. Let $\pi:G\to G/H$ be the canonical coset projection. If the natural left action of $G$ on $G/H$ is not effective, replace $G$ by $G/N$, where $N\subset G$ is the closed, normal subgroup $N$ consisting of the elements that act trivially on $G/H$. -Let $\omega = (\omega^i)$ (where $1\le i\le n$) be a basis for the left-invariant $1$-forms on $G$ such that $\omega = 0$ defines the foliation of $G$ by the left cosets of $H$. Then exists a unique non-degenerate, symmetric $n$-by-$n$ matrix of constants $Q$ such that $\pi^*g = Q_{ij}\,\omega^i\circ\omega^j$. -Since $Q$ is constant, there exists a unique $n$-by-$n$ matrix $\theta = (\theta^i_j)$, whose entries are left-invariant $1$-forms on $G$, such that -$$ -\mathrm{d}\omega = {}-\theta\wedge\omega -\qquad\text{and}\qquad -Q\theta + {}^t(Q\theta) = \mathrm{d}Q = 0. -$$ -(This is just the Fundamental Lemma of (pseudo-)Riemannian Geometry in this -context.) Note that applying $\mathrm{d}$ to both sides of this equation gives -$0 = \Theta\wedge\omega$, where, of course, $\Theta = \mathrm{d}\theta + \theta\wedge\theta$, is the curvature of the connection $\theta$ and hence can be written in the form $\Theta = R(\omega\wedge\omega)$, where the coefficients in $R$ are constants, since $R$ is left-invariant as a function on $G$. -Suppose now that a vector field $Y$ on $G$ be $\pi$-related to a $g$-Killing vector field $Z$ on $G/H$, and let $\omega(Y) = a$. Then the $g$-Killing equation for $Z$ implies that -$$ -\mathrm{d}a = {}-\theta\ a + b\ \omega -$$ -where $b$ is an $n$-by-$n$ matrix of functions that satisfies $Qb + {}^t(Qb)=0$. -Taking the exterior derivative of this equation gives -$$ -0 = {} -\Theta\ a + (\mathrm{d}b +\theta\ b - b\ \theta)\wedge\omega. -$$ -Now, by counting dimensions to show that the corresponding linear algebra problem always has a unique solution, it is easy to see (and, in any given case, explicitly compute) that there exists a matrix $\rho = (\rho^i_j)$ of $1$-forms such that $\Theta\ a = \rho\wedge\omega$ and such that $Q\rho + {}^t(Q\rho) = 0$. In fact, one has $\rho^i_j = r^i_{jkl}\,a^k\omega^l$, where the $r^i_{jkl}$ are constants determined by the curvature form $\Theta$. (The exact formula is not important for the following argument.) Thus, the above equation can be written as -$$ -\mathrm{d}b = -\theta\ b + b\ \theta + \rho(a,\omega), -$$ -where I have written the term $\rho$ as $\rho(a,\omega)$ to emphasize that this is some constant-coefficient bilinear pairing of $a$ and $\omega$ taking values in the Lie algebra ${\frak{so}}(Q)$. -The above equations for $\mathrm{d}a$ and $\mathrm{d}b$ are then a total (linear) differential system whose solutions give the Lie algebra of $g$-Killing vector fields on $G/H$. -Now, this system is not Frobenius unless $(G/H,g)$ is a (pseudo-)Riemannian space form, so, usually, one must differentiate these equations. The derivative of the $\mathrm{d}a$-equation won't give anything new, so one must differentiate the $\mathrm{d}b$-equation. This yields equations of the form -$$ -0 = \mathrm{d}(\mathrm{d}b) = B, -$$ -where $B = (B^i_j)$ and $B^i_j = (u^i_{jklm}a^m + v^{ip}_{jklq}b^q_p)\,\omega^k\wedge\omega^l$ for some explicit constants -$u^i_{jklm}= -u^i_{jlkm}$ and $v^{ip}_{jklq} = - v^{ip}_{jlkq}$. -It follows that the $a^i$ and $b^i_j$ are subject to the constant-coefficient linear relations -$$ -u^i_{jklm}a^m + v^{ip}_{jklq}b^q_p = 0 -$$ -in addition to the linear relations on $b$ already known: $Qb + {}^t(Qb) = 0$. Differentiating these new linear relations and using the $\mathrm{d}a$ and $\mathrm{d}b$ formulae to express the results in terms of left-invariant forms with coeffcients that are constant linear combinations of the $a$- and $b$-components, one might get further constant coefficient linear relations among the $a$- and $b$-components. Repeat this process with the new relations (if any) until no new linear relations are found. -At that point, the linear relations between the $a$- and $b$-components will have some solution space of dimension $n{+}s{+}r$ for some $r\ge0$ (but, necessarily, $r\le n(n{-}1)/2 - s$). It will then follow that the space of $g$-Killing vector fields on $G/H$ has dimension $n{+}s{+}r$. Moreover, one can compute the Lie algebra structure on this space by using the formula -$$ -\omega\bigl([Y_1,Y_2]\bigr) - = Y_1(a_2) - Y_2(a_1) - \theta\wedge\omega\bigl(Y_1,Y_2\bigr) -$$ -and the formulae for $\mathrm{d}a_1$ and $\mathrm{d}a_2$. Thus, the algebra structure of the $g$-Killing fields will follow directly by algebraic operations from the structure of the algebras $\frak{g}$ and $\frak{h}$ and $Q$. -In any given instance, this algorithm can be implemented on a computer without difficulty.<|endoftext|> -TITLE: Fermat-like equation $c^n=a^{2n}+a^n b^n + b^{2n}$ -QUESTION [13 upvotes]: Hello, -Is there a solution to the following equation: -$c^n=a^{2n}+a^n b^n + b^{2n}$ -where $a,b,c \in \mathbb{N}^*$, and $n$ is an integer $\geq 2$. -The problem is due to Antoine Balan. -Thanks in advance. - -REPLY [12 votes]: Yes, it's a nice question. It does seem that there are no solutions in -positive integers to A. Balan's equation, and indeed no integer -solutions at all other than those with $a=0$, $b=0$, or (when $n$ is -odd) $a+b=0$. As usual the problem is equivalent to finding all -rational points on an algebraic curve $B_n$; here $B_n$ is the -"superelliptic" curve with affine equation $y^n = x^{2n} + x^n + 1$. -We want to show that the only rational points are those with -$x=\infty$, $x=0$, and (when $n$ is odd) $x=-1$. Herewith: -1) Verification that $ab$ and $c$ may be assumed coprime; by -M.Bennett's clever rewriting and the results he quotes from DMP = -Darmon, Merel, and Poonen, this reduces the problem to $n=2$ and $n=3$. -2) $n=2$, worked out in more detail than in my comment to the original -question. -3) $n=3$; fortunately a genus-2 quotient $C_3$ of $B_3$ has torsion Jacobian. -4) A bit more on the analogy JSE suggested: -Balan is to $\lbrace0,\infty,\rho,\rho^2 \rbrace$ -as Fermat is to $\lbrace 0, 1, \infty \rbrace$ -(where $\rho = e^{2\pi i/3}$), and why the proof suggested -by this analogy seems to barely fail. -5) Where Bennett's approach fits in this picture. -(1) Mike finished his answer with the sentence "The condition on -coprimality might be a problem, but it's too early in the morning for -me to be sure!"; by now he's probably had the chance to check this, but -I don't see it yet as an edit or a comment, so here goes. To apply DMP -to $c^n + (ab)^n = (a^n+b^n)^2$ we need $c$ coprime to $ab$. If $p$ -divides $c$ and (say) $a$, then taking the equation mod $p$ yields -$p|b^n$, and then $p|b$, so $p^{2n} | a^{2n} + a^n b^n + b^{2n} = c^n$, -whence $p^2 | c$. At this point we may replace $(a,b,c)$ by the equivalent -solution $(a/p, b/p, c/p^2)$. Unless $(a,b,c)=(0,0,0)$, in finitely -many steps we reach an equivalent solution with $\gcd(ab,c)=1$ as -desired. If $n\geq 4$, it follows from DMP that at least one of $ab$, -$a^n+b^n$, and $c$ vanishes, and we're done. -It remains to deal with $n < 4$. -(2) $B_2: y^2 = x^4 + x^2 + 1$ is an elliptic curve. It has -rational points at infinity, so we put it in extended Weierstrass form -by the usual centuries-old technique. Choose a point $P$ at infinity, -and expand $y$ about $P$ in a Laurent series, truncating before the -constant term to get a quadratic polynomial; here it's just $x^2$. -Hence $t := y+x^2$ has a double pole at $P$ and nowhere else on the curve, -so it can be used as the abscissa. Writing $y=t-x^2$ in $y^2 = x^4 + x^2 + 1$ -gives a quadratic in $x$, namely $(2t+1)x^2 = t^2-1$, which has solutions -iff $4(2t+1)(t^2-1)=u^2$ for some $u$; -dividing $t$ by $2$ yields $(t+1)(t-2)(t+2) = u^2$, which is -the curve of conductor $48$ with coefficient vector $[0,1,0,-4,-4]$. -Thanks to rational $2$-torsion (and trivial Sha[2] obstruction), -Fermat's descent technique, now automated in Cremona's mwrank, -suffices to prove the curve has rank zero and no rational points -other than the four trivial ones we already knew of. -(3) $B_3: y^3 = x^6 + x^3 + 1$ has genus $4$. There's an obvious -quotient curve of genus $1$, namely -$E: Y^{\phantom.3} = X^{\phantom.2} + X + 1$, -which is already in the usual extended Weierstrass form -(with the roles of $X$ and $Y$ reversed). Unfortunately the -obvious points with $Y=1$ are of infinite order; indeed plugging the -coefficient vector $[0,0,1,0,-1]$ into mwrank we find that -$(X,Y\phantom.) = (0,1)$ (or equivalently $(-1,1)$) -generates the group of rational points. (We could have known in advance -that the rank cannot exceed $1$ because its conductor is $3^5 = 243 < 389$...) -So we cannot use this quotient to prove that $B_3$ has no more rational points. -Fortunately there is a genus-2 quotient $C_3: Y^{\phantom.3} = X^4 + X^3 + X^2$ -that does work. [To get this quotient, first write $B_3$ as -$(x^2 y)^3 = x^{12} + x^9 + x^6$, then take $(X,Y\phantom.) = (x^2 y, x^3)$.] -To put this curve in hyperelliptic form, divide through by $X^2$ -to get $z^3 X = X^2 + X + 1$ where $z = Y/X$, then find the discriminant -of this quadratic in $X$ to get $z^6 - 2z^3 - 3 = w^2$. The three -known points correspond to the Weierstrass point $(z,w) = (-1,0)$ -and the two points at infinity, call them $P_\pm$. -The subgroup of the Jacobian $J(C_3)$ represented by degree-zero -divisors supported on these three points is isomorphic with -${\bf Z} / 6 {\bf Z}$: twice the Weierstrass point is equivalent with -$P_+ + P_-$, and $3P_+ \sim 3 P_-$ because the difference is the -divisor of $z^3 - 1 - w$. [An analogous result holds for all odd $n$, -but the next step might not.] It turns out that this torsion group -is all of $(J(B_1))({\bf Q})$: I asked magma -_ := PolynomialRing(Rationals()); -C1 := HyperellipticCurve(z^6 - 2*z^3 - 3); -J := Jacobian(C1); -TorsionSubgroup(J); -RankBound(J); - -(adapting one of the examples in -http://magma.maths.usyd.edu.au/magma/handbook/text/1375), -and found that the Jacobian $J(C_3)$ has $6$ torsion points and rank zero. -It follows that our list of rational points on $C_3$, and thus -also on $B_3$, is complete. -[The Jacobian of the genus-$4$ curve $B_3$ is isogenous with -$E \times E \times J(C_3)$: in general the Jacobian of a superelliptic -curve of the form $y^m = P(x^m)$ is isogenous with the product of the -Jacobians of the $n$ curves $Y^{\phantom.m} = X^k P(X)$ with $k=0,1,2,\ldots,m$, -and here $k=0$ and $k=1$ both yield $E$, while $k=2$ yields $C_3$.] -(4) Recall the following geometrical description of the Frey curves -associated with a putative point on the Fermat curve $F_p: x^p + y^p = z^p$, -with $p>3$ prime. -The rational function $(x/z)^p$ on $F_p$ realizes $F_p$ as a $\mu_p^2$ -cover of ${\bf P}^1$, branched only above $\lbrace 0, 1, \infty \rbrace$. -Use the modular function $\lambda$ to identify ${\bf P}^1$ with -the modular curve ${\rm X}(2)$, whose three cusps are at $\lambda=0,1,\infty$. -The image of a putative Fermat counterexample yields a Frey curve -$E: Y^{\phantom.^2} = X (X \pm x^p) (X \mp y^p)$, -with the sign chosen to make the curve semistable even at $2$ -(this is where $p>3$ is needed), and thus modular. -But the cover ${\rm X}(2p) \rightarrow {\rm X}(2)$ of modular curves -has the same ramification points and orders as our map from $F_p$. -Once $E$, and thus the Galois representation on $E[p]$, has been proved modular, -it then follows that this mod-$p$ representation comes from a cuspform -of level $2$, which is impossible because fortunately ${\rm X}_0(2)$ -has genus zero. -Now the Balan curve $B_p$ is likewise a $\mu_p^2$ cover of ${\bf P}^1$, -but with four branch points, at $(a/b)^p = 0$, $\infty$, $\rho$, and $\rho^2$. -These don't seem to be the cusps of any modular curve. But for our purpose -it's enough for some subset to be the cusps, and JSE suggested the only -possibility: twist $\Gamma(2)$ by ${\bf Q}(\sqrt{-3})$ to -put two of its cusps at $\rho$ and $\rho^2$ and the third at $\infty$ -(or equivalently at $0$). Then the Frey curve becomes -$$ -E': Y^{\phantom.2} = X \bigl(X^2 + (a^p+2b^p) X + (a^{2p}+a^pb^p+b^{2p}) \bigr), -$$ -with discriminant $-48 a^{2p} (a^{2p}+a^pb^p+b^{2p})^p = -48 a^{2p} c^p$. -So we must deal with reduction at $3$ as well as at $2$. Now the -good news is that $E'$ is semistable at $3$: while there are coprime -integers $A,B$ such that -$Y^{\phantom.2} = X \bigl(X^2 + (A+2B) X + (A^2+AB+B^2)\bigr)$ -has additive reduction at $3$, this can only happen if $A\equiv B \mod 3$, -but then $A^2 + AB + B^2 \equiv 3 \bmod 9$ so it cannot be a $p$-th power. -The bad news is that $E'$ is not in general semistable at $2$, -and even after applying the available symmetries $(a,b) \leftrightarrow (b,a)$ -and $(a,b) \leftrightarrow (-a,-b)$ the power of $2$ in the conductor -might be as large $2^3$ $-$ this happens exactly when $a \equiv b \bmod 4$. -In this case, all that we can say is that the Galois representation -on $E'[p]$ is isomorphic to the the $p$-torsion of an elliptic curve -of conductor 24, such as $y^2 = x^3 - x^2 + x$ (it is not entirely -coincidental that this curve is a quadratic twist of the one we -encountered above for exponent $2$). At this point I'm stuck: -there might be a way to push this argument further or modify it -to finish off the proof, but all I can obtain this way is the -unsatisfying conclusion that in any primitive solution $a \equiv b \bmod 4$ -and $E'[p]$ has the same Galois structure as the $p$-torsion of -$y^2 = x^3 - x^2 + x$. The first condition follows from the computation -that in all other cases $E'[p]$ has conductor dividing $12$ for at least one -of the four equivalent choices of $(a,b)$, and then there's no possible -cuspform because ${\rm X}_0(12)$ is rational. This condition is peculiar -because the curve $B_n$ has good reduction at 2 (it's the construction of -the auxiliary curve $E'$ introduces problems at this prime). The -condition on $E'[p]$ seems rather hard to exploit... -(5) Fortunately there's another way, which comes down to what -Bennett found: compose the map, call it -$t = (a/b)^p: B_p \rightarrow {\bf P}^1$, -with the map $s: t \mapsto t + 1/t$, giving the quotient map -${\bf P}^1 \rightarrow {\bf P}^1$ under the involution -$t \leftrightarrow 1/t$ that switches $0$ with $\infty$ and -$\rho$ with $\rho^2$. This map has double points at $s = \pm 2$ -(images of the fixed points $t = \pm 1$ of the involution), -and takes $t=0,\infty$ to $s=\infty$ and $t = \rho,\rho^2$ -to $s = -1$. So now, in place of ${\rm X}(2)$ we need a -modular curve with cusps at $s=-1$ and $s=\infty$ -and an elliptic point of order $2$ at $s=2$ or $s=-2$ -(or possibly both). The modular curve ${\rm X}_0(2)$, -with two cusps and one elliptic point of order $2$, -does the trick, and if we choose to put the elliptic point -at $s=-2$ then there's no problem with reduction at $3$ -(putting it at $s=+2$ would cause such a problem because -$2 \equiv -1 \bmod 3$). This yields the elliptic curve -$E'': Y^2 = X \bigl(X^2 \pm 2(a^p+b^p) X + c^p \bigr)$. -That's still not the end of the game: the smallest -valuation at $2$ of any quadratic twist of $E''$ is $5$, -and the modular curve ${\rm X}_0(2^5)$ has genus $1$, -so there is a modular form of level $2^5$ that must be -dealt with. Fortunately this form, unlike the one of level $24$, -is CM. Thus the corresponding Galois representation mod $p$ -is well understood, and with some further cleverness and much effort -Darmon and Merel were able to dispose of this final case.<|endoftext|> -TITLE: Who coined the name tensor and why? -QUESTION [15 upvotes]: Who coined the name "tensor" and why? What does the word "tensor" really mean, not the mathematical definition? - -REPLY [18 votes]: I think the OP referes to the modern meaning of the word, in which case, according to that website, it first appeared in german physicist Woldemar Voigt's paper Die fundamentalen physikalischen Eigenschaften der Krystalle in elementarer Darstellung published in 1898. (I do not have access to this paper, but probably this deals with deformation tensors in crystals). - -REPLY [11 votes]: A tensor muscle is a muscle that stretches some part of the body, e.g. the tensor veli palatini or tensor tympani. The word -ultimately derives from the Latin tendere meaning "to stretch", see Douglas Harper's etymonline. -Hamilton first introduced the term to mathematics; see its entry in "Earliest Known Uses of Some of the Words of Mathematics".<|endoftext|> -TITLE: Brauer group of projective space -QUESTION [6 upvotes]: I've read that $\text{Br} \mathbb{P}^n_k$ (here $\text{Br}$ is the cohomological Brauer group, i.e. $H^2_{ét}(-,\mathbb{G}_m)$) is just isomorphic to $\text{Br} k$. As proof of this fact seems to be not so easy in the general case, but there should be a simple and conceptual proof when $k$ has characteristic zero. Does anyone know this simple proof? - -REPLY [4 votes]: Here is a proof that works in all characteristics. The idea comes from Colliot-Thélène ["Formes quadratiques multiplicatives et variétés algébriques: deux compléments", Bull. Soc. Math. France, 108(2), 213–227]. -We'll use several times that the Brauer group of a smooth variety injects into the Brauer group of its function field. -First we prove $\mathrm{Br}\,\mathbb{P}^1 = \mathrm{Br}\,k$. -By Tsen's theorem, this is true if $k$ is algebraically closed. -By Grothendieck ["Le Groupe de Brauer III", Corollaire 5.8], it is true if $k$ is separably closed, and so for general $k$ we have an isomorphism $\mathrm{Br}\, \mathbb{P}^1 = \mathrm{Br}_1\,\mathbb{P}^1$. -But $\mathrm{Br}_1\,\mathbb{P}^1$ is contained in $\mathrm{Br}_1\,\mathbb{A}^1$, and Auslander–Goldman ["The Brauer group of a commutative ring", Theorem 7.5] have proved $\mathrm{Br}_1\,\mathbb{A}^1 = \mathrm{Br}\,k$. -Now we prove $\mathrm{Br}\,\mathbb{P}^n = \mathrm{Br}\, k$ by induction. -Pick any point $P \in \mathbb{P}^n(k)$. Resolving the projection away from $P$ gives a morphism $\pi \colon X \to \mathbb{P}^{n-1}$, where $X$ is the blow-up of $\mathbb{P}^n$ at $P$. -By induction we have $\mathrm{Br}\,\mathbb{P}^{n-1} = \mathrm{Br}\,k$. -Since the injective map $\mathrm{Br}\,\mathbb{P}^n \to \mathrm{Br}\,k(\mathbb{P}^n)$ factors through $\mathrm{Br}\,X$, it is enough to prove that $\pi^* \colon \mathrm{Br}\,\mathbb{P}^{n-1} \to \mathrm{Br}\,X$ is surjective. -Let $\eta\colon \mathrm{Spec}\,K \to \mathbb{P}^{n-1}$ be the generic point. We have $X_\eta \cong \mathbb{P}^1_K$, so the natural map $\pi^* \colon \mathrm{Br}\,K \to \mathrm{Br}\,X_\eta$ is an isomorphism. -Now let $\sigma \colon \mathbb{P}^{n-1} \to X$ be a section (just take any hyperplane in $\mathbb{P}^n$ not containing $P$). Let $\alpha$ lie in $\mathrm{Br}\,X \subset \mathrm{Br}\,X_\eta$. We've just shown that $\alpha$ lies in the image of $\mathrm{Br}\,K$, that is, there exists $\beta \in \mathrm{Br}\,K$ such that $\alpha = \pi^* \beta$. Applying $\sigma^*$ shows that $\beta = \sigma^* \pi^* \beta = \sigma^* \alpha$ lies in $\mathrm{Br}\,\mathbb{P}^{n-1}$, that is, $\mathrm{Br}\,\mathbb{P}^{n-1} \to \mathrm{Br}\,X$ is surjective. -I've not been able to find this proof given explicitly in the literature.<|endoftext|> -TITLE: Elementary proof of the equidistribution theorem -QUESTION [15 upvotes]: I'm looking for references to (as many as possible) elementary proofs of the Weyl's equidistribution theorem, i.e., the statement that the sequence $\alpha, 2\alpha, 3\alpha, \ldots \mod 1$ is uniformly distributed on the unit interval. With "elementary" I mean that it does not make use of complex analysis in particular the Weyl's criterion. -Thank you very much. - -REPLY [4 votes]: Here is an elementary proof for the equidistribution of orbits of irrational rotations. Let $\alpha$ be any irrational number and fix $0\le a \frac{p}{q}$. Then $0 < j\alpha - j\frac{p}{q} < \frac{1}{q}$ for every $0\leq j0$. By Dirichlet's theorem there are arbitrarily large rational approximations to $\alpha$ that satisfy \eqref{eq:best}. Choose $p$ and $q$ coprime that satisfy \eqref{eq:best} with $q$ large enough ( such that $\frac{1}{q}<\frac{\varepsilon}{3}$ ), and choose $N$ to be large enough ( such that $\frac{q}{N}<\frac{\varepsilon}{3}$ ). -Then, for any $n\ge N$, write $n = sq+r$ with $0\le r < q$. The following estimate holds: -\begin{align*} -\# \left\{0\le j < n\,:\,\{j\alpha\} \in[a,b]\right\} & \ge -\sum_{t=1}^{s} \# \left\{(t-1)q\le j < tq\,:\,\{j\alpha\} \in[a,b]\right\} \\ -& \ge s\left(q(b-a) - 2\right) \\ -& = n\left(b-a\right) - 2s - r(b-a). -\end{align*} -In the second inequality the number of subintervals from a partition of the unit interval into $q$ subintervals of length $1/q$ that intersect $[a,b]$ bounds the number of orbit points in $[a,b]$. The error in this count is at most two, the reason being that in any partition of the interval into subintervals there are at most two subintervals that intersect $[a,b]$ but are not contained in it. Since $\frac{s}{n} \leq \frac{1}{q} < \frac{\varepsilon}{3}$ and $\frac{r}{n} \leq \frac{q}{n} < \frac{\varepsilon}{3}$, this proves that -$$ -\frac{\# \left\{0\le j < n\,:\,\{j\alpha\} \in[a,b]\right\}}{n} \ge b-a - \varepsilon. -$$ -The other inequality is proved similarly: -\begin{align*} -\# \left\{0\le j < n\,:\,\{j\alpha\} \in[a,b]\right\} & \le -\sum_{t=1}^{s+1} \# \left\{(t-1)q\le j < tq\,:\,\{j\alpha\} \in[a,b]\right\} \\ -& \le (s+1)\left(q(b-a) + 2\right) \\ -& = n\left(b-a\right) + 2(s+1) + (q - r)(b-a), -\end{align*} -so, -$$ -\frac{\# \{0\le j < n\,:\,\{j\alpha\} \in[a,b]\}}{n} \le b-a + \varepsilon. -$$ -Since $\varepsilon$ is arbitrary, this proves \eqref{eq:equidistribution}.<|endoftext|> -TITLE: Is there a non-Hopfian lacunary hyperbolic group? -QUESTION [21 upvotes]: The question's in the title and is easily stated, but let me try to give some details and explain why I'm interested. First, a disclaimer: if the answer's not already somewhere in the literature then it could be rather hard; I'm asking this question here because MO is lucky enough to have some of the foremost experts on lacunary hyperbolic groups as active participants. -Definitions - -A group $\Gamma$ is non-Hopfian if there is an epimorphism $\Gamma\to\Gamma$ with non-trivial kernel. -A group $\Gamma$ is lacunary hyperbolic if some asymptotic cone of $\Gamma$ is an $\mathbb{R}$-tree. - -To motivate this second definition, note that a group is word-hyperbolic if and only if every asymptotic cone is an $\mathbb{R}$-tree. -Lacunary hyperbolic groups were defined and investigated in a paper of Ol'shanskii, Osin and Sapir (although examples of lacunary hyperbolic groups that are not hyperbolic already existed---I believe the first one was constructed by Simon Thomas). They construct examples that exhibit very non-hyperbolic behaviour, including torsion groups and Tarski monsters. -Question -Once again:- - -Is there a non-Hopfian lacunary hyperbolic group? - -Motivation -My motivation comes from logic, and the following fact. -Proposition: A lacunary hyperbolic group is a direct limit of hyperbolic groups (satisfying a certain injectivity-radius condition). -That is to say, lacunary hyperbolic groups are limit groups over the class of all hyperbolic groups. (Note: the injectivity-radius condition means that there are other limit groups over hyperbolic groups which are not lacunary hyperbolic. I'm also interested in them.) Sela has shown that limit groups over a fixed hyperbolic group $\Gamma$ (and its subgroups) tell you a lot about the solutions to equations over $\Gamma$. For instance, his result that a sequence of epimorphisms of $\Gamma$-limit groups eventually stabilises (which implies that all $\Gamma$-limit groups are Hopfian) has the following consequence. -Theorem (Sela): Hyperbolic groups are equationally Noetherian. That is, any infinite set of equations is equivalent to a finite subsystem. -In the wake of Sela's work we have a fairly detailed understanding of solutions to equations over a given word-hyperbolic group $\Gamma$. But it's still a matter of great interest to try to understand systems of equations over all hyperbolic groups. -Pathological behaviour in lacunary hyperbolic groups should translate into pathological results about systems of equations over hyperbolic groups. A positive answer to my question would imply that the class of hyperbolic groups is not equationally Noetherian. And that would be quite interesting. -Note: This paper of Denis Osin already makes a connection between equations over a single lacunary hyperbolic group and equations over the class of all hyperbolic groups. - -This question attracted great answers from Yves Cornulier and Mark Sapir, as well as some excellent comments from Denis Osin. Let me quickly clarify my goals in answering the question, and try to summarise what the state of knowledge seems to be. I hope someone will correct me if I make any unwarranted conjectures! -My motivation came from the theory of equations over the class of all (word-)hyperbolic groups. For these purposes, it is not important to actually find a non-Hopfian lacunary hyperbolic group; merely a non-Hopfian limit of hyperbolic groups is enough. (That is, the injectivity radius condition in the above proposition can be ignored.) Yves Cornulier gave an example of a limit of virtually free (in particular, hyperbolic) groups which is non-Hopfian. From this one can conclude that the class of word-hyperbolic groups is not equationally Noetherian, as I had hoped. -[Note: I chose to accept Yves's answer. Mark's answer is equally worthy of acceptance.] -Clearly, the pathologies of Yves's groups derive from torsion---the class of free groups is equationally Noetherian---and there are some reasons to expect torsion to cause problems, so I asked in a comment for torsion-free examples. These were provided by Denis Osin, who referred to a paper of Ivanov and Storozhev. Thus, we also have that the class of all torsion-free hyperbolic groups is not equationally Noetherian. -Let us now turn to the question in the title---what if we require an actual lacunary hyperbolic group that is non-Hopfian. First, it seems very likely that such a thing exists. As Mark says, 'A short answer is "why not?"'. More formally, Denis claims in a comment that the subspace of the space of marked groups consisting of lacunary hyperbolic groups is comeagre in the closure of the subspace of hyperbolic groups. This formalises the idea that lacunary hyperbolic groups are not particularly special among limits of hyperbolic groups. -Mark also suggested two possible approaches to constructing a non-Hopfian lacunary hyperbolic group; however, in a comment, Denis questioned whether one of these approaches works. In summary, I feel fairly confident in concluding that a construction of a non-Hopfian lacunary hyperbolic group is not currently known, although one should expect to be able to find one with a bit of work. - -REPLY [6 votes]: Contrary to I think the expectations of the earlier answers, in a recent paper, Coulon--Guirardel showed that the answer to the title question is 'no'! - -Theorem (Coulon--Guirardel): Every lacunary hyperbolic group is Hopfian. - -The proof is remarkably short, and although it of course uses the fact that hyperbolic groups are Hopfian, it doesn't need the details of the proof.<|endoftext|> -TITLE: Where does the definition of "Tower of Algebras" come from? -QUESTION [5 upvotes]: A tower of algebras is a sequence of algebras -$$A_0 \hookrightarrow A_1 \hookrightarrow \cdots \hookrightarrow A_n \hookrightarrow \cdots$$ -with embeddings $A_n \otimes A_m \hookrightarrow A_{n+m}$ satisfying an associativity condition. -I found this definition recently in papers in representation theory such as: -"Algebraic structures on Grothendieck groups of a tower of algebras" -by Huilan Li and Nantel Bergeron -http://arxiv.org/pdf/math/0612170 -"Combinatorial Hopf algebras and Towers of Algebras - Dimension, Quantization, and Functoriality" -by Nantel Bergeron, Thomas Lam and Huilan Li -http://arxiv.org/pdf/0903.1381 -"Representation theories of some towers of algebras related to the symmetric groups and their Hecke algebras" -by F. Hivert and N. Thiéry -http://garsia.math.yorku.ca/fpsac06/papers/75.pdf -Also, I found this definition in a homotopy theoretic (and operadic) paper: -"On Quillen homology and a homotopy completion tower for algebras over operads" -by J. Harper and K. Hess. - -My question is: where does the definition of tower of algebras come from and what was the original motivation to introduce this object? - -REPLY [4 votes]: I don't know what the original motivation was (or where the original definition was). But one reasonable conceptual explanation for the definition is this: let $\mathcal C$ be a category with objects $\mathbb N$, endomorphisms $A_n$, and other morphism spaces 0. Then the embeddings $A_m \otimes A_n \to A_{m+n}$ (and the associativity conditions) are what is needed to provide $\mathcal C$ with a strict monoidal structure, where $m \otimes n := m + n$. Then one reason towers of algebras tend to be interesting is that monoidal categories tend to be interesting.<|endoftext|> -TITLE: Mathematics needed for higher dimensional category theory? -QUESTION [5 upvotes]: I'm a undergrad(third year, Manchester uni and want to do a PhD) that is thinking of doing a PhD in this area or category theory in general.(Sorry for asking it here, Maths exchange stack didn't help as asked twice last week and then today, I spend like an hour a day worrying about what I'm studying and what I should study). Imagine a chess board and you need to move a piece, how you know you are moving the best piece?, now times that by a thousand. -Just wondering, what branches of Maths should I focus on? As I've been told that topology particular Algebraic topology is the main area for this. A lecturer told me I should focus on getting down Algebraic topology before thinking of doing category theory as most of the examples of category theory are from algebraic topology. -However, another lecturer told me I should be doing logic and particular model theory. I can take a course in it this year, but would mean not doing commutative algebra. -So before you delete this can you please tell me what I should be studying? As it would save me countless hours of worrying if I'm doing the right subjects. -P.S. I can do most of the stuff in Conceptual Mathematics category book. I don't understand topoi through. - -REPLY [20 votes]: It seems to me that category theory (whether higher or not) may be the worst part of mathematics to approach with a narrow viewpoint --- just topology, or just algebraic geometry, or just logic, or just any one area. To me, much of the value and beauty of category theory lies in how it exhibits connections and similarities among many areas of mathematics.<|endoftext|> -TITLE: In an arbitrary abelian category, does chain complex homology commute with coproduct? -QUESTION [13 upvotes]: On page 55 of Weibel's Introduction to homological algebra the following passage appears: - -Here are two consequences that use the fact that homology commutes with arbitrary direct sums of chain complexes - -I understand why homology commutes with arbitrary direct sums when the direct sum of a collection of monics is a monic (i.e the direct sum functor is exact) but I was under the impression that there were abelian categories where the direct sum functor is not exact. After a bit of thought, I realised that I don't know an example of an abelian category in which the coproduct functor is not exact. -Sheaves of abelian groups on a fixed topological space give an example of an abelian category in which the product functor is not exact. - -Question 1: Is the passage from Weibel's book correct? If so, then why? -Question 2: Is there an example of an abelian category where the direct sum functor is not exact? - -REPLY [28 votes]: This was an error in the original book, and I added a correction to the -errata in 2007. Homology does not commute with direct sums unless (AB4) -holds, as Sam points out. --Chuck Weibel<|endoftext|> -TITLE: Berry Esseen type result for probability density functions -QUESTION [13 upvotes]: Let $X_1, X_2, \cdots$ be i.i.d. random variables with $E(X_1) = 0, E(X_1^2) = \sigma^2 >0, E(|X_1|^3) = \rho < \infty$. -Let $Y_n = \frac{1}{n} \sum_{i=1}^n X_i$ and let us note $F_n$ (resp. $\Phi$) the cumulative distribution function of $\frac{Y_n \sqrt{n}}{\sigma}$ (resp. of the standard normal distribution). -Then, Berry Esseen theorem states that there exists a positive constant $C$ such that for all $x$ and $n$, -$$|F_n(x)-\Phi(x)| \leq \frac{C \rho}{\sigma^3 \sqrt{n}}.$$ -Are there known conditions on the distribution of $X_1$ that allow to derive a similar statement for probability density functions instead of cumulative distribution functions? - -REPLY [9 votes]: The magic words are "local limit theorem".<|endoftext|> -TITLE: Is there an easy description of the structure of this infinite group? -QUESTION [21 upvotes]: Let $S_\infty$ denote the full symmetric group on countably many points and index the points by $\mathbb{N}$. For any weight (for lack of a better name) function $w:\mathbb{N}\rightarrow\mathbb R^+$, define a subgroup $S_\infty(w)\subset S_\infty$ as follows: -$$S_\infty(w) := \{\pi\in S_\infty|\sum_{\pi(k)\neq k} w(k)<\infty \}$$ -The intuitive idea is that given a weight function $w$ a permutation $\pi\in S_\infty$ is in $S_\infty(w)$ if the sum of the weights of all points moved by $\pi$ is finite. -There are two well-known examples of such subgroups: - -$S_\infty(1)$ is the group of finitely supported permutations of $\mathbb{N}$ or equivalently the direct limit of the finite symmetric groups. -$S_\infty(\frac{1}{n^2})$ is simply $S_\infty$ since the series $\{\frac{1}{n^2}\}$ is convergent. - -So in this context, I have been thinking about the group $G := S_\infty(\frac{1}{n})$. It is clear $S_\infty(1)\not\cong G$ since $G$ contains many subgroups isomorphic to $S_\infty$ (for example the set of all permutations of the points indexed by $b^i$ for $b$ fixed and $i\in\mathbb{N}$). -On the other hand, $G$ contains a proper subgroup isomorphic to the direct sum of countably many copies of $S_\infty$, so in particular it is easy to see $S_\infty\not\cong G$ since in $G$ elements of infinite order always have nontrivial centralizers while in $S_\infty$ this is not always the case. - - -Question: Aside from the description of $G$ as $S_\infty(\frac{1}{n})$, is there any easy to describe/understand presentation of this group? - - -Primarily I am torn on whether $G$ is itself isomorphic to a direct sum of countably many copies of $S_\infty$; every time I attempt to describe such an isomorphism, there are elements of $G$ not accounted for. - -REPLY [5 votes]: Not an answer, just a series of observations which will hopefully be of some use, and too long for a comment. The last observation might be relevant to your question about whether $G$ is the direct sum of countably many copies of $S_\infty$, unless I've over-simplified something. - -For each weight $w$, we may associate the collection -$$I_w = \left\{A \subseteq \omega\ |\ \sum_{n \in A}w(n) < \infty\right\}$$ -Then $I_w = P(\omega)$ iff $\sum_{n \in \omega} w(n) < \infty$. If the series is divergent, then $I_w$ is a non-principal ideal extending the ideal $\mathrm{Fin}$ of finite sets, and moreover this ideal is not maximal, i.e. its dual filter is not an ultrafilter. -$S_{\infty}(w) = \bigcup _{A \in I_w}S_A$ where $S_A$ is the subgroup of $S_{\infty}$ which fixes $\omega \setminus A$ pointwise, i.e. it's the set of permutations of $A$. -The map $(I_w,\subseteq) \to ($subgroups of $S_\infty,\leq)$ defined by $A \mapsto S_A$ is an injective homomorphism of posets. We're essentially interested in the union of the range of this map, but unfortunately this is not equal to the image of the union of the domain of this map. -Also note that $(I_w,\subseteq)$ is a lattice and a directed set, as is the range of the above map. -If $\lim_{n\to\infty}w(n) = 0$ but $\sum_{n\in\omega}w(n) = \infty$, then there is no partition of $\omega$ into countably many infinite sets $A_n (n \in \omega)$ such that $S_\infty(w)$ is isomorphic to the direct sum of the $S_{A_n}$ in the "obvious" way. That is, $S_\infty(w)$ doesn't consist merely of those permutations which can be expressed as a finite product of elements from finitely many of the $S_{A_n}$. [Proof: For each $n$, pick some $m_n \in A_n$ such that $w(m_n) < 2^{-n}$. Set $A = \{m_n : n \in \omega\}$. Then $S_A \leq S_{\infty}(w)$ but there are elements of $S_A$ which can't be written as a finite product of elements of the $S_{A_n}$.]<|endoftext|> -TITLE: Applications of the GCD metric -QUESTION [14 upvotes]: In the pre-MO era, I once realized that on the integers, the function -$$ - d(m, n) := \sqrt{\log \frac{\sqrt{mn}} {\text{gcd}(m,n)}}\ , -$$ -is a metric (all properties are easily verified; in fact this is a Hilbertian metric). -Now that I got reminded of it, I wanted to ask - -Has anyone seen this metric before? If so, in what context? What could be potential applications of this metric? - - -Added Given the formulae mentioned by Emil and Quid, an additional thing that I am wondering about is the following (I hope I'm not being obtuse): - -Does the fact that $d(m,n)=\|\phi(m)-\phi(n)\|$, where $\phi(m)$ and $\phi(n)$ lie in some Hilbert space, have any interesting ramifications? - -REPLY [5 votes]: The closest thing I am aware of would be a word metric. More precisely the word metric on the group of positive rationals (with multiplication) relative to the natural generating set formed by the prime numbers and their reciprocals. -The distance between two (positive) integers $m,n$ in this metric is then -$$\sum_p |v_p(m) - v_p(n)|$$ -where the sum goes over the primes and $v_p$ is the p-adic valuation. -As per Emil's comment your metric is then closely related to a 'weighted version' of this.<|endoftext|> -TITLE: Why SU(3) is not equal to SO(5)? -QUESTION [6 upvotes]: I am asking in the sense of isometry groups of a manifold. SU(3) is the group of isometries of CP2, and SO(5) is the group of isometries of the 4-sphere. Now, it happens that both manifolds are related by Arnold-Kuiper-Massey theorem: $\mathbb{CP}^2/conj \approx S^4$; one is a branched covering of the other, the quotient being via complex conjugation. -Now, for the case of a manifold and a lower dimensional submanifold, it is not rare to find that the corresponding isometry groups are subgroups one of the other. But here, which is the equivalent result? is SO(5) an "enhanced SU(3)" in some way? -The context of the question comes from 11D Kaluza Klein, more particularly from the classification of Einstein metrics in compact 7-manifolds. It is easy to produce from the 7-sphere metric a "squeezed sphere" whose isometry group is, instead of SO(8), just the one of $S^4 \times S^3$. But it is not known if there is some relationship between the 7-sphere and the "Witten manifolds" of the kind $CP^2 \times S^3$. -EDIT: to add more context, some dynkin diagrams. -o====o SO(5), isometries of the sphere S4 -o----o SU(3) are the isometries of CP2 -o o SU(2)xSU(2), isometries of S2xS2. Also SO(4), so isometries of S3 - -So it seems that the quotient under conjugation has implied, or is compensated by, some change in the angles between roots, but not in the number of roots. -For isometries of 7-manifolds we have also some similarities. - o o o - / - / -o----o SO(8) o----o SU(3)xSO(4) o====o SO(5)xSO(4) - \ - \ - o o o - -where the first diagram is the [isometry group of] the seven sphere, the last is the squashed sphere, and the intermediate is the one I am intrigued about, as it contains the physicists standard model gauge group. -By the way, the last drawing makes one to ask about how triality survives in the representations of these product groups, but that is other question :-) - -REPLY [8 votes]: $SU(3)$ has center of order 3 and $SO(5)$ has center reduced to the identity. In fact they are not even locally isomorphic: $SU(3)$ is of Cartan type $A_2$ and $SO(5)$ is of type $B_2$.<|endoftext|> -TITLE: Are there large cardinals for $n$-elementarity? -QUESTION [8 upvotes]: In July, Asaf Karagila asked three questions about elementary substructures of the universe of sets. The latter two were answered, the upshot being that the hypothesis $V_\kappa \prec V$ doesn't alone endow $\kappa$ with any large cardinal properties. However, Asaf's first question went unanswered: - -For given $n$, is there a large cardinal property $\Phi_n$ such that, if $\kappa$ is a $\Phi_n$ cardinal, then $V_\kappa\ \prec_n\ V\ $? - -As he noted, for $n = 1, 2, 3$ we can take $\Phi_n$ to be inaccessibility, supercompactness, and extendibility, respectively. (See Kanamori, pp. 299 and 319.) What of larger $n$? -I'm curious about this in connection with Zermelo's "Grenzzahlen" picture, which suggests a view of set existence on which there is no fixed universe of sets, but rather an indefinitely extensible sequence of natural universes, each contained as a set in all its successors. To salvage the separate claim that each set theoretic statement has a determinate truth value, someone with this "no fixed universe" view might argue that, for each $n$, there are arbitrarily large $\Phi_n$ cardinals; faced with a $\Sigma_n$ sentence $\varphi$, she could then declare that $\varphi$ is true simpliciter iff $V_\kappa \vDash \varphi$ for some (or each) $\Phi_n$ cardinal $\kappa$. -Of course, someone with a "no fixed universe" view wouldn't formulate things as I did in the displayed line above, since on her view there is no such object as $V$. Rather, she might wonder: - -For given $n$, is there a large cardinal property $\Psi_n$ that "transcends" $\Psi_m$ for all $m < n$ and is such that, if $\kappa < \lambda$ are $\Psi_n$ cardinals, then $V_\kappa\ \prec_n\ V_\lambda\ $? - -Given a positive answer, she could define truth simpliciter much as above, so that the truths are those propositions that hold in all sufficiently rich natural models. - -REPLY [11 votes]: The $\Sigma_n$ reflecting cardinals, of course, have -precisely the property that you mention, by definition. A -cardinal $\kappa$ is $\Sigma_n$-reflecting if -$V_\kappa\prec_{\Sigma_n} V$. One sometimes sees this term -defined in such a way to insist also that $\kappa$ is -inaccessible, but to my way of thinking, these should -simply be called the inaccessible $\Sigma_n$-reflecting -cardinals, as the concept is coherent and useful without -the extra inaccessibility hypothesis. -One can prove in ZFC, for any particular $n$, that there is -a closed unbounded class of $\Sigma_n$-reflecting -cardinals, and this is essentially the content of the -Reflection Theorem. If one adds the hypothesis that -$\kappa$ is inaccessible, then for $n\geq 2$ it follows -that there is a proper class of inaccessible cardinals and -more, but the strength is bounded below a Mahlo cardinal -(that is, very low in the large cardinal hierarchy), -because if $\delta$ is Mahlo, then $V_\delta$ has a club in -$\delta$ of $\kappa$ with $V_\kappa\prec V_\delta$, and -this club will then contain many inaccessible cardinals. -One may easily combine the reflecting idea with other large -cardinal concepts to obtain new stronger large cardinal -concepts that also exhibit your desired property. For -example, rather than considering an inaccessible -$\Sigma_n$-reflecting cardinal, one may consider a -measurable $\Sigma_n$ reflecting cardinal, or a -supercompact $\Sigma_n$-reflecting cardinal, and so on. -These large cardinal concepts will again exhibit the -property you request, but be much stronger in large -cardinal consistency strength. -Perhaps the right perspective is that the -$\Sigma_n$-reflecting idea is a way of adding "epsilon" to -a large cardinal concept. The $\Sigma_5$-reflecting -measurable cardinal hypothesis, for example, is definitely -stronger than mere measurability, but it is not as strong -as a cardinal $\delta$ that is stationary for -measurability---meaning that $\delta$ is regular and the measurable cardinals below $\delta$ are a stationary subset of $\delta$---since if the set of measurable cardinals -$\kappa$ below $\delta$ is stationary in $\delta$, then -since there is a club subset of $\delta$ of $\alpha$ with -$V_\alpha\prec V_\delta$, there will be many measurable -$\kappa$ that are fully reflecting in $V_\delta$. -In particular, if $\kappa$ has nontrivial Mitchell rank, -then the set of measurable cardinals below $\kappa$ is -stationary, and this hypothesis is satisfied. So for -example, if $\kappa$ is only $(\kappa+2)$-strong (also -known as the $P^2(\kappa)$-hypermeasurable cardinals), then -this is plenty strong enough. -The concept of a large cardinal having property $X$ and -also being $\Sigma_n$-reflecting is weaker in consistency -strength than a cardinal such that the $X$-cardinals below -are stationary, which is another common way of adding -"epsilon", in other words, of making a mild increase in -consistency strength over a given large cardinal concept. -For any fixed large cardinal concept $X$, the notion $\Phi_n(\kappa)=\kappa$ has property $X$ and is $\Sigma_n$-reflecting has the properties that you request in the second part of your question, while also being stronger than $X$. In most cases, however, this property is at the same time weaker than the typical next cardinal $Y$ above $X$, since as I mentioned, insisting on this level of reflection on top of $X$ is only adding a little, and is usually swamped by the next larger large cardinal concept. Thus, the following sequence provides a way of strengthening (in consistency strength) any given large cardinal concept: - -$\kappa$ has large cardinal property $X$. (e.g. inaccessibility, measurability, supercompactness, etc.) -There are a proper class of cardinals with property $X$. -There is a $\Sigma_2$-reflecting cardinal with property $X$. -There is a $\Sigma_3$-reflecting cardinal with property $X$, or $\Sigma_n$-reflecting, etc. -There is a stationary class of cardinals with property $X$. -There is a regular cardinal $\delta$ with the $X$-cardinals below $\delta$ being stationary. - -Each step increases strictly in consistency strength (for the reflecting case, you have to get above the complexity of the $X$ notion to get strict increases). So this shows that asking for reflection along with a given large cardinal notion is something in between asking for a proper class of those cardinals and asking for a stationary proper class of those cardinals. This is the same as the move from inaccessible cardinals to Mahlo cardinals, and the move from Mahlo to $1$-Mahlo, or hyper-Mahlo, and so on. But ultimately, that move is subsumed by the moves to higher levels of the large cardinal hierarchy. -It is traditional in set theory to consider individual large cardinal properties, rather than hypotheses implying a proper class or a stationary class of cardinals with the property, but these extra hypotheses are merely strenthenings of a given notion that typically have strength less than the next higher large cardinal notion.<|endoftext|> -TITLE: Lie $2$-groups and differential equations -QUESTION [11 upvotes]: I was reading the abstract of a recent preprint (Division Algebras and Supersymmetry III by Juhn Huerta), and I wondered if something much simpler than what he was talking about had been worked on: have Lie $2$-groups been applied to the resolution of differential equations, in the same manner that Lie groups originated from the study of differential equations? -In other words, do Lie 2-groups arise as symmetries for (certain kinds of) differential equations, and can these in turn be used for the integration/resolution of those same differential equations? If they do not, then in what setting can a 2-group be understood as a symmetry (if any), and to what 'use' can this information be put? -My motivation here is to expand the toolset I can use to solve problems in classical analysis (like differential equations), and not to explore the other areas where Lie groups have developed in to (like Lie algebras and their classification, etc). For the purposes of this question, these issues are out-of-scope. In-scope are applications (to classical analysis) of generalizations going all the way to $\infty$-Lie groupoids. - -REPLY [11 votes]: A well-studied special case of higher symmetries of differential equations is that of differential equations that arise as Euler-Lagrange equations of local action functionals. The symmetries and symmetries-of-symmetries and symmetries-of-symmetries-of-symmetries of such a system of equations form an $\infty$-groupoid whose infinitesimal version is encoded by the corresponding BRST complex -- which is the Chevalley-Eilenberg algebra of the corresponding L-∞ algebroid. In simple cases (or else locally) this is the global quotient by a smooth ∞-group: the "ghosts" in the BRST complex are the cotangents to the local symmetries, the "ghosts-of-ghosts" are the cotangents to the local symmetries-of-symmetries, and so on. -For instance - -for the action functional of the Yang-Mills field the symmetries form an ordinary Lie group; -for the action functional of the Kalb-Ramond field the symmetries form the circle 2-group $\mathbf{B}U(1) = (U(1) \to 1)$, (or rather the 2-group of functions with values in the circle 2-group); -for the action functional of the supergravity C-field the symmetries are governed by the circle 3-group $\mathbf{B}^2 U(1) = (U(1) \to 1 \to 1)$; -the higher abelian Chern-Simons theory in dimension $4k+3$ has the circle (2k+1)-group $\mathbf{B}^{2k} U(1)$ as its gauge group; -the symmetries of full string field theory form a general $\infty$-group (not an $n$-group for any finite $n$) the structure of which nobody really understands, I think. -every ∞-Chern-Simons theory (or equivalently its Euler-Lagrange equations) has a higher group of symmetries. In general, this is not just a higher gauge group, but even a higher gauge groupoid . - -the gauge groupoid of the Poisson sigma-model is controled by the Lie integration of a [Poisson Lie algebroid](http://ncatlab.org/nlab/show/Poisson+Lie algebroid), which is a [symplectic+groupoid](http://ncatlab.org/nlab/show/symplectic groupoid); -the gauge 2-groupoid of the Courant sigma-model is controled by the Lie integration of a Courant Lie 2-algebroid, which is a symplectic Lie 2-algebroid; -the gauge $n$-groupoid of a grade $n$ AKSZ sigma-model is similarly controled by a symplectic Lie n-groupoid. -the 7-dimensional "fivebrane Chern-Simons theory" has string 2-group-symmetries<|endoftext|> -TITLE: Definition of relativization of complexity class -QUESTION [14 upvotes]: Is there any general definition, for a class $C$ of languages, what is the relativized class $C^A$ for an oracle $A$? -Usually, these classes and their relativizations seem to be defined in an ad-hoc way. For example, $P$ is the class of languages decided by poly Turing machines, $P^A$ is the class of languages decided by poly Turing machines with access to $A$. There is not any logic, as far as I can see, to tell you how to add $A$ to the definition of $P$. -Obviously, the definition of a relativization of $C$ depends on more than simply the set of languages that comprise $C$. For example, $P = NP$ is unknown, but there exists an oracle $A$ such that $P^A \neq NP^A$. So, the definition of the relativization should somehow have access to the "defining property" of $C$. But, is there any rigorous way to do this? - -REPLY [15 votes]: [EDIT: I wrote my original answer when I was in a bit of a hurry. I have now expanded my answer.] -The short answer is no. The simplest way to see that $C^A$ cannot possibly depend only on $C$ and $A$ as sets of strings is the following spurious argument that has confused generations of students. Assume that $P=NP$. Then for all oracles $A$, $P^A = NP^A$. But by Baker–Gill–Solovay, we know that there exists an oracle $A$ such that $P^A \ne NP^A$. This is a contradiction. Hence $P\ne NP$. Q.E.D., and I await my $1 million check. -The standard notation $C^A$ is an abuse of notation. A class per se cannot be relativized; one must specify a model of computation and provide that model of computation with access to an oracle. Choosing different models of computation, or different oracle access mechanisms, may give you different relativizations. For example, space-bounded classes are notoriously tricky to relativize, as explained in this paper by Hartmanis et al. The trouble is that it's not obvious what the "right" oracle access mechanism is in certain cases. -Having said that, I think it is a fascinating open-ended problem to try to "formalize" the concept of relativization in a way that would allow one to rigorously prove statements of the form, "such-and-such a type of argument relativizes and therefore cannot (for example) separate $P$ from $NP$." Though this is a very tempting idea, there is still no fully satisfactory theory along these lines. Probably the best attempt in this direction so far is the unpublished manuscript Relativizing versus nonrelativizing techniques: the role of local checkability, by Arora, Impagliazzo, and Vazirani, but this is really just a first attempt. For example, in his paper The role of relativization in complexity theory, Fortnow argues that there are certain important limitations to the Arora–Impagliazzo–Vazirani model. Fortnow's paper is recommended reading if you want to work on this open problem yourself.<|endoftext|> -TITLE: Minimum cardinality of a difference set in $R^n$ -QUESTION [7 upvotes]: Cross-posted from https://math.stackexchange.com/questions/65195/minimum-cardinality-of-a-difference-set-in-mathbb-rn. -Given a finite set $S$ of $m$ points in $\mathbb R^n$ that do not all lie in the same $(n-1)$-dimensional hyperplane, consider the set of difference vectors: -$\{x-y \, | \, x,y \in S\}$ -What is the minimum cardinality of this set, as a function of $m$ and $n$? -(The sets that minimize this should be "small" subsets of a lattice, but I don't know what specific shapes minimize it.) -What is the status of exact results for this problem for small $n$ (say $n = 2$ or $3$)? - -REPLY [8 votes]: A basic inequality proved in 1987 by Freiman, Heppes, and Uhrin ("A lower estimation for the cardinality of finite difference sets in $R^n$", Number theory, Vol. I (Budapest, 1987), 125–139, Colloq. Math. Soc. János Bolyai, 51, North-Holland, Amsterdam, 1990) is that $|S-S|\ge(n+1)|S|-n(n+1)/2$. A number of improvements have been obtained since then; in particular, Stanchescu ("On finite difference sets", Acta Math. Hungarica 79 (1998), no. 1-2, 123–138) showed that for $n=3$ one has $|S-S|\ge 4.5|A|-9$, with an explicit description of those sets $S$ for which equality is attained. -You can recover much more starting with these two papers and their MathReviews.<|endoftext|> -TITLE: What is the tensor product for the Eilenberg-Moore category of a commutative monad? -QUESTION [7 upvotes]: In Linear logic, monads and the lambda calculus (DRAFT), Proposition 3.0.2 says that the Eilenberg-Moore category for a commutative monad has the structure of a symmetric monoidal closed category. My question is: How do you construct the tensor product for the EM-category? When I followed the reference to Kiegher's paper "Symmetric monoidal closed categories generated by commutative adjoint monads", that paper only seems to give the construction for monads of the form A ⊗ -. I don't see how to generalize the construction to arbitrary commutative monads. - -REPLY [13 votes]: Caveat: this construction only works if your category of algebras has coequalizers of reflexive pairs, i.e. coequalizers of parallel pairs of arrows with a common right inverse. -Let $T \colon C \to C$ be your monad. Being commutative, it comes with maps $\mathrm{dst} \colon T(A) \otimes T(B) \to T(A \otimes B)$. -Let $\phi \colon TA \to A$ and $\psi \colon TB \to B$ be algebras. Then $\phi \otimes \psi$ is the coequalizer in $\mathrm{Alg}(T)$ of $T(\phi \otimes \psi)$ and $\mu \circ T(\mathrm{dst})$ (which is a reflexive pair of morphisms from the free algebra on $T(A) \otimes T(B)$ to the free algebra on $A \otimes B$). The unit $I$ in $\mathrm{Alg}(T)$ is the free algebra $\mu \colon T^2(I) \to T(I)$. Moreover, the free functor $C \to \mathrm{Alg}(T)$ preserves monoidal structure. -A good example to keep in mind is where $T$ is the free vector space monad on the category of sets. The coequalizers then is pretty much directly the usual tensor product construction with bilinear maps. -This goes back to Anders Kock, see his papers -"Closed categories generated by commutative monads" (J. Austr. Math. Soc. 12:405-424, 1975), and -"Monads on symmetric monoidal closed categories" (Archiv der Mathematik, 21:1-10, 1970).<|endoftext|> -TITLE: The space of valuations of a function field -QUESTION [7 upvotes]: Hello, I'm looking for someone who can help me to understand Zariski's theory of valuations. -First I outline the theory: we take a field $K$ which is a finitely generated transcendent extension of another field $k$. We only consider the case $k=\mathbb{Q}$ or $\mathbb{C}$. By definition, A model of $K$ is a variety $V\subset \mathbb{CP}^n$ defined over $k$, such that the rational function field of $V$ over $k$ is isomorphic to $K$. We define the underlying topological space of $V$ to be a space, whose points are irreducible subvarieties of $V$, endowed with Zariski topology. -Now comes the interesting thing: Zariski gave an homeomorphism between the space of valuations on $K/k$ and the inverse limit of underlying topological spaces of all models of $K$. -Question: Plz give me some concret examples of the above correspondence. -The only example I know is that, given an irreducible hypersurface of a model $V$, one can count the order of rational functions on $V$ over the hypersurface. This gives a discret rank one valuation. -Is there some other easily-described points in the inverse limit, whose corresponding valuations are non-discret, or of higher rank? -Other comments are welcome! - -REPLY [3 votes]: Simply put (more or less already said above). Let $K=C(x,y)$ and let $xi(t)$ be a generalized power series (a power series with well-ordered exponents) where the exponents are non-negative real numbers. Assume (this is important) that $P(t,\xi(t))\neq 0$ for any $P\in K$. Then the map: $\nu(P(x,y))=ord_{t}(t,\xi(t))$ is a valuation (the "order of contact of P with $(t,\xi(t))$"). -Notice that if $\xi(t)=t^{\pi}$, for example, the valuation has rational rank $2$. If $\xi(t)$ corresponds to an analytic branch of a curve (non-algebraic), the valuation has rational rank $1$, etc. -By the way, here you may find something useful. That book may be of help. -You should read something about point blowing-ups and then you understand the projective limit thing. But without it, it gets somewhat too algebraic.<|endoftext|> -TITLE: Is there a relation between 4-dimensional general relativity and exotic smooth structures on $\mathbb{R}^4$? -QUESTION [17 upvotes]: Let's say General Relativity is the study of the Einstein equation on smooth Lorentzian manifolds, i.e. pseudo-Riemannian manifolds of signature $(n-1,1)$. -I've heard more than once people say that the reason why there are many non-diffeomorphic smooth structures on $\mathbb{R}^n$ only for $n=4$ would be somehow "related to the fact that the spacetime in which we live is $4$-dimensional". Or, on the other hand, that there are exotic smooth structures only on $\mathbb{R}^4$ because "that's the dimension of spacetime". - -So, first of all - -1) Are there properties or phenomena of general relativity that are "specific" to $\mathbb{R}^{3,1}$, that is, the equivalent version on $\mathbb{R}^{n-1,1}$ does not hold for $n\neq 4$, or more generally on any other Lorentzian manifold homeomorphic to $\mathbb{R}^4$? - -I know string theorists proved that if you cross $\mathbb{R}^{n-1,1}$ by a Calabi-Yau manifold then $n=4$ would result in a "critical dimension" for some properties (supersymmetry?) of string theory on the whole $\mathbb{R}^{3,1}\times\mathrm{CY}$, but I'm wondering if something special happens already at the level of GR. - -And - -2) Has anybody seriously considered the relationship between "the spacetime being $4$-dimensional" (i.e. $n=4$ being a "critical dimension" for GR/ string theory on $\mathbb{R}^{n-1,1}$) and the existence of exotic $\mathbb{R}^n$'s only for $n=4$? To which conclusions did they arrive? Which explanations did they provide? - -REPLY [5 votes]: I assume by general relativity you mean classical general relativity as opposed to any quantum version of general relativity. With that in mind I will attempt to answer your questions: - -Are there properties or phenomena of general relativity that are "specific" to $\mathbb{R}^{3,1}$, that is, the equivalent version on $\mathbb{R}^{n-1,1}$ does not hold for $n \ne 4$. - -There are many such properties. The most obvious of which is the count of local degrees of freedom. (For a reference as to what "local degrees of freedom" in this context means refer to the the end of section 10.2 in Wald's General Relativity.) For $n < 4$ general relativity has no local degrees of freedom. For $n = 4$ general relativity has 2 local degrees of freedom per space-time point. For $n > 4$ general relativity has more than 2 local degrees of freedom per space-time point. - -...or more generally on any other Lorentzian manifold homeomorphic to $\mathbb{R}^4$ - -Here I assume when you say homeomorphic you mean homeomorphic and not diffeomorphic. In other words your question is what properties of general relativity change when one changes the smooth structure on $\mathbb{R}^4$? -Beyond the obvious answer, a previously smooth metric is no longer smooth, I think this question is not answered in the current literature. -The closest one comes to an answer is in section III of Witten's Global Gravitational Anomalies. There he argues that in more than $4$ dimensions taking the connected sum of a manifold $M$ with an exotic sphere $S$ is equivalent to placing a gravitational instanton on $M$. In four dimensions he is silent, waiting on the resolution of the smooth 4-dimensional Poincare conjecture. However, instantons are only really relavent for quantum general relativity. - -Has anybody seriously considered the relationship between "the spacetime being 4-dimensional" and the existence of exotic $\mathbb{R}^n$'s only for $n=4$? - -I'd say no one has seriously considered this relationship. There has been lots of speculation. For example a series of articles by Carl Brans an collaborators seem to raise a lot of questions in this direction but not reach any solid conclusions. -My guess is that one would have to prove, in a physicist's sense of the word, that the path-integral is only well-defined, whatever that might mean, due to the inclusion of gravitational instantons if $n=4$. I don't think we are anywhere near being able to do that yet.<|endoftext|> -TITLE: Estimates for the diameter of a (nice) surface? -QUESTION [5 upvotes]: The Question -Let $M$ be a compact, connected, orientable surface without boundary of bounded genus smoothly embedded in $\mathbb{R}^3$; define the diameter $d_M$ of $M$ as the maximum minimal (geodesic) distance between any pair of points $x,y \in S$. My question is: - -What's a good estimate for $d_M$? - -To make this question more interesting, let's try to avoid pathological examples. Because this is MathOverflow I'm a little insecure about the best way to say this, but let me try the following: suppose that the surface has fixed area and bounded curvature. For instance, one might require that the Willmore energy $\int_M H^2 dA$ is bounded, where $H$ is the mean curvature. This way we avoid examples like an incredibly long, thin (capped) cylinder which has nearly zero surface area but enormous diameter, or similarly an infinitely thin infinitely long spiral. These types of examples tend to confound estimates that otherwise work pretty well. -To understand the scope of the question it may help to know that I really want to be able to evaluate this estimate (using a 5-year-old laptop, say) for a simplicial approximation of $S$. For this reason, I prefer to work with quantities that can be computed efficiently like surface area, enclosed volume, mean/Gaussian curvature, etc. I'm willing to compute a few eigenvalues of the Laplacian (e.g., the smallest and largest), but I refuse to compute all of them. I'm not terribly excited about estimates that bound $d_M$ from only one side. -One Approach -Here's one idea I had based on some rough intuition about Laplacian eigenfunctions. It works surprisingly well in practice (e.g., within 20% of the actual diameter for a wide variety of example surfaces), but I have trouble saying anything concrete about it. -Let $\phi$ be an eigenfunction of the scalar Laplace-Beltrami operator $\Delta$ on $M$, i.e., $\phi$ has unit $L^2$ norm and satisfies $\Delta \phi = \lambda \phi$ for some eigenvalue $\lambda \in \mathbb{R}$. (I'll adopt the convention that $\Delta$ is positive-semidefinite so that $\lambda \geq 0$.) Then we have -$$ \int_M |\nabla\phi|^2 dA = \int_M \Delta \phi \cdot \phi = \int_M \lambda \phi^2 dA = \lambda \int_M \phi^2 = \lambda,$$ -hence $\|\nabla \phi\|_{L^2} = \sqrt{\lambda}$. -Now consider the "first" nontrivial eigenfunction $\phi$, i.e., the one corresponding to the smallest nonzero eigenvalue $\lambda$. I'm going to make a fairly crude approximation and replace $|\nabla \phi|$ at every point with the value -$$ g := \frac{\phi_\max-\phi_\min}{d_M}, $$ -where $\phi_\max$ and $\phi_\min$ are the maximum and minimum (pointwise) values of $\phi$. -Where does this idea come from? The idea is best illustrated in 1D -- consider the circle $S^1$, which we can identify with the interval $[0,2\pi) \subset \mathbb{R}$. In this case the first eigenfunction $\phi$ looks like $\sqrt{\pi}\cos(x)$ -- just one full oscillation as we go from zero to $2\pi$: - (source) -In this case the value $g$ is an overestimate of $|\nabla \phi|$ in some places and an underestimate in others. So, we substitute $g$ for $|\nabla \phi|$ in the first relationship above and get -$$ A g^2 \approx \lambda, $$ -where $A$ is the total surface area. Solving for $d_M$ gives us an approximation -$$ d_M \approx (\phi_\max - \phi_\min)\sqrt{A/\lambda}. $$ -You'll notice my liberal use of "$\approx$" here, because I haven't been able to establish any hard bounds! But again, keep in mind that however crude the argument may be, this approximation appears to work unreasonably well for "reasonable" surfaces. -Still, it would be nice to be able to say something concrete about this approximation (maybe with additional restrictions on $M$) or to have an alternative approximation with established bounds. -Thanks! - -REPLY [4 votes]: This paper of Peter Topping doesn't really meet all of your criteria, but may be of interest to you. For surfaces it bounds the diameter from above in terms of the $L^1$ norm of the mean curvature of the surface.<|endoftext|> -TITLE: Indecomposable modules over preprojective algebras -QUESTION [5 upvotes]: Would you please give some references concerning the number of indecomposable modules over preprojective algebras of type $A_n$? -More precisely, I need references about the following claim: The number of indecomposable modules over the preprojective algebra of type $A_n$ is: -1) only one for $n=1$, -2) four for $n=2$, -3) $12$ for $n=3$, -4) $40$ for $n=4$, and -5) an infinite number of indecomposable modules for $n\geq5$. - -REPLY [7 votes]: This is stated in Rigid modules over preprojective algebras by Geiss, Leclerc and Schröer, Sections 8.1-8.3. As noted in my comment a proof of the $n=5$ statement can be found in The module theoretic approach to quasi-hereditary algebras. (These are of course not the original references, I doubt that there is one original reference) -Note also that you can split up the $n\geq 5$ case in $n=5$, which is the tame case, i.e. there is a classification of the infinitely many indecomposable modules, and the $n>5$ case, where a classification is not possible. -EDIT: There is a proof in Semicanonical bases and preprojective algebras by Geiss, Leclerc, Schröer, Section 9, via the theory of Galois coverings. The Auslander-Reiten quiver of these algebras can also be found there. The statement was long before and is probably part of the folklore in the representation theory of quivers.<|endoftext|> -TITLE: A Family of Bases for a Vector Space -QUESTION [9 upvotes]: Let $V$ be a multiset of $kn$ nonzero vectors in $\mathbf{R}^n$. Suppose that for $1 \leq d \leq n$, each $d$-dimensional subspace of $\mathbf{R}^n$ contains at most $kd$ members of $V$. Then $V$ can be partitioned into $k$ bases of $\mathbf{R}^n$. -A proof or a reference would be appreciated. - -REPLY [5 votes]: Apparently the question was answered in the comments: Edmonds proves a more general theorem for a family of matroids. If the matroids are all the same and if that matroid is a vector matroid, then it is exactly the question stated.<|endoftext|> -TITLE: Distinct simple zeros of Dirichlet L-functions -QUESTION [18 upvotes]: Given a finite set of distinct primitive Dirichlet characters, $\chi_1, \dots, \chi_r$, is it known that the product of the L-functions, $$L(s):=\prod_{i=1}^r L(s,\chi_i),$$ has a simple zero? It's conjectured that all of the zeros of the $L(s,\chi_i)$ are distinct and simple, but I don't know what is known unconditionally except in the case that $r\leq 2$. -It's known that each $L(s,\chi)$ has infinitely many simple zeros (in fact, a positive proportion of its zeros are simple and lie on the half-line), which immediately answers the question in the case that $r=1$. If $r=2$, the answer to my question seems to be provided by work of Conrey, Ghosh, and Gonek (Simple zeros of the zeta function of a quadratic number field, I. Invent. Math., MR0860683). They prove that the Dedekind zeta function associated to a quadratic field has $\gg T^{6/11}$ simple zeros with imaginary part up to $T$, all arising from $\zeta(s)$. It appears that their method can be adapted to consider the product of any two Dirichlet L-functions, and this is confirmed by a statement of Bombieri and Perelli (Distinct zeros of L-functions. Acta Arith., MR1611193), who additionally write that $r=2$ is the limit of the Conrey-Ghosh-Gonek method. -I have not been able to find any work which applies to my question in the case that $r\geq 3$. The paper of Bombieri and Perelli referenced above discusses counting distinct zeros of more general L-functions, but it is not obvious to me how to isolate the simple zeros in their argument. -I also don't know if the fact that I'm only looking for a single simple zero of $L(s)$ saves me anything. That is, I don't know of techniques that detect the existence of such a zero without proving that there are an infinite number. Nevertheless, this could prove to be useful, since it seems entirely possible that showing that $L(s)$ has infinitely many simple zeros when $r\geq 3$ could be quite difficult. - -REPLY [10 votes]: You probably guessed this from the lack of responses, but the answer is that there is no hope of making progress on this question using current methods. -One measure of the complexity of an L-function is its degree, where the Riemann zeta function and Dirichlet L-functions have degree 1, the L-function of a holomorphic cusp form has degree 2, the standard L-function of a GL(n) automorphic form has degree n, etc. The precise definition of degree is the number of $\Gamma$-factors in the functional equation, where a $\Gamma(s+A)$ counts as two $\Gamma$-factors. -These days we have very few tools for dealing with degree 3 and higher. Your product of Dirichlet L-functions is like one degree $r$ L-function, and so you are stuck once $r$ is bigger than 2. In particular, the fact it is a product doesn't seem to help much. -It also doesn't help that you want only one simple zero, unless you have a particular explicit case in mind, in which case you can show it by direct calculation.<|endoftext|> -TITLE: Is de Rham cohomology of affine schemes over discrete valuation rings finitely generated (modulo torsion)? -QUESTION [13 upvotes]: Let $U$ be a open affine subscheme of a smooth, proper scheme $X$ over $\mathbf{Z}_p$. Over $\mathbf{Q}_p$ we know that $\mathrm{H}^i(U \times \mathbf{Q}_p/\mathbf{Q}_p)$ is finite-dimensional (where $\mathrm{H}^i(\cdot/R) = \mathbf{H}^i(\Omega_{\cdot/R}^\bullet)$ is the $i$-th de Rham cohomology of a scheme over a ring $R$). My question is: - -Is the image of $\mathrm{H}^i(U/\mathbf{Z}_p)$ in $\mathrm{H}^i(U \times \mathbf{Q}_p/\mathbf{Q}_p)$ finitely generated as $\mathbf{Z}_p$-module? - -I am particularly interested in the case of (smooth, geometrically intgeral) curves and $i=1$. For simplicity, let $Z = X \setminus U$ be a single $\mathbf{Z}_p$-rational point. Here are my partial results: - -If $X$ is of genus $0$, then the answer is obviously 'yes' as $\mathrm{H}^1(U \times \mathbf{Q}_p/\mathbf{Q}_p)$ vanishes. -If $X$ is ordinary and of genus $1$, then the answer is 'no'. -The answer can also be 'no' if $X$ is of genus $1$ and supersingular. (An example is: $U$ defined by $y^2 = x^3 + 1$ over $\mathbf{Z}_5$. The calcuations are somewhat lengthy.) - -Let me give you the starting point for the proof/example: The image of $\mathrm{H}^1(U/\mathbf{Z}_p)$ is finitely generated if and only if there exists an integer $\ell$ such that $\mathrm{H}^0(\Omega_X(\ell Z))$ surjects onto the image. This happens if and only if for every differential $\omega \in \mathrm{H}^0(\Omega_U)$ there exists a function $f \in \mathrm{H}^0({\cal{O}}_U)$ such that $p^r\omega - df \in p^r\mathrm{H}^0(\Omega_X(\ell Z))$. (Note that reducing to $\mathrm{H}^0(\Omega_X(\ell Z))$ is not enough.) In particular, one must be able to reduce differentials with Laurent tail $t^{-p^r-1}dt + O(t^{-p^r})dt$ (The function $t$ is such that $\{ t, p \}$ are local coordinates for the stalk of ${\cal{O}}_X$ at the special point of $Z$. In particular, $t$ is a local uniformizer over $\mathbf{Q}_p$ as well as over $\mathbf{F}_p$.) This is only possible if the following holds: - -For every $r \in \mathbf{N}$, is there a function $f \in \mathrm{H}^0({\cal{O}}_X(p^rZ))$ with Laurent tail $t^{-p^r} + O(t^{-p^r +1})$ and $df \in p^r\mathrm{H}^0(\Omega_X((p^r+1)Z))$? - -Update: If $X$ is ordinary of genus $1$, here's how to show that such a function cannot exist. The main ingredients are reduction modulo $p$ and the residue theorem. A $\bar{\cdot}$ will denote reduction modulo $p$ of the object in question. Since $p^r$ divides $df$, the Laurent expansion of $\bar{f}$ with respect to $\bar{t}$ is $\bar{t}^{-p^r} + O(\bar{t})$ (we can always assume that the coefficient of $\bar{t}^0$ is $0$). Choose a generator $\bar{\vartheta}$ of $\mathrm{H}^0(\Omega_{\bar{X}})$ and let $\sum_{i \ge 0}\bar{\vartheta}_i\bar{t}^id\bar{t}$ be its Laurent expansion. By the residue theorem $0 = \mathrm{res}(\bar{f}\bar{\vartheta}) = \mathrm{res}_{\bar{Z}}(\bar{f}\bar{\vartheta}) = \bar{\vartheta}_{p^r-1}$ (all other summands contributing to the residue are $0$). But $\bar{\vartheta}_{p^r-1} = \bar{c}\bar{\vartheta}_0 = \bar{c}$ (the trace of the Frobenius modulo $p$, see [1]). Since $X$ is ordinary, $\bar{c}$ cannot be $0$. This is the contradiction. -[1] Lang, Elliptic functions, Appendix 2. - -REPLY [5 votes]: After more studying, I found a solution. An $\overline{\cdot}$ will denote the reduction modulo $p$ of the object in question. Let us assume that $\overline{X}$ is a geometrically integral curve of genus $g$ and that $Z = X \setminus U$ is a (relative, reduced) normal crossing divisor. Then the following holds: - -If $\gamma$ is the Hasse-Witt invariant of $\overline{X}$, i.e. the rank of the - $p$-torsion of the Jacobian of $\overline{X}$, then $$ \mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion}) \cong \mathbf{Q}_p^{2g-\gamma} \oplus \mathbf{Z}_p^{\deg{\overline{Z}}-1+\gamma}.$$ - -The module $\mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion})$ embeds into $\mathrm{H}_{\mathrm{dR}}^1(X_{\mathbf{Q}_p} \setminus Z_{\mathbf{Q}_p})$ and this space is known to be isomorphic to $\mathbf{Q}_p^{2g + \deg{Z} - 1}$. Thus $\mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion})$ is isomorphic (as $\mathbf{Z}_p$-module) to $\mathbf{Q}_p^r \oplus \mathbf{Z}_p^s$ with $r+s = 2g + \deg{Z}-1$. Tensoring with $\mathbf{F}_p$ gives $\mathbf{F}_p^s$. Since $\deg{Z} = \deg{\overline{Z}}$, we need to show that the dimension of $\mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion}) \otimes \mathbf{F}_p$ is $\deg{\overline{Z}}-1+\gamma$. -Let $d{\cal{O}}_{X}(X \setminus Z) \colon (p^\infty)$ be the $p$-saturation of -$d{\cal{O}}_{X}(X \setminus Z)$, i.e. it contains all differentials $\omega$ -such that $p^r\omega = df$ for some $r \in \mathbf{N}$ and $f \in {\cal{O}}_X(X -\setminus Z)$. Then there is the exact sequence $d{\cal{O}}_{X}(X -\setminus Z) \colon (p^\infty) \to \Omega_X(X \setminus Z) \to \mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion}) \to 0$. Since tensoring with $\mathbf{F}_p$ is right-exact, we get $$d{\cal{O}}_{X}(X \setminus Z) \colon (p^\infty) \otimes \mathbf{F}_p \to \Omega_{\overline{X}}(\overline{X} \setminus \overline{Z}) \to \mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion}) \otimes \mathbf{F}_p \to 0.$$ -Write $\Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$ for -the image of $d{\cal{O}}_{X}(X \setminus Z) \colon (p^\infty) \otimes \mathbf{F}_p$ in $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})$. Its elements are called pseudo-exact differentials [1, Sec. 2]. What we want to prove follows from the two claims - -$$\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z}) \big/ - \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) \cong - \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \big/ \left( - \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \cap - \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) \right)$$ - -and - -$$\dim{\Omega_{\overline{X}}(\overline{Z})(\overline{X}) \big/ - \left( \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \cap - \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) \right)} = \deg{\overline{Z}} - 1 + \gamma.$$ - -Concerning the second claim, note that pseudo-exact differentials are -residue-free [1, Sec. 3] and that $\overline{Z}$ is reduced since $Z$ is a reduced normal crossing divisor. Thus, $\dim{\Omega_{\overline{X}}(\overline{Z})(\overline{X}) / ( \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \cap - \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})} )$ is the sum of $\dim{\Omega_{\overline{X}}(\overline{Z})(\overline{X}) / \Omega_{\overline{X}}(\overline{X})}$ and $\dim{\Omega_{\overline{X}}(\overline{X}) / -\Omega_{\overline{X}}(\overline{X}) \cap \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})}$. The former dimension is $\deg{\overline{Z}}-1$ and the latter is $\gamma$ [4, Cor. 1]. -Let $C$ be the Cartier operator on the differentials of $\overline{X}$. If $\overline{x}$ is a separating element for the function field $K(\overline{X})$, then every differential may be written as $(\sum_{i=0}^{p-1} \lambda_i\overline{x}^i)d\overline{x}$ with $\lambda_i \in k$ and -$C$ maps this differential to $\lambda_i^{1/p} d\overline{x}$. The Cartier operator restricts to a map (also called $C$) on $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})$ and [3, Sec. 1] $$ - \Omega_{\overline{X}}(\overline{X} \setminus \overline{Z}) = \bigcup_{i \in \mathbf{N}} C^{-i}(\Omega_{\overline{X}}(\overline{Z})(\overline{X})) \hspace{10pt} \text{and} \hspace{10pt} \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) = \bigcup_{i \in \mathbf{N}} C^{-i}(\{ 0 \}).$$ -Let $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i)} := C^{-i}(\Omega_{\overline{X}}(\overline{Z})(\overline{X}))$ and $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i, \mathrm{p-ex})} := \Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i)} \cap \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$. Then $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i)} / \Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i, \mathrm{p-ex})}$ is isomorphic to $\Omega_{\overline{X}}(\overline{Z})(\overline{X}) / \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \cap \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$. But $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i)} / \Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i, \mathrm{p-ex})}$ is also isomorphic to $\Omega_{\overline{X}}(\overline{X} \setminus -\overline{Z})^{(i)} + \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) / \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$. Varying $i$, these spaces form a direct system whose limit is $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z}) / \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$ and whose maps are isomorphisms. This finishes the proof. (Compare with [2, Satz 3].) - -[1] Lamprecht "Zur Klassifikation von Differentialen in Körpern von Primzahlcharakteristik. I" 1958 -[2] Kunz "Einige Anwendungen des Cartier-Operator" 1962 -[3] Kodama "Residuenfreie Differentiale und der Cartier-Operator algebraischer Funktionenkörper" 1971 -[4] Kodama "On the Rank of Hasse-Witt Matrix" 1984<|endoftext|> -TITLE: Symbol of pseudodiff operator -QUESTION [10 upvotes]: Hello, -I am trying to understand the calculus of pseudodifferential operators on manifolds. All the textbooks I could put my hand on define the principal symbol of a pseudodifferential operator locally, then prove that it transforms well, hence becomes a "global" object. Is there any good way to define the the principal symbol without coordinate patches? Am I asking too much here :)? - -REPLY [2 votes]: I think that Hormander beautiful short paper - -Pseudo-differential operators, Comm. - Pure Appl. Math. 18 1965, 501–517 - -is always a good place to start. In this paper he gives a coordinate free definition of a (scalar) pseudo-differential operator. This paper did it for me.<|endoftext|> -TITLE: Restrictions for presenting groups with cyclic quotient -QUESTION [7 upvotes]: Let $H$ be a group, $\phi$ an automorphism of $H$ of order n and fix $h_0 \in H$. I wonder, what the restrictions are, such that -$$G:= \lt H,g \mid g^n=h_0,\quad \forall h \in H: ghg^{-1}=\phi(h) \gt$$ -defines a group which has $H$ as normal subgroup such that $G/H$ is cyclic of order n. -There are two obvious restrictions: -(1) $h_0$ has to be in the center of $H$. For, $h_0hh_0^{-1} = g^nhg^{-n}=\phi^n(h) = h$, since $\phi$ has order n. -(2) $\phi(h_0) = h_0$. For, $\phi(h_0) = gh_0g^{-1}=gg^ng^{-1} = g^n = h_0$. -But I can't figure out, if there some more restrictions. -I tried to apply the classification of extensions with non-abelian kernel (Kenneth Brown, Cohomology of Groups, chapter IV, §6), but that requires to consider $H^3(Out(H),C)$ ($C$ the center of $H$) and I'm unable to do so, because I have no information about $Out(H)$ and $C$. -Any help is appreciated. - -REPLY [3 votes]: Your conditions are sufficient. Indeed, consider the $H$-by-cyclic group $G_0=\langle H, g\mid ghg^{-1}=\phi(h), h\in H\rangle$. By (1) and (2), both $g^n$ and $h_0$ are central in that group. Hence the element $u=h_0^{-1}g^n$ is central. Now factor out the central subgroup $\langle u\rangle$: $G=G_0/\langle u\rangle$. That group $G$ is what you need.<|endoftext|> -TITLE: what's the contragredient of induced representation -QUESTION [5 upvotes]: Let $G$ be a real reductive Lie group, $P=MN$ its parabolic subgroup with Levi decomposition. Suppose $\sigma$ is a smooth admissible irreducible representation of $M$, extend this to $P$ by letting $N$ act trivially. Form the unitarily induced representation $Ind_P^G(\sigma)$. -My question is what is the contragredient representation (smooth admissible dual) of $Ind_P^G(\sigma)$ in terms of $\sigma$ ? In particular, is it equal to $Ind_P^G(\sigma')$, where $\sigma'$ is the smooth admissible dual of $\sigma$? - -REPLY [8 votes]: Yes. The point is that $Ind_P^G(\sigma)$ is by definition equal to the space of -sections of a certain $G$-equivariant vector bundle $E_{\sigma}$ on $G/P$ and $Ind_P^G(\sigma')$ -is equal to the sections of the corresponding bundle $E_{\sigma'}$. Now the point -is that because you are using unitary induction there is a natural map -$E_{\sigma}\otimes E_{\sigma'}\to \Omega_{G/P}$ where $\Omega$ is the bundle -on differential forms of top degree (more precisely, it has to be tensored -with the corresponding orientation sheaf which we can trivialize if we choose a -$G$-equivaruiant orientation of $G/P$ - let me for simplicity assume that -we can do that). This gives a map -$Ind_P^G(\sigma)\otimes Ind_P^G(\sigma')$ to differential forms which we can integrate -(since I assumed that we have chosen an orientation on $G/P$). This gives a pairing -between the two induced representations and the fact that it is a perfect pairing -is easy.<|endoftext|> -TITLE: Formalizing Euclid's proof of the infinitude of primes -QUESTION [10 upvotes]: Euclid's proof of the infinitude of primes requires me to take an arbitrary finite set of primes and multiply them together. If I want to formalize this proof in Peano Arithmetic, I need to know that any finite set of numbers has a product. Carl Mummert's comment on Russell O'Connor's answer to this question points out that this is trickier than it sounds, since "finite" means "indexed by a natural number", which is not the same thing as "indexed by a standard natural number". So one needs an inductive definition of the product to prove it exists. -The obvious inductive definition is: -$$\prod_{k=1}^{S(n)}a_k = a_{S(n)}\prod_{k=1}^n a_k$$ -where I write $S$ for "successor". Is this enough to fill the "gap" in Euclid's proof, or am I missing something? (I ask because, in the discussion linked above, there seems to be a hint that it's not quite this simple.) - -REPLY [14 votes]: The definition of iterated product is the easy part of the proof, the bottleneck is elsewhere. Even quite weak theories of bounded arithmetic, such as PV (this is an open theory with function symbols for polynomial-time functions), can prove that the product of a sequence of natural numbers exists (and is divisible by each element of the sequence), by formalizing the usual inductive definition in a straightforward way. -The next thing you need is to show that every number greater than one is divisible by a prime. PV is not known (and not expected) to prove this, but it is provable in the slightly stronger theory $S^1_2$. -Thus, $S^1_2$ can formalize Euclid’s proof as the statement “there is no sequence containing all primes”. -As Carl noted, this is not a very natural rendering of the statement “there are infinitely many primes” (especially in weak theories, where sequences can only have logarithmic length). The statement one really wants to prove is “the set of primes is unbounded”. In order to infer this from Euclid’s proof, you need the principle that any bounded set can be put in a sequence, and this is equivalent to the axiom that exponentiation is total. Thus, formalization of Euclid’s proof in this sense can be done in $I\Delta_0+\mathrm{EXP}$ (and a fortiori in very strong theories like PRA). -It is possible to prove that there exist unboundedly many primes in theories weaker than $I\Delta_0+\mathrm{EXP}$, but this requires a much more elaborate proof than Euclid’s. Paris, Wilkie and Woods have shown the unboundedness of primes in the theory $I\Delta_0+\Omega_1$ (where $\Omega_1$ roughly says that $x^{\log x}$ exists for every $x$), and more precisely, in its subtheory $I\Delta_0+\mathrm{WPHP}(\Delta_0)$ (the latter is the weak pigeonhole principle: there is no $\Delta_0$-definable injection from $[0,2x]$ to $[0,x]$). It can be seen that their proof actually goes through in the theory $S^1_2+\mathrm{rWPHP}(\Theta)$ (itself included in $T^3_2$ and $S^4_2$), where $\Theta$ is the set of provably total $\Sigma^b_2$-definable functions of $S^1_2$ (which coincides with functions from the complexity class $\mathrm{FP}^{\mathrm{NP}[\mathrm{wit},\log n]}$), and $\mathrm{rWPHP}(\Theta)$ is a version of WPHP where the injection is supplied with an explicit retraction, both coming from $\Theta$. To the best of my knowledge, this is the weakest theory known to prove the unboundedness of primes.<|endoftext|> -TITLE: Two ways of generalizing factorials via symmetric groups -QUESTION [7 upvotes]: By the Bruhat decomposition of $GL(n, \mathbb{F}_q) / B_n$ we know that $$[n]! = \sum_{ \sigma \in S(n)} q^{l(\sigma)}$$ where $[n]! = \prod_{j=1}^n (1+q + \cdots + q^{j-1})$ and $l(\sigma)$ is the length of the permutation $\sigma \in S(n)$ (also known as the number of involutions of $\sigma$). -We also know that $$\theta^{(n)} = \sum_{\sigma \in S(n)} \theta^{[\sigma]}$$ where $[\sigma]$ is the number of cycles of the permutation $\sigma \in S(n)$ and $\theta^{(n)} = \theta(\theta+1) \cdots (\theta+n-1)$. Notice that $$\lim_{q \rightarrow 1} \ [n]! = n! = \lim_{\theta \rightarrow 1} \ \theta^{(n)}.$$ Is there a way to write $$\sum_{\sigma \in S(n)} q^{l (\sigma)} \theta^{[\sigma]}$$ explicitly as a function of $q$ and $\theta$? - -REPLY [8 votes]: It seems like there is no hope for a nice closed form for $F(q,\theta)=\sum_{\sigma\in S_n}q^{l(\sigma)}\theta^{[\sigma]}$. When the sum is restricted to $\sigma\in S_n$ which are involutions, the computation can be found in I. Gessel's paper "A q-analog of the exponential formula". -For the general case, the study of $F(q,\theta)$ is the main topic of P.H. Edelman's paper "On inversions and cycles in permutations" (Europ. J. Combinatorics,(1987) vol 8, 269-279). he proves a bunch of properties of this bivariate generating function, yet, according to this paper even computing the number of permutations which achieve the minimum number of cycles with a fixed number of inversions hasn't been carried. It gives a bunch of open problems about $F(q,\theta)$. -If on the other hand you let $l(\sigma)$ denote the number of inversions of $\sigma$ written in cycle notation (i.e. the number of inversions in $(a_1\dots a_{k_1})(a_{k_1+1}\dots a_{k_2})\cdots(a_{k_{r}+1}\dots k_{r+1})$) then the sum is $\prod _{i=1}^n [i] _{q} ^{\theta}$, where -$$[i]_{q}^{\theta}= 1+q+\cdots+q^{i-2}+\theta q^{i-1}.$$ -This is proved in "Cycles and patterns in permutations" by R. Parviainen.<|endoftext|> -TITLE: Are there non-reflexive modules isomorphic to their bi-dual? -QUESTION [15 upvotes]: Let $M$ be an $R$-module. We say that $M$ is reflexive if the natural map $M\rightarrow M^{**}$ is an isomorphism. -I'd like to know if there exists a module isomorphic to its bi-dual but not reflexive, do you know an example? - -REPLY [13 votes]: Yes, there are such examples. In the case of Abelian groups for example, one can have a group A which is not reflexive, but which is isomorphic to its double dual. The book "Almost Free Modules" by Eklof and Mekler (North Holland) contains much of what is known. -As a specific example, take E to be a stationary and costationary subset of $\omega_1$, and let $X=\omega_1 + 1\backslash E$, given the subspace topology (of the space $\omega_1 +1$ with the interval topology). Then $C(X, Z)$ (continuous functions to the integers with the discrete topology) is such a group. In fact for this group $C(X,Z)^{**} = \sigma[C(X,Z)] \oplus Z$ where $\sigma$ is the natural map, so that $C(X,Z)$ is not reflexive, but $C(X,Z)\oplus Z \cong C(X,Z)$ because for example $C(X,Z)$ has $Z^\omega$ as a summand.<|endoftext|> -TITLE: Why is the Fast Fourier Transform efficient? -QUESTION [9 upvotes]: Is there a conceptual way to understand where the Fast Fourier Transform is avoiding redundant computation and thus achieving $O(n\log n)$ instead of $O(n^2)$. -Consider a standard example of the FFT to multiply two polynomials faster. Its not obvious to me conceptually why the FFT should yield a faster way to multiply two polynomials. - -REPLY [12 votes]: Conceptually the FFT takes advantage of a shortcut similar to the distributive law for multiplication. To compute $$(x_1 + x_2)(x_3 + x_4)$$ on could either add first (twice) and then multiply (once), or one could expand $$sx_1x_3 + x_1x_4 + x_2x_3 + x_2x_4$$ and multiply (four times) and then add (three times). This idea has been spelled out in the paper The Generalized Distributive Law.<|endoftext|> -TITLE: How much can a diagonal matrix change the eigenvalues of a symmetric matrix? -QUESTION [6 upvotes]: Suppose that we have a symmetric matrix ${\bf S}$ with eigenvalue decomposition ${\bf S} = {\bf Q}{\bf \Lambda}{\bf Q}^T$. Assume that we have two diagonal matrices ${\bf D}_1$ and ${\bf D}_2$ that are multiplying ${\bf S}$ from the left and right, i.e. ${\bf A} = {\bf D}_1{\bf S}{\bf D}_2$. -Can we relate the eigenvalues of ${\bf S}$ to the ones of ${\bf A}$? -How about the case where ${\bf S}$ is not symmetric? - -REPLY [10 votes]: In general there is no relation: for example consider the simplest case $S$ itself is diagonal and invertible. Letting $D_1=S^{-1}$ then $A$ can be any diagonal matrix $D_2$. The only considerations you can do are related to the presence of the zero eigenvalues using Binet formula for determinants. Notice also that in general $A$ itself can be nonsymmetric, and its eigenvalues can be complex. However small perturbations, i.e. small $D_1$ and $D_2$, result in a small perturbation of the eigenvalues of $S$ in the complex plane.<|endoftext|> -TITLE: Grauert's criteria for ample line bundles -QUESTION [9 upvotes]: In their book "Compact complex surfaces", W.P. Barth, K. Hulek, C.A.M. Peters and A. Van de Ven refer to the following theorem: -Let $X$ be a compact complex space and $L$ a holomorphic line bundle on $X$. Then $L$ is ample if and only if the folowing holds: given any irreducible analytic subset $Y$ of strictly positive dimension on $X$, there exist an $n=n(Y)$ such that $L^{\otimes n} |_Y$ has a section which has at least one zero, but does not vanish identically. -They don't give a proof in their book. Instead they refer to Grauert's original paper Über Modifikationen und exzeptionelle analytische Mengen which was published in Math. Ann. 1962. -I do not know German. So my question is: Can I find the proof of this theorem somewhere else? Or instead, some comments on the idea of proof will also be very helpful. -Finally, this paper of Grauert is among one of the papers I want to read with greatest enthusiasm. Is there a translation? Or can I find some books or papers which give an explanation of the results of this paper? - -REPLY [3 votes]: Suppose that $H^0(L^{\otimes n})|_ Y$ -is non-zero for all $Y$ and $n$ sufficiently big, and has a section which vanishes somewhere on $Y$. Then it follows that the base -set of $L$ is trivial: indeed, $L$ has a non-zero -section on any complex subvariety, which includes -the base set. This implies that the natural map -$P_n:\; X\rightarrow {\mathbb P}(H^0(X, L^{\otimes n}))$ -is holomorphic, for $n$ sufficiently big. -Also from this assumption it follows that -$P_n$ does not map any irreducible, positive-dimensional -subvariety to a point (again, for $n$ sufficiently big). -This implies that $P_n$ is a finite, proper map -to a projective variety, hence $X$ is a -ramified covering of a projective variety. -A ramified covering of a projective variety -is projective, which can be seen from vanishing -of cohomology of powers of $L$ (a finite -map is acyclic on coherent sheaves, hence -the cohomology of $L^{\otimes n}$ on $X$ are the -same as cohomology of ${\cal O}(1)$ -on its image). -However, this works only when $H^0(L^{\otimes n} |_Y)\neq 0$; -the implication $H^0(L^{\otimes n} |_Y)\neq 0$ -$\Rightarrow$ $H^0(L^{\otimes n})|_Y\neq 0$ is not that easy. -If $Y$ does not lie in a zero divisor of $L^{\otimes n}$, -we are done, otherwise we replace $Y$ with the zero -divisor of $L^{\otimes n}$. Using induction on dimension, -we may already assume that the restriction $L|_Y$ is ample. -Consider the exact sequence -$$ -0\rightarrow H^0(L^{\otimes k}) -\rightarrow H^0(L^{\otimes n+k}) \stackrel r \rightarrow H^0(Y, -L^{\otimes n+k} |_Y) \rightarrow H^1(L^{\otimes k}) -\rightarrow H^1(L^{\otimes n+k}) \rightarrow 0. -$$ -The arrow $r$ of this sequence is the restriction map; -we need to prove that $r$ does not vanish. -If it vanishes, we have $\dim H^0(Y, L^{\otimes n+k}|_Y) \leq -\dim H^1(L^{\otimes k})$ for all $n\gg 0$ and $k$. -The last term of this exact sequence actually -implies that $\dim H^1(L^{\otimes n+k})\leq \dim H^1(L^{\otimes k})$, -for all $k$, and all $n\gg 0$, -hence $H^1(L^{\otimes n+k})$ is bounded by a universal constant. -This implies that $\dim H^0(Y, L^{\otimes n+k} |_Y)$ -is also bounded, whenever $r=0$, which is -impossible, because $L|_Y$ is ample.<|endoftext|> -TITLE: Smooth convergence of minimizing varifolds -QUESTION [10 upvotes]: Dear all, -a question came to me when I read the paper "Complete three dimensional manifolds with positive Ricci curvature and scalar curvature/ R. Schoen and S.-T. Yau, 1982". -The question is as follows. -suppose we have a sequence of complete non-compact smooth submanifolds $\Sigma_k\subset M$ and each $\Sigma_k$ is area-minimizing (in its isotopy class) with respect to any compactly supported deformation. Then by treating them as Radon measures, we can get a convergent subsequence of $\Sigma_k$, say $\Sigma_k\to\Sigma$. The convergence is of the sense of Radon measure, or is called "in the sense of varifold". As I know, it is merely stronger than Cheeger-Gromov convergence. However, in Schoen-Yau's article, they say that the convergence is smooth. (page 217, the last paragraph in the proof of Lemma 3.) I cannot see the reason for this convergence to be smooth. -Maybe they have used some other properties in that paper (I guess not), but it's difficult for me to provide all the details here.... Anyway, thank you all for any comments. - -REPLY [14 votes]: The reason for this is Allard's regularity theorem. -Roughly speaking Allard's theorem says that if near a point of the support of a stationary varifold the varifold has unit density and area close to that of the ball of the appropriate dimension (for a 2-varifold it would be area of a disk) then the support of the varifold is smooth at that point. -More precisely, there is an $\epsilon>0$ and $r_0>0$ (depending on the ambient geometry and dimension of the varifold). So that if $\Sigma$ is an stationary $m$-varifold in $M$ , a Riemannian manifold, and a point $p\in spt \Sigma$ satisfies -$$\mathcal{H}^m(B_r(p)\cap \Sigma) \leq \omega_m r^m(1+\epsilon)$$ -for $r\leq r_0 $ then $spt \Sigma$ is smooth near $p$. Here $B_r(p)$ is the $r$-ball in $M$ and $\omega_m$ is the volume of the unit ball in $\mathbb{R}^m$. -Allard's proof is unfortunately quite technical -- a good reference is Leon Simon's (sadly) hard to find book "Lectures on Geometric Measure Theory". Luckily, for your purpose there is a simpler version with a very easy proof due to Brian White (see here for the paper). -Specifically, suppose you have instead of being a stationary varifold you know that $\Sigma$ is a smooth minimal surface. If $p$ is a point of $\Sigma$ so that $$\mathcal{H}^m(B_r(p)\cap \Sigma) \leq \omega_m r^m(1+\epsilon)$$ - then one has -$$|A|(p)\leq r^{-1}$$ - here $|A|$ is the norm of the second fundamental form. In other words you obtain a quantitive bound on curvature. -How does this relate to your question? Well you have that $\Sigma_k$ converge to $\Sigma$ as Radon measures. Let $p\in \Sigma$ this convergence implies that -$$\mathcal{H}^m(\Sigma_k\cap B_r(p))\to \mathcal{H}^m(\Sigma\cap B_r(p))$$ -(it is worth noting that we are using that the surfaces are area minimizing to ensure the convergence is with multiplicity one). -Now since $\Sigma$ is smooth near $p$ it is locally modelled on a flat plane. In other words, there is a scale $r$ so that -$$\mathcal{H}^m(B_r(p)\cap \Sigma) \leq \omega_m r^m(1+\frac{1}{2}\epsilon)$$ -hence by the convergence -$$\mathcal{H}^m(B_r(p)\cap \Sigma_k) \leq \omega_m r^m(1+\epsilon)$$ -and so since the $\Sigma_k$ are smooth (either a priori or by Allard's full theorem) -and the point $p$ was not important we deduce that for some $\delta>0$ we have -$$\sup_{B_\delta(p)\cap\Sigma_k} |A|\leq r^{-1}.$$ -In other words there is a uniform curvature bound on the $\Sigma_k$. The result then follows from "Standard elliptic PDE" and the Arzela-Ascoli theorem.<|endoftext|> -TITLE: Undecidable problems in geometry -QUESTION [8 upvotes]: Are there any (many) algorithmically undecidable problems in computational (combinatorial/discrete) geometry? -Update: the Wang tiles answer the question with "any". (I have somewhat overlooked to count them when I was browsing the answers to general undecidable problems.) But I still suspect that there are not many examples naturally arising in combinatorial/discrete geometry. And I am of course interested in any such example. I do not count artificial reformulations of problems stated (by the authors) in different settings. -For instance, it was open for quite some time whether STRING graphs (intersection graphs of curves in the plane) are recognizable. However, it turned out that they indeed are. If the answer was opposite, it would be an example of problem I seek for. -(Let me also exclude problems very similar to Wang tiles if there are any.) - -REPLY [6 votes]: The problem to determine whether two 4-manifolds, given as simplicial complexes, are homeomorphic. This was shown to be undecidable by Markov. (Some theories of physics involve a sum over such manifolds, one additive term for each homeomorphism class, and this lead to speculation that physics was noncomputable in some sense [Geroch and Hartle 1986]. I am not sure what the current status of that is.)<|endoftext|> -TITLE: Question about the definition of hamiltonian group action. -QUESTION [6 upvotes]: So I'm reading the part in Ana Cannas da Silva's book "Lectures on Symplectic Geometry" available (on her website) about hamiltonian group actions on a symplectic manifold. She starts by defining $\mathbb R$-actions and $\mathbb S^1$ actions by saying that the vector field on $M$ that they generate must be hamiltonian. Then she defines hamiltonian $\mathbb T^n$-actions (p.154) by the requirement that the restriction of the action to each circle $\{1\} \times\ldots\times\mathbb S^1\times\ldots\times\{1\}$ be hamiltonian (plus the requirement that each of the $n$ corresponding hamiltonian functions be invariant under the action of the rest of $\mathbb T^n$) -And then, finally, she defines a hamiltonian action of a general Lie group G as one having a "moment map". This is a natural generalisation because the existence of a moment map is equivalent (I believe!) to the fact that for each $X\in\operatorname{Lie}(G)$, the vector field $X^* $ on $M$ induced by $X$ is hamiltonian (i.e. $ X^*_p=\frac{d}{dt}(\exp(tX)\cdot p)(0)$). -For indeed, if that it that case, then on can just define the moment map $\mu : M \rightarrow\operatorname{Lie}(G)^* $ by setting $\langle\mu(p),X\rangle:=\mu^X(p)$, where $\mu^X$ is a hamiltonian function for $X^* $ chosen so that $\mu$ is G-equivariant with respect to the coadjoint action on $\operatorname{Lie}(G)^* $. (Note that in the case where $G$ is commutative such as $G=\mathbb T^n$, this last condition boils down to $\mu$ being $G$-invariant.) -The question: I am trying to prove that the ad-hoc definition implies the general definition in the $\mathbb T^n$ case. The problem I am having is that we only know that n vector fields $ X_1^*,\ldots,X_n^* $ (one for each subcircle $ 1\times\ldots\times\mathbb S^1\times\ldots\times 1 \subset\mathbb T^n $) are hamiltonian. Knowing that the $X_i$'s form a basis of $\operatorname{Lie}(\mathbb T^n)$, does this imply that every $X\in\operatorname{Lie}(\mathbb T^n)$ induces a hamiltonian $X^* $. Does something like $\color{red}{(X+Y)^* =X^* +Y^*}$ hold? -More generally, given a Lie group $G$ acting on a symplectic manifold $M$, is it necessary to check that each $X\in\operatorname{Lie}(G) $ induce a hamiltonian vector field on $M$, or is it sufficient to check this for a basis of $\operatorname{Lie}(G) $ ? -Thanks. - -REPLY [3 votes]: Just so you're aware, not every author insists that a momentum map be infinitesimally equivariant (Prof. Figueroa-O'Farrill's condition 2), although it is part of da Silva's definition (edit: actually, on checking, da Silva requires the slightly stronger condition of equivariance, i.e. $\mu(g\cdot p)=\mu(p)\circ\mathrm{Ad}_{g^{-1}}$). The literature isn't uniform - for example, Marsden just requires the first condition. I'm biased towards this definition since Jerry Marsden, who unfortunately passed away a year ago today, was my advisor. -To answer your main question, yes the map $X\in\mathfrak{g}\mapsto X^* \in\mathfrak{X}(M)$ is always linear, regardless of whether the action is Hamiltonian. To see this explicitly, let $\Phi^p:G\rightarrow M$ be the map $g\mapsto g\cdot p$. By definition $X^* _p$ is precisely $T_e\Phi^p(X)$ (you can see this agrees with your definition by writing $\exp(tX)\cdot p$ as $\Phi^p(\exp(tX))$, and using the chain rule to calculate $\frac{d}{dt}$), and derivative maps $T_xf$ are always linear. So yes, it's enough to check condition 1 on a basis.<|endoftext|> -TITLE: Categories presented with Arrows only, no objects: partial monoids -QUESTION [9 upvotes]: Hi, -I received an answer to a question a while back. The question was about how we can present a category as a collection of arrows and a large list of algebraic relations between them. One of the answers I got was about Freyd's "Categories, Alegories", and here it is: -products in a category without reference to objects or sources and targets -Can anyone (Wouter maybe?), give much more detail about this presentation. Can anyone give the precise definition of these "kinds" of partial monoids a la Freyd? -As a side note, could someone suggest a good way to define a dcpo of such partial monoids? - -REPLY [4 votes]: One reference that might be of interest to you is C. Ehresmann's Catégories et structures (Dunod, 1965). It starts Chapter 1 with the following definition (my translation) - -Let $C$ be a class; a (partially defined) law of composition on $C$ is a - function $\kappa$ from some subclass $K$ of the product class $C\times C$ - into $C$; the class $K$ is called the class of composables; if $(g,f)\in K$, - we call $\kappa(g,f)$ the composite of $g$ and $f$. The pair $(C,\kappa)$ - of a class $C$ and a law of composition on $C$ is called a multiplicative class. - -By page 5, we've reached the definition of a category, as a multiplicative class satisfying four further axioms, being (paraphrased) - -Existence of identities (at notional source and target) -Proper domains and codomains of composites (in terms of identities) -Associativity -`Enough composites', that is, if $\mathrm{dom}\ g = \mathrm{cod}\ f$, then $g$ and $f$ are composable. - -The book goes on to cover most of basic category theory, as far as I can tell.<|endoftext|> -TITLE: An isoperimetric problem on the hypercube -QUESTION [10 upvotes]: Fix integer $n\ge 1$, and let $E=\{e_1,...,e_n\}$ denote the standard basis of the vector space ${\mathbb F}_2^n$. Thus, for a set $A\subset{\mathbb F}_2^n$, the sumset $A+E:=\{a+e\colon a\in A,\ e\in E\}$ consists of all those elements of ${\mathbb F}_2^n$ which are at Hamming distance $1$ from an element of $A$. Now I wonder, - -How small can $|A+E|$ be in terms of $|A|$? How to choose a set $A$ of prescribed size to minimize the size of $A+E$? - -I would expect the answer to be well-known -- any reference? -Somewhat closer to what I actually need is the situation where $A$ consists of even vectors only; that is, of vectors orthogonal to $e_1+\dotsb+e_n$. How to choose $A$ (of prescribed size) under this additional constrain to minimize $|A+E|$? - -REPLY [6 votes]: When A consists of even vectors only, the problem was probably solved by Korner and Wei [Odd and even Hamming spheres also have minimum boundary, Discrete Math. 51 (1984), 147–165]. See also Lemma 1.10 in [D. Galvin, On homomorphisms from the Hamming cube to Z, Israel J. of Math 138 (2003), 189-213].<|endoftext|> -TITLE: Continuity of a convolution (Version 2) -QUESTION [6 upvotes]: Hello, -This problem bothers me for some time. Suppose that - -$\mu$ is a non-negative Radon measure (or positive linear functional of the space of continuous functions with compact support); -$\psi$ is a continuous function, vanishing at infinity and integrable, i.e., $\psi\in C^0_0(R)\cap L^1(R)$; -$\sup_{x \in R}|(\psi*\mu)(x)|<+\infty$. - -Then, we would like to prove that the function $x\mapsto (\psi*\mu)(x)$ is continuous. -Thank you very much for your help and any hints! -Anand - -Version 2. If we add an additional property, - -$\sup_{x\in R} |(G*\mu)(x)|<+\infty$, where $G(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$. - -Then, is it possible to prove that the function $x\mapsto (\psi*\mu)(x)$ is continuous? -Thanks -Anand - -REPLY [5 votes]: Here's a counter-example. -Let $\mu$ be counting measure supported on $\mathbb Z$; so $\int f(x) \ d\mu(x) = \sum_{m\in\mathbb Z} f(m)$ for $f$ continuous with compact support. -Choose a very rapidly decreasing sequence of positive reals $\delta_n$. -Let $\psi$ be the piecewise linear function which is $1$ at $n+1/n$ for $n\geq 10$ (say), and is $0$ at $n+1/n \pm \delta_n$ (and is zero at all $x<10+1/10-\delta_{10}$). By definition, $\psi$ is continuous (and could be made smooth by the use of a bump function) and $\psi$ is Lebesgue integrable (if $\delta_n$ decreases fast enough). -Then set -\[ \alpha(x) = (\psi*\mu)(x) = \int \psi(x-y) \ d\mu(y) = \sum_{m\in\mathbb Z} \psi(x+m). \] -A priori, this sum might diverge, but only to $+\infty$. Clearly $\alpha$ is periodic in that $\alpha(x+k)=\alpha(x)$ for any $k\in\mathbb Z$. -Fix $x\in[-1/2,1/2)$, and consider which $m\in\mathbb Z$ are such that $\psi(x+m)>0$. This occurs iff there is $n\geq 10$ with $1/n-\delta_n < x+m-n < 1/n+\delta_n$. As $m-n\in\mathbb Z$ and $n\geq 10$ and $|x|\leq 1/2$, this can only occur if $m=n$. So $1/n-\delta_n < x < 1/n+\delta_n$. Choosing $(\delta_n)$ suitably, we can arrange that $1/(k+1)+\delta_{k+1} < 1/k-\delta_k$ for all $k$, and then $n$ is unique for any given $x$. -We conclude that for every $x$, there is at most one $m\in\mathbb Z$ with $\psi(x+m)>0$. In particular, $\alpha(x)\in[0,1]$ for all $x$. -However, clearly $\alpha(1/n)\geq 1$ for all $n\geq 10$, but $\alpha(0) = 0$ as if $\psi(m)>0$ for some $m\in\mathbb Z$, then there is $n\geq 10$ with $-\delta_n < m-n-1/n < \delta_n$ which forces $m=n$, but then $-\delta_n <-1/n<\delta_n$, which we can avoid by a suitable choice of $(\delta_n)$. So $\alpha = \psi*\mu$ is not continuous at $0$.<|endoftext|> -TITLE: Spin structures and quadratic forms on surfaces -QUESTION [7 upvotes]: In his paper "Spin structures and quadratic forms on surfaces", Johnson constructs a bijection between the set of spin strucutres on a smooth closed orientable surface $S$ and the set of quadratic forms on $H_1(S,\mathbb{Z}_2)$. -It seems to me that this result (and all the proofs) extends to the case of a not closed orientable surface. Is it true ? - -REPLY [12 votes]: It is not quite true. There is an algebraic gadget one can produce from a Spin structure on a surface with boundary (where we fix the Spin structure along the boundary), and one gets a bijection from Spin structures to such gadgets, but the process is horribly non-canonical. It does however let you prove that there are precisely two isomorphism classes of Spin surfaces of each topological type (for genus $\geq 2$). -This is discussed in Section 2.3 of my paper "Homology of the moduli spaces and mapping class groups of framed, r-Spin and Pin surfaces" (http://arxiv.org/abs/1001.5366), albeit it more generality (so-called $r$-Spin structures). -Briefly, the point is this: one ought to have some sort of algebraic structure on $H_1(S, \partial S ;\mathbb{Z}/2)$, and Johnson's construction works fine for elements represented by unions of simple closed curves, but for arcs between boundary components there is all sorts of indeterminacy and choices to be made.<|endoftext|> -TITLE: Finding minimum (or maximum) element of a low rank matrix. -QUESTION [14 upvotes]: Let $A\in\mathbb{R}^{n\times n}$ and suppose that $A$ is of rank $m\leq n$. Moreover suppose we know $u_1,\ldots, u_m \in\mathbb{R}^{n\times 1}$ and $v_1,\ldots, v_m \in\mathbb{R}^{n\times 1}$ such that -$ -A = \sum_{k=1}^m u_kv_k^T -$ -Is there a faster way than $\mathcal{O}(n^2)$ for finding the minimum (or maximum) element of $A$? Here I'm thinking of $m$ being small in comparison to $n$. -For instance, if $m=1$ it is easy to find the minimum (or maximum) element just by finding the minimum of $u_1$ and then of $v_1$, taking $\mathcal{O}(n)$. -For instance, if $m=1$ then to compute the maximum you can do, -$ -\max\{\min u_1\cdot \min v_1,\min u_1\cdot\max v_1, \max u_1\cdot\min v_1,\max u_1\cdot\max v_1\} -$ -which is $\mathcal{O}(n)$. -Thanks. - -REPLY [2 votes]: We can reformulate the problem as follows: given vectors $u_1,\ldots,u_n,v_1,\ldots,v_n \in \mathbb{R}^m$, find $\max | \langle u_i,v_j \rangle |$ over all $i,j \in [n]$. -Now, if an approximate answer is good enough, I think that for small $m$ an approximate answer with multiplicative error of $1+\epsilon$ can be found in time $\approx n m + (1/\epsilon)^m$. Take an $\epsilon$-net of the $m$-dimensional sphere (that is, a set of points on the sphere such that each other point on the sphere is $\epsilon$-close to one of them). Such a net has size $ (1/\epsilon)^{O(m)}$. Now, hash the vectors $u_1,\ldots,u_n$ to the sphere. Let $u'_i=u_i / \|u_i\|$ be the image of $u_i$. Find the point on the net closest to $u'_i$ (trivially this can be done in time linear in the size of the net, but probably there are some geometric algorithms to do this is constant or near constant time). For each point in the net store the vector close to it with maximal norm. Do the same for $v_1,\ldots,v_n$. To finish, go over all pairs of points in the net and compute the inner products of their associated vectors.<|endoftext|> -TITLE: Dynamical Systems for undergraduate students -QUESTION [6 upvotes]: Hello, -I'm searching for a good subject of Dynamical Systems theory in which I can propose a theme for a undergraduate research opportunity program. As I'm a undergraduate student, I have had only be exposed to subjects such as linear and abstract algebra, real analysis (calculus, basic topology of $\mathbb{R}^{n}$, metric spaces, and a "informal" view of measure theory). -I have been looking at subjects such as unidimensional dynamics and circle homeomorphisms, and reading An Introduction to Chaotic Dynamical Systems by R. Devaney. --Thanks. - -REPLY [3 votes]: I see that you have already received suggestions about books, so I am not going to suggest any books, but I will suggest a topic, if that's what you are looking for. I think the notion of entropy and its role in dynamical systems could potentially be a good research theme. You can find out about different notions of entropy, their connections and their importance in dynamical systems.<|endoftext|> -TITLE: Topological Characterisation of the real line. -QUESTION [39 upvotes]: What is a purely topological characterisation of the real line( standard topology)? - -REPLY [8 votes]: Take the category of completely regular spaces (subspaces of compact Hausdorff spaces or uniformisable spaces, you do not need the real numbers) with continuous maps. The cogenerators in this category are precisely all spaces containing the real numbers as a subspace. Now you can choose all connected cogenerators which are contained in every cogenerator. You get (up to homeomorphisms) $[0,1]$, $[0,1)$ and $\mathbb{R}$. Only one of these objects can be embedded densely into the other two objects: $\mathbb{R}$. Notice that you can state this definition without refering to anything else than the morphisms in the category of completely regular spaces. This definition is quite amusing in my opinion, although it might not be very practical.<|endoftext|> -TITLE: Del pezzo surfaces in positive characteristic -QUESTION [6 upvotes]: For me a Del Pezzo surface $X$ over an algebraically closed field of characteristic $p$ is an algebraic surface where the anticanonical bundle $\omega^{-1}_X$ or $-K_X$ is ample. (I prefer the second notation, although it is not very correct), -In characteristic 0, as far as I know, there is a classification. $X$ has to be isomporphic either to $\mathbb{P}^1\times \mathbb{P}^1$ or a blow-up of $\mathbb{P}^2$ at $9-K_X^2\geq 0$ points in general position, i.e. not 3 points in the same line and not 6 points in the same conic of $\mathbb{P}^2$. -I would like to know if there is such a classification in characteristic $p\geq 2$. As far as I know the classification is definitely the same for $p>3$, and probably even for $p=3$ and there are 'extra' surfaces for $p=2$. -The perfect answer would confirm whether these assertions are true, describe the extra cases (in particular in terms of which curves can live in them or minimality) and/or give a reference. -But anything is better than nothing, so even if you know a bit of this I would like to know. - -REPLY [10 votes]: The classical classification of (smooth) Del Pezzo surfaces as blow-ups relies on the Kodaira vanishing theorem in characteristic zero, but is actually true over any algebraically closed field. See for example Kollar's Rational curves on algebraic varieties book, section III.3. (This paper by Xie on the Kawamata vanishing theorem on rational surfaces in characteristic $p$ might also be useful.) -There is also an interesting classification of Del Pezzo surfaces as complete intersections in weighted projective spaces, which holds over any base field (not neccessarily algebraically closed). Again, the reference is Kollar's book, chapter III.3.<|endoftext|> -TITLE: Probability of Generating a Connected Graph -QUESTION [9 upvotes]: $N$ points are generated randomly within a unit square, with a uniform distribution. -What is the probability that the points form a connected graph, given that two points are connected if the distance between them is less than or equal to $d$? -(this should obviously be some function of $N$ and $d$). -If you don't know the answer, but have an idea that may (or may not) lead me a step forward, please let me know as well. -Moreover, if you know for sure that this problem is yet unsolved, that's also good news for me. I can then do it through a Monte-Carlo simulation, but my approach would be justified. -Thanks, -Melvin - -REPLY [9 votes]: Just a few more comments to the answers and references already posted. I will denote your graph by $G(n,d(n))$. I'm not sure if this is satisfactory enough, but with fairly standard methods one can prove that if $\mu=ne^{-\pi nd(n)^2}\to 0$ as $n\to \infty$ then the graph is aas connected. In general the probability that $G$ is connected is $\sim e^{-\mu}$. References include the monograph by Penrose mentioned in the comments, the paper by Gupta and Kumar, see also the paper by Penrose "The longest edge of the random minimal spanning tree" The Annals of Applied Probability (1997) 7, 340–361, and M.J. Apple, R.P. Russo "The connectivity of a graph on uniform points on $[0,1]^d$". -Note that the situation in $[0,1]^d$ with metric $\ell_p$ is similar. In fact Goel, Rai, and Krishnamachari proved this for all monotone properties of graphs (like connectivity, non-planarity etc.) "Monotone properties of random geometric graphs have sharp thresholds". (Monotone property here means that it is preserved by addition of edges.)<|endoftext|> -TITLE: Acyclic models via model categories? -QUESTION [19 upvotes]: Recall the acyclic models theorem: given two functors $F, G$ from a "category $\mathcal{C}$ with models $M$" to the category of chain complexes of modules over a ring $R$, a natural transformation $H_0(F) \to H_0(G)$ induces a natural transformation $F \to G$ (unique up to natural chain homotopy) if $F$ is free on the models and $G$ is acyclic. (This has many useful applications, such as showing that things like the Alexander-Whitney maps exist and are unique up to homotopy.) -The proof is a bit of homological algebra, but I suspect that there may be a fancier homotopical way to think about it. Namely, I want to say something of the form that the category of functors from $\mathcal{C}$ to chain complexes has a model structure on it, the free functors are cofibrant in some sense, and the map $G \to H_0(G)$ should be an acyclic fibration. Thus the lifting of $F \to H_0(F) \to H_0(G)$ to $F \to G$ would just be a lifting argument in a general model category (as would be the existence of a natural chain homotopy). However, I'm not sure what the model structure should be. The projective model structure doesn't seem to be the right one because $G \to H_0(G)$ is not a weak equivalence of chain complexes (it only is for the models $M$). So I guess the model structure relevant here would be a hybrid of the projective model structure, where instead of weak equivalences and fibrations levelwise, one would just want them to hold on $M$: that is, a morphism $A \to B$ is a weak equivalence (resp. fibration) if and only if $A(m) \to B(m)$ is one for each $m \in M$. It seems that the lifting property needed is essentially the proof of the acyclic models theorem itself. -I strongly suspect this can be done. Am I right? Can this be pushed further? - -REPLY [7 votes]: Akhil's answer works proposes a model structure, but it doesn't have the property required by the original question, i.e. that free functors are cofibrant. Also, proving factorization is much more involved than Akhil's answer would lead you to believe (in fact, this is where all the hard work is done). In this answer I'll fill in the details of that proof and at the bottom address the non-cofibrancy of free functors. -The proof that the model structure exists mimics Hirschhorn's proof that the projective model structure exists, and that proof takes about 6 pages. I'll summarize it here, but those who know it can just skip the next section where I fill in the details in Akhil's setting (i.e. with fibrations and weak equivalences defined based on $M\subset K$). The last section brings it back to the acyclic models theorem. -Projective Model Structure -The reason factorization works in the standard projective model structure is that you transfer the model structure from the cofibrantly generated $C^{K^{disc}} = \prod_{ob(K)} C$ along a Quillen adjunction $(F,U)$, and you apply the small object argument. This transfer principle is Hirschhorn's Theorem 11.3.2. Let $I$ and $J$ be the generating cofibrations and generating trivial cofibrations for $C$. For every $k\in K$ let $I_k= I \times \prod_{k'\neq k} id_{\phi}$, where $\phi$ is the initial object of $C$. This is how to view $I$ in just one component of $C^{K^{disc}}$. The generators for $C^{K^{disc}}$ are $I_{ob(K)} = \cup_{k\in K} I_k$ and similarly $J_{ob(K)} = \cup_{k\in K} J_k$. -The existence of the projective model structure (and the fact that it's cofibrantly generated) is Hirschhorn's Theorem 11.6.1 and the proof verifies that the hypotheses of 11.3.2 hold, namely: - -$F(I_{ob(K)})$ and $F(J_{ob(K)})$ permit the small object argument. -All maps in $U(F(J_{ob(K)})-cell)$ are weak equivalences. - -For Hirschhorn, the functor $F$ takes a discrete diagram $X$ to a diagram $\coprod_{k \in K} F^k_{X(k)}$ in $C^K$ where $F^k_A = A \otimes K(k,-)$. The functor $U$ is the forgetful functor. Hirschhorn's proof of (2) requires Lemma 11.5.32, which analyzes pushouts of $F_A^k \to F_B^k$, since for any $g:A\to B$, $F(g\otimes \prod_{k'\neq k} id_\phi)$ is the map $F_A^k\to F_B^k$. The point of Lemma 11.5.32 is that if $X\to Y$ is a pushout of this map then $X(k)\to Y(k)$ is a pushouts in $C$ of a giant coproduct of copies of $g$. This coproduct is a trivial cofibration, so pushouts of it are trivial cofibrations and hence $X\to Y$ is a weak equivalence as desired. -Lemma 11.5.32 gets used again (along with the Yoneda lemma and basic facts about smallness) to prove (1). The point is that the domains of the generators are $F_A^k$ and $A$ is small relative to $I$-cof, so one can use adjointness to pass between maps in $C^K$ and maps in $C$ (i.e. one component of $C^{K^{disc}}$) and use smallness there to get the desired commutativity of colimit and hom. -Modified Projective Model Structure -To prove Akhil's proposed model structure has factorization you need to define the correct functor $F$ AND you have to verify (1) and (2). It turns out that Akhil has the correct $F$ (if you soup it up so it goes from $C^{K^{disc}}\to C^K$ rather than $C\to C^K$), but verifying (1) and (2) are non-trivial because this $F$ doesn't automatically have Lemma 11.5.32. Anyway, the correct $F$ takes $X$ to $\coprod_{m\in M}F^m_{X(m)}$ in $C^K$. Note that it doesn't land in $C^M$ because $K(m,-)\neq M(m,-)$. This is good, since if it landed in $C^M$ you'd just end up with the projective model structure on $C^M$. -To make a long story short, you can prove an analog of Lemma 11.5.32 and follow Hirschhorn to get the modified projective model structure. A better way is to notice that on the coordinates in $K - M$ all maps are trivial fibrations, the cofibrations are isomorphisms, and both the generating cofibrations and the generating trivial cofibrations are $\lbrace id_\phi \rbrace$. So if $M$ is empty you get the trivial model structure (Hovey's example 1.1.5) on $C^K$, and if $M$ is a one-point set you get a product of $C$ with a bunch of copies of that trivial model structure. Of course if $M=K$ you get the usual projective model structure. -With this observation, a better way to verify existence of this modified projective model structure is to take a product where in coordinate $k\in M$ one uses the projective model structure and in coordinate $k\not \in M$ one uses the trivial model structure. Note that this modified projective structure is done in Johnson and Yau's "On homotopy invariance for algebras over colored PROPs" for the case where $K$ is a group acting on $C$ and $M$ is a subgroup. Unfortunately, I only found this reference after working the above out for myself. -Connection to Acyclic Models Theorem -The main reason why I don't think this model structure helps with the acyclic models theorem is what I alluded to in my comment to Akhil's answer. You really need $K$ to be small, and this rules out all the interesting examples to which the acyclic models theorem applies. The other reason is the observation above about getting the trivial model structure in coordinates $k\notin M$. With this, you can't have the "free functors being cofibrant" that the OP asks for. Recall that a free functor here means there are models $M_\alpha \in M$ and $m_\alpha \in F(M_\alpha)$ such that for all $X\in K$ the set $\lbrace F(f)(m_\alpha)\rbrace$ is a basis for $F(X)$. Because cofibrations are isomorphisms in the coordinates $k\notin M$, the only cofibrant object there is the initial object, i.e. the empty functor. These free functors are not of that form because their behavior is not restricted in that way on $k\notin M$.<|endoftext|> -TITLE: Geodesics in $\mathbb{R}^2 \times \mathbb{S}^1$ under "segment" metric -QUESTION [13 upvotes]: Represent the position of a unit-length, oriented segment $s$ in the plane -by the location $a$ of its basepoint and -an orientation $\theta$: $s = (a,\theta)$. So $s$ can be -viewed as a point in $\mathbb{R^2} \times \mathbb{S^1}$. -Now I'll define a metric on this space. -Define the distance $d(s_1,s_2)$ between two positions of -unit-length segments as the average distance between their corresponding -points: - -      - - -Above the distances are about 0.31, 0.61, and 0.53, left-to-right. -So if the endpoints of $s_i$ are $a_i$ and $b_i$, -then $d(s_1,s_2)$ is the average of the Euclidean distances -between $(1-t) a_1 + t b_1$ and $(1-t) a_2 + t b_2$ as $t$ varies in $[0,1]$. -This is indeed a metric, I believe, because the triangle inequality -holds between corresponding points in three positions of the segment. -This metric is intended to capture the intuitive notion of how much -work is required to move $s_1$ to $s_2$. -My question is: What are the geodesics in this space under this metric? -Certainly a pure translation of $s$ is a geodesic. -It seems that a pure rotation by at most $\pi$ of $s$ about a -point $p \in s$ should also be a geodesic, but even this is -not so clear to me. Certainly a rotation about a point not on $s$ -is (generally) not a geodesic. Of course the main interest would be -in geodesics that mix translation and rotation, showing (locally) optimal -repositioning paths. -I investigated this long ago when working on motion-planning algorithms -("moving a ladder"), but got quite blocked on this natural question. -This superficially seems related to the -Kakeya needle problem, -but the metric I propose does not measure swept area. -Perhaps it has been studied in some guise previously. -If so, a pointer would be appreciated. Thanks! -Addenda. (26Sep11.) I just ran across this book, by V. A. Dubovit͡s︡kiĭ, -which seems relevant: -The Ulam problem of optimal motion of line segments, -Translation Series in Mathematics and Engineering, Optimization Software, 1985. -It may take some time for me to locate a copy... -(11Nov11). I finally have this book in my hands. -The Preface by Hestenes says, - -Dubovitskij has succeeded in solving in closed form a generalization of a problem - of S[.] Ulam..: - Among all continuous motions of an oriented line segment $S$ in $\mathbb{E}^n$ from one - position to another, which preserves its length [...], find one for which the sum - of the lengths of the paths swept by its endpoints is minimal. - -The concentration here on the motion of the endpoints—in contrast to -the average distance metric I proposed—seems to render these results as not directly relevant, although nevertheless quite interesting. - -REPLY [10 votes]: In addition to Anton Petrunin's answer, here is a trick to simplify (and in some sense solve) the geodesic equation. -Since the metric has three-dimensional group of isometries (generated by rigid motions of the plane), the corresponding Noether's integrals make the geodesic flow completely integrable. More precisely, the first derivative is uniquely determined by the position and the values of the integrals. Actually a two-dimensional group is sufficient because for the third integral one can always take the speed (= the norm of the velocity vector) of a geodesic. -So let's use invariance under parallel translations only. We have a manifold of all unit segments in the plane, and a tangent vector to that manifold can be thought of as the vector field along a segment (representing velocities of all its points). This vector field has a form -$$ - v(t) = tx + (1-t)y, \qquad t\in[0,1] -$$ -where $x,y\in\mathbb R^2$ are such that the vector $x-y$ is orthogonal to the segment. The norm of this tangent vector is $\int |v(t)| dt$ (as in Anton's answer), and the Noether integral corresponding to translations (if my quick computation is correct) boils down to the following: if $v$ is a velocity vector of a geodesic represented in the above form, then -$$ - \int_0^1 \frac{v(t)}{|v(t)|} \ dt = V_0 -$$ -where $V_0$ is a constant (for each geodesic) vector in the plane. This vector $V_0$ (and, say, the assumption that the geodesic is unit-speed) determines the parameters $x$ and $y$ in the formula for $v(t)$ uniquely (as a function of the segment's current direction), so the geodesic equation is reduced to a 1st order ODE. Furthermore, the above integral can be found as an explicit (but weird) function of $x$ and $y$, so the equation becomes easy to solve, at least numerically. -By the way, I second Jean-Marc Schlenker's proposal to consider quadratic mean of the distances (i.e. $L^2$ norm rather than $L^1$), especially if you have any physics-related application in mind. In this case the metric is Riemannian, the energy is just the standard kinetic energy of the moving segment, so the Noether intergrals are just the standard conservation laws: the linear momentum and the angular momentum. And the geodesics are very simple: the segment rotates at a fixed angular speed while its midpoint (the barycenter) moves along a straight line with a constant speed.<|endoftext|> -TITLE: is there any fibration $\mathbb{R}^n\to \mathbb{S}^n$? -QUESTION [12 upvotes]: It is probably a trivial question. But I don't see the answer. -Is there any Hurewicz fibration $\mathbb{R}^n\to \mathbb{S}^n$ ? -Is there any fibration $X\to \mathbb{S}^n$, when $X\subset \mathbb{R}^n $? -I appreciate any help. Thank you very much! - -REPLY [12 votes]: Edit: The following simplifies the original answer (which unnecessarily used singular cohomology). -If $f:\Bbb R^n\to S^n$ is a fibration, then as Mark noted, a fiber $F$ of $f$ is weak homotopy equivalent to $\Omega S^n$ (using the 5-lemma, see Prop. 4.66 in Hatcher). I claim that $F$ is in fact homotopy equivalent to $\Omega S^n$. -Indeed, $F$ is homotopy equivalent to the corresponding homotopy fiber of $f$ (Prop. 4.65 in Hatcher). The homotopy fiber consists of pairs $(x,p)$ where $x\in\Bbb R^n$ and $p$ is a path in $S^n$ connecting $f(x)$ and the basepoint. It is homotopy equivalent to the space $X$ of maps $[0,1]\to MC(f)$ (the mapping cylinder) sending $0$ into $\Bbb R^n$ and $1$ into the basepoint of $S^n$. Milnor showed that $X$ and $\Omega S^n$ are homotopy equivalent to CW-complexes. Hence, being weak homotopy equivalent to each other, by Whitehead's theorem they are homotopy equivalent to each other. -Now $F$ is finite-dimensional, so its Cech cohomology is eventually zero. Cech cohomology is a homotopy invariant, so we get that the cohomology of $\Omega S^n$ is eventually zero, -contradicting Mark's comment. (It does not matter which cohomology of $\Omega S^n$, they are all isomorphic since $\Omega S^n$ is homotopically a CW-complex.)<|endoftext|> -TITLE: What's tropical about tropical algebra? -QUESTION [16 upvotes]: Please allow me to ask a potentially dumb question (or maybe more precisely, a question floating on clouds of ignorance): - -Why is a max-plus algebra called a tropical algebra? - -REPLY [20 votes]: A lot of sources mention that the adjective "tropical" is given in honor of Imre Simon, but it seems hard to find who precisely coined the term. I found some sources which attribute this to some French mathematicians. Here is what Bryan Hayes writes on the topic: - -For starters, what is that word “tropical” supposed to mean? Speyer and Sturmfels explain: “The adjective tropical was coined by French mathematicians, including Jean-Eric Pin, in honor of their Brazilian colleague Imre Simon.” Pin, in a 1998 paper (.pdf), deflects the credit to another French mathematician, Dominique Perrin, again noting that the name honors “the pioneering work of our brazilian colleague and friend Imre Simon.” Simon himself, in a 1988 paper (.ps), attributes the term to yet a third French mathematician, Christian Choffrut. Apparently, no one wants to lay claim to the word, and I can’t entirely blame them. Speyer and Sturmfels go on: “There is no deeper meaning in the adjective ‘tropical’. It simply stands for the French view of Brazil.”<|endoftext|> -TITLE: Gauss Theorem and Weil Conjectures for elliptic curves -QUESTION [7 upvotes]: It is known (by Gauss) that for a prime $p \equiv 1 \pmod 3$ there is a "unique" writing of $4p=A^2+27B^2$ where $A=1+p-M_p$ and $M_p$ is the number of solutions of $X^3+Y^3+Z^3=0$ in the projective plane $\mathbb{P^2(\mathbb F_p)}$. -It's also known that the points on the cubic $y^3=x^3+1$ are in 1-1 correspondence with the points on the elliptic curve $y^2=x^3-432$. -I noted that $A$ is the same (up to its sign) of the Frobenius trace in the Weil conjectures for elliptic curves; and the number $A^2-4p$ is actually the discriminant of the characteristic polynomial of the map the Frobenius automorphism induces on the Tate module. -So, I've the following question to you: -Is there a proof of $p \equiv 1 \pmod 3 \Rightarrow 4p=A^2+27B^2, \ A=1+p-M_p$ using Weil conjectures? - -REPLY [8 votes]: I am going to interpret "the Weil conjectures" as a synecdoche for "thinking about counting points in terms of eigenvalues of Frobenius operators." In this case, the answer is a resounding yes! However, the specific facts about eigenvalues of Frobenius which which are being used here are easier than the ones discussed in the Weil conjectures. Silverman's The Arithmetic of Elliptic Curves might be a good reference for you. -Let $E$ be the genus one curve $X^3+Y^3+Z^3$ over $\mathbb{F}_p$. This curve has an automorphism $\Phi: (x:y:z) \mapsto (z^p:y^p:z^p)$. The map $\Phi$ sends every point with values in $\mathbb{F}_p$ to itself. However, points with values in $\mathbb{F}_{p^n}$ for higher values of $n$ are permuted. Our goal is to find the number of fixed points of $\Phi$. -The Weil conjecture way to do this is to associate to $E$ certain vector spaces $H^0(E)$, $H^1(E)$ and $H^2(E)$. The map $\Phi$ acts functorially on cohomology, and we have -$$\#(\mbox{Fixed points of }\Phi) = Tr(\Phi^*:H^0 \to H^0) - Tr(\Phi^*:H^1 \to H^1) + Tr(\Phi^*:H^2 \to H^2).$$ -Now, in this case, you can really get quite explicit about what these vector spaces are. $H^0$ and $H^2$ are both $1$-dimensional, and $\Phi$ acts by $1$ and $p$ respectively. $H^1$ the Tate module of $E$ (Silverman III.7); a two dimensional space. -So the number of fixed points is $p+1-a$ where $a$ is the trace of $\Phi^*$ acting on the Tate module. -One can show that the endomorphism ring of $E$ is $\mathbb{Z}[\omega]$ where $\omega$ is a primitive cube root of unity (see Silverman V.3). So we must have $\Phi = c+d \omega$ for some integers $c$ and $d$. The fact that $\Phi$ has degree $p$ (see Silverman II.2) means that $N(c \omega +d) = c^2 - cd + d^2 =p$. And one computes that $a = Tr(\Phi) = 2c - d$. -So, in short, the number of points on $E$ is $p+1-a$, where $a=2c-d$ with $(c,d)$ such that $N(c+d \omega)=p$. Converting that into your statement, and getting the signs right, is left to you. - -Focusing on the more specific question of using the Weil conjectures. The fact that $N(c+d \omega)$ is $p$ comes from the Riemann hypothesis part of the Weil conjectures. But you also need to know that $\mathbb{Z}[\omega]$ is the full ring of automorphisms, which is not particularly related to the Weil conjectures. And you need to figure out which of the $6$ elements of $\mathbb{Z}[\omega]$ which have norm $p$ is the correct one, which is also not a Weil computation. So I would say that this is not specifically a Weil conjecture result, although definitely can be proven using the same sort of ideas as in the Weil conjectures.<|endoftext|> -TITLE: Is the Crandall, Dilcher and Pomerance heuristic concerning Wall-Sun-Sun primes still state of the art? -QUESTION [7 upvotes]: This is a question about the open problem Fibonacci divisibility from the Open Problem Garden. -The problem, originally stated in 1960 by D.D. Wall, has several equivalent formulations one of which is: -Find a prime $p$ with $p^2|a_{p-\left(\frac{p}{5}\right)}$, -where $a_n$ is the $n$-th Fibonacci number. Such primes are often called Wall-Sun-Sun primes. -However in his paper his hypothesis was that Fibonacci-wieferich primes do not exist. He stated that one might conjecture something similar to the Wieferich question, because a lack of proof otherwise. He could not prove that Fibonacci Wieferich primes do not exist. The Sun brothers later followed his open question with the Wall-Sun-Sun prime conjecture. -Since always $p|a_{p-\left(\frac{p}{5}\right)}$ Crandall, Dilcher and Pomerance assumed uniform distribution of the residues and proposed the heuristic $\log \log y -\log \log x$ for the number of such primes in the interval $[x,y]$. Several authors did computer aided searches and no such prime was found up to $9.7 \times 10^{14}$. That confuses me and therefore my question. - -Q: Is the heuristic of Crandall, Dilcher and Pomerance (maybe in its patched version by Klaska) still considered state of the art? If not, are there other approaches? - -Edit: For me the question is sufficiently answered. That is more that I could hope for. Thank you very much! - -REPLY [8 votes]: As quid mentioned, Klyve and I have done some computational investigations on Fibonacci-Wieferich/Wall-Sun-Sun primes. In particular, we collected all primes $p < 9.7\times10^{14}$ such that $F_{p-(p/5)} \equiv Ap \pmod{p^2}$ with $|A| < 2\times10^6$. I've just crunched our data for primes in the range from $6.5\times10^{14}$ to $9.5\times10^{14}$ to see if the incidence of small values of $A$ matches the Crandall–Dilcher–Pomerance heuristic. Here are the results: -$$\begin{matrix} -p \in [6.5\times10^{14},7.0\times10^{14}) & : & 2112 & 2085 & 2083 & 2155 & 2170.39 \cr -p \in [7.0\times10^{14},7.5\times10^{14}) & : & 1905 & 1915 & 2021 & 1953 & 2016.36 \cr -p \in [7.5\times10^{14},8.0\times10^{14}) & : & 1867 & 1854 & 1781 & 1870 & 1882.50 \cr -p \in [8.0\times10^{14},8.5\times10^{14}) & : & 1768 & 1779 & 1707 & 1669 & 1765.11 \cr -p \in [8.5\times10^{14},9.0\times10^{14}) & : & 1598 & 1561 & 1650 & 1686 & 1661.35 \cr -p \in [9.0\times10^{14},9.5\times10^{14}) & : & 1568 & 1592 & 1519 & 1556 & 1568.96 \cr -\end{matrix}$$ -The first four columns report the count of primes $p$ in the given interval whose corresponding $A$ values lie in the respective intervals $(-2\times10^6,-10^6)$, $(-10^6,0)$, $(0,10^6)$, and $(10^6,2\times10^6)$. The last column represents the value predicted by the Crandall–Dilcher–Pomerance heuristic (namely $10^6\log\left(\frac{\log y}{\log x}\right)$ for the interval from $x$ to $y$). -The agreement between experimental data and theoretical values is pretty good. If I understand Klaška's adjustment correctly, he proposes an expected count of roughly half that proposed by the Crandall–Dilcher–Pomerance heuristic. Thus, the data does not appear to support Klaška's modified heuristic. However, note that Klaška's argument is specifically for the special value $A = 0$, so the above data does not invalidate his proposed estimate.<|endoftext|> -TITLE: Polynomial-time complexity and a question and a remark of Serre -QUESTION [9 upvotes]: My question is about the theory of complexity, but let me first explain my motivation, which comes from number theory or more precisely from trying to understand a question/conjecture of Serre and a remark he made about it. Yesterday at Harvard, Jean-Pierre Serre gave a wonderful colloquium on "Variation with $p$ of the number of solutions mod $p$ of a system of polynomial equations". There was many things in it, most of them I knew (but beautifully explained nevertheless), but some lines of thought seemed completely new (and at least they surely where for me): -Let $P(X)$ be a fixed non-zero polynomial in $\mathbb{Z}[X]$. Let us denote by $N(p)$ the number of solution of $P(X) \equiv 0 \pmod{p}$. Serre noticed that there was a very simple, efficient and old algorithm, already present in a paper by Galois and probably older (since Galois refers to it as well-known), to compute $N(p)$ for big values of $p$: compute by Euclid's method the gcd of -$X^p-X$ and $P(X)$ in $\mathbb{Z}/p\mathbb{Z}[X]$. The degree of this polynomial is the number of solutions $N(p)$. The algorithm, Serre said, runs in $O(\log(p)^{2+\epsilon})$. -Now if it happens that all roots of $P(X)$ are in a cyclotomic fields (for example, a quadratic polynomial) then $N(p)$ is determined by some congruences, that is there is an $m$ -such that $N(p)$ depends only on $p \pmod{m}$ (For $P$ quadratic, it is just the quadratic -reciprocity law), and in this case, determining $N(p)$ can be made in $O(\log p)$ steps -obviously. -Serre asked (or conjectured, I don't remember for sure the term he used) if conversely, the existence of an algorihhm in linear time $O(\log p)$ to compute $N(p)$ would imply -that all roots of $P$ are in a cyclotomic field. -After that he made a remark that I didn't fully understand, that he had asked some friends -that were specialists of complexity theory and and that they told him there were not -theorem proving statement like "for a given problem, there is no algorithm solving it in linear time", but I am not sure if this was really that, nor if he was talking only of this specific problem or of all computable problems. -Well, my question just aims to understand Serre's remark - -Do we know of a computable question (with answer yes or no) depending of a parameter $n$ - such that the answer can be given in $O((\log n)^r)$ steps but provably not in - $O((\log n)^s)$, where $1 \leq s < r$ are real numbers? If yes, how do we do that? If no, are there conjecture that for some problem it is not possible? - -(Remark:the answer to the question of Serre is a natural integer, not yes or no, -but since this integer is less than the degree of $P$, obviously it can be transformed into a finite set of "yes or no" questions). -I am an absolute beginner in complexity theory. I made some research on the web this morning and yesterday, but I didn't find an answer or anything close. My question may be very basic -and a reference to a text discussing this kind of questions would be a perfectly nice answer. - -REPLY [12 votes]: Yes, if $1\le r< s$, there exists a set of integers computable in time $O((\log x)^s)$, but not in time $O((\log x)^r)$, where $x$ is the input number written in binary. This is an instance of the time hierarchy theorem. However, the only known such problems are obtained by diagonalization. There is no known natural problem (say, from number theory or combinatorics) that would be computable in polynomial time, but provably required more than linear time. -You can find a discussion of the time hierarchy theorem in any textbook on complexity theory. Two are mentioned in the Wikipedia article, another popular one is by Arora and Barak.<|endoftext|> -TITLE: Understanding Penrose diagrammatical notation -QUESTION [6 upvotes]: I arrived to Penrose's paper Applications of negative dimensional Tensors after reading some bits of Baez's Prehistory (link) and the first two chapters of Turaev's Quantum invariants of knots and 3-manifolds (link). The main result in the last one is a presentation of $\text{Rib}$ (the category of ribbon graphs) by a set of generators and relations (braiding, twist and their inverses, duality morphisms...). -On the other hand Penrose seems to add to the natural braided ribbon structure also a differential one finding generators also for the (anti)symmetrization and covariant-derivative operations. Are there some easily-found references for a structure theorem similar to Lemma 3.1.1 in the book of Turaev? -Edit: I'm sorry I forgot the second question: I have some problems in figuring out the "contraction" in Penrose notation, because in Turaev exposition the juxtaposition of "coupons" seems to be reserved to the composition of morphisms (contraction contrarily implicates some sort of Einstein summation, am I right?) -This is the last edit, I promise: the following convention to contract the sum of two tensors - -doesn't make any sense to me.. How can the "f" leg of $\chi^b_{fde}$ split into two? - -REPLY [6 votes]: To the best of my knowledge, no abstract formulation and soundness theorem of the full Penrose notation (with symmetriser, antisymmetriser, covariant derivative, etc) exists. There is, however, a veritable zoo of graphical languages for various kinds of monoidal categories (with a regrettably small subset of them having published soundess/completeness proofs). The best place to start is Selinger's survey paper from 2011: -http://www.mathstat.dal.ca/~selinger/papers.html#graphical<|endoftext|> -TITLE: First Passage Percolation on Trees -QUESTION [5 upvotes]: Let $T$ be a rooted Galton-Watson random tree generated accordingly to a probability distribution $\mu$. Now assign to each edge $e$ a random non-negative weight $w_e$ distributed a accordingly to a distribution $\nu$. We also assume that the weights are independent for different edges. -Let $T_{n}$ be the collection of nodes at (hop) distance $n$ from the root. For each $v\in T_{n}$, let $P_{v}$ denote the path from the root to $v$. Define -$$ -Z_{v}=\sum_{e\\,\in P_{v}}{\\,w_e}. -$$ -Now for each $n$ let $Y_{n}=\min_{v\in T_{n}}Z_{v}$. It was proved in Limit distributions for minimal displacement of branching random walks that the sequence of random variables -$$ -\{Y_{n}-\mathbb{E}(Y_{n})\}_{n\geq 1} -$$ -is tight. -My question are: - -Is it known what is the behavior of $\mathbb{E}(Y_{n})$ as $n$ increases in terms of $\mu$ and $\nu$? -Is it known for the case $\mu=\delta_{k}$, i.e. when $T$ is a $k+1$ regular tree? - -REPLY [6 votes]: This paper by Dekking and Host is quite old and much has been done in this area since. Today we know that under reasonable assumptions, there are constants $a\in\mathbb{R}$, $b\ge 0$, such that $E(Y_n) = an + b \log n + O(1)$. How to get the constant $a$ was known for quite a long time, see -Biggins, J. D. (1977). Chernoff’s Theorem in the Branching Random Walk. Journal of Applied Probability, 14(3), 630. doi:10.2307/3213469 -For the second term and for almost sure behaviour of $Y_n$, see -Hu, Y., & Shi, Z. (2009). Minimal position and critical martingale convergence in branching random walks, and directed polymers on disordered trees. The Annals of Probability, 37(2), 742-789. doi:10.1214/08-AOP419 -For the definite answer for the law of $Y_n$ in the non-lattice case, see -Aïdékon, E. (2011). Convergence in law of the minimum of a branching random walk. Retrieved from http://arxiv.org/abs/1101.1810 -Note that all of this was already known long before for branching Brownian motion, see the references in the respective articles. -UPDATE: I forgot to add the important reference -Addario-Berry, L., & Reed, B. (2009). Minima in branching random walks. The Annals of Probability, 37(3), 1044-1079. doi:10.1214/08-AOP428 -Here, the authors show the above-mentioned result for $E[Y_n]$ in almost complete generality, and exponential tails for $Y_n−E[Y_n]$ as well<|endoftext|> -TITLE: Random bipartite graphs -QUESTION [10 upvotes]: Consider the following situation: I have a set $A$ of $n$ vertices and a set $B$ of $N = n^2$vertices. I consider the bipartite graph $(A, B)$ and put at random $M = n^{1 + \varepsilon}$ edges (or I could put each edge independently with probability $p$ such that $pnN = M$, this shouldn't make a big difference). Then I remove the isolated vertices from $B$, so effectively I get vertex sets of size $n$ and $\Theta(n^{1 + \varepsilon})$. - -Are there any references on how in general such bipartite random graphs look like (their degree sequence, connectivity etc.)? The model considered above is rather specific, however, I'd be happy with any references on bipartite graphs on $(n, N)$ vertices, where $N$ depends on $n$ (or information on how can one tackle them with techniques similar to ordinary random graphs; in this context it isn't clear to me whether we should treat the graph as a "sparse" or a "dense" one). -Ultimately I'm interested in spectral properties of such a graph (or rather a slight modification of it). What can be said about the second largest eigenvalue of its adjacency matrix or Laplacian? Now suppose that we take a union of this graph and a "good" graph (possibly also random) on the set $A$ only (by "good" I mean it has good spectral properties, I'm not trying to be very specific here). What can be said about eigenvalues of this graph? - -REPLY [9 votes]: Take the case of choosing edges independently with probability $p=n^{-2+\epsilon}$. As you say, it won't make much difference compared to choosing $n^{1+\epsilon}$ edges. Assume $\epsilon<\frac12$. -Consider a particular vertex on the left. The number of possible 2-edge paths from that vertex to any other vertex on the left is less than $n^3$, and each has probability $p^2$, so the expected number of them is $O(n^{-1+2\epsilon})$. So almost certainly most vertices on the left are the centre of a star and nothing bigger. -By similar calculation, the probability that there are any cycles at all is $O(n^{-2+4\epsilon})$. -So the graph is with high probability a forest consisting of $n-O(n^{2\epsilon})$ stars and a few larger components. You can work out the distribution of the star sizes and hence get most of the eigenvalues.<|endoftext|> -TITLE: Is $ord(xy)$ independent of $ord(x)$ and $ord(y)$ in a finite group? -QUESTION [13 upvotes]: Let $r,s,t>1$ be positive integers. Must there exist a finite group $G$ with elements $x$ and $y$ such that $ord(x)=r$, $ord(y)=s$, and $ord(xy)=t$? -The answer is probably "yes." Is there a nice description of such a $G$? - -REPLY [17 votes]: Let $a$ and $b$ be elements of a group $G$. If $a$ has order $m$ and $b$ has -order $n$, what can we say about the order of $ab$? The next theorem shows -that we can say nothing at all. -THEOREM: For any integers $m,n,r>1$, there exists a finite group $G$ with -elements $a$ and $b$ such that $a$ has order $m$, $b$ has order $n$, and $ab$ -has order $r$. -PROOF: We shall show that, for a suitable prime power $q$, there exist elements $a$ -and $b$ of $SL_{2}(F_{q})$ such that $a$, $b$, and $ab$ have -orders $2m$, $2n$, and $2r$ respectively. As $-I$ is the unique element of -order $2$ in $SL_{2}(F_{q})$, the images of $a$, $b$, $ab$ in -$SL_{2}(F_{q})/\{\pm I\}$ will then have orders $m$, $n$, and $r$ -as required. -Let $p$ be a prime number not dividing $2mnr$. Then $p$ is a unit in the -finite ring $\mathbb{Z}/2mnr\mathbb{Z}$, and so some power of it, $q$ say, -is $1$ in the ring. This means that $2mnr$ divides $q-1$. As the group -$F_{q}^{\times}$ has order $q-1$ and is cyclic, -there exist elements $u$, $v$, and $w$ of $F_{q}^{\times}$ having -orders $2m$, $2n$, and $2r$ respectively. -Let -$$ -a=\left( -\begin{array}{cc} -u & 1\\ -0 & u^{-1} -\end{array} -\right)$$ - -and $$b=\left( -\begin{array}{cc}% -v & 0\\ -t & v^{-1}% -\end{array} -\right)$$ -(elements of $SL_{2}(F_{q})$), where $t$ has been chosen so that -$$ -uv+t+u^{-1}v^{-1}=w+w^{-1}. -$$ - - -The characteristic polynomial of $a$ is $(X-u)(X-u^{-1})$, and so $a$ is -similar to $diag(u,u^{-1})$. Therefore $a$ has order $2m$. Similarly $b$ has -order $2n$. The matrix -$$ -ab=\left( -\begin{array}{cc} -uv+t & v^{-1}\\ -u^{-1}t & u^{-1}v^{-1}% -\end{array} -\right) , -$$ -has characteristic polynomial -$$ -X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})\text{,} -$$ -and so $ab$ is similar to $diag(w,w^{-1})$. Therefore $ab$ has order -$2r$. - -I don't know who found this beautiful proof. Apparently the -original proof of G.A. Miller is very complicated; see MO24940. - -REPLY [15 votes]: Here's my comment as an answer: -Take the $r,s,t$--(ordinary) triangle group $T(r,s,t)=\langle x,y \ | \ x^r=y^s=(xy)^t = 1 \rangle$, in which $x$, $y$, and $xy$ have the correct orders. See the section on ``von Dyck" groups here. -As Anton mentions in his comment, $T(r,s,t)$ is infinite when $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} \leq 1$. However, $T(r,s,t)$ is residually finite. The easiest way to see this is to use the facts that finitely generated linear groups are residually finite (due to Malcev, as Steve D mentions), and the fact that $T(r,s,t)$ is linear. -To see that $T(r,s,t)$ is linear, note that when $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = 1$, it is a discrete subgroup of the affine group $\mathbb{R}^2 \rtimes \mathrm{SL}_2(\mathbb{R})$, and when $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} < 1$, it is a discrete subgroup of $\mathrm{Isom}^+(\mathbb{H}^2) \cong \mathrm{PSL}_2(\mathbb{R}) \cong \mathrm{SO}_0(2,1)$, where $\mathrm{SO}_0(2,1)$ is the identity component of $\mathrm{SO}(2,1)$. See this again. -Now, since $T(r,s,t)$ is residually finite, there is a quotient $G$ in which $$x, x^2, \ldots, x^{r-1}, y, y^2, \ldots, y^{s-1}, (xy), (xy)^2, \ldots, (xy)^{t-1}$$ are all nontrivial. This is the $G$ you seek. -Also see Steve D's answer here.<|endoftext|> -TITLE: Biholomorphism between Neighborhood of a complex submanifold and a Neighborhood of zero section of its normal bundle -QUESTION [5 upvotes]: Let X be a compact kahler manifold, $M\subseteq X$ be a complex submanifold, is there a biholomorphism between a neighborhood of the zero section of the normal bundle $NM$ of $M$ and a neighborhood $U(M)\subseteq X$? - -REPLY [7 votes]: The answer is "no" because in general there does not even exist a neighbourhood $U$ of your submanifold that holomorphically retracts to the submanifold. -In fact Finnur Lárusson has proved the following (warning: I have kept his notations, so as to faithfully quote him. In particular $[X]$ is $\mathcal O(X)$, the line bundle on $M$ associated to the divisor $X$) - -Theorem Let $X$ be a connected hypersurface in a compact complex manifold - M of dimension at least $2$. If - (1) $H^0 (M, TM) = 0$ - (2) $dim H^0(M; [X]) \geq 2$, and - (3) $H^1(M; [X]^{-1} \otimes TM) = 0$, - then no neighbourhood of $X$ retracts holomorphically onto $X$. - -Note that the hypotheses are very easy to satisfy: (1) holds for $K3$ surfaces for example and (2), (3) will follow from $X$ being sufficiently ample. -Here is a freely downloadable version of Lárusson's article, from his homepage.<|endoftext|> -TITLE: What is so "plactic" about the plactic monoid? -QUESTION [11 upvotes]: The plactic monoid is the monoid consisting of all words from the alphabet $\mathbb{Z}^+$ modulo certain relations. It is important mainly because its elements enumerate semistandard Young tableaux. -I believe the plactic monoid was introduced by Knuth, but without that name. Lascoux and Schützenberger named it "le monoïde plaxique" in a French paper (1981) of the same name. (DISCLAIMER: I have never seen that paper; perhaps my second question is answered in it.) -Several questions: -1) How did plaxique $\rightarrow$ plactic? (This isn't the most obvious Anglicization; note that the MathSciNet entry for the original Lascoux/Schützenberger paper translates the title as "Plaxic'' monoids.) Who introduced the latter form of the word and why? -2) What is plactic/plaxique supposed to mean? As far as I know neither was a word in their respective languages before being applied to the word monoid/monoïde. I am entertaining an etymology from Greek $\pi \lambda \alpha \xi$ "flat surface," but I don't find it very compelling. - -REPLY [4 votes]: The English translation of Symmetric Functions, Schubert Polynomials and Degeneracy Loci by Laurent Manivel contains the following footnote on the phrase "plactic ring": - -From the Greek $\pi \lambda \alpha \xi$, flat place, stone plate, tablet. This terminology is due to Lascoux and Sch\"utzenberger. - -There is no evidence provided for this assertion, but the fact that the author is French makes me suspicious that he has insider information. There is no discussion of why this name was chosen or what it is meant to suggest.<|endoftext|> -TITLE: Random Sampling a linearly constrained region in n-dimensions... -QUESTION [8 upvotes]: Hi, -So here is my problem: -Given a nonlinear, discontinous, cost function $f(x_1,x_2,..,x_N)$ along with linear constraints $x_n \ge 0, \forall n$ -$x_n \le c_n$ - and $\sum_{n=1}^N x_n = 1$ find an optimal (local) solution by randomly sampling the feasible region. $c_n$ are just constants. -The issue I am having is indexing the space, since it is not a simple n-dimensional cube but rather a polytope(i believe its convex). Discretizing and enumerating all possible combinations of points to sample is much too hard since N =20. Approximating the polytope with a n-dimensional cube and sampling from the cube, only about 1% of the samples fall within the feasible region...which is inefficient if I'm trying to generate many samples. -I've tried finding the volume of the space analytically, however the complexity of computing the integral gets overwhelming for many dimensions. -I was wondering if anyone has come across this type of problem and has any recommendation as to different methods I could try to sample this space. Essentially, I need a good way to estimate the volume...am I looking at this the correct way? -any help would be greatly appreciated... - -REPLY [2 votes]: If the $c_n$ are sufficiently small (say, all smaller than $1/\sqrt{n}$) or sufficiently large (say, all at least $1$) the polytope is a simplex, so estimating the volume and sampling are both easy as pointed out in @Mike's answer. On the other hand, trying to find a local optimum by uniform sampling is doomed in $20$-dimensional space, since the distance between the samples will be very large for any reasonable number of samples. (for example, if your polytope were a cube of side one, generating $2^{40}$ points will have inter point distances around $1/4.$ Needless to say, generating $2^{40}$ points might take a while.<|endoftext|> -TITLE: Any nice examples of small cancellation theory appearing in applied mathematics? -QUESTION [9 upvotes]: Are there any nice discussions of applications of small cancellation theory, or other cases of the word problem, in applied mathematics or algorithms for seemingly non-group theoretic problems? -I suppose two candidates are Anshel–Anshel–Goldfeld key exchange and braid group based cryptography. - -REPLY [6 votes]: The group-based cryptography is a very active area now. There are books (eg Group-based Cryptography) and a special journal, Groups Complexity Cryptology. -As another application, one can reformulate the problem P=NP as "Is it true that every finitely presented group with polynomial Dehn function has word problem solvable deterministically in polynomial time?" (see the paper Isoperimetric and isodiametric functions of groups arXiv:math/9811105). -Small cancelation itself is not used much in any of these areas, but many ideas are inspired by the classical small cancelation theory.<|endoftext|> -TITLE: Exact consistency-strength of "all projective sets are Ramsey" -QUESTION [8 upvotes]: I wonder if the exact consistency strength of -"All projective sets have the Ramsey property" -is still open. -In Solovay's model, all sets have the Ramsey property, so the consistency strength of this is below an inaccessible. As far as I know, there are some implications under forcing axioms. -And all sets up to a certain level of the projective hierarchy having the Ramsey property is still do-able from just ZFC. -But I can't seem to find anything which shows Solovays inaccessible is really necessary in case of the Ramsey property (like Raisonnier/Shelahs result for Lebesgue measure), or oppositely, that you can get a model where "All projective sets have the Ramsey property" from just ZFC (like for the Baire property). - -REPLY [9 votes]: Hi David, -This is my first foray onto MathOverflow as well, so this answer is an experiment to see if I can get things to work, rather than an attempt to convey a lot of serious information. -As Andres said, the problem is still open as far as I know. I worked on this with Shelah a bit in the late 90s and we generated many things that led nowhere. Some related material: -1) Roslanowski and Shelah investigated "sweetness" and "sourness" (properties of ccc posets motivated by the constructions in Shelah's "Can you take Solovay's Inaccessible Away") in a series of quite technical papers early in the 2000s. This was partially motivated by the problem of getting all nicely definable sets to be Ramsey without using an inaccessible. -2) CH + "every set of real in L(R) is Ramsey with respect to every Ramsey ultrafilter" is equiconsistent with the existence of Mahlo cardinals. Mathias got the consistency result assuming a Mahlo, and the other direction is in a paper of mine from 1999 or so. The trick I used didn't seem to shed any light on whether or not the inaccessible is needed when we drop the reference to ultrafilters. -Best, -Todd<|endoftext|> -TITLE: Existence of definite symmetric matrices satisfying affine linear constraints -QUESTION [5 upvotes]: Suppose I have a generic 3-dimensional affine linear subspace of the 6-dimensional space of symmetric $3 \times 3$ matrices. Does such a space necessarily intersect the set of definite (either positive or negative definite) symmetric matrices? -Genericity is important here; this is certainly not necessarily true for a linear subspace, or for spaces parallel to coordinate planes, for example. But will a small perturbation of the affine subspace always suffice to obtain a nontrivial intersection? - -REPLY [7 votes]: No. Consider the 3-dimensional affine subspace consisting of all matrices having $1,1,-1$ on the diagonal (non-diagonal entries are arbitrary). A small perturbation of this subspace cannot intersect the cone of definite matrices. Indeed, a matrix from a perturbed subspace either has diagonal entries close to $1,1,-1$, or the maximum non-diagonal entry is much larger than the diagonal ones (and in this case, the respective $2\times 2$ minor is negative). -Even 4-dimensional subspace is not enough. Indeed, let $C$ denote the set of nonnegative matrices. It is a sharp (i.e. not containing straight lines) closed convex cone in the 6-dimensional space of all symmetric matrices. Therefore it is a cone over a compact convex set $K$ lying in a 5-dimensional affine subspace (e.g. in the subspace of matrices with trace 1). In this affine subspace, there is a 4-dimensional subspace $L$ which does not intersect $K$ (for example, the set of matrices with trace 1 and the upper-left entry equal to 2). Consider the distance from the set $C\cup(-C)$ as a function on $L$. It is positive and have a linear growth rate at infinity, and these properties are preserved under small perturbations of $L$.<|endoftext|> -TITLE: SQ-universality in the class of amenable groups -QUESTION [11 upvotes]: This question arises from HNN Embedding Theorem for Amenable Groups? -Recall that a group $G$ is called SQ-universal if every countable group is isomorphic to a subgroup of a quotient of $G$. The first non-trivial example of an SQ-universal group was provided by Higman, Neumann and Neumann in 1949. They proved that the free group of rank $2$ is SQ-universal, which is equivalent to the statement that every countable group embeds into a $2$-generated one. Presently many other examples of $SQ$-universal groups are known (e.g., hyperbolic and relatively hyperbolic groups). -It is straightforward to see that any SQ-universal group contains a non-abelian free subgroup and hence is non-amenable. However the following problem seems open. -Problem 1. Does there exist a finitely generated amenable group $A$ such that every countable amenable group embeds into a quotient of $A$? -I believe, the answer is "no". One way to disprove it would be to use the Folner functions, defined by Vershik in 70's. Recall that for a finitely generated amenable group $A$, $Fol_A\colon \mathbb N\to \mathbb N$ is defined by $Fol_A(n)$ = the size of a smallest finite subset $S \subseteq A$ satisfying $|\partial S|/|S|\le 1/n$. The asymptotic growth of $Fol_A(n)$ is independent of the choice of a finite generating set of $A$ up to a natural equivalence. -It is not hard to show that, when we pass to subgroups and quotient groups, this function does not decrease in the sense of the natural relation -$$ -f\preceq g \; {\rm iff}\; \exists\, C>0\; {\rm such\; that}\; f(n) \le Cg(Cn)\; \forall\, n. -$$ -Thus to answer Problem 1 negatively it would be sufficient to prove the following. -Conjecture 2. For any function $f\colon \mathbb N\to \mathbb N$, there exists a finitely generated amenable group $A$ such that $f\preceq Fol_A$. -Erschler [On isoperimetric profiles of finitely generated groups, Geom. Dedicata 100 (2003), 157–171] showed the existence of amenable groups with $Fol$ growing faster than any iterated exponential function. She also announced the proof of Conjecture 2 there, but I did not find it in her later papers. -Final remark: Problem 1 also makes sense if we replace "finitely generated" with "countable". - -REPLY [10 votes]: Here's what you were looking for: -MR2254627 (2007k:20086) -Erschler, Anna(F-PARIS11-M) -Piecewise automatic groups. (English summary) -Duke Math. J. 134 (2006), no. 3, 591–613. -20F65 (20F69 43A07 57M07) -"The main result of the paper under review is stated as follows: For any function f:N→N there exists a finitely generated group of an intermediate group (and thus amenable) whose Følner function satisfies FølG,S(n)≥f(n) for all sufficiently large n."<|endoftext|> -TITLE: Real analytic function, injective, non surjective and preserving the rationals ? -QUESTION [9 upvotes]: I'd like to prove the non-existence of a real analytic function, injective, non-surjective -that sends rationals to rationals. -Is it a classical result ? If not, any hints on how to prove it ? -Thanks in advance for you help. - -REPLY [14 votes]: The statement in the question is not true. Given any two enumerable and dense sets in open intervals of the reals, there is a (complex) analytic1 function giving a bijection between them. See the following paper: Analytic Transformations of Everywhere Dense Point Sets, Philip Franklin, Transactions of the American Mathematical Society, Vol. 27, No. 1 (Jan., 1925), pp. 91-100. -The analytic functions can be constructed along similar lines to the method outlined in this MO answer. Is a real power series that maps rationals to rationals defined by a rational function? Also see this mathforum.org discussion on the subject with plenty of links. A question on real-analytic functions -1 The analytic function can be chosen to be entire on the complex plane except for the obvious case where either of the ends of the real interval in the domain is bounded but the corresponding end in the range is unbounded, where the function must have a singularity.<|endoftext|> -TITLE: A K3 over $P^1$ with six singular $A_1$- fibers? -QUESTION [11 upvotes]: Hirzebruch, in the paper 'Arrangements of Lines and Algebraic Surfaces' -constructs a special $K3$ surface out of a 'complete quadrilateral' in - $CP^2$. A complete quadritlateral consists of - 4 points in general position and the $6$ lines joining them. -Over each line Hirzebruch forms the local 2:1 fold cover to get -a new surface which comes as a branched -covering of $CP^2$, branched over the $6$ lines. -Away from the lines the covering has degree $2^{6-1}$. -This surface has singularities of conical type at the original $4$ points -(At these points 3 lines are coincident. ) -Blow up the singularities coming from these 4 points.. The resulting smooth surface is Hirzebruch's $K3$. -Viewed from a different perspective, I believe that I can get this same -$K3$ has an elliptic surface over $CP^1$ with $6$ singular fibers. -I also believe that each of the singular fibers are of $A_1$ type, meaning two $CP^1$'s intersecting transversally -(as in $xy = 0$), but am less sure of this. -The corresponding singular points on $CP^1$ can be taken to be the vertices of the octahedron. -And I believe that the manifest symmetry group of order $4! = 24$ seen in Hirzebruch's construction (permute the -original 4 points) agrees with the symmetry group of the octahedron. -Questions. Do you know this second K3? If so, could you give me a reference for it? -Have you seen a place which shows that the second $K3$ is the same as Hirzebruch's? -More generally, what are the first few 'simplest' elliptic $K3$'s? -By 'elliptic' I mean expressed as elliptic surface $f: X \to CP^1$, over -$CP^1$. By 'simplest' I mean a small number of singular fibers whose singularities -are as 'simple' as possible. For example, if all singular fibers are of $A_1$-type, -what is the fewest number of fibers? Must this number be $6$? -(I have looked in Barth-Hulek-Peters-van de Ven's 'Compact Complex Surfaces', - esp. ch. V, sec. 2 and suppose this information is buried there somehow -or other, but is rather beyond me to untangle it from there. Neither did I -find this 2nd $K3$ in Gompf and Stipsicz's book ) - -REPLY [3 votes]: As explained by Noam Elkies, the fibers you want are of Kodaira type: - -$I_2$ : two rational curves intersecting at two distinct points -$III$ : two rational curves meeting at a double point - -Since these are the only fibers you want, I will recommand not using a Weierstrass equation but rather a Jacobi quartic form for your elliptic fibration. -Indeed, a smooth Weierstrass model admits only singular fibers of type $I_1$ (nodal curves) and type $II$ (cuspidial curve). So if you want to have singular fibers of type $I_2$ and $III$, you will have to introduce singularities in the total space of your fibration and resolve them following Tate's algorithm. -You can avoid dealing with the resolution of singularities if you start with the Jacobi quartic form. Consider a weighted projective plane $\mathbb{P}^2_{1,2,1}$ bundle, a smooth quartic equation will define an elliptic curve as you can see by using the adjunction formula. The canonical form of such a curve is -$$ -\text{Jacobi form}: \quad y^2=x^4+ e x^2 z^2+ f x z^3+ g z^4 -$$ -where $[x:y:z]$ are the projective coordinates of weight $1$, $2$ and $1$ respectively. -For a fibration $Y\rightarrow B$, the weighted projective plane is replaced by a weighted projective bundle $\mathbb{P}_{1,2,1}[\mathscr{O}\oplus\mathscr{L}\oplus\mathscr{L}^2]$ with weight $1,2,1$, where $\mathscr{L}\ $ is a line bundle over the base $B$ of your fibration. In that equation $e,f,g$ are now sections of $\mathscr{L}^2$, $\mathscr{L}^3$ and $\mathscr{L}^4$ respectively. The projective coordinates $x$, $z$, $y$ are respectively sections of $\mathscr{L}\otimes \mathscr{O}(1)$, $\mathscr{O}(1)$ and $\mathscr{L}^2\otimes\mathscr{O}(2)$, where $\mathscr{O}(1)$ is the tautological line bundle of the weighted projective bundle. - -Using the adjuction, you can see that you have a $K3$ elliptic fibration iff $c_1(\mathscr{L})=c_1(B)$. - -In case you actually want a Weierstrass model, the Jacobian fibration will give you a birationally equivalent Weierstrass model: $y^2 =x^3+ Fx + G$ with $F=-4 g +e^2/3$ and $G=f^2-8/3 eg +2e^3/27$. You can compute the birational transformation using Maple. -Going back to the Jacobi form, it is not difficult to see that a smooth elliptic fibration will have fibers of type $I_1$, $II$, $I_2$ and $III$. -In particular, the fibers are: - -type $I_1$ at a general point of $4 - F^3+27 G^2=0$ -type $II$ at $F=G=0$ -and $e\neq 0$ -type $I_2$ iff $f= - e^2-4 g=0$ and $g\neq 0$ -type $III$ iff $f=e=g=0$. -With a $\mathbb{P}^1$ base and -$\mathscr{L}=O(2H)$, you have a $K3$ -surface with fibers $I_1,II, I_2, - III$. - -You can avoid the fiber $I_1$ and $II$ by manipulating the equation. For example if you take the equation with a $\mathbb{P}^1$ base and $\mathscr{L}=O(2H)$, you can write the following K3 elliptic surface - -$$y^2=x^4+ g z^4\Longrightarrow \ \text{8 fibers of type}\ III $$ - -You need $g$ to be a smooth polynomial of degree 8 in $P^1$. For example $g=x_1^8+x_0^8$ with $[x_0:x_1]$ the projective coordinates of $\mathbb{P}^1$. You will have $8$ fibers of type $III$ at the zero of $g$.Note that with that choice, the singular fibers are at the vertices of an octagon on $\mathbb{P}^1$ defined by $x_1^8+x_0^8=0$. -These types of fibrations based on the Jacobi quartic form are known as $E_7$ elliptic fibration in the string theory literature. They have some nice topological properties and transitions. You can read more about them in this paper.<|endoftext|> -TITLE: Genus of smooth varieties with small Chow group -QUESTION [11 upvotes]: Let $X$ be a smooth projective variety over $\mathbb C$ with $d = \dim X \geq 1$. Let $CH(X)$ denotes the total Chow group of (cycles modulo rational equivalences of) $X$ and $CH(X)_{\mathbb Q} = CH(X)\otimes_{\mathbb Z} \mathbb Q$. My question is - -Suppose $CH(X)_{\mathbb Q}$ is finite-dimensional as a $\mathbb Q$-vector space. When can we conclude that $H^d(X, \mathcal O_X) =0$? - -I suspect that the answer might be always yes: in dimension one, such varieties have to be rational. In dimension two, I think it follows from a paper of Mumford. Also, it is true for flag manifolds. -More vaguely, I am also quite interested in: - -Suppose $CH(X)_{\mathbb Q}$ is finite-dimensional as a $\mathbb Q$-vector space. What can we say about the geometry of $X$? - -Any partial answers would be appreciated. Thanks! - -REPLY [7 votes]: For a smooth proper complex variety $X$, the condition that $CH^*(X)_{\mathbb Q}$ is a finite dimensional vector space is equivalent to the Chow motive of $X$ being a direct sum of Tate motives. In particular, the singular cohomology of $X$ (with $\mathbb Q$ coefficients) coincides with $CH^*(X)_{\mathbb Q}$ (and thus the Hodge decomposition is rather trivial). This can be proved using the technique of decomposition of the diagonal. -A version of this argument (for an arbitrary Chow motive) appears in an article of Kimura: -MR2562457 (2010j:14014), Surjectivity of the cycle map for Chow motives. in ``Motives and algebraic cycles'', 157–165, Fields Inst. Commun., 56, Amer. Math. Soc., Providence, RI, 2009. -Actually, one can even make do without the ``supported correspondences'', by reasoning as in Kapil Paranjape's article -MR1283872 (95g:14008), Cohomological and cycle-theoretic connectivity. Ann. of Math. (2) 139 (1994), no. 3, 641–660. -But they do give a clean way to write the proof. -There is a follow-up by Charles Vial (again, there is no real need to use ``birational motives'' in the proofs, for this result): -MR2738925 (2012c:14010), Pure motives with representable Chow groups, C. R. Math. Acad. Sci. Paris 348 (2010), no. 21-22, 1191–1195.<|endoftext|> -TITLE: Can you hide a letter without losing information? -QUESTION [7 upvotes]: Consider the following game between Alice and Bob. -$\Sigma$ is a finite nonempty alphabet, $\Delta \notin \Sigma$ denotes -a special symbol, and $k > 0$ is a positive integer constant representing -the length of contexts. - -Alice gets a message $w \in \Sigma^{\omega}$. -She then finds an appropriate index $i > k$ and passes the following message: -$w[i - k \ldots i - 1]\ \Delta\ w[i + 1 \ldots i + k]$ to Bob. -Bob receives the message $l\ \Delta\ r$, where $l, r \in \Sigma^k$. -Based on the context $(l, r)$ he then recovers the original hidden letter $w[i]$. - - -The question is, whether there exists any such protocol enabling - the above game? Is it possible for Alice to always find an appropriate - index $i$, such that Bob is then able to recover the hidden letter? - -Background/Motivation -If such protocol existed it would enable to encode any information into -a sufficiently long word only by means of replacing appropriate letters -by the $\Delta$ symbol. It would be also possible to recover the original -word only by looking at the limited context surrounding these $\Delta$ symbols. -Similar, but a little more involved coding was used in our paper: -P. Cerno, F. Mraz: Delta-Clearing Restarting Automata and CFL, -Proceedings of the DLT 2011 15th International Conference on Developments in Language Theory (Milano, Italy), Springer, Berlin, 2011, LNCS, Vol. 6795, 153-164. -Partial Solution -I was able to find a correct protocol only for a two-letter alphabet. -Let $\Sigma = \{a, b\}$. Then there exist integers $B, k > 0$, -and a table $T$ of triples $(x, z, y)$, $x, y \in \Sigma^k$, -$z \in \Sigma$, such that: - -For each context $(x, y) \in \Sigma^k \times \Sigma^k$ there exists -at most one $(x, z, y) \in T$. -For each $w \in \Sigma^B$ there exists at least one -$(x, z, y) \in T$, such that $xzy$ is a subword of $w$. - -Proof. -Let us set $B = 8$, $k = 2$, and $T$ to be the following set: -{(aa, a, aa), (ab, a, aa), (ba, b, aa), (bb, a, aa), - (aa, a, ab), (ab, a, ab), (ba, b, ab), (bb, a, ab), - (aa, b, ba), (ab, a, ba), (ba, b, ba), (bb, b, ba), - (aa, b, bb), (ab, a, bb), (ba, b, bb), (bb, b, bb)} - -Note that $T$ can be shortly described as -$T = \{ (xx, y, y?) \} \cup \{ (xy, x, ??) \mid x \neq y \}$, where -$x, y \in \{a, b\}$, and the symbol $?$ represents an arbitrary letter. -The intuition why this works is following: -Consider a sufficiently long word $w \in \{a, b\}^*$. -Then there are only two cases: either each (internal) letter in $w$ is -doubled, i.e. each (internal) letter $x$ in $w$ has a neighbor $x$, -or there exists an (internal) letter $x$ in $w$ which does not have -a neighbor $x$. In the first case the pattern $xxyy$ will occur, -and in the second case the pattern $xyx$, where $x \neq y$, will occur. - -The question is, whether there are similar protocols also for larger alphabets? - -REPLY [5 votes]: We say that Alice catches the word if she can make the desired move. We prove that a protocol exists by the induction on $d=|\Sigma|$. -Your example states the base for $d=2$. Assume that we know Alice's strategy for $d-1$ letters; let $k'$ be the length of the words in the catching triples. Note that in fact Alice catches all the finite words of some length $N'$; otherwise by Koenig's lemma there exists an infinite word which is not caught. -We may assume that $k'\geq 2d$. Now fill an alphabet with $d$th letter $x$. We claim that we can set $k=N'$. -We will use the "catching words" of length $\leq k$ instead of exactly $k$; this means that we extend all the left words by arbitrary letters from the left, and all right words --- to the right to obtain the words of desired length. -0) First, take all Alice's words for the alphabet $\Sigma\setminus\{x\}$. Then she will catch any word which contains a subword of length $N'$ not containing $x$ (this subword should stand far enough from the beginning of the word). -We say that a word is valid if it does not contain $x$. -Now we need to catch all other words; they consist of subwords of the form $xWx$, where $0\leq |W|\leq N'$, and $W$ is a valid word (these subwords overlap by letters $x$). We will consider only the words $W$ standing far enough from the beginning. -We distinguish three cases: 1) maximal length of $W$ is at least $2d$; 2) this maximal length is in $[d,2d-1)$, and 3) this maximal length is less than $d$. -To catch 1), assign to each $a\in \Sigma$ its value $\nu(a)$ from 1 to $d$ bijectively; let $\nu(W)$ be the sum of values of letters of $W$ (with repetitions) modulo $d$. -Now we use the catching triples of the form $(xU,a,Vx)$ where $U,V$ are valid, $0\leq |U|\leq d-1$, $d\leq |V|\leq N'$, $2d-1\leq |U|+|V|\leq N'-1$, and $\nu(UaV)=|U|$. Now, if we have the subword $xWx$ with $|W|\geq 2d$, then we set $U$ to be the beginning of $W$ of length $\nu(W)$, $a$ to be the next letter of $W$, and $V$ to be the remaining tail. Our word is caught since $a$ is determined uniquely by $U$ and $V$. -To catch 2), we choose the pairs of the form $(xU,x,?)$ where $d\leq |U|\leq 2d-1$ and $U$ is a valid word. -Finally, to catch 3), we take all the triples of the form $(xU,x,Vx)$ where $0\leq |U|,|V|\leq d-1$, and $U,V$ are valid words. -It is easy to note that our triples do not contradict to each other. To see this, in each triple consider the tails $L_d$ and $L_{2d}$ of the left word and the beginnings $R_d$ and $R_{2d}$ of the right word, of lengths $d$ and $2d$ respectively; recall that $k'\geq 2d$. In 0), $L_{2d}$ does not contain $x$. In 2), $L_{2d}$ contains $x$, while $L_d$ does not. In 1), $L_d$ contains $x$ while $R_d$ does not. Finally, in 3) both $L_d$ and $R_d$ contain $x$. -So, the algorithm for Bob is as following. By the rules above, he determines which type of triple he has received. In 0), he finds the letter by the algorithm for $d-1$ letters; in 2) and 3), he simply answers $x$. In 1), he determines $U$ and $V$ (as the longest tail and beginning not containing $x$) and then finds $a$ from the condition $\nu(UaV)=U$.<|endoftext|> -TITLE: Is $O(10^{-6})$ an acceptable notation in numerical analysis? -QUESTION [6 upvotes]: The following question has been on math.SE for several days. Without having a satisfying answer, I'd like to ask the experts here. -In mathematics, the big $O$ notation is used to describe the limiting behavior of a function. It is abuse of notation to say -$$ -f(x)=O(g(x)). -$$ -But this is understandable. However, in the class of numerical analysis, I found that the teacher used the big $O$ notation as the following: - -If $\kappa = O(10^{-6})$, $\epsilon_{machine} = O(10^{-16})$, then we can only expect $O(10^{-10})$ accuracy. - -I am surprised that they regard $O(10^{-6})$, $O(10^{-16})$, $O(10^{-10})$ as different things. Since according to the definition, they are nothing but $O(1)$. -I guess this is another kind of abuse of notions: when one says $O(10^{-6})$, s/he actually means $c\times 10^{-6}$ where $1 -TITLE: Elastostatics and homotopy type -QUESTION [5 upvotes]: In perfect elastostatics, the unknown is the displacement $x\mapsto y$, where $x\in\Omega\subset{\mathbb R}^3$ is the reference configuration, and $y\in{\mathbb R}^3$. It obeys to an 2nd-order PDEs. When we rewrite the PDE as a 1st-order system, the unknown becomes the deformation gradient $F(x):=\nabla y$. -A fundamental constraint is that the matter cannot interpenetrate. In particular, one must have $\det F>0$ everywhere. This is highly non-trivial, in the sense that the set $P$ of $3\times3$ real matrices with positive determinant has a non-trivial homotopy type. Thanks to the polar factorization, $P$ is homeomorphic to the product of ${\bf SPD}_3$ (a convex cone, something trivial) with ${\bf SO}_3$ whose fundamental group is ${\mathbb Z}_2$. -This raises the following questions. Given a domain $\Omega\subset{\mathbb R}^3$, how many connected components are there in $C(\partial\Omega;P)$ ? How can we determine if an $f\in C(\partial\Omega;P)$ can be lifted as an $F\in C(\Omega;P)$ ? How can we determine if an $f\in C(\partial\Omega;P)$ can be lifted as an $F\in C(\Omega;P)$ such that $F$ is homotopic to the identity ? - -REPLY [5 votes]: Complementing Dmitri's answer: one can see $SO(3)\cong \mathbb{P^3}(\mathbb{R})$ as the 3-skeleton of $K(\mathbb{Z}/2,1)\cong \mathbb{P^\infty}(\mathbb{R})$. Assuming $\Omega$ and its boundary are non-pathological, they will have the homotopy type of 2-dimensional $CW$-complexes. If $X$ is a 2-dimensional $CW$-complex, one can deform any mapping $X\to K(\mathbb{Z}/2,1)$ into a mapping to the 2-skeleton and any homotopy between such mappings into a mapping to the 3-skeleton. So $H^1(X,\mathbb{Z}/2)\cong [X,K(\mathbb{Z}/2,1)]\cong [X,SO(3)]$.<|endoftext|> -TITLE: Principal congruence subgroups in higher rank -QUESTION [8 upvotes]: I don't seem to have seen any explicit generators for the principal congruence subgroups of $SL(n, \mathbb{Z}),$ for $n>2,$ although it is known (Sury+Venkataramana) is that the number of generators is bounded independently of the modulus. It is certainly possible to write a program to generate the generators (since the subgroup is given as a kernel of a map to a finite group, if we fix a generating set, the kernel is generated by the loops in the induced Cayley graph of the image group), but this seems painful (and will probably produce really awful generating sets without much further work). -For the self-proclaimed obtuse The question is: is there a known way to produce, given $n, p,$ a generating set of the level $p$ principal congruence subgroup in $SL(n, \mathbb{Z}).$ Particularly a nice, small generating set. -EDIT The Sury and Venkataramana paper is actually available for free. and having overcome my laziness somewhat, I see that they do actually construct a generating set using an idea somewhat related to @Andy's, but the generating set is quite large. Since $SL(n, \mathbb{Z})$ itself can be generated by three elements, one is tempted to make a rash conjecture that the minimal size of a generating set of $\Gamma_n(p)$ (with the obvious notation) is bounded independently of $n$ and $p.$ BUT (EDIT), as pointed out by @Andy Putman in his answer and his comment, this would be rash indeed, since the rank of the abelianization of $\Gamma_n(p)$ is $n^2-1.$ - -REPLY [13 votes]: I'm pretty sure that no such explicit generating set exists in the literature. However, at some point I spent some time thinking about this and I have a conjectural generating set. -Normal generators. Let $\Gamma_n$ denote $SL_n(\mathbb{Z})$ and let $\Gamma_n(p)$ denote the level $p$ congruence subgroup of $\Gamma_n$. For $1 \leq i,j \leq n$ such that $i \neq j$, let $e_{ij}$ be the $(i,j)$-elementary matrix. Define $S = \{e_{ij}\ |\ 1 \leq i,j \leq n, i \neq j\}$ and $S(p) = \{e_{ij}^p\ |\ 1 \leq i,j \leq n, i \neq j\}$. Clearly we have $S(p) \subset \Gamma_n(p)$. Moreover, the proof of the congruence subgroup theorem (by Mennicke and Serre) shows that $\Gamma_n(p)$ is the normal closure in $\Gamma_n$ of the subgroup generated by $S(p)$. -Abelianization of congruence subgroup. -However, $S(p)$ does not generate $\Gamma_n(p)$. This follows from the computation of the abelianization of $\Gamma_n(p)$ by Lee and Szczarba. Briefly, they defined a homomorphism $$\phi : \Gamma_n(p) \rightarrow \mathfrak{sl}_n(\mathbb{Z}/p).$$ -Here $\mathfrak{sl}_n(\mathbb{Z}/p)$ is the abelian group of $n \times n$ matrices of trace $0$ with entries in $\mathbb{Z}/p$. The definition of $\phi$ is as follows. An element of $\Gamma_n(p)$ is of the form $1 + p A$ for some matrix $A$. Define $\phi(1+pA) = A$ modulo $p$. It is easy to see that the trace of $A$ mod $p$ is $0$. Moreover, this map is a homomorphism due to the identity -$$(1+pA)(1+pB) = 1+p(A+B)+p^2 AB.$$ -Lee and Szczarba proved two things about $\phi$. - -They proved that $\phi$ is surjective. I'll say more about this below. -They proved that the kernel of $\phi$ (which, by the way, is clearly $\Gamma_n(p^2)$) is the whole commutator subgroup of $\Gamma_n(p)$ for $n \geq 3$. - -This implies that the abelianization of $\Gamma_n(p)$ is $\mathfrak{sl}_n(\mathbb{Z}/p)$. -They only normally generate. -Now, above I claimed that $S(p)$ does not generate $\Gamma_n(p)$. If $G$ is the subgroup generated by $S(p)$, it is easy to see that $\phi(G)$ is exactly the group of matrices whose diagonals are $0$. The problem thus is that you don't get the diagonal matrices. -Hitting the diagonal. -We thus need generators that hit the diagonal matrices. For $1 \leq i < n$, let $f_i$ be the result of inserting the $2 \times 2$ matrix $\left( \begin{array}{ll} 1+p & -p \\ p & 1-p\\ \end{array} \right)$ into the $n \times n$ identity matrix with the upper left at position $(i,i)$. Define $T = \{f_i\ |\ 1 \leq i < n\}$. -I conjecture that $S(p) \cup T$ generates $\Gamma_n(p)$. Here is my evidence for this. - -Let $H$ be the subgroup generated by $S(p) \cup T$. Then $\phi(H)$ is all of $\mathfrak{sl}_n(\mathbb{Z}/p)$. -I can prove this by hand for $p=2$ and $p=3$. - -Conjectural proof sketch. -My proof for $p=2$ and $p=3$ should work in general, so let me briefly describe it. By the congruence subgroup property business discussed above, the normal closure in $\Gamma_n$ of $H$ is $\Gamma_n(p)$. To prove that $H = \Gamma_n(p)$, therefore, it is enough to prove that $H$ is normal. To do this, it is enough prove the following. Let $e_{ij} \in S$ be an elementary matrix (a generator for $\Gamma_n$). Then for $s \in S(p) \cup T$, it is enough to prove that $e_{ij} s e_{ij}^{-1}$ and $e_{ij}^{-1} s e_{ij}$ can be written as words in $S(p) \cup T$. For $p=2$ and $p=3$, this is fairly easy, but I haven't managed to do the computations for higher $p$.<|endoftext|> -TITLE: Is there a connected non-affine scheme $S$ such that it is the union of rings of integers of number fields? -QUESTION [9 upvotes]: I was woolgathering about the notion of a scheme, and it occurred to me that I know of no non-affine scheme $S$ that is the union of $Spec(O_K)$'s of some number field $K$ (I allow $K$ to vary - so that $S$ might be $Spec(O_K)\cup Spec(O_L)$ for example). -It would an interesting notion if one could patch rings of integers together to form some non-affine $1$-dimensional normal scheme $S$. The fact that I've never seen an example makes me think it's impossible. -Question -Is there a connected non-affine scheme $S$ such that it is the union of open subschemes of it that are $Spec$'s of rings of integers of number fields? -More pointedly, if $Spec(O_K)$ (the ring of integers of some number field $K$) is an open subscheme of a normal scheme $S$ then is it equal to it? - -REPLY [28 votes]: If $i: \mathrm{Spec}(O_K)\to S$ is an open immersion into a connected separate scheme $S$, then $i$ is an isomorphism. Indeed, the canonical morphism $\pi : \mathrm{Spec}(O_K)\to \mathrm{Spec}(\mathbb Z)$ is finite (hence proper) and can be decomposed into $i$ followed by the canonical morphism $S\to \mathrm{Spec}(\mathbb Z)$. As the latter is separated, this implies that $i$ is also proper, hence closed. The connectedness of $S$ implies that $i$ is onto.<|endoftext|> -TITLE: Relationship between free probability and deterministic graphs? -QUESTION [9 upvotes]: Consider the $N\times N$ matrix $$ -M = \left(\begin{array} \\ -0 & 1 & & 0 \\ -1 & \ddots & \ddots & \\ - & \ddots & \ddots & 1 \\ -0 & & 1 & 0 \\ -\end{array}\right) -$$ -which comes from the adjacency matrix of a graph corresponding to a one-dimensional chain of $N$ nodes with dangling ends. A cartoon of this graph is $$\circ -\circ -\circ -\circ -\cdots-\circ -\circ$$ -It turns out that if you plot a histogram of its eigenvalues, it appears to fit exactly with an arcsine distribution $$f(x) = \frac{1} {\pi \sqrt{4-x^2}}, \vert x \vert < 2 $$ which is exactly what one would expect from the free convolution of the binomial distribution $$ p(x) = \frac 1 2 \left( \delta\left(x-1\right) + \delta \left(x+1\right)\right)$$ with itself. - -Is this mere coincidence, or evidence of something deeper? I feel like this must be some example of a known result out there. - -I've gotten as far as figuring out how $\pm 1$ shows up; you can write $M$ as the sum of two pieces -$$ M = A + B $$ -$$ A = \left(\begin{array}{cccccc} -0 & 1\\ -1 & 0\\ - & & 0 & 1\\ - & & 1 & 0\\ - & & & & \ddots\\ - & & & & & \ddots -\end{array}\right) = \sigma_x \oplus \sigma_x \oplus \cdots $$ -$$ B = \left(\begin{array}{cccccc} -0\\ - & 0 & 1\\ - & 1 & 0\\ - & & & 0 & 1\\ - & & & 1 & 0\\ - & & & & & \ddots -\end{array}\right) = [0] \oplus \sigma_x \oplus \sigma_x \oplus \cdots $$ -where $\sigma_x$ is the Pauli sigma matrix which of course has eigenvalues $\pm 1$. It must be that these two matrices are freely independent in the $N\rightarrow \infty$ limit, and possibly even for finite $N$ also, so that this reduces to the free convolution described above. -I may be reading too much into this, but it's interesting to me that this is a completely deterministic matrix problem with free probabilistic characteristics. I'm not at all familiar with the algebraic aspects of free probability theory, let alone what the graph theoretic relationships would be. - -REPLY [3 votes]: The relation between free probability and graphs you are looking for comes from the free product of (rooted) graphs, see for example the paper of Accardi, Lenczewski and Salapata https://arxiv.org/abs/math/0609329, where a decompostion of this prodcut as a sum of free operators is given in terms of the individual adjacency operators of the factors. -Now, there are actually two small differences between the free product of K2 (corresponding to $1/2\delta_{-1}+1/2\delta_1$) with itself and the graphs you are considering: First, the free product is actually infinite: $$⋯∘-∘-∘-∘-∘-∘-⋯$$ and second, one needs to consider the vector state $$, where $e$ is root of the graph, instead of the normalized trace $tr$. However, both differences "vanish" in the limit since almost all the vertexes look locally as belonging to the actual free product.<|endoftext|> -TITLE: Orthogonal group of quadratic form -QUESTION [12 upvotes]: Orthogonal group of the quadratic form over fields, somehow, is well-studied. Indeed -E. Cartan has proved for quadratic forms over the reals or complexes that any -orthogonal transformation is a product of at most $n$ symmetries, where $n$ is -the dimensionality of the underlying vector space. This result was generalized -by Dieudonne to quadratic forms over arbitrary base fields. -One can try to understand the integral orthogonal group of an integral quadratic form, more precisely: -Question: Let $q(x_1,\dots,x_n)$ be an integral quadratic form, can we say the integral orthogonal group, denoted as usual by $O_\mathbb{Z}(q)$, is finitely generated. If this would be the case, what can we say about the number of generator? -Let me pick the following special example: -$$ -q(x,y,z)=x^2+y^2-z^2 -$$ -One can show $O_\mathbb{Z}(q)$ acts transitively on -$$ -\{(x,y,z)\in \mathbb{Z}^3: q(x,y,z)=0\} -$$ -So understanding the number of generators of $O_\mathbb{Z}(q)$, could gives us the space of integral solutions of $q(x,y,z)=0$. For instance, Keith Conrad has a wonderful note entitled with "Orthogonal group of $x^2 + y^2 - z^2$", who proved $O_\mathbb{Z}(q)$ is generated by five elements (I think it was known before but Conrad exposition is great). -This example shows, that the above question can be interesting. What do we know bout the about question? - -REPLY [14 votes]: Yes, these groups are arithmetic lattices, and are therefore finitely generated. -I believe Selberg showed that they are cofinite volume (with respect to the -discrete action on the appropriate symmetric space). -When the form is definite, it is a finite group. When the form is Lorentzian, -the group may be shown to have a nice Ford domain with respect to its action -on hyperbolic space, which shows that it is -finitely generated. In principle, using the volume computation, one could -give an upper bound on the number of generators in this case. For small -examples, I think that the number of generators grows linearly with the -dimension. Also, these groups are not generated by reflections for high enough dimensions -by a result of Nikulin (even up to finite-index). -When the form has rank $>1$, then the group has property -(T), and therefore is finitely generated by a result of Kazhdan. The original -proof though appears to be due to Borel-Harish Chandra. I think this -may also be proved using the Borel-Serre compactification. For information -about arithmetic groups, check out the book (in progress) by Dave Witte-Morris. -In this case, the groups are generated by reflections up to finite index. -Venkataramana has some results on the number of generators of such -lattices and finite-index subgroups.<|endoftext|> -TITLE: What is the best homology/cohomology theory for the Lefschetz fixed point theorem? -QUESTION [20 upvotes]: Short version: One can define a version of the Lefschetz fixed point theorem using any homology or cohomology theory. All versions will be true on some topological spaces, since they agree on some topological spaces, but some might be true more generally than others. If two versions have the same generality, one might be more exact, and make stronger statements. Under this criterion of goodness, which one is best? -Potential answer: Is the the theorem true for Cech cohomology on all T3 compact spaces? -Intuition: Suppose there is a continuous map from a topological space X to itself. Does this have a fixed point? -This is an extraordinarily elementary question. It makes sense to ask this question about any topological space. Yet the tools that are typically used to answer it are usually very specific, referring to special topological spaces like $\mathbb R^n$. -The Lefschetz fixed point theorem is a powerful result in this area. It is usually defined in terms of singular homology, which is not an elementary construction at all. But one could easily use another homology or cohomology theory to calculate the matrix chases and, thereby, the number of fixed points. -In particular Cech cohomology seems extraordinarily elementary, being defined entirely in terms of open sets and their intersections. Therefore, one would expect Cech cohomology to be the best one. -Consider, as an example, the topologist's sine curve. All maps from it to itself have a fixed point. But under singular homology it has characteristic 2, not 1. Cech gives characteristic 1 and is thus more exact. -Edit: I'm more interested in the weak than the strong version of the theorem, because the weak version is presumably more generalizable. I'm interested in any sort of information that compares the theorem in different theories even if it doesn't fall into a strict/better worse dichotomy. - -REPLY [20 votes]: To answer your specific question about compact T3 spaces: First of all, every compact Hausdorff space (T2 space) is automatically a T4 space (a Hausdorff normal space). In the literature one usually says "compact Hausdorff space". A continuum is a compact, connected, metrizable Hausdorff space. There is a famous example of Kinoshita, described in another MO question, of a contractible continuum that has a fixed-point-free self-map. It even embeds in $\mathbb{R}^3$. So, no Lefschetz fixed point theorem in this generality. As that MO question also explains, Borsuk earlier found Cech-acyclic examples. -One possible answer for the natural generality of the Lefschetz fixed point theorem is the version that Lefschetz himself proved: For compact spaces that are ANRs (absolute neighborhood retracts). ANRs are nicely approximable by simplicial complexes in a certain sense, and actually for compact ANRs Cech and singular cohomology always agree. If you accept Lefschetz's level of generality, then the quest to generalize his fixed point theorem beyond singular cohomology looks a bit self-defeating. Still, there is something in the literature of fixed-point theory called a "Lefschetz space" which is simply directly assumed to satisfy some version of his theorem. -I also wouldn't call Čech cohomology "extraordinarily simple". Like many other definitions of (co)homology, it really comes down to constructing a simplicial complex or a simplicial set, or in this case an inverse system of them, and then using simplicial (co)homology. That's not really all that simple. Also using open covers to build simplicial complexes is a brilliant idea, but not a perfect idea. Actually the message of the examples of Kinoshita and Borsuk and others (including one by my mother, that there can't be a Čech Hurewicz theorem) is that there can't be one perfect cohomology theory in topology, not even for compact, connected Hausdorff spaces. Maybe Čech cohomology is still arguably the best, but it isn't even dual (in the sense of universal coefficients) to any reasonable homology theory. -One of the beautiful aspects of de Rham cohomology, even though it is defined only for smooth manifolds and real coefficients (or generalizations of real coefficients), is that it truly doesn't use simplicial complexes or simplicial sets, only calculus. In fact only that partial derivatives commute. Generality is not the virtue that trumps all others.<|endoftext|> -TITLE: Lifting to Characteristic 0 not over W -QUESTION [15 upvotes]: I thought of this several months ago and forgot about it. Now I rethought of it again and I just can't find it anywhere in the literature, so I'll ask here. -Is it known whether or not there exists a (smooth, proper, ...) variety over a field $k$ (perfect? alg. closed?) of positive characteristic that lifts to characteristic $0$ over some ramified extension of $W(k)$ and also lifts to $W_2(k)$, but does not lift over $W(k)$ itself? -In other words, if it lifts to characteristic $0$ (in a way related to $W(k)$) and it lifts to $W_2(k)$ must it lift via $W(k)$ itself? -I have looked at examples that lift to $W_2(k)$ but not to $W_3(k)$ and hence not over $W(k)$, but they don't seem to lift to char $0$ at all. I've also looked examples that lift over a ramified extension but not $W$, but these can easily be shown to not lift to $W_2$ as well. - -REPLY [10 votes]: Theorem 1.1, M1a of the following article of Ravi Vakil, -"Murphy's Law in algebraic geometry: Badly-behaved deformation spaces", Invent. Math. 164 (2006), 569--590, available at http://math.stanford.edu/~vakil/files/Mjul0705.pdf. -Apply this theorem to the "germ" of $\text{Spec}(\mathbb{Z}[x]/\langle p(x^2-p) \rangle)$ over $\text{Spec}(\mathbb{Z})$.<|endoftext|> -TITLE: A meta-mathematical question related to Hilbert tenth problem -QUESTION [6 upvotes]: I am reading Bjorn Poonen's very nice survey on Hilbert's Tenth problem -(http://www-math.mit.edu/~poonen/papers/uniform.pdf), and while I believe I understand the mathematics well, I have widespread difficulties with the meta-mathematics of these questions. To illustrate them, and to ask a question that is answerable and whose answer might be helpful for me, let me focus on one little passage of this paper, that concerns not directly Matthiasevich's theorem that the tenth problem has a negative solution -but an older, weaker version : -"[...] the work of K. Gödel, A. Church, and A. Turing in the 1930s made it clear that there was no algorithm for solving the [...] problem of deciding the truth of first-order sentences over $\mathbb{Z}$". (page 6) - -What does that assertion exactly mean? - -I understand well what an algorithm is and what a first-order sentence in arithmetic is. -The difficult word in the quoted sentence is "truth". -Here is my tentative interpretation. Define a "platonist" as someone -who believes that natural integers actually exist and that first order sentences about them are either absolutely true or absolutely false. I am such a person. So for a platonist, -the passage quoted above would mean: "there is no Turing machine that take a first-order sentence as input and produces the output TRUE or FALSE according to wether the sentence is absolutely true or absolutely false." Ok. The problem is that this interpretation makes -sense only for a platonist. I am not going to name names here, but I know very good mathematicians that are not platonists in the above sense. - -Is there another (weaker) interpretation of the quoted sentence, that would make sense for pretty all mathematicians? - -Or, is the statement from Poonen's paper simply rejected as non-sensical by those -non-platonist mathematicians ? -I have in mind Gödel's incompleteness theorem itself, that comes in two versions: one for everyone, that says that there is a first-order arithmetical sentence -that can not be proved nor disproved in, say, PA; and one stronger version -for platonists that says that there is a first-order arithmetical sentence, that cannot be proved in, say, PA, but that is nevertheless true. But for the theorem of Gödel-Church-Turing -quoted by Poonen I don't see what would be the version acceptable by everyone. -Edit: Many people seem to have great difficulties to understand my question. I am not sure I understand why. Let me try to explain it more from the "philosophical" point of view. -I think anyone would agree that it is not self-evident that a first-order statement -about numbers makes sense, and is either true or false independently of the system of axioms -we choose. (Surely, a statement about sets does not necessary make sense, like -"is the set of all sets an element of itself".) Actually, I do believe that any first-order statement about numbers makes sense, but for me this is like a religious belief, not something -I would feel authorized to use in a serious mathematical theorem. Basically, like most number theorists I suppose, I work with ZF with the enumerable axiom of choice, and I feel pain in the stomach when occasionally I need to use the full axiom of choice or Grothendieck's axiom of universes (and in general, we convince ourselves that they are just used "to simplify the exposition", and that they could be avoided at the cost of just a lost in elegance). -So when I see a statement like the one in boldface above, as a platonist I understand what it means, but I wonder if it is a reasonable statement that I can agree with with -non-platonist colleagues. - -REPLY [13 votes]: There's a general "trick" for handling all issues of this sort. Take any mathematical theorem that a platonist regards as meaningful. Formalize it as a formal theorem T in ZFC. The formalist will now accept the sentence, "ZFC proves T." -Here, the only potentially confusing concept is that of truth. But to say that some first-order sentence of arithmetic is true just means that it is satisfied by the structure $\mathbb N$. The satisfaction relation, like all ordinary mathematics, is readily defined set-theoretically, as you can see in any textbook on logic. So the nonexistence of the algorithm in question can be expressed as a first-order sentence of set theory, and the formalist will agree that this sentence is a theorem of ZFC. -For some kinds of finitistic statements, the formalist doesn't have to do this little dance of translating "true" into formal set-theoretic terms and replacing "T" with "ZFC proves T." For example, in the sentence, "It is true that ZFC proves T," the formalist can use his "native" understanding of the word "true" and doesn't have to convert "ZFC proves T" into an arithmetic statement S and use the set-theoretic definition of truth to get a set-theoretic assertion whose ZFC-theoremhood he can agree with. But the little dance is always available as an option. -EDIT: Reading various comments to the original question and to other answers, I see that something more may need to be said about the satisfaction relation, even though it is standard textbook material. To say that a first order sentence $\phi$ is true, or that it belongs to $\mathrm{Th}(\mathbb N)$, means that it is satisfied by $\mathbb N$, where satisfiability is defined inductively. For example, $\exists x: \phi(x)$ is satisfied by $\mathbb N$ if there exists $x\in \mathbb N$ such that $\phi(x)$ is satisfied by $\mathbb N$. Further details may be found here. -Now, you might complain that in order to "make sense" of the satisfiability relation, you have to "make sense" of $\mathbb N$. However, you don't have to believe in $\mathbb N$ as some kind of platonically existing thing in order to correctly manipulate sentences about $\mathbb N$. Any sufficiently powerful set-theoretic meta-theory will suffice to carry out the definition of $\mathbb N$ and the satisfaction relation. ZFC is the standard choice but you could use something else if you prefer. A way to assert the existence of $\mathbb N$ in the first-order language of set-theory is as follows: $$\exists x:(\emptyset \in x \wedge \forall y\in x: (y\cup\lbrace y\rbrace\in x))$$ -Here I've used various abbreviations, e.g., $\emptyset\in x$ expands formally to $\exists z : (z\in x \wedge \neg \exists w: (w\in z))$. Similar but more complicated formalizations can be produced for "set of first-order sentences of arithmetic" and "$\mathbb N$ satisfies $\phi$." As long as you know the axioms and rules of inference for ZFC, you can verify that the existence of $\mathrm{Th}(\mathbb N)$ is provable in ZFC. (Note: This is NOT the same as saying that every true sentence of arithmetic is provable in ZFC, which is absolutely false!) And once you have $\mathrm{Th}(\mathbb N)$, you can simply interpret "x is true" as $x\in \mathrm{Th}(\mathbb N)$. In particular, there is nothing mysterious about truth; it is just a mathematical concept formalizable in ZFC like any other mathematical concept.<|endoftext|> -TITLE: tr(ab) = tr(ba)? -QUESTION [33 upvotes]: It is well known that given two Hilbert-Schmidt operators $a$ and $b$ on a Hilbert space $H$, their product is trace class and $tr(ab)=tr(ba)$. A similar result holds for $a$ bounded and $b$ trace class. -The following attractive statement, however, is false: -Non-theorem: -Let $a$ and $b$ be bounded operators on $H$. If $ab$ is trace class , then $ba$ is trace class and $tr(ab)=tr(ba)$. -The counterexample is $a=\pmatrix{0&0&0\\0&0&1\\0&0&0}\otimes 1_{\ell^2(\mathbb N)}$, $b=\pmatrix{0&1&0\\0&0&0\\0&0&0}\otimes 1_{\ell^2(\mathbb N)}$. -I'm guessing that the following is also false, but I can't find a counterexample: -Non-theorem?: -Let $a$ and $b$ be two bounded operators on $H$. If $ab$ and $ba$ are trace class, then $tr(ab)=tr(ba)$. - -REPLY [20 votes]: This follows from proposition 7.3 in "Trace-class operators and commutators" by N.J. Kalton. The theorem actually proves something stronger, that for $AB-BA$, if you arrange the eigenvalues with algebraic multiplicity so that $|\lambda_n|\geq |\lambda_{n+1}|$, then $$\sum_{n=1}^{\infty}\frac{\lambda_1+\cdots+\lambda_{n}}{n}<\infty$$ -which implies trace zero, but the converse is false.<|endoftext|> -TITLE: One-line proof of the Euler's reflection formula -QUESTION [9 upvotes]: A popular method of proving the formula is to use the infinite product representation of the gamma function. See ProofWiki for example. -However, I'm interested in down-to-earth proof; e.g. using the change of variables. As the formula being connected to the beta function, there could be one-line proof for it. -Could anyone help me? - -REPLY [3 votes]: It can be shown (from the Beta function) that -\begin{eqnarray} - \Gamma(1-x) \Gamma(x) = \mathrm{B}(x, 1 -x) - = \int_0^{\infty} \frac{s^{x-1} d s}{s+1} \quad \quad (1) -\end{eqnarray} -Now we show that -\begin{eqnarray*} - \int_0^{\infty} \frac{s^{x-1} d s}{s+1} = \frac{\pi}{\sin \pi x} -\end{eqnarray*} -We use the contour shown in Figure below - -The singularities are a pole at $s=-1$ and a branch point at $s=0$ where -the function is multivalued. We can use the positive $x$ axis as a branch -cut. The contour encloses the pole, so the integral along the contour -$C=C_1 \cup C_2 \cup C_3 \cup C_4$ is given by -\begin{eqnarray*} - \int_C \frac{s^{x-1} ds}{s+1} = 2 \pi \mathrm{i} (-1)^{x-1} - = 2 \pi \mathrm{i} \, \mathrm{e}^{ \mathrm{i} \pi (x-1)}. -\end{eqnarray*} -since $(-1)^{x-1}=\mathrm{e}^{- \mathrm{i} \pi (x-1)}$ is the only residue of the -integrand. -We now evaluate the four individual integrations for the four paths in the figure. -Let us call $I_i = \int_{C_i} s^{x-1}/(1+s) ds$. -We start with the integrals along the circular paths. For the small circle -we can write $s= \epsilon \mathrm{e}^{\mathrm{i} \theta}$ where $\theta \in[ \delta, -2 \pi - \delta]$ with $\epsilon$ the radius of the disk, and $\delta$ the initial -angle of integration. We then change variables from $s$ to $\theta$ with -$ds = \mathrm{i} \epsilon \mathrm{e}^{\mathrm{i} \theta}$. That is, -\begin{eqnarray*} - |I_4| = - \left | \int_ - {2 \pi - \delta_{\epsilon}}^ - {\delta_{\epsilon}} - \frac{\mathrm{i} \epsilon^x \mathrm{e}^{\mathrm{i} \theta (x+1) } d \theta } - {\epsilon \mathrm{e}^{\mathrm{i} \theta} -1 } \right | - \le - \int_{\delta_{\epsilon}}^{2 \pi - \delta_{\epsilon}} - \frac{|\epsilon^x| d \theta } - {1 - \epsilon } - = (2 \pi - 2 \delta_{\epsilon} ) \frac{|\epsilon^x| } - {1 - \epsilon } -\end{eqnarray*} -which goes to $0$ as $\epsilon, \delta_{\epsilon} \to 0$, since $0 < x < 1$. -Likewise along the big circle $s = R \mathrm{e}^{\mathrm{i} \theta}$ and -$ds = \mathrm{i} R \mathrm{e}^{\mathrm{i} \theta} d \theta$, and so -\begin{eqnarray*} - |I_2| = - \left | \int_{\delta_{R}}^{2 \pi - \delta_{R}} - \frac{\mathrm{i} R^x \mathrm{e}^{\mathrm{i} \theta (x+1) } d \theta } - {R \mathrm{e}^{\mathrm{i} \theta} -1 } \right | - \le - \int_{\delta_{R}}^{2 \pi - \delta_{R}} - \frac{|R^x| d \theta } - {R-1 } - = (2 \pi - 2 \delta_{R} ) \frac{R^x } - {R-1 } -\end{eqnarray*} -We use the L'H\^{o}pital rule to find that -\begin{eqnarray*} - \lim_{R \to \infty} |I_2| = \lim_{R \to \infty} 2 (\pi - \delta_R) \frac{x R^{x-1}}{1} = - \lim_{R \to \infty} \frac{x}{R^{1-x}} = 0 -\end{eqnarray*} -since $1 > 1-x > 0$. -We are left with the integrals along $C_1$ and $C_3$. If in the integral along $C_1$ -we take the limit as $\epsilon \to 0 , R \to \infty$, is the original -integral (1). On the other hand, the integral over $C_3$ has the -argument shifted by $2 \pi$ with respect to the original integral. That is, -\begin{eqnarray*} - \lim_{\epsilon \to 0, R \to \infty} \int_{C_3} \frac{s^{x-1} ds}{(s+1)^{x+1}} - = \int_{\infty}^0 \frac{ \mathrm{e}^{2 \pi \mathrm{i} (x-1)} s^{x-1} ds}{(\mathrm{e}^{2 \pi - \mathrm{i}} s - + 1)^{x+1}} = -\mathrm{e}^{2 \pi \mathrm{i} ( x-1)} \int_0^{\infty} - \frac{s^{x-1} ds }{(s+1)^{x+1}} -\end{eqnarray*} -Putting all integrals together we find that -\begin{eqnarray*} - (1 - \mathrm{e}^{2 \mathrm{i} \pi ( x-1) }) - \int_0^{\infty} \frac{s^{x-1} ds}{(s+1)^{x+1}} = - 2 \pi \; \mathrm{i} \; \mathrm{e}^{ \mathrm{i} \pi ( x-1)}. -\end{eqnarray*} -Hence -\begin{eqnarray*} - \int_0^{\infty} \frac{s^{x-1} ds}{ (s+1)^{x+1}} = - \frac{2 \pi \mathrm{i} \mathrm{e}^{ \mathrm{i} \pi ( x-1)}}{ - 1 - \mathrm{e}^{2 \mathrm{i} \pi(x-1)}} - = \frac{\pi}{( -\mathrm{e}^{- \pi x} + \mathrm{e}^{\mathrm{i} \pi x})/2 i} - = \frac{\pi}{\sin \pi x} -\end{eqnarray*} -We then showed that Euler's reflection formula<|endoftext|> -TITLE: $Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$ -QUESTION [14 upvotes]: Observe that we have $Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$. -More generally, assume that $K$ is a finite extension of Q. Is there any $\alpha \in K$ such that $K=Q(\alpha^n)$ for every $n \in N$? - -REPLY [21 votes]: In a more general setting, the following is true : let $K$ be an infinite field and $L/K$ be a finite separable extension whose Galois closure $M$ contains only finitely many roots of unity (this assumption is true for number fields). Then there exists $\alpha \in L$ such that $L=K(\alpha^n)$ for every $n \geq 1$. -Proof : let $d=[L:K]$ and $\sigma_1,\ldots,\sigma_d$ be the distinct $K$-embeddings of $L$ into $M$. Given $\alpha \in L$, we have $L=K(\alpha)$ if and only if $\sigma_1(\alpha),\ldots,\sigma_d(\alpha)$ are pairwise distinct. If $i \neq j$ then the equation $\sigma_i(\alpha)=\sigma_j(\alpha)$ determines a strict $K$-subspace of $L$. Moreover, the equation $\sigma_i(\alpha^n)=\sigma_j(\alpha^n)$ is equivalent to $\sigma_i(\alpha)= \zeta \cdot \sigma_j(\alpha)$ for some $\zeta \in \mu_n(M)$. The set of $\alpha$ satisfying the last condition is again a strict $K$-subspace of $L$ because $\sigma_i(1)=\sigma_j(1)=1$. Since the union of finitely many strict subspaces of $L$ cannot be equal to $L$, the result follows. -EDIT. Note that this proof is based on the same idea as in Denis's answer : if $K$ is a quadratic field, the embeddings of $K$ are just the identity map and the map $\alpha \mapsto \overline{\alpha}$.<|endoftext|> -TITLE: Isometric embedding a convex cap to render its boundary planar -QUESTION [6 upvotes]: I would like to know if there is a polyhedral analog to this beautiful -theorem of Hong: - -Theorem 11.0.1. - Any smooth positive disk $(\bar{D},g)$ with a positive geodesic - curvature along $\partial D$ admits a smooth isometric - embedding in $\mathbb{R}^3$ that maps $\partial D$ to a planar curve. - Moreover, the embedding is unique within rigid motions. - -This is proved in the book by -Qing Han and Jia-Xing Hong, -Isometric Embedding of Riemannian Manifolds in Euclidean Spaces, - -American Mathematical Society Monograph, Volume 130, 2006, p.225. -I would like to replace the disk by a convex polyhedral cap $C$ -with a convex boundary $\partial C$. $C$ is a connected subset of the -surface -of a convex polyhedron, and $\partial C$ has at every point $p \in \partial C$ at most $\pi$ angle -incident to $p$ within $C$: - -           - - -Alexandrov's book -Convex Polyhedra -includes an article in the Appendix written by L.A. Shor -"On Flexibility of Convex Polyhedra with Boundary," -which describes (rather complex) conditions under which flexing can -occur, -but does not seem to address flexing to achieve planarity of $\partial C$. -It appears that Alexandrov's and Shor's results imply that -it cannot always be possible, because (for example) if $\partial C$ -contains no vertices of $C$, then it must be rigid -(but not necessarily planar). -But perhaps there are conditions under which -some polyhedral analog of Hong's theorem holds? -Thanks for pointers! -Addendum. My concentration on the conditions for "flexing" above was misplaced, -as Sergei Ivanov's answer demonstrates: the analog isometrically embeds a convex cap $C$ with -$\partial C$ planar, but not by hinging $C$'s faces as rigid plates, -but rather through an embedding which -in general alters the facial structure of $C$ while maintaining isometry. - -REPLY [12 votes]: Yes the polyhedral analog is true. Just consider the doubling of $C$, i.e., attach an isometric copy $C'$ of $C$ along the boundary, and apply Alexandrov's embedding theorem to the doubling. The common boundary of $C$ and $C'$ will go to a plane automatically. -Indeed, Alexandrov's theorem says that the doubling $S=C\cup C'$ (with its natural intrinsic metric) admits a unique (up to a rigid motion) isometric embedding $f:S\to\mathbb R^3$ as a surface of a convex polyhedron. The intrinsic metric of $S$ has a self-isometry $i$ which swaps $C$ and $C'$ and whose set of fixed points is their common boundary. Since $f$ is unique up to a rigid motion, $f\circ i$ extends to a rigid motion of $\mathbb R^3$. The set of fixed points of this rigid motion is an affine subspace, and $f(\partial C)$ is contained in this subspace. Hence $f(\partial C)$ is contained in a plane (and that rigid motion is the reflection in that plane). -The uniqueness also follows from the uniqueness part of Alexandrov's theorem as long as you require that the image is a convex half-polyhedron. -The usual warning attached to Alexandrov's embeddings applies here: you get an isometric embedding of the intrinsic metric, but its facial structure may differ from the original one.<|endoftext|> -TITLE: Torsion subgroups in families of twists of elliptic curves -QUESTION [15 upvotes]: Here is the short version: - -Fix an elliptic curve $E/\mathbb{Q}$. How does the torsion structure $E_d(\mathbb{Q})_{tors}$ vary, as $E_d$ runs through the quadratic twists of $E$? - -Here is the longer version: -I have been playing with SAGE this morning. I inserted the elliptic curve ('11a1') $$E : y^2 + y = x^3 - x^2 - 10x - 20$$ which has rational torsion subgroup isomorphic to $\mathbb{Z}/5\mathbb{Z}$. I then computed its quadratic twist $E_d$ for all squarefree $d$ up to 2000, and observed $E_d(\mathbb{Q})_{tors}$ was always trivial. - -Can it be that, in this particular family of quadratic twists, all but one of the curves have trivial torsion? Is this a general phenomenon? - -(I ran this experiment for several other curves $E$ and got the same impression; that all but one of the curves in a family of twists have the same torsion structure.) - -REPLY [10 votes]: The maximum number of quadratic twists with $n$-torsion (n odd) that an elliptic curve over a number field $K$ can have is 2, and here is an easy proof. -You can see this by asking how many twists with $n$-torsion can an elliptic curve which already has $n$-torsion have? If this number is $k$, then clearly an elliptic curve without $n$-torsion can have $k+1$ twists with $n$-torsion. -Suppose now that $E/K,\ E(K)\supset \mathbb Z/ n\mathbb Z$, has 2 twists $E_{d_1}$ and $E_{d_2}$ with $n$-torsion. It follows from what Chris wrote that -$$E(K(\sqrt{d_1},\sqrt{d_2}))[n]\supset E(K)[n] \oplus E_{d_1}(K)[n] \oplus E_{d_2}(K)[n]\supset (\mathbb Z/ n\mathbb Z)^3,$$ -which is clearly impossible. -Note that the answer depends a lot on the number field $K$ and the number $n$ you choose. The possibility of 2 twists exists only over number fields $K$ such that the complete $n$-tosion can be defined over a quadratic extension of $K$. For example, it follows that over $\mathbb Q$, an elliptic curve can have at most 2 twists with 3-torsion, 1 twist with 5 or 7-torison and 0 twists with $p$ torsion for all primes $p>7$. -So, Giuseppe, your curve $E$ with 5-torsion, as any other curve with 5-torsion, cannot have a twist with 5-torsion over $\mathbb Q$.<|endoftext|> -TITLE: size of smallest generating set of a group -QUESTION [21 upvotes]: Suppose I have a nice (e.g., word-hyperbolic? bi-automatic? automatic?) group and I want to know how big the smallest generating set is. Is that tractable (or, to put it more optimistically, what is the biggest class of groups for which it is tractable)? I am actually most interested in the question of whether there is a generating set of cardinality $2,$ but I suspect that is as hard as the general question. -EDIT What I really want to know is the answer for lattices (e.g., $SL(n, \mathbb{Z}),$) but that's probably not in any tractable class. -UPDATE It is, in fact, known that $SL(n, \mathbb{Z})$ itself is generated by $2$ elements (Hua+Reiner wrote down a generating set with three elements in 1949, as did M. Conder et al in 1992, for $SL(3, \mathbb{Z})$, but Stanton M. Trott did it with two generators in 1962). The generators are: -$\begin{pmatrix} -1 & 0 & 0 & \dots & 0 \\ -1 & 1 & 0 & \dots & 0 \\ -0 & 0 & 1 & \dots & 0 \\ -. & . & . & \dots & .\\ -0 & 0 & 0 & \dots & 1 -\end{pmatrix}$ -and -$\begin{pmatrix} -0 & 1 & 0 & \dots & 0\\ -0 & 0 & 1 & \dots & 0\\ -. & . & . & \dots & .\\ -0 & 0 & 0 & \dots & 1\\ -(-1)^n & 0 & 0 & \dots & 0 -\end{pmatrix}$ - -REPLY [5 votes]: There is no upper bound for finite index subgroups of $SL_n({\mathbb Z})$. That is, given any integer $k$, there exists a finite index subgroup of $SL_n({\mathbb Z})$ which needs at least $k$ generators. I do not know to whom this is originally due, but it is a remark of Rapinchuk that if you pull the centre of, say $SL_{2n}({\mathbb Z}/m {\mathbb Z})$ where $m$ is the product of $k$ distinct primes back to $SL_{2n}({\mathbb Z}) $, you get a finite index subgroup of $SL_{2n}({\mathbb Z})$ whose abelianisation has $({\mathbb Z}/2{\mathbb Z})^k$ as a quotient and hence is at least $k$ generated. -On the other hand, a finite index subgroup of $SL_n({\mathbb Z})$, contains a smaller finite index subgroup which is generated by 3 elements; this result is true for any higher rank non-uniform arithmetic group in a semi-simple linear group (and is due to Ritumoni Sarma and Venkataramana)<|endoftext|> -TITLE: Forcing over models without the axiom of choice -QUESTION [17 upvotes]: In the vast majority of papers forcing is always developed over ZFC. -Not surprisingly too, since infintary combinatorial principles are often used to prove results based on properties such as chain conditions, closure, and so on. -I am looking for a good start on forcing over models of ZF. I have before me two papers which I have yet to read thoroughly, however may not be as useful for this purpose as I am hoping. - -Grigorieff, S. Intermediate Submodels and Generic Extensions in Set Theory. The Annals of Mathematics, Second Series, Vol. 101, No. 3 (May, 1975), pp. 447-490 -Monro, G. P. On Generic Extensions Without the Axiom of Choice. The Journal of Symbolic Logic, Vol. 48, No. 1 (Mar., 1983), pp. 39-52 - -While I do intend to read them either way, it seems that neither develops the theory of forcing in the absolute absence of choice. I am currently looking for references which deal with such situation, or with the relation between forcing theorems proved in ZFC and the amount of choice needed for them to hold. -Edit: I probably should have mentioned that I am quite familiar with permutation models of ZFA+embedding theorems and transfer theorems (Jech-Sochor, Pincus' theorem) as well with symmetric extensions. -I am not looking for ways to develop forcing extensions of ZF without the axiom of choice; rather I am looking for theorems such as c.c.c forcing does not collapse cardinals and similar theorems extended to the choiceless contexts if possible, or the strength of choice needed for these theorems to hold. -Consider two examples: - -Suppose a model of ZF in which the axiom of choice does not hold. Can we, by set forcing add the axiom of choice? If not, can it be done using a machinery similar to a symmetric extension? If we can in fact find such extension, does that mean the model without choice is a symmetric extension between two larger models? -Suppose A is an infinite Dedekind-finite set, what can we say on a forcing poset based on A (either domain of functions are partial to A or the range is in A)? Can we "collapse" amorphous sets onto ordinals? Can we collapse one amorphous set onto another? And so on. - -REPLY [2 votes]: Here is an equivalence of a forcing principle to the Axiom of Choice, courtesy of Arnold Miller, found in his preprint, "The maximum principle in forcing and the axiom of choice": - -(Abstract) In this paper we prove that the maximum principle in forcing [ $p$ $\Vdash$ $\exists$$x$ $\theta$($x$) iff there exists a a name $\tau$ such that $p$ $\Vdash$ $\theta$($\tau$)--my quote from Miller's preprint, found on his homepage at www.math.wisc.edu/~miller/] is equivalent to the axiom of choice. We also look at some specific partial orders in the basic Cohen model [his model in which the axiom of choice fails]. - -This paper also discusses partial orders in Cohen's original model for which the maximum principle fails and one (at the end of this paper) for which the maximum principle succeeds. -It would be interesting to find out what difficulties (if any) the absence of the maximum principle causes for forcing in $ZF$ (and if these difficulties could be overcome).<|endoftext|> -TITLE: Non-trivial algebraic consequence of an elementary geometric theorem -QUESTION [5 upvotes]: A well-known theorem in projective geometry states that the three Pascal lines of an arbitrary hexagon inscribed in a quadric intersect in one point. I found an algebraic reformulation, which states that for any six lines $l_i=a_ix+b_iy+c_iz$, if the quadrics $l_i^2$ are linearly dependent, then $l_1l_3$, $l_3l_5$, $l_1l_5$, $l_2l_4$, $l_2l_6$, $l_4l_6$ are linearly dependent. But I can not understand the algebraic meaning of this fact. Is it just an accident or a special case of some duality? - -REPLY [4 votes]: Pascal's theorem is a special case of the purely algebraic Cayley-Bacharach theorem. For a brief discussion see e.g. http://terrytao.wordpress.com/2011/07/15/pappuss-theorem-and-elliptic-curves/<|endoftext|> -TITLE: p-adic Hodge theory for varieties defined over \C _p ? -QUESTION [13 upvotes]: I have a question on p-adic Hodge theory: -When e.g. $X$ is a smooth proper scheme over a finite extension $K$ of $\mathbf{Q}_{p}$ then e.g. one variant of $p$-adic Hodge theory says that there is a functorial isomorphism -$$ B_{\mathrm{HT}}\otimes_K\mathrm{gr}H^\ast_{\mathrm{dR}}(X/K) \cong B_{\mathrm{HT}}\otimes_{\mathbf{Q}_p} H^\ast_{\mathrm{\acute{e}t}}(X\times_\overline{K},\mathbf{Q}_p).$$ -To my knowledge there it is not known whether such an isomorphism exists also for schemes defined over -$\mathbf{C}_p$ -Of course there may not be a Galois action on the left hand side, but one may still ask, whether there exists a canonical functorial isomorphism of $\mathbf{C}_{p}$-vector spaces: -$$ H^\ast_{\mathrm{dR}}(X) \cong H^\ast_{\mathrm{\acute{e}t}}(X) $$ -I am thinking whether it may be possible to find such an isomorphism by approximating it modulo $p^n$ and by using $p$-adic hodge theory over bigger and bigger finite extensions of $\mathbf{Q}_p$? - -REPLY [21 votes]: In general, you cannot expect that for a proper smooth scheme $X$ over $\mathbb{C}_p$, you have a canonical decomposition -$$ -H^i_{\mathrm{\acute{e}t}}(X)\otimes \mathbb{C}_p\cong \bigoplus_j H^{i-j}(X,\Omega_X^j)\ . -$$ -However, there is a natural filtration on $H^i_{\mathrm{\acute{e}t}}(X)\otimes\mathbb{C}_p$ with associated gradeds $H^{i-j}(X,\Omega_X^j)$. For abelian varieties over $\mathbb{C}_p$, see Theoreme II.1.1 (and Remarque II.1.2) in Fargues' book on the isomorphism of the Lubin-Tate and Drinfeld tower.<|endoftext|> -TITLE: Applications of Hilbert's metric -QUESTION [10 upvotes]: Among the fascinating constructions in mathematics is the Hilbert metric on a bounded convex subset of ${\mathbb R}^n$. - -Where, within mathematics, is it used ? I know at least a proof of the Perron--Frobenius Theorem for non-negative matrices. -What are its applications in other sciences ? - -REPLY [6 votes]: This might not be considered an application, but Hilbert metrics have been studied geometrically and dynamically. Here are several examples of questions that have been partially or totally answered: - -when is a convex domain endowed with its Hilbert metric $\delta$-hyperbolic? -what is the volume entropy of Hilbert metrics? -does there exist convex sets that admit a cocompact group of isometries (relative to their Hilbert metric)? (the answer is yes!) - -I guess that amoung these, the last part can be considered an application: Hilbert metrics yields interesting subgroups of $\mathrm{PSL}(n,\mathbb{R})$. -For more details you can look at the works of Yves Benoist (in particular the "convexes divisibles" series and "Convexes hyperboliques et fonctions quasisymétriques", in french), Constantin Vernicos, Ludovic marquis and Mickaël Crampon.<|endoftext|> -TITLE: G = [G,G] with two generators -QUESTION [6 upvotes]: Is it true that groups $\langle a,b \mid a^n b^k=b^ka^{n+1}, b^la^s=a^sb^{l+1}\rangle$ are non-trivial for almost all (in any sense:))) $n,k,l,s\in\mathbb N$? - -REPLY [16 votes]: The group is always trivial. Indeed, $a^{-ns}b^{l^n}a^{ns}=b^{(l+1)^n}$. On the other hand, -$a^{ns}=b^ka^{(n+1)s}b^{-k}$. Substitute $a^{ns}$ from the second equality to the first. You will get that $a^{(n+1)s}$ conjugates $b^{l^n}$ to $b^{(l+1)^n}$. Since $a^{ns}$ does the same, you get that $a^s$ commutes with $b^{l^n}$. Hence $b^{l^n}=b^{(l+1)^n}$ and $b$ is torsion, $b^p=1$. Similarly, $a$ is torsion, $a^q=1$. Notice also that $p=(l+1)^n-l^n$ is coprime with $l$ and $q$ is coprime with $n$. Now we have $b^{-kp}a^{n^p}b^{kp}=a^{(n+1)^p}$. Hence $a^{n^p}=a^{(n+1)^p}$. We already know that $a^{n^l}=a^{(n+1)^l}$. Suppose that the order of $a$ is $t$. Then $t$ divides both $(n+1)^l-n^l$ and $(n+1)^p-n^p$ and is co-prime with $n$. Then $t$ divides $$(n+1)^p-n^p-(n+1)^p+n^l*(n+1)^{p-l}=-n^p+n^l*(n+1)^{p-l}.$$ Since $t$ and $n$ are co-prime, $t$ divides $(n+1)^{p-l}-n^{p-l}$. Proceed by using the Euclidean algorithm. Since $p$ and $l$ are co-prime, we get that $t$ divides $(n+1)-n=1$, so $a=1$. Similarly $b=1$ and the group is trivial.<|endoftext|> -TITLE: An example of an affine variety with non-zero Chow groups -QUESTION [5 upvotes]: Are there any examples known of an affine variety $A$ over an algebraically closed field such some Chow group (say, of codimension at least $2$) of $A$ with coefficients in $\mathbb{Z}/n\mathbb{Z}$ ($n$ is prime to the residue field characteristic) is non-zero? - -REPLY [4 votes]: I think there will be many such varieties. -For example, let $Q$ be a smooth $4$-dimensional projective quadric, $Q'$ a hyperplane section of $Q$ and $A = Q \backslash Q'$. For any $i$, we have the localisation sequence -$$ CH^{i-1}(Q') \to CH^i(Q) \to CH^i(A) \to 0 \ . $$ -Since $CH^1(Q') = \mathbb{Z}/n\mathbb{Z}$ and $CH^2(Q) = (\mathbb{Z}/n\mathbb{Z})^{\oplus 2}$ (with $\mathbb{Z}/n\mathbb{Z}$ coefficients) it follows that $CH^2(A)$ is nonzero.<|endoftext|> -TITLE: Functoriality of adjugate matrix -QUESTION [5 upvotes]: i'm interested in geometric interpretations of many linear algebra notions (check also related geometric interpretation of matrix minors). it came to me recently that geometric description of adjugate matrix (for example in case 3×3-matrix) might be quite hard—feel welcome to fill the gap!—but what catched my attention is functoriality of adjugate matrix ($\scriptstyle \mathbf I^\mathrm D = \mathbf I$ and $\scriptstyle (\mathbf{AB})^\mathrm D = \mathbf B^\mathrm D \mathbf A^\mathrm D$); my question is: - -what kind of functor is the adjugating (for linear endomorphisms)? - -it seems to have strong relationship with (hermitian) adjoint but has slightly different properties (it commutes with transpose). thanks in advance! - -REPLY [3 votes]: Copied from math.stackexchange: Another way to approach this avoids the use of duality via inner-products. Functorality should be clear. It may be that one requires finite rank free modules here so that the bilnear pairing below is perfect. -The bilinear pairing is, -$$ -V \times \Lambda^{n-1} V \to \Lambda^n V -$$ -sending -$$ -(v, \eta) \to v \wedge \eta -$$ -and written -$$ -\langle v, \eta \rangle = v \wedge \eta. -$$ -Then given $T : V \to V$, we define, -$$ -\Lambda^{n-1} T : \Lambda^{n-1} V \to \Lambda^{n-1} V -$$ -on indecomposable elements by -$$ -\Lambda^{n-1} T (v_1 \wedge \cdots v_{n-1}) = T(v_1) \wedge \cdots \wedge T(v_{n-1}) -$$ -and extend to all of $\Lambda^{n-1} V$ by alternating multilinearity as usual. -The adjugate $\operatorname{adj}(T) : V \to V$ is the adjoint of $\Lambda^{n-1} T$ with respect to the pairing: -$$ -\langle \operatorname{adj}(T) (v), \eta \rangle = \langle v, \Lambda^{n-1} T (\eta) \rangle, -$$ -or using the definition of the pairing, -$$ -\operatorname{adj} (T) (v) \wedge \eta = V \wedge \Lambda^{n-1} T \eta -$$ -Now one observes that, -$$ -\begin{split} -\langle \operatorname{adj} (T) \circ T (v), \eta\rangle &= \langle T(v), \Lambda^{n-1} T(\eta)\rangle \\ -&= T(v) \wedge \Lambda^{n-1} T (\eta) \\ -&= \Lambda^{n} T (v \wedge \eta) \\ -&= \det T v \wedge \eta \\ -&= \langle \det T v, \eta \rangle -\end{split} -$$ -If the pairing is perfect, this implies that, -$$ -\operatorname{adj} (T) \circ T = \det T \operatorname{Id}. -$$ -In particular, if $\det T$ is an invertible element of the underlying ring, then $\det T \operatorname{Id}$ is invertible and $\operatorname{adj} (T)$ commutes with $T$ (which both are invertible) as well as satisfying, -$$ -\operatorname{adj} (T) = \det T T^{-1} -$$ -which is the usual formula. -This all explained (sections 5 to 8) here: -http://people.reed.edu/~jerry/332/27exterior.pdf<|endoftext|> -TITLE: Existence of non-principal ultrafilters on sets -QUESTION [7 upvotes]: Is it known to be consistent with ZF that there is no non-principal ultrafilter on any infinite set? (Feel free to use your favorite interpretation of "infinite" in this context. -If infinite just meant infinite ordinals, that would be fine, too. You may use all kinds of large cardinals.) - -REPLY [12 votes]: Yes, it is consistent with ZF that every ultrafilter is principal. This is a result of Andreas Blass, A model without ultrafilters, Bull. Acad. Polon. Sci. 25 (1977), 329–331.<|endoftext|> -TITLE: Schur-Weyl duality -QUESTION [7 upvotes]: I have a question about the interpretation of multiplicities and dimensions using Schur-Weyl duality. -$V$ is an n-dimensional complex vector space. Then $V$ $\otimes$ $V$ $\otimes$ $V$ decomposes as: -$V$ $\otimes$ $V$ $\otimes$ $V$ = $Sym^3$ $V$ $\oplus$ $\wedge^3$ $V$ $\oplus$ $S_{(2,1)}V$ $\oplus$ $S_{(2,1)}V$ -where $S_{(2,1)}V$ is a Schur module for the partition (2,1). (Fulton-Harris, Chapter 6). -Then Schur-Weyl duality says that the multiplicity of $S_{(2,1)}V$ in this decomposition (it is 2) should be the dimension of the irrep of $S_3$ labelled by the partition (2,1) - which is correct. -My question is about the other half of this duality: the dimension of $S_{(2,1)}V$ in this decomposition (it is 8) should correspond to some sort of multiplicity for the irrep of $S_3$ labelled by the partition (1,2) - but I am unable to see exactly what... -Any help is most welcome. - -REPLY [3 votes]: Let $V$ be the vector representation of $GL_n(\mathbb{C})$, and let $d\leq n$. -You want to see the multiplicities of a given irreducible $S_d$ module in $V^{\otimes d}$ in terms of the dimension of an associated irreducible $GL_n(\mathbb{C})$-module. To do this, observe that $S_d$ acts on $V^{\otimes d}$ by permuting the tensor factors and this action commutes with the diagonal action of $GL_n(\mathbb{C})$ on $V^{\otimes d}$. Therefore, for a partition $\lambda=(\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_n\geq 0)$ (equivalently, a polynomial weight for $GL_n(\mathbb{C})$) we can consider the space $W_\lambda$ of highest weight vectors in $V^{\otimes d}$ of highest weight $\lambda$. Since the action of $S_d$ commutes with the action of $GL_n(\mathbb{C})$ (and, in particular, the action of the standard maximal torus, and standard Borel), $W_\lambda$ is an $S_d$-module. By character considerations, one can show that this space is the irreducible Specht module, $W_\lambda\cong S^{\lambda}$. -More generally, if $X\in GL_n(\mathbb{C})$, then $X\cdot W_\lambda$ is another $S_d$-module since, again, the action of $S_d$ commutes with the action of $GL_n(\mathbb{C})$. Moreover, this module is obviously isomorphic to $S^\lambda$. This means that there are $\dim L(\lambda)$ copies of $S^\lambda$ in $V^{\otimes d}$, where $L(\lambda)$ is the irreducible $GL_n(\mathbb{C})$-module of highest weight $\lambda$. -This proves half of the statement that, as a $(GL_n(\mathbb{C}),S_d)$-bimodule, -$V^{\otimes d}\cong \bigoplus_{\lambda\vdash d}L(\lambda)\otimes S^\lambda.$ From your question, I assume you understand the other half.<|endoftext|> -TITLE: Is "normal scheme" defined in EGA? -QUESTION [6 upvotes]: In answering a question on this site yesterday, I was led to introduce a non-separated scheme, which I said was normal. -However when I tried to check the definition of "normal scheme" in EGA , in order to make sure that "normal" didn't have "separated" in its definition, I was surprised to find that I couldn't find the definition, although "normal prescheme" is used in many places (which seems to indicate that normal doesn't imply separated, else the authors would have said normal scheme: schemes were supposed to be separated in contradistinction to preschemes). -Another ambiguity: it is not clear to me whether for Grothendieck-Dieudonné, "normal" only depends on the local rings of the scheme : do they consider that the spectrum of the product of two copies of a field $k$ is normal, even though $k\times k$ is not a domain, hence not an integrally closed domain ? -[My guess is "yes", but I don't want to guess] -So, my question is:Do the EGA contain a definition of "normal (pre)scheme" ? -I'm interested because I have never used a definition at odds with EGA, and I have no intention to start now... - -REPLY [10 votes]: Normality for schemes (as opposed to general ringed spaces) seems not to be treated in EGA until $\S$5 in Chapter IV. It is discussed in the context of properties R$_n$ and S$_n$. Serre's normality criterion (Normal is equivalent to R$_1$ and S$_2$) is proved in EGA IV$_2$ (5.8.6). The definition is briefly recalled at the beginning of the proof.<|endoftext|> -TITLE: In which situations can one see that topological spaces are ill-behaved from the homotopical viewpoint? -QUESTION [36 upvotes]: In the eighties, Grothendieck devoted a great amount of time to work on the foundations of homotopical algebra. -He wrote in "Esquisse d'un programme": "[D]epuis près d'un an, la plus grande partie de mon énergie a été consacrée à un travail de réflexion sur les fondements de l'algèbre (co)homologique non commutative, ou ce qui revient au même, finalement, de l'algèbre homotopique." (Beginning of section 7. English version here: "Since the month of March last year, so nearly a year ago, the greater part of my energy has been devoted to a work of reflection on the foundations of non-commutative (co)homological algebra, or what is the same, after all, of homotopic[al] algebra.) -In a letter to Thomason written in 1991, he states: "[P]our moi le “paradis originel” pour l’algèbre topologique n’est nullement la sempiternelle catégorie ∆∧ semi-simpliciale, si utile soit-elle, et encore moins celle des espaces topologiques (qui l’une et l’autre s’envoient dans la 2-catégorie des topos, qui en est comme une enveloppe commune), mais bien la catégorie Cat des petites caégories, vue avec un œil de géomètre par l’ensemble d’intuitions, étonnamment riche, provenant des topos." [EDIT 1: Terrible attempt of translation, otherwise some people might miss the reason why I have asked this question: "To me, the "original paradise" for topological algebra is by no means the never-ending semi-simplicial category ∆∧ [he means the simplex category], for all its usefulness, and even less is it the category of topological spaces (both of them imbedded in the 2-category of toposes, which is a kind of common enveloppe for them). It is the category of small categories Cat indeed, seen through the eyes of a geometer with the set of intuitions, surprisingly rich, arising from toposes."] -If $Hot$ stands for the classical homotopy category, then we can see $Hot$ as the localization of $Cat$ with respect to functors of which the topological realization of the nerve is a homotopy equivalence (or equivalently a topological weak equivalence). This definition of $Hot$ still makes use of topological spaces. However, topological spaces are in fact not necessary to define $Hot$. Grothendieck defines a basic localizer as a $W \subseteq Fl(Cat)$ satisfying the following properties: $W$ is weakly saturated; if a small category $A$ has a terminal object, then $A \to e$ is in $W$ (where $e$ stands for the trivial category); and the relative version of Quillen Theorem A holds. This notion is clearly stable by intersection, and Grothendieck conjectured that classical weak equivalences of $Cat$ form the smallest basic localizer. This was proved by Cisinski in his thesis, so that we end up with a categorical definition of the homotopy category $Hot$ without having mentionned topological spaces. (Neither have we made use of simplicial sets.) -I personnally found what Grothendieck wrote on the subject quite convincing, but of course it is a rather radical change of viewpoint regarding the foundations of homotopical algebra. -A related fact is that Grothendieck writes in "Esquisse d'un programme" that "la "topologie générale" a été développée (dans les années trente et quarante) par des analystes et pour les besoins de l'analyse, non pour les besoins de la topologie proprement dite, c'est à dire l'étude des propriétés topologiques de formes géométriques diverses". ("[G]eneral topology” was developed (during the thirties and forties) by analysts and in order to meet the needs of analysis, not for topology per se, i.e. the study of the topological properties of the various geometrical shapes." See the link above.) This sentence has already been alluded to on MO, for instance in Allen Knutson's answer there or Kevin Lin's comment there. -So much for the personal background of this question. -It is not new that $Top$, the category of all topological spaces and continuous functions, does not possess all the desirable properties from the geometric and homotopical viewpoint. For instance, there are many situations in which it is necessary to restrict oneself to some subcategory of $Top$. I expect there are many more instances of "failures" of $Top$ from the homotopical viewpoint than the few I know of, and I would like to have a list of such "failures", from elementary ones to deeper or less-known ones. I do not give any example myself on purpose, but I hope the question as stated below is clear enough. Here it is, then: - -In which situations is it noticeable that $Top$ (the category of general topological spaces and continuous maps) is not adapted to geometric or homotopical needs? Which facts "should be true" but are not? And what do people usually do when encountering such situations? - -As usual, please post only one answer per post so as to allow people to upvote or downvote single answers. -P.S. I would like to make sure that nobody interpret this question as "why should we get rid of topological spaces". This, of course, is not what I have in mind! - -REPLY [9 votes]: My answer is in agreement with Grothendieck that topological spaces may be seen as inadequate for many geometric, and in particular, homotopical purposes. Round about 1970, I spent 9 years trying to generalise the fundamental groupoid of a topological space to dimension 2, using a notion of double groupoid to reflect the idea of ``algebraic inverse to subdivision'' and in the hope of proving a 2-dimensional van Kampen type theorem. In discussion with Philip Higgins in 1974 we agreed that: -1) Whitehead's theorem on free crossed modules, that $\pi_2(X \cup \{e^2_\lambda\},X,x)$ was a free crossed $\pi_1(X,x)$-module, was an instance of a 2-dimensional universal property in homotopy theory. -2) If our proposed theories were to be any good, then Whitehead's theorem should be a corollary. -However we observed that Whitehead's theorem was about relative homotopy groups. So we tried to define a homotopy double groupoid of a pair of pointed spaces, mapping a square into $X$ in which the edges go to $A$ and the vertices to the base point, and taking homotopy classes of such maps. This worked like a dream, and we were able to formulate and prove our theorem, published after some delays (and in the teeth of opposition!) in 1978. -We could then see how to generalise this to filtered spaces, but the proofs needed new ideas, and were published in 1981; this and subsequent work has evolved into the book ``Nonabelian algebraic topology'' published last August. -Contact with Loday who had defined a special kind of $(n+1)$-fold groupoid for an $n$-cube of spaces led to a more powerful van Kampen Theorem, with a totally different type of proof, published jointly in 1987. This allows for calculations of some homotopy $n$-types, and has as a Corollary an $n$-ad connectivity theorem, with a calculation of the critical (nonabelian!) $n$-ad homotopy group, as has been made more explicit by Ellis and Steiner, using the notion of a crossed $n$-cube of groups. -Thus we could get useful strict homotopy multiple groupoids for kinds of structured spaces, allowing calculations not previously possible. -In this way, Grothendieck's view is verified that as spaces with some kind of structure arise naturally in geometric situations, there should be advantages if the algebraic methods take proper cognisance of this structure from the start. That is, one should consider the data which define the space of interest.<|endoftext|> -TITLE: What is a coalgebra intuitively? -QUESTION [47 upvotes]: How to think about coalgebras? Are there geometric interpretations of coalgebras? -If I think of algebras and modules as spaces and vectorbundles, what are coalgebras and comodules? What basic examples of coalgebras should one keep in mind? -Anything that helps to think about coalgebras without headache is welcome ;) - -REPLY [6 votes]: a pretty general kind of coalgebra is the path coalgebra of a quiver (directed graph).<|endoftext|> -TITLE: How big are the ultrapowers of the hyperfinite $II_1$-factor? -QUESTION [7 upvotes]: Let $R$ be the hyperfinite $II_1$-factor. It is well-known that it is the smallest $II_1$-factor, in the sense that every $II_1$-factor contains a copy of $R$. -Now, let $\omega$ be a free ultrafilter on the natural numbers and construct the tracial ultrapower $R^\omega$. It is well-known, that this is a quite big factor. Indeed -1) Its predual is not separable. Even, just to make a bit of self-advertisement, its unitary group contains an orthogonal copy of the free group on uncountable many generators (see Capraro V. and Paunescu L. Product of ultrafilters and applications to Connes Embedding Problem, to appear in J. Oper. Theory) -2) A well-known conjecture, Connes' Embedding Conjecture, states that every $II_1$-factor with separable predual should be embeddable into $R^\omega$. Namely, all known $II_1$-factors with separable predual are known to embed into $R^\omega$ (or we do not know if they embed) -On the other hand, $R^\omega$ is constructed starting from the smallest $II_1$-factor using a quite natural construction that should not increase to much. So my question is: - -** Question: ** Is it true that any $II_1$-factor with non-separable predual contains a copy of (some) $R^\omega$? - -Actually I don't think this is true, since for instance it is not clear how $R^\omega$ could be contained in the group factor of the free group on uncountable many generators. In some sense, the latter has very few relations and $R^\omega$ has lots of relations, but I don't understand quite well how this difference can be reflected in the group factors. -My interest in this problem comes from the following result by Taka Ozawa. He showed that there is indeed no way to find such a universal factor among the ones with separable predual. See N.Ozawa,* There is no separable universal $ II_1$-factor*, Proc. Amer. Math. Soc., vol. 132, iss. 2, pp. 487-490, 2004. -So, a way to reformulate my question is: Is Connes' embedding conjecture the best conjecture one can formulate? -Maybe, it's not the best conjecture, but it's the best one can hope for. -Thanks in advance for any comment. -Valerio - -REPLY [9 votes]: This is not true. If we denote by $\mathbb F_{\mathbb R}$ the free group with generators indexed by $\mathbb R$ then any separable von Neumann subalgebra $N \subset L\mathbb F_{\mathbb R}$ must necessarily be contained in the von Neumann subalgebra $L\mathbb F_I$ for some countably infinite subset $I \subset \mathbb R$. (This is a direct consequence of applying Parseval's identity to vectors in $L^2N \subset \ell^2 \mathbb F_{\mathbb R}$.) -The separable $II_1$ factor $L ({\rm PSL}(3, \mathbb Z))$ embeds into $R^\omega$ since the group is residually finite, but does not embed into $L\mathbb F_I$ by a result of Connes and Jones since it has property (T). Hence, $L\mathbb F_{\mathbb R}$ cannot contain $R^\omega$.<|endoftext|> -TITLE: Radical of group algebra -QUESTION [6 upvotes]: Let $G$ be a finite group of size $p^a\cdot r$. -Does there exist a simple way to calculate radical of the group algebra $F_p[G]$? - -REPLY [2 votes]: I think the answer is "no, in general", though the terms in the question are not really well-defined. For example, if you knew the radical of the group algebra, then you could easily compute the dimension of the absolutely irreducible modules in characteristic $p.$ When $G$ is the symmetric group, this is already a notoriously difficult (and as yet unsolved) problem.<|endoftext|> -TITLE: What do the stable homotopy groups of spheres say about the combinatorics of finite sets? -QUESTION [114 upvotes]: The Barratt-Priddy-Quillen(-Segal) theorem says that the following spaces are homotopy equivalent in an (essentially) canonical way: - -$\Omega^\infty S^\infty:=\varinjlim~ \Omega^nS^n$ -$\mathbb{Z}\times ({B\Sigma_\infty})_+$, where $\Sigma_\infty$ is the group of automorphisms of a countable set which have finite support, and $+$ is the Quillen plus-construction. -The group completion of $B\left(\bigsqcup_n \Sigma_n\right)$, where $\Sigma_n$ is the symmetric group on $n$ letters, and $B(\sqcup_n \Sigma_n)$ is given the structure of a topological monoid via the block addition map $\Sigma_n\times \Sigma_m\to \Sigma_{n+m}$. -$\Omega|S^\bullet.\operatorname{FinSet}|$, where $S^\bullet$ is the Waldhausen $S$-construction, and $\operatorname{FinSet}$ is the category of pointed finite sets, given the structure of a Waldhausen category by declaring cofibrations to be injections and weak equivalences to be isomorphisms. I don't want to define this since it's complicated (for a reference, see Chapter IV of Weibel's K Book), but it should be thought as a homotopical version of the Grothendieck ring of finite sets, where addition is given by disjoint union and multiplication is given by the cartesian product. Clark Barwick's answer here makes this more precise. - -Now, the homotopy groups of the first space are manifestly the stable homotopy groups of spheres; on the other hand, the last two spaces clearly encode some information about the combinatorics of finite sets. So my question is: - -Is there a concrete combinatorial interpretation of the higher stable homotopy groups of spheres in terms of the combinatorics of finite sets or symmetric groups? - -For example it is easy to see via (3) or (4) that $\pi_{0+k}(S^k)=\mathbb{Z}$ corresponds to the Grothendieck ring of finite sets. Similarly, (2), or with some theory (4), make it clear that $\pi_{1+k}(S^k)=\mathbb{Z}/2\mathbb{Z}$ corresponds to the abelianization of $\Sigma_n$ (via the sign homomorphism). I am interested in concrete interpretations of the higher stable homotopy groups in this style. -A good answer would be, for example, a direct combinatorial interpretation of $\pi_{2+k}(S^k)=\mathbb{Z}/2\mathbb{Z}$ and $\pi_{3+k}(S^k)=\mathbb{Z}/24\mathbb{Z}$; a not-so-good answer would be a statement like "the sphere spectrum is a categorification of the integers," which is not the sort of concrete thing I'm looking for. -EDIT: So with the exception of Jacob Lurie's comment on $\pi_{2+k}(S^k)$ below (interpreting it as the Schur multiplier $H_2(\Sigma_\infty, \mathbb{Z})$ of $\Sigma_\infty$), it seems like it might be too much to hope for any reasonably complete combinatorial interpretation of the stable homotopy groups. So I'd settle for something like the following: namely, a sequence of groups $G_n$ defined in some combinatorial way, and maps $f_n: G_n\to \pi_{n+k}(S^k)$ or $g_n: \pi_{n+k}(S^k)\to G_n$ such that - -$f_n$ or $g_n$ are nontrivial for infinitely many $n$, -The maps are related in some way to the constructions 2-4 above, and -The $G_n$ are combinatorially interesting. - -One such example is $G_n:=H_n(\Sigma_\infty, \mathbb{Z})$ with $g_n$ the Hurewicz map (whence the interpretation of $\pi_{2+k}(S^k)$). But even in this case, the combinatorial meaning is sort of mysterious (to me at least) for large $n$. - -REPLY [24 votes]: Write $S = \bigsqcup_n BS_n$ for the symmetric monoidal category of finite sets and bijections under disjoint union, and write $\mathbb{S}$ for the sphere spectrum, thought of as a symmetric monoidal $\infty$-groupoid. There is a natural map -$$S \to \mathbb{S}$$ -of symmetric monoidal $\infty$-categories. I'll interpret "combinatorics of finite sets or symmetric groups" to mean $S$, and so I'll interpret the question to mean "what does this map tell us about $S$?" The guiding principle for everything that follows is the following pair of universal properties, which in particular tell us where the above map comes from: - -$S$ is the free symmetric monoidal $\infty$-category on a point, while $\mathbb{S}$ is the free symmetric monoidal $\infty$-groupoid with inverses on a point. - -The punchline is the following: - -$S$ naturally acts on all symmetric monoidal $n$-categories, and on symmetric monoidal $n$-groupoids with inverses, this action naturally factors through the $n^{th}$ truncation $\mathbb{S}_{\le n}$ of the sphere spectrum. - -$n$ is, of course, allowed to take the value $\infty$. The $n$-truncated map $S \to \mathbb{S}_{\le n}$ can be thought of as a "higher sign character" following e.g. Ganter-Kapranov. -I apologize in advance for the extreme length of this answer, but I'm making some very formal statements and they won't count for much if I can't make them at least a little concrete. - -Let's start with the case $n = 1$. (The case $n = 0$ is an exercise.) Any symmetric monoidal category $(C, \otimes)$ admits a natural family of endofunctors -$$k : V \mapsto V^{\otimes k}$$ -which themselves admit a natural family of natural automorphisms given by the symmetric groups -$$S_k \ni \sigma : V^{\otimes k} \to V^{\otimes k}.$$ -A concise way of summarizing the compatibilities these things satisfy is that any symmetric monoidal category $C$ is naturally a right module category over the free symmetric monoidal category $S$ on a point (with a second monoidal structure). This is because, by the universal property, there is a natural identification -$$[S, C] \cong C$$ -(where here $[S, C]$ denotes the category of symmetric monoidal functors $S \to C$), and so $[S, S]$ naturally acts on $C$ by precomposition. But by the universal property we have $[S, S] \cong S$, with composition giving us a second monoidal structure on $S$ naturally making it a monoid object in symmetric monoidal categories (with respect to a suitable tensor product). -If $C$ is a symmetric monoidal category, let $C^{\times}$ denote the subcategory of invertible objects in $C$ and isomorphisms between these, and let's restrict our attention to this subcategory. For invertible $V$, the endofunctors $k : V \mapsto V^{\otimes k}$ have natural pointwise inverses -$$-k : V \mapsto (V^{-1})^{\otimes k}$$ -and so the natural action of $\pi_0(S) \cong \mathbb{Z}_{\ge 0}$ naturally extends to an action of $\mathbb{Z}$. This $\mathbb{Z}$ is, of course, $\pi_0(\mathbb{S})$. -By invertibility, the natural map $\text{Aut}(1) \to \text{Aut}(V^{\otimes k})$ is an isomorphism. By the Eckmann-Hilton argument, $\text{Aut}(1)$ is an abelian group, and so the natural action of $S_k$ on $V^{\otimes k}$ must factor through its abelianization, and hence through the sign character $S_k \to \mathbb{Z}_2$. Moreover, these sign actions on $V^{\otimes k}$ determine and are determined by a sign action on $1$, and naturally give rise to sign actions on $(V^{-1})^{\otimes k}$ by dualizing. -This $\mathbb{Z}_2$ is, of course, $\pi_1(\mathbb{S})$, and a concise summary of the above pair of observations, together with all of the compatibilities relating them, is that the action of $S$ on symmetric monoidal groupoids with inverses factors through the natural map to the $1$-truncation $\mathbb{S}_{\le 1}$ of the sphere spectrum. This says a bit more than just that the natural actions of $S_k$ factor through their abelianizations: it also says that the sign character is compatible with the natural maps $S_k \times S_{k'} \to S_{k + k'}$, as well as that the data of the sign actions on $V^{\otimes k}$ can be packaged into the data of an operation -$$\pi_0(C^{\times}) \ni V \mapsto \sigma \in \text{Aut}(V^{\otimes 2}) \cong \text{Aut}(1) \cong \pi_1(C^{\times})$$ -where $\sigma$ is the nontrivial element of $S_2$ acting on $\text{Aut}(V^{\otimes 2})$. This operation can be thought of as assigning every invertible object in $C$ its dimension, and the fact that $\pi_1(\mathbb{S}) \cong \mathbb{Z}_2$ implies in particular that the dimension of an invertible object is always $2$-torsion. -Example. Let $C$ be the symmetric monoidal category of super vector spaces over a field $k$ equipped with the tensor product. This operation assigns to the even invertible super vector space $1 \in \text{Aut}(1) \cong k^{\times}$ and assigns to the odd invertible super vector space $-1$. - -The story for general $n$ (including $n = \infty$) is formally identical but takes more work to unpack. Again any symmetric monoidal $n$-category admits a natural action of $S$, and for the same reason. If $(C, \otimes)$ is a symmetric monoidal $n$-category, let $C^{\times}$ denote the subcategory of invertible objects in $C$, isomorphisms between these, isomorphisms between isomorphisms, etc. -Now let's try to run the above arguments again for the action of $S$ on $C^{\times}$. It is again true that for invertible $V$, the natural map $\text{Aut}(1) \to \text{Aut}(V^{\otimes k})$ (now a map of $n$-groups) is an isomorphism. The analogue of Eckmann-Hilton here asserts that the group operation on $\text{Aut}(1)$ is symmetric monoidal, so the natural action of $S_k$ on $V^{\otimes k}$ factors through a map -$$S_k \to \text{Aut}(V^{\otimes k}) \cong \text{Aut}(1)$$ -to a symmetric monoidal $n$-group. The "symmetric monoidal $n$-abelianization" of $S_k$ is determined by taking the $n$-truncation of the answer for $n = \infty$, which is the loop space of the free connective spectrum on $BS_k$, or $\Omega \Sigma^{\infty} BS_k$. -I don't know much about that space, but fortunately we don't have to deal with it because we haven't used all of the available structure: we're talking about the individual symmetric groups instead of talking about $S$, so let's talk about $S$. The identification $[S, C] \cong C$ says that symmetric monoidal functors $S \to C$ can be identified with objects of $C$. If we restrict our attention to $C^{\times}$, then symmetric monoidal functors $S \to C^{\times}$ naturally factor through the "group completion" of $S$ (by which I mean the left adjoint to the inclusion of symmetric monoidal $\infty$-groupoids with inverses into symmetric monoidal $\infty$-categories), which Barratt-Priddy-Quillen-Segal tells us is the sphere spectrum $\mathbb{S}$. If $C$ is an $n$-groupoid then we can further factor through the $n$-truncation $\mathbb{S}_{\le n}$. Note that if you believe that $S$ and $\mathbb{S}$ have the universal properties that I said they did, then BPQS is a formal consequence of those universal properties. On the other hand, while it's not hard to show that $S$ is the free symmetric monoidal category on a point, it's less clear why $S$ is the free symmetric monoidal $\infty$-category on a point. - -Okay, but that was a lot of formal junk. What am I actually saying for $n = 2$? Here $C$ is a symmetric monoidal $2$-category and we're considering the action of the symmetric groups $S_k$ on the tensor powers $V^{\otimes k}$ of some invertible object in $C^{\times}$. Write -$$\varphi : S_k \to \text{Aut}(V^{\otimes k})$$ -for the corresponding morphism of $2$-groups. The starting observation is that part of what it means to be a morphism of $2$-groups is that $\varphi$ does not in any sense just "strictly preserve" the group operation on both sides, in the sense that we don't just have -$$\varphi(\sigma_1 \sigma_2) = \varphi(\sigma_1) \varphi(\sigma_2).$$ -What we have instead is a family of $2$-morphisms -$$\eta_{\sigma_1, \sigma_2} : \varphi(\sigma_1 \sigma_2) \to \varphi(\sigma_1) \varphi(\sigma_2)$$ -satisfying the coherence condition that the two ways of going from $\varphi(\sigma_1 \sigma_2 \sigma_3)$ to $\varphi(\sigma_1) \varphi(\sigma_2) \varphi(\sigma_3)$ coincide. As previously explained here, in the special case that the only element of $\text{Aut}(V^{\otimes k})$ is $\text{id}_{V^{\otimes k}}$ (this is an evil condition but it's easier to say things this way) this family of $2$-morphisms is precisely a $2$-cocycle on $S_k$ with values in $\text{Aut}(\text{id}_{V^{\otimes k}})$, and the coherence condition is precisely the $2$-cocycle condition. -$\pi_2(\mathbb{S})$ describes how nontrivial these cocycles can be subject to the condition that the natural action of $S_k$ on $\text{Aut}(V^{\otimes k}) \cong \text{Aut}(1)$ is coherently compatible with the symmetric monoidal structure $S_k \times S_{k'} \to S_{k + k'}$. These compatibilities imply, in particular, that we can package all of the morphisms $S_k \to \text{Aut}(V^{\otimes k})$ into a single morphism -$$S_{\infty} \to \text{Aut}(1)$$ -of $2$-groups. We can trivialize this map on $\pi_0$ by passing from $S_{\infty}$ to the infinite alternating subgroup $A_{\infty}$, which has the effect of letting us ignore the non-identity elements of $\text{Aut}(1)$. We then get a natural map -$$A_{\infty} \to B \text{Aut}(\text{id}_1) \cong B \pi_2(C^{\times})$$ -of $2$-groups, and the fact that $\pi_2(\mathbb{S}) \cong \mathbb{Z}_2$ says that the universal map from $A_{\infty}$ to a connected $2$-group (a $2$-group of the form $BA$, for $A$ an abelian group) is a map -$$A_{\infty} \to B \mathbb{Z}_2.$$ -By universal coefficients, this is in turn equivalent to Lurie's assertion in the comments that -$$\pi_2(\mathbb{S}) \cong H_2(BA_{\infty}, \mathbb{Z}) \cong \mathbb{Z}_2$$ -is the Schur multiplier of $A_{\infty}$. In the same way that passing from $S_{\infty}$ to $A_{\infty}$ universally trivializes the action of $S_{\infty}$ at the level of $\pi_0$, passing from $S_{\infty}$ to $A_{\infty}$ to the universal central extension $\widetilde{A}_{\infty}$ universally trivializes the action of $S_{\infty}$ at the level of $\pi_0$ and $\pi_1$. - -The generator of $\pi_2(\mathbb{S})$ can also be thought of as describing an operation -$$\pi_0(C^{\times}) \to \pi_2(C^{\times})$$ -which assigns to every invertible object of $C$ its "$2$-dimension." This is, in a very precise sense, the dimension of the dimension: the ring spectrum structure on $\mathbb{S}$ induces in particular a multiplication -$$\pi_1(\mathbb{S}) \times \pi_1(\mathbb{S}) \to \pi_2(\mathbb{S})$$ -and it's known that the generator of $\pi_2(\mathbb{S})$ is the square of the generator of $\pi_1(\mathbb{S})$. This means the above operation factors as a composite -$$\pi_0(C^{\times}) \to \pi_1(C^{\times}) \to \pi_2(C^{\times})$$ -where in the first step we take the dimension of an invertible object $V$, which is itself an invertible object in $\text{Aut}(1)$, and then take the dimension again. -Example. Let $C$ be the symmetric monoidal $2$-category of $k$-algebras, $k$-bimodules, and $k$-bimodule homomorphisms, where $k$ is a commutative ring. It turns out to be possible to describe the dimension of any object of $C$, not just the invertible ones, and it turns out to be the zeroth Hochschild homology -$$H_0(A) = A / [A, A].$$ -$H_0$ respects the tensor product of $k$-algebras in the sense that $H_0(A \otimes_k B) \cong H_0(A) \otimes_k H_0(B)$, so in order for a $k$-algebra to be invertible (which here means Morita invertible) it must in particular have invertible $H_0$. The sequence of abelian groups given by taking $H_0$ and then taking the dimension -$$\pi_0(C^{\times}) \to \pi_1(C^{\times}) \to \pi_2(C^{\times})$$ -can in fact be identified with a natural sequence of abelian groups -$$\text{Br}(k) \to \text{Pic}(k) \to k^{\times}.$$ -Subexample. To see a case where the composite of the two maps above is nontrivial, let $k = \mathbb{R}$, but now consider super $k$-algebras, super $k$-bimodules, and super $k$-bimodule homomorphisms. Then -$$\pi_0(C^{\times}) \cong \text{SBr}(\mathbb{R}) \cong \mathbb{Z}_8$$ -$$\pi_1(C^{\times}) \cong \text{SPic}(\mathbb{R}) \cong \mathbb{Z}_2$$ -$$\pi_2(C^{\times}) \cong \mathbb{R}^{\times}.$$ -where the $S$ stands for "super." The super Brauer group is generated by a Clifford algebra -$$\text{Cl}(1) \cong \mathbb{R}[e]/(e^2 = -1)$$ -where $e$ has degree $1$. The zeroth Hochschild homology of this Clifford algebra is -$$H_0(\text{Cl}(1)) \cong \mathbb{R}[e]/(e^2 = -1, [e, e] = 2 e^2 = 0) \cong \mathbb{R} e;$$ -that is, it's a $1$-dimensional odd vector space, and hence its image in both $\text{SPic}(\mathbb{R})$ and $\mathbb{R}^{\times}$ is $-1$ as above. - -Okay, but so far I've more or less repackaged things that other people have already said. To really justify having written all of that let's do $n = 3$. -As far as combinatorics goes, we could talk about coherence data for compatible families of maps from the symmetric groups $S_k$ to symmetric monoidal $3$-groups, and we'd eventually learn (I think) that $\pi_3(\mathbb{S}) \cong \mathbb{Z}_{24}$ can be identified with the third homology -$$H_3(B \widetilde{A}_{\infty}, \mathbb{Z})$$ -of the universal central extension of $A_{\infty}$, but at this point it seems prudent to introduce a better language for extracting something understandable from all of the structure available here. I'll use the language of framed cobordism. The natural map $S \to \mathbb{S}$ naturally factors through a third object -$$S \to \text{Bord} \to \mathbb{S}$$ -where $\text{Bord}$ is the symmetric monoidal $\infty$-category of framed cobordisms. That is, it has objects framed points, morphisms framed $1$-cobordisms between points, etc. Like $S$ and $\mathbb{S}$, it has a universal property explaining where the above maps come from, which is a variant of the cobordism hypothesis: - -$\text{Bord}$ is the free symmetric monoidal $\infty$-category with duals on a point. - -The point is that a symmetric monoidal $\infty$-groupoid with inverses is in particular a symmetric monoidal $\infty$-category with duals. -The natural map $\text{Bord} \to \mathbb{S}$ induces a map -$$\pi_n(\text{Bord}) \to \pi_n(\mathbb{S})$$ -which is precisely the Pontryagin-Thom isomorphism from the group of cobordism classes of closed framed $n$-manifolds to the $n^{th}$ stable homotopy group of spheres. A generalization of the claims I made about dimensions and "$2$-dimensions" above is that if $C$ is any symmetric monoidal $\infty$-category, then there is a natural map -$$\pi_n(\mathbb{S}) \times \pi_0(C^{\times}) \to \pi_n(C^{\times})$$ -but since the generator of $\pi_1(\mathbb{S})$ does not cube to a generator of $\pi_3(\mathbb{S})$ and its higher powers vanish, it is not possible to interpret these operations as arising from iterating the construction of the dimension, and we need something new. That something new is the observation that the same argument that shows that every symmetric monoidal $\infty$-groupoid with duals is naturally a module over $\mathbb{S}$ shows that every symmetric monoidal $\infty$-category with duals is naturally a module over $\text{Bord}$. Hence if $C$ is any symmetric monoidal $\infty$-category, then the above map refines to a natural map -$$\pi_n(\text{Bord}) \times \pi_0(C^{fd}) \to \pi_n(C^{fd})$$ -where $C^{fd}$ denotes the subcategory of fully dualizable objects, isomorphisms between these, etc. In terms of the usual cobordism hypothesis, this map comes from taking a fully dualizable object $V$, making it the value at a point of an $n$-dimensional fully extended topological field theory, and then evaluating the TFT on a framed $n$-manifold $M$ representing an element of $\pi_n(\text{Bord})$. If $C$ is chosen to be a suitable Morita category of $E_n$ algebras then this operation recovers factorization homology. For lack of a better term I'll call it factorization homology in general and use factorization homology notation for it: -$$\pi_n(\text{Bord}) \times \pi_0(C^{fd}) \ni (M, V) \mapsto \int_M V \in \pi_n(C^{fd}).$$ -Hence (with the appropriate framings) $\int_{S^1} V$ is the dimension / Hochschild homology and $\int_{S^1 \times S^1} V$ is the $2$-dimension. This is all we get when $n = 1, 2$ because $\pi_1(\mathbb{S}) \cong \mathbb{Z}_2$ is generated by $S^1$ with the Lie group framing and $\pi_2(\mathbb{S}) \cong \mathbb{Z}_2$ is generated by $S^1 \times S^1$. But $\pi_3(\mathbb{S})$ is not generated by $(S^1)^3$: instead it's generated by $S^3 \cong \text{SU}(2)$ with the Lie group framing, and so the operation we get from a generator of $\pi_3(\mathbb{S})$ is taking factorization homology on $S^3$: -$$\pi_0(C^{fd}) \ni V \mapsto \int_{S^3} V \in \pi_3(C^{fd}).$$ -Since $\pi_3(\mathbb{S}) \cong \mathbb{Z}_{24}$ and the cube of the generator of $\pi_1(\mathbb{S})$ represents the class $12 \in \mathbb{Z}_{24}$, factorization homology of a fully dualizable object on $S^3$ is $24$-torsion and moreover represents a distinguished $12^{th}$ root of the $3$-dimension $\int_{S^1 \times S^1 \times S^1} V$, which is $2$-torsion. -Warning. "Fully dualizable" above is an incredibly strong condition, and in particular it implies being $n$-dualizable for all $n$. Aside from invertible objects, the only examples I can cook up even heuristically are $E_{\infty}$ algebras in a suitable Morita $\infty$-category. -On the other hand, the upshot of the above digression into the cobordism hypothesis is that the operations induced by elements of $\pi_n(\mathbb{S})$, thought of as closed framed $n$-manifolds, make sense on a much more general class of objects than just invertible objects, but in fact on $n$-dualizable objects. So to understand how these operations work we can relax our hypotheses substantially: for example, dimension $\int_{S^1} (-)$ makes sense on $1$-dualizable objects, and these are much easier to find. -Example. Let $C$ be the symmetric monoidal $3$-category of conformal nets in the sense of Bartels-Douglas-Henriques. Then $\pi_3(C^{\times}) \cong U(1)$, and so factorization homology on $S^3$ defines an invariant -$$\pi_0(C^{\times}) \ni V \mapsto \int_{S^3} V \in \pi_3(C^{\times}) \cong U(1).$$ -of invertible conformal nets taking values in the $24^{th}$ roots of unity. There is an invertible conformal net called the free fermion $\text{Fer}(1)$, which (conjecturally) is to tmf what the Clifford algebra $\text{Cl}(1)$ is to K-theory, and in Topological modular forms and conformal nets, Douglas and Henriques show that $\int_{S^3} \text{Fer}(1)$ is in fact a primitive $24^{th}$ root of unity. - -One last comment on in what sense we can connect the above back to combinatorics, or $S$. Roughly speaking, I want to argue that this question doesn't respect the symmetries of the situation. What I mean is that starting from the action of $S$, we can at best write down some $0$-morphisms, some $1$-morphisms, and some coherence data, and out of this data we can try to say something about $\pi_n(\mathbb{S})$ if we can use this data to put together a closed framed $n$-manifold $M$. But the invariant $\int_M V$ itself doesn't depend on a choice of presentation of $M$, and presenting special cases of it this way blinds us to that: for example, it blinds us to a potentially very rich action of the framed diffeomorphisms of $M$ on $\int_M V$.<|endoftext|> -TITLE: How to prove the Hahn-Banach constructively -QUESTION [6 upvotes]: I am just wondering, how to prove the Hahn-Banach theorem constructively for a finite dimensional normed vector space. -Thanks in advance for any helpful answers. - -REPLY [2 votes]: The idea is to show that one can extend a linear functional from an $n$-dimensional space to a space of dimension $n+1$ without increasing its norm. See, for instance, my notes (Lemma E.2) -In fact, by doing so, you can prove THBT constructively for any separable space.<|endoftext|> -TITLE: pushforward of locally free sheaf is locally free? -QUESTION [9 upvotes]: Hi, -Is there an example of a proper smooth map of schemes $f:X\to Y$ and a vector bundle $E$ on $X$ -such that $f_*E$ is not locally free on $Y$? -Thanks - -REPLY [6 votes]: There is even a more simple example (although essentially the same) than in the Anand's answer. Take $C = P^1 = P(V)$, where $\dim V = 2$, let $Y = S^2V$ and $X = C\times Y$ with the map $f:X \to Y$ being just the projection. Take $E$ to be the universal extension of $O(2)$ by $O(-2)$. This means that $E$ fits into exact sequence -$$ -0 \to p^*O(-2) \to E \to p^*O(2) \to 0, -$$ -where $p$ is the projection $X \to C$, and the extension is given by the canonical element in $S^2V\otimes S^2V^* \subset S^2V\otimes k[S^2V] = Ext^1(p^*O(2),p^*O(-2))$. Applying the pushforward via $f$ to the above sequence one gets -$$ -0 \to f_*E \to S^2V^*\otimes O_Y \to O_Y \to R^1f_*E \to 0, -$$ -and it is clear that the middle map is given by the canonical embedding $S^2V^* \to k[S^2V]$. Thus $R^1f_*E$ is the structure sheaf of the point $0 \in Y = S^2V$ and $f_*E$ is the second syzygy sheaf of a point on a 3-dimensional variety, which is the simplest example of a reflexive non-locally-free sheaf. -By the way, the pushforward of a vector bundle under a smooth morphism is always reflexive. This is why the above example is the simplest possible.<|endoftext|> -TITLE: Linear orders with only short cuts -QUESTION [7 upvotes]: Here's a question from Shelah's book Classification Theory. Given an order $I$, we consider cuts of the form $(A,B)$ where for all $a \in A, b \in B$ we have $a < b$ and $A \cup B = I$. The cofinality of a cut $(A,B)$ is the pair $(\lambda, \kappa)$ where $\lambda$ is the cofinality of $A$ and $\kappa$ is the cofinality of $B$ in the reverse order. -The reader is asked to show in Exercise VII.1.7 that given any order $J$ with $|J| \leq \kappa$ there is an order $I$ with $J \subseteq I$ and $|I| = \kappa$ so that if $(A,B)$ is a cut of $I$ then the cofinality of $(A,B)$ is $(\aleph_{0}, \aleph_{0})$. In other words, given any order of size at most $\kappa$, it can be expanded to an order of size $\kappa$ so that every cut has countable cofinality. -Does anyone know how to show this? In the absence of any hypothesis on $\kappa$ it is difficult to see how one could kill off all uncountable cuts in an order of size $\kappa$ without increasing the cardinality. - -REPLY [4 votes]: I don´t think the statement is true as it is. If $J=\omega_1+1$ then for any $I \supseteq J$ (of whatever cardinality) you can define a cut as: $x \in A$ iff $x \in I$ and $\exists \alpha \in \omega_1 (x<\alpha) $ and $B=I \setminus A$. This cut can't have cofinality $(\omega, \omega)$.<|endoftext|> -TITLE: tr(ab)=tr(ba), part 2. -QUESTION [34 upvotes]: This is a Banach space version of Andre Henriques' question - Trace Question -for Hilbert spaces. Let $a:X\to Y$ and $b:Y\to X$ be bounded linear operators between Banach spaces s.t. $ba$ and $ab$ are both nuclear. Assume whatever approximation properties on $X$ and $Y$ that you want (say, assume that both $X^*$ and $Y^*$ have the bounded or even metric approximation property), so that the trace of $ab$ and of $ba$ are well defined. Then must $tr(ab)=tr(ba)$? -When $X$ and $Y$ are Hilbert spaces, you can find three correct proofs and one interesting but incomplete proof at the above link. None of these generalize immediately to the Banach space setting. -Caveat: I have not done a literature search or thought much about this problem, but it is natural to consider it after reading Andre's question. - -REPLY [21 votes]: My question has a negative answer. -Lemma. Suppose $X$ has the approximation property (AP), $Y$ is a subspace of $X$, and $X/Y$ fails the AP. Then there is a nuclear operator $T$ on $X$ s.t. $TX\subset Y$, $T^2=0$, and $tr(T)=1$. -Suppose you have $X$, $Y$, $T$ as in the lemma and $Y$ has the AP. Define $a:X\to Y$ to be $T$ considered as an operator into $Y$ and let $b:Y\to X$ be the inclusion map. Then $ba=T$ has trace one but $ab=0$. -Experts will see immediately that you can realize the situation in the previous paragraph by letting $Z$ be a James-Lindenstrauss space s.t. $Z^{**}/Z$ fails the AP while $Z^{**}$ and $Z$ have Schauder bases. More remarkable is that you can even have $X=\ell_p$ with $1 -TITLE: Jet bundles and partial differential operators -QUESTION [30 upvotes]: A geometric way of looking at differential equations -In the literature for the h-principle (for example Gromov's Partial differential relations or Eliashberg and Mishachev's Introduction to the h-principle), we often see the following (all objects smooth): - -Give a fibre bundle $\pi:F\to M$ over some manifold $M$, denote by $F^{(r)}$ the associated $r$-jet bundle. A partial differential relation $\mathcal{R}$ is a subset of $F^{(r)}$. -$\mathcal{R}$ is said to be a partial differential equation if it is a submanifold of $F^{(r)}$ with codimension $\geq 1$. -A solution $\Phi$ to the relation $\mathcal{R}$ is a holonomic (in the sense that $\Phi = j^r\phi$ for some section $\phi$ of $F$) section of $F^{(r)}$ that lies in $\mathcal{R}$. - -This description is very powerful in the context of the topologically motivated techniques of the h-principle. And for partial differential inequalities where $\mathcal{R}$ is an open submanifold, this allows the convenient setting for the Holonomic Approximation Theorem. -A geometric way of looking at differential operators -Here I recall the famous theorem of Peetre. - -Let $E\to M$ and $F\to M$ be two vector bundles. Let $D$ be a linear operator mapping sections of $E$ to sections of $F$. Suppose $D$ is support-non-increasing, then $D$ is (locally) a linear differential operator. - -where - -A linear differential operator $D:\Gamma E\to\Gamma F$ is a composition $D := i\circ j$ where $j: E\to J^RE$ is the $R$-jet operator, and $i: J^RE \to F$ is a linear map of vector bundles. - -Most of the time in applications, $F$ can be taken to be the same bundle as $E$. This way of phrasing things is also convenient for analysis. For example, we can easily define the principal symbol of a linear differential operator in the following way. -Let $D_1$ and $D_2$ be two linear differential operators of order $r$ (that is, $r$ is the smallest natural number such that if $j^r\phi = j^r\psi$, then $D\phi = D\psi$). Let $\pi^r_{r-1}: J^rE \to J^{r-1}E$ the natural projection. We say that $D_1$ and $D_2$ have the same principal part if their corresponding $i_1 - i_2|_{\ker \pi^{r}_{r-1}} \equiv 0$. (In words, their difference is a l.d.o. of lower order. This defines an equivalence relation on linear differential operators of order $r$. Each equivalence class defines a unique linear map of vector bundles $P: \ker \pi^{r}_{r-1} \to F$. -Now, it is known (sec 12.10 in Kolar, Michor, Slovak's Natural operations...) that $\ker \pi^{r}_{r-1}$ is canonically isomorphic to $E\otimes S^r(T^*M)$, where $S^r$ denotes the $r$-fold symmetric tensor product of a space with itself. So we have naturally an interpretation of $P$ as a section of $F\otimes E^* \otimes S^r(TM)$. In the case where $E$ and $F$ are just, say, the trivial $\mathbb{R}$ bundle over $M$, we recover the usual description of the principle symbol of a pseudo-differential operator being (fibre-wise) a homogeneous polynomial of the cotangent space. -Question -Given the above, another way of looking at partial differential equations is perhaps the following. -Let $\pi_X: X\to M$ and $\pi_Y: Y\to M$ be fibre bundles. A system of partial differential operators of order $r$ is defined to be a map $H: J^rX \to Y$ that commutes with projection $\pi_X \circ \pi^r_0 = \pi_Y \circ H$. And a system of partial differential equations is just the statement $H(j^r\phi) = \psi$, where $\phi\in\Gamma X$ and $\psi \in\Gamma Y$. Observe that by considering $H^{-1}(\psi) \subset J^rX$, we clearly have a partial differential relation the sense defined in the first section. If we require that $H$ is a submersion, then $H^{-1}(\psi)$ is also a partial differential equation in the sense defined before. -On the other hand, if $\mathcal{R}$ is an embedded submanifold of $J^rX$, at least locally $\mathcal{R}$ can be written as the pre-image of a point of a submersion; though there may be problems making this a global description. So perhaps my definition is in fact more restrictive than the one given in the first part of this discussion. -My question is: is this last "definition" discussed anywhere in the literature? Perhaps with its pros and cons versus the partial differential relations definition given in the first part of the question? I am especially interested in references taking a more PDE (as opposed to differential geometry) point of view, but any reference at all would be welcome. -Note: For the reference-request part of this question, I would also appreciate pointers to whom I can ask/e-mail on these matters. - -REPLY [12 votes]: If $\mathcal{R}\subset J^rX$ is closed, then there's a smooth function $f:J^rX\to\mathbb R$ with $\mathcal{R}=f^{-1}(0)$. So you can construct a differential operator $H:J^kX\to M\times \mathbb{R}$ by $H(\theta):=(\pi_X(\pi^r_0(\theta)),f(\theta))$ and the equation $\mathcal{R}$ will be given by $H(j^r\phi)=0$. -So there is no big difference between the two definitions. If you are only interested in the "space" of solutions of the differential equation, then i'd say that the set $\mathcal{R}$ is enough, or put differently, you could choose the differential operator which suits you best to represent the equation. -Edit In response to Willies comment: Here's a counterexample to what you are asking for: recall that there's no submersion from $\mathbb{RP}^2$ to something, which has $\mathbb{RP}^1\subset \mathbb{RP}^2$ as a fiber. So take $M=\mathbb{R}$, $X=M\times \mathbb{RP}^2$ and $\mathcal{R}=M\times \mathbb{RP}^1\subset X$. Then there's no fiber bundle $Y\to M$ allowing a submersion $H:X\to Y$ with $\mathcal{R}=H^{-1}(\psi)$ for any $\psi\in \Gamma(Y)$. This is probably a silly example since the PDE is of order zero, but I'm sure one can come up with examples in higher order. -Anyway: if what you are interested in is an intrinsic notion of overdeterminedness of a PDE you might want to take a look at Bryant and Griffiths Characteristic Cohomology of Differential systems. Roughly the codimension of the characteristic variety serves as such a measure. And the characteristic variety can be defined completely without referring to an operator describing the equation. As Deane says, much of this can be found in the book Exterior differential systems. -There are also the books by Vinogradov, Krasil'shchik and Lychagin.<|endoftext|> -TITLE: Growing random trees on a lattice $\rightarrow$ Voronoi diagrams -QUESTION [29 upvotes]: Imagine growing trees from $k$ seeds on a square $n \times n$ region -of $\mathbb{Z}^2$. -At each step, a unit-length edge $e$ between two points of -$\mathbb{Z}^2$ is added. -The edge $e$ is chosen randomly among those that have one endpoint -touching an existing tree and the other endpoint not touching -any tree. Edge $e$ then extends the tree $T$ it touches, maintains -$T$ a tree, and does not join to another tree. -The process is repeated until no such edges remain. -Here is a view of (a portion of) the trees grown from three seeds: - -                    - - -I expected the trees to intertwine and tangle with one another -as they grew, -but instead they appear to be approximating the Voronoi -diagram of the seeds. -Here is an example (of which the above is a detail) -of $k=10$ trees grown within a $100 \times 100$ square: - -      - - -I have overlaid the Voronoi diagram of the seeds. -My primary question is: - -Q1. Is it known that a tree-growth process like - mine approximates the Voronoi diagram in some sense? - -Short of this, I would like to correct my faulty intuition: - -Q2. Why do the trees not entangle more significantly - where they meet? - -Finally: - -Q3. Are there any results known on the structure of - the individual grown trees? For example, their expected height, perhaps - as a function of their Voronoi region shape? - (Not surprisingly, the tallest tree ($h=110$) above is the mauve one - with root $(61,41)$ that encompasses the upper right corner.) - -Although the book Percolation has a chapter on -"Random Voronoi Percolation," -I have not been successful in matching their models to mine -closely enough to draw any definite conclusions. - -                    - - -Thanks for any pointers to relevant literature! - -REPLY [9 votes]: Update: -I think your model for growth has been studied before and goes by the name of the Eden growth model (See section 4 of the (linked) original paper by Murray Eden). It seems what is usually studied is a "site addition" model, whereas your model is a "bond addition" model, but I bet the main results will be the same. I haven't managed to find any recent reviews focusing just on the Eden model yet (and the Wikipedia page is sadly very sparse), but I found a description of it in the first pages of this book chapter of Jean-François Gouyet's "Physics and Fractal Structures". -Famously, the interface of the Eden model was the motivation of Kardar, Parisi and Zhang when they defined the universality class now known as KPZ. Here's a nice review of the KPZ universality class by Ivan Corwin. -Thus a lot is (conjecturally) known about this -- in particular, the size of the fluctuations at the perimeter of a single tree with $N$ bonds should scale as $N^{1/6}$ (fluctuations go as radius$^{1/3}$, and I am pretty sure the radius goes like $N^{1/2}$). Because of this site versus bond addition discrepancy, I haven't found anything about colliding Eden clusters, but I think these ideas can justify that you do get Voronoi cells. I would be delighted to hear criticism or details from an expert! - -I don't have anything rigorous to say yet, but some post-facto intuition for Q2 is as follows. -Consider the case of a single seed. I'll tell a story about why the growth from a single seed ought to look more or less like a disk with a wiggly edge, rather than something with a lot of branchy and spread-out fingers (like something one gets out of diffusion-limited aggregation, for instance). If you buy my story, then it should be clearer why there isn't that much "entanglement" at the interfaces between two such growing bodies. -At a given time step $i$, we have a tree $T_i$; let the set of edges that connect $T_i$ to $\mathbb{Z}^2\setminus T_i$ be $G_i$. Then we may form $T_{i+1}$ by choosing uniformly an edge in $G_i$ and adding it to $T_i$. So far, this is just your process in different terms. Here comes some handwaving. Any "protrusions" on $T_i$ will not get too long, because to grow a subset of $T_i$ which sticks out a significant distance from the rest of $T_i$, we'd have to had chosen edges near the "tip" of this protrusion repeatedly. But of course the tip of a long protrusion has a small perimeter compared to the sides of the protrusion, and thus it's much more likely that added edges will smooth out any such features instead of extending them. -Why does this end up with something disk-like? Your process is basically one where you add a "protrusion" to the tree centered somewhere uniformly random on the "boundary". Thus consider the following "off-lattice" version of your process. Let $T_0$ be a disk of radius 1 centered at the origin. $T_i$ is the union of $T_{i-1}$ with a disk centered at a point $p_i$ chosen uniformly from the boundary of $T_{i-1}$. I hope you can see that this is a stochastic version of normal evolution of the boundary. And as is well-known to doodlers, normal evolution tends towards to circular disks, so "intuitively" a stochastic version will tend to a wiggly disk. (One complication here is that your process isn't allowed to form loops in the tree, whereas the continuum version I have in mind does).<|endoftext|> -TITLE: expectation of supremum -QUESTION [6 upvotes]: Hello, -Suppose $(X_{n}(t))_{n\geq 1}$ is a sequence of real valued stochastic processes, and $T>0$ a fixed number. -Do we have the following implication ? -$\displaystyle{ \lim_{n \to \infty} \sup_{t\in[0,T]}} \mathbb{E}[|X_n(t)|] =0$ implies $\displaystyle{ \lim_{n \to \infty} \mathbb{E}[\sup_{t\in[0,T]}}|X_n(t)|] =0$ -If not, what are the weakest conditions on $X_n(t)$ such that the above implication is true ? - -Edit 2 : is the implication true if -\begin{equation} -\mathbb{E}\left[\displaystyle{\sup_{n>0}}\ |X_n(t+h)-X_n(t)|\right]\leq c(h) -\end{equation} -with $\displaystyle{\lim_{h\to 0}}\ c(h)=0$ - -Edit 1 : is the implication true if \begin{equation} -\displaystyle{\sup_{n>0}}\ \mathbb{E}\left[|X_n(t+h)-X_n(t)|\right]\leq c(h) -\end{equation} -with $\displaystyle{\lim_{h\to 0}}\ c(h)=0$. Proven false by Jeff Schenker (cf below). - -REPLY [2 votes]: This is false even with your edit. Here is a counter example with $T=1$. -Let $j$ be a random integer chosen uniformly from $\{0,\ldots,n-1\}$. Let $X_n(t)$ be a piecewise linear function on $[0,1]$ as follows: - -$X_n(t)=0$ if $t\not \in J_n$ where $J_n=[\frac{j}{n},\frac{j+1}{n}]$. -If $t\in J_n$ then the graph of $X_n(t)$ has a "tent shape": it vanishes at each endpoint and increases linearly with slope $2n$ as we move toward the midpoint so that it takes value $1$ at the midpoint. - -The resulting function $X_n(t)$ is piecewise linear on $[0,1]$, bounded by $1$ and the slope of any linear segment is bounded by $2n$. Then - -For each $t$, $\mathbb{E}[|X_n(t)|]\le \frac{1}{n}$ since $X_n(t)\neq 0$ only with probability $\frac{1}{n}$ and $0\le X_n(t)\le 1$. -$\mathbb{E}[\sup_t |X_n(t)|]=1$ since $\sup_t |X_n(t)|=1$ for every outcome. -$ \mathbb{E}[|X_n(t+h)-X_n(t)|]\le C h $ where $C$ is a constant independent of $n$, since $|X_n(t+h)-X_n(t)|\neq 0$ only with probability less than $\frac{c}{n}$ and is bounded by $2 n h$ for every outcome.<|endoftext|> -TITLE: Vanishing of $\hat{A}$ genus and positive scalar curvature -QUESTION [12 upvotes]: Classicly, for a spin Riemannian manifold $M$, the $\hat{A}(M)$ genus will be $0$, if the scalar curvature is positive. -The proof is to use the Lichnerowicz formula. we have the index of the Dirac operator will be $0$, i.e., -$$Ind(D_+)=0.$$ -On the other hand, by the index theorem of Atiyah and Singer, we have -$$Ind(D_+)=\hat{A}(M).$$ -So we get -$$\hat{A}(M)=0.$$ -My question is "Can we have a different method to proof this result? without using the Lichnerowicz formula or without using the index theorem" Maybe a formal proof or explanation. - -REPLY [24 votes]: After thinking about this question for two hours, my belief is strengthened that there cannot be a proof of the vanishing of the A-genus -for spin manifolds with positive scalar curvature without using both the index theorem for the Dirac operator and the Lichnerowicz formula (or other analytic techniques). -In the comments to the question, a proof using the integrality of the $\hat{A}$-genus and Chern-Weil theory has been suggested. I wish to argue -that there is absolute no reason to believe such a proof might work out. -Objections to the use of the integrality theorem. -The manifold $CP^2$ has positive scalar curvature and $\hat{A}$-genus $1 /16$. It is of course not spin. Take the connected sum -of 16 copies of $CP^2$ and call it $M$. The $\hat{A}$-genus of $M$ is $1$; $M$ is still not spin. The famous surgery lemma by Gromov and Lawson -implies that $M$ has a metric of positive scalar curvature, because $M$ is obtained from the manifold $\coprod_{16} CP^2$ -by a sequence of surgeries of codimension $4 \geq 3$. So there are connected pscm manifolds with integral but nonzero $\hat{A}$-genus. More generally, -it can be shown that any simply connected, high-dimensional, oriented and non-spin manifold admits a metric of positive scalar curvature. This also -goes back to Gromov and Lawson, -and Sven F\"uhring worked out the details in his thesis: -http://www-m10.ma.tum.de/bin/view/Lehrstuhl/SvenFuehring#EigenePublikationen -This result shows that the integrality of the $\hat{A}$-genus has nothing to do with positive scalar curvature and that the spin hypothesis has to be used in a more subtle way. -Objections to the use of Chern-Weil theory: -At a first glance, it seems tempting to use Chern-Weil theory for such questions, because it expresses characteristic classes in terms of curvature. Moreover, -a conceptual proof of the Gauss-Bonnet theorem can be given with help of this theory. But an examination of the argument quickly shows that this -is a lucky accident. The first accident is that the Euler class has a direct -geometric-topological meaning, which allows to relate it to another topological invariant (the Euler number), while the complicated rational polynomial in the Pontrjagin classes has no such meaning - (Paul Siegel alluded to this point in his comment). -The second accident is that the integrand of the Gauss-Bonnet-Chern theorem only depends on the curvature tensor and not on the metric itself -(though to define the integrand, -one needs to know that the connection preserves some metric). -The problem with scalar curvature is that it depends on both, the connection and the metric, while the Chern-Weil forms only depend on the connection and not on the metric. -Of course there is a universal formula for the $\hat{A}$-form of any Riemann manifold $(M,g)$ of the form -$\hat{A} = f(jet^2 g) \cdot vol_g$, where $f$ is an algebraic function of the $2$-jet of the metric. I have not worked out the concrete formula for $f$; -for $4$-dimensional manifolds, this should not be too complicated and might be a worthy exercise. I predict that you won't find a formula for $f$ -that depends on the scalar curvature in a controllable way. Another remark is that if such a formula gives a result for positive scalar curvature, -it should give a result for negative scalar curvature as well. On the other hand, any manifold of large dimension has a metric of negative scalar curvature (Lohkamp). -The conclusion I draw is that Chern-Weil theory is insensitive to the sign of scalar curvature. The next problem is that Chern-Weil theory is -insensitive to the spin condition as well: the map $Sym^{\ast} (\mathfrak{so}_n)^{SO(n)}\to Sym^{\ast} (\mathfrak{spin}_n)^{Spin(n)}$ -on the space of invariant polynomials is an isomorphism. In other words, the spin condition does not yield any relation between Chern-Weil classes. The integrality of the $\hat{A}$-genus cannot be obtained -by a local consideration; this would equate it to another manifestly integral class, which is not the case. -The view that positivity should enter the proof in a more global way is furthermore supported by Gromov's h-principle, a special case of which states that -any open manifold (no compact component) has a metric of positive scalar curvature. This implies -that any obstruction to positive scalar curvature has to arise in an essential global way. -My conclusion of the above discussion is that the integrality result forgets the relevant information from the spin condition and that -Chern-Weil theory is not the -correct recipient for the information coming from the positivity of the curvature. Philosophically, integrality is too global to capture the spin condition and Chern-Weil theory is too local to capture the sign of the curvature. -It is a pity that Misha Gromov is not active on this site..... -EDIT: here is an example which shows that any hope to use Chern-Weil theory to obtain information on scalar curvature is in vain. -Fact 1: Let $\nabla_g$ be the Levi-civita connection of the Riemann manifold $(M,g)$, then for all $c>0$, $\nabla_{cg} = \nabla_g$. -Fact 2: the scalar curvature of $(M,cg)$ is $c^{-1} scal_g$. Fact 3: the scalar curvature of a product is the sum of the scalar curvature. -Now take $S^2$, $g_0$ the standard metric with $scal_{g_0}=1$; $F$ a genus $\geq 2$ surface with metric $g_1$ and $scal_{g_1}=-1$. -Then the product $M=S^2 \times F$ with product metric $(c_0 g_0 )\times (c_1 g_1)$ has scalar curvature $c_{0}^{-1} - c_{1}^{-1}$. Suitable choices of $c_i$ yield any real value for the scalar curvature. But by fact 1, the Levi-Civitta connection and hence the Chern-Weil invariants do not depend on $c_i$.<|endoftext|> -TITLE: Is [mD] very ample if D is ample? -QUESTION [10 upvotes]: Let D be an ample R-divisor, is the round down [mD] very ample for any sufficiently divisible number m? -I think it's true. But I do not know how to arrange an argument. - -REPLY [5 votes]: I am not sure if this is the shortest proof, but I think that it is a proof. -Let $A=$ very ample line bundle. After replacing D by a multiple, you may assume that -$$C=D - K_X - (n+1) A$$ is ample where $n=\dim X$. -By Angehrn and Siu, we know that $$K_X+(n+1)A + \text{(ample line bundle)}$$ is very ample. -Now -$$[mD] = K_X+ (n+1) A + C + [mD] - D$$ -and you just need to make sure that $[mD]+C-D$ is ample if $m\gg 0$. This is easy to check by Diophantine approximation.<|endoftext|> -TITLE: A book about model theory -QUESTION [7 upvotes]: I am looking for a good book about model theory. As this is obviously too vague, let me -explain what I am looking for and why. -First I am interested about the basics and foundations of model theory. Right now I am not interested in their applications (like proving things in mathematics or even independence results like Cohen's -- but of course it is not a problem if a book deals with some of these applications if it does not only that). -Second, until a few days ago I believed I knew well enough what was a model. -But since two days I am not so sure. I have a problem with the notion of model inside ZF (or ZFC, or any formalized set theory) of a theory, and in particular, with the meaning of the satisfaction relation in this case. I would like a book which treats that aspect. -These problems arose with my trying to understand the answers to my question -A meta-mathematical question related to Hilbert tenth problem -I am currently having endless discussions in comments about that notion what I am said doesn't make sense to me (and the converse is clearly also true). A good book will certainly help me and save time for my respectable interlocutors. Thanks... - -REPLY [2 votes]: I might be completely mistaken, but I dare say you should take a look at set theory first, maye that could clarify your questions. -I'd recommend the first four chapters of Kenneth Kunen's book, at least they helped my understanding, but maybe that's not the thing you had in mind. -In particular: -Chapter 1, §14: "Formalizing the metatheory" -Chapter 4, §9: "Model theory in the metatheory" -Chapter 4, §10: "Model theory in the formal theory"<|endoftext|> -TITLE: When are constructible points closed? -QUESTION [5 upvotes]: Let $X$ be a scheme. What technical hypotheses must be imposed on $X$ to assure that a point $p \in X$ is closed if and only if the 1-point set $\{p\}$ is constructible? - -REPLY [12 votes]: Let $X$ be locally noetherian. Then $\{x\}$ is constructible if and only if $\{x\}$ is locally closed (for non-noetherian schemes the notion of constructibility is more complicated and all kind of terrible things can happen, e.g. there exist closed points $x$ such that $\{x\}$ is not constructible). Moreover it is a nice exercise that this is the case if and only if the canonical morphism ${\rm Spec} \kappa(x) \to X$ is of finite type. On the other hand, $\{x\}$ is closed if and only if the canonical morphism ${\rm Spec}\kappa(x) \to X$ is finite (or, equivalently, a closed immersion). Thus we are looking for a scheme $X$ that has a covering by open affine subschemes $U = {\rm Spec}A$ such that every finitely generated $A$-algebra which is a field is already a finite $A$-algebra. Such rings are called Jacobson rings. They are also characterized by the property that every prime ideal is the intersection of (not necessarily finitely many) maximal ideals. -Upshot: For a locally noetherian scheme $X$ the following properties are equivalent. -(i) For all $x \in X$ the set $\{x\}$ is closed if and only if it is constructible. -(ii) $X$ is Jacobson. -Every scheme locally of finite type over a Jacobson ring is a Jacobson scheme. Examples for Jacobson rings are of course fields but also integral domains of dimension $1$ with infinitely many prime ideals. Thus every scheme locally of finite type over a field is Jacobson (as already remarked by Damian) but also every scheme locally of finite type over ${\mathbb Z}$ is Jacobson.<|endoftext|> -TITLE: What is the limit of the "knight" distance on finer and finer chessboards? -QUESTION [14 upvotes]: Consider the "infinite chessboard" on the plane. Think of it as the lattice $X_1:=\mathbb{Z}^2$, and also finer chessboards $X_n$ corresponding to $\frac{1}{n}\cdot \mathbb{Z}^2$, $n\geq 1$. Given two squares (i.e. vertices) $u,v$ of $X_n$ one can define the "knight distance" $d_n(u,v)$ as the minimum number of moves that a "knight" (moving as in the usual game of chess) must do to get from $u$ to $v$, divided by $n$. -Now take two points $a,b$ on the plane $\mathbb{R}^2$, and define their "knight distance at step $n$" to be the minimum of $d_n(u,v)$ for $u,v\in X_n$ such that $d_E(u,a)$ and $d_E(v,b)$ are minimal (where $d_E$ is the Euclidean distance on $\mathbb{R}^2$). -Define the "knight distance" on the plane by $d_K(a,b):=\lim_{n\to\infty}d_n(a,b)$. - -Does it define an actual distance (metric) on $\mathbb{R}^2$? Assuming it does, how does the spheres look like for this metric? Any interesting properties? Is there a self-homeomorphism of $\mathbb{R}^2$ that pullbacks $d_K$ to $d_E$? - -(This was a question I happened to ask myself at high school -and never thought really to answer- of which I was reminded of by just reading this MO thread by Joseph O'Rourke. Provided it makes sense, it still looks like a legitimate question to me...) - -REPLY [13 votes]: Let $(x,y)$ and $(x+2a,y+a)$ be points in space. Then clearly the distance between these two points is $a$. Therefore, the unit ball around 0 must contain the octagon with vertices $(2,1)$, $(1,2)$, $(-1,2)$ and so on. -I argue that this is all it contains. To see this, construct linear invariants showing how far you can get with $k$ knight's moves of length $1/k$. For instance, each knight's move increases $x$ by no more than $2/k$, so if $|x_1-x_2|>2$ then $d(x_1,x_2)>1$. Similarly, it increases $x+y$ by no more than 3. With the 6 other linear functionals, one can restrict the unit ball to that octagon. -This metric is a kind of taxicab metric. It is not like the Euclidean metric because there are an infinite number of geodesics of length 1 between $(0,0)$ and $(2,0)$. -So, answers: Yes, octagons, I can't think of any other than the properties of the taxicab metric, no.<|endoftext|> -TITLE: Counting copies of a BA within a BA: arbitrarily many vs infinitely many -QUESTION [8 upvotes]: Informally, I am wondering if a Boolean algebra $\mathcal{B}$ contains infinitely many disjoint copies of a Boolean algebra $\mathcal{A}$ whenever it contains arbitrarily many disjoint copies of $\mathcal{A}$. -More formally, fix Boolean algebras $\mathcal{A}$ and $\mathcal{B}$. Assume that, for each $n \in \omega$, there are $b_1,\dots,b_n \in B$ with $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ for $1 \leq i \leq n$ and $b_i \wedge b_j = 0$ if $i \neq j$. Is there necessarily a sequence $\{ b_i \}_{i \in \omega} \in B$ with $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ for all $i \in \omega$ and $b_i \wedge b_j = 0$ if $i \neq j$? -Though I am curious about the question in the general setting, my primary interest is when $\mathcal{A}$ and $\mathcal{B}$ are both countable. - -Edit (YCor Nov 2019): topological reformulation: let $X,Y$ be Hausdorff compact, totally disconnected spaces (the OP is mostly interested in the case when $X,Y$ are metrizable). Suppose that for every $n$, there are pairwise disjoint clopen subsets $Y_1,\dots,Y_n$ of $Y$, each homeomorphic to $X$. Does it follow that there exists a sequence $(Y_i)$ of pairwise disjoint clopen subsets of $Y$, each homeomorphic to $X$? - -REPLY [2 votes]: This is a nice question from a while ago, and this is only a very partial answer, hoping it will motivate more complete ones. -I'll address the topological equivalent reformulation. It has a positive answer in a few cases described below. First, say that $X$ is good if the question has a positive answer for this given $X$, for every $Y$. -1) If $X$ has a clopen subset homeomorphic to $X\sqcup X$ then $X$ is good. (In particular the answer to the question is positive when $Y=X$). Indeed, in this case, define by induction pairwise disjoint clopen subsets $Y_1,\dots,Y_{n-1},Y'_{n}$ of $Y$ each homeomorphic to $X$, and then choose two disjoint clopen subsets $Y_n,Y'_{n+1}$ of $Y'_n$, each homeomorphic to $X$. -Similarly, the answer to the question is positive if $Y$ has a clopen subset homeomorphic to $Y\sqcup Y$ (or to $Y\sqcup X$). -2) If $X$ embeds as a clopen subset in a finite disjoint union $nX'$ and vice versa, then $X$ is good iff $X'$ is good. -3) If $X$ is countable then $X$ is good. Indeed, the case $X$ empty being trivial, suppose $X$ nonempty and let $\alpha=\alpha_X$ be the largest Cantor-Bendixson rank of a point in $X$; it is achieved by $n$ points, where $n=n_X$ is a positive integer, the pair $(\alpha,n)$ characterizing $X=X(\alpha,n)$ up to homeomorphism. Note that $X$ is then homeomorphic to the disjoint union of $n$ copies of $X(\alpha,1)$, so, in view of (2), let us focus on the case $n_X=1$. -So the condition that $Y$ has $k$ disjoint clopen copies of $X$ means that $Y$ has $\ge k$ points with a countable neighborhood and of Cantor-Bendixson rank $\alpha$. Hence the condition that it has this for arbitrary large $k$ means that $Y$ has infinitely many points with a countable neighborhood and of Cantor-Bendixson rank $\alpha$. Let $(x_n)$ be an injective sequence of such points. By induction let $Y_n$ be a countable clopen neighborhood of $x_n$ disjoint from $\bigcup_{in$). Then $Y_n$ is homeomorphic to $X$ for each $n$, and these are pairwise disjoint clopen subsets. -4) Call $X$-point a point in a topological space a point with a basis of neighborhoods homeomorphic to $X$. If $X$ possesses an $X$-point that is isolated among $X$-points, then then $X$ is good (call this isolated $X$-point). This is actually a generalization of (3) since countable spaces $X$ with $n_X=1$ have this property (the point of maximal Cantor-Bendixson rank being an isolated $X$-point). The argument follows that of (3). An example not covered by the previous items is the case of the space obtained as union of a Cantor space and a discrete countable set accumulating at a single point of this Cantor space. -5) I don't know if $X,X'$ good imply $X\sqcup X'$ good; however this is clear if $X,X'$ have no nonempty homeomorphic subsets. This applies to $X$ Cantor and $X'$ countable. -Actually the few metrizable Stone spaces I can think of, so far, seem to satisfy close variants of the latter arguments. For the classification of metrizable Stone spaces, see notably - Reichbach, M. -The power of topological types of some classes of 0-dimensional sets. -Proc. Amer. Math. Soc. 13 1962 17-23 (Open link).<|endoftext|> -TITLE: Space of metrics with positive sectional curvature -QUESTION [5 upvotes]: Hello; -We know that the space of riemannian metrics on a compact manifold is an open cone in the space of symmetric 2-tensors. -Is it reasonable to think that metrics with positive sectional curvature (even positive at a specific point $x \in M$) also form a convex cone? -This is a question about the local behaviour of metrics, so I am not imposing the condition that the sectional curvature be positive everywhere. -Also, due to certain corollary of this statement for a class of metrics, I am quite certain that this cannot hold for metrics of negative curvature. - -REPLY [18 votes]: No, the formula for curvature is nonlinear with respect to metric tensor in a very essential way. -In particular, -a convex combination of two positively curved metrics can have negative curvature. -In fact, arbitrary large negative sectional curvature. -For example, the induced metric on any embedding $\mathbb{S}^2\hookrightarrow\mathbb{S}^2\times \mathbb{S}^2$, -such that both projections are diffeomorphisms is a convex combination of metrics with constant curvature 1/2. -But it is not hard to make such metric arbitrary bad.<|endoftext|> -TITLE: Determine if circle is covered by some set of other circles -QUESTION [20 upvotes]: Suppose you have a set of circles $\mathcal{C} = \{ C_1, \ldots, C_n \}$ each with a fixed radius $r$ but varying centre coordinates. Next, you are given a new circle $C_{n+1}$ with the same radius $r$ as the previous circles but with a new centre coordinate. -How can you determine whether the area covered by the $C_{n+1}$ is fully covered by $\mathcal{C}$? -How to do this if the circles can have varying radii? -Note: I couldn't yet work out a nice mathematical solution for this. Coming from computer science, the best I could come up with is solving this in a nasty brute force way using some sort of Monte Carlo sampling, i.e. draw a large number of random points from the area of $C_{n+1}$ and then checking for each point if it is enclosed by at least one circle in $\mathcal{C}_{\text{intersecting}}$ (subset of $\mathcal{C}$ with circles that are within $2r$ of $C_{n+1}$). - -REPLY [6 votes]: The problem (for discs of arbitrary radii) is essentially equivalent to the following: given a convex polyhedral set in $\mathbb R^3$ (represented by a system of linear inequalities), determine whether it intersects the sphere $x^2+y^2+z^2=1$. To solve this, one can check if the set is compact and enumerate its vertices to see whether it lies inside the sphere, and minimize $x^2+y^2+z^2$ over the set to find out whether it is completely outside. I am not an expert in complexity but I believe that the former can be done in $O(n^2)$ time and the latter is even faster. -The reduction goes as follows. Construct a stereographic projection from the plane to the sphere in such a way that the main disc $C_{n+1}$ is mapped to the upper hemisphere. Every other disc is mapped to spherical cap which can be represented as the set of solution of a linear inequality (whose coefficients are easy to compute) intersected with the sphere. Add the inequality $z\le 0$ for the lower hemisphere. Now we need to find out whether the union of corresponding half-spaces cover the sphere. Or, equivalently, whether the intersection of their complement half-spaces have an empty intersection with the sphere. Q.E.D. -The fact that the spherical caps arise from discs in the plane (i.e., not from complements of discs) means that the complement half-spaces except the last one contain the south pole $(0,0,-1)$. However this does not help much. If the original radii are equal, then the complement half-spaces also contain the center $(0,0,0)$. This rules out the case when the intersection is outside the sphere and thus simplifies the problem a little bit.<|endoftext|> -TITLE: What is the size of a largest antichain in this poset? -QUESTION [11 upvotes]: Let $[n]:=\lbrace 1, \dots, n \rbrace$. We define a partial ordering on the set of subsets of $[n]$ as follows. We say that $X \preceq Y$ if there is an injective map $f:X \to Y$ such that $x \leq f(x)$ for all $x \in X$. This is a pretty standard construction in poset theory. -The motivation for this question comes from a subset sum problem I've been playing with. Let us regard $[n]$ as the set of indices of a set $A:=\lbrace a_1, \dots, a_n \rbrace$ of numbers (indexed so that $a_1 < \dots < a_n$). If $X \preceq Y$, then the sum of the elements in $A$ corresponding to $X$ is at most the sum of the elements in $A$ corresponding to $Y$. If $X$ and $Y$ are incomparable, then we don't know which sum is bigger (without additional information about $A$). -I would like to cover this poset with as few chains as possible, so it is natural to apply Dilworth's Theorem and then ask - -What is the size of a largest antichain in this poset? - -One natural candidate is to take all subsets of $[n]$ with the same sum $s$. To maximize the size of this antichain, we should take $s$ to be halfway between $0$ and $1+\dots + n$. I'd guess that this is optimal. Any references or thoughts would be much appreciated. - -REPLY [15 votes]: Calling this poset $M(n)$, the fact that it has the Sperner property was conjectured in B. Lindström, "A conjecture on a theorem similar to Sperner's", Combinatorial Structures and Their Applications, p. 241. -It turns out that $M(n)$ has the $k$-Sperner property for all $k$, see R. Stanley's paper "Weyl groups, the hard Lefschetz theorem, and the Sperner property", in the section on the type $B_n$ the properties of $M(n)$ are shown to follow from the general theorems about algebraic posets, the main ingredient being the hard Lefschetz theorem. See also Stanley's article "Some applications of algebra to combinatorics". The papers are available at his website.<|endoftext|> -TITLE: A question on the set of element orders of a finite group -QUESTION [10 upvotes]: Let $G$ be a finite group of order $n$ and denote by $\pi_e(G)$ the set of element orders of $G$. What can be said about $G$ if $\pi_e(G)$ forms a sublattice of the lattice of divisors of $n$? - -REPLY [4 votes]: The obvious conjecture following Mark Sapir's post is that $\mathcal{C}$ consists just of the finite nilpotent groups. That is false. Let $P$ be a nonabelian group of order $p^3$ and exponent $p$, for $p$ an odd prime. Then the groups defined by the presentation below have element orders $\{1,2,p,2p\}$ but are not nilpotent. -$$\langle x,y,z,t \mid x^p=y^p=z^p=t^2=(xt)^2=(yt)^2=1, yx=xyz, xz=zx, yz=zy \rangle$$ -Further question: are there any non-solvable groups in $\mathcal{C}$?<|endoftext|> -TITLE: Linking the residual finiteness of $G$ with $Aut(G)$ or $Out(G)$ -QUESTION [7 upvotes]: There is a classic result of Baumslag which states, -Thm: If $G$ is residually finite then so is $\operatorname{Aut}(G)$. -While Grossman proved the (essentially) analogous result for $\operatorname{Out}(G)$, -Thm: If $G$ is conjugacy separable and every conjugating automorphism is inner then $\operatorname{Out}(G)$ is residually finite. -(A conjugating automorphism is an automorphism $\delta: g\mapsto g^{w_g}$ where $w_g$ is dependent on the $g\in G$ - so every element is sent to a conjugate of itself). -I was wondering if it was possible to "go the other way", so to speak, - -What conditions, if any, can we put on $\operatorname{Aut}(G)$ or $\operatorname{Out}(G)$ to ensure that $G$ is residually finite? - -One obvious condition is that if $\operatorname{Aut}(G)$ is residually finite and $G$ is centerless then $G$ is residually finite. However, you have the added stipulation that $G$ is centerless, which is a condition on $G$ not on $\operatorname{Aut}(G)$. That said, this is a relatively harmless condition on $G$ (checking $Z(G)=1$ is often easier than checking $G$ is residually finite). So I suppose my question can be taken up to "relatively harmless conditions on $G$". - -REPLY [10 votes]: First, Baumslag's result is for finitely generated groups only. HW already says that essentially the Out of a residually finite group can be ``arbitrary". Now if you take a Tarski monster with trivial Out, then the direct product of it with the residually finite group above gives a non-residualy finite group with an arbitrary Out. Another way is to use Minasyan's result from Minasyan, Ashot, Groups with finitely many conjugacy classes and their automorphisms. Comment. Math. Helv. 84 (2009), no. 2, 259–296. He constructed a 2-generated group with two conjugacy classes (hence simple and not residually finite) with arbitrary countable Out. So there is no connection between properties of Out and residual finiteness of the group. -With Aut the situation is more complicated since if the group is not res. finite and does not have center then Aut is not residually finite. On the other hand, there are non-residually finite f.g.groups with residualy finite Aut. For example, Anna Erschler, I think, constructed such a group as a central extension of a Grigorchuk group. See Erschler, Anna -Not residually finite groups of intermediate growth, commensurability and non-geometricity. -J. Algebra 272 (2004), no. 1, 154–172.<|endoftext|> -TITLE: Classification compact Riemann Surfaces -QUESTION [7 upvotes]: I know that all compact Riemann surfaces with the same genus are topologically equivalent. Moreover they are diffeomorphic. But are they biholomorphic, too? -In other words, is the complex structure conserved? - -REPLY [2 votes]: The answers here are great, but I think anyone drawn to this question should find the keyword "Teichmüller space" somewhere and maybe some references. Here is the wikipedia page as a start. -And Chapters 10-12 in A primer on mapping class groups serves as a great introduction. See also here for more suggestions. -Note: with the top of page 10 (ibid) in mind, in general complex smooth varieties that are biholomorphic need not be biregular (see here).<|endoftext|> -TITLE: Gamma spaces and monoidal categories -QUESTION [11 upvotes]: In his paper "Categories and cohomology theories" Graeme Segal gives examples how to construct a Gamma category and therefore also a Gamma space from a strict monoidal category like finite chain complexes of complex fin.-dim. vector bundles with the alternating sum over the dimensions equal to 1 and chain homotopy equivalences as morphisms. This yields a nice model for $BU_{\otimes}$ and proves that this is an infinite loop space. My question is - -Can you still construct a Gamma category if the monoidal structure is not strict, for example if there is a non-trivial associator? If such a Gamma category $C$ exists, what would $C(3)$ and the induced morphisms look like? - -REPLY [6 votes]: If your monoidal category is not strict you can first form a multicategory out of it. This process involves some choices (how to bracket higher tensor products) but they are not essential (see e.g. "Tom Leinster - Higher Categories, Higher Operads", chapter 3.3 for discussion). In nature symmetric monoidal categories very often come from multicategoires, like the category of vector bundles etc. So in some sense the underlying multicategory is the even more natural objects. -As a next step you can extract a gamma category out of a multicategory, see "Permutative categories, multicategories, and algebraic K-theory" by Elmendorf and Mandell. This is a direct generalization of the Gamma category of Segal associated to a permutative category. In particular if your category from the start was permutative it gives back the old construction of Segal.<|endoftext|> -TITLE: Is the sum of two prime ideals in different polynomial rings, K[X_i] and K[Y_i] a prime ideal in K[X_i Y_i]? -QUESTION [11 upvotes]: Let $P$ be a prime ideal in the polynomial ring $K\left[x_1,...,x_m\right]$ and $Q$ be a prime ideal in the polynomial ring $K\left[y_1,...,y_n\right]$. -Is $P+Q$ a prime ideal in $K\left[x_1,...,x_m,y_1,...,y_n\right]$ ? -For example for $Q=\left(y_1,...,y_n\right)$, it is easy to prove that. But how about general cases? - -REPLY [5 votes]: The geometric analogue to darij grinberg's answer is that you are effectively asking whether the product of two particular irreducible algebraic varieties is also irreducible. Here is how the translation works. -The two polynomial rings $K[x_1, \dots, x_m]$ and $K[y_1, \dots, y_n]$ are the coordinate rings of the affine spaces $\mathbb{A}^m_K$ and $\mathbb{A}^n_K$, and their prime ideals $P$ and $Q$ are, respectively, irreducible subvarieties $X$ and $Y$ in these. The combined ring $K[x_1, \dots, x_m, y_1, \dots, y_n]$ is the ring of $\mathbb{A}^{n + m}_K$, which is isomorphic to the product $\mathbb{A}^n_K \times_K \mathbb{A}^m_K$. Considering $P$ and $Q$ as prime ideals in the combined ring is the same as forming the varieties $X \times_K \mathbb{A}^m_K$ and $\mathbb{A}^n_K \times_K Y$, and their intersection is itself isomorphic to $X \times_K Y$. You want to know whether this is necessarily irreducible. -Over an algebraically closed field the answer is "yes". This formally follows from the theorem darij cited, but you can also understand it geometrically in that if, for example, $K = \mathbb{C}$, you can cut the ${}_K$ from the product and convince yourself that since the slices of $X \times Y$ over $Y$ are irreducible (isomorphic to $X$) and of equal dimension, the product itself is also irreducible. (To understand the second condition, think about the reducible variety $\{0\} \times \mathbb{A}^1 \cup \mathbb{A}^1 \times \{0\}$ and its map to $\mathbb{A}^1$ in, say, the second factor.) -Over a non-closed field you have darij's example. It's hard to give a totally geometric picture for this sort of thing, since geometry takes place over a closed field, but you can think about it in a sort of quasi-pictorial way. Namely, the real affine line $\mathbb{A}^1_\mathbb{R}$ (with coordinate ring $\mathbb{R}[x]$) has two kinds of points: those in $\mathbb{R}$, corresponding to prime ideals $(x - a)$, and conjugate pairs of those in $\mathbb{C}$, corresponding to prime ideals $(x^2 + 2bx + c)$ with $b^2 < c$. This means that $\mathbb{A}^1_\mathbb{R}$ is like $\mathbb{A}^1_\mathbb{C}$ with conjugate pairs of points being grouped together into one point. -In $\mathbb{A}^2_\mathbb{R}$, things are more complicated because of dimension, but for closed points, it's similar: each coordinate is like a pair of conjugate complex numbers grouped together. -Suppose you have points like $\{z, \bar{z}\}$ and $\{w, \bar{w}\}$ in $\mathbb{A}^1_\mathbb{R}$. Their product, taken in the naive sense, is going to look like four pairs: -$$ \{ (z, w), (\bar{z}, \bar{w}), (z, \bar{w}), (\bar{z}, w) \}. $$ -The first two and the last two are conjugate pairs, but (in the absence of coincidences like one of them being real) not conjugate to each other, so you will get two distinct points in $\mathbb{A}^2_\mathbb{R}$, which is not an irreducible set. -This corresponds to the fact that, as darij computed, the coordinate ring of this product is $\mathbb{C} \otimes_\mathbb{R} \mathbb{C}$, which is isomorphic to $\mathbb{C} \oplus \mathbb{C}$ since $\mathbb{C}$ is Galois of degree 2 over $\mathbb{R}$. And in fact, everything I've said about conjugates in the last few paragraphs has used this fact implicitly.<|endoftext|> -TITLE: New results on Chow's notion of closed-form numbers? -QUESTION [11 upvotes]: In an interesting article (available here), Timothy Chow proposes that a closed-form number be defined as an element of the smallest subfield of $\mathbb{C}$ that is closed under $\exp$ and a chosen branch of $\log$. It is fun to check that pretty much any number that you might accept as closed-form answer to a calculus problem belongs to this field. -He writes, "My hope is that this definition of closed-form expression for a number will become standard, and that many readers will be lured into working on the many attarctive open problems in this area." The bulk of the article relates his notion of closed-form numbers to standard conjectures in transcendental number theory, most notably Schanuel's conjecture. -My questions are - -To what extent has this notion become accepted as standard? -Are there new results since the time of his writing? - -There was a rekindling of interest in Schanuel's conjecture after Boris Zil'ber's categoricity results on algebraically closed exponential fields in characteristic zero. In what way has this changed the status of problems mentioned in Chow's article (if it has)? - -REPLY [6 votes]: There is also the recent paper by Borwein and Crandall, Closed Forms: What they are and why we care", to appear in the Notices of the AMS. He gives 7 different methods via which one can approach closed forms. Chow's notion is #4. For some strange reason, I am rather a fan of the approach they baptized diffeoclosed...<|endoftext|> -TITLE: Quanitative de Moivre–Laplace theorem (reference request) -QUESTION [6 upvotes]: The classical de Moivre-Laplace theorem states that we can approximate the normal distribution by discrete binomial distribution: -$${n \choose k} p^k q^{n-k} \simeq \frac{1}{\sqrt{2 \pi npq}}e^{-(k-np)^2 / (2npq)}.$$ -My question is: are there more precise, quantitative versions of this theorem in the literature? Are there good estimates how to measure the error? I am unfortunately not familiar with the subject but need a result of this type. -Of course there is always the option of going through existing proofs and checking the details, and turning them from "soft" to "hard", but I suspect this has to be already done. And maybe this is not optimal, maybe there are good accessible ways. -Can someone point me a good reference in this direction? - -REPLY [2 votes]: You might be interested in this paper (a very precise estimate, apparently overlooked by most people!) - -J. E. Littlewood, On the probability in the tail of a binomial - distribution, Adv. Appl. Prob. 1 (1969) 43–72. - -revisited and corrected by McKay - -Brendan D. McKay, On Littlewood's Estimate for the Binomial - Distribution, Advances in Applied Probability, Vol. 21, No. 2 (Jun., - 1989), pp. 475-478<|endoftext|> -TITLE: What is a category of sets? -QUESTION [7 upvotes]: One knows that many models of set theory exist. In topos theory,"the" category of sets is to play the role of the point. Since many models of set theory are around, I believe one of the following to be true. - -There is one category of sets and the model determines what is true about the category of sets. -There are many different categories that can be called the category of sets, one for each model. - -The question is: is 1 or 2 correct? Perhaps neither one is correct. In either case,what would a good reference be? - -REPLY [16 votes]: Here is a thoroughly Platonist answer to your question: Both 1 and 2 are true. There is one category of sets. Its objects are all of the sets, and its morphisms are all of the functions between them. But there are many other categories that can be (and in fact have been) called the category of sets (by abuse of language). In fact, I often think of an arbitrary elementary topos as the category of sets. -Most non-Platonist mathematicians would say that 1 is false and 2 is true, or perhaps that one shouldn't speak of the category of sets at all but rather of a category of sets. -A few mathematicians, who don't believe the consistency of any sort of set theory, might say that there are no categories that can be called the category of sets because there are no models of set theory. -In an effort to clarify these positions, their respective advocates would assert them more loudly.<|endoftext|> -TITLE: A question about J.H. Conway's SURREAL NUMBERS -QUESTION [9 upvotes]: My quesion is: What set theory are the mathematicians who are developing the theory of -these numbers working in-or are they, in fact, working outside any of the standard set -theories?. Each surreal number is a mapping of an ordinal number into the pair (+,-) so -that the collection S of all these numbers is a proper class. Moreover S is a real closed -(ordered) field containing sub-collections which are ordinally similar to the class of -ordinal numbers and to the set of real numbers (in their usual order). Since S is densely -ordered but not order-complete, there exists an order-complete ordered collection C -(constructed from the Dedekind cuts of S), which contains a dense sub-collection that is -ordinally similar to S. Now the elements of C are proper classes and if we are going to -have theorems about sub-collections of C (such as closed intervals), then the underlying -set theory (if any) must be one that allows some proper classes to be elements of collections. - -REPLY [6 votes]: Why on Earth do you feel the need to coerce Conway's system into some form of Set Theory? -Please go and read the Appendix to Part Zero of On Numbers and Games, which is a manifesto for the Mathematician's Lib Movement. -Conway's games are already formulated in much the same way as set theories are, the only difference being that they have left- and right-handed members. -Why do you think that an official one-handed set theory is any more suitable as a foundation than Conway's system free-standing? -Anyway, if you have coerced it into your preferred set theory, it becomes useless for anybody else's. For example, Mike Shulman recently solved the problem of eliminating the double negations from every level, as part of the presentation of Conway Numbers in Homotopy Type Theory (in the final chapter of the book).<|endoftext|> -TITLE: Embedding of riemannian manifolds into space forms -QUESTION [5 upvotes]: Hi; -By a celebrated theorem of J.Nash, we know that any riemannian manifold with smooth enough metric tensor can be realised as an embedded submanifold of $\mathbb{E}^N$ for some $N$. -Can one hope to be able to embed (compact) manifolds into some space with constant curvature? If the answer is yes and we have the isometric embedding $(M,g) \hookrightarrow Q(k)$, where $Q(k)$ is a space form of constant curvature $k$, do we have any estimates on the largest possible $|k|$? One possible bound comes from boundedness of diameter of positive space forms in terms of their curvature, although, the embedded submanifold might have longer geodesics than the ambient space. -Also, if we let the embedding be merely conformal, that is the induced metric belong to the same conformal class as that of $(M,g)$, what are the possible obstructions? - -REPLY [8 votes]: There is no bound on curvature, only on the dimension. -(The bound for the diameter of ambient space gives no bound for the intrinsic diameter of embedded space.) -In fact, it is straightforward to modify the proof of Nash's theorem to prove the following: - -Any $n$-dimensional Riemannian manifold can be embedded into any $m$-dimensional Riemannian manifold if $m\ge N(n)$.<|endoftext|> -TITLE: A question about quotient singularity -QUESTION [6 upvotes]: If a finite group G acts on a smooth variety X over complex number field and the fixed locus of G is smooth subvariety of codimension 1, will the resulting quotient variety be smooth? What will happen if the fixed locus has lager codimension ? thanks. - -REPLY [9 votes]: In general, for a finite group $G$ acting faithfully on a smooth variety $X$, whether or not the quotient is smooth is determined by the -Chevalley-Shephard-Todd theorem: -For $x \in X$, let $G_x\subset G$ be the stabilizer of $x$. Then a necessary and sufficient condtion for the quotient to be smooth is that each $G_x$ should be generated by pseudoreflections i.e. elements which fix pointwise a codimension $1$ subvariety of $X$ containing $x$. -In particular, one cannot just look at the fixed locus of $G$ to determine whether the quotient is smooth. It could well be empty but the quotient could still be singular.<|endoftext|> -TITLE: A Diophantine Equation -QUESTION [8 upvotes]: Does the Diophantine equation $2(x - \frac{1}{x}) = y - \frac{1}{y}$ have only trivial rational solutions, i.e, $x=\pm1, y = \pm1$? - -REPLY [7 votes]: Since at least one answer mentions the historical pedigree of this particular problem to some work in the 80's, I thought I would mention that the problem goes back much further. As various people have remarked in the comments, one first observes that -$$(y + 1/y)^2 = (y-1/y)^2 + 4 = 4(x^2 + 1/x^2 - 1).$$ -In particular, if $u = x^2 + 1/x^2$, then $u-1$ is a square. Similarly, -$u-2 = (x-1/x)^2$ and $u+2 = (x+1/x)^2$ are both squares. If follows that the product of $u-1$, $u-2$, and $u+2$ is also a square, -leading to a point on the curve -$$v^2 = (u-2)(u-1)(u+2).$$ -Now Fermat proved many years ago that there do not exist four rational squares in a non-trivial arithmetic progression. This is the same as asking that $1$, $1+r$, $1+2r$, and $1+3r$ are all squares, leading to a point on the curve -$$s^2 = (1+r)(1+2r)(1+3r).$$ -Yet this is the same equation as the one above, with $u = 6r+4$ and $v = 6s$. -Like in all cases of elliptic curves with rational $2$-torsion and trivial $2$-part of Sha, (edit and rank zero) there is an elementary proof that there are no rational points by infinite descent. However, these arguments, although classical, can sometimes be a little tricky to find. -There are, however, several articles one can find on the web giving "elementary" proofs of Fermat's theorem, for example: -http://www.maths.mq.edu.au/~alf/SomeRecentPapers/183.pdf -gives an argument which involves finding pairs $A$, $B$ with $A^2+B^2$ and $A^2+4B^2$ are both squares, then using pythagorean triples and descent (something that came up in the comments). Google searches will reveal many more roughly equivalent arguments, which all essentially show that the elliptic curve above has rank zero, and all the rational torsion points correspond to "trivial" or "degenerate" solutions.<|endoftext|> -TITLE: Reference request: a locally cyclic group is isomorphic to a section of the rational numbers -QUESTION [9 upvotes]: A group $G$ is locally cyclic if whenever $H \le G$ is a finitely generated subgroup then $H$ is cyclic. If $G$ is a locally cyclic group then $G$ is isomorphic to a quotient of a subgroup of the rational numbers under addition. An online proof of this fact appears at groupprops. However, despite some searching, I have been unable to find a proof in the published literature on abelian groups. - -Is there a published paper or textbook that has a proof that every locally cyclic group is isomorphic to a quotient of a subgroup of the rational numbers? - -REPLY [9 votes]: For torsion-free groups it is proved in Kurosh, "Group theory", See 3d edition, Chapter VIII, Section 30 (of course the result can be found in the 1st edition as well). Oroginally it was proved in Reinhold Baer, "Abelian groups without elements of finite order". Duke Math J. 3 (1): 68–122, 1937. For groups with torsion, it follows from old results as well but I am not sure anybody specifically mentioned it somewhere. Of course you need to look at Fuchs, "Infinite Abelian groups" (both volumes). If it is not there, it is probably not anywhere else.<|endoftext|> -TITLE: What is known about zero-sets of Schur polynomials? -QUESTION [8 upvotes]: Consider a set S of partitions not containing the empty partition (I would be happy with, say, all the partitions of size less than k, except for the empty one). -Let $U_\lambda^{(r)}$ be the zero-set in $\mathbb{C}^r$ of the Schur polynomial $s_\lambda(x_1,\cdots,x_r)$. -What is known about $\cap_{\lambda \in S} U_\lambda^{(r)}$, beyond the fact that it is symmetric under the action of $S_r$? -(I am having trouble finding information about this: all the hits are about the different question of the Schur stability of univariate polynomials, a concept based on the location of the roots of those polynomials). - -REPLY [2 votes]: For $k :=|\lambda| \ge r$, the statement that all $s_\lambda(x_1, \ldots, x_r)$ vanish is equivalent to all the elementary polynomials $e_j(x_1, \ldots, x_r) := \sum_{i_1 < \ldots < i_j} x_{i_1} \ldots x_{i_j}$, $j \le r$, vanish, since the latter form a basis of the algebra $\Lambda_r$. But this is true if and only if all the $x_1, \ldots, x_r$ are zero, since $e_j$ is the $y^{r-j}$ coefficient of a degree $r$ polynomial, i.e., -$$ \prod_{i=1}^r (y - x_i) = \sum_{j=0}^r (-1)^j e_j(x_1, \ldots, x_r) y^{r-j}.$$ -If all the $e_j$'s are zero, except $e_0 \equiv 1$, then it must be $y^r$, hence all the $x_i$'s vanish. -For $k < r$, the zero set consists of roots of polynomials of the form -$y^r + c_{k+1} y^{r-k-1} + c_{k+2} y^{r-k-2} + \ldots + c_r$.<|endoftext|> -TITLE: Polynomials are dense in weighted $L^2$ space -QUESTION [5 upvotes]: Hi, -It seems to be a common knowledge that the polynomials $x^n$ are dense in $L^2$ spaces with various probability weights, such as the gamma distribution weight $x^{\alpha-1}e^{-x}/\Gamma(\alpha)\;dx$. - -Is there any reference to this fact preferrably including the condition which property of the weight implies the density of polynomials? - -My guess is that if -$$ -\int_{-\infty}^{+\infty}e^{-\lambda|x|}P(dx)<\infty -$$ -for some $\lambda>0$, then the polynomials are dense in $L^2(\mathbb{R},P)$. - -REPLY [8 votes]: Sure, the beautiful book by N.I.Akhiezer The Classical Moment Problem and Some Related Questions in Analysis, where in particular you can find a thorough discussion of the property of the density of the polynomials in $L^1$ and $L^2$ for measures with finite moments of all order (together with sufficient conditions and counter-examples to the density.)<|endoftext|> -TITLE: Convex hull on a Riemannian manifold -QUESTION [15 upvotes]: Let $M$ be a complete Riemannian 2-manifold. -Define a subset $C$ of $M$ to be convex -if all shortest paths between any two points -$x,y \in C$ are completely contained within $C$. -For a finite set of points $P$ on $M$, define -the convex hull of $P$ to be -the intersection of all convex sets containing $P$. -It is my understanding that this definition is due to Menger. -In the Euclidean plane, the convex hull of $P$ coincides -with the minimum perimeter polygon enclosing $P$. -This does not hold on every $M$. -For example, the convex hull of four points on a sphere that do not -fit in a hemisphere is the whole sphere (this is Lemma 3.4 in the book below), -different from the minimum perimeter geodesic polygon: - -           - - -The shortest path connecting $a$ and $b$ goes around the back of the sphere, -but the illustrated quadrilateral is (I think!) the minimum perimeter polygon enclosing -$\lbrace a,b,c,d \rbrace$. -My specific question is: - -Q1. - Under what conditions on $M$ and on $P$ will the convex hull of $P$ - coincide with the minimum perimeter geodesic polygon enclosing $P$? - -I am teaching the (conventional, Euclidean) -convex hull now, and it would be enlightening to say something -about generalizing the concept to 2-manifolds. -More generally: - -Q2. - Which properties of the convex hull in $\mathbb{R}^d$ are retained - and which lost when generalizing to the convex hull in a - $d$-manifold? - -(The earlier MO question, -Convex Hull in CAT(0), -is related but its focus is different.) -I recall reading somewhere in Marcel Berger's writings that some questions about -convex hulls of just three points in dimension $d > 3$ are open, but I cannot find the -passage at the moment, and perhaps he was discussing a different concept of hull... -Added: I found the passage, in Berger's -Riemannian geometry during the second half of the twentieth century (American Mathematical Society, Providence, 2000), p.127: - -A most naive problem is the following. What is the convex envelope of $k$ points in a Riemannian - manifold of dimension $d \ge 3$? Even for three points and $d \ge 3$ the question is - completely open (except when the curvature is constant). A natural example to look at would be - $\mathbb{C P}^2$, because it is symmetric but not of constant curvature. - -(Caveat: These quoted sentences were published in 2000.) -Thanks for pointers and/or clarification! -C. Grima and A. Márquez, - -Computational Geometry on Surfaces: -Performing Computational Geometry on the Cylinder, the Sphere, the Torus, and the Cone, -Springer, 2002. - -REPLY [6 votes]: Convex hulls are extremely useful in hyperbolic/negatively curved geometry (in particular, I disagree with Anton's bold accessment in his answer that convex hulls have few applications in Riemannian world). Search on the phrases "convex core" and "hyperbolic manifold" to see how convex hulls help to work with Kleinian groups and (non-simply-connected) hyperbolic manifolds. -In the simply-connected case there is the following basic fact: if two convex sets $A$, $B$ in the hyperbolic $n$-space intersect, then the convex hull of $A\cup B$ is contained in the $\Delta$-neighborhood of $A\cup B$, where $\Delta$ is the thin triangle constant for the hyperbolic plane. (See e.g. Lemma 2.12 of this paper by Baker-Cooper). This illustrates fundamental difference of hyperbolic and Euclidean geometry. -One place where convex hulls appear prominently is the paper "The Dirichlet problem at infinity for manifolds of negative curvature" by Michael Anderson, JDG (1983), 701-721. -My personal favorite application of this work (derived by Bowditch) is that in a simply-connected manifold of pinched negative curvature the convex hull of any quasi-convex set $Q$ is within bounded Hausdorff distance of $Q$. Surprisingly, this result requres that the curvature is pinched negative and not merely negative.<|endoftext|> -TITLE: Optimal pants decompositions of a hyperbolic surface -QUESTION [5 upvotes]: Let $S$ be a hyperbolic surface, which is not the punctured torus or $4$-holed sphere. I am interested in finding a ``geometrically optimal'' pants decomposition on $S$. -Here is a candidate definition. Given a pants decomposition $P$, order the curves of $P$ from longest to shortest (in the hyperbolic metric). Then, pants decompositions $P$ and $P'$ can be compared by comparing the lengths of their curves lexicographically. That is: if the longest curve $c_1$ of $P$ is shorter than the longest curve $c'_1$ of $P'$, then $P$ is better. Or, if $\ell(c_1) = \ell(c'_1)$ and $\ell(c_2) < \ell(c'_2)$, where $c_2, c'_2$ are the second-longest curves of $P$ and $P'$, then $P$ is better. And so on, lexicographically. -With this definition, the induced ordering on pants decompositions becomes a well-ordering. More precisely: given a fixed pants decomposition $P$, there are finitely many curves shorter than the longest curve of $P$, hence finitely many better pants decompositions. In particular, there exists an ``optimal'' decomposition, whose longest curve is no longer than the Bers constant. -It is clear that optimal decompositions are not necessarily unique (otherwise, the optimal decomposition would never change as we move in Teichmuller space). But if $P$ and $P'$ are both optimal on a given surface, what can be said about how far apart they are? For example: are their shortest curves necessarily disjoint? - -REPLY [4 votes]: A pair of optimal pants decompositions $A, B$ need not have disjoint shortest curves. Here is an example. -Let $S$ be the genus two hyperbolic surface built from four equilateral right-angled hexagons by "doubling". That is, let $H$ be such a hexagon. Let $a = 2 \cosh^{-1}(\sqrt{3/2}) = 1.31695\ldots$ denote the side-length of $H$. (See Beardon.) Let $\alpha$ be the union of three non-adjacent sides of $H$ and let $\beta$ be the other three non-adjacent sides. Doubling $H$ across $\alpha$ gives a pants $P$. So $\alpha$ gives the seams of $P$ and $\beta$ doubles to give $\partial P$, each component having length $2a$. Now double across $\beta$ to get $S$. -Let $B = \partial P$ be the double of $\beta$ in $S$. So $B$ is a pants decomposition of $S$, each curve having length $2a$. Likewise $A$, the double of $\alpha$, is a pants decomposition. -Claim: All curves of $A$ and $B$ are systoles of $S$. -Proof: Any geodesic in a genus two surface is preserved by the hyperelliptic and so double covers an arc or a loop in the orbifold $O = S^2(2,2,2,2,2,2)$. In our situation $O$ is the double of $H$ across $\partial H$. The shortest arc is an edge of $H$. The shortest loop divides $O$ in half and has length $4\cosh^{-1}(\sqrt{2}) = 3.525494\ldots$. QED -Thus both $A$ and $B$ are optimal, yet every curve of $A$ crosses every curve of $B$.<|endoftext|> -TITLE: Unprovable sentence about integers -QUESTION [7 upvotes]: Is there any natural* statement S about the natural integers such that if PA contains no contradictions then neither PA+S nor PA+not S contains a contradiction? -If unknown, where can I read about the philosophical views on it? -*By natural I mean not a logical trick or non-constructive existence proof such as Rosser's sentence, but a clear statement such as the Collatz conjecture. - -REPLY [5 votes]: Let $A$ be any set of natural numbers that is computably enumerable, but not decidable. There are numerous natural instances of such sets, such as the set of all finite presentations of the trivial group, the set of all finite tile families that cannot tile the plane, and so on. -Every such set $A$ is saturated with infinitely many instances of the type you request. That is, there must be infinitely many candidates $a$, which are not in $A$, but such that this assertion is neither provable nor refutable in PA, or in whichever fixed background theory you wish to work, such as ZFC or ZFC + large cardinals. The reason is that otherwise, we would be able to decide membership in $A$ as follows: given a candidate $a$, wait for $a$ to show up in the enumeration of elements of $A$ and simultaneously search for a proof that $a$ is not in $A$. If non-membership was always provable for large candidates, then this would show that $A$ is decidable, contrary to assumption. -Thus, for every such $A$, there must be infinitely many particular non-members $a$ such that $a$ is not in $A$, but this is neither provable nor refutable from the fixed axioms. -To anticipate an objection, let me say that when $A$ is a "naturally" defined set, and $a$ is a particular candidate instance, then the assertion that $a$ is not in $A$ would seem to be a perfectly clear statement, not directly involving any logical "tricks". For example, we would have the assertion that a particular finite group presentation is not a presentation of the trivial group, or the assertion that a particular finite set of tiles does not tile the plane. -The logical trickery magic comes into play, of course, in the details of the explanation that a particular set $A$ is not decidable, where one might show a reduction from the halting problem or something similar. My opinion, however, is that consistency statements themselves are highly natural, and should not be considered a form of trickery. There is a sense in which every $\Pi^0_1$ statement is a consistency statement.<|endoftext|> -TITLE: What is the generator of $\pi_9(S^3)$? -QUESTION [19 upvotes]: I have Toda's book Composition methods in homotopy groups of sphere. -I have not found the generator of $\pi_9(S^3)=\mathbb{Z}_3$. -What is it? - -REPLY [10 votes]: Here's another way to describe the generator of $\pi_9(S^3)\cong\mathbb{Z}/3$. -By Proposition 13.6 in Toda's book, and the remarks preceding it, a generator is given by the composition -$$ -S^9 \stackrel{\alpha_1(6)}{\longrightarrow} S^6 \stackrel{\alpha_1(3)}{\longrightarrow} S^3 -$$ -where $\alpha_1(3)\in \pi_6(S^3)\cong\mathbb{Z}_{12}$ is a generator of the $3$-primary component and $\alpha_1(6)=E^3\alpha_1(3) \in \pi_9(S^6)\cong\mathbb{Z}_{24}$ is its third suspension. -The element $\alpha_1(3)$ has a very nice geometric representative described in my answer here. Explicitly, let $\mathbb{H}$ denote the quaternions, and represent the $6$-sphere as -$$ -S^6 = \{(p,w)\in \mathbb{H}\times\mathbb{H} \mid \mathfrak{Re}(p)=0\mbox{ and } |p|^2+|w|^2=1\}. -$$ -The map $b:S^6\to S^3\subseteq \mathbb{H}$ is given by -$$ -b(p,w) = \left\{\begin{array}{ll} \frac{w}{|w|} e^{\pi p} \frac{\overline w}{|w|}, & w\neq 0 \\ --1, & w=0, \end{array}\right. -$$ -where $e^{\pi p} = \cos(\pi |p|) + \sin(\pi|p|) \dfrac{p}{|p|}$ is the quaternionic exponential. -This description appears in the paper -Abresch, U.; Durán, C.; Püttmann, T.; Rigas, A., Wiedersehen metrics and exotic involutions of Euclidean spheres, J. Reine Angew. Math. 605, 1-21 (2007). ZBL1125.57017.<|endoftext|> -TITLE: Dodecahedral K3? -QUESTION [12 upvotes]: In pondering -this -MO question and in particularly its 1st answer, and answers to -this one recently posed, I realized there ought to be a dodecahedral K3 surface $X$. -This $X$ would fiber as an elliptic surface over $CP^1$ -with 12 singular fibers, each of type $I_2$. The corresponding singular points on $CP^1$ -would form the vertices of the icosahedron (centers of the dodecahedron). The automorphism group of -this $K3$ will then project onto the symmetry group of the dodecahedron. Do you know this -$K3$? Do you have a reference? - -REPLY [17 votes]: There's a pretty K3 with icosahedral symmetry and 12 singular fibers each of type II (so a double root of the discriminant but with additive reduction); would that do? It's $y^2 = x^3 + P(t)$ where $P$ has icosahedral symmetry. Explicitly one can take $P(t) = t^{11} - 11 t^6 - t$. This surface is isotrivial (i.e. with constant $j$-invariant, here $j=0$), and is closely related with the isotrivial surface of Chahal, Meijer, and Top, which also has $j=0$ but with $a(t) = t^{12} - 11 t^6 - 1$. Both surfaces attain the maximum of $18$ for the Mordell-Weil rank of an elliptic K3 surface over ${\bf C}(t)$. The Chahal-Meijer-Top paper is on the arXiv (9911274), and is published in Comment. Math. Univ. St. Pauli 49 (2000), 79–89. My recent student G.Zaytman studied the icosahedral surface and its Mordell-Weil lattice in the course of his thesis research. -EDIT Come to think of it, this is up to isomorphism the unique elliptic K3 surface over ${\bf C}(t)$ with icosahedral symmetries acting on the base: for the narrow Weierstrass form $y^2 = x^3 + a(t) x + b(t)$ to have icosahedral symmetry, the same must be true of the coefficients $a$ and $b$; but $a$ has degree $8$, so must vanish identically, while $b$ has degree 12 and is thus determined up to a scalar factor.<|endoftext|> -TITLE: A "dimension" for Tychonoff spaces -QUESTION [20 upvotes]: It's well-known that any Tychonoff space $X$ can be embedded in $[0,1]^k$ for some cardinal $k$. It's natural to ask what the smallest such $k$ is (let's call it $k(X)$). However, this probably probably doesn't deserve to be called dimension, since it fails to satisfy some desirable properties. For instance, although $k(\text{point}) = 0$, we have $k(\text{2 points}) = 1$. This leads me to consider a local version: -If $X$ is Tychonoff and $x \in X$, let $D(X, x)$ be the smallest cardinal $k$ such that some neighbourhood of $x$ can be embedded in $[0,1]^k$, and let $D(X) = \sup_{x \in X} D(X, x)$. This satisfies some obvious properties: - -If $\{U_\alpha\}$ is an open cover of $X$, then $\dim(X) = \sup D(U_\alpha)$. -If $A$ is a subspace of $X$, then $D(A) \le D(X)$. Equality holds if, for instance, $A$ contains a neighbourhood of a point $x$ with $D(X,x) = D(X)$. -$D(X) = n$ if $X$ is a $n$-dimensional manifold. -$D(X \times Y) \le D(X) + D(Y)$. - -This last inequality may be strict; for instance, if $X$ is the Cantor set, then $D(X) = 1$ and $X \times X \cong X$. -If $\dim$ denotes the Lebesgue covering dimension, then for $X$ compact, we have $\dim(X) \le \dim([0,1]^{D(X)}) = D(X)$. I have no idea when equality holds (it would if $\dim(X) = n$ implied that $X$ could be locally embedded in $\mathbb{R}^n$, but I don't know if that's true). -Is there a name for this $D$, or has such an invariant been studied before? How is this related to other notions of dimension for a topological space? In particular, are there classes of nice spaces (for instance, compact metrizable) on which they agree? - -REPLY [10 votes]: It is a classic result that a separable metrizable space $X$ can be embeded into $\mathbb{R}^{2n+1}$, where $n=\dim X$; thus $D(X)\le2\dim X+1$. For the universal spaces of Menger and Nöbeling this is optimal and since they are locally homeomorphic to themselves we find that $D(X)=2\dim X+1$ is possible.<|endoftext|> -TITLE: Supremum amongst Kolmogorov-Sinai entropies: ergodic or just invariant measures. -QUESTION [12 upvotes]: Cases where -$sup_{\mu \in E(T)} h_\mu(T) -\neq -\sup_{\mu \in M(T)} h_\mu(T)$. -Background -For a topological space $X$, -let $T: X \to X$ be a continuous application. -Then, call the set of $T$-invariant probability measures -$M(T)$, and call the set of $T$-ergodic (probability) measures -$E(T)$. -It is evident that $E(T) \subset M(T)$. -But it may happen that $M(T) = \emptyset$. -For example, take $X = \mathbb{R}$ and $T(x) = x+1$. -Since an ergodic measure is invariant, it is immediate that -$ - \begin{equation*} - \sup_{\mu \in E(T)} h_\mu(T) - \leq - \sup_{\mu \in M(T)} h_\mu(T). - \end{equation*} -$ -The question is whether equality holds or not. -When $X$ is compact, it is well known that equality holds. -In this case, it is a consequence of Jacobs' Theorem, which -states that for any $\mu \in M(T)$, there exists a measure -$\tau$, over the set $E(T)$, such that -$ - \begin{equation*} - h_\mu(T) - = - \int_{E(T)} h_m(T) d\tau(m). - \end{equation*} -$ -When $X$ is compact (locally compact, in fact), -the above equation is a consequence of -Choquet Representation Theorem -and the Krein-Milman Theorem. -(See, for example, Theorem 8.4 from -Walters, P. An Introduction to Ergodic Theory) -Now, when $X$ is not necessarily compact, but it is a Borel subset of -a compact metrizable set $\widetilde{X}$, Pesin and Pitskel' -argue in their -Topological Pressure and the Variational Principle for Noncompact Sets, -at the end of page 310: -(I will rename the spaces and applications in order -to conform to this post's notation.) - -We may assume that measure $\mu$ is ergodic. - In fact, consider the partition $\eta$ of $X$ into - ergodic components $X_s,\, s \in S$, of measure $\mu$. - Denote by $\mu_s$ the measures on $X_s$ - (then $T * \mu_s = \mu_s$), and by $\nu$ the measure on the - quotient space $X / \eta$. - Then $h_\mu(T) = \int_{Y/\eta} h_m(T) d\nu(m)$. - -As far as I understand, $Y/\eta$ is just the same as $E(T)$, -since each ergodic component is associated with an ergodic measure. -And for the same reason, $\nu$ is just our $\tau$. -So, what is being stated is the validity of -$ - \begin{equation*} - h_\mu(T) - = - \int_{E(T)} h_m(T) d\tau(m), - \end{equation*} -$ -which in turns implies the equality -$ - \begin{equation*} - \sup_{\mu \in M(T)} h_\mu(T) - = - \sup_{\mu \in E(T)} h_\mu(T). - \end{equation*} -$ -In Pitskel' and Pesin's paper, $T$ is not even supposed to be the restriction -to $X$ of a continuous transformation -$\widetilde{T}: \widetilde{X} \to \widetilde{X}$. -Questions - -How do I prove that when $X$ is a Borel subset of a compact -metrizable space $\widetilde{X}$ and $T$ is a continuous application -$T: X \to X$, then for any $\mu \in M(T)$, there exists -a measure $\tau$ over $E(T)$ such that -$h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m)$? -In case the answer to question "1" is negative, -is there a prove for the specific case where $T$ is the restriction -of a continuous application -$\widetilde{T}: \widetilde{X} \to \widetilde{X}$? -Do you know nice examples of transformations -of measurable spaces where $E(T) = \emptyset$ while -$M(T) \neq \emptyset$? - - -PS: This is my first post to MathOverflow. -This is really exciting! :-) - -REPLY [4 votes]: You make a pretty common mistake here - your question has nothing to do with continuity, compactness etc. and belongs entirely to the measure category. What matters here is that you have a measure preserving transformation of a probability Lebesgue space (sometimes these spaces are also called standard probability spaces - there is a nice wikipedia article about them) and a measurable partition of this space pointwise invariant with respect to the transformation (particular, and, essentially, the only interesting case: the partition into ergodic components of the transformation). Then the entropy of the transformation is the integral of the entropies of its restrictions to the elements of the partition (the formula which appears in Pesin and Pitskel'). In fact, it should not be attributed to them - a much better reference (actually, for the whole entropy theory as well) is Rokhlin's lecture notes on entropy theory in Russian Math Surveys 1967 (the formula we discuss is in section 8.11 there). It takes care of your questions (1) and (2). As for (3), any invariant probability measure has a unique (mod 0) decomposition into ergodic invariant measures (once again, this is an entirely measure category property), so that there are no (nice or not so nice) examples of transformations $T$ with $E(T)=\emptyset$ while $M(T)\neq\emptyset$.<|endoftext|> -TITLE: Is the nc torus a quantum group? -QUESTION [9 upvotes]: The non-commutative n-torus appears in many applications of non-commutative geometry. To stay in the setting $n=2$: it is a C$^\ast$-algebra generated by unitaries $u$ and $v$, satisfying $u v = e^{i \theta} v u$. It is the deformation of the 2-torus, i.e. a group. -So my question is: besides viewing the nc torus as a 'non-commutative space', is it also a compact quantum group? That is, is there Hopf algebraic structure in it? - -REPLY [4 votes]: Despite the negative result quoted by MTS, there have been some attempts to put a Hopf-like structure on the quantum torus. -One of these attemps, which seems orthogonal to the one mentioned by Pierre in his answer, is via Hopfish algebras. To be short, Hopfish algebras (after Tang-Weinstein-Zhu) are unital algebras equipped a coproduct, a counit and an antipode that are morphisms in the Morita category (they are bimodules,rather than actual algebra morphisms). -The Hopfish structure on the quantum torus has been studied in details in this paper. -To be complete, let me emphazise the following point (taken from the above paper): - -It is important to note that, although - the irrational rotation algebra may be - viewed as a deformation of the algebra - of functions on a 2-dimensional torus, - our hopfish structure is not a - deformation of the Hopf structure - associated with the group structure on - the torus. Rather, the classical limit - of our hopfish structure is a second - symplectic groupoid structure on - $T^∗\mathbb{T}^2$ (...), whose - quantization is the multiplication in - the irrational rotation algebra. We - thus seem to have a symplectic double - groupoid which does not arise from a - Poisson Lie group.<|endoftext|> -TITLE: Grothendieck-Riemann-Roch interpretation of a calculation -QUESTION [5 upvotes]: Let $X$ be some smooth projective variety over $\mathbb{C}$ and let -$$\mathscr{H}_{q}(X):=ch(\Omega_{X}^{q})Td(X).$$ -For $Y$ a certain elliptic fibration -$$\varphi:Y\to B$$ -where $B$ is of arbitrary dimension, I have been computing -$$\varphi_{*}\mathscr{H}_{q}(Y)$$ -where $\varphi_{*}$ is the proper pushforward. The total space $Y$ is a subvariety of a projective bundle -$$\mathbb{P}(\mathscr{O}\oplus \mathscr{L}\oplus \mathscr{L}\oplus \mathscr{L})$$ -where $\mathscr{L}$ is a line bundle on $B$. I have computed that -$\varphi_{*}Td(Y)=(1-e^{-L})Td(B)$ -$$\varphi_{*}\mathscr{H}_{1}(Y)=(1-e^{-L})\mathscr{H}_{1}(B)+(-4-e^{-L}+3e^{-2L}+2e^{-3L})Td(B),$$ -where $L=c_1(\mathscr{L})$. What I would like is an explanation of the result of these calculations in terms of Grothendieck-Riemann-Roch as I have only recently acquainted myself with GRR. Thanks everyone. - -REPLY [6 votes]: Let $T\phi$ be the relative tangent bundle. So -we have an exact sequence $0\to T\phi\to TY\to \phi^*TB\to 0$. Now GRR and the projection formula gives -$$ -\phi_*{\rm Td}(Y)=\phi_*({\rm Td}(T\phi){\rm Td}(\phi^*TB))= -{\rm Td}(TB)\phi_*({\rm Td}(T\phi))={\rm ch}(1-R^1\pi_*({\cal O}_Y)){\rm Td}(TB) -$$ -which suggests that ${\cal L}=R^1\pi_*({\cal O}_Y)^\vee=\pi_*(\Omega_\phi):=\pi_*(T\phi^\vee)$ -(by Grothendieck duality). I will take this for granted; up to $\otimes$ by a torsion bundle, it is forced upon you by the equation; if $\cal L$ and $\pi_*(\Omega_\phi)$ differ by a torsion line bundle, the calculations below still work. -Furthermore, applying GRR and the projection formula again, we may compute -$$ -\phi_*{\cal H}_1(Y)=\phi_*({\rm ch}(\Omega_Y){\rm Td}(TY))= -\phi_*(\ [\phi^*{\rm ch}(\Omega_B)+{\rm ch}(\Omega_\phi)]{\rm Td}(T\phi)\phi^*{\rm Td}(TB)\ )= -$$ -$$ -{\rm Td}(TB){\rm ch}(\Omega_B)\phi_*({\rm Td}(T\phi))+{\rm Td}(TB)\phi_*({\rm Td}(T\phi) -{\rm ch}(\Omega_\phi))= -$$ -$$ -{\rm Td}(TB){\rm ch}(\Omega_B)\phi_*({\rm Td}(T\phi))+{\rm Td}(TB){\rm ch}({\cal L}- -R^1\pi_*(\Omega_\phi))= -$$ -$$ -{\rm Td}(TB){\rm ch}(\Omega_B)\phi_*({\rm Td}(T\phi))+{\rm Td}(TB){\rm ch}({\cal L}-1)= -$$ -$$ -{\rm Td}(TB){\rm ch}(\Omega_B)(1-{\rm ch}({\cal L}^\vee))+{\rm Td}(TB){\rm ch}({\cal L}-1)= -(1-e^{-L}){\cal H}_1(B)+(e^{L}-1){\rm Td}(TB)\,\,\, (*) -$$ -Now use the fact that $\cal L$ is actually a torsion bundle, because the discriminant modular -form will trivialise ${\cal L}^{\otimes 12}$ (or possibly a higher power, if one needs to introduce -level structures). This last fact is also a consequence of GRR, since -$$ -\phi_*({\rm Td}(T\phi))=\pi_*({\rm ch}(1))\phi^*\phi_*({\rm Td}(T\phi))=0={\rm ch}(1-R^1\pi_*({\cal O}_Y)) -$$ -(because $T\phi=\pi^*\pi_* T\phi)$. -Hence, one gets, all in all, that -$$ -\phi_*{\rm Td}(TY)=0 -$$ -and in view of (*), that -$$ -\phi_*{\cal H}_1(Y)=0 -$$ -which is equivalent to the two equations you are considering, since ${\cal L}$ is a torsion line bundle (observe that the degree $0$ part of $−4−e^{-L}+3e^{−2L}+2e^{−3L}$ vanishes).<|endoftext|> -TITLE: Kähler Structure for Projective Varieties over a Finite Field -QUESTION [8 upvotes]: (i) In 1960 Serre proved a famous analogue of the Weil conjectures for Kähler manifolds. This poses an obvious question: Does there exist an analogue of a Kähler structure for (non-singular) projective varieties over a finite field. That is, do there exist things like almost complex structures, Lefschetz operators, Kähler identities, etc? -(ii) Moreover, the study of generalised Hodge structures is a well-studied field. Does there exist a subfield of generalised Kähler structures? - -REPLY [10 votes]: This may not be quite the answer you want, but from my point of view, smooth projective -varieties are the analogues of compact Kähler manifolds, where the metric corresponds to a -choice of ample line bundle. Over $\mathbb{C}$, the first Chern class of an ample line bundle -is precisely the Kähler class for a Fubini-Study metric. This is, however, an imperfect analogy; some things translate and some thing don't. For example, -Lefschetz operators make sense in any Weil cohomology, and Deligne even gave a proof of the hard Lefschetz theorem for projective varieties by reducing to finite fields. There is -a definition of sorts for the $\Lambda$ and $*$ operators within algebraic geometry (see Kleiman, Algebraic cycles and the Weil conjectures). But these -are all at the cohomology level, not at the differential form level. So you shouldn't -expect Kähler identities in the conventional sense (as far as I can tell). -I'm not sure how to interpret (ii).<|endoftext|> -TITLE: How to disjoint two cycles with zero intersection? -QUESTION [13 upvotes]: Suppose that $M^n$ is a smooth connected orientable manifold and $Z^k$ with $Z^{n-k}$ are two real cycles in $M^n$ with zero index of intersection $Z^k\cdot Z^{n-k}=0$ (these cycles are submanifolds if this helps). Is it true that the cycles can be replaced by homologous ones that do not intersect at all? -PS. I would like to find a reference for this statement, if it exists, hence the bounty - -REPLY [13 votes]: I think that the answer is yes and that we don't really need the full Whitney's trick for this since being homologous is a much coarser relation than being isotopic, which is what Whitney's trick gives. So, rather than using the full trick, one can use just a half of it. -Let $M$ be an oriented connected smooth manifold, and let $Z_1,Z_2$ be oriented pseudo-manifolds representing two cohomology classes (recall that a pseudo-manifold is a stratified space that has no codimension 1 strata and such that each connected component is the closure of a single connected codimension 0 stratum; any homology class can be represented by a pseudo-manifold). Assume $\dim Z_2\geq 1$ [upd: and $\mathop{\mathrm{codim}}_M Z_2>1$; this assumption excludes the cases when one of the cycles is 0-dimensional and the other is of the maximal dimension, in which case the statement we're after is clearly true, and when $\dim Z_1=\dim Z_2=1$; this case has to be considered separately]. -First, let's make $Z_2$ connected by joining the connected components with tubes. [upd: as Bruno Martelli points out in the comments, some care is needed here. However, if we have a tube that induces the wrong orientations of one of the components it's supposed to connect, we can always twist the tube since we assume $\mathop{\mathrm{codim}}_M Z_2>1$.] While doing this we may introduce new intersection points, but after a small isotopy these will be all transversal and their signs will add up to 0. Second, take two intersection points $P,Q$ with opposite signs and join them with a non-self-intersecting path $\gamma\subset Z_2$ that does not pass through the singularities [upd: and through other intersecrtion points; some care is needed here as well when $\dim Z_2=1$: in this case we take $P$ and $Q$ to be neighbors on $Z_2$]. -Now equip $M$ with a Riemannian metric and let's modify $Z_1$ by taking out two small balls around $P$ and $Q$ in $Z_1$ and inserting a thin tube $T$ instead where $T$ is obtained by exponentiating the sphere subbundle of $N_M Z_2|\gamma$ of sufficiently small radius. More precisely, some work is needed to identify the spheres in $N_MZ_2$ at $P$ and $Q$ with the boundaries of the balls, but this should be no problem. -The result will be homologous to $Z_1$: the fact that $P$ and $Q$ have different signs ensures that the small balls around them and the tube together form the boundary of the exponential of a a ball subbundle of $N_{M}Z_2|\gamma$. [I wish I could draw a picture here but don't know how to do that.] Notice that when $Z_1$ is a loop around $(0,0)$ and $(1,0)$ in $\mathbb{R}^2$ and $Z_2$ is a loop around $(-1,0)$ and $(0,0)$, as in Simon Rose's example, then this procedure cuts $Z_1$ into two loops, one around $(0,0)$, the other around $(1,0)$. -In this way one can eliminate every pair of intersection points with opposite signs. Notice that we haven't done anything to $Z_2$ in the process, apart from making it connected. -[upd: in the case when $M$ is a surface and $Z_1,Z_2$ are 1-dimensional, then one could try to adapt the above argument for possibly self-intersecting $Z_2$, but this would require some work. In a sense, this makes sense: if 4 dimensions are not enough to perform Whitney's trick, then it's not too surprising that 2 dimensions are not enough to perform half of it.]<|endoftext|> -TITLE: What motivates modern algebraic geometry for a combinatorial/constructive algebraist? -QUESTION [40 upvotes]: This is, basically, me trying to generalize "Why should I care for sheaves and schemes?" into a reasonable question. Whether successfully, time will tell, but let me hope that if not the question, then at least the answers might be of use not just for me. -To differentiate this from some questions already asked, let me clarify: - -I am talking only about modern algebraic geometry, as in: everything that is better dealt with in terms of sheaves and schemes rather than varieties and curves. I know well enough that classical ("Italian") algebraic geometry has lots of applications; I am interested in knowing a reason to study (and a golden thread to follow in that) the kind of algebraic geometry that started with Serre, Leray, Grothendieck. -A "combinatorial/constructive algebraist" is a notion I cannot really formalize, but I mean an algebraist who is interested in actual computable things and their "fine structure" rather than topological abstracta and their "crude structure"; for example, actual polynomial identities rather than equality of zero-sets; actual isomorphisms instead of isomorphy; "for every point not on the zero-set of some particular ideal" rather than "for almost every point". The "combinatorial/constructive algebraist" (himself an abstraction) is fine with abstraction and formalism as long as he knows how to transform the abstract results into concrete equations and algorithms in case of need. He is not fine with nonconstructive existence results, although he is wary of declaring proofs unconstructive at first sight merely due to their formulation... - -I believe I know of one example of this kind, a problem on matrix factorization solved using cohomology of sheaves somewhere on MathOverflow (any help with finding it is appreciated). There is also the interpretation of commutative Hopf algebras as coordinate Hopf algebras of affine schemes - but affine schemes are not really what I consider to be modern algebraic geometry; they correspond 1-to-1 to rings and are more frequently considered as functors than as locally ringed spaces in Hopf algebra theory. I would personally be more convinced by applications to invariant theory (viz., results from classical invariant theory proved with geometric methods) or the combinatorial kind of representation theory. I used to think that Swan's paper I linked in question 68071 is another application of scheme theory, but after understanding Seiler's proof it seems rather unnecessary to me. - -REPLY [16 votes]: If you are interested in actual computations using modern algebraic geometry, there are plenty -to be had in Gromov-Witten theory and enumerative geometry. For example, Kontsevich's formula -counting rational plane curves is a famous example. The proof itself does not use any scheme theory, but it was based on the structure of a very delicate object called moduli space of stable maps, which could not be constructed without using schemes. Basically, counting problems -in enumerative geometry are usually transformed into intersection theory on moduli spaces -of the objects being counted. We want the moduli space to be compact so we can use the "invariant -of numbers" principle (example : two lines intersect at one point in projective plane, but not necessarily so in -affine line). Compactifying the moduli space meaning you allow your objects to have limits, thus even if your objects of interested are smooth varieties, their limits may be deformed and have unreduced structures (read : schemes and sheaves). For example, a conic may be denegrate to a double line which only makes sense as a scheme. -You can read a nice exposition of Kontsevich's formula in : http://arxiv.org/abs/alg-geom/9608011. -As another example, in http://arxiv.org/abs/alg-geom/9612004 , Getzler computed the intersection pairing matrix of $\mathcal M_{1,4}$ to obtain relation between cycles and then use it to compute, for example, the number of elliptic curves of degree $5$ passing through $20$ lines in $\mathbb P^3$ to be $2,583,319,387,968$, among other thing.<|endoftext|> -TITLE: Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature -QUESTION [35 upvotes]: A curve in the plane is determined, up to orientation-preserving -Euclidean -motions, by its curvature function, $\kappa(s)$. -Here is one of my favorite examples, from -Alfred Gray's book, - -Modern Differential Geometry of Curves and Surfaces with Mathematica, -p.116: - -               - - - -Q1. - Is there an analogous theorem stating that a surface in $\mathbb{R}^3$ - is determined (in some sense) by its Gaussian curvature? - -I know such a reconstruction path (curvature $\rightarrow$ surface) is needed in -computer vision, and so there are approximation algorithms, -but I don't know what is the precise theorem underlying this work. - -Q2. - Are there higher-dimensional generalizations, determining a - Riemannian manifold by its curvature tensor? - -I have no doubt this is all well known to the cognoscenti, -in which case a reference would suffice. Thanks! -Addendum (4Oct11). Permit me to augment this question with a relevant reference -which loosens the notion of "determines" and answers my Q1 with that notion replaced -by "find some." -The paper by Gluck, Krigelman, and Singer, entitled "The converse to the Gauss-Bonnet Theorem in PL," -J. Diff. Geom, 9(4): 601-616, 1974, poses this question: - -Suppose that a closed smooth two-manifold $M$ and a smooth real-valued function $K \;:\; M \rightarrow \mathbb{R}$ are given, and that one is asked to find a Riemannian metric for $M$ having $K$ as its Gaussian curvature. [...] With these restrictions on $K$ [just elided], the problem has been completely solved for all closed smooth two-manifolds by: Melvyn Berger [...], Gluck [...], - Moser [...], Kazdan and Warner [...]. Recently Kazdan and Warner have obtained a uniform solution. - The problem for compact two-manifolds with boundary, however, seems not to have been addressed in the smooth category. - -The MathSciNet review of this paper was written by Gromov. - -REPLY [3 votes]: For $\bf{Q1}$, (local) congruence of sub-manifolds under a continuous group action can be determined by the method of moving (co)frames (inspired by Elie Cartan and rigorously formulated by Olver and Fels): see Section 10 in http://www.math.umn.edu/~olver/mf_/mcII.pdf. -For an application of this to image recognition and computer vision (since you mentioned it) see the section on image processing here: http://www.math.umn.edu/~olver/paper.html<|endoftext|> -TITLE: ``Nice'' metrics for a Morse gradient field: counterexample request -QUESTION [17 upvotes]: Ralph Cohen (professor at Stanford) is teaching a class on algebraic topology and moduli spaces this quarter, beginning by reviewing his perspective of Morse theory. He defined "nice" metrics, proved that they are dense in the $L^2$ space of metrics on $\mathbb R^n$, proved one result using that, and doesn't need them anymore. But given the previous assumptions, I and some classmates want to know: are not all metrics "nice"? -The setup: Assume a real-valued $C^\infty$ function $f$ on a closed smooth manifold $M$ of dimension $n$ is Morse: critical points are nondegenerate in the sense of having full-rank Hessians. Given some critical point $p$ of index $k$, find a small neighborhood $U$ and a diffeomorphism $U\cong\mathbb R^n$ such that $f$ becomes a function $\sum_{i=k+1}^nx_i^2-\sum_{i=1}^kx_i^2$. -A "nice" (smooth Riemannian) metric on $M$ is defined to be one that, when restricted to $U\cong\mathbb R^n$, gives $f$ a gradient field that after some diffeomorphism of $\mathbb R^n$ can be written as $\sum c_ix_i\partial_i$ for nonzero constants $c_i$. (Edited to add the missing $x_i$s.) -Can anyone make an illuminating example of a non-nice metric on $\mathbb R^n$? In fact (because $U$ is bounded) I might prefer one in the setting where $U$ is mapped an open ball rather than all of $\mathbb R^n$. - -REPLY [25 votes]: Assuming Giuseppe's suggested correction is right, here's what you have to worry about: Consider the function $f(x,y) = \tfrac12 x^2 + y^2 + x^2 y$ and the metric $g = (1+2y)\ dx^2 + dy^2$ on the half-plane $y > -\tfrac12$. You can compute that -$$ -\nabla f = x\ \frac{\partial\ }{\partial x} + (2y+x^2)\ \frac{\partial\ }{\partial y} -$$ -and this is one of those vector fields that cannot be linearized smoothly near $(x,y) = (0,0)$, i.e., it is not equivalent, under any smooth change of variables near this point, to the vector field -$$ -x\ \frac{\partial\ }{\partial x} + 2y\ \frac{\partial\ }{\partial y} -$$ -(There is a large literature about conditions that obstruct or allow a vector field to be linearized near a singular point.) -One way you can see this is that the flow lines of the first vector field are $x=0$ and those of the form $y = (c + \ln |x|)\ x^2$ where $c$ is a constant, while the flow lines of the second vector field are $x=0$ and those of the form $y = c\ x^2$. (Of course, $(x,y) = (0,0)$ is fixed by both vector fields.) Since the flow lines of $\nabla f$ aren't smooth curves, no smooth change of variables will convert them to smooth curves.<|endoftext|> -TITLE: Red-blue alternating Menger's theorem -QUESTION [8 upvotes]: Suppose we have a graph where every edge is colored red or blue. We say that a path is alternating if the red and blue edges alternate in it. Our goal is to find many edge/vertex-disjoint alternating paths from a given vertex $s$ to another given vertex $t$. Has this problem been studied before? -Update: It DOES NOT seem to me ANYMORE that the following, Menger-type theorem is true, as pointed out by Ilya in the comments: -Lemma: -We say that a partition of the vertices into $S, T, R, B$ is a colored cut if $s\in S, t\in T$, there are no edges between $S$ and $T$ and, except for the edges between $R$ and $B$, the vertices of $R$ are only adjacent to red edges and the vertices of $B$ are only adjacent to blue edges. There are $k$ edge/vertex-disjoint paths from $s$ to $t$ if and only if after deleting any $k-1$ edges/vertices, there is no colored cut. -Update: Gyula Pap told me that if we consider the directed version and edge-disjoint paths, then doubling every vertex (to redin-blueout and bluein-redout) reduces the problem to the monochromatic Menger's theorem from which (if I see well) the above lemma follows. So now I am mainly interested in the vertex-disjoint version. - -REPLY [5 votes]: It seems that the problem of determining whether there exist $2$ vertex disjoint red-blue alternating paths joining vertices $s$ and $t$ is NP-complete. Thus, unless NP $=$ co-NP, there exist no efficient characterization of obstructions to existence of such paths, similar to the one you propose in the lemma. -Below is a reduction to the classical result of Fortune, Hopcroft and Wyllie that the DIRECTED $2$-LINKAGE problem is NP-complete. ( Given a digraph $D$ and four distinct vertices $u_1,v_1,u_2,v_2$; does $D$ contain a pair of vertex-disjoint paths $P_1, P_2$, so that $P_i$ is a directed path from $u_i$ to $v_i$ for $i=1,2$?) -Given a digraph $D$, we replace every directed edge $e=xy$ of $D$ by a vertex $w_e$ joined to $x$ by a red edge and to $y$ by a blue edge. Then we add vertex $s$ joined by a blue edge to -$u_1$ and by a red edge to $v_2$, and a vertex $t$ joined by a blue edge to $u_2$ and a red edge to $v_1$. It is not hard to see that a pair of vertex disjoint paths from $s$ to $t$ in the new graph corresponds exactly to the $2$-linkage as described above.<|endoftext|> -TITLE: Crystalline realization of mixed Tate motives -QUESTION [15 upvotes]: Deligne and Goncharov, in their article of 2005, mention that the crystalline realization functor has yet to be worked out. What's the current state of the literature on this? And how big of an issue is it? - -REPLY [9 votes]: Apparently this issue was worked out shortly after, during Go Yamashita's stay at the IHES in 2006, and after some discussion with Deligne. -For any Tate motive $M$ unramified at $v$, its crystalline realization is defined as $D_\mathrm{crys}(M_p)$, where $D_\mathrm{crys}$ is Fontaine's functor $(B_\mathrm{crys} \otimes_{\mathbb{Q}_p}-)^{G_{k_v}}$, and $M_p$ is the p-adic realization of $M$. -This realization is functorial, and has the comparison isomorphism: -$$k_v \otimes_{k_0,v}D_\mathrm{crys}(M_p) \cong k_v\otimes_k M_\mathrm{dR}$$ -All of this is perfectly explained and proved in section 4 of: - -Go Yamashita, Bounds for the Dimensions of p-adic Multiple L-Value Spaces (2009)<|endoftext|> -TITLE: Chord arrangement that avoids confining small or large disks -QUESTION [10 upvotes]: These two questions are two-dimensional variations on this recent MO question, -"Threading pinholes in the wall of cylinder to pass through an internal coordinate." -Noam Elkies suggested that even a 2D version might lead to interesting mathematics. -Here I've taken the liberty to pose two specific 2D questions directly inspired by the -original 3D question: - -P1. Arrange $n$ points on a circle to maximize the smallest radius of a chord-confined disk. - -The $n$ points determine $\binom{n}{2}$ chords, which partition the disk into cells. -Among cells entirely bounded by chords (as opposed to bounded by chords and a circle arc), -the question asks to maximize the size of the inscribed disks, -i.e., to avoid any small disk. -In the example below, the green disk is the smallest; I do not know if another $n=5$ arrangement -results in a larger smallest disk. -           - -Let $r(n)$ be this max-min radius. -It seems that finding the exact value of $r(n)$ for small $n$ could be interesting, -as well as determining its asymptotic behavior. -Another version is: - -P2. Arrange $n$ points on a circle to minimize the largest radius of an interior disk - not crossed by any chord. - -Call this min-max quantity $R(n)$. -Note (updated) that, unlike in P1, here disks confined by chords and the outer circle -are included. -This is perhaps closer to the spirit of the 3D question, -for then no point in the disk is ever far from a chord. Again I do not know if, for $n=5$, -the blue disk above is the minimum of the largest confined disks over all 5-point arrangements. -Update. Certainly not, as Simon shows in his answer. And here is a better -arrangement for P2 -with $n=5$, following Gerhard's suggestion (the three blue disks have the same radius): - -REPLY [4 votes]: The blue disk given is not the minimum size of the largest disk. By symmetry, there are only three sized involved, and the yellow disk is strictly smaller than the blue one. -So if we pick one of the five points and perturb it a little, the blue disk will shrink while one of the yellow ones next to it will grow---however, there is a non-zero amount that you can move the point before the yellow one becomes larger than the blue one.<|endoftext|> -TITLE: Can most 3 dimensional hyperbolic orbifolds with finite volume be covered by a hyperbolic manifold? -QUESTION [6 upvotes]: If G is a discrete cofinite volume subgroup of PSL(2,C),then G acts on H3, H3/G is a 3-dim hyperbolic orbifold N with finite volume, my question is : Is it right in most situations that we can find a hyperbolic 3 manifold M as a finite covering space of N? -This question is equivalent to the following : -do most dicrete cofinte volume klein groups have a finit order torsion free subgroup? - -REPLY [8 votes]: Yes, this is true for all of them. Any finitely generated matrix group has a torsion-free subgroup of finite index; this is the so-called "Selberg's lemma". A canonical source is Ratcliffe's Hyperbolic Manifolds book (you can probably find the relevant section on google books for free, or on gigapedia.com if you are so inclined).<|endoftext|> -TITLE: Polynomial reducible modulo every integer -QUESTION [6 upvotes]: Hi, -let $f\in\mathbb{Z}[X]$ be a monic polynomial. Assume that the reduction of $f$ modulo $m $ is reducible for all integers $m\geq 2$. -Q1: Is $f$ reducible in $\mathbb{Z}[X]$ ? -I've thought about this question without making substantial progress. If $p$ is a prime number , $p\nmid disc(f)$, then $f$ mod $p$ is separable, so if there is a nontrivial factorisation, we may find a factorisation into product of coprime monic polynomials. By Hensel's lemma, $f$ is reducible in $\mathbb{Z}_p[X]$. -If $p\mid disc(f)$, bad things could happen. For example, $X^4+1$ is reducible mod $2$, but irreducible mod $4$. -So I'm stuck here, and may be this approach will be not the right one but I have several related questions : -Q2: Assume that $p\mid disc(f)$. Does the fact that $f$ is reducible modulo $p^r$ for all $r\geq 1$ implies that $f$ is reducible in $\mathbb{Z}_p[X]$ ? -Q3: Let $f\in\mathbb{Z}[X]$ be a monic polynomial. Assume that $f$ is reducible in $\mathbb{Z}_p[X]$ for all prime integers $p$. Does $f$ is reducible in $\mathbb{Z}[X]$ ? -Thanks! -Greg - -REPLY [17 votes]: The polynomial $x^4-72x^2+4$ is irreducible over $\mathbb{Z}$, but reducible modulo every integer. -See E. Driver, P. A. Leonard and K. S. Williams Irreducible Quartic Polynomials with Factorizations modulo $p$ -The American Mathematical Monthly -Vol. 112, No. 10 (Dec., 2005), pp. 876-890 - -REPLY [5 votes]: The answer to Q2 is yes, while the answer to Q1 and Q3 (which are equivalent by Q2 and CRT) is no. -Q2 is essentially formal. The set of non trivial factorizations into monics of $f$ mod $p^r$ form an inverse system, and the inverse limit is nonempty (why?) So take your compatible system of factorizations mod $p^r$ and take inverse limits coefficient by coefficient and you get a factorization over $\mathbb{Z}_p[x]$. -As for Q1/Q3, $f\in\mathbb{Z}[X]$ irreducible monic defines a number field $K=\mathbb{Q}[x]/(f)$ and $f$ is irreducible over $\mathbb{Z}_p$ if and only if there is only one prime of $K$ lying over $p$. So to find a counterexample, you need to find a number field with more than one prime above every rational prime $p$. This can easily be arranged, for example with $K$ a suitable biquadratic extension.<|endoftext|> -TITLE: Why are some axioms preserved in generic extensions? -QUESTION [6 upvotes]: It is a known theorem that for a model of $ZF$, $M$, if $M\models AC$ and $G$ is a $P$-generic filter over $M$, for some $P\in M$, then $M[G]\models AC$. -On the other hand, it is long known that other axioms, for example $GCH, CH, V=L, ...$ are not preserved by such extensions. -In a paper of Monro [1] the first paragraph speaks on a question by Dana Scott about why do some axioms get preserved by generic extensions and other not; it also says that there is an [almost obvious] equivalence between preservation in generic extensions, and preservation in Boolean-valued models. -The paper goes on to prove that some restricted versions of the axiom of choice are not preserved by generic extensions (and in a recent phone call it was explained to me how to break weak choice principles such $DC_\kappa$ quite easily in relatively simple generic extensions of models of $ZF$). -Lastly the paper says nothing about an answer to the question above, nor it cites any resource for a possible answer. However, the paper is quite old now, and in the past three decades some progress might have occurred. -Is there an answer for Scott's question: which axioms are preserved by all generic extensions? -Edit: If there is no "simple" and uniform answer to the above question, is there a possible answer to why is the Axiom of Choice preserved in generic extensions, while restricted versions are not? - -Bibliography: - -G. P. Monro, On Generic Extensions Without the Axiom of Choice. The Journal of Symbolic Logic Vol. 48, No. 1 (Mar., 1983), pp. 39-52 - -REPLY [7 votes]: In the comments you consider the more general question of -which sentences are necessarily preserved by forcing. One -could cast the question as: what are the most general -relations between set-theoretic truth and forceability? -This more general setting is the theme of my work with -Benedikt Loewe on The modal logic of forcing, Trans. AMS, -vol. 360, -2008. -The idea here is to investigate the two operators: - -$\varphi$ is forceable or possible, written -$\lozenge\varphi$, if $\varphi$ holds in some forcing -extension. -$\varphi$ is necessary, written $\square\varphi$, if -$\varphi$ holds in all forcing extensions. - -These are modal operators, but they are expressible in the -language of set theory. One can view a set-theoretic -universe in the context of all its forcing extensions as an -enormous Kripke model of possible worlds, with each -accessing its forcing extensions. Your question was, when -can we expect $\varphi\to\square\varphi$? -Our inquiry was what are the most general valid principles -of forcing. For example, one can easily verify the -following validities, under the forcing interpretation: - -(K) $\ \ \square(\varphi\to \psi)\to (\square\varphi\to - \square\psi)$ -(Dual) -$\ \ \neg\lozenge\varphi\leftrightarrow\square\neg\varphi$ -(S) $\ \ \square\varphi\to\varphi$ -(4) $\ \ \square\varphi\to\square\square\varphi$ -(.2) $\ \ \lozenge\square\varphi\to \square\lozenge\varphi$ - -The last axiom (.2) uses product forcing. The modal theory -obtained from these axioms is known as S4.2. We define that -a modal assertion $\varphi(p_0,\ldots,p_n)$ in the language -of modal logic expresses a valid principle of forcing if -$\varphi(\psi_0,\ldots,\psi_n)$ holds with the forcing -interpretation of the modal operators for all set-theoretic -assertions $\psi_0,\ldots,\psi_n$. The main theorem of our -paper is: -Theorem. (Hamkins+Loewe) If ZFC is consistent, then -the ZFC-provably valid principles of forcing are exactly -those in the modal theory S4.2. -The proof uses the concept of buttons and switches, where a -set theoretic sentence $\psi$ is a switch if both $\psi$ -and $\neg\psi$ are necessarily possible, that is, forceable -over any forcing extension, and $\varphi$ is a button if -it is possibly necessary, that is, if it can be forced in -such a way that it remains true in any forcing extension. -Thus, switches can be turned on and off, but once you have -pushed a button, you cannot unpush it. The arguments are a -nice blend of easy forcing and modal logic. -Meanwhile, it is consistent that a model can exhibit more -than merely the S4.2 validities. For example, the -maximality principle is the scheme asserting: -$$\lozenge\square\varphi\to\square\varphi$$ -for all sentences $\varphi$. (See J. D. Hamkins, A simple -maximality principle, JSL, vol. 68, -2003, -also independently investigated by Stavi and Vaananen.) -Thus, the maximality principle asserts that any sentence -that could be forced in such a way that it remains true in -all subsequent forcing extensions, is already necessarily -true in this way. Thus, it asserts that all buttons have -been pushed. The maximality principle is equiconsistent -with ZFC, but it is not true that every model of set theory -has a forcing extension with MP. If one allows parameters -into the scheme, the strength rises, and the necessary -maximality principle with real parameters implies -$\text{AD}^{L(\mathbb{R})}$. -The fact that different models of set theory can exhibit -different valid principles of forcing implies that even -when one looks at the general modal form of a set-theoretic -sentence, the question of which sentences are invariant by -forcing will depend on the model of set theory. -Incidently, there are numerous open questions in the modal -logic of forcing. For example, if one restricts to ccc -forcing or to proper forcing or any of several other -natural forcing classes, we don't yet know the exact modal -theory of validites. What are the ZFC-provably valid -principles of ccc forcing?<|endoftext|> -TITLE: Furstenberg-Zimmer Theorem: non-invertible systems. -QUESTION [5 upvotes]: Questions - -Is there a version of Furstenber-Zimmer Theorem for -non-invertible measure preserving systems? -Where can I find it? -What is the precise statement? - -Background -In many works that reference the Furstenberg-Zimmer Theorem, -the theorem itself is not stated. -Authors usually cite the works of Furstenberg -(The structure of distal flows -and/or -Ergodic behavior of diagonal measures and a -theorem of Szemerédi on arithmetic progressions) -and Zimmer -(Extensions of ergodic group actions -and/or -Extensions of ergodic actions and generalized discrete spectrum). -The point is that in many places, the theorem is being used -for non-invertible systems. -This happens, for instance in -On Li-Yorke Pairs, -where the systems are assumed to be surjective, but not -necessarily invertible. -In this paper, for the proof of Theorem 2.1, the authors -use Furstenber-Zimmer Theorem. -As far as I understood, Zimmer's work deals with -group actions. -That is, invertible systems. -And for Furstenberg's Ergodic behaviour of diagonal measures [...], -he deals with regular measure preserving systems. -Unfortunately, Furstenberg and Zimmer (obviously) did not call their result -the Furstenberg-Zimmer Theorem. -In fact, it seems to me that -Furstenberg didn't even call it a theorem. :-P -I could find a precise statement of the theorem -for the invertible case at a -Terry Tao's post. -But I could not find any precise statement for the non-invertible case. - -REPLY [3 votes]: Posted as requested - consult the book by Manfred Einsiedler and Tom Ward - "Ergodic Theory with a view towards number theory" - published in GTM, especially in ch 7.<|endoftext|> -TITLE: Mathematicians failing to solve problems despite having all methods required -QUESTION [20 upvotes]: On this wikipedia page, there is the following quote by Anil Nerode: - -Being attached to a speculation is not a good guide to research planning. One should always try both directions of every problem. Prejudice has caused famous mathematicians to fail to solve famous problems whose solution was opposite to their expectations, even though they had developed all the methods required. - -What are some good examples of this? - -REPLY [2 votes]: The history of the first version of Poincaré's essay submitted to the competition sponsored by King Oscar II of Sweden, could be representative of this situation but at the same time of the capacity of overcome previous errors. -The problem of the stability of a planetary system was central from the dawn of newtonian mechanics. A father of Analytical mechanics such as Dirichlet thought to have proved the stability for the n-body problem, but he died suddenly before to write it. -The prize of King Oscar was aimed to obtain such a proof of the stability and infact in the first version of his essay, Poincaré claimed the stability of the restricted 3-body problem. This essay won the prize, but just after the publication of the paper in the Acta he realized the presence of a serious error for the presence of homoclinic orbits. Consequently the published issues were recalled and the second version of the essay was printed. -The existence of a first version of Poincaré's essay has been discovered only in 1994 by June Green-Barrow. The second version is known as the starting point of the qualitative geometric methods in mechanics. -For more information a possible source is Diacu, F., The solution of the n-body problem, Math. Intelligencer 18 (3) 66-70, 1996.<|endoftext|> -TITLE: Krull-Schmidt Analogue for Complete / Graded Rings -QUESTION [5 upvotes]: Over the ring $\mathbb{Z}$, all finitely-generated modules decompose uniquely as a direct sum of indecomposable submodules; that's the Krull-Schmidt theorem. -I'm given to understand that if a (commutative, Noetherian) ring $R$ is $\mathbb{N}$-graded over a field $k$ (and the degree zero part of $R$ is equal to $k$), then $R$ satisfies this same Krull-Schmidt condition. I'm told that the same holds if $R$ is a complete local ring over a field. -On the other hand, I can't find a good reference for either of these facts. (I can find references to the statements of both facts, but I dislike the notion of citing an unsupported assertion...) So: can anyone point me to a good proof of a Krull-Schmidt theorem for graded or complete local rings? Thanks! - -REPLY [4 votes]: Self-advertisement alert: In Chapter 1 of my book with Roger Wiegand we give a complete proof for complete local rings. It follows from a more general fact about additive categories in which every idempotent splits, that the key property is that endomorphism rings of indecomposable modules must be local (in the non-commutative sense). We don't have much use for graded rings, so don't address them,<|endoftext|> -TITLE: Is the function $e^{x^2/2} \Phi(x)$ monotone increasing? -QUESTION [11 upvotes]: Hello, -Here is an interesting problem. It looks elementary, but it has taken me some efforts without solving it. Let -$$ -h(x) = e^{x^2/2} \Phi(x),\qquad \text{with}\quad \Phi(x):=\int_{-\infty}^x \frac{e^{-y^2/2}}{\sqrt{2\pi}} dy. -$$ -The question is whether the function $h(x)$ is monotone increasing over $R$? Are there some work dealing with such function? -It seems a quite easy problem. By taking the first derivative, we need to prove that -$$ -h(x)' = h(x) x + \frac{1}{\sqrt{2\pi}} \ge 0. -$$ -which again, not obvious (for $x<0$). Some facts, that might be useful, are: -$$ -\lim_{x\rightarrow -\infty} h(x) =0, \quad \lim_{x\rightarrow -\infty} h(x)' =0. -$$ -Thank you very much for any hints! -Anand - -REPLY [2 votes]: A useful inequality is -\begin{equation} -\tag{1} -\frac{1}{\sqrt{2 \pi}} \frac{x}{x^2+1} \mathrm{e}^{-x^2/2} \le \frac{1}{\sqrt{2\pi}} \int_x^\infty \mathrm{e}^{-y^2/2} \, \mathrm{d} y \le \frac{1}{\sqrt{2 \pi}} \frac{1}{x} \mathrm{e}^{-x^2/2} -\end{equation} -for $x >0$, which can be shown by elementary arguments (see below). For example, one can prove the law of the iterated logarithm (for the Browian motion) by means of this inequality. By symmetry, we have for $x <0$ that -$$|x \mathrm{e}^{x^2/2} \Phi(x)| = \frac{|x|\mathrm{e}^{x^2/2} }{\sqrt{2\pi}} \int_{|x|}^\infty \mathrm{e}^{-y^2/2} \, \mathrm{d} y \le \frac{1}{\sqrt{2\pi}}$$ -and this proves $h'(x) \ge 0$. -The second inequality in $(1)$ can be proven by using $\exp(-y^2) \le y/x \exp(-y^2)$ for $y \ge x$ and the first one in $(1)$ follows by applying partial integration in the form -\begin{equation*} -\frac{1}{x^2} \int_x^\infty \mathrm{e}^{-y^2/2} \, \mathrm{d} y \ge \int_x^\infty \frac{1}{y^2} e^{-y^2/2} = \frac{1}{x} \mathrm{e}^{-x^2/2} - \int_x^\infty \mathrm{e}^{-y^2/2} -\end{equation*} -and properly rearranging.<|endoftext|> -TITLE: A result of Shelah about the nonstationary ideal -QUESTION [8 upvotes]: Suppose that $\kappa$ is a regular cardinal and let $NS$ be the ideal of its nonstationary subsets. One can consider the Boolean algebra $P(\kappa) /NS$ and say that (if $\lambda$ is another cardinal) $NS$ is $\lambda$ saturated iff there are no antichains in $P(\kappa) / NS$ of length $\lambda$. It is an elegant result of Gitik and Shelah that $NS$ cannot be $\kappa^+$ saturated for every regular $\kappa > \aleph_1$, on the other hand Foreman Magidor and Shelah could show that assuming a supercompact cardinal it is consistent that $NS$ of $\omega_1$ is $\aleph_2$ saturated. These results are all well known and one can find them for example in Jech's book. However it is stated there that Shelah eventually found that even a Woodin cardinal suffices to obtain the consistency of the statement "$NS$ on $\omega_1$ is $\aleph_2$ saturated". -Do you know where I can find a proof of this result? -Thank you - -REPLY [7 votes]: I have a writeup Shelah's proof of the consistency of "NS is saturated" relative to "there is a Woodin cardinal" on my web page: https://ivv5hpp.uni-muenster.de/u/rds/sat_ideal_better_version.pdf<|endoftext|> -TITLE: Turing machines that always halt -QUESTION [8 upvotes]: Needed for this paper: -Here is a possibly more clear version of my question. A Turing machine (with $1$ tape) has sets of tape letters $Y$, state letters $Q$, two symbols $\alpha$ and $\omega$ that mark the ends of the tape and a set of commands $\Theta$. A configuration is any word of the form $\alpha uqv\omega$ where $u,v$ are words in $Y$, $q\in Q$. Usually we distinguish the stop state $q_0$ and the input state $q_1$. The input configuration is any configuration of the form $\alpha uq_1\omega$. A command of the Turing machine is a substitution $aqb\to a'q'b'$ where $a,a', b, b'$ are either empty or letters or symbols $\alpha,\omega$ (with natural restriction: if, say, $a=\alpha$, then $a'=\alpha$, etc.).The command is applicable to a configuration $\alpha uqv\omega$ if the configuration contains a subword equal to $aqb$. -The machine can start working with any configuration $\alpha u q v\omega$. If it starts working with an input configuration $\alpha u q_1\omega$ and ends with a configuration containing $q_0$ we say that the machine accepts $u$.The machine can stop without accepting by arriving to a configuration where no command from $\Theta$ is applicable. -The language of all words accepted by a Turing machine $M$ is denoted by $L(M)$. Any recursive set $L$ of words is accepted by a (deterministic) Turing machine which stops on any input word (but accepts only words from $L$). That machine may never stop when started with some non-input configuration. For every $L$ it is possible to construct a Turing machine with $L=L(M)$ and which stops starting with every configuration (not necessarily input). - -Question: Where in the literature can I find a construction of such a Turing machine (for every recursive $L$)? - -REPLY [2 votes]: [I know this has already been answered satisfactorily, but for the sake of future readers, here's some additional terminology and an additional reference.] -These are also known as mortal Turing machines. See, e.g., this answer. Also discussed in - -Hooper, P. K. The Undecidability of the Turing Machine Immortality Problem. J. Symb. Logic, 1966. - -The key to the proof in the above paper is avoiding infinite loops starting from unreachable (finite) configurations, which is the same as the key to the construction alluded to in the question. -(This paper also contains the great line: "Most of the detailed work has been banished to the Appendices, to give - the casual reader an opportunity to sample the flavor of the construction - without choking on its bones.") -Note, however, that some authors use the term "mortal" differently, e.g. Hughes in "Undecidability of finite convergence for concatenation, insertion and bounded shuffle operators" uses it to mean a TM that halts on all configurations including infinite ones, and he proves that any such TM runs in $O(1)$ time, so this version of mortality is obviously too strong for most purposes.<|endoftext|> -TITLE: Formality of de Rham algebra for two-dimensional closed surfaces -QUESTION [5 upvotes]: Is it possible to embed de Rham cohomology of a two-dimensional closed surface of genus $g\geq 2$ into the differential graded algebra of differential forms (with de Rham differential and wedge product) on the surface as a differential graded subalgebra (in a way that is compatible with canonical projection from closed forms to cohomology, associating to a closed form its cohomology class)? - -REPLY [14 votes]: The answer is no. Suppose that $\alpha_1,\ldots,\alpha_g,\beta_1,\ldots,\beta_g$ were closed $1$-forms on $M$ such that their cohomology classes were a basis of $H^1(M)$ and they satisfied $\alpha_i\wedge\alpha_j = \beta_i\wedge\beta_j = 0$ while $\alpha_i\wedge\beta_j = \delta_{ij}\ \gamma$ where $\gamma$ is a single $2$-form whose cohomology class is nonzero. Then $\gamma$ cannot vanish identically. -Let $U\subset M$ be the open set on which $\gamma$ is nonzero. Then none of the $\alpha_i$ or $\beta_i$ can vanish on $U$. Since $\alpha_1\wedge\alpha_2 = \beta_1\wedge\alpha_2=0$ while $\alpha_2\not=0$ on $U$, it follows that $\alpha_1$ and $\beta_1$ are multiples of $\alpha_2$ on $U$. But this implies that $\alpha_1$ and $\beta_1$ must be linearly dependent on $U$ as well, which implies that $\alpha_1\wedge\beta_1 = 0$.<|endoftext|> -TITLE: partial Derivatives of Eigen value decomposition or Singular value decomposition -QUESTION [5 upvotes]: Hi All, -Suppose I've a symmetric matrix $A_{N \times N} = (A_{ij})$ which has a eigen value decomposition $A = UDU'$. I would like to know under what conditions $\frac{\partial U}{\partial A_{ij}}$ exists for all $i,j = 1,2, \ldots, N$. I found the following paper which talks about estimating the Jacobian of the SVD transformation -http://www.ics.forth.gr/cvrl/publications/conferences/2000_eccv_SVD_jacobian.pdf -But its not very clear regarding the conditions that the matrix $A$ would need to satisfy. Any help is much appreciated. -Thanks -Ashin - -REPLY [5 votes]: To make Igor's more precise, Kato's book tells us that - -if an eigenvalue of a matrix $A$ is simple, then it extends as an analytic function $M\mapsto\lambda(M)$ defined in a neighbourhood of $A$, such that $\lambda(M)$ is an eigenvalue of $M$. -if $s\mapsto A(s)$ is an analytic, one-parameter, family of real symmetric matrices, their eigenvalues $\lambda_1(s),\ldots,\lambda_n(s)$ can be arranged so that they are analytic functions. Mind that they are not in increasing order in general, because their order can change at values of $s$ for which $A(s)$ has a multiple eigenvalue. -The previous result becomes false when the family depends upon several parameters. A typical example with two parameters is -$$A(s,t)=\begin{pmatrix} s & t \\\\ t & -s \end{pmatrix},$$ -for which the eigenvalues $\pm\sqrt{s^2+t^2}$ are even not $C^1$-functions. -However, Weyl inequalities tell us that for real symmetric matrices, ${\rm dist}({\rm Sp}(B),{\rm Sp}(A))\le\|B-A\|$. Hence the eigenvalues are Lipschitz function, with unit Lipschitz constant.<|endoftext|> -TITLE: Questions on Thurston's earthquake flow -QUESTION [9 upvotes]: Here are some questions about the earthquake deformation of hyperbolic surface that I can't answer or find references. -I briefly recall the settings. Let's fix a closed surface $S$ with genus $g\geq 2$. A point $h$ in the Teichmuller space $\mathscr{T}$ of $S$ may be thinked of -either (a) as a marked hyperbolic structure on $S$ -or -(b) as a conjugate class of representations $\pi_1(S)\rightarrow PSL(2,\mathbb{R})$. -For any simple closed curve $\alpha$ on $S$, the Fenchel-Nielson twist along $\alpha$ gives rise to a flow $\phi_\alpha^t$ on $\mathscr{T}$. Wolpert proved that $\phi_\alpha^t$ is an Hamiltonian flow with respect to the Weil-Petersson symplectic structure. We can describe the deformation in the representation level. Namely, take a representation $h:\pi_1(S)\rightarrow PSL(2,\mathbb{R})$ which defines a point in $\mathscr{T}$, if $\alpha$ is seperating, then $\pi_1(S)$ is the amalgamated product of two groups, and $\phi_\alpha^t(h)$ is a representation in which we modify restriction of $h$ on one of the two groups by conjugation. -Thurston's eathquake deformation is a generalization of the above construction where we take a geodesic lamination instead of simple closed curve. -Question 1: Is there any example of an explicit family of representations $h_t:\pi_1(S)\rightarrow PSL(2,\mathbb{R})$ which gives an earthquake deformation supported on some non-simple lamination? -Question 2: Is the earthquake flow Hamiltonian, say, generated by the "generalized length function" of laminations? -Question 3: Can one describe the limit of $\phi_\alpha^t(h)$, as a projectived measured lamination, when $t\rightarrow\pm\infty$ (at least when $\alpha$ is simple closed)? For example, is it a measured lamination supported on $\alpha$? -Remark to question 3: I'm not sure of this, it seems that the review of a paper of Bonahon suggests $\phi_\alpha^t$ extends to a non-trivial action on the Thurston boundary of $\mathscr{T}$. If it is true, then the limit in question 3 does not always exist. - -Addendum: Now I realized that I made some conceptual mistakes about Question 3. Once we think of the earthquake flow intuitively as "horocycle flow" like on the upper half plan, the picture would be clear. - -REPLY [5 votes]: On Q1 I would guess that you can find explicit deformations corresponding to an earthquake path on a non-simple lamination in the quite special case of the punctured torus. You might find some help for instance in -MR0697067 (85d:32047) -Waterman, Peter; Wolpert, Scott -Earthquakes and tessellations of Teichmüller space. -Trans. Amer. Math. Soc. 278 (1983), no. 1, 157–167. -The answer to Q2 is yes, the earthquake flow is the Hamiltonian flow of the length function. I'm not sure where this was first proved but you could check in Kerckhoff's paper on the Nielsen realization problem. -On Q3 I'm convinced that the earthquake path $\phi^t_\alpha(h)$ limits to $\alpha$, I believe you could prove it by understanding the asymptotic behavior of the length of closed curves on this 1-parameter family of metrics, using the tools in the Kerckhoff paper mentioned above. -As for your remark, as Igor Rivin mentiond, I don't believe the Bonahon paper that you cite states that the earthquake flow extends to the boundary as nicely.<|endoftext|> -TITLE: Two-cardinal diamond principles and saturation of the nonstationary ideal -QUESTION [11 upvotes]: In the paper "Stationary reflection and the club filter", the author Masahiro Shioya says that the club filter on $P_{\omega_1}(\lambda)$ cannot be $2^\lambda$-saturated for $\lambda > \omega_1$, citing Shelah's book "Nonstructure Theory" (in preparation). I have three questions: -1) Is there a published reference for this result? -2) Does the theorem apply to $P_{\omega_1}(\lambda) | S$ for an arbitrary stationary set $S$? -3) Does the proof go through a two-cardinal diamond principle? I.e., did Shelah prove (in ZFC) that $\lozenge_{\omega_1,\lambda}$ holds for $\lambda > \omega_1$? What about $\lozenge_{\omega_1,\lambda}(S)$ for arbitrary stationary $S$? -I am particularly interested in the case $\lambda = 2^{\omega} = \omega_2$. In this case $\lozenge_{\omega_1,\lambda}(S)$ was proved by Donder and Matet in the paper "Two cardinal versions of diamond" for stationary sets $S$ of the form $\lbrace a \in P_{\omega_1}(\lambda) : \sup a \in B\rbrace$ where $B \subset \lambda$ is a stationary set consisting of points of cofinality $\omega$. Does this hold for arbitrary stationary $S$? - -REPLY [3 votes]: I just came across this, but probably you resolved this question long ago. In any case, to address your questions in order as asked, -1) Shioya has an article on this. Let me know if you'd like me to ask him for a pdf. -2) I believe the consistency of local $\lambda^+$-saturation of $NS_{\omega_1,\lambda}$ is still open for $\lambda > \omega_1$. -3) Yes, the proof does go through proving in ZFC $\Diamond_{\omega_1,\lambda}$ for all $\lambda > \omega_1$.<|endoftext|> -TITLE: Grothendieck group for projective space over the dual numbers -QUESTION [10 upvotes]: Fix a field $k$. For a singular variety $X$, I understand that the Grothendieck group $K^0(X)$ of vector bundles on $X$ is not necessarily isomorphic to the Grothendieck group $K_0(X)$ of coherent sheaves on $X$. -I am curious to learn what is known about these two groups in one family of examples: $\mathbb P^n_{D}$, where $D$ is the dual numbers $D=k[\epsilon]/(\epsilon^2)$. -References would be especially appreciated, as I know very little about K-theory. - -REPLY [9 votes]: If $X$ is a noetherian separated scheme and $X_{red}$ its reduction , we have $K_0(X)=K_o(X_{red})$: in other words $K_o$ doesn't see nilpotents . -Much more generally and profoundly, Quillen has proved that for all his $K$-theory groups, $K_i(X)=K_i(X_{red})$. -In your particular case you thus have (in the following $T$ is an indeterminate) -$$K_0(\mathbb P^n_D)=K_0(\mathbb P^n_k)=\mathbb Z[T]/(T^{n+1})$$ -As for $K^0$, a special case of a theorem of Berthelot (SGA 6, Exposé VI, Théorème 1.1, page 365) -states that, for any commutative ring $A$, we have $K^0(\mathbb P^n_A)=K^o(A)[T]/(T^{n+1})$. -If $A=D=k[\epsilon]$, we have $K^0(D)=\mathbb Z$, since projective modules over local rings (like $D$) are free. -So here too $$K^0(\mathbb P^n_D)=\mathbb Z[T]/(T^{n+1})$$ -Bibliography -Srinivas has written this nice book on $K$-theory. -And as an homage to the recently sadly departed Daniel Quillen, let me refer to his groundbreaking paper - "Higher algebraic $K$-theory I", published in Springer's Lecture Notes LNM 341.<|endoftext|> -TITLE: Several questions on semi infinite flag manifold -QUESTION [6 upvotes]: Let $G$ be a connected simply-connected Lie group correspondence to simple Lie algebra $g$,consider the loop algebra $g((t))$ and so called "Natural Borel subalgebra" $n[t,t^{-1}]\oplus h[t]$,denoted by $\mathfrak{b}$ and consider ind-group $G((t))$ associated to $g((t))$ and ind group $N_{-}((t))$ and $H[[t]]$ correspondence to completion of $n_{-}[t,t^{-1}]$ and $h[t]$ in $g((t))$. -Look at the "Natural Borel subgroup" $B$ correspondence to "natural borel subalgebra" $\mathfrak{b}$ -It is equal to $N_{-}((t))$$H[[t]]$. Now consider the quotient $X:= G((t))/B$. Frenkel and Ben zvi claimed that $X$ can not be given a scheme or an ind-scheme structures. -My first question is How to see this? For example, if $G=SL_2(\mathbb{C})$, how to see $X$ can not be given a scheme structure? Can one use analogue of Birkhoff decomposition to give a scheme structure(the $w$-translate of big "semi-inifnite" cell form an open affine cover and algebra of regular functions on these big cell should identify with contragradient Wakimoto modules) -Then they claimed that $X$ should be viewed as formal loop space of finite dimensional flag variety $G/B_{-}$, which is $Hom(Spec\mathbb{C}((t)),G/B_{-})$. I wonder whether this statement is equivalent to say they are "isomorphic" to each other(in what sense?) and how to prove this statement? -Any hint and related comments is welcome. -Thanks! - -REPLY [6 votes]: About defining the (ind)scheme structure: working with particular strata is basically never a good way to do this. What you need in order to define an algebro-geometric object is to -define a functor from $Schemes$ to $Sets$ that it represents (it is enough to do it for affine schemes, i.e. it is enough to say what is an $R$-point of your space when $R$ is a ring). This is easy to do for semi-infinite flags. After you have done this, you can ask whether this functor is representable by a scheme or an ind-scheme (but I want to emphasize -that this question doesn't make sense before you define the functor).<|endoftext|> -TITLE: Are non-algebraic stacks useful in algebraic geometry? -QUESTION [13 upvotes]: The title is a bit vague. What I want to know is if there is any geometric application of non-algebraic stacks. I know e.g. the category of coherent sheaves is an example. But I want to ask if people think about some of the general stacks geometrically (instead of merely thinking about them as fibered categories). -If the paragraph still doesn't explain my purpose, let me be more specific: people talk about "locally around a curve in M_g" something happens. But they don't say "locally near a coherent sheaf in Coh", right? -I read Vistoli's notes on general stacks before I saw people talk about algebraic stacks geometrically, and I always feel like for me there is a gap between these two ways of thinking. I feel like e.g. when a birational geometer uses stacks as a tool he probably don't put general fibered categories in his mind, (although abstract nonsense shouldn't be a big problem for him) right? -[A related question is, is category over Schemes fibered in something other than groupoid useful in geometry? Why people only care about categories fibered in groupoids?] - -REPLY [9 votes]: The answer is a qualified yes, and it depends on how broadly you define "algebraic geometry". Here is a class of examples: -A sheaf with flat connection on a smooth variety is equivalent to a sheaf on the de Rham stack of that variety, and de Rham stacks of positive dimensional varieties are not algebraic. You are basically taking a quotient with respect to the action of an infinitesimal symmetry, and the corresponding covering morphism satisfies only a formal version of smoothness. More generally, there are variations on the notion of sheaf with connection, e.g., action of an algebroid, that can be viewed functorially using sheaves on non-algebraic stacks which are stackifications of reasonably nice formal groupoids. -There is a sketch of a theory of such stacks in sections 2.9.10-2.9.12 of Chiral Algebras by Beilinson and Drinfeld, but in fact, the entire book is secretly about them, restricted to the language of $D$-modules. This is a situation where you can do some kind of "geometry", but it is a more degenerate environment than most algebraic geometers tend to consider. For example, you may end up using "compound tensor structures" because ordinary tensor products on quasicoherent sheaves, with their nice properties, are unavailable.<|endoftext|> -TITLE: Smoothness of $f(\sqrt x)$ -QUESTION [17 upvotes]: I found that I need to use the following facts in a paper that I am writing. -Let $f\in C^\infty(\mathbb R)$, then - -If $f(0)=0$, then $f(x)=x g(x)$ for some $g\in C^\infty(\mathbb R)$. -If $f$ is even, i.e. $f(-x)=f(x)$ for all $x$, then $f(x)=g(x^2)$ for some $g\in C^\infty(\mathbb R)$. - -These facts are trivial for analytic functions (just look at the Taylor series). In the smooth case, one can prove them by analyzing the derivatives of $f(x)/x$ and $f(\sqrt x)$, respectively, and showing that they have certain limits at 0. However this is somewhat cumbersome (especially if one wants to analyze how $g$ depends on $f$). This is not a problem for me because the facts fall into the category "you should be able to prove this yourself if you are reading my paper". But I would like to know if there is a nicer proof. -For the first statement, I know the following trick (which can be found in textbooks): define -$$ - g(x) = \int_0^1 f'(tx) \ dt -$$ -and observe that $f(x)=xg(x)$, and $g\in C^\infty$ since the function $t\mapsto f'(tx)$ under the integral is smooth in the parameter $x$. As a bonus, this argument also shows easily that $g$ (as a point of $C^\infty$) depends smoothly on $f$. (This is another fact that I need to use.) -Is there a similarly nice proof of the second statement? And, by the way, is there a textbook reference for it? -Added. Here is a more precise mathematical question that more or less formalizes what I mean. The function $g$ such that $f(x)=g(x^2)$ is uniquely defined only on $\mathbb R_+$, and its extension to negative arguments involves some choice. -Is there a canonical way to associate $g$ to $f$? Even more precisely, can one make a mapping $f\mapsto g$ which is linear, preserves the pointwise multiplication, and is continuous as a map from $C^\infty$ to $C^\infty$? - -REPLY [3 votes]: There are general results on $C^\infty$ functions of several variables which are invariant under the action of a group, many of them due to Georges Glaeser. One of most beautiful is the following: take a $C^\infty$ function $f(x_1,\dots,x_n)$ which is symmetric, i.e. such that -for all permutations $\sigma$ of $\{1,\dots,n\}$, -$$ -f(x_{\sigma(1)},\dots,x_{\sigma(n)})=f(x_1,\dots,x_n). -$$ -Then it is a $C^\infty$ function $F$ of the standard symmetric functions -$$ -\sum x_j,\dots,\prod x_j. -$$ -This result is elementary for polynomials ($f$ polynomial, then $F$ polynomial) but the above result is highly non-trivial for smooth non-analytic functions.<|endoftext|> -TITLE: Does the Weierstrass function have a point of increase? -QUESTION [8 upvotes]: Problem -The Weierstrass function $W(x)$ is given by -$W(x)=\sum_{n\geq 0} a^n \cos(b^n \pi x)$ -where $0< a <1$ and $b$ is an odd integer such that $ab > 1+3\pi/2$. -A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is said to have a point of increase if there exists a $t \in \mathbb{R}$ and $\delta>0$ such that -$f(t-s)\leq f(t) \leq f(t+s) \quad \forall s \in [0,\delta]$. -So my question is does the Weierstrass function have a point of increase? -Motivation -In Burdzy's paper there is a proof that a Brownian motion does not have a point of increase. There are examples of nowhere differentiable functions which have a point of increase that one could construct but I have been having difficulty seeing if the Weierstrass function does. -I would be grateful for any references or heuristics regarding this problem, or any comments as to the difficulty. - -REPLY [10 votes]: The original proof of Weierstrass (see pages 4 to 7 in Elgar (ed.): Classics on Fractals, Westview Press, 2004) constructs, for any $x_0\in\mathbb{R}$, two sequences $(x'_n)$ and $(x''_n)$ such that -$$x'_n < x_0 < x''_n,\qquad x'_n\to x_0,\qquad x''_n\to x_0,$$ -but -$$\frac{W(x'_n)-W(x)}{x'_n-x}\qquad\text{and}\qquad -\frac{W(x''_n)-W(x)}{x''_n-x}$$ -are of opposite signs and their absolute values tend to infinity. This shows that $W(x)$ has no point of increase and no point of decrease.<|endoftext|> -TITLE: Convex Analytic or linear algebraic proof that a certain psd matrix is a sum of rank 1 psd matrices -QUESTION [5 upvotes]: Can you prove the following using techniques from convex analysis or linear algebra? I was originally seeking an elementary proof, but I think it is better to broaden the scope for this bounty question. - -A psd matrix of the form $\left(\begin{smallmatrix} -a & b & c\\ -b & c & d \\ -c & d & e -\end{smallmatrix}\right)$ can be written as the sum of finitely many rank 1 matrices of the form $\left(\begin{smallmatrix} -x^4 & x^3 y & x^2 y^2\\ -x^3 y & x^2 y^2 & x y^3 \\ -x^2 y^2 & x y^3 & y^4 -\end{smallmatrix}\right)$? - -Edit Thank you Greg for your answer. In our comments, we observe that every psd matrix of the form $\left(\begin{smallmatrix} -a & b & c & d\\ -b & c & d & e\\ -c & d & e & f\\ -d & e & f & g -\end{smallmatrix}\right)$ is a finite sum of rank 1 and rank 2 matrices. Can one prove the following in a convex analytic manner? - -A psd matrix of the form $\left(\begin{smallmatrix} -a & b & c & d\\ -b & c & d & e\\ -c & d & e & f\\ -d & e & f & g -\end{smallmatrix}\right)$ can be written as the sum of finitely many rank 1 matrices of the form $\left(\begin{smallmatrix} -x^6 & x^5 y & x^4 y^2 & x^3 y^3\\ -x^5 y & x^4 y^2 & x^3 y^3 & x^2 y^4\\ -x^4 y^2 & x^3 y^3 & x^2y^4 & x y^5 \\ -x^3y^3 & x^2y^4 & x y^5 & y^6 -\end{smallmatrix}\right)$? - - - Motivation The first question is a convex analytic proof to the (known fact) that $P_{2,4}=\Sigma_{2,4}$, while the second question is to prove $P_{2,6}=\Sigma_{2,6}$. Below, I describe the origin of the problem for the former. -I came upon this problem as a convex analytic approach to the (known fact) that $P_{2,4}=\Sigma_{2,4}$. Here $P_{2,4}$ is the cone of nonnegative-valued binary quartics. $\Sigma_{2,4}$ is the cone of binary quartics that are sums of squares. (A binary quartic is a homogeneous polynomial in 2 variables of degree 4). Obviously $P_{2,4}\subseteq \Sigma_{2,4}$. The standard proof that equality holds is by dehomonogizing and applying the Fundamental theorem of algebra. -Since the cone $\Sigma_{2,4}$ is closed in the vector space ${\mathbb{R}}[x,y]_4$ of homogeneous polynomials in 2 variables of degree 4, a separation theorem in convex geometry provides a necessary and sufficient condition for a binary quartic $f$ to lie in $\Sigma_{2,4}$. This condition is that, for any linear functional $T$ which spans an extremal ray of the dual cone $\Sigma_{2,4}^{\vee}$, we have $T(f)\ge 0$. -The cone of positive semidefinite matrices of the form $\left(\begin{smallmatrix} -a & b & c\\ -b & c & d \\ -c & d & e -\end{smallmatrix}\right)$ is isomorphic to $\Sigma_{2,4}^{\vee}$. Under this isomorphism, point evaluations correspond to rank 1 matrices of the form $\left(\begin{smallmatrix} -x^4 & x^3 y & x^2 y^2\\ -x^3 y & x^2 y^2 & x y^3 \\ -x^2 y^2 & x y^3 & y^4 -\end{smallmatrix}\right)$. The above proof of which I'm seeking a convex analytic proof is equivalent to the assertion $P_{2,4}=\Sigma_{2,4}$. - -REPLY [6 votes]: $\newcommand{\R}{\mathbb{R}}$ -Let $K$ be a compact convex body in $\R^n$, or some other $n$-dimensional vector space or affine space. Then every point $p \in K$ has an extremality rank, which is the largest dimension of a flat open ball $B$ such that $p \in B \subset K$. The 0-extremal points are thus the usual extremal points, while the $n$-extremal points are the interior points. Also, the finite-dimensional Krein-Milman theorem says that $K$ is the convex hull of its extremal points. Also, if we intersect $K$ with a hyperplane $H \subset \R^n$, then the extremality rank of a point $p \in H \cap K$ is either the same or one less than its extremality rank in $K$. In particular, the extremality rank of $p$ cannot decrease by more than 1. If $K$ has no 1-extremal points, then the extremal points of $K \cap H$ are all extremal points of $K$ as well. -Let $K_n \subset S^2(\R^n)$ be the convex body of positive, semidefinite symmetric matrices with trace 1. Since you can canonicalize $p \in K_n$ as a symmetric form, its extremality rank can only depend on its rank as a matrix. (Well, an arbitrary change of basis won't preserve the trace, but that doesn't matter since it still yields a projective transformation on the trace 1 affine space. It may have been better to do this without the trace 1 condition, with closed cones instead, but the compact version is easier to see.) If it has matrix rank $r$, then it lives in the interior of an extremal copy of $K_r$, so its extremality rank is $\binom{r}{2}-1$. In particular, $K_n$ has no 1-extremal points. The extremal points are those of the form $v \otimes v$. After that are the 2-extremal points, which are rank 2 matrices $v \otimes v + w \otimes w$. Actually, you can see things most clearly by recognizing $K_2$ as a round 2-dimensional disk, which is a convex set that have "vertices" and interior points but no edges. Anyway, it means that if you impose any linear condition on PSD matrices represented by a hyperplane $H$, the extremal points in $K_n \cap H$ are still rank 1 matrices (that satisfy the same condition). -Your question is a special case of this general result. You are looking at $K_3 \cap H$, where $H$ is the condition that the middle entry of the matrix equals the northeast or southwest entry. The extremal points are all of the form $v \otimes v \in H$, which forces $v$ to have the form $(x^2,xy,y^2)$. -(Note that the complex version of $K_n$ is an important object in quantum information theory; it's convex body of mixed states on an $n$-state qudit. That is how I learned about this.)<|endoftext|> -TITLE: Conjugacy in $GL(n,\mathbb Z)$ -QUESTION [10 upvotes]: How can I determine whether $A_1,A_2\in GL(n,\mathbb Z)$ conjugate in $GL(n,\mathbb Z)$ and if they are, how can I find a $P\in GL(n,\mathbb Z)$ for which $A_2 = P^{-1}.A_1.P$ ? -In $GL(n,\mathbb Q)$ one could achieve this by checking if the Frobenius normal forms (FNF) are equal and if they are -$\quad\quad FNF_2 = FNF_1$ -$\Leftrightarrow P_2^{-1}.A_2.P_2=P_1^{-1}.A_1.P_1$ -$\Leftrightarrow A_2=M^{-1}.A_1.M\quad\quad\quad M=P_1.P_2^{-1}$ -I found an algorithm which gives the FNF of a matrix with P a matrix of integers. Is there an way of performing subsequent elementary similarity transformations on $P_i$ (and hence also on $P_i^{-1}$) until $P_i\in GL(n,\mathbb Z)$ while also checking whether it is even possible to arrive at such a $P_i$? - -REPLY [10 votes]: The Conjugacy Problem in $\operatorname{GL}(n, \mathbb{Z})$ by Eick, Hofmann and O'Brien gives an algorithm for solving this problem, which has been implemented in Magma.<|endoftext|> -TITLE: Increasing union of contractible CW complexes -QUESTION [6 upvotes]: Let X be CW complex. I'm trying to prove (using Zorn's lemma) that there is maximal contractible subcomplex. Problem is that I'm not able to show that increasing union of contractible subcomplexes has to be contractible itself. - -REPLY [14 votes]: By various standard lemmas, a CW complex $X$ is contractible if and only if every map $u:S^{n-1}\to X$ (for any $n>0$) can be extended over $B^n$. In this context $u(S^{n-1})$ is compact and therefore (by another standard lemma) contained in some subcomplex with only finitely many cells. If $X$ is the union of some totally ordered family of subcomplexes $X_\alpha$, it follows that $u(S^{n-1})\subseteq X_\alpha$ for some $\alpha$. This is enough to prove what you want.<|endoftext|> -TITLE: The geometrical meaning of the common value in the law of sines in hyperbolic geometry -QUESTION [5 upvotes]: What is the geometrical meaning of the common value in the law of sines, $\frac{\sin A}{\sinh a} = \frac{\sin B}{\sinh b} = \frac{\sin C}{\sinh c}$ in hyperbolic geometry? I know the meaning of this value only in Euclidean and spherical geometry. -EDIT, Will Jagy. The OP is looking for some fourth fairly natural real number that can be calculated from a triangle, that gives the same answer as the common value in the Law of Sines. The original question is at https://math.stackexchange.com/questions/69345/the-law-of-sines-in-hyperbolic-geometry - -REPLY [3 votes]: There is a meaning, though whether it is geometric is up to you to decide. You can see it in: -http://mathworld.wolfram.com/GeneralizedLawofSines.html -which is fairly incomprehensible without -http://mathworld.wolfram.com/HyperbolicPolarSine.html -To see how you might derive such a thing, see this: -http://arxiv.org/abs/math/0211261<|endoftext|> -TITLE: Is the following a sufficient condition for asphericity? -QUESTION [19 upvotes]: I recently came across the following question while working on some problems on manifolds with lower Ricci curvature bounds. -Given $n$ does there exist a large $R>0$ with the following property: -Suppose $M^n$ is a closed Riemannian manifold of diameter $=1$ such that for any $p$ in the universal cover $\tilde M$ the ball $B(p,R)$ is contained in a homeomorphic copy of $\mathbb R^n$ which in turn is contained in $B(p,R+\frac{1}{10})$. Then $M$ is aspherical. -In our situation we had extra geometric assumptions which allowed us to prove asphericity but I've been wondering if the above holds as is. I suspect not but I could not construct a counterexample. - -REPLY [4 votes]: Sorry, I realized that this is not an answer. I am constructing a Riemannian 3-manifold $M$ with small diameter and nontrivial $\pi_2 M$ such that for any point $p$ in the univesal cover $\widetilde M$ there is a sequence of open embeddings -$$B_R(p)\hookrightarrow\mathbb R^3\hookrightarrow B_{10\cdot R}(p),$$ -and its composition coinsides with the inclusionn $B_R(p)\hookrightarrow B_{10\cdot R}(p)$. -I hope that it still might be interesting. -Take a the surface of an $(2{\cdot}R+\tfrac1{100})$-long and $\varepsilon$-thin cylinder $C$ with caps in $\mathbb R^3$ (further $C$ is called sausage). - Think of it as a surface of revolution around $X$-axis. -Idetify points on $C$ along the folloing equivalence relation -$$x\sim y\ \ \ \text{if}\ \ \ x-y=(\tfrac12,\varepsilon,0).$$ -This way you obtain a $2$-dimensional CW-complex, say $W=C/\sim$ with $\pi_1 W=\mathbb Z$ and nontrivial $\pi_2 W$. -If you equip $W$ with the induced intrinsic metric then then $\mathop{\rm diam} W\approx \tfrac12$ and any $R$-ball in the universal cover $\widetilde W$ is contractible in a ball of radius $R+\tfrac1{10}$. -(A rough reason: $\widetilde W$ glued from a sequence of sausages. If a ball of radius $R$ intersects one sausage then it can not contain it all, but the ball of radius $R+\tfrac1{10}$ with the same center containa at least one of the ends, which makes possible to shrink the intrsection to a point.) -Now $W$ can be embedded into $\mathbb R^3$, it seems that thickening and then doubling produces a $3$-dimensional manifold $M$ with the property described above. -(Fortunately or unfortunately, any ball in $\widetilde M$ contains a closed curve such that to shrink it one has to go about $R$-far out of the ball.)<|endoftext|> -TITLE: Taking "Zooming in on a point of a graph" seriously -QUESTION [65 upvotes]: In calculus classes it is sometimes said that the tangent line to a curve at a point is the line that we get by "zooming in" on that point with an infinitely powerful microscope. This explanation never really translates into a formal definition - we instead approximate the tangent line by secant lines. -I seem to have found a way to obtain tangent lines (and more) by taking "zooming in" seriously. -Example 1 -Take the curve $y = x(x-1)(x+1)$. -I want to find an equation for the tangent line to this curve at the origin. So I zoom in on the origin with a microscope of magnification power $c$ (i.e. I stretch both vertically and horizontally by a factor of $c$) to obtain -$\frac{y}{c} = \frac{x}{c}(\frac{x}{c} - 1)(\frac{x}{c}+1)$. -Multiplying through by $c$ I have -$y = x(\frac{x}{c} - 1)(\frac{x}{c}+1) $ -Now letting my magnification power go to infinity I have -$y = -x$ -Which is the correct answer. -Example 2 -Take the curve $y = x^2$. -I want to find an equation for the tangent line to this curve at the point (3,9). I first rewrite the equation as -$(y-9) + 9= ((x-3) + 3)^2$ -so that I am focusing on the appropriate point. To zoom on this point with magnification $c$ I have -$\frac{y-9}{c} + 9 = (\frac{x-3}{c} + 3)^2$. -$\frac{y-9}{c} + 9 = \frac{(x-3)^2}{c^2} + 6\frac{x-3}{c} + 9 $ -Multiplying through by $c$ I have -$y - 9 = \frac{(x-3)^2}{c} + 6(x-3) $ -Now letting my magnification power $c$ go to infinity I have -$y - 9 = 6(x-3)$ -Which is the correct answer. -Example 3 -Here is the example which actually motivated me to consider this at all: -Take the curve $y^2 = x^2(1 - x)$. -This is a cubic curve with a singularity at the origin, and so it doesn't really have a well defined tangent line. It sort of looks like it should have two tangent lines (y = x, and y = -x), but it is a little bit tricky to formalize this. Let's see what "zooming in" does: -$\frac{y^2}{c^2} = \frac{x^2}{c^2}(1 - \frac{x}{c})$ -$y^2 = x^2(1 - \frac{x}{c})$ -Letting $c$ go to infinity I have -$y^2 = x^2$, or $(y-x)(y+x) = 0$, which is the pair of lines I desired. -My Questions - -Do any books take this approach when developing the derivative? -I would imagine that algebraic geometers do this kind of thing formally. Is there a more rigorous analogue of the prestidigitation I engage in above? Where would I look to read up on such things? - -p.s. It would be nice to illustrate each of these examples with a little movie of the "zooming in" process, but I am not sure how to put such things on MO. Any hints? - -REPLY [5 votes]: Excellent question! One of the earliest advocates of the "zooming in" approach is the mathematician and education specialist David Tall of Warwick; see e.g. http://www.tallfamily.co.uk/david/papers/2.embodied-calculus -His latest related paper is a reevaluation of Cauchy's legacy here. The basic idea is that ideas of zooming and informal infinitesimals should be introduced before any formalisation takes place. Once the student has a grasp of the key concepts, the course can branch out either into standard or non-standard calculus.<|endoftext|> -TITLE: quasinilpotence and finite spectrum -QUESTION [6 upvotes]: Let A be a quasinilpotent operator on a Hilbert space and let $A^{*}A$ have finite spectrum. -Does then follow, that A is nilpotent ? - -REPLY [2 votes]: I found a counterexample : -Let $e_{1},e_{2},...$ be ON basis of the Hilbert space and define A by -$Ae_{2n-1} = \sqrt{1-\frac{1}{n^{2}}} \ e_{2n} \ + \ \frac{1}{n} -\ e_{2n+1}$ , $\ \ $n=1,2,3,... , -$Ae_{2n} = 0$ , $\ \ $n=1,2,3,... -Then A is a partial isometry and therefore $A^{*}A$ a projection. -Furthermore A is quasinilpotent but not nilpotent.<|endoftext|> -TITLE: Is the ideal of a closure of a Bruhat cell generated by generalized minors? -QUESTION [10 upvotes]: Let $G$ be your favorite complex semi-simple algebraic group, and let $B\supset T$ be your favorite Borus. For any $w\in W$, we have the Bruhat cell $BwB$, and its closure $\overline{BwB}$. -Now, it's very easy to write down some functions that cut out this variety. Let $V$ be any finite-dimensional representation of highest weight $\lambda$, and let $v$ be highest weight vector, and $\delta$ a non-zero functional killing all but the highest weight space. Then the generalized minors $\omega_{w'\delta,v}(g)=w'\delta(gv)$ $\Delta_{w'\delta,v}(g)=w'\delta(gv)$ for all $w'$ with $w'\lambda > w\lambda$ all vanish on $BwB$ (since $BwBv$ in contained in the sum of weight spaces $\leq w\lambda$), and on no Bruhat cells $Bw'B$ with $w'>w$. -That is, the radical of the ideal generated by these functions is all functions vanishing on $\overline{BwB}$. In fact, it's enough to just consider $\lambda$ fundamental to get an ideal with the correct radical. So, my question is: - -Is the ideal generated by these generalized minors already radical? - -REPLY [3 votes]: Exactly as Alexander said, this will fail for $G/P$ nonminuscule. The smallest example is the closed orbit $SO(5)/P$ of $SO(5)$ acting on ${\mathbb P}({\mathbb C}^4)$. The $T$-weight diagram of this representation is -.1. -111 -.1. -The representation arises as the space of sections of ${\mathcal O}(1)$ on $SO(5)/P$. If we take the space of sections over the Schubert point $P/P$, we get just the top $1$. The extremal weight vectors you want to kill correspond to the left, right, and bottom $1$s. But to get the Schubert point on the nose, you have to kill the $1$ in the middle too.<|endoftext|> -TITLE: How has modern algebraic geometry affected other areas of math? -QUESTION [82 upvotes]: I have a friend who is very biased against algebraic geometry altogether. He says it's because it's about polynomials and he hates polynomials. I try to tell him about modern algebraic geometry, scheme theory, and especially the relative approach, things like algebraic spaces and stacks, etc, but he still thinks it sounds stupid. This stuff is very appealing for me and I think it's one of the most beautiful theories of math and that's enough for me to love it, but in our last talk about this he asked me well how has the modern view of algebraic geometry been useful or given cool results in math outside of algebraic geometry itself. I guess since I couldn't convince him that just studying itself was interesting, he wanted to know why else he'd want to study it if he isn't going to be an algebraic geometer. But I found myself unable to give him a good answer that involved anything outside of algebraic geometry or number theory (which he dislikes even more than polynomials). He really likes algebraic topology and homotopy theory and says he wants to learn more about the categorical approaches to algebraic topology and is also interested in differential and noncommutative geometry because of their applications to mathematical physics. I know that recently there's been a lot of overlap between algebraic topology/homotopy theory and algebraic geometry (A1 homotopy theory and such), and applications of algebraic geometry to string theory/mirror symmetry and the Konstevich school of noncommutative geometry. However, I am far from qualified to explain any of these things and have only picked up enough to know they will be extremely interesting to me when I get to the point that I can understand them, but that's not a satisfactory answer for him. I don't know enough to really explain how modern algebraic geometry has affected math outside of itself and number theory enough to spark interest in someone who doesn't just find it intrinsically interesting. -So my question are specifically as follows: -How would one explain how the modern view of algebraic geometry has affected or inspired or in any way advanced math outside of algebraic geometry and number theory? How would one explain why modern algebraic geometry is useful and interesting for someone who's not at all interested in classical algebraic geometry or number theory? Specifically why should someone who wants to learn modern algebraic topology/homotopy theory care or appreciate modern algebraic geometry? I'm not sure if this should be CW or not so tell me if it should. - -REPLY [17 votes]: My favorite such example is Stanley's proof of McMullen's conjecture characterizing the number of faces that a simplicial polytope can have. -I summarize the proof below (although Stanley's article is not much longer) in order to illustrate the extent to which "modern algebraic geometry" is used: -(1) Build a toric variety from the dual polytope -(2) Count points over a finite field. Since the toric variety is built by replacing each face of the dual polytope by an algebraic torus of this dimension, the number of points over a field with $q$ elements is -$$\sum_{faces} (q-1)^{\mbox{dimension of face}}$$ -where we include the enclosed face "inside" the polytope -(3) The original polytope being simplicial, hence the dual polytope being simple, means the toric variety is rationally smooth. In particular, its cohomology = its intersection cohomology -(4) Purity of intersection cohomology, hence of cohomology = intersection cohomology, together with the point count above, implies (by the Weil conjectures) that the variety has no odd cohomology and that its $2i$'th Betti number is the coefficient of $q^i$ above. -(5) Intersection cohomology obeys Hard Lefschetz, giving a condition on this Poincare polynomial, translating into a condition on the faces. (polytopes with arbitrary face numbers obeying this condition had already been constructed.) -Example: For a polygon, we have $$(q-1)^2 + (\#faces)(q-1) + (\#vertices) = q^2 + (\#faces-2) + 1$$ So by the Hard Lefschetz theorem, every lattice polygon has at least 3 sides.<|endoftext|> -TITLE: Finding Generators of O( Z^3,x^2 + xy + y^2 - z^2) and integer solutions -QUESTION [5 upvotes]: All pythagorean triples can be generated from $(3,4,5)^T$ and $(5,4,3)^T$ by the matrices: -\[ A = \left(\begin{array}{ccc} -1 & 2 & 2 \\\\ -2 & 1 & 2 \\\\-2 & 2 & 3 \end{array} \right) \hspace{0.25in} -B = \left(\begin{array}{ccc} 1 & 2 & 2 \\\\ 2 & 1 & 2 \\\\2 & 2 & 3 \end{array} \right) \hspace{0.25in} -C = \left(\begin{array}{ccc} 1 & -2 & 2 \\\\ 2 & -1 & 2 \\\\2 & -2 & 3 \end{array} \right) \] -Giving pythagorean triples the structure of a ternary tree. These correspond to generators of $\mathrm{O}(\mathbb{Z}^3, x^2+y^2-z^2)$ and we're looking for integer points on the ``light cone" where the norm is zero. -I remember seeing there being 5 generators for the quadratic $x^2 + xy + y^2 = z^2$. Does anyone have the reference? - -REPLY [7 votes]: I wouldn't call it GL, it is the orthogonal group of the lattice we are discussing. References, as i said, include Lattices and Codes by W. Ebeling, Rational Quadratic Forms by Cassels, these two being available and inexpensive. -We double the quadratic form to get an integral Gram matrix: -$$ G \; = \; - \left( \begin{array}{rrr} - 2 & 1 & 0 \\\ - 1 & 2 & 0 \\\ - 0 & 0 & -2 -\end{array} - \right). - $$ -Your solutions, the light cone, are column vectors $v$ such that $ v^T G v = 0.$ A root, since we have doubled everything to get an "even" lattice, is a vector $r$ with norm 2, $ r^T G r = 2.$ The general definition of reflection (Cassels calls this a symmetry, page 19) in any vector $w$ is that -$$ x \mapsto \; \; x \; - \; \frac{2 \, x^T G w}{w^T G w} \; w.$$ -As a result, when we take $w$ to be a root, the factors of 2 cancel and we are taking lattice points to other lattice points. In your original form, a root $(x,y,z)$ solves $x^2 + x y + y^2 = 1 + z^2.$ Then the reflection in the root is just a linear map, determinant $-1,$ and is therefore given by a square matrix with respect to the original basis. Finally, the reflection is an isometry, part of the orthogonal group of the quadratic form, and if we call the matix $A,$ it solves $$A^T G A = G.$$ -I think it wise to include reflection in the root $(1,0,0)^T$ to get some negative values taken care of, -$$ A_0 \; = \; - \left( \begin{array}{rrr} --1 & -1 & 0 \\\ -0 & 1 & 0 \\\ -0 & 0 & 1 -\end{array} - \right). - $$ -We also include reflection in the root $(0,1,0)^T,$ -$$ A_{00} \; = \; - \left( \begin{array}{rrr} -1 & 0 & 0 \\\ --1 & -1 & 0 \\\ -0 & 0 & 1 -\end{array} - \right). - $$ -Note that we have already included the automorph that interchanges the first two items, $$ (x,y,z) \mapsto (y,x,z). $$ -$$ A_0 A_{00} A_0 \; = \; - \left( \begin{array}{rrr} -0 & 1 & 0 \\\ -1 & 0 & 0 \\\ -0 & 0 & 1 -\end{array} - \right). - $$ -Although it may not be called a root, having norm $-2,$ we include reflection in $(0,0,1)^T,$ or -$$ A_{000} \; = \; - \left( \begin{array}{rrr} -1 & 0 & 0 \\\ -0 & 1 & 0 \\\ -0 & 0 & -1 -\end{array} - \right). - $$ -Also, as is usual, we explicitly include negation, -$$ (x,y,z) \mapsto (-x,-y,-z), $$ so -$$ A_{0000} \; = \; - \left( \begin{array}{rrr} --1 & 0 & 0 \\\ -0 & -1 & 0 \\\ -0 & 0 & -1 -\end{array} - \right). - $$ -It occurred to me that we would get some reflections with quite small matrix entries by taking other vectors of norm $-2,$ for example $(1,1,2)^T.$ -$$ A_{-1} \; = \; - \left( \begin{array}{rrr} -4 & 3 & -4 \\\ -3 & 4 & -4 \\\ -6 & 6 & -7 -\end{array} - \right). - $$ This still solves, as with all the $A'$s, both $A^T G A = G$ and $A^2 = I.$ -Taking the root $(3,4,6)^T$ gives the reflection -$$ A_1 \; = \; - \left( \begin{array}{rrr} - -29 & -33 & 36 \\\ --40 & -43 & 48 \\\ --60 & -66 & 73 -\end{array} - \right). - $$ -If we take the initial triple in the light cone to be $c = (3,5,7)^T,$ we get $A_1 c = (0,1,1)^T$ which is indeed another solution, and shows that care must be used in constructing the "tree." -Taking the root $(5,15,18)^T$ gives the reflection -$$ A_2 \; = \; - \left( \begin{array}{rrr} --124 & -175 & 180 \\\ --375 & -524 & 540 \\\ --450 & -630 & 649 -\end{array} - \right). - $$ -we get $A_2 c = (13,35,43)^T$ -Taking the root $(8,19,24)^T$ gives the reflection -$$ A_3 \; = \; - \left( \begin{array}{rrr} --279 & -368 & 384 \\\ --665 & -873 & 912 \\\ --840 & -1104 & 1153 -\end{array} - \right). - $$ -we get $A_3 c = (11,24,31)^T$ -Taking the root $(7,32,36)^T$ gives the reflection -$$ A_4 \; = \; - \left( \begin{array}{rrr} --321 & -497 & 504 \\\ --1472 & -2271 & 2304 \\\ --1656 & -2556 & 2593 -\end{array} - \right). - $$ -we get $A_4 c = (80,357,403)^T$ -Taking the root $(12,47,54)^T$ gives the reflection -$$ A_5 \; = \; - \left( \begin{array}{rrr} --851 & -1272 & 1296 \\\ --3337 & -4981 & 5076 \\\ --3834 & -5724 & 5833 -\end{array} - \right). - $$ -we get $A_5 c = (159,616,709)^T$ -One typically includes $\pm 1$ anyway. Ian Agol would know how many reflections are enough, but I suspect this will do. Nothing really wrong with finding too many. -Note that the squarefree parts in $1+z^2$ in the five nontrivial roots I chose are $37, 13, 577, 1297, 2917.$<|endoftext|> -TITLE: dimension of a real affine variety -QUESTION [5 upvotes]: Let $V$ be a real affine variety in $\mathbb R^n$, i.e. the zero set of a real polynomial $p(x_1,\dots,x_n)$. Consider the following three definitions of the dimension of $V$, $dim(V)$. - -Definition 1: if $I$ is the ideal of - polynomials vanishing on $V$, then - $dim(V)$ is the maximum dimension of a - coordinate subspace in $V(\langle -> LT(I)\rangle)$ with a given graded - order $>$ on the monomials - ($\sum\alpha_i>\sum\beta_i$ implies - $x^\alpha>x^\beta$) (see the book - "Ideals, Varieties, and Algorithms" - chapter 9) - -Definition 2: $dim(V)$ is the largest $d$ such that there exists an injective semi-algebraic map from $(0, 1)^d$ to $V$ (see "Algorithms in Real Algebraic Geometry" chapter 5) - -Definition 3: $dim(V)$ is the largest $d$ such that there exists a - subset of $V$ homeomorphic to $(0, -> 1)^d$ - -Are these three definitions, for the case of real affine varieties, pairwise equivalent, or pairwise different, or something else? - -REPLY [4 votes]: They are all equivalent, including definition 1, to the Krull dimension of $S/I$, where $S=\mathbb{R}[x_1,\ldots,x_n]$ is the polynomial ring $I$ lives in. This is very good news for people like me who want to apply algebraic geometry to statistics, where numbers are mostly real. -Here's how it goes: -Definition 1 is a way of computing the Krull dimension of $S/I$ via Groebner bases. See, e.g., p. 250 of Computing in algebraic geometry: a quick start using SINGULAR by Wolfram Decker and Christoph Lossen -Definition 2 is shown equivalent to Krull dimension in Corollary 2.8.9 of Real Algebraic Geometry by Bochnak, Coste and Roy. Note that they define the "dimension" of a real variety to be Krull dimension of its coordinate ring. I recommend reading the whole chapter. -Definition 3 is equivalent to Defintion 2 because any semialgebraic set admits a decomposition into finitely many pieces homeomorphic to $(0,1)^d$ (see Theorem 2.3.6 in RAG), and a finite union of semialgebraic sets of dimension less than $d$ cannot contain a set of dimension $d$. I.e., the only way a real variety can contain an open set homeomorphic to $(0,1)^d$ is by containing a semialgebraic set homeomorphic to $(0,1)^d$ in its semialgebraic cell decomposition.<|endoftext|> -TITLE: prime numbers and Pythagorean triplets -QUESTION [8 upvotes]: What are the necessary conditions for two of the terms in the Pythagorean triplet $a^2 = b^2 + c^2$ to be prime numbers? - -REPLY [20 votes]: There is a well-known parametrization of Phythagorean triples as $k(m^2 - n^2)$, $2kmn$ , $k(m^2 + n^2)$ with positive integers $k,m,n$ and $m$ greater $n$. -Now, if two are prime we get $k=1$. And also the middle term is never prime. -So the question is when are $m^2 - n^2$ and $m^2 + n^2$ both prime. -The former factors as $(m-n) (m+n)$. For this to be prime we need $m-n = 1$. -So we get two of three are prime if and only if $2n+1$ and $2n^2 + 2n + 1$ are prime for some positive integer $n$.<|endoftext|> -TITLE: Quotient of flat module is flat - a property in Mumford's Red book -QUESTION [8 upvotes]: Hi, -In Mumford's "Red Book of Varieties and Schemes", Chapter III, Paragraph 10 (entitled "Flat and smooth morphisms"), the following property is stated: -Let $M$ be a $B$-module, and $B$ an algebra over $A$. Let $f\in B$ have the property that for all maximal ideals $m \subset A$, multiplication by $f$ is injective in $M / m \cdot M$. Then $M$ flat over $A$ implies $M / f \cdot M$ flat over $A$. -My question is whether this statement is true as stated, without some finiteness assumptions (it is not clear, for example, what is the role of $B$ here if no finiteness is assumed), and, secondly, can someone indicate a proof, or an exact reference for a proof. -Thank you, -Sasha - -REPLY [2 votes]: The purpose of this post is to analyze the situation while ignoring $B$ altogether. Fix a maximal ideal $m$ of $A$. By tensoring the exact sequence $0 \to M \to M \to M/fM \to 0$ with $A/m$ one deduces immediately that -$$Tor_i^A(M/fM, A/m) =0 $$ -for $i>0$. -Flatness can be checked locally on maximal ideals, so we may as well let $R=A_m, k = R/mR$ and considering: - -If $Tor_i^R(N,k) =0$ for $i>0$, when is $N$ flat? - -In general, $N$ may not be flat, as a-fortiori's example show. However, it will be flat in many situations. The finiteness assumptions in EGA suffice. You can also check out the paper "A local flatness criterion ..." by Hans Schoutens (available on his website). For example, there he showed that $N$ will be flat if we assume it has projective dimension at most one. So if we know $M$ is projective to begin with, then $M/fM$ will be flat.<|endoftext|> -TITLE: What is the "right" definition of the homology(cohomology) of an orbifold? -QUESTION [11 upvotes]: What is the "right" analog in the orbifold case of a singular homology of a topological space? -We can not just take the homology of the underlying space, because it does not contain much information. -For example, is there any kind homology of orbifolds such that the first homology group is the abelianization of the fundamental group of the orbifold? And such that an $n$-dim orbifold will have all homology group higher than $n$-dim equal to zero? And if there is such a homology, will there be Poincare duality in the orbifold case??? -Thanks very much! - -REPLY [8 votes]: If your orbifold is given to you as a proper etale Lie/topological groupoid, you can: - -take the nerve of this groupoid, to get a simplicial manifold/space, -take the levelwise singular set to get a bisimplicial set, -take the diagonal to get a simplicial set, -take geometric realisation to get a topological space, -and then take (co)homology. - -This recipe at the very least gives you the homotopy type of the orbifold as a space (to which you can apply the (co)homology functor you want), but it perhaps too big and unwieldy for your purposes.<|endoftext|> -TITLE: When is a holomorphic tangent bundle stable? -QUESTION [9 upvotes]: Hello, I was thinking about the stability condition in terms of Mumford and I have a question: - -If $S$ is a compact Kahler surface (complex 2D), when is the holomorphic tangent bundle $T^{1,0}S$ stable? - -When $S$ is Kahler-Einstein, $T^{1,0} S$ is stable by the Donaldson-Uhlenbeck-Yau theorem. But Kahler-Einstein examples are rather restrictive as it requires the hermitian metric $h$ on $TS$ to be induced by the riemannian metric $g$. Would there be more examples other than Kahler-Einstein with stable holomorphic tangent bundles? -Thank you in advance. -===== -The following has been added to clarify some points. -Appendix 1: What I meant by Kahler-Einstein being "restrictive". -Hermitian-Yang-Mills connections or equivalently Hermite-Einstein metrics can be thought as a generalization of Kahler-Einstein metrics. $T^{1,0}S$ admits a Hermite-Einstein metric with the hermitian structure induced from $(S, J, g)$ when $S$ is a Kahler-Einstein. However, in principle $T^{1,0}S$ can admit a Hermite-Einstein metric even when $S$ is not Kahler-Einstein. In this case, the hermitian structure on $T^{1,0}S$ will be different from the one naturally induced from $(S, J, g)$. -Appendix 2: What I meant by stable bundle. -The stability condition introduced by Mumford is the following. - -A vector bundle $V$ is stable if for all coherent sub-sheaves $U$ -$$\frac{\deg (U)}{\mathrm{rank} (U)}<\frac{\deg (V)}{\mathrm{rank} (V)}.$$ - -Here the degree is computed using the Kahler form $\omega$ (polarization). By the Donaldson-Uhlenbeck-Yau theorem, on Kahler manifolds stability is equivalent to $V$ admitting a Hermitian-Yang-Mills connection. This theorem was later generalized by Li and Yau assuming stability to non-Kahler manifolds. - -REPLY [6 votes]: What do you mean by stable? usually stable is with respect to a polarization $H$. If $K_X$ is ample and you choose $H=K_X$ then we know by Aubin-Yau's theorem that there exists a Kahler-Einstein metric. Thus, in this case, the existence of such a metric is not restrictive. -If you choose any ample line bundle $H$, then Donaldson has proved that if a holomorphic vector bundle $E$ over a compact Kahler manifold $M$ admits an approximate $\omega$-Einstein-Hermitian structure if and only if $E$ is $H$-semistable, where $\omega$ is a Kahler form in the class of $H$. -You can find many details in the book by Kobayashi -http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183554744<|endoftext|> -TITLE: A topological space for which having the ccc is independent of ZFC? -QUESTION [9 upvotes]: It is well known that a generalized Cantor space $2^A$ is separable if and only if $|A| \leq 2^{\aleph_0}$. This means that one cannot decide in $ZFC$ whether the space $2^{\omega_2}$ is separable or not. On the other hand it is also well known that every $2^A$ has the ccc. - -Does anyone know a topological space - for which its ccc status cannot be - decided in $ZFC$? - -(A topological space has the ccc if any pair-wise disjoint collection of open sets is at most countable. A space is separable if it has a countable dense subset) -Note: I´m not looking for ad hoc examples like: "let $X$ be a dense subset of $2^{\omega_2}$ of minimum size and give $X$ the discrete topology". - -REPLY [8 votes]: Let $T$ be the $L$-least Suslin tree of $L$, with the usual cone topology. Thus, in $L$, this is a Suslin tree, a tree of height $\omega_1$ wtih all countable levels, and satisfying the countable chain condition. As a topological space, it is c.c.c. in $L$. -This tree is absolutely definable, in the sense that the definition, "the $L$-least Suslin tree in $L$," picks out exactly the same object in the universe as it does in all inner and outer models of the universe. -But meanwhile, it is independent of ZFC whether this space is ccc or not, since if $V=L$, it definitely is c.c.c., but in the forcing extension where we have forced over this tree, then this tree is no longer c.c.c. -There are many other examples in the same vein. For example, - -the partial order $\text{Coll}(\omega,\omega_1^L)$, consisting of finite partial functions from $\omega$ to $\omega_1^L$. In $L$, this partial order is not c.c.c., but in a universe where $\omega_1^L$ has become countable, then the poset is countable and hence c.c.c. And the partial order is abolutely definable. -More generally, for any absolutely definable ordinal $\theta$, you may consider the partial order of all finite partial functions from $\omega$ to $\theta$. If $\theta$ is uncountable, then this partial order is not c.c.c., but if it is, then the whole partial order is countable and hence c.c.c. And it is absolutely definable. - -Perhaps you will object that these examples are ad hoc in the sense you mention, but because these examples involve absolutely definable partial orders, I think you will have a hard time to cache out a robust concept of ad hoc that excludes them. The absolutely definable nature of these posets would seem to make them even more non-ad-hoc than the sample spaces you mention, which are not literally the same space in a model of set theory as in all its forcing extensions.<|endoftext|> -TITLE: Ideals of the ring of smooth functions -QUESTION [18 upvotes]: The ring $C^\infty(M)$ of smooth functions on a smooth manifold $M$ is a topological ring with respect to the Whitney topology and the usual ring operations. Is it possible to describe, maybe under some conditions on $M$, the ideals and the closed ideals of $C^\infty(M)$? - -REPLY [20 votes]: In 1948 Whitney proved his ideal (spectral) theorem [1] describing the closed ideals -Let $M$ be an $n$-dimensional manifold. -for each point $p \in M$ and each natural $k$ we define $N(k)$ to be the number of (up to $n$) tuples $m$ such that $|m| \leq k$. Define the map $J_p^k: C^\infty(M) \rightarrow \mathbb{R}^{N(k)}$ by assigning to $f$ the $m$-jets of $f$ at $p$ up to $|m|=k$. -If $I$ is an ideal of $C^\infty(M)$ then its closure is the ideal of functions $f$ such that for each $p$ in $M$ and $k \geq 0$ then $J^k_p f \in J^k_p(I)$. -So in some sense the closed ideals are like $I_\infty$ in Neil's answer. -[1] H.Whitney. On ideals of differentiable functions. American Journal of Mathematics. Vol. 70, No. 3, pp. 635-658 (1948)<|endoftext|> -TITLE: Example for Busemann function is not an exhaustion when Ricci $\ge 0$ -QUESTION [5 upvotes]: For an open complete Riemannian manifold $M$ with non-negative sectional curvature, the Busemann function defined below is a convex exhaustion function (by Cheeger-Gromoll's proof of soul theorem) -The Buesemann function: -$$b(x)=\sup_{\gamma} b_{\gamma}(x)$$ -where the $sup$ is taken over all rays from a given point and $b_{\gamma}$ is the Busemann function associated with ray $\gamma$: -$$b_{\gamma}(x)=\lim_{t\to \infty}(t-d(x, \gamma(t))$$ -A function $f:M\to \mathbb [a,\infty)$ is called an exhaustion function on $M$ if its sublevel set $\Omega_c:=f^{-1}((-\infty, c])$ is compact for all $c$ and $M=\cup_c \Omega_c$. -I am wondering that when we assume Ricci curvature is non-negative, is there any example where the Busemann function $b$ is not an exhaustion? (In this case the Busemann function is subharmonic.) - -REPLY [4 votes]: Sorry it's not an answer. I just came across a similar question recently and found a reference. -If the manifold has Ricci nonnegative and Euclidean volume growth, then the Busemann function is an exhaustion function. This is a result due to Zhongmin Shen. -See P400 Lemma 3.4 in Shen, Zhongmin, Complete manifolds with nonnegative Ricci curvature and large volume growth. Invent. Math. 125 (1996), no. 3, 393–404. -But I did not check the details of this result.<|endoftext|> -TITLE: classification of irreducible admissible (g,K)-module for GL(3,R) -QUESTION [6 upvotes]: classification of irreducible admissible (g,K)-module for GL(3,R) -Is there a classification of irreducible admissible (g,K)-module for GL(3,R)? -For GL(2,R) we have principal series, discrete series and etc. Is there such a result for GL(3,R) or GL(n,R)? - -REPLY [9 votes]: For a general real reductive group, all irreducible admissible $({\mathfrak g},K)$-modules are quotients of parabolically-induced discrete series (or limits thereof) representations (where we allow "trivial" parabolic induction ($P=G$) for discrete series on the group). See Theorem 14.92 in Knapp's Representation Theory of Semisimple Groups. This is a refinement of the Langlands Classification (which replaces "discrete series" with "tempered"). Knapp's paper "Local Langlands Correspondence: the archimedean case", in volume 2 of Motives, PSPM 55, gives an explicit classification for $GL_n$ (over $\mathbb R$ and $\mathbb C$). Also see Moeglin's article "Representations of GL(n) over the Real Field" in Representation Theory and Automorphic Forms, PSPM 61. -For $GL_n(\mathbb R)$, we can say that given an irreducible admissible $({\mathfrak g},K)$-module $V$, there exists a parabolic subgroup $P=MN$ of $GL_n$ with block sizes either $1$ or $2$ (since $GL_n$ only has discrete series for $n=1$ or $2$), and a discrete series representation $\sigma$ of $M$, such that $V$ is isomorphic to the unique quotient of the $({\mathfrak g},K)$-module underlying ${\rm Ind}_P^G(\sigma,s)$, where $s$ is a tuple of complex parameters, one for each block in $M$. Further analysis can tell you when two induced representations give you the same $({\mathfrak g},K)$-module, and when the induced representation is irreducible.<|endoftext|> -TITLE: Introductory text on Galois representations -QUESTION [40 upvotes]: Could someone please recommend a good introductory text on Galois representations? In particular, something that might help with reading Serre's "Abelian l-Adic Representations and Elliptic Curves" and "Propriétés galoisiennes des points d'ordre fini des courbes elliptiques". - -REPLY [59 votes]: Kevin Ventullo's suggestion of Silverman's book is a very good one. The first examples -of Galois representations in nature are Tate modules of elliptic curves, and if you haven't -read about them in Silverman's book, you should. -If you have read Silverman's book, a nice paper to read is Serre and Tate's "On the good reduction of abelian varieties". It is a research paper, not a text-book, and is at a higher level than Silverman (especially in its use of algebraic geometry), but it has the merit of being short and beautifully written, and uses Galois representation techniques throughout. -One fantastic paper is Swinnerton-Dyer's article in Lecture Notes 350. Here he -explains various things about the Galois representations attached to modular forms. -The existence of the Galois representations is taken as a black box, but he explains -the Galois theoretic significance of various congruences on the coefficients of the -modular forms. Reading it is a good way to get a concrete feeling of what Galois -representations are and how you can think about and argue with them. -Another source is Ken Ribet's article "Galois representations attached to modular forms -with nebentypus" (or something like that) in one of the later Antwerp volumes. It presupposes some understanding of modular forms, but this would be wise to obtain anyway if you want to learn about elliptic curves, and again demonstrates lots of Galois representation techniques. It would be a good sequel to Swinnerton-Dyer's article. -Yet another good article to read is Ribet's "Converse to Herbrand's criterion" article, which is a real classic. It is reasonably accessible if you know class field theory, know a little bit about Jacobians (or are willing to take some results on faith, using your knowledge of elliptic curves as an intuitive guide), and something about modular forms. Mazur recently wrote a very nice article surveying Ribet's, available here on his web-site. -One problem with reading Serre is that he uses $p$-adic Hodge theory in a strong way, -but his language is a bit old-fashioned and out-dated (he was writing at a time when this -theory was in its infancy); what he calls "locally algebraic" representations would -now be called Hodge--Tate representations. To learn the modern formulation of and perspective on $p$-adic Hodge theory you can look at Laurent Berger's various exposes, available on his web-site. (This will tell you much more than you need to know for -Serre's book, though.) -For a two page introduction to Galois representation theory, -you could read Mark Kisin's What is ... a Galois representation? for a two-page -introduction. -Yet another source is the Fermat's Last Theorem book (Cornell--Silverman--Stevens), which -has many articles related to Galois representations, some more accessible than others. -The article of Taylor that Chandan mentioned in a comment is also very nice, although it -moves at a fairly rapid clip if you haven't seen any of it before. -Serre's article in Duke 54, in which he explains his conjecture about the modularity -of 2-dimensional mod p Galois representations, is also very beautiful, and involves -various concrete computations which could be helpful -One last remark: if you do want to understand Galois representations, you will need to have -a good understanding of the structure of the Galois groups of local fields (as described e.g. in Serre's book "Local fields"), in particular the role of the Frobenius element, -of the inertia subgroup, and of the significance of tame and wild inertia.<|endoftext|> -TITLE: When and where did the term "module" enter commutative algebra? -QUESTION [9 upvotes]: Bruns/Herzog "Cohen-Macaulay-Rings" has a note in the notes for Chapter 1, saying roughly that after the influx of homological algebra into commutative ring theory, modules became popular objects (instead of ideals). They cite Gröbner's 1949 book "Moderne algebraische Geometrie" as the birthplace of "Vektormoduln", which are submodules of free modules. When did the term "module" (with its current definition) appear first, and why would the word "module" be chosen for this concept? - -REPLY [8 votes]: From the website Chris Dionne mentioned in the comments: -MODULE. A JSTOR search found the English term in E. T. Bell’s “Successive Generalizations in the Theory of Numbers,” American Mathematical Monthly, 34, (1927), 55-75. Bell was describing the work of Dedekind, basing his account on Dedekind’s French article, “Sur la Théorie des Nombres entiers algébriques” (1877) Gesammelte mathematische Werke 3 pp. 262-298. Dedekind used the French word module to translate his German term Modul. Stillwell writes in the Introduction to his English translation, Theory of Algebraic Integers (1996, p. 5), “Dedekind presumably chose the name ‘module’ because a module M is something for which ‘congruence modulo M’ is meaningful.” Curiously le module had once before been translated into English but then it went into English as the MODULUS of a complex number. [John Aldrich]<|endoftext|> -TITLE: Statistics for Haar measure of random matrices? -QUESTION [10 upvotes]: Let's say I have $M$ samples of $N\times N$ real orthogonal matrices. What statistics can I calculate to test if they could have been drawn from a distribution consistent with Haar measure over $O(N)$? -This question is related to this previous question. - -REPLY [4 votes]: The ensemble of real orthogonal matrices (uniformly distributed with respect to the Haar measure) is the socalled Circular Real Ensemble (CRE) of random-matrix theory. (Not the Circular Orthogonal Ensemble, COE, which confusingly enough contains symmetric complex unitary matrices.) -The probability distribution of the eigenvalues in the CRE is known, and you could use that to test whether your set of matrices is indeed drawn from that ensemble. (For example, by comparing moments of the eigenvalues or by comparing the spacing distribution.) -You can find the eigenvalue probability distribution in Section 2.9.2 of P.J. Forrester's book "Log-gases and Random matrices". You will have to distinguish the cases of determinant equal to +1 or -1, and even or odd N. The formulas are a bit lengthy, but if the book is not available to you, let me know and I will try to record them here. -Alternatively, you could just generate matrices from the CRE and compare the eigenvalue statistics of those matrices with your M matrices from an unknown ensemble. Generating matrices from the CRE is easy: They contain the eigenvectors of random real symmetric matrices with a Gaussian distribution of the matrix elements (the Gaussian Orthogonal Ensemble --- yes, here the word orthogonal is appropriate).<|endoftext|> -TITLE: Hall's treatment of algebraic operations -QUESTION [5 upvotes]: Marshall Hall, in his famous book Theory of Groups, does not always require a binary operation be "well-defined", i.e. an operation is a relation instead of a function (there might be more than one $c$ s.t. $ab=c$). For example, in the discussion of "quasi-group with the inverse property", he use the inverse property to prove that the binary operation is well-defined. This is, I think, different from modern treatment. Could you recommend other references about such treatment? - -REPLY [6 votes]: I am looking at page 7 of Hall's book and I do not see that he is using binary operation in -any non-standard fashion. He defines a "quasigroup with the inverse property" to be a set -with a binary product and unary inverse such that -$$ - a^{-1}(ab) = b = (ba)a^{-1}. -$$ -He then wants to show that this is a quasigroup, by which he means a set with a binary operation such that if $ab=c$, then any two of $a$, $b$, $c$ determine the third. This means -he must show that $ax=c$ and $yb=c$ have unique solutions, and that is exactly what he does. That $a$ and $b$ determines $c$ is a consequence of his definition of binary operation. -I can see that Hall's wording is slightly misleading, but he is not using binary operation -in any sense that differs from what we do now. (In his earlier presentation of the axioms for a group he is quite explicit that a product takes a unique value.)<|endoftext|> -TITLE: Subexpressions of reduced words in Coxeter groups -QUESTION [9 upvotes]: Let $\underline{w} = [s_1, s_2, \dots ,s_n]$ be a reduced expression in a Coxeter group $W$. Given $x$ in $W$ one can consider the set $\Pi(\underline{w},x)$ consisting of all subexpressions of $\underline{w}$ with product $x$. (A subexpression of $\underline{w}$ is a sequence $[t_1, t_2, \dots, t_n]$ such that $t_i$ is either the identity or $s_i$ for all $i$. Given a subexpression $[t_1, \dots, t_n]$ its product is (obviously) $t_1 \dots t_n$. -There are lots ways of thinking about subepxpressions. Probably one of the nicest is as a path in the Coxeter complex, where at time $= i$ one either chooses to either stay put or to cross a wall coloured by $s_i$. In this interpretation the product is the end-point of the path. -My question is the following: -How should one go about calculating the set $\Pi(\underline{w},x)$? -Of course there is the obvious answer: enumerate all $2^n$-subsequences and calculate all products (or perhaps take some obvious subset given, for example by length restrictions). However, this obviously gets difficult quickly. -I am wondering if anybody has studied this problem from a computational point of view. In small examples it seems like there are a lot of tricks that give an answer much faster than the brute-force method above. -(My motivation is that I have recently been writing software to do calculations in the Hecke algebra, where this problem is the bottleneck. Often I have a long reduced expression (length 40 or so) and think that one should be able to calculate $\Pi(\underline{w},x)$ for elements which are not too much smaller than w (say length 30). However either it's hard or I'm missing something!) -Myself and my computer would like to thank you in advance for any labour saving tips! - -REPLY [4 votes]: I have found a much more efficient way of solving this problem on computer. Having asked the question I guess I should provide a brief account. However I feel like the algorithm is technical and not very enlightening, and so I'll be brief. Please leave comments if you would like more detail. -First we consider an important special case: Is there an efficient algorithm to find all subexpressions of a fixed (reduced) expression $\underline{w} = [s_1, \dots, s_m]$ whose product is the identity? In the language of the question, how does one calculate $\Pi(\underline{w}, id)$? -Of course, there is one canonical such subexpression: $(id,id,…,id)$. Moreover, given any subexpression for the identity we can apply "cancellation moves" to get to this expression. -By a "cancellation move" I mean the following: write $\pi[i,j]$ for the product $t_it_{i+1}…t_j$. Suppose that -$s_i \pi[i+1,j-1] = \pi[i+1,j-1] s_j$ -then a "cancellation move" is one of the following: -$\pi = [\dots ,s_i, \dots, id, \dots] \mapsto \pi ' = [\dots, id,…,s_j, \dots]$ -$\pi = [\dots ,s_i, \dots, s_j, \dots ]\mapsto \pi' = [\dots ,id, \dots, id, \dots]$ -where $\pi$ remains unchanged except at the $i^{th}$ and $j^{th}$ place. Obviously, applying a cancellation move to a subexpression does not change the product. Also, cancellation moves either increase the number of $id$'s or increase the number of $id$'s to the left. It follows that by repeatedly applying cancellation moves to any subexpression for the identity we end up at the canonical subexpression. -For example, consider the Coxeter group with simple reflections $s$, $t$ and $u$ and only braid relation $(st)^3 = id$ (so that $su$ and $tu$ have infinite order). Let $\underline{w} = [s,t,s,u,t,s,t]$. Then the following is a sequence of cancellation moves: -$[s,t,s,id,t,s,t] \mapsto [id,t,s,id,id,s,t] \mapsto [id,t,id,id,id,id,t] \mapsto [id, id, id, id, id, id, id]$. -Now one can reverse this, and define "reverse cancellation moves". Applying all possible reverse cancellation moves to the canonical subexpression yields all subexpressions for the identity. (This is easily programmed.) -We now turn to the general case. The above "cancellation moves" make sense for any subexpression and one can show that again there is a canonical subexpression. (It is characterised by the fact that the subexpression is reduced, and the $t_i \ne id$ occur as far to the right as possible.) Again, once one has this canonical subexpression then it is a simple matter to reconstruct all subexpressions by reverse cancellation moves. -Of course this begs the question: can one find this canonical subexpression efficiently? Yes. Firstly, consider $R(x)$ the right descent set of $x$, and choose $s \in R(x)$ such that s occurs as far as possible to the right in $\underline{w}$. Now, delete everything to the right and including the right-most occurrence of $s$ and repeat with $x$ replaced by $xs$. (This also gives a reasonably efficient algorithm to decide whether $x \le y$ in the Bruhat order.) -(One can ask what this canonical expression "means". Here is one possible explanation: because Schubert varieties are normal the fibre of the Bott-Samelson resolution over point corresponding to $x \le w$ is connected and hence its homology is one-dimensional in degree zero. Now subexpressions also index BB-cells, and hence a basis for the cohomology of the fibre. This "canonical subexpression" corresponds to a generator of the one-dimensional degree zero part.) -For example, take $W = S_7$ (with simple reflections = simple transpositions) and take the following reduced expression for the longest element: -$w_0 = 121321432154321654321$ -then there are 6408 subexpressions with product the identity. On my laptop the above algorithm takes 3.33 seconds to find them. A brute force attack takes 93 seconds. -Here is another example which I actually cared about in my calculations. Take the reduced expression -$\underline{w} = 13572613574352461357$ -in $S_8$. There are 80 subexpressions with product the longest element in the standard parabolic subgroup generated by all simple transpositions except 4. The above algorithm takes 0.03 seconds to find all of them, whereas the naive approach takes roughly 15 seconds. (So here the above algorithm is 500 times as fast.) -I imagine that the differences become (even) more pronounced with longer words.<|endoftext|> -TITLE: What is the identity class of the set of equivalence classes of binary cubic forms of discriminant $D$ ? -QUESTION [5 upvotes]: Let $\Delta = \sigma + 4 m$ be the fundamental discriminant of a quadratic field, where $\sigma \in \{ 0, 1 \}$. The binary quadratic form $Q(x, y) = A x^2 + B x y + C y^2$ of discriminant $\Delta$ belongs to the identity class (principal) of the narrow class group of forms, under substitution by $(x,y) \mapsto M (x', y')$ with $M \in \text{SL}_2(\mathbb{Z})$, if there exist rational integers $x, y$ satisfying $Q(x, y) = 1$, or equivalently, if $Q(x, y)$ is equivalent to $Q_0(x, y) = x^2 + \sigma x y - m y^2$. Davenport and Heilbronn, On the density of discriminants of cubic fields (I), Bull. Lond. Math. Soc. 1, (1969), 345--348, for example, considered binary cubic forms $a x^3 + b x^2 y + c x y^2 + d y^3$ of discriminant $D = 18 a b c d + b^2 c^2 - 4 a c^3 - 4 b^3 d - 27 a^2 d^2$, and their quadratic covariant binary quadratic forms $(b^2 - 3 a c) x^2 + (b c - 9 a d) x y + (c^2 - 3 b d) y^2$, of disc $\Delta = - 3 D$. M. Bhargava has described the composition of binary cubic forms in which $3 \mid b, c$. I would have thought, probably mistakenly, that a group law on the former binary cubic forms would consist of composing quadratic covariants and finding a binary cubic form for which the covariant belongs to the class of the composed form. I would like to know which class of binary cubic forms to consider as the neutral element of a class group of forms. For example, the binary cubic form $C = (39820040, 28889459, 6986439, 563185)$ of discriminant $D = -3299$ has quadratic covariant $x^2 - 99 x y - 24 y^2$ of discriminant $9897$. Does $C$ belong to an identity class of a class group of binary cubic forms? \\ -Lemma : Let $\mathcal{P} : x^2 + \sigma x y - m y^2 = 1$ be a Pell conic with $\Delta = - 3 D = \sigma + 4 m$, $\sigma \in \{ 0, 1 \}$, and let $(x, y) = (b, - 3 a) \in \mathcal{P}(\mathbb{Z})$ be the least non-trivial point such that $3 \mid y$. Let $c = \frac{b^2 - 1}{3 a}$, and $d = \frac{b c - \sigma }{9 a}$. Then the binary cubic form $(a, b, c, d)$ of discriminant $D$ has quadratic covariant equal to $Q_0(x, y)$. \\ -Proof : Since $b^2 - 3 \sigma a b - 9 m a^2 = 1$, $c = \frac{b^2 - 1}{3 a} = \sigma b + 3 a m \in \mathbb{Z} $. Also, -$$d = \frac{b c - \sigma }{9 a} = \frac{b (\sigma b + 3 a m ) - \sigma }{9 a} = \frac{\sigma (b^2 - 1) + 3 a b m }{9 a} = \frac{3 \sigma a b + 9 \sigma a^2 m + 3 a b m }{9 a} = \frac{\sigma - D}{4} b + \sigma a m ,$$ -which is clearly an integer. The following identity shows that $(a, b, c, d)$ is a binary cubic form of discriminant $D$. $$18 a b c d + b^2 c^2 - 4 a c^3 - 4 b^3 d - 27 a^2 d^2 = \frac{4 \sigma b - 3 \sigma a - 4 \frac{b^2 - 1}{3 a}}{9 a} = -\frac{\sigma + 4 m }{3} = D .$$ The quadratic covariant of $C(x ,y) = (a, b, c, d)$ is $$( b^2 - 3 a c , b c - 9 a d , c^2 - 3 b d ) = ( b^2 - b^2 + 1, b c - b c + \sigma , \frac{(b^2 - 1)^2 - b^2 (b^2 - 1) + 3 \sigma a b}{9 a^2} ),$$ equal to $(1, \sigma , - m )$. Should $C(x, y)$ be called principal? When $\Delta > 0$, what is a different $\text{SL}_2(\mathbb{Z})$ class of binary cubic form with principal quadratic covariant? \\ -References : \\ -Lemmermeyer, Conics A poor mans elliptic curves http://www.rzuser.uni-heidelberg.de/~hb3/ \\ -Bhargava, High composition laws and applications, http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_13.pdf - -REPLY [3 votes]: I'm not sure that I will be answering your question, so let me first recall the background. The equivalence classes of primitive binary cubic forms with discriminant $\Delta$ correspond to $SL_2(\mathbb Z)$-equivalence classes of triply symmetric Bhargava cubes. There is a natural homomorphism from this group to the strict class group of binary quadratic forms with discriminant $\Delta$, and in fact the image has order dividing three for (almost trivial) reasons. The kernel of this homomorphism has order $(U:U^3)$, where $U$ is the unit group of the quadratic order with discriminant $\Delta$. -The group structure on the set of binary cubic forms is defined via the group structure -on the group of $\Gamma$-equivalence classes of Bhargava cubes, where $\Gamma$ is the direct product of three copies of $SL_2(\mathbb Z)$. Given a binary quadratic form $Q$ with discriminant $\Delta$ it is easy to see that there is exactly one $\Gamma$-equivalence class of triply symmetric cubes belonging to $(Q,Q,Q)$. -Your problem seems to come from confusing the $\Gamma$-equivalence classes of cubes in the second paragraph with the $SL_2(\mathbb Z)$-equivalence classes of cubes from the first one, and I think that this is what you asked in your comment - observe that an element -$S \in SL_2(\mathbb Z)$ acts on a cube via the action of $(S,S,S)$. -I do not understand your question in the numerical example; perhaps you can clarify this part by editing the question.<|endoftext|> -TITLE: Records for low-height points on elliptic curves over number fields -QUESTION [12 upvotes]: Elkies maintains a list of nontorsion points of low height on elliptic curves over Q; does anyone know of anything similar for curves over number fields? -Everest and Ward give examples of points of height 0.01032... and 0.009721... on curves over Q(w) for w a cube root of unity or the golden ratio respectively. I have made a modest improvement in the latter case, recovering a point of height 0.009128... . -In the context of the elliptic Lehmer problem the aim is to minimise dh(P) for d the degree of the number field, so working over quadratic extensions a point would have to have height less than 0.005 to be competitive with the examples in Elkies' table. Are there any examples? - -REPLY [6 votes]: Using PARI, I recently extended Taylor's computations based on Elkies' parameterizations and found four examples in quadratic fields of non-torsion points whose heights are smaller than those on Taylor's website. -As in Elkies' parameterization, $P=(0,0)$ for each example. -a, c and u are also as in his parameterization. -${\mathbb K}={\mathbb Q}(\sqrt{d})$ and -hgtRatio=2hgt(P)/$\log \left( {\mathcal N}_{{\mathbb K}/{\mathbb Q}} \left( \Delta_{E} \right) \right)$. -w=quadgen(d), if d is congruent to 1 mod 4 -and w=quadgen(4*d) otherwise. -hgt=0.000681, hgtRatio=0.00001794, d=10, a=3 - w, u=1 + 1/2*w, c=8/9, -a1=-118 - 53*w, a2=2520 + 1044*w, a3=-755640 - 239040*w -hgt=0.001466, hgtRatio=0.00003851, d=7, a=2 + w, u=1/7*w, c=343, -a1=-16422 - 6207*w, a2=43423359 + 16412487*w, a3=-1276637821704 - 482523741504*w -hgt=0.001481, hgtRatio=0.00003735, d=33, a=1, u=-3 + w, c=1/3, -a1=-123 + 35*w, a2=-471018 + 139674*w, a3=435016512 - 128997696*w -hgt=0.001597, hgtRatio=0.00005050, d=7, a=1/3 + 1/3*w, u=-1/2, c=288, -a1=443 + 167*w, a2=-178080 - 67308*w, a3=6681612 + 2525412*w -The example above with the smallest height also has the smallest hgtRatio, 0.00001794. -The second smallest value of hgtRatio occurs for the following example: -hgt=0.001822, hgtRatio=0.00002038, d=7, a=1/7*w, u=-5/6 + 1/6*w, c=13608, -a1=4081 + 1423*w, a2=-33595908 - 12371892*w, a3=63747980088 + 23913434976*w<|endoftext|> -TITLE: Cohomology of fixed point subspaces -QUESTION [6 upvotes]: Suppose $M$ is a smooth manifold and $\phi : M \to M$ is a homeomorphism whose fixed point set is a smooth submanifold $M_{\phi}$. Is there any relation between the cohomology ring of $M_{\phi}$ and the cohomology ring of $M$ augmented by its natural $\mathbf{Z}[\phi]$-algebra structure? Is it too much to try to compute the former in terms of the latter? I am sorry for the vagueness of this; I am not sure precisely what question I want to be asking, but I am posting this to find out if I am completely wasting my time by looking for such interactions. - -REPLY [7 votes]: Suppose $G$ is a finite group on a space $X$. Then [Swan, Richard G. A new method in fixed point theory. Bull. Amer. Math. Soc. 65 1959 128--130. MR0107238 (21 #5963); Swan, Richard G. A new method in fixed point theory. Comment. Math. Helv. 34 1960 1--16. MR0115176 (22 #5978)] constructs a spectral sequence which has $$E_2^{i,j}=\hat H^i(G,H^j(X))$$ converging to a certain filtered graded group $J^\bullet(X)$. The spectral sequence degenerates at $E_2$ if the action of the group acts trivially. Here cohomology of spaces is Cech cohomology, with any coefficients you like, and $\hat H$ is Tate cohomology of groups. -Now, suppose $G$ is cyclic, that $X=M$ is a manifold, and that each point in $M$ with non-trivial stabilizer is actually fixed by the whole of $G$. Then Swan shows that $J^\bullet(X)$ is isomorphic to $J^\bullet(X^G)$. -Over a field, an under the above conditions, this gives information about $H^\bullet(X^G)$ from information on $H^\bullet(X)$ and its $G$-module structure. -Swan constructs a multiplicative structure on the spectral sequence, and this gives information on the cohomology ring of the invariants.<|endoftext|> -TITLE: Convex Hull of Path Connected sets -QUESTION [10 upvotes]: This is a pretty easy question to ask, but haven't seen it anywhere. - -Suppose I have some continuous path $X$ in $\mathbb{R}^n$ and I want to get the convex hull of $X$, $\operatorname{co}(X)$. -Is it enough to consider only pairwise convex combinations of points in $X$ to generate $\operatorname{co}(X)$? I.e., -$$\left(\forall z\in \operatorname{co}\left(X\right) \right) \left(\exists\lambda \in \left[0,1\right] \land x_{0},x_{1}\in X \right) \left(z=\lambda x_{0}+(1-\lambda)x_{1} \right)$$ -Also, if this is true, is it generalizable to more general topological spaces? Thanks! - -REPLY [3 votes]: This is just a computational footnote to the thrust of this old question, but -I wanted to mention that finding the convex hull of a polygonal path is -computationally easier in $\mathbb{R}^2$ that finding the hull of unconnected -points: It can be computed in $O(n)$—linear time—compared to the $\Omega(n \log n)$ lower bound for unconnected points: - -Melkman, Avraham A. "On-line construction of the convex hull of a simple polyline." Information Processing Letters 25, No. 1 (1987): 11-12. - (ACM link.) - -           - - -           - - Image from Joe Mitchell's course notes: - PDF download.<|endoftext|> -TITLE: What is the series expression for (1+1/x)^x about x = \infty? -QUESTION [8 upvotes]: This seems like it must have been addressed somewhere already, but I cannot find it in any standard series tables. -I have the equation: -$f(z) = \left(1 + \frac{1}{z}\right)^z$. -What is the general form for the $n$th term of the series? That is, if I have -$f(z) \sim \sum_{n=0}^{\infty} \frac{c_n}{z^n}$ -near $z = \infty$, what is the form of $c_n$? - -REPLY [14 votes]: Markus Brede proves the following formula in the paper "On the convergence of the sequence defining Euler’s number". Let $$\left(1+\frac{1}{z}\right)^z=\sum_{n\geq 0} \frac{a_n}{z^n}$$ -then we have -$$a_n=e\sum_{v=0}^n \frac{S(n+v,v)}{(n+v)!}\sum_{m=0}^{n-v}\frac{(-1)^m}{m!}$$ -where $S(a,b)$ are Stirling numbers of the first kind. This shows that all coefficients are rational multiples of $e$. I found the article through OEIS. - -REPLY [8 votes]: $\log f(z) = z \log(1+1/z) = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} z^{-k}$ as $z \to +\infty$, so $f(z)$ is the exponential of this sum. See http://oeis.org/A055505 for the numerators and http://oeis.org/A055535 for the denominators of the coefficients.<|endoftext|> -TITLE: reverse mathematics strength of "Lipschitz functions are somewhere differentiable" -QUESTION [6 upvotes]: What is the reverse mathematics strength of - -"For all Lipschitz functions $\; f : \mathbb{R} \to \mathbb{R} \;$, $\;$ there exists a real number $x$ such that $f$ is differentiable at $x$." ? - -(defined using epsilon-delta, so not requiring that there exist a function witnessing the convergence) -Since Lipschitz functions are differentiable almost everywhere, - -I would guess the answer is "is equivalent to WWKL$_0$ (over RCA$_0$)". - -REPLY [4 votes]: Here it is, the newest version -http://dl.dropbox.com/u/370127/papers/randomnessanalysisjuly_2011.pdf -What jason says is right, even for nondecreasing functions. It is in our paper, Subection 5.1 -Hence there is (I believe) a real $y$ computable from the function $f$ for which $f$ is differentiable at $x$. -Hence your theorem holds in every $\omega$-model of RCA0. -This implies that your theorem is true in RCA0 or RCA0 plus some first order principle. -Not sure what these fo principles do, I guess it's standard stuff?<|endoftext|> -TITLE: How did "Ore's Conjecture" become a conjecture? -QUESTION [21 upvotes]: The narrow question here concerns the history of one development in group theory, but the broader context involves the sometimes loose use of the term "conjecture". This goes back to older work of Oystein Ore, with whom I had only a slight interaction in 1963 as an entering graduate student at Yale (he generously found my reading knowledge of mathematical German adequate). -In 1951 his short paper Some remarks on commutators appeared in Proc. Amer. Math. Soc. -here. -He raised the question of expressing every element of a given group as a commutator, noting at the outset (but without a specific example): "In a group the product of two commutators need not be a commutator, consequently the commutator group of a given group cannot be defined as the set of all commutators,but only as the group generated by these." In the paper he proves that every element of a (nonabelian) finite or infinite symmetric group is in fact a commutator, while each element of an alternating group is a commutator in the ambient symmetric group. He notes that for the simple groups $A_n$ ($n \geq 5$) the proof can be adapted to show that each element is a commutator within the group, adding: "It is possible that a similar theorem holds for any simple group of finite order, but it seems that at present we do not have the necessary methods to investigate the question." -There was a lot of subsequent progress stimulated in part by work on the classification of finite simple groups, culminating recently in a definitive treatment: Liebeck-O’Brien-Shalev-Tiep, The Ore conjecture. J. Eur. Math. Soc. (JEMS) 12 (2010), no. 4, 939–1008. (See a related discussion in an older MO question 44269.) -But there is some distance between Ore's remark and the designation "conjecture", so I'm left with my question: - -How did "Ore's Conjecture" become a conjecture? - -This is not an isolated instance in mathematics, where terms such as "problem", "question", "(working) hypothesis" tend to morph into "conjecture". (This has happened to me, which is of course fine with me when my tentative suggestion turns out to be true.) An example I encountered as a reviewer occurs in a 1985 Russian paper by D.I. Panyushev with a title translated into English as A question concerning Steinberg's conjecture. The Russian word resembling "hypothesis" is also used in the sense of "conjecture", as it seems to be here; but Panyushev is providing counterexamples. As I noted in my review, Steinberg was explicitly stating a "problem" in his ICM address, not a conjecture, though he probably hoped for a positive solution. -ADDED: Igor's comments prompt me to mention that another 1951 paper dealt more directly than Ore's with commutators in the finite simple alternating groups: Noboru Ito, A theorem on the alternating group $\mathfrak{A}_n (n \geq 5)$. Math. Japonicae 2 (1951), 59–60. Both papers were reviewed by Graham Higman, who added a reference "Cf. O. Ore ...." to his one-line summary of Ito's paper. Since Ito went on to make substantial contributions to finite group theory, maybe one should refer to "Ito's conjecture" (even if he didn't formulate it any more explicitly than Ore did). -AFTERWORD: I asked the question partly out of curiosity after revisiting Ore's old paper, hoping there might be some further indication in the literature or at least in individual recollections of how to justify the transition to "conjecture". One person I should have asked years ago was one of my teachers Walter Feit, who joined the Yale faculty around 1962 and brought finite group theory into the department during the later years of Ore's career. Ore himself was never much involved with finite simple groups and in that period concentrated on graph theory. His speculative remark isn't at all strong, and he makes no mention of other simple groups (like the Mathieu groups or finite classical groups) known at the time. In the absence of evidence, I'm more inclined than Igor is to withhold judgment. The literature suggests a rather casual decision by other people to refer to "Ore's conjecture". Fortunately for his reputation it eventually turned out to be true, but this couldn't have been foreseen by him (or others) around 1950. -My other reason for asking for question is to emphasize that there is still no conjectural classification of which finite nonabelian groups consist of commutators and which don't. Everything known so far amounts to study of individual groups. In particular, why does simplicity contribute to a positive answer? - -REPLY [9 votes]: Not so much an answer, as a collection of comments: - -Ore's question became a conjecture quite early (there are some papers published in Chinese(!) in the early sixties where the math review refers to the "Ore conjecture"). My conjecture is that Ore (who was a very well-known mathematician then, and a professor at Yale, reasonably central then as now) must have talked about this result, and conjectured it with somewhat greater conviction than in his paper. -@Gil confounds two different Thompsons -- Robert C (who proved some results on linear groups) and John G the Fields medalist who made the "Thompson conjecture". -In fact, most of the statement @Gil attributes to RC Thompson goes back to Shoda, who showed that in a complex semi-simple lie group every element is a commutator; this was extended by various people. -It is a conjecture I attribute to myself, but probably goes back to the ancients, that in every reasonable simple group every element is a commutator. In particular, this ought to be true in $Homeo(M^n)$ and in $Diffeo(M^n).$ For homeo and diffeo it is known that everything is a product of "few" commutators (I think from some early work of Armstrong you can get six for Diffeo; von Neumann and Ulam claimed something like four for $Homeo(S^2).$ You can look at my paper with Manfred Droste ("on extension of coverings") for more examples of when this commutator statement is known to be true, and why one might care. -The reason I mention the stuff above is that the question was probably going around already in the late forties, and Ore, very likely, decided to show it in a relatively simple interesting case (most of his paper is actually on the infinite symmetric group) and made a conservative conjecture (being a conservative guy, I guess), but probably he already guessed something like what I am saying above.<|endoftext|> -TITLE: Topple height of randomly stacked bricks -QUESTION [40 upvotes]: What is the expected height of a stack of unit-length bricks, each one -stacked on the previous with a uniformly random shift within $\pm \delta$? -The stack topples if the center of gravity of the top $k$ does -not rest on the brick under that top group. For example, -here I used $\delta = \frac{1}{3}$, and the stack was stable after -the 21st brick was placed (the green top-group center-of-gravity dots), -but the 22nd brick caused the centers of gravity to shift (red dots), -with the cg of the top four bricks (arrow marked) now overhanging -the 18th brick beneath: - -           - - -Let $N(\delta)$ be the expected height (number of bricks) of such a stack. -Computing this is elementary in some sense, -but I am not arriving at a satisfactory formulation -to let me see the overall growth clearly. -Perhaps this has been studied, given the vast literature -on brick-stack overhangs? If so, a pointer would be appreciated. -I set out to explore random stacks of disks, but even the above simpler -rectangular bricks case is more intricate than I anticipated. - -           - - -Apologies if all this is too elementary! -Added. To provide a data point, here is one simulated topple-height histogram: - -           - - -Update. Here is a plot of Aaron Golden's data, $N(\delta)$ vs. $\delta^{-2}$, -which certainly supports Ori's $\delta^{-2}$ conclusion, as JSE noted: - -REPLY [16 votes]: The distribution should be roughly geometric with expectation roughly $\delta^{-2}$. To bound from above, in every $2\delta^{-2}$ steps there is a constant (independent of $\delta$) probability that the location of the $\delta^{-2}$ brick will be at least offset of 3 from the first brick. Then there is a constant probability for the next $\delta^{-2}$ bricks to be all at offset 2 or more, thus causing collapse. This event can happen independently for every batch of $\delta^{-2}$ bricks, so we get domination by geometric distribution with expectation $O(\delta^{-2})$. -On the other hand, the probability that the center of the brick never strays more than $1/4$ away from the center for $t$ bricks should be roughly $e^{-ct\delta^2}$ so we get bound from below by a geometric distribution with expectation $\Omega(\delta^{-2})$ too. -The coins case should be similar, still $\delta^{-2}$, but with a different constant.<|endoftext|> -TITLE: Topological rigidity of compact manifolds in dimension three -QUESTION [9 upvotes]: The Borel Conjecture asserts that homotopy equivalent aspherical closed manifolds are homeomorphic, which is still open in general. -But, for three-dimensional manifolds, this conjecture holds (I read this in Bessieres-Besson-Boileau), whose proof depends on the geometrization theorem (Perelman). -Question: -Does the relative version of the Borel conjecture also hold for compact 3-manifolds with boundary (by the geometrization)? -The relative version: If there is a homotopy equivalence between two compact aspherical manifolds that is a homeomorphism between their boundaries, are those manifolds homeomorphic? - -REPLY [12 votes]: Yes. -When the manifolds are Haken this is a theorem of Waldhausen. See Ian Agol's answer here. -Since your manifolds are aspherical, they are irreducible by the Poincaré conjecture. Since they have boundary and are irreducible, they are Haken.<|endoftext|> -TITLE: What's the Lipschitz constant of the exponential map for $\mathrm{SO}(n,R)$? -QUESTION [14 upvotes]: $\DeclareMathOperator\SO{SO}\DeclareMathOperator\so{\mathfrak{so}}$Consider the Lie algebra $\so(n)$ equipped with the metric $\langle e_i \wedge e_j, e_k \wedge e_l \rangle = \delta_{i,k} \delta_{j,l}$. Similarly equip the tangent space at other points of $\SO(n)$ by left translation. My question is, is the exponential map $\exp: \so(n) \to \SO(n)$ 1-Lipschitz? I can show it's Lipschitz using the following result from Wikipedia: -$$ \| e^{X+Y} - e^Y \| \le e^{\|X\|} e^{\|Y\|} \|X\|$$ for any matrix norm $\|\bullet \|$. -But this yields an optimal Lipschitz constant of 2, by rescaling the Hilbert Schmidt norm. I really need the constant to be $1$. Maybe it's not even true? - -REPLY [20 votes]: This is also true for the Riemannian metric described in the original question. I'm sure there is an elementary argument to prove it - it should certainly follow from the formula for the differential of the matrix exponential for example but here is a quick argument which involves no computations at all (assuming you know basic Riemannian geometry). -Since $SO(n)$ is compact and the metric in question is bi-invariant, the Riemannian exponential map coincides with the matrix exponential and sectional curvature is nonnegative. By Rauch comparison this right away implies that exponential map is $1$-Lipschitz on a neighborhood of $0$. To get the same statement on the whole $so(n)$ one can observe that -$SO(n)$ is a symmetric space which means that the curvature tensor is parallel and Jacobi equations along any radial geodesic giving differential of the exponential map essentially reduce to the trivial form $J(t)=f(t)X(t)$ where $X$ is a unit parallel vector field and $f(t)$ satisfies $f''+kf=0, f(0)=0, f'(0)=1$ with $k\ge 0$. This of course is easily solved and gives $f(t)=\frac{1}{\sqrt{k}}\sin (\sqrt k t)$ (and $f(t)=t$ when $k=0$) which certainly satisfies $|f(t)|\le |t|$ which means that the exponential map is $1$-Lipschitz. -John Jiang, -To be clear, the argument about the 1-Lipschitz part near $0$ in the above can be skipped because the Jacobi field computation reproduces it not only locally but globally. It doesn't work on noncompact Lie groups because most of them don't have bi-invariant metrics. -Also, there is a nice closed formula for the differential of the Lie group exponential which I was referring to above but was too lazy to look up when I wrote my first response. You can find it in most books on Lie groups. See Brian Hall's An Elementary Introduction to Groups and Representations for a completely elementary proof of it. -The formula is this -$d\exp_X(Y)=L_{\exp(X)}(\int_0^1Ad_{\exp(-tX)}Ydt)$. For matrix groups this becomes -$d\exp_X(Y)=\exp(X)\int_0^1 \exp(-tX)Y\exp(tX)dt$. -It easily implies the fact that you want since $X$ is skew-symmetric and $\exp(tX)$ is orthogonal.<|endoftext|> -TITLE: Failures that lead eventually to new mathematics -QUESTION [23 upvotes]: Possible Duplicate: -Most interesting mathematics mistake? - -In the 25-centuries old history of Mathematics, there have been landmark points when a famous mathematician claimed to have proven a fundamental statement, then the proof turned out to be false or incomplete, and finally the statement became a nice problem playing an important role in the development of Mathematics. Two immediate examples come to me: - -Fermat's last theorem, -the existence of a minimizer of the Dirichlet integral in calculus of variations, which led Weierstrass to introduce the notion of compactness. - -This must have happened in almost all branches of Mathematics. - -What are the best examples of such an evolution? How did they influence our every mathematical life? - -REPLY [9 votes]: In the early 1960's Smale published a paper containing a conjecture whose consequence was that (in modern language) "chaos didn't exist". He soon received a letter from Norman Levinson informing him of an earlier work of Cartwright and Littlewood which effectively contained a counterexample to Smale's conjecture. Smale "worked day and night to resolve the challenges that the letter posed to my beliefs" (in his own words), trying to translate analytic arguments of Levinson and Cartwright-Littlewood into his own geometric way of thinking. This led him to his seminal discovery of the horseshoe map, followed by the foundation of the field of hyperbolic dynamical systems. For more details, see Smale's popular article "Finding a horseshoe on the beaches of Rio", Mathematical Intelligencer 20 (1998), 39-44.<|endoftext|> -TITLE: What is known about Ulam's problem of metric spaces with isometric squares? -QUESTION [15 upvotes]: Background -In the book Problems in Modern Mathematics, S. Ulam asks the following question: -Suppose $A$ and $B$ are metric spaces, such that $A^2$ and $B^2$ equipped with the 2-metric $d((x_1, y_1),(x_2, y_2)) = \sqrt{d(x_1, x_2)^2 + d(y_1, y_2)^2}$ are isometric. Does it follow that $A$ and $B$ must also be isometric themselves? -The answer to this general question is negative. Take for example the space of the rational numbers $\mathbb{Q}$ and the space $\mathbb{Q} \sqrt{2}$ of rational multiples of $\sqrt{2}$. These two spaces are obviously not isometric, but their squares are, isometry being just the rotation by $\frac{\pi}{4}$. See http://www.ams.org/journals/proc/1971-029-03/S0002-9939-1971-0278262-5/S0002-9939-1971-0278262-5.pdf for the proof. -Question -What is the status of this problem for slightly less general metric spaces? I am especially interested in the following cases: - -What is known about this problem for complete metric spaces? -What is known for finite metric spaces? - -A good survey if it exists would be more than welcome. - -REPLY [7 votes]: This is known for geodesic metric spaces of finite affine dimension by a beautiful result of Alexander Lytchak and Thomas Foertsch. The de Rham decomposition theorem for metric spaces. -Geom. Funct. Anal., Vol. 18 (2008), 120-143 -Here a metric space is called affine if it is isometric to a convex subset of a normed vector space. Given a metric space $X$ its affine rank, $rank_{aff}(X)$ is defined as the supremum over all topological dimensions of affine spaces that admit an isometric embedding into $X$. Note that $rank_{aff}(X)$ is bounded above by the topological dimension of $X$. -Lytchak and Foertsch proved that a geodesic metric space of finite affine dimension can be uniquely written as a product $X=Y_0\times Y_1\times\ldots\times Y_k$ where $Y_0$ is a Euclidean space (possibly a point) and $Y_i$ are irreducible (i.e. can not be written as products) and different from $\mathbb R$ or a point. This of course implies the result you are asking about.<|endoftext|> -TITLE: Homological computations -QUESTION [8 upvotes]: Suppose I have a group acting on some Hadamard manifold, and I want to understand as much as possible about the (co)homology of the quotient. In my case I can find a fundamental domain for the action and the side-pairing transformations. Is there some canned piece of software which will compute the homology groups (I know how to do this in principle, but rather not actually do all the work) and the cohomology ring structure (in my ignorance, I don't know how to do this even in principle, so references would certainly be appreciated)? -*EDIT -To elaborate slightly: think of the projective model of $\mathbb{H}^n,$ and a group (discrete, but not known a priori to act without fixed points), and construct the Dirichlet domain, which is a convex polytope, for which we have the side-pairing transformations. Now, clearly by triangulating the fundamental domain enough we get a simplicial decomposition of the quotient, but since all this is happening in high dimensions (five or higher), this is already a pain to program.... - -REPLY [7 votes]: What is the fundamental domain of the action? In case you can easily create a cubical or simplicial decomposition of this region, there is tons of software out there to help you with the grunt-work. -I would recommend the Computational Homology Project (CHomP) which is run by Konstantin Mischaikow's group at Rutgers. I believe he co-wrote the book on efficient homology computation. Here is a link to the CHomP website and some documentation, I think you will want to use the program called "homsimpl" in case of a simplicial decomposition and "homcubes" in the (highly!?) unlikely case that your fundamental domain can be represented as a union of axis-parallel cubes with integer vertices. -It's been a while since I have personally used this code, but I think it is dimension-independent. I know for a fact that it is written in C++, so you will need a compiler such as gcc to build it from source. That being said, binaries are available here for most non-obscure operating systems. -As far as the cup product computation is concerned, I suspect that you will run into some difficulties. For cubical complexes, there are results (pdf) of Kaczynski (also a co-author of the Computational Homology book) which give pseudo-code (see page 6) but I must confess that I have been unable to find a software implementation. For simplices the situation is considerably harder because in general the product of two simplices is not a simplex, and so there are technical difficulties in constructing the Kunneth map. It is difficult to see how software will bypass this issue for arbitrary complexes, but I would love to test-drive an implementation if one exists. -All the best with your computations!<|endoftext|> -TITLE: The "binary" product preserves pushouts? -QUESTION [5 upvotes]: In the category Set of sets and functions, consider the functor F(X) = X * X where * is the product (its action on arrows is just F(f) = f * f). Does this functor preserve pushouts? Or at least pushouts of pairs of epimorpisms? - -REPLY [14 votes]: General pushouts, no; pushouts of pairs of epis, yes. In fact, yes to pushouts of pairs where only one of the arrows of the pair is epi. -It's very easy to see the answer is no for general pushouts: since a coproduct $A + B$ is the pushout of a pair $A \leftarrow 0 \to B$, and since the squaring functor preserves $0$, the squaring functor would preserve this pushout only if it preserved the coproduct. But we all know that for finite cardinalities, $a^2 + b^2$ is generally not equal to $(a+b)^2$; therefore squaring cannot preserve coproducts. -But in general, we can say that for the category of sets, given a pair of functions -$$A \stackrel{f_1}{\leftarrow} P \stackrel{f_2}{\to} B,$$ -the canonical arrow -$$\phi: A^2 +_{P^2} B^2 \to (A +_P B)^2$$ -is monic. (Lemma to be proved.) Assuming this, suppose for example that $f_2: P \to B$ is epic. Then the pushout of $f_2$ along $f_1$ is also epic (well-known fact); let $A \to C$ denote this epi (so $C$ is shorthand for $A +_P B$). Then $A^2 \to C^2$ is also epic, and since this obviously factors as -$$A^2 \to A^2 +_{P^2} B^2 \stackrel{\phi}{\to} C^2$$ -we see $\phi$ is also epi. But it is monic by the lemma, hence $\phi$ is an isomorphism, as we wanted to show. -To prove the lemma, it helps to have a clear picture of how pushouts are formed in $Set$. The pushout $C$ is the set of equivalence classes on $A + B$ where $x \in A + B$ is deemed equivalent to $x' \in A + B$ iff there is a zig-zag path -$$x = x_0 \stackrel{f_{i_1}}{\leftarrow} p_0 \stackrel{f_{i_2}}{\to} x_1 \leftarrow \ldots x_{n-1} \stackrel{f_{i_{n-1}}}{\leftarrow} p_{n-1} \stackrel{f_{i_n}}{\to} x_n = x'$$ -where for each $k$, either $p_k$ belongs to $P$ and the arrows out of $p_k$ alternate between $f_1$ and $f_2$, or we are in a "holding pattern" where $p_k$ belongs to $A$ or $B$ and the two arrows out of $p_k$ are both identities. Now it is not hard to convince yourself that given $(x, y)$ in $A^2$ or $B^2$, and $(x', y')$ in $A^2$ or $B^2$, if there is a zig-zag path from $x$ to $x'$, and a zig-zag path from $y$ to $y'$, then there is a zig-zag path from $(x, y)$ to $(x', y')$ with respect to the pair of maps -$$A \times A \stackrel{f_1 \times f_1}{\leftarrow} P \times P \stackrel{f_2 \times f_2}{\to} B \times B;$$ -all we do is pair together zig-zag paths in the separate $x$- and $y$-components. (Note that if the zig-zag to get from $y$ to $y'$ is longer than the zig-zag from $x$ to $x'$, we can always insert a holding pattern in the $x$-component so that the zig-zag in the $y$-component can "catch up", i.e., so that the lengths of the zig-zags match up.) But this means precisely that the map $\phi$ is monic. -Edit: In answer to vincenzoml's questions in the comments, I wrote up a proof of the more general desired result which applies to a general $\infty$-pretopos, here.<|endoftext|> -TITLE: Why the quarter in the $\frac{1}{4}$-pinched sphere theorem? -QUESTION [14 upvotes]: Is there any hope of a high-level explanation of why the fraction $\frac{1}{4}$ -plays such a prominent role as a -sectional curvature -bound in Riemannian geometry? -My (dim) understanding is that the idea is that if the sectional curvature of a manifold is -constrained to be close to 1, then the manifold must be topologically a (homeomorphic to the) -sphere $S^n$. -Conjectured by Hopf and Rauch, proved by Berger and Klingenberg, and strengthened by -Brendel and Schoen to establish diffeomorphism to $S^n$. -Here is the definition of "local" $\frac{1}{4}$-pinched from -Brendel and Schoen's paper "Manifolds with $1/4$-pinched curvature -are space forms" -(J. Amer. Math. Soc., 22(1): 287-307, January 2009; -PDF): - -We will say that a manifold $M$ has pointwise $1/4$-pinched sectional curvatures if $M$ has positive sectional curvature and for every point $p \in M$ the ratio of the maximum to the minimum sectional curvature at that point is less than 4. - -I know the "$\frac{1}{4}$" in the -$\frac{1}{4}$-pinched sphere theorem -is optimal, -and perhaps that is the answer to my question: $\frac{1}{4}$ appears because the theorem -is false otherwise—punkt! -But I am wondering if there is a high-level intelligble reason for the appearance of $\frac{1}{4}$, -rather than, say, $\frac{3}{8}$, or $\frac{e}{\pi^2}$ for that matter? -I am aware this is a "fishing expedition," and a fair response is: -Study the Brendel-Schoen proof closely, and enlightenment will follow! - -REPLY [11 votes]: The new proofs add a layer of complication that is unnecessary for your question. From Cheeger and Ebin, Comparison Theorems in Riemannian Geometry first paragraph of Chapter 6: - -The symmetric spaces of positive - curvature are known to admit metrics - such that $1 \geq K_M \geq -> \frac{1}{4}$; see Example 3.38. In - fact, we will prove that any - riemannian manifold with $1 \geq K_M -> \geq \frac{1}{4},$ which is not a - sphere, is isometric to one of these - spaces. - -Back in chapter 3, page 73, - -A complete classification of symmetric - spaces is available (Helgason [1962]). - In particular, the only simply - connected symmetric spaces having - positive curvature are the spheres of - constant curvatures, complex and - quaternionic projective spaces, and - the Cayley plane. These are sometimes - referred to as the rank one - symmetric spaces, and except for the - spheres they have canonical metrics - varying between $\frac{1}{4}$ and $1.$ - As an example, we will compute the - curvature of complex projective space. - The calculations for the other rank one spaces are similar. - -Then they begin the example, which is numbered Example 3.38. The punchline is that $\alpha$ and $\beta$ are orthogonal unit vectors, $J(\alpha)$ and $\beta$ are unit real vectors, so their real inner product lies between $-1$ and $1$ by Cauchy-Schwarz, followed by an identity about the sectional curvature that I shall write as -$$ K_{\alpha \beta} = \frac{1}{4} + \frac{3}{4} \langle J(\alpha), \beta \rangle^2. $$ - -REPLY [6 votes]: This was too long for a comment: -Certainly 1/4 is optimal, as Will Jagy points out, for e.g. the Fubini-Study metric on $CP^n$. The totally real tangent planes have curvature 1 (tangent to a totally geodesic $RP^2$), and the complex-invariant tangent planes to a totally geodesic $CP^1$ have curvature $1/4$. This is usually computed via the Riemannian submersion $S^{2n+1}\to CP^n$, but I think one can get a feel for this geometrically, by thinking about a quotient of $S^{2n+1}/U(1)=CP^n$. If you have $S^3/U(1)$, you get $CP^1$, and the map is the Hopf fibration. $S^3$ is the 3-sphere of radius 1, and when you quotient by $U(1)$, you get a sphere of half the diameter, since the circles of the Hopf fibration are distance $\pi/2$ apart, and thus the curvature is $1/4$. On the other hand, the $RP^n\subset CP^n$ subspaces are just a 2-fold quotient of a totally real $S^n\subset S^{2n+1}$ by the antipodal map, so have curvature $1$. -I saw Brendle give a talk about their theorem. One thing to note is that they have a local pinching theorem (pointwise), which is much stronger than a global pinching theorem. I think they actually show that they get positive isotropic curvature, which has something to do with complexifying the curvature tensor (this was introduced by Micallef-Moore I think), and is important for studying the stability of minimal 2-spheres. But I don't recall the details to be able to explain where $1/4$ comes in, but it has to do something with obtaining positive eigenvalues after complexifying the curvature tensor appropriately. - -REPLY [6 votes]: The number $1/4$ allows them to use a lemma by Berger to show that $M \times \mathbb{R}^2$ has non-negative isotropy curvature for a pointwise quarter-pinched manifold $M$. Non-negative isotropy curvature condition is preserved by Ricci flow. -Berger's lemma says that if the sectional curvatures at a point $p$ are bounded by $\kappa_1$ and $\kappa_2$, then the Riemann curvature tensor satisfies $R(e_1, e_2, e_3, e_4) \leq \frac{2}{3}(\kappa_2 - \kappa_1)$ for all orthonormal four-frames $\{e_1, e_2, e_3, e_4\}$ at $p$. The reference for this lemma is M. Berger, Sur quelques vari\'{e}t\'{e}s riemanniennes suffisamment pinc\'{e}es, Bull. Soc. Math. France 88, 57-71(1960). -A related result is that Yau and Zheng showed that a negatively $1/4$-pinched manifold has non-positive complex sectional curvature from this lemma. See S. T. Yau and F. Zheng, Negatively $1/4$-pinched riemannian metric on a compact K\"{a}hler manifold, Invent. Math., 103, 527-535(1991).<|endoftext|> -TITLE: A route towards understanding Shimura varieties? -QUESTION [22 upvotes]: I'm in the embarrassing situation that I want to ask a question that -was already asked, but (for complicated reasons) never answered. I'd -like to try with a blank slate. -Shimura varieties show connections to a lot of interesting -mathematical subjects. They're a topic of active research and have -been of importance in number theory and the Langlands program. -However, the theory has a bit of a reputation: for heavy -prerequisites; for a large and difficult-to-penetrate body of -literature; for seminar talks that spend a minimum of half an hour -getting past the definitions. ("Aren't you assuming that the -polarizations are principal here?" "I don't see why that has -cocompact center.") -Let's suppose that a poor graduate student doesn't have the best -access to the experts, but has gone to lengths to make themselves -familiar with "the basics" on modular curves and Shimura curves. -There is still a bewildering abundance of new material and new -ideas to absorb: - -Abelian schemes. -Reductive algebraic groups and the switch to the adelic perspective. -Representation theory and the switch in perspective on -modular/automorphic forms. -$p$-divisible groups and their various equivalent formulations. -Moduli problems and geometric invariant theory. -Deformation theory. -Polarizations. (Yes, I think this deserves its own bullet point.) -(This is a placeholder for any and all major topics that I forgot.) - -Obviously there is a lot to learn, and there's no magic way to obtain -enlightenment. -But for an outsider, it's not clear where to start, what a good place -to read is, what really constitutes the "core" of the subject, or even -if one might cobble together a basic education while learning things -that will prove useful outside of this specialty. -Is a route from modular curves to Shimura varieties that will help -both with understanding the basics of the subject, and with getting an -idea of where to learn more? -Thank you (and sorry for the side commentary). -(Often these kinds of questions ask for "roadmaps"; but "roadmap" -seems like it presupposes the existence of roads.) - -REPLY [3 votes]: I think the general wisdom is that Deligne's Travaux de Shimura and Milne's Introduction to Shimura Varieties are the most comprehensive references, with the latter being somewhat lighter on prerequisites (but heavier on examples). -I've heard it suggested by people who work in the area that the best way to learn the theory is via special cases and examples, motivated by focused research problems. I suppose this is true of many things, though. -It might also help to learn about quaternionic Shimura curves first, assuming that you know a bit about modular curves.<|endoftext|> -TITLE: The number of conjugacy classes and the order of the group -QUESTION [9 upvotes]: In my response to the OP in Embedding $S_3$ into $Aut(F_2)$ (I continue notation from my response in that thread), I indicated the possibility that lifting elements from $GL_{2}(\mathbb{Z}/(p))$ to $Aut(B(2,p))$ could extend the ability of character-free methods in proving congruences relating the order of a group and its number of conjugacy classes. I can now report a small bit of concrete progress in this direction. Specifically, what I can prove is: -Theorem. If $G$ is a finite group of exponent 3, then $|G| \equiv c(G) \mod{16}$. -Proof. We regard the Burnside group $B(2,3) = < x,y|(*)^{3}=1 >$ (here $ * $ ranges over every word in the generators) as $< x,y>$ and specify elements of $Aut(B(2,3))$ by their action on $x$ and $y$. Specifically, consider the following elements of $Aut(B(2,3))$: -$\alpha: (x,y) \to (xy,x^{-1}y)$ -$\beta: (x,y) \to (x,y^{-1})$ -By checking that $\beta^{2} = 1$ and $\beta\alpha\beta = \alpha^{3}$, we see that $<\alpha,\beta>$ is a quotient of the semidihedral group of order $16$. By checking that $\alpha^{4}$ sends $(x,y)$ to $(x^{-1},xy^{-1}x^{-1})$ so that $\alpha^{4} \neq 1$, we see that $<\alpha,\beta>$ is semidihedral of order $16$. -Now let $<\alpha,\beta>$ act on non-commuting pairs of elements of our group $G$ of exponent 3. We wish to show that every non-identity element of $<\alpha,\beta>$ has no fixed points in its action on the set of non-commuting pairs of elements of $G$. Every non-identity element of $<\alpha,\beta>$ has a power equal to $\alpha^{4}$ or is conjugate to $\beta$, so it suffices to check this property for just $\alpha^{4}$ and $\beta$. -If $(x,y)$ is a fixed point of $\alpha$, then $x = x^{-1}$ so $x^{2} = 1$ and, since $x^{3} = 1$, $x = 1$ and $(x,y)$ is not a non-commuting pair. -If $(x,y)$ is a fixed point of $\beta$, then $y = y^{-1}$ so $y^{2} = 1$ and, since $y^{3} = 1$, $y = 1$ and $(x,y)$ is not a non-commuting pair. -Therefore every orbit for the action of $<\alpha,\beta>$ on the set of non-commuting pairs of elements of $G$ has exactly 16 elements and we conclude that 16 divides $|G|(|G| - c(G))$ . Since $|G|$ is odd, we conclude that $|G| \equiv c(G) \mod{16}$. -I do not know how to give a character-free proof of this congruence (either for general $p$-groups or just those of exponent $p$) for any prime $p$ with $p \equiv 3 \mod{4}$ and $p > 3$. Complicating hopes of extending this is the fact that $B(2,n)$ is not known to be finite for any larger odd value of $n$. -(i) I'm probably just not thinking clearly enough right now, but how does one use this to prove the congruence when $G$ is an arbitrary 3-group? -(ii) Is it hopeless to expect to extend this to larger primes congruent to 3 mod 4? - -REPLY [8 votes]: Burnside's classical formula $|G| \equiv c(G) \text{ mod } 16$ has been generalized by Hirsch in the paper -"On a theorem of Burnside, Quart. J. Math. Oxford 1 (1950), 97-99". -By elementary methods (without using characters or other techniques from representation theory) he shows in the main theorem: - -Let $p_1,...,p_k$ be the prime-divisors of $|G|$ and let $d$ be the greatest common divisor of the numbers $p_i^2-1$ $(i=1,...,k)$. Then - $$\begin{array}{ll} -|G| \equiv c(G) \text{ mod } 2d, & \text{ if } |G| \text{ is odd} \newline -|G| \equiv c(G) \text{ mod } 3, & \text{ if } |G| \text{ is even and } 3, |G| \text{ are coprime } -\end{array}$$ - -This also generalizes Geoff Robinson's note for $p$-groups with $p \equiv \pm 1(8)$.<|endoftext|> -TITLE: Optic fibers after Joseph O'Rourke -QUESTION [30 upvotes]: Let $\gamma\colon[a,b]\to \mathbb R^3$ be a smooth curve with curvature $< 1$. -Consider a tube, formed by one parameter family of unit circles with center at $\gamma(t)$ in the plane orthogonal to $\dot\gamma(t)$. -A light ray which is comming into tube from one end and bouncing with perfect reflection from the interior walls will emerge from the other end with probability 1; see this question. Let us call a tube with this property an optic fiber. -(Note that I want an optic fiber to be bidirectional.) -One can construct an optic fiber along the same lines using any simple close smooth plane curve $(x(\theta),y(\theta))$ instead of circle. -To do this one has to choose a parallel normal frame $e_1,e_2$ along $\gamma$ -(i.e., such that $\dot e_i(t)\parallel\dot\gamma(t)$ for all $t$) -and consider the tube $[a,b]\times\mathbb S^1\hookrightarrow\mathbb R^3$ defined as -$$(t,\theta)\mapsto \gamma(t)+x(\theta){\cdot}e_1(t)+y(\theta){\cdot}e_2(t)$$ -(The condition that the frame is parallel implies that any normal plane to $\gamma$ cuts tube at right angle.) -This way we get an optic fiber with congruent ends. - -Question 1. Are there any constructions of optic fibers different from the one described above? -In other words, is it always possible so slice an optic fibers by planes which cut the walls at right angle? - -In particular, - -Question 2. Is there an optic fiber with noncongruent ends? - -Comments - -I feel that the answer is "NO", but have no idea "WHY". -From Liouville's theorem, it is clear that the ends must have the same area. -I realized that if the walls are only piecewise smooth then one can make an optic fiber with a pair of equidecomposable figures at the ends. (The construction is the same, but one splits tube into few on the way and then rearrange them back together.) -In dimension 2, a line passing through focal points cuts from confocal ellipses an optic fiber. (I learned it from Arseniy Akopyan.) I do not know smooth 3D examples of that type. [One might also impose an additional condition on optic fibers that random ray which starts inside leaves it with probability 1. The described confocal-ellipses-example does not have this property.] - -An extract from the answer of Marcos Cossarini: Note that if one can cut an optic fiber in two pieces in such a way that almost all rays pass the cut at most once then the cut has to be flat and orthogonal to the boundary. After such cut, one gets two optical fibers. Applying a bit of differential geometry the problem can be reformulated in an equivalent way: is it true that any optic fiber admits such a cut. - -REPLY [2 votes]: I don't know why this question appeared yesterday in my main mathoverflow screen, since the last comment appears to be from March 1st. I also think the conjecture is true, but my argument has plenty of holes. Here's what I thought. -Let $\Omega$ be an optic fiber that is the interior region of a compact connected $\mathcal C^1$ surface in $\mathbb R^3$, that is the union of a reflective surface $R$, a starting surface $F_0$ and an ending surface $F_1$. I assume (0) all optic fibers are like this. -Assume (1) that you can foliate $\Omega$ by a uniparametric family of plane surfaces $(F_s)_{s\in[0,1]}$, such that every ray starting at $F_0$ with positive initial $\dot s$ keeps a positive $\dot s$, until it reaches $F_1$. Let's call this a foliated optic fiber. Assume (2) that all the $F_s$ are plane surfaces and call it planely foliated, and let's call the fiber Petruninean if each $F_s$ is orthogonal to the reflective surface $R$ along its border. - -Claim (3): Every planely foliated optic fiber is Petruninean. - -Let my say why I believe it. -Let $m$ be the last value such that the fiber is Petruninean for $s\in[0,m]$. By finding a foliation of curves that is orthogonal to $F$ we can identify the points of $F_s$ at different values of $s$, and think of $F_s$ as a uniparametric family of regions in $\mathbb R^2$. -Conjecture (4): $F_s$ has constant area. -I don't know why this should be true, but it seems to be clear to Anton (if I understood well his comment concerning the Liouville theorem), and I would like to know why. Is it evident after studying billiard problems? I'll assume it true. -Can $F_s$ be nonconstant? It is constant until $s=m$, and then it starts to change. Since the area is conserved, it must advance at some points of its border and recede at other points. Find a point where it has just started receding. It represents a point $P\in R$. Fire a ray from $P$ into $\Omega$, with initial velocity orthogonal to $R$. Because of how we chose $P$, it will have negative initial $\dot s$, and by adjusting our choice of $P$ I would like (5) to be able to ensure that it reaches $F_m$. After that the optical fiber is Petruninean, and the ray makes its way to $F_0$. By reversing this ray, we obtain a ray that starts at $F_0$, reaches $P$, and bounces back to $F_0$, so $\Omega$ was not an optic fiber, after all. -The holes in the argument suggest more questions: - -Question 0: Is every foliated optic fiber planely foliated? - -If $\Omega$ is foliated, the phase space is the disjoint union of $T\Omega_+=\{\dot s>0\}$, $T\Omega_-\{\dot s<0\}$, and their zero-measure common border $T\Omega_0=\{\dot s=0\}$. Does every phase in $T\Omega_+$ correspond to a ray that came from $F_0$ and is going to $F_1$? If this is true (and also the analogous statement for $T\Omega_-$), then it is easy to see (by shooting rays between points of the same $F_s$) that every $F_s$ must be locally convex, and hence plane. - -Question 1: Is every optic fiber foliated? - -We can express the phase space as a union of the set $T\Omega_+$ of phases corresponding to rays that go from $F_0$ to $F_1$, the set $T\Omega_-$ defined analogously (so that $\dot x\in T\Omega_+$ iff $-\dot x\in T\Omega_-$), and the set $T\Omega_0$ of phases corresponding to rays that remain inside $\Omega$ eternally (in both directions of time). Observe how for each $x\in\Omega$, the classification of phases partitions $T_x\Omega$ into two opposite cones $T_x\Omega_+$, $T_x\Omega_-$ and a selfopposite cone $T_x\Omega$. -Does $T\Omega_0$ have measure zero? Are the cones convex? I have no idea. -Can we totally order the phases of $T\Omega_+$, so that it is then possible to assign them a scalar parameter, that makes possible to apply supremum arguments to prove things? At least we can partially order it. Do supremum arguments work in posets?<|endoftext|> -TITLE: 3x3 submatrix with only $0$ or $1$ entries -QUESTION [6 upvotes]: I decided to cross-post the question here from math.stackexchange.com because I got no answer from there. -It is a quick question on bipartite Ramsey numbers (I'm not an expert on the subject, so perhaps the question is trivial). -What is the least positive integer $r$ such that, any $r \times r$ 0-1 matrix contains at least one $3 \times 3$ submatrix filled with only 0 or only 1 entries ? -I found some articles with upper/lower bounds, but not a clear chart with the particular values I need. - -REPLY [7 votes]: According to this: -MR1622032 (99c:05139) -Hattingh, Johannes H.(SA-RAND); Henning, Michael A.(SA-NTL2) -Bipartite Ramsey theory. (English summary) -Util. Math. 53 (1998), 217–230. -the answer is $17$, not $15.$ -Addition A proof is contained in Irving's paper "A bipartite Ramsey problem and the Zarankiewicz numbers" (available on line). In the later paper with an almost identical title: -Bipartite Ramsey Numbers and Zarankiewicz Numbers (by Goddard et al) they seem to indicate that the Irving paper is slightly buggy, but that the result holds anyhow.<|endoftext|> -TITLE: Balls in spaces of operators -QUESTION [12 upvotes]: I am interested in some geometrical aspects of spaces $L(E)$, of bounded operators on a given Banach space $E$. I am unable to estimate if my problem deserves to be asked at MO, but let me try. -Is there an infinite-dimensional Banach space (non-separable preferably) $E$ such that for some non-zero -$T\in L(E)$ -the set -$$\{S\in L(E)\colon \|S-T\|=\|S+T\|\}$$ -contains an open ball? In fact, I am more interested in the negation: -Is there a Banach space such that for none non-zero $T\in L(E)$ this can happen? -I cannot (dis)prove it even if $E$ is a Hilbert space. - -REPLY [9 votes]: In what follows I show that such an operator exists if $E$ can be written (isometrically) as the $\ell_\infty$-direct sum of two (nonzero) subspaces (I have not tried the Hilbert space case, but I started writing my answer before the edits were made to the question.) -Let $E = X\oplus_\infty Y$, where $X$ and $Y$ are nonzero (infinite dimensional, if you like). Each $V\in L(E)$ satisfies $\Vert V \Vert = \max ( \Vert P_X V \Vert, \Vert P_Y V\Vert )$, where $P_X$ and $P_Y$ denote the projections onto the complemented subspaces $X$ and $Y$. -Let $T= P_X$ and $S=3P_Y$, so that $\Vert T-S\Vert =3=\Vert T+S\Vert $. To construct the desired example, we show that if $\Vert R-S\Vert <1$, then $\Vert T-R\Vert = \Vert T+R\Vert $. So take such $R$ and note that then $\Vert P_YR \Vert >2$ and $\Vert P_XR\Vert<1$. It follows that -$$ -\Vert T-R\Vert = \max (\Vert P_X(T-R) \Vert ,\Vert P_Y(T-R)\Vert ) = \max (\Vert T-P_XR \Vert ,\Vert P_YR\Vert ) = \Vert P_Y R\Vert -$$ -(since $\Vert T-P_XR \Vert \leq \Vert T\Vert + \Vert P_XR \Vert$<2 and $\Vert P_Y R\Vert >2$). -Similarly, we conclude that $\Vert T+R\Vert = \Vert P_Y R\Vert$, hence $\Vert T+R\Vert = \Vert T-R\Vert $. -Edit: Note that since each $U\in L(X\oplus_1 Y)$ satisfies $\Vert U\Vert = \max (\Vert UP_X\Vert , \Vert UP_Y\Vert )$, a similar construction gives an example of such a ball for spaces isometrically isomorphic to $X\oplus_1 Y$ for nonzero $X$ and $Y$.<|endoftext|> -TITLE: Legendre and sums of three squares -QUESTION [19 upvotes]: The Three-Squares-Theorem was proved by Gauss in his Disquisitiones, and this proof was studied carefully by various number theorists. Three years before Gauss, Legendre claimed -to have given a proof in his Essais de theorie des nombres. Dickson just says that Legendre -proved the result using reciprocal divisors. -I am a little bit surprised that I can't seem to find a discussion of Legendre's proof -anywhere. Reading the original is not exactly a pleasure since the material in Legendre's -book is organized, if I dare use this word, in a somewhat suboptimal manner. So before I'll start reading Legendre's work on three squares I'd like to ask whether anyone knows a discussion of his proof (or its gaps). -Edit. I have meanwhile found a thoroughly written thesis from Brazil on -Legendre's work in number theory by Maria Aparecida Roseane Ramos Silvia: -Adrien-Marie Legendre (1752-1833) e suas obras em teoria dos Números, which has, however, preciously little on Legendre's "proof". -One gets a better idea of what Legendre was doing by reading Simerka's article -Die Perioden der quadratischen Zahlformen bei negativen Determinanten, although Simerka is doing his own thing; in particular, he uses Gauss's definition of equivalence and composition of forms and describes parts of Legendre's work in Gauss's language. -BTW I have only recently learned about the existence of this Czech number theorist, who discovered Carmichael numbers long before Korselt or Carmichael, and who factored (10^17-1)/9 using an algorithm based on the class group of quadratic forms discovered long after by Shanks, Schnorr, and others. - -REPLY [8 votes]: There is a discussion of this by Andre Weil in "Number theory from Hammurapi to Legendre", where seems to imply that Legendre's proof might have been a bit problematic (p. 332). Weil also gives some nice proofs of the $n$-squares theorems (appendix 2 to the Euler chapter). You can find a link to the Weil book here: -http://dl.dropbox.com/u/5188175/WeilNumbers.pdf<|endoftext|> -TITLE: Schemes over ℤ with a “graded existence over ₁” -QUESTION [13 upvotes]: Let $X$ be a (separated, and with whatever other tameness conditions are appropriate) scheme over the integers $\mathbb{Z}$. (If you don't like schemes much, imagine that $X$ is described by algebraic equations over the integers, so that it is possible to make the algebraic variety $X(k)$ for any field $k$.) I am interested in the case in which $X(k)$ is "the same" for all fields $k$ so that it is possible to discuss its properties over the "field with one element $\mathbb{F}_1$". (Actually I think that $\mathbb{F}_1$ is a scam to some extent, but I wanted to make the question catchy.) What I mean is examples such as when $X$ is a Grassmannian. In this case the Poincaré-Hilbert polynomial of the topological cohomology $H^*(X(\mathbb{C}),\mathbb{Q})$ is the same up to the change of variables $q^2 \mapsto q$ as that of the cohomology $H^*(X(\mathbb{R}),\mathbb{Z}/2)$, and that's the same as the polynomial expression of the number of $\mathbb{F}_q$-rational points of $X(\mathbb{F}_q)$, and probably a lot of other things are the same. They're all the same because the Schubert cell decomposition is perfect in the sense that it has a vanishing differential over either $\mathbb{C}$ or $\mathbb{R}$, and it has equally nice properties over any other field. -Is there a name for a scheme with this type of good behavior? The question is not entirely rigorous, except in the primitive form of reducing only to the fields listed above. Even so, most schemes clearly don't share these stability properties possessed by Grassmannians. For instance, the picture fails badly for an elliptic curve. - -I got two good answers to the question (and a third one that was not bad). In the interest of wrapping things up, I would like to accept one of them. The answers are basically tied in mathematical value, but I particularly like Jordan Ellenberg's prose. The picture from both answers taken together is that people have worked both on upper bounds --- all schemes that could possibly qualify for the question --- and lower bounds --- schemes that by design must be allowed as examples. The former are called "polynomial count" varieties; the latter are sometimes offered as varieties defined over $\mathbb{F}_1$. Pure Tate cohomology is another type of upper bound. It seems that the discrepancies between the upper and lower bounds are not well understood. - -REPLY [2 votes]: The following paper http://ens.math.univ-montp2.fr/~toen/souz.pdf of Toen and Vaquie may be interesting.<|endoftext|> -TITLE: Area of distance sphere in manifold with Ricci $\ge 0$. -QUESTION [19 upvotes]: Let $M$ be a open complete manifold with Ricci curvature $\ge 0$. -By a theorem of Calabi and Yau, the volume growth of $M$ is at least of linear. -I am wondering whether the following statement is true: -Let $p$ be any fixed point in $M$ and $B(p, r)$ be the distance ball of radius $r$ in $M$. Then for any given $R>0$, there exists a constant $c=c(p,R)>0$ such that $Area(\partial B(p, r))\ge c(p,R)$ for any $r>R$. - -REPLY [2 votes]: The answer is YES. -Let $b\colon M\to \mathbb R$ be a Busemann function for a ray $\gamma$ from $p$, so that $b(p)=0$ and $b(x)\le 0$ for any $x\in \gamma$. -Set -$$L_t=b^{-1}(t)\ \ \text{and}\ \ L^-_t=b^{-1}(-\infty,t].$$ -Note that the sublevel sets $L_t^-$ are mean curvature concave for all $t$. -In particular, any area minimizing hypersurface in $L_t^-$ with the boundary in $L_t$ lies in $L_t$. -Fix $t<0$. -From above, -$$\mathop{\rm area}\partial B(p,r)\ge\mathop{\rm area}(\partial B(p,r)\cap L_t^-)\ge \mathop{\rm area}( B(p,r)\cap L_t)\ge \mathop{\rm area}( B(p,R)\cap L_t);$$ -i.e., the inequality holds for $c(p,R)=\mathop{\rm area}( B(p,R)\cap L_t)$. -It remains to choose $t$ and $R$ so that $\mathop{\rm area}( B(p,R)\cap L_t)>0$; $R=2$ and $t=-1$ will do the job.<|endoftext|> -TITLE: sheaf valued functors: how much can you prove about them just using category theory -QUESTION [7 upvotes]: This question is related to my earlier question. I have learnt a lot since I asked that question and I think I can phrase my thoughts more clearly now. -To someone who is comfortable with category theory, it seems reasonable to define sheafification as the left adjoint to the forgetful functor from sheaves to presheaves (assuming such a functor exists). Intuitively, sheafification associates to the presheaf $\mathcal{F}$, a sheaf $\mathcal{F}^{+}$ which has isomorphic stalks. However, this is not obvious from the definition of sheafification as an adjoint. - Question: Is it possible to prove that the unit associated to the adjunction (sheafification, forgetful) induces isomorphisms on the stalks without explicitly constructing sheafification? -Note that I am not saying that defining sheafification as an adjoint is a good idea. Indeed, the explicit construction of the sheafification of a presheaf is a good exercise, but the fact that the unit $\mathcal{F} \to \mathcal{F}^+$ induces isomorphisms on the stalks seems like it should follow from properties of adjoint functors. -I have similar questions about the inverse image functor but I think that if i can find out how to solve the above question, then I should be able to work out the details for the inverse image functor aswell - -REPLY [10 votes]: The stalk of a (pre)sheaf $\mathcal{F}$ at a point $x$ is the colimit of the directed system of (pre)sheaves $\mathcal{F}_U = j_* j^* \mathcal{F}$, where $U$ runs over all open sets containing $x$ and $j$ is the inclusion map of $U$. (This is the sheaf version of the definition of a stalk; it realizes the stalk as a skyscraper sheaf concentrated at $x$.) Since sheafification is a left adjoint, it is right exact, i.e. preserves colimits.<|endoftext|> -TITLE: Is the free product of arbitrarily many copies of `${\mathbb{Z}}$` and `${\mathbb{Z}}/2$` residually nilpotent? -QUESTION [8 upvotes]: A group is residually nilpotent if the intersection of the terms in its lower central series is the trivial group. -Is the free product of arbitarily (possibly infinite) many abelian groups residually nilpotent? In particular, this would imply that the free product of abelian groups with any free group is residually nilpotent. - -Later: Thank you for all the comments and answers. It is good to know that the free product of abelian groups is residually solvable. -The group I am interested in is the free product $F*G$ where $F$ is a free group and $G$ is a free product of arbitrarily many ${\mathbb{Z}}/2$. In the special case ${\mathbb{Z}}*{\mathbb{Z}}/2$, I know that this group is residually nilpotent. This follows from a result of Mal'cev on adjoint groups of residually nilpotent algebras. - -REPLY [2 votes]: Mark Sapir has already given you a definitive answer, but I think it is interesting to note that his argument generalises to show something stronger. A finitely-generated non-trivial free product of groups is residually nilpotent only if it is $p^{\prime}$-torsion-free, for some prime $p$. (Here, $p^{\prime}$ denotes the set of primes not equal to $p$ so, in this case, we are allowed only $p$-torsion.) For, in a finitely-generated residually nilpotent group, elements of finite coprime order commute. And, if $G = A\ast B$ is such a free product (with $A\neq 1\neq B$), and $G$ is not $p^{\prime}$-torsion-free, for any prime $p$, there are distinct primes $p$ and $q$, and elements $u$, of order $p$, and $v$, of order $q$ in $G$. These may be taken (up to conjugation) to be members of $A\cup B$. Now, if $u$ and $v$ belong to different free factors ($A$ or $B$) then they cannot commute, so it must be that $u$ and $v$ belong to the same factor, say $A$. But then, taking a non-trivial element $b$ in $B$ and forming the conjugate $w = b^{-1}vb$, we get an element of order $q$ which again does not commute with $u$. (The commutator $[u,w] = u^{-1}b^{-1}v^{-1}bub^{-1}vb\neq 1$, by the normal form theorem for free products.) This again contradicts the fact that elements of coprime order commute.<|endoftext|> -TITLE: Use of games to approximate solutions to Partial Differential Equations -QUESTION [10 upvotes]: Hi there, -Hopefully the mathematics community can help me out this one, I'm currently studying my senior capstone at my college, and decided to do some research on a chapter in Stanley Farlow's book "Partial Differential Equations for Scientists and Engineers". Basically in one of the chapters he describes a way one can create a game to approximate solutions to PDE's by using the various difference formula's. -Farlow gives a few examples in his book. Namely the dirichlet problem on a square with the laplace equation, however I'd like to extend this result to parabolic and hyperbolic PDE's, and then eventually extend those results to 3-dimensional games. My problem is that I'm unable to find any past research using game theory in this manner other than this one chapter in this book. And I'm not sure if my own foundation in numerical solutions is good enough to do this thing on my own. In other words, I need help. -To give an example of what the game is, start with an nxm lattice, then pick an interior point and run a Monte Carlo simulation that computes random walks from any point $u_{i,j}$ on the lattice to a bound. The game ends when the player hits the bound, and is then given a value(prize) that depends on the boundry/initial conditions. It turns out that the solution to the pde using finite difference equations at that point is the average of the four neighboring points, and so on. Here are some of the questions I NEED to answer, and it would be of great help if I could get some direction on them: - -Do you think the random walks on the bounded lattice are self avoiding? I.e. Can the player intersect with the path they've already made? This would definitely change the outcome of the probabilities of a player reaching the bound. -Are there any other good undergraduate-ish texts/resources that give examples of employing finite difference methods on nonlinear PDE's and PDE's with variable coefficients such as $u_{xx} + \sin(x)u_{yy}=0$? (with appropriate boundary/initial conditions) Or just good undergrad resources in general for numerical methods for PDE's? I've gone to my library to check some books out, and some of them are way way way over my head, so I've got to be careful. -What would be the best way to code such a program to compute these walks? I'm pretty sure it's doable in both Mathematica and C++, as to what kind of plan I would make to write it..I've no idea yet.(Making the program would answer definitively if the walks are self avoiding or not) - -Google books has a slice of the chapter online, http://books.google.com/books?id=DLUYeSb49eAC&pg=PA347&lpg=PA347&dq=%22tour+du+wino%22&source=bl&ots=vmsz2z6uAP&sig=rZiqwWJ8ESgokmYmBga8qtGVG8U&hl=en&ei=dEaOTp26EeTKsQLg1LymAQ&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDIQ6AEwAw#v=onepage&q=%22tour%20du%20wino%22&f=false -I wish there was someone like Erdos who could just have insane memory to remember past papers on a topic. - -REPLY [12 votes]: The connection between random walks, diffusion, and the heat equation is an amazing example of "the unreasonable effectiveness of mathematics." However, it's important to understand that this doesn't extend to other PDE's. -I'd encourage you to make the effort to dig a little deeper so that you understand why a Monte Carlo simulation of a random walk gives solutions to the heat equation. The keys to this are understanding how the binomial distribution applies to the random walk, understanding that the limiting case of the binomial distribution is the normal distribution, seeing that the normal distribution (in particular its probability density function) is a solution to the heat equation, understanding that the linearity of the heat equation allows you to use superposition of solutions, and finally understanding how boundary conditions for your random walk and your heat equation can be made equivalent. -Once you understand all of this, it will become obvious that you don't want to use a self-avoiding random walk in your Monte Carlo simulation- the distribution of the regular random walk is normal and does the right thing for this problem. -In terms of implementing this on the computer, I would suggest that you start with an expample in one space dimension rather than 2 or 3 dimensions. I would also suggest doing this in a computational environment (such as Maple, Mathematica, MATLAB, R, or Python) that already has facilities for generating pseudo-random numbers and plotting the results of your simulation. -A separate, equally large project would be to learn about finite difference methods for the solution of the heat equation boundary value problem. It turns out that this is not as simple as just plugging in finite difference approximations for the derivatives- you also have to work carefully to come up with a stable numerical scheme. Analyzing the stability of your scheme will involve Fourier analysis- it gets complicated. -To answer your specific questions: - -You don't want self-avoiding random walks. -Finite difference methods for the heat equation are addressed in many introductory textbooks on numerical analysis. The discussion in Burden and Faires (which has been through many editions and now has new coauthors...) might be suitable for you. Methods for nonlinear PDE's are a more advanced topic. -As I've suggested above, you'll want to use an environment with built-in tools for random number generation and plotting.<|endoftext|> -TITLE: Is the degree of a finite morphism stable by base change -QUESTION [10 upvotes]: Let $f:X\longrightarrow Y$ be a finite morphism of schemes of degree $n$. Let $S\to Y$ be a morphism of schemes. -Is the degree of the finite morphism $X\times_Y S \longrightarrow S$ equal to $n$? -If not, what conditions should be put on $X$ and $Y$? -If it helps, you can assume all the schemes to be integral. - -REPLY [19 votes]: I shall assume that $X,Y$ are integral, locally noetherian schemes and that $f$ is dominant. Then the degree of $f$ is the degree of the corresponding extension of fields, namely -$$deg(f)=[Rat(X):Rat(Y)]$$. -We have for the fibers $X_y \; (y\in f(X))$ of $f$ the interesting result: -$$dim_{\kappa (y)} \mathcal O(X_y)\geq deg(f)$$ -with equality for all fibers -$$ dim_{\kappa (y)} \mathcal O(X_y)= deg(f) \quad (\star)$$ -if and only if $f$ is flat (cf. Qing Liu's book, page176). -So non flat morphisms will give you counterexamples by taking for $S$ a point of $Y$. -For an explicit counterexample, consider the case where $Y$ is a node, $X$ the affine line (both over a field $k$) and $f$ the normalization morphism. This is a finite morphism of degree one, but the fiber of the singular point has degree $2$ over $k$. -More generally, normalizations of non-normal varieties are never flat and will yield any number of countereamples. -Also if $f$ is flat the criterion will tell you, since flatness is preserved under base-change, that the degree of $f$ will be preserved under some reasonable assumptions on the morphism $S\to Y$, the most obvious one being that $S$ should be locally noetherian and integral too. -A well-known formula Here is an arithmetically flavoured illustration of the above. - Let A be a Dedekind domain with fraction field $K$ and $L$ a separable field extension of $K$ of degree $[L:K]=n$. Let $B$ be the ring of elements in $L$ integral over $A$. -That ring $B$ is flat over $A$ (because for Dedekind rings flat=without torsion) and is a Dedekind domain, finite over $A$ (Krull-Akizuki). - We can apply the considerations above above to the associated morphism $f:Spec(B)=X\to Y=Spec(A)$. -Take a nonzero prime $\mathfrak p =y \in Y $ and write ${\mathfrak p}B=\prod {\mathfrak P}_i^{e_i}$. -Since $X_y=Spec(B/{\mathfrak p}B) $, the formula $(\star )$ translates into the very classical formula of algebraic number theory (where $f_i=[B/{\mathfrak P}_i: A/ \mathfrak p]$): -$$n=\sum e_if_i$$<|endoftext|> -TITLE: For any $n$, does there exist a number field with at least $n$ solutions to the unit equation -QUESTION [5 upvotes]: Let $n$ be a positive integer. -Does there exist a number field $K$ such that the number of solutions of the unit equation $$a+b =1, \quad a,b\in O_{K}^\ast$$ is at least $n$? Can we write down such a number field explicitly? -I know that the number of solutions is always finite in a fixed number field. - -REPLY [6 votes]: The unit equation is discussed in detail in Chapter 5 of Bombieri and Gubler's book Heights in diophantine geometry. In particular, they note (see Example 5.2.6) that in the cyclotomic field ${\Bbb Q}(e^{2\pi i/p})$ there are exactly $(p-1)(p-2)$ non-real solutions to the unit equation $x+y=1$. This observation is attributed to H.W. Lenstra. -In Example 5.2.7 it is observed that there seem to be many more real solutions to the unit equation in cyclotomic fields. For example the maximal real subfield of ${\Bbb Q}(e^{2\pi i/19})$ contains $28 398$ solutions to the unit equation. -Another intriguing example is 5.2.8. Let $\alpha$ be the real root $\alpha>1$ of the Lehmer polynomial -$$ -x^{10}+x^{9}-x^7-x^6-x^5-x^4-x^3+x+1. -$$ -Then the field ${\Bbb Q}(\alpha)$ (with unit group of rank $5$) contains $2532$ solutions to the unit equation.<|endoftext|> -TITLE: Serre's Analogue of the Weil Conjectures for Non-Compact Kahler Manifolds -QUESTION [6 upvotes]: The classical Riemann Hypothesis concerns the locations of zeroes of the Riemann zeta-function, or more generally the Dedekind zeta-functions of number fields. Its analogue for varieties defined over finite fields is part of the Weil conjectures, which concerns the absolute values of the zeroes and poles of the zeta-functions of these varieties. -The zeta-functions for varieties have analogues for complex manifolds. When a complex manifold is compact Kahler, the ablosute values of the zeroes and poles of its zeta-functions then satisfy similar properties as the zeta-functions for varieties. This theorem is due to Serre, and his proof is based on Hodge theory on compact Kahler manifolds. -Does anyone know if this can be extended to non-compact Kahler manifolds? - -REPLY [13 votes]: Here is a link to Serre's paper: Analogues Kahleriennes de Certaines Conjectures de Weil. There seems to be some misunderstanding about what Serre actually proves implicit in the question, so I'll clarify it a bit, explain an easy analogue for some non-compact varieties, and indicate what should be true in general. -First of all, Serre's result is not for arbitrary compact Kahler manifolds. Rather, he starts with a smooth complex projective variety $X$, and an endomorphism $f: X\to X$, with an ample divisor $E$ so that $f^{-1}(E)$ is algebraically equivalent to $qE$ for some integer $q>0$. Then he shows that the eigenvalues of $f^*$ acting on $H^r(X, \mathbb{C})$ have absolute value $q^{r/2}$. Contrary to the statement of the question, the algebraicity of $X$ is built into the very result, since it requires the existence of an ample divisor. -Serre doesn't explicitly define the zeta function of $(X, f)$, but by analogy to the Weil conjectures, one may define $$Z(X, f, t)=\prod_i \det(1-f^*t\mid H^i(X, \mathbb{C}))^{(-1)^{i+1}}.$$ Then this zeta function satisfies the desired "Riemann hypothesis," by the result above. -When looking for a non-compact analogue, the motto one should have in mind is that zeta functions should behave well under "cutting-and-pasting." For example, if $Z(X, t)$ is the zeta function from the Weil conjectures, where $X/\mathbb{F}_q$ is an arbitrary variety, and $Y\subset X$ is a closed subscheme, then $$Z(X, t)=Z(X\setminus Y, t) \cdot Z(Y, t).$$ -So here's an analogue of this fact in the setting of smooth quasi-projective varieties. Suppose $X$ is a smooth projective variety over $\mathbb{C}$, and $f: X\to X$ is a self-map satisfying the conditions of Serre's result. Suppose $Y\subset X$ is a smooth closed subvariety, so that $f(Y)\subset Y$. Then $Y, f|_Y$ also satisfy the conditions of the theorem, and so $Z(X, f, t)$ and $Z(Y, f|_Y, t)$ satisfy the "Riemann Hypothesis." Now suppose $f(X\setminus Y)\subset X\setminus Y$ as well, and $f|_{X\setminus Y}$ is proper. Then we may define $$Z(X\setminus Y, f|_{X\setminus Y}, t)=\prod_i \det(1-f|_{X\setminus Y}^*t\mid H^i_c(X\setminus Y, \mathbb{C}))^{(-1)^{i+1}}.$$ -Here $H^i_c(X, \mathbb{C})$ denotes the cohomology of $X$ with compact support. Does this zeta function satisfy some kind of Riemann hypothesis? Well, from the long exact sequence relating the compactly supported cohomology of $X\setminus Y$ to that of $X$ and $Y$, we have that $$Z(X\setminus Y, f|_{X\setminus Y}, t)=Z(X, f, t)/Z(Y, f|_Y, t).$$ -So certainly all of the poles and zeroes of $Z(X\setminus Y, f|_{X\setminus Y}, t)$ have absolute value $q^{-r/2}$ for some integer $r$. This gives a sort of "Riemann hypothesis" for quasi-projective varieties admitting a particularly nice compactification. What you'll notice immediately though is that the $r$ does not match up with the cohomological degree, as it does in the compact case---there is some "slippage" coming from the boundary map in the long exact sequence. Rather, the $r$ is related to the "weight filtration" on the cohomology of $X\setminus Y$. -Now suppose $U$ is an arbitrary quasiprojective variety, and $f: U\to U$ is a proper map. $U$ might not have the ridiculously nice compactification we need to run the above argument, but the zeta function defined using cohomology with compact support still makes sense. I doubt that it will in general satisfy a "Riemann hypothesis" of the type you're looking for, since the condition on ample divisors may not makes sense. What is true, though, is that if $U$ admits a smooth compactification $X$ so that $f$ extends to a map $g: X\to X$ satisfying the conditions of Serre's result, so that $g(X\setminus U)\subset X\setminus U$, (where $X\setminus U$ is not necessarily smooth!) this zeta function will satisfy a "Riemann hypothesis." Namely, the zeroes and poles of $Z(U, f, t)$ will be of the form $q^{-r/2}$, with the $r$ coming from the weight filtration on the compactly supported cohomology of $U$. -To see this, one may imitate Serre's argument using the mixed Hodge structure on the cohomology of $X\setminus Z$. I don't immediately see how to give an analogue of Serre's condition on the existence of the ample divisor $E$ for $U$ without reference to a compactification, though there should be a way to do so; then one should be able to imitate Serre's argument, just working with the mixed Hodge structure on the compactly supported cohomology of $U$. -$ $<|endoftext|> -TITLE: D-modules and Algebraic Solutions of PDEs -QUESTION [20 upvotes]: I am not certain if this is a complete question and I fear it might be shot down. Anyway, I try to pose it. My question is in connection to using D-modules to study PDEs (and systems of PDEs). When I was doing a perusal on "A primer of algebraic D-modules by S. C. Coutinho" the justification on the importance of D-modules; they provide an algebraic tool towards the solution of differential equations. This is the story I always hear! Do someone have a reference or more information about D-modules and algebraically solution of PDEs ?. - -REPLY [12 votes]: My guess is that this is a reference to Bernshtein's proof using the theory of D-modules that every constant coefficent partial differential operator $D$ has a fundamental solution, i.e. there is a distributional solution to the PDE $Du = \delta$ where $\delta$ is the Dirac distribution. This was an important theorem in analysis due to Malgrange and Ehrenpreis in the 1950's, and I think it came as a bit of a surprise that it can be done purely algebraically - all of the analytic input is encapsulated by a few basic facts about distributions. Everyone knows that the fundamental theorem of algebra is proved in analysis class; maybe this means the fundamental theorem of analysis will one day be proved in algebra class! -Here is a link to Bernshtein's paper: http://www.math.tau.ac.il/~bernstei/Publication_list/publication_texts/bernstein-mod-dif-FAN.pdf. This was written in the 70's, and I think since then the argument has been cleaned up and made even more algebraic.<|endoftext|> -TITLE: Simple basis for Barnes-Wall lattices in dimension `$2^n$` -QUESTION [6 upvotes]: I'm searching for a "simple" description of the basis of the Barnes-Wall lattices -in (real) dimension $2^n$, if possible in a basis of minimal vectors, so that I can -do some calculations. -Can anyone tell me where to find such a description ? -Note : I'm not looking for examples in fixed dimensions, like the D4, or E8 -lattices. I know where to look up those. I'm looking for a formula / systematic -description in all dimensions of powers of 2. - -REPLY [10 votes]: Henry Cohn cited a very nice definition of the Barnes-Wall lattices, but in my opinion, this definition that I just found in a paper by Micciancio and Nicoli is even better. (Although the two definitions are similar.) The Barnes-Wall lattice in $\mathbb{C}^{2^{n-1}}$ is a lattice over the Gaussian integers generated by the rows of the matrix -$$\begin{pmatrix} 1 & 1 \\ 0 & 1+i \end{pmatrix}^{\otimes {n-1}}.$$ -Then of course to get the generator matrix in $\mathbb{R}^{2^n}$, you use the replacement -$$a+ib \mapsto \begin{pmatrix} a & b \\ -b & a \end{pmatrix}.$$ -This isn't a basis of minimal vectors. However, if you change the Gaussian generator matrix to -$$\begin{pmatrix} 1 & 1 \\ 1 & i \end{pmatrix}^{\otimes {n-1}},$$ -then it is a basis of minimal vectors, and the associated real basis is also minimal.<|endoftext|> -TITLE: Do indiscernibility embeddings exist for an initial segment of an inner model of many measurable cardinals? -QUESTION [11 upvotes]: Background -I am interested in elementary embeddings from a model of set theory into itself. One way of producing such elementary embeddings is when the model is generated by indiscernibles; this idea is very closely related to the existence of sharps. Jech and Kanamori discuss $0^{\#}$ and $0^\dagger$ in detail but don't tell me much about other sharps. More advanced resources are difficult to understand without a lot of background knowledge. -Hypotheses -Let $\theta$ be an inaccessible cardinal, and suppose that some set $A$ of measurable cardinals below $\theta$ is a stationary subset of $\theta$. For each $\kappa \in A$, let $\mu_\kappa$ be a normal measure on $\kappa$, and let $\mathcal{U} = \{ \langle \kappa, \mu_\kappa \rangle : \kappa \in A \}$. Let $L[\mathcal{U}]_\theta$ denote those elements of $L[\mathcal{U}]$ of rank less than $\theta$. -Question statement -Do there exist large cardinal assumptions which imply the existence of a closed unbounded set of ordinal indiscernibles for $L[\mathcal{U}]_{\theta}$ such that every order-preserving map of these indiscernibles extends to an elementary embedding $j:L[\mathcal{U}]_{\theta} \to L[\mathcal{U}]_{\theta} \, \, ?$ -Remarks -The large cardinal assumptions may be on $\theta$, the elements of $A$, or some other large cardinal. The values of $\theta$, $A$, and the $\mu_\kappa$ may be chosen in whatever way you like subject to the hypotheses above -- I just want this to work in some example, not in every example. -In The Core Model, Dodd mentions double mice, a generalization of $0^\dagger$. Maybe some version of these can be used to answer the question affirmatively, but I know nothing about them. - -REPLY [9 votes]: My reading of this question was different from Andreas', because Norman asked for order preserving maps of the indiscernibles to extend to embeddings -$j:L[U]_\theta \rightarrow$ $L[U] _\theta$ -i.e. the indiscernibles should be below $\theta$ as the ordinal height of the structures mentioned is $\theta$? -In that case the measurable or Ramsey above $\theta$ only guarantees indiscernibles above $\theta$ and so "missing the target"? -In any case there are generalisations of "double mice" that you surmise that provide a positive answer. Let $M$ be the "least" in a certain canonical well-ordering of all such iterable structure that have a measurable cardinal $\kappa$ which is, in $M$, the limit of measurables cardinals below $\kappa$. (Such structures are called "mice".) By Loewenheim Skolem we may assume that $M$ is countable. `Iterability' here means we may form -all iterated ultrapowers of the first structure $M=M_0$ using this top measure repeatedly; call the ultrapower structures $M_\tau$ for all ordinal $\tau$; then all the $M_\tau$ will be wellfounded. In the $\tau$'th model let $\kappa_\tau$ be the top-most measurable cardinal. Then $\tau ,\mu \rightarrow \kappa_\tau<\kappa_\mu$ and the class of all such $\kappa_\tau$ forms a closed unbounded class of indiscernibles for the model $W$ "left behind" by these iterates. -[Fornally $W = \bigcup _ \tau H(\kappa_\tau)^{M_\tau}$.] -Then $W$ is the `least' inner model with a proper class of measurables cardinals, -(because as $\tau$ increases the order type of the measurables in the models $M_\tau$ increases; but these measures are left behind in the lower $H(\kappa_\tau)^{M_\tau}$ part -of the model that goes into $W$). The $\kappa_\tau$ are indiscernibles for it, just as the Silver indiscernibles are for $L$. -The large cardinal assumption is then that "This iterable $M$ exists" just as a parallel -to "$0$-sharp exists" is to $L$.<|endoftext|> -TITLE: What exactly is the relation between string theory and conformal field theory? -QUESTION [67 upvotes]: Maybe it would be helpful for me to summarize the little bit I -think know. A 2D CFT assigns a Hilbert space ${\cal H}$ to a circle and -an operator -$$A(X): {\cal H}^{\otimes n}\rightarrow {\cal H}^{\otimes m}$$ -to a Riemann surface $X$ with $n$ incoming boundaries and $m$ -outgoing boundaries. This data is subject to natural conditions -arising from the sewing of surfaces. -Here is how I understand the relation to string theory. The -Hilbert space ${\cal H}$ might be the space of functions on the -configuration space of a string sitting in a manifold $M$. So -${\cal H}=L^2(Maps(S^1,M))$ with some suitable restrictions on the maps. -It is natural then that the functions on the configuration space -of $n$ circles is ${\cal H}^{\otimes n}$. Now we consider $n$ strings -evolving into $m$ strings. There are many ways to do this, one for -each Riemann surface $X$ as above. When $X$ is fixed, $A(X)$ -is the evolution operator, usually described in terms of some path -integral over maps from $X$ into $M$ involving a conformally -invariant functional. -All this makes a modicum of sense. So ${\cal H}$ is -the Hilbert space arising from quantization of the cotangent -bundle of $Map(S^1,M)$, while $A$ describes time evolution. So in -this sense, such a conformal field theory appears to be the -quantization of the classical string. I guess what is missing in -my description up to here is the prescription for turning -functions on $T^*Map(S^1,M)$ into operators on ${\cal H}$. To my -deficient understanding, maybe this situation corresponds to -having quantized only the Hamiltonian. -Now, what I was really wondering was this: When I was in graduate -school, I remember frequently hearing the phrase: -CFT theory is the space of classical solutions to string -theory. -Does this make some sense? And if so, what does it mean? This phrase has been hindering my understanding of conformal field theory ever since, making me feel like my grasp of physics is all wrong. -According to the paragraphs above, my naive formulation would have -been: -Quantization of a string theory gives rise to a CFT. -What is wrong with this naive point of view? If you could provide -some enlightenment on this, you'll have resolved a long-standing -cognitive itch in the back of my mind. -Thanks in advance. - -Added: -As Jose suggests, I could simply be remembering incorrectly, or misunderstood before what I heard. That, in fact, is what I had hoped to be the case. But read, for example, the first page of Moore and Seiberg's famous paper "Classical and quantum conformal field theory": -http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.cmp/1104178762 - -Added again: -To quote Moore and Seiberg more precisely, the second sentence of the paper reads 'Conformal field theories are classical solutions of the string equations of motion.' Now, I might attempt to understand this as follows. When the Riemann surface is $$S^1\times [0,t]$$(with the conformal structure induced by the standard metric) -one interprets -$$A(S^1\times [0,t])$$ -as $$e^{itH}.$$ -Thus, when applied to a vector $\psi_0\in {\cal H}$, the theory would generate a solution to Schroedinger's equation -$$\frac{d}{dt}\psi =iH \psi$$ -with initial condition $\psi_0$ as $t$ varies. So one might think of the various $A(X)$ as $X$ varies as being 'generalized solutions' to Schroedinger's equation for a quantized string. I suppose I could get used to such an idea (if correct). But then, the question remains: why do they (and others) say classical solutions? Is there some kind of second quantization in mind with this usage? - -Added, 11 October: -Even at the risk of boring the experts, I will have one more go. Jeff Harvey seems to indicate the following. We can think of $Map(X, M)$ as the fields in a non-linear sigma model on $X$, provisionally thought of as 1+1 dimensional spacetime. However, it seems that one can also associate to the situation a space of fields on $M$ (the string fields?). If we denote by ${\cal F}$ this space of fields, it seems that there is a functional $S$ on ${\cal F}$ with the property that the extrema of $S$ (the 'string equations of motion') can be interpreted as the $A(X)$. From this perspective, my main question might then be 'what is ${\cal F}$?' Since I think of fields on $M$ as being sections of some bundle on $M$, I can't see how to get such a thing out of maps from $X$ to $M$. -Thank you very much for your patience with these ignorant questions. - -Final addition, 11 October: -Thanks to the kind guidance of Jose, Aaron, and especially Jeff, I think I have some kind of an understanding of the situation. -I will attempt to summarize it now, superficial as my knowledge obviously is. I don't wish to waste more of the experts' time on this question. However, I am hoping that truly egregious errors will offend their sensibilities enough to elicit at least a cry of outrage, enabling me to improve my poor understanding. I apologize in advance for putting down even more statements that are either trivial or wrong. -As far as I can tell, the sense of Moore and Seiberg's sentence is as in my second addition: it is referring to second quantization. Recall that in this process, the single particle wave functions become the classical fields, and Schroedinger's equation is the classical equation of motion. Now the truly elementary point that I was missing (as I feared), is that -quantization of a 'single particle' string theory cannot give you a conformal field theory. -At most, a single string will propagate though space, giving us exactly the operators $A(S^1\times [0,t])$. If we want operators -$$A(X):{\cal H}^{\otimes n}\rightarrow {\cal H}^{\otimes m}$$ -corresponding to a Riemann surface with many boundaries, then we are already requiring a theory where particle numbers can change, that is, a quantum field theory coming from second quantization. WIth such a theory in place, of course, the $A(S^1\times [0,t])$ are exactly the solutions to the classical equations of motion, while the general $A(X)$ can be viewed either as 'generalized classical solutions' (I hope this expression is reasonable) or contributions to a perturbation series, as in the field theory of a point particle. So this, I think. already answers my original question. To repeat, because of the changing 'particle number' -the operators of conformal field theory cannot be the quantization of a 'single particle' theory. They must be construed as classical evolution operators of some kind of quantum (string) field theory. -The part I'm still far, far from understanding even superficially is this: The classical fields in the case of strings would be something like functions on $Map(S^1, M)$. I haven't the vaguest idea of how to get from this to fields on spacetime. The difficulty surrounding this issue seems to be discussed in the beginning pages of Zwiebach's paper referred to by Jeff, which is quite heavy reading for a pure mathematician like me. Some mention is made of infinitely many fields arising from the situation (alluded to also by Jeff), which perhaps is some way of turning the data of a function on loop space to fields on space(-time). - -REPLY [6 votes]: This is reiterating a lot of what Jeff said, but maybe I can explain from a different perspective. -There are two things going on here (as there always are in string perturbation theory.) The first is the string worldsheet, and the second is what is going on in spacetime. -The string worldsheet is a non-linear sigma model into spacetime. Here, spacetime is a Riemannian manifold (with plenty of other structure depending on the exact string model you're using.) The "nonlinear sigma model" on the string worldsheet (a surface potentially with multiple punctures/boundaries) has a metric (different from the metric on the target manifold) and a map from the worldsheet into the spacetime manifold -- there are other fields in fancier versions of string theory, but I'll neglect them. In string perturbation theory, you integrate over the moduli space of metrics and embeddings. The resulting theory is invariant under conformal transformations, and because metrics in two dimensions don't have a huge amount information in them, an essential part of the theory on the worldsheet ends up being a conformal theory. There are various other conditions which ensure that the CFT gives rise to a full theory of 2D quantum gravity, meaning that you really can integrate over the space of the metrics. If those conditions hold, using the CFT, you can compute string amplitudes corresponding to your punctured Riemann surface. The can be thought of as scattering string in spacetime. -The important thing is that the amplitudes computed above are supposed to be terms in an asymptotic expansion of, er, something. This is why it's called string perturbation theory: in analogy to quantum field theory, combining individual string amplitudes of higher and higher genus in a formal power series (where the parameter is called the "string coupling") is supposed to be an expansion arising from some "nonpertubative" theory. What this theory is in complete generality is still unknown (although we know a lot in various special cases). -We can try to ask what this all looks like from the point of view of spacetime. Now, a basic fact about perturbation theory is that it only really makes sense (or, at least, makes the best sense) when you're perturbing around a solution. All of this is a roundabout way of saying that the spacetime only makes sense when the target manifold and its various structures give rise to a good perturbation expansion which means that the two dimensional theory is a conformal field theory. -This is what people mean when they say that a 2D CFT is a solution to the equations of motion of string theory. In fact, you can drop the requirement that your 2D theory is a "non-linear sigma model", ie, that it has the structure of maps into a manifold. Then you get into the "moduli space" of two dimensional field theories. Which, as far as I know, is completely undefined. But, even in this case (the world of string field theory), the "classical solutions" are the ones where you can define a good perturbation expansion around, and those are the conformal field theories. -Added 10/12: -I wouldn't go too far with the entire "first quantization"/"second quantization" thing. You could imagine a free string field theory where the string isn't allowed to interact, but the nice thing about string perturbation theory is that the interactions and the propagation are different aspects of the same thing. This is in contrast to the perturbation theory of quantum field theories where the interactions and the propagators are different things (one is pointlike, the other is lifelike). The CFT (really, a theory of 2D quantum gravity) is what you start with, and you automatically get both the "free" string theory and the interactions. The question of "second quantization" (a term I really hate) is whether or not you can derive the formal power series resulting from adding the various amplitudes associated with Riemann surfaces of varying general as the perturbation expansion of another theory. -To answer your questions about how you go from fields on the worldsheet to fields in spacetime, you quantize the theory on the cylinder, and each vector in the Hilbert space corresponds to a spacetime field (because you can Fourier expand fields on the cylinder, this isn't as crazy as it sounds). However, because you really are doing 2D quantum gravity, you have to deal with the gauge invariances. The nice way to do this is using BRST quantization, and the actually physical fields are the cohomology of the BRST operator acting on the CFT Hilbert space. -This is pretty standard material in a first course on string theory. I don't have them on me, but I'd expect Eric D'Hoker's lectures in the IAS volumes on QFT and strings for mathematicians would do this.<|endoftext|> -TITLE: Are there nonobvious cases where equations have finitely many algebraic integer solutions? -QUESTION [12 upvotes]: Let $X$ be a scheme of finite type over $\mathbb{Z}$. Let $R$ be the ring of algebraic integers. My intuition is that $X(R)$ is practically always infinite. -More specifically, suppose that $X$ is faithfully flat over $\mathbb{Z}$, of relative dimension $\geq 1$, and the generic fiber is geometrically irreducible. Is that enough to guarantee infinitely many algebraic integer points? -This question is inspired by this one; I have no application in mind. - -REPLY [12 votes]: This is basically true, in view of a density theorem due to Robert Rumely -(Arithmetic over the ring of all algebraic integers, J. reine u. angew. Math. 368, 1986, p. 127-133). It relies on Rumely's capacity theory, and his extension of the theorem -of Fekete-Szegö. -For a generalization, and an algebraic proof, see also Laurent Moret-Bailly, Groupes de Picard et problèmes de Skolem. II. Annales scientifiques de l'École Normale Sup��rieure, Sér. 4, 22 no. 2 (1989), p. 181-194. (Numdam, http://www.numdam.org/item?id=ASENS_1989_4_22_2_181_0) -The hypothesis of Moret-Bailly's Theorem is that $X$ be irreducible, surjective and of positive relative dimension over ${\rm Spec}\mathbf Z$, and that its generic fiber be geometrically irreducible. Then, he proves that $X$ has $\overline{\mathbf Z}$-points which can be chosen arbitrarily close to a given $p$-adic point (end even more...).<|endoftext|> -TITLE: Pairs of Permutations up to Simultaneous Conjugation -QUESTION [6 upvotes]: The conjugacy classes of $S_n$ are the cycle types since if $\tau = (\dots)(\dots)\dots(\dots)$, the conjugation $\tau \mapsto \sigma \tau \sigma^{-1}$ permutes the labels in the cycles of $\tau$. -Has anyone studied pairs of permutations up to simultaneous conjugation $(\tau_1,\tau_2) \mapsto (\sigma \tau_1 \sigma^{-1}, \sigma \tau_2 \sigma^{-1})$? -These are related to branched covers of a once-punctured torus since $\pi_1(\mathbb{T}-\{ pt\}) = \langle a,b| \text{ no relations }\rangle = \mathbb{F}_2$ we need two generators, $\tau_1, \tau_2$. - -REPLY [4 votes]: Another way to say what's in oeis.org/A110143 is the following. Let -$\lambda$ be a partition of $n$, denoted $\lambda\vdash n$, and let -$z_\lambda$ be the number of permutations in $S_n$ commuting with a -fixed permutation $w$ of cycle type $\lambda$, so -$z_\lambda=1^{m_1}m_1!2^{m_2}m_2!\cdots$, where $\lambda$ has $m_i$ -parts equal to $i$. Thus the size of the conjugacy class containing -$w$ is $n!/z_\lambda$. It is then immediate from the Cauchy-Frobenius -lemma (aka Burnside's lemma) that the number of classes is - $$ \frac{1}{n!}\sum_{\lambda\vdash n} \frac{n!}{z_\lambda}z_\lambda^2 = - \sum_{\lambda\vdash n} z_\lambda. $$ -Similarly the number of $k$-tuples of permutations up to simultaneous -conjugation is $\sum_{\lambda\vdash n}z_\lambda^{k-1}$.<|endoftext|> -TITLE: Expressing a element of a Matrix subgroup in terms of subgroup generators -QUESTION [5 upvotes]: I'm no (computational) algebraist, and my searches have been pretty unyielding (probably due to the vast amounts written on the key words), but perhaps someone may know if this is possible, and if so, lead me to some solution. -Consider the subgroup $S$ of $GL(n,\mathbb{Z})$ which is generated by elements $s_1,...,s_k$. If $x\in S$, then $x$ has a representation as a word in the form $x=\Pi_{i}s_i$. Is it possible to find such a representation? -If this is possible, and computable, is there any efficient software out there? My current problem is a 10x10 matrix and I'm trying to fit this in to a subgroup generated by 9 10x10 matrices. - -REPLY [5 votes]: If you know in advance that $x$ can be expressed as a word in $s_1,\ldots,s_k$, then clearly it is possible to find such a word: just try all words until you find one that works! -But the generalized word problem is known to be unsolvable for finitely generated subgroups of ${\rm GL}(n,\mathbb{Z})$ for $n \ge 4$. This means that, there is no algorithm to decide whether an arbitrary element $x \in {\rm GL}(n,\mathbb{Z})$ can be written as a word over $s_1,\ldots,s_k$. To find references for that result, it is probably easiest just to do a search for "generalized word problem GL(n,Z)". There are good survey papers by C.F. Miller on decision problems in group theory. -So there cannot be an efficient algorithm to solve your problem that is guaranteed to work on all inputs, because if there were then we could use it to solve the generalized word problem: just run it for the prescribed time, and if it fails to solve the problem, then that must be because $x$ is not in the subgroup. Of course, for specific problems, then there might happen to be a special purpose method that works in that case.<|endoftext|> -TITLE: Isoperimetric-like inequality for non-connected sets -QUESTION [17 upvotes]: The classical isoperimetric inequality can be stated as follows: if $A$ and $B$ are sets in the plane with the same area, and if $B$ is a disk, then the perimeter of $A$ is larger than the perimeter of $B$. -There are several ways to define the perimeter. Here is a unusual one: if $A \subset \mathbb{R}^2$ is a convex set, the -Cauchy-Crofton formula says that the perimeter of $A$ equals the measure of the set of lines that hit $A$, or -$$ p(A) = \frac{1}{2} \int_{S^1} \lambda(P_{\theta} A) d\theta, $$ -where $P_\theta$ is the orthogonal projection in the direction $\theta \in S^1$, and $\lambda$ the Lebesgue measure on any line. -Now, this definition of $p(A)$ makes sense for non-necessarily convex sets, excepts that it is not the usual notion of perimeter, so let's call it rather "mean shadow". My question if whether the isoperimetric inequality holds for the mean shadow instead of perimeter: if $A,B$ are (open, say) subsets of the plane with equal area, and if $B$ is a disk, is the mean shadow of $A$ larger that the mean shadow of $B$ ? -The inequality is true if $A$ is convex, and we can assume that $A$ is a disjoint union of convex sets (since taking the convex hull of a connected set does not change the mean shadow). - -REPLY [3 votes]: I believe that the answer is positive. If $A$ is connected, then it has the same mean shadow as its convex hull $CH(A)$ so the isoperimetric inequality for $CH(A)$ shows that the mean shadow of $A$ is larger than the mean shadow of $B$. -If $A$ is not connected I believe the same inequality holds. I'll sketch a proof when $A$ has finitely many connected components $A_1, \cdots, A_n$, the general case then follows by an approximation argument. Choose a point $x\in CH(A)$ and define a 1-parameter family of -deformations of $A$ by making a parallel translate of each connected component $A_i$ so that its barycenter moves towards $x$, say at constant speed to reach it at time $t=1$. Stop this 1-parameter family of deformation as soon as a contact occurs between two connected components, then merge those two connected components and repeat. -The point is that the area of $A$ does not change under this deformation, however the mean shadow is non-increasing -- actually the size of the shadow is non-increasing in every direction. At the end of this deformation one obtains a connected set to which the usual isoperimetric inequality can be applied, and the same inequality then also applies to $A$.<|endoftext|> -TITLE: implicit function theorem for algebraic sets -QUESTION [5 upvotes]: We know by the standard Implicit Function Theorem that - -If $f:\mathbb R^4\rightarrow\mathbb -> R^2$ is a polynomial (or in fact any - continuously differentiable function), - then there is a point $a\in\mathbb -> R^2$ such that $f^{-1}(a)$ is at least - two-dimensional. - -Now imagine that instead of $\mathbb R^2$ we have an algebraic surface $S\subset\mathbb R^3$, i.e. the zero set of a trivariate polynomial. It's reasonable that the statement still holds. A general statement along these lines would be something like - -If $A,B$ are two algebraic sets such - that $\dim(A)>\dim(B)$, and if - $f:A\rightarrow B$ is a polynomial - mapping, then there is a point $b\in -> B$ such that $f^{-1}(b)$ is of - dimension at least $\dim(A)-\dim(B)$. - -Is it a well-known theorem? Any reference for it? - -REPLY [3 votes]: This is a well-known theorem. I imagine it has been routinely used for many years, so tracking down a historical reference would be hard. The following argument will work for any $f$ which is definable over the real field, i.e. whose graph is a semi-algebraic set. -The graph $\Gamma \subset A\times B$ of your mapping has dimension $\dim(A)$, since the restriction to $\Gamma$ of the projection $\pi_A: A\times B \to A$ is 1-to-1. Now do a cylindrical algebraic decomposition of $A\times B$ adapted to $\Gamma$ and the projection $\pi_B$. Then, any cell in $f(B)$ is the image of a cell of $\Gamma$. In particular, this is true for all the $\dim(A)$-dimensional cells that appear in $\Gamma$; if $C$ is such a cell, any point $b$ in the image $\pi_B(C)$ verifies $\dim(f^{-1}(b)) \geq \dim(A)-\dim(\pi_B(C)) \geq \dim(A)-\dim(B)$. -So not only can you find such a $b$, but there should be quite a few of them.<|endoftext|> -TITLE: Cohomology of representation varieties -QUESTION [11 upvotes]: Perhaps this question is too general then I am sorry about this. -My question is the following. -Let $\pi$ be the fundamental group of a compact surface of genus $g$ (with if necessary $n$ punctures) and $G$ a Lie group. I make no assumptions on the Lie group but if it interesting to do so then it is ok. -Are there any results known about the cohomology of the representation variety $Hom(\pi,G)$ if it is not a manifold. -So I know the calculations for $Hom(\pi,G)/G$ by Atiyah and Bott where it has a smooth structure. But here I am only interested in the cohomology of the topological space that is in more general results. - -REPLY [6 votes]: If $G=U(n)$ then I know a fair amount. Letting $n$ tend to infinity, Hom$(\pi_1 M^g, U)$ is homotopy equivalent to $U^{2g} \times BU$, so stably one can write down the cohomology using standard facts about the infinite unitary group. There is also a stability range for the inclusions Hom$(\pi_1 M^g, U(n))\to$ Hom$(\pi_1 M^g, U(n+1))$ (they are $(2n-2)$-connected maps). Roughly this range is the stability range for the unitary groups minus 2, if I remember correctly. -The way to prove these things is to follow Atiyah-Bott and think about the representation space as the space of flat connections modulo the based gauge group. If this is the sort of information you're looking for, and you'd like to know more details, I can say more. I think I wrote some notes a few years ago that go through this carefully. But since you mentioned Atiyah-Bott already, you may be looking in a different direction. -If you puncture the surface, the fundamental group becomes free. Then you might be interested in Tyler Lawson's paper about simultaneous similarity of unitary matrices (Math. Proc. Camb. Phil. Soc. 2008 or http://arxiv.org/abs/0809.0466 ). Or you might be interested in keeping track of some additional structure related to the punctures, in which case some of Tom Baird's work may be of interest to you.<|endoftext|> -TITLE: Dimension of certain subgroup of isometry group of positively curved manifold -QUESTION [5 upvotes]: Let $M$ be a closed $n$-dimensional Riemannian manifold with positive sectional curvature. -Let $G$ be a close subgroup of isometry group ${\rm Iso}(M)$. Suppose the action of $G$ on $M$ is not transitive, hence $M/G$ has dimension at least $1$. -By a theorem of Grove and Searle the symmetry rank $${\rm symran}(M)\le [\frac{n+1}{2}]$$ -I am wondering is there any upper bound for the dimension of $G$ mentioned above? - -REPLY [10 votes]: Forgetting positive curvature, if $\dim M^n/G=k$ then by looking at the transitive action of $G$ on the principal orbit one gets a trivial bound $\dim G\le \dim O(n-k)=\frac{(n-k+1)(n-k)}{2}$. This bound is realized for $k=1$ on a round $S^n$ and $G=O(n-1)$. As the sphere is positively curved this bound is sharp. -Addressing the comment below, the assumption of $M/G$ being a manifold is not a natural one in this context. It hardly ever happens when $M$ has positive curvature. In particular, by a result of Wilking (Lemma 5 in "Positively curved manifolds with symmetry") based on his connectedness principle, if the principal isotropy group $H$ is not trivial and $M/G\ne pt$ then $M/G$ has a boundary. If $H=1$ and $M/G$ is a smooth manifold then the $G$-action is free and hence $rank G\le 1$ by Berger's vanishing theorem. -I should add that there is large literature on the subject of isometric group actions on positively curved manifolds (mostly by Wilking, Grove, Ziller, Searle and Rong) and I suggest you study it if you want to pursue these kind of questions. A good place to start is a survey by Wilking "Nonnegatively and positively curved manifolds".<|endoftext|> -TITLE: $S^1$-action in three dimensions -QUESTION [6 upvotes]: Let $M$ be a closed, orientable 3-manifold with a non-trivial differentiable $S^1$-action. -What does this imply for $M$? What are examples except for (products of) spheres? - -REPLY [8 votes]: If the action has no fixed points, that class of manifolds has a name: Seifert-fibred 3-manifold with an orientation on the fiber. Seifert classified a slightly larger class of manifolds (Seifert-fibered ones -- no orientability constraint on the fibers). This is available in either the Jaco or Hatcher 3-manifolds lecture notes, also Orlik's book on Seifert-Fibered spaces. -Such examples include things like lens spaces, and relatively complicated 3-manifolds like the double of a torus knot complement. -When you allow fixed points, the orbit decomposition theorem gives you a stratification of the manifold into the fixed-point set (a link) together with its Seifert-fibred complement. So you could view these manifolds as certain Dehn Fillings on Seifert-fibred manifolds with boundary. As the references above point out, they're abundant.<|endoftext|> -TITLE: Can local duality for elliptic curves be proven with "big rings"? -QUESTION [11 upvotes]: From Exercise 5.14, Ch. V of Silverman's "Advanced Topics in the Arithmetic of Elliptic Curves", I learned that the local duality for elliptic curves over $p$-adic fields can be proven for Tate curves by a relatively easy argument in Galois cohomology. Essentially, when the elliptic curve is $E_q = G_m / q^Z$ over a $p$-adic field $K$, one can find various long exact sequences connecting the Galois cohomology $H^1(K, E_q(\bar K))$ to the cohomology of $G_m$, $Q/Z$, etc., which are well-known by class field theory. -Without being an expert in $p$-adic cohomology, Hodge theory, etc., I know that by passing to a big ring ($B_{dR}$ will work), one can find a pair of periods for an elliptic curve over $K$ with good reduction. There might not be any interesting nontrivial uniformization of such elliptic curves, but the periods carry the information instead. -So can one exhibit (some of) the duality between $H^1(K, E(\bar K))$ and $Hom(E(K), Q/Z)$ when $E$ has good reduction, by using a big ring like $B_{dR}$? -When I see period rings, they are always used as linear algebraic gadgets. But since $B_{dR}$ is a $K$-algebra with Galois action, might someone consider $H^1(K, E(B_{dR}))$? In other words, rather than taking a linear algebraic gadget over $\bar K$, and tensoring up to $B_{dR}$, might one study a variety over $\bar K$ and base change to $B_{dR}$ or at least take $B_{dR}$-points? Might $H^1(K, E(B_{dR}))$ pick up the Weil-Chatelet group in the spirit of my first question? -Any references including $B_{dR}$-points of varieties would be greatly appreciated, as well as answers to the questions. - -REPLY [5 votes]: If I've understood correctly what you want, this is in the Bloch–Kato article in volume 1 of the Grothendieck Festschrift. The local duality result (I think) you are talking about is at the top of page 353 and the generalization is Proposition 3.8. Basically, you can show that if you have an abelian variety $A$ over $K$ with Tate module $T$, then the image of the Kummer map -$$A(K)\rightarrow H^1(K,T)$$ -is the Bloch–Kato Selmer group $H^1_f(K,T)$ (which is defined using $B_{\text{cris}}$; basically a derived functor of tensoring with $B_\text{cris}$ and taking Galois invariants). For a general $p$-adic Galois representation on a finite free $O_K$-module $T$, one can still define $H^1_f(K,T)$. One shows that under the usual local Tate duality -$$H^1(K,T)\times H^1(K,T^\vee\otimes\mathbf{Q}_p/\mathbf{Z}_p)\rightarrow H^2(K,\mathbf{Q}_p/\mathbf{Z}_p(1))\cong\mathbf{Q}_p/\mathbf{Z}_p$$ -the annihilator of $H^1_f(K,T)$ is $H^1_f(K,T^\vee\otimes\mathbf{Q}_p/\mathbf{Z}_p)$ (and vice versa) (where $T^\vee$ denotes the Tate dual). That the latter thing is the $H^1(K,E(\overline{K}))$ that you want it to be is equation (3.3) of the Bloch–Kato paper. -For example, when trying to generalize some construction using Kummer theory for elliptic curves to the case of higher weight modular forms, people use $H^1_f(K,T)$. Of course, the Bloch–Kato Selmer groups pretty much permeate a lot of things right now.<|endoftext|> -TITLE: Devlin's "Constructibility" as a resource -QUESTION [21 upvotes]: It is fairly well-known among set-theorists that Keith Devlin's 1984 book "Constructibility" has flaws in its initial development of fine structure theory. (See Lee Stanley's review 1 of the text for the Journal of Symbolic Logic, for example.) -I've had the book on my shelf for twenty years now, and although there is much in there that I find interesting, the fact that I know there are some errors in it means that I've been reluctant to invest a lot of time working through it. So, this brings me to my questions for the experts: -How badly do these flaws mar the rest of the book? Is the damage localized to the initial development of properties of the the J-hierarchy, or is it much more widespread? -Of particular interest to me are the following questions: -1) Is Devlin's treatment of the Covering Lemma for L on solid ground? -2) What about his treatment of morasses? -I know that there are other sources for this material, but I've always appreciated Devlin's writing style. - -REPLY [24 votes]: Mathias has a paper where he corrects the flaws that occur in Devlin's theory BS (= Basic Set Theory). The theory has to be only slightly strengthened to be correct. (It is more than sufficent to add an axiom that asserts for any set and any $n \in \omega$ there is a set of all its sized $n$-subsets.) It is really only in dealing with syntax and showing that certain straightforward concepts are $\Delta_1$ that BS comes unstuck. -I think the book can be safely read beyond a certain point. It is true that Devlin has not correctly proved that the satisfaction relation for $\Delta_0$ formulae is $\Delta_1$, but one can just take the attitude that the result is correct (as Jensen showed), it is just that that particular development in BS failed. BS needs another axiom and all would be well. -Thus, the constructions of the $J$-hierarchy, the fine structural concepts of projecta, mastercodes these are all fine, -as are the constructions of trees, $\Box$, Morasses etc, and the Covering Lemma can all be safely read since there one is past the point where these delicate matters are being considered. -Mathias: Weak Systems of Gandy, Jensen and Devlin, Centre de Recerca Matem`atica, Barcelona, 2003--2004, Birkh\"auser Verlag, 2006, Eds: Bagaria, Joan and Todorcevic, Stevo, Trends in mathematics Series.<|endoftext|> -TITLE: Elliptic curves over the complex numbers: everything "well known"? -QUESTION [5 upvotes]: This may be a very naive question. We always hear about elliptic curves over the rational numbers, or over other arithmetically significant fields or rings. -But, are there open problems or recent fertile theories related to elliptic curves over the complex numbers or is everything considered "classical" and well known? -For example, what about moduar parametrizations: are they only interesting as far as they involve curves defined over "arithmetic bases"? If yes, why? -What about elliptic cohomology, $\mathrm{tmf}$ and the like: if I'm not mistaken, the moduli spaces considered in that theory are defined over $\mathbb{Z}$; anything relevant/interesting happens over $\mathbb{C}$? -(I asked this question after having read this question -which, I must say, I'm not able to understand due to my ignorance of arithmetic geometry- in which one is lead to consider even points valued in -I think- de Rham differential forms...) - -REPLY [3 votes]: Your question is very vague. I don't know of any open problem about elliptic curves over the complex numbers per se, although one could come up with some unproven identity among elliptic functions or modular functions and say it's "about" elliptic curves. Then again, I am a number theorist. -As for your more specific question about modular parametrizations. If an elliptic curve admits a non-constant map from a modular curve (in the usual sense) to it, then the elliptic curve is defined over a number field, since there are only finitely many such elliptic curves for each modular curve (which is an easy consequence of Poincaré's complete reducibility or de Franchis's theorem or whatever). -Finally, about the other MO question you link to, the ring $B_{dR}$ has very little to do with de Rham differentials directly. It's one of Fontaine's rings of $p$-adic periods.<|endoftext|> -TITLE: Is there an infinite-dimensional Banach space with a compact unit ball? -QUESTION [19 upvotes]: A popular pair of exercises in first courses on functional analysis prove the following theorem: - -The unit ball of a Banach space $X$ is compact if and only if $X$ is finite-dimensional. - -My question is, is the "only if" part of this (i.e., that the unit ball of an infinite-dimensional Banach space is noncompact) necessarily true without some form of the axiom of choice? The usual proof uses the Hahn--Banach theorem, which may reasonably be regarded as a weak form of the axiom of choice (see this answer, and other answers to the same question, for some interesting points related to this). - -REPLY [2 votes]: I am a bit surprised. Perhaps I have misunderstood the question, but the standard proof of the OP's question does not rely upon the HB Theorem, but rather simply on Riesz' Lemma - in whose proof in turn I do not see any application of the AoC.<|endoftext|> -TITLE: Bounds on $\max \mathrm{pcf}(A)$ if $\Pi A$ is big -QUESTION [6 upvotes]: For concreteness, let $A = \{\aleph_n : n < \omega\}$. We know $\max \mathrm{pcf}(A) \in [\aleph_{\omega+1},\Pi A]$. My question is, if $\Pi A$ is big (say, $\aleph_{\omega_1+1}$), then which cardinals in that interval can $\max \mathrm{pcf}(A)$ really be? -My second question, which motivates the first, is: Let $E = \{\aleph_{2n} : n < \omega\}$ and $O = \{\aleph_{2n+1} : n < \omega \}$; then is it possible for $\max \mathrm{pcf}(A) = \max \mathrm{pcf}(E) \gg \max \mathrm{pcf}(O)$ (where $\gg$ means "much larger than" in any way you care to make precise)? -If the answer to the first question is that $\max \mathrm{pcf}(A)$ can never really be that big, then the second question doesn't matter. But if it can be big, and the answer to the second question is that the even alephs and the odd alephs can have very different $\max\mathrm{pcf}$'s, I would find that somewhat disturbing. - -REPLY [4 votes]: Shelah proved that it is consistent that GCH holds below $\aleph_\omega$, while $2^{\aleph_\omega}=\aleph_{\omega+\alpha+1}$ for any countable ordinal $\alpha$ you care to choose. (See Theorem 36.5 of Jech's book, for example). -In such a model, ${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$ as well. Now if you add $\aleph_{\omega_1+1}$ Cohen reals (which has no effect on the pcf structure) you end up with a model where - -$|\prod A| = \aleph_{\omega_1+1}$, and -${\rm max pcf}(A)=\aleph_{\omega+\alpha+1}$. - -So ${\rm max pcf} (A)$ could potentially be any successor cardinal below $\aleph_{\omega_1}$. -(Of course, it's still unknown if $\aleph_{\omega_1}\leq{\rm max pcf }(A)$ is possible, so this is the best answer we can hope for given our current knowledge.) -I don't know the answer to your "evens and odds" question, but certainly you can split $A$ up into two disjoint pieces whose "gap" is as large as possible: -Let $\tau$ denote ${\rm max pcf}(A)$, and suppose $\aleph_{\omega+1}<\tau$. -We know there exists an unbounded $B\subseteq A$ such that $\prod B$ contains a scale (mod finite) of length $\aleph_{\omega+1}$. This implies ${\rm max pcf}(B)=\aleph_{\omega+1}$. -The set $A\setminus B$ cannot be in the ideal $J_{<\tau}[A]$ (otherwise, we contradict ${\rm max pcf}(A)=\tau$), and so we must conclude ${\rm max pcf}(A\setminus B)=\tau$ as well.<|endoftext|> -TITLE: Generalizations of pcf theory -QUESTION [5 upvotes]: Does anyone know of generalizations of pcf theory where we might consider products of the form: -$$\aleph_1 \times (\aleph_2 \times \aleph_2) \times (\aleph_3 \times \aleph_3 \times \aleph_3) \dots$$ -$$(\aleph_1 \times \aleph_2 \times \dots) \times (\aleph_1 \times \aleph_2 \times \dots) \times \dots$$ -or, more abstractly: -$$P_1 \times P_2 \times P_3 \times \dots$$ -where each $P_i$ is a $\mathrm{cof}(P_i)$-directed partial order. My motivation is that I'm interested in what the relationship between -$$\max\mathrm{pcf}\langle \aleph_1,\aleph_2,\aleph_2,\aleph_3,\aleph_3,\aleph_3,\dots\rangle$$ -and -$$\max\mathrm{pcf}\langle\aleph_n\rangle_{0 < n < \omega }$$ -might be because it might help me understand the relationship between -$$\max \mathrm{pcf} \langle \aleph_n \rangle_{0 < n <\omega }$$ -and -$$\max \mathrm{pcf} \langle \aleph_{2n} \rangle_{0 < n <\omega }$$ - -Claim: The $\mathrm{pcf}$ structure on $(\aleph_1 \times \aleph_2 \times \dots) \times (\aleph_1 \times \aleph_2 \times \dots) \times \dots$ gives nothing new. -Proof: For notational convenience, let $A : \omega \to \mathrm{Reg}$ be defined by $A(n) = \aleph_n$ and let $B : \omega \cdot \omega \to \mathrm{Reg}$ be defined by $B(\omega\cdot m + n) = \aleph_n$. Define -$$\mathrm{pcf}(A) = \{\mathrm{cf}(\Pi_{n<\omega}A(n)/U)\ :\ U \in \beta \omega\}$$ -$$\mathrm{pcf}(B) = \{\mathrm{cf}(\Pi_{\alpha<\omega\cdot\omega}B(\alpha)/U)\ :\ U \in \beta (\omega\cdot\omega)\}$$ -Where $\beta X$ denotes the set of all ultrafilters on $X$. It's not hard to see that $\mathrm{pcf}(A) \subseteq \mathrm{pcf}(B)$ and since $\mathrm{pcf}(A)$ is an interval of regular cardinals, it suffices to show that $\max \mathrm{pcf}(B) = \max \mathrm{pcf}(B)$. We know that we can find an everywhere-pointwise-dominating family on $\Pi A$ of size $\lambda := \max\mathrm{pcf}(A)$. If we can find a dominating family on $\Pi B$ of that same size, we'll be done. -So, given a dominating family $\mathcal{F}$ on $\Pi A$ of size $\lambda$, just let $\mathcal{F}^\ast \subseteq \Pi B$ consist of functions of the form: -$$f^\ast (\omega\cdot m + n) = f(n)$$ -for each $f \in \mathcal{F}$. Now given $g \in \Pi B$, define: -$$g'(\omega\cdot m + n) = \sup_{m' \in \omega}g(\omega\cdot m' + n)$$ -We see that $g' \geq g$ everywhere pointwise, and $g' \in \Pi B$ since we're always taking countable suprema within uncountable regular cardinals. Since $g'$ has the same value at any of its coordinates that correspond to the same $\aleph_n$, it's clear that there's some $f \in \mathcal{F}$ such that $f^\ast \in \mathcal{F}^\ast$ dominates $g'$ everywhere, and hence $g$ everywhere. - -REPLY [6 votes]: I don't know if anyone has looked at such things systematically, but I know Shelah has made use of structures of this form at various times. The examples which follow are just what I can remember off-hand; I know there's more buried in his work, but this is where I remembered seeing such a construction: -1) Clause $(\gamma)$ on page 1641 of [Sh:589]: -Applications of PCF theory. J. Symbolic Logic 65 (2000), no. 4, 1624–1674. -(It's available on JSTOR here; the Arxiv version looks like an older iteration) -2) The second proof of his Revised GCH theorem in [Sh:460], in particular Claim 2.6. -The generalized continuum hypothesis revisited. Israel J. Math. 116 (2000), 285–321. -(Available on SpringerLink here, again the Arxiv version is a bit dated) -I don't think that you get anything really new as far as pcf theory by doing this. Rather, it's just that sometimes such a structure is the right way to organize things.<|endoftext|> -TITLE: Can an operator have Exp(z) as its characteristic "polynomial"? -QUESTION [28 upvotes]: Let $\mathcal{H}$ be a Hilbert space, and let $T: \mathcal{H} \rightarrow \mathcal{H}$ be a trace-class operator. Define -$$ f_T(z) = \sum_{i=0}^\infty \mbox{Tr}(\wedge^k T) \cdot z^k, $$ -the ordinary generating function for traces of exterior powers of $T$. Expressed another way, -$$ f_T(z) = \mbox{Det}(I + zT) $$ -where $\mbox{Det}$ is the Fredholm determinant. This function is entire, and can be considered a generalization of the characteristic polynomial. -I am wondering if -$$f_T(z) = e^z $$ -for some natural choice of $T$ on some nice incarnation of Hilbert space. -The motivation for this problem comes from representation theory, where finite minors of such a $T$ provide a formula for dimensions of irreps of $S_n$, the symmetric group on $n$ letters. ---edit-- -In order to provide slightly more background on the sort of application I have in mind, here is an attractive identity which might pique your interest: -If $\beta : \mathbb{C} \rightarrow \mathbb{C}$ is any function, define $$\gamma(z) = \frac{1}{\Gamma(z+1)}, \hspace{.4in} \delta(z) = \frac{\beta(z)}{\Gamma(z+1)}$$ -Now -$$ -\left| -\begin{pmatrix} -\gamma(3) & \gamma(4) & \gamma(5) \\\\ -\gamma(-1) & \gamma(0) & \gamma(1) \\\\ -\gamma(-2) & \gamma(-1) & \gamma(0) -\end{pmatrix} -\begin{pmatrix} -\delta(3) & \delta(4) & \delta(5)\\\\ -\delta(-1) & \delta(0) & \delta(1)\\\\ -\delta(-2) & \delta(-1) & \delta(0) -\end{pmatrix} -\right| -+ $$ -$$ -\left| -\begin{pmatrix} -\gamma(2) & \gamma(3) & \gamma(4) \\\\ -\gamma(0) & \gamma(1) & \gamma(2) \\\\ -\gamma(-2) & \gamma(-1) & \gamma(0) -\end{pmatrix} -\begin{pmatrix} -\delta(2) & \delta(3) & \delta(4)\\\\ -\delta(0) & \delta(1) & \delta(2)\\\\ -\delta(-2) & \delta(-1) & \delta(0) -\end{pmatrix} -\right| -+ $$ -$$ -\left| -\begin{pmatrix} -\gamma(1) & \gamma(2) & \gamma(3) \\\\ -\gamma(0) & \gamma(1) & \gamma(2) \\\\ -\gamma(-1) & \gamma(0) & \gamma(1) -\end{pmatrix} -\begin{pmatrix} -\delta(1) & \delta(2) & \delta(3)\\\\ -\delta(0) & \delta(1) & \delta(2)\\\\ -\delta(-1) & \delta(0) & \delta(1) -\end{pmatrix} -\right| = \frac{\delta(1)^3}{3!} -$$ - -REPLY [15 votes]: Here's a proof that $\exp(z)$ is not a characteristic function using the product expansion for the determinant, which is essentially equivalent to Lidskii's theorem stating that the trace of a trace class operator is the sum of its eigenvalues. - -If $T$ is a trace class operator on a Hilbert space $\mathcal{H}$ with eigenvalues $\lambda_n$ (counted up to algebraic multiplicity), then $\sum_n\vert\lambda_n\vert < \infty$ and - $$ -{\rm det}(1+zT)\equiv\sum_{k=0}^\infty{\rm Tr}(\Lambda^kT)z^k=\prod_n(1+z\lambda_n) -$$ - for all $z\in\mathbb{C}$. - -See Trace Ideals and Their Applications, Theorem 3.7. Actually, they use this result to prove Lidskii's theorem, that ${\rm Tr}(T)=\sum_n\lambda_n$. -Then, ${\rm det}(1+zT)$ cannot be equal to $\exp(z)$ everywhere. This is because $\exp$ has no zeros, so $\lambda_n$ would have to be all zero, giving ${\rm det}(1+zT)=1$.<|endoftext|> -TITLE: The Wronskian of sin(kx) and cos(kx), k=1...n -QUESTION [19 upvotes]: What is the determinant of the Wronskian of the functions $\{\cos\ x, \sin\ x, \cos\ 2x, \sin\ 2x,\ldots, \cos\ nx, \sin\ nx\}$? This determinant seems to be an integer, and the sequence starts with 1, 18, 86400, 548674560000... It is not in the Encyclopedia. - Question What is this sequence? I guess it is enough to prove that it consists of integers (constants), because then to compute it, one can simply put $x=0$. - Update 1. The fact that the determinant of the Wronskian matrix is a constant is obvious. Take the derivative of the determinant. It is a sum of determinants of matrices each of which has two proportional columns. - Update 2. The determinant is equal to the square of the Vandermonde determinant of $1, 2^2,\ldots, n^2$ times $n!$ (alternatively see Felipe Voloch's answer below). It is interesting that for $n=1$ we get just the equality $\cos^2 x+\sin^2 x=1$ (the equation of the circle). So the equality for $n > 1$ can be considered as the generalization of this equation. What is the geometry behind this identity? Of course the parametrization $(\cos x,\sin x,\ldots, \cos nx, \sin nx)$ defines some curve in $\mathbb{R}^{2n}$. What is known about that curve? - Update 3. Here is an easier formula for the determinant. It is equal to -$$(1! 3!\ldots (2n-1)!)^2/n!$$ - Update 4 I found a related paper: -Larsen, Mogens Esrom, Wronskian harmony. -Math. Mag. 63 (1990), no. 1, 33–37. He considers the Wronskian of $\sin x, \ldots, \sin nx$. - Update 5. The sequence is in the Encyclopedia now. - -REPLY [10 votes]: If you only need that it doesn't depend on $x$ consider the following argument. For any $\pi \in S_{2n}$ written as $\pi(1)\pi(2)\cdots \pi(n)$, let $[\pi]$ be the set of all permutations $\sigma \in S_{2n}$ which satisfy $\lbrace\sigma(2k-1),\sigma(2k)\rbrace =\lbrace\pi(2k-1),\pi(2k)\rbrace$ as sets for all $k$. Then $S_{2n}$ can be written as a disjoint union of such classes $[\pi]$ and so the terms in the determinant can be grouped accordingly. One can see that for each $[\pi]$ the correpsonding terms in the determinant add up to zero if $\pi(2j-1)+\pi(2j)$ is even for some $j$, and to -$$ (-1)^{\text{something}}\prod_{j=1}^{n}\left( j^{\pi(2j-1)+\pi(2j)}(\sin^2(jx)+\cos^2(jx))\right)$$ -otherwise, which makes it clear that the determinant doesn't depend on $x$. - -REPLY [9 votes]: These functions are solutions of the homogeneous differential equation with characteristic equation $(r^2+1)(r^2+4)(r^2+9)\dots(r^2+n^2)$. By Abel's identity, the Wronskian of a fundamental set of solutions is its value at zero times the integral $\int_0^x -p_{2n-1}(t) dt$, where $p_{2n-1}(t)$ is the coefficient at $y^{(2n-1)}$. But in our case $p_{2n-1}(t)=0$, so we have that the Wronskian is constant. - -REPLY [5 votes]: Is something similar to $(i/2)^n$ times the Vandermonde of $i,2i,\ldots,ni,-i,-2i,\ldots,-ni$.<|endoftext|> -TITLE: Number of the Reidemeister moves needed to transform one diagram into another one -QUESTION [11 upvotes]: A recent question Random Reidemeister moves to unknot contains a link to the paper http://www.ams.org/journals/jams/2001-14-02/S0894-0347-01-00358-7/S0894-0347-01-00358-7.pdf, in which J. Hass and J. Lagarias show that one can transform any unknot diagram with $n$ crossings into the standard unknot diagram using not more than $2^{cn}$ Reidemeister moves, with $c=10^{11}$. -[As an aside: this is quite a large bound, so the first thing that comes to mind when one looks at it is a computer falling apart with all its atoms decaying long before it manages to untie a diagram with a single crossing. As far as I understand, for those diagrams the algorithm works faster, but still it is probably impractical for untying knots that can't be untied by trial and error.] -It seems plausible that the methods of Hass and Lagarias can be adapted to give a similar explicit upper bound for the number of the Reidemeister moves needed to transform two diagrams representing isotopic links into one another. I would like to ask whether this is indeed the case, and if so, whether there is a reference for that. -A related question: given a nonnegative integer $n$, is it possible to estimate from above the minimal $m$ such that any two link diagrams with $\leq n$ crossings that represent isotopic links can be connected by a sequence of diagrams with $\leq m$ crossings such that each is obtained from the preceding one by a Reidemeister move? - -REPLY [10 votes]: Lackenby updated his arXiv paper today (12Dec2014) in which he proves a polynomial bound -on the moves to uncross the unknot. I am uncertain if this also implies -a polynomial bound on moving from one knot to another. - -Marc Lackenby. -"A polynomial upper bound on Reidemeister moves." -2014. (62 pages.) -(arXiv abs link.) - - -"We prove that any diagram of the unknot with $c$ crossings may be reduced to the trivial diagram using at most $(236 c)^{11}$ Reidemeister moves. -Moreover, every diagram in this sequence has at most $(7 c)^2$ crossings." - - -(19Feb2021). See update at "Are there any very hard unknots?". -for Lackenby's Feb 2021 "Unknot recognition in quasi-polynomial time."<|endoftext|> -TITLE: Existence of a non-trivial zero (in the rational cyclotomic field) of a form -QUESTION [5 upvotes]: It is well known that if a field K is quasi-algebraically closed (i.e. all forms with coefficients in K of degree d in n > d variables have a non-trivial zero in K) then it has no central divison algebras. It is also known that there are no central division algebras over the rational cyclotomic field (obtained by adjoining to Q all roots of unity). -Is it known whether the rational cyclotomic field is quasi-algebraically closed? -For instance, I know that the answer is yes for p-adic cyclotomic fields and that the converse of the first result above is false. - -REPLY [4 votes]: This is an old conjecture of Artin and, as far as I know, it is still open. It is mentioned as such on page 477 of this 2007 article on the work of Lang: -http://www.ams.org/notices/200704/fea-lang-web.pdf<|endoftext|> -TITLE: ASCII prime plots and prime-rich quadratic polynomials -QUESTION [6 upvotes]: This is a series of questions inspired by the MathOverflow question -Find the least prime so that p-1 has two factors greater than $m$ and $n$ posted by Aaron Sterling. -I suggested plotting primes by marking the status of the number $(nm+1)$ at coordinate $(n,m)$. Using commutativity, I have combined two ASCII art plots -for $1 \leq n,m \leq 50$ in a figure-ground contrast. (Perhaps Joseph O'Rourke -will be inspired to provide some similar but nicer looking plots for other ranges of $n$ and $m$.) In the plots below, + indicates $nm+1$ is prime, and other characters (after a shift in one coordinate) indicates whether $nm + 1$ has a factor of two, three or five. -+ .o.Oo .oO o .O. o Oo. oO.o. O .o.Oo .oO o .O. o Oo -++ O o O o Oo oO o O o Oo oO o O o Oo oO o - + O . .O. . O . .O. . O . .O. . O . .O. . O . .O. -+ ++ oO o O o Oo oO o O o Oo oO o O o Oo o - + . o .o. o .o. o .o. o .o. o .o. o .o. o .o. o -+++ ++ O O O O O O O O O - + + Oo. oO.o. O .o.Oo .oO o .O. o Oo. oO.o. O .o - + + O o O o Oo oO o O o Oo oO o O o Oo - + + + . O . .O. . O . .O. . O . .O. . O . .O. . -+ ++ ++ + o o o o o o o o o o o o o o - + + + . oO.o. O .o.Oo .oO o .O. o Oo. oO.o. O -+ + ++ ++ O O O O O O O O - + + + + Oo. oO.o. O .o.Oo .oO o .O. o Oo. oO.o - ++ + ++ + O o Oo oO o O o Oo oO o O o - + + + + + . . . . . . . . . . . . . . . . . . -+ ++ + ++ o Oo oO o O o Oo oO o O o Oo - + + + O o .O. o Oo. oO.o. O .o.Oo .oO o -++ + ++ +++ + + O O O O O O O - + + .oO o .O. o Oo. oO.o. O .o.Oo .o - ++ + + + + + o o o o o o o o o o - + + + + + + . .O. . O . .O. . O . .O. . O -+ ++ + ++ ++ + Oo oO o O o Oo oO o O o - + + + + O o .O. o Oo. oO.o. O .o.Oo - ++ + + ++ +++ + O O O O O - + + + + + .o. o .o. o .o. o .o. o .o - ++ ++ + + ++ + + oO o O o Oo oO o O - + + + + + + + O . .O. . O . .O. . O . -+ + + + + ++ + ++ + Oo oO o O o Oo oO o - + + + + . O .o.Oo .oO o .O. o -++ +++++ + + ++++++ + + - + + + .o.Oo .oO o .O. o Oo - + + + + + + ++ + + O o O o Oo oO o - + + + + + + + + + O . .O. . O . .O. - ++ + + ++ ++ ++ + + oO o O o Oo o - + + + + + + + + + . o .o. o .o. o -+++ + ++ +++ + + + + ++ + O O O - + + + + Oo. oO.o. O .o - ++ ++ + + + + + O o O o Oo - + + + + + + + + + + + . O . .O. . -+ ++ + + ++ + + + ++ + + o o o o - + + + + + . oO.o. O -+ + + ++++ + ++ + + ++++ + ++ + + O O - + + + + + + + Oo. oO.o - + ++ ++ + + ++ ++ O o - + + + + + + + + + + . . . -+ + + + + + + ++ + + ++ + ++ o Oo - + + + + + + + + O o - + ++ + + + + + + ++++ + + + + ++ O - + + + + + + .o - ++ + + + ++ + ++ + + + + ++ + + - -Based on this plot, I suspect my conjecture about the prime "nearest" to and greater than $n*m$ being at most $4nm$ not only holds (as a sort of 2-dimensional Bertrand's -conjecture), but that this prime differs in taxicab distance by $O(\log(nm)^2)$. In other words, there is an absolute constant $C$ such that there is a prime $p$ with $p-1 = n'm'$, and with $n \leq n' \leq n + C\log(nm)^2$ and also $m \leq m' \leq m + C\log(nm)^2$. I am interested in information supporting or refuting my suspicion (and I suspect Aaron Sterling shares this interest), but that is incidental to what follows. -The primary question is a reference request: has anyone seen a plot like this before in the literature? I know of Ulam Spirals ( http://en.wikipedia.org/wiki/Ulam_spiral ) -and it seems that artefacts in the plot might be related to a conjecture of Hardy and Littlewood regarding primes of the form $ax^2 +bx +c$. What I find striking are the diagonals that occur in the plot, especially those starting at $(a,a)$ and continuing in the direction $(2,-1)$. In particular, the sequence 101,109,113,113,109,101,89,73,53,29 appears as such a diagonal. Is it possible that primes of the form $a^2 + 1$ lead to prime rich polynomials of the form $(a+2t)(a-t)+1$? The secondary question series is: what is known about prime rich quadratic polynomials, and does such knowledge follow naturally from studying plots like those above? -Gerhard "Ask Me About System Design" Paseman, 2011.10.11 - -REPLY [3 votes]: joro's plots are much more detailed and informative, but here is a plot closer to your original ASCII art: red is prime, green is composite with no prime factor in $\lbrace 2,3,5,7,11 \rbrace$, -and the other colors indicate composites with smallest factor among those primes -(blue, purple, magenta, orange, yellow). -The range here is $n,m \in [1,300]$, with $(1,1)$ in the lowerleft corner. -I find this display relatively unrevealing, -only repeating the grid-like pattern evident in joro's plots. -Edit: I saved the image through a different process and the grid-like structure -disappeared, pretty much confirming that the grid-pattern was an imaging artifact. -This less-deceiving image is, alas, not very interesting... -            - - -To respond to joro's question, here is just primes vs. not-primes (red and white), $n,m \in [1,1000]$, -which should be directly comparable to his last plot.<|endoftext|> -TITLE: Hirsch length and cohomological dimension -QUESTION [6 upvotes]: It seems to be known that Hirsch length and cohomological dimension agree for (torsion-free, finitely generated) polycyclic groups. -If we drop the assumption "torsion-free", then cd is of course infinite. But, is it still true (as one might expect) that the rational cohomological dimension is bounded above by the Hirsch length? -More generally, are there known conditions on a group G such that cd(G)≥cd(H) if there is a surjective homomorphism G-->H? (For the Hirsch length this inequality is immediate from the definition.) - -REPLY [8 votes]: Hillman extended the notion of Hirsch length to elementary amenable groups and proved that it is bounded above by the rational cohomological dimension. The reference is - -Jonathan A. Hillman, Elementary amenable groups and 4-manifolds with Euler characteristic 0, J. Austral. Math. Soc. (Series A) 50 (1991), 160-170. - - -To answer the question in the comments about nilpotent groups, theorem 5, section 8.8 in Gruenberg's book "Cohomological topics in group theory" (Springer LNM 143) says that for a torsion free nilpotent group $G$ with finite Hirsch length, one has $cd(G)=h(G)$ when $G$ is finitely generated and $cd(G)=h(G)+1$ otherwise.<|endoftext|> -TITLE: Uniqueness of splitting field for linear representations of finite groups -QUESTION [7 upvotes]: If $F$ is any field, $\bar{F}$ its algebraic closure, then it is well-known that all irreducible (indecomposable) $\bar{F}$-representations of a finite group $G$ are realizable over some finite extension $E$ of $F$, and we call $E$, a splitting field for $G$. Any extension of $E$ is also a splitting field of $G$. By finiteness of $[E\colon F]$, we can have a minimal extension of $F$ which is splitting field for $G$. -I puzzled by the following question: -Is such minimal extension unique? -(For example, consider a finite group $G$, $\mathbb{F}_p$ a field of order $p$ with $(|G|,p)=1$. Then there is $n\in \mathbb{N}$ such that $\mathbb{F}_q$ ($q=p^n$) is a splitting field field of $G$. -Is it possible that there are subfields of $\mathbb{F}_q$ of order $p^{m_1}$ and $p^{m_2}$, $m_1\nmid m_2$, and $m_2\nmid m_1$ such that they are splitting fields of $G$, but their subfields are not splitting fields?) - -REPLY [16 votes]: You need two conditions for a field to be a splitting field for a specific irreducible representation (in characteristic zero to begin with): It must contain the character values of the representation. For this there is of course a minimal field, the field generated by those values. However, a splitting field must also split a division algebra and for that there is no unique minimal field. -As a specific example we can take the quaternion group of order $8$ over the rational numbers. The component of the group algebra corresponding to the only faithful representation is a quaternion algebra over $\mathbb Q$ ramified at $\infty$ and $2$ and hence is split by any quadratic imaginary field for which $2$ is either ramified or non-split. -To be more concrete about the quaternion representation, the ordinary real quaternion algebra makes sense over the rationals, denoted $\mathbb H_{\mathbb Q}$. It has $\mathbb Q$-basis 1,i,j,k and the usual multiplication table. Then $\pm\{1,i,j,k\}$ is a multiplicative group that is a copy of the quaternion group $Q$ and therefore we have an algebra map $\mathbb Q[Q]\to\mathbb H_{\mathbb Q}$. It is evidently surjective so that $\mathbb H_{\mathbb Q}$ is one of the factors in the group algebra (more precisely $\mathbb Q[Q]=\mathbb Q^4\times\mathbb H_{\mathbb Q}$, where the first four factors correspond to the four one-dimensional representations). Now, for a field $K$ of characteristic zero $\mathbb H_K$ has a two-dimensional irreducible representation exactly when it is split, i.e., when it is isomorphic to the algebra of $2\times2$-matrices. That is thus exactly the condition for a two-dimensional irreducible representation to exist over $K$. Now, it is well-known that the algebra is split precisely when the reduced norm form ($N(\alpha)=\alpha\overline{\alpha}$) restricted to the purely imaginary quaternions has a non-trivial zero. As $N(ai+bj+ck)=a^2+b^2+c^2$ we get the condition that David mentions in the comments. -The situation in positive characteristic is different. As the group algebra can be defined over the prime field and the Brauer group of finite fields is trivial, one only needs for (the mod $p$) character values to be in the field so in that case there is a minimal field. -Addendum: The question of a minimal field for all irreducible representations has essentially the same answer. First the field has to contain all character values, then there are still division algebras to split (in characteristic $0$). The example of the quaternion group still illustrates the problem, all characters are rational-valued and one must still kill the quaternion algebra for which there is no unique minimal field.<|endoftext|> -TITLE: Do these matrix rings have non-zero elements that are neither units nor zero divisors? -QUESTION [14 upvotes]: First, a disclaimer: This is a repost of a question I asked on stackexchange (no answer there). -Let $R$ be a commutative ring (with $1$) and $R^{n \times n}$ be the ring of $n \times n$ matrices with entries in $R$. -In addition, suppose that $R$ is a ring in which every non-zero element is either a zero divisor or a unit [For example: take any finite ring or any field.] My question: - -Is every non-zero element of $R^{n \times n}$ a zero divisor or a unit as well? - -We know that if $A \in R^{n \times n}$, then $AC=CA=\mathrm{det}(A)I_n$ where $C$ is the classical adjoint of $A$ and $I_n$ is the identity matrix. -This means that if $\mathrm{det}(A)$ is a unit of $R$, then $A$ is a unit of $R^{n \times n}$ (since $A^{-1}=(\mathrm{det}(A))^{-1}C$). Also, the converse holds, if $A$ is a unit of $R^{n \times n}$, then $\mathrm{det}(A)$ is a unit. -I would like to know if one can show $0 \not= A \in R^{n \times n}$ is a zero divisor if $\mathrm{det}(A)$ is zero or a zero divisor. -Things to consider: -1) This is true when $R=\mathbb{F}$ a field. Since over a field (no zero divisors) and if $\mathrm{det}(A)=0$ then $Ax=0$ has a non-trivial solution and so $B=[x|0|\cdots|0]$ gives us a right zero divisor $AB=0$. -2) You can't use the classical adjoint to construct a zero divisor since it can be zero even when $A$ is not zero. For example: -$$A=\begin{pmatrix} 1 & 1 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \end{pmatrix} \qquad \mathrm{implies} \qquad \mathrm{classical\;adjoint} = 0 $$ -(All $2 \times 2$ sub-determinants are zero.) -3) This is true when $R$ is finite (since $R^{n \times n}$ would be finite as well). -4) Of course the assumption that every non-zero element of $R$ is either a zero divisor or unit is necessary since otherwise take a non-zero, non-zero divisor, non-unit element $r$ -and construct the diagonal matrix $D = \mathrm{diag}(r,1,\dots,1)$ (this is non-zero, not a zero divisor, and is not a unit). -5) This is somewhat related to the question: Rings in which every non-unit is a zero divisor -6) This is definitely true when $n=1$ and $n=2$. It is true for $n=1$ by assumption on $R$. To see that $n=2$ is true notice that the classical adjoint contains the same same elements as that of $A$ (or negations): -$$ A = \begin{pmatrix} a_{11} & a_{12} \cr a_{21} & a_{22} \end{pmatrix} \qquad \Longrightarrow \qquad \mathrm{classical\;adjoint} = C = \begin{pmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{22} \end{pmatrix} $$ -Thus if $\mathrm{det}(A)b=0$ for some $b \not=0$, then either $bC=0$ so that all of the entries of both $A$ and $C$ are annihilated by $b$ so that $A(bI_2)=0$ or $bC \not=0$ and so $A(Cb)=\mathrm{det}(A)bI_2 =0I_2=0$. Thus $A$ is a zero divisor. -7) Apparently strange behavior can occur when $R$ is non-commutative (not surprising). Like a matrix can be both a left inverse and left zero divisor. [The determinant keeps this from happening in the commutative case.] - -REPLY [10 votes]: The answer given by David Speyer can be strengthened as follows. If $A$ is a non-invertible $n\times n$ matrix with entries in $R$ as described in the problem, then the linear maps $R^n \to R^n$ defined by either left or right multiplication are non-injective. In particular, $A$ is both a left-zero-divisor and a right-zero-divisor. -This is a consequence of McCoy's rank theorem. You can find a nice, brief account of this in Section 2 of this paper by Kodiyalam, Lam, and Swan. One consequence of the theorem is that for any commutative ring $R$, a square matrix $A$ over $R$ has linearly independent columns if and only if its determinant is not a zero-divisor, if and only if its rows are linearly independent. -So if every element of $R$ is either invertible or a zero-divisor, this means that every square matrix over $R$ defines a linear transformation that is either invertible or non-injective.<|endoftext|> -TITLE: generators of Out(F_n) and homology -QUESTION [8 upvotes]: It is known that the mapping class group of a closed (orientable) surface is generated by elements of finite order. Is this also known to be true for $Out(F_n)?$ A related question is the following: The mapping class group is known to be a perfect group for $g\geq 3,$ that is, it is equal to its commutator. Is this known for $Out(F_n)?$ I can find references to the effect that the rational homology is trivial in low dimensions, but that's not quite the same... - -REPLY [6 votes]: In Bridson's "A condition that prevents groups from acting nontrivially on trees" -it is shown that for $n\geq 3$, $Aut (F_n)$ is generated by subgroups $A_1,A_2,A_3$ such that $\langle A_i,A_j \rangle$ is finite for $i,j=1,2,3$. This answers your question and moreover implies (via Helly's theorem) that for $n\geq 3$, any action of $Aut (F_n)$ on a tree fixes a point. -Beautiful stuff!<|endoftext|> -TITLE: God's number for the $n \times n \times n$-cube -QUESTION [13 upvotes]: This is a question about Rubik's Cube and generalizations of this puzzle, such as Rubik's Revenge, Professor's cube or in general the $n \times n \times n$ cube. -Let $g(n)$ be the smallest number $m$, such that every realizable arrangement of the $n \times n \times n$ cube can be solved with $m$ moves. In other words, this is the "radius" of the Cayley graph of the $n \times n \times n$ cube group with respect to the canonical generating system. -We have $g(1)=0$, $g(2)=11$ and - quite recently - in 2010 it was proven that $g(3)=20$: God's number is 20. -Question. Is anything known about $g(4)$ or $g(5)$? -I expect that the precise number is unkwown since the calculation for Rubik's cube already took three decades. Nevertheless, is there any work in progress? Are any lower or upper bounds known? -I would like to ask the same question about $g(n)$ for $n>5$, or rather: -Question. Is anything known about the asymptotic value of $g(n)$? - -REPLY [5 votes]: Based on this discussion from 2015, God's number for the 4x4x4 cube lies beween 35 and 55 (inclusive) for the outer block turn metric, with the corresponding bounds being 32 and 53 for the single slice turn metric, and 29 and 53 for the block turn metric. Scroll down to the comments for the optimal bounds. -The choice of metric is irrelevant for the asymptotic results of Demaine, Demaine, Eisenstat, Lubiw and Winslow which gives $\Theta(n^2/\log(n))$ as the answer to question two. -Since this question was first asked, cube20.org claims to have also shown that God's number for the quarter turn metric is 26. -As for the 5x5x5, one can find this post claiming an upper bound of 130 for the outer block turning metric.<|endoftext|> -TITLE: Axiom of class collection -QUESTION [9 upvotes]: One version of the Axiom of Collection says that any surjection $A\to B$ from a class $A$ to a set $B$ is factored through by some surjection $C\to B$ where $C$ is a set. -Note that assuming $B$ is a set, the axiom of replacement ensures that $C$ is a set if and only if each fiber of $C\to B$ is a set (i.e. the map $C\to B$ is "small" in the sense of algebraic set theory). Thus it is natural to consider the following "class-collection axiom": any surjection $A\to B$ of classes is factored through by some surjection $C\to B$ whose fibers are all sets. -Has this "class-collection axiom" been studied at all? Assuming classical logic, it seems to follow from the axiom of foundation in the same way that collection follows from replacement and foundation: let the fiber of $C$ over $b\in B$ consist of all those $a\in A$ lying over $b$ and of minimal rank. Can it be proven in any intuitionistic context? - -REPLY [2 votes]: Maybe in -A. Joyal, I. Moerdijk: A completeness theorem for open maps, Annals of Pure and Applied Logic 70 (1994) 51-86 -an axiom related to this occurs - but its study is not the main purpose of this paper. So I'm looking for a more extensive treatment of it myself.<|endoftext|> -TITLE: partial differential equation for ruled surfaces -QUESTION [14 upvotes]: We say that a surface $f(x,y,z)=0$ is ruled if for each point $p$ in the surface there is a line that passes through $p$ and is contained in the surface. See http://en.wikipedia.org/wiki/Ruled_surface for more information. -Does anybody know if there is a partial differential equation whose solutions are all ruled surfaces and only them? And what is the equation? Some reference would be helpful, too (different from Salmon's old book about surfaces which I unfortunately find unreadable). - -REPLY [2 votes]: First of all, for bivariate functions $z=u(x,y)$, let's write down the following notation -\begin{gather*} -r=\dfrac{\partial^2u}{\partial\,\!x^2}\qquad\,s=\dfrac{\partial^2u}{\partial\,\!x\partial\,\!y}\qquad\,t=\dfrac{\partial^2u}{\partial\,\!y^2}\\ -\\ -\lambda_1=\dfrac{-s+\sqrt{s^2-rt}}{t}\qquad\,\lambda_2=\dfrac{-s-\sqrt{s^2-rt}}{t}\\ -\\ -\Diamond\,\!u=\dfrac{\partial\,\!\lambda_1}{\partial\,\!x}+\lambda_1\dfrac{\partial\,\!\lambda_1}{\partial\,\!y} -\\\\ -\bar\Diamond\,\!u=\dfrac{\partial\,\!\lambda_2}{\partial\,\!x}+\lambda_2\dfrac{\partial\,\!\lambda_2}{\partial\,\!y} -\end{gather*} - -If the smooth surface $z=u(x,y)$ is ruled, also required $t\ne0$, then the bivariate function $z=u(x,y)$ satisfies the third order partial differential equation $\Diamond\,\!u=0$ or $\bar\Diamond\,\!u=0$; -If the bivariate function $z=u(x,y)$ satisfies the third-order partial differential equation $\Diamond\,\!u=0$ or $\bar\Diamond\,\!u=0$, also satisfies the inequality $s^2-rt\ge0$ and $t\ne0$, the surface represented by it is ruled. - -Cf. [Monge 1780] Gaspard Monge, “Mémoire sur les Propriétés de plusieurs genres de -Surfaces courbes, particulièrement sur celles des Surfaces développables, -avec une Application à la Théorie des Ombres et des Pénombres”, Savans -Étrangers 9 (1780), pp.382-440. -https://archive.org/details/mmoiresdemath09acad/page/434 -Cf. [J. Ockendon, S. Howison, A. Lacey, A. Movchan] Applied Partial Differential Equations (2003), pp.380-382. -Example: $u(x,y)=\dfrac{xy}{x^2+y^2}$, i.e. $z=\dfrac{xy}{x^2+y^2}$; Assume that $x>y>0$, then -\begin{gather*} -\begin{split} -r&=+\dfrac{2(x^2-3y^2)xy}{(x^2 +y^2)^3}\\\\ -s&=-\dfrac{(x^2 - 2 x y - y^2) (x^2 + 2 x y - y^2)}{(x^2 +y^2)^3}\\\\ -t&=-\dfrac{2(3x^2-y^2)xy}{(x^2 +y^2)^3}\\\\ -\sqrt{s^2-rt}&=\dfrac{\left|x^2-y^2\right|}{(x^2 +y^2)^2}=\dfrac{x^2-y^2}{(x^2 +y^2)^2}>0\\ -\end{split}\\ -\\ -\lambda_1=-\dfrac{(x^2-3y^2)x}{(3x^2-y^2)y}\qquad\,\lambda_2=\dfrac{y}{x}\\ -\\ -\begin{split} -&\color{red}{\Diamond\,\!u=-\dfrac{3(x-y)(x+y)(x^2+y^2)^3}{(3x^2-y^2)^3y^3}}\\ -&\color{red}{\phantom{\Diamond\,\!u}<0}\\ -\\ -&\color{blue}{\bar\Diamond\,\!u\equiv0} -\end{split} -\end{gather*}<|endoftext|> -TITLE: K-theory and K-theory pushforward in topology vs. in algebraic geometry -QUESTION [8 upvotes]: Let $f : X \to Y$ be a [fill in the blank] morphism of [fill in the blank] complex varieties. Then we have the pushforward $f_! : K(X) \to K(Y)$ which is defined by $f_!(E) = \sum_i (-1)^i [R^i f_\ast E]$, the alternating sum of the higher direct images. Here we take $K(X)$ to mean the $K$-group of coherent sheaves. -On the other hand we can also define $K(X)$ as the $K$-group of $C^\infty$ complex vector bundles on $X$ considered as a real manifold. Then we can define a Gysin map $f_!$ using the Thom isomorphism theorem for $K$-theory. - - -Which adjectives do I need to fill in the blanks with to make the two notions of $K(X)$ agree? - -Which adjectives do I need to fill in the blanks with to make the two notions of $f_!$ agree? - - - -If $X$ is smooth and projective, then any coherent sheaf has a finite resolution by locally free sheaves, so we have a map $K^{alg}(X) \to K^{top}(X)$. On the other hand, I don't think it's true that any $C^\infty$ complex vector bundle has a holomorphic structure, so I don't think there is a map $K^{top}(X) \to K^{alg}(X)$ ... - -REPLY [5 votes]: I found a paper which I think answers #2: -Baum, Fulton, MacPherson Riemann-Roch and topological K-theory for singular varieties. -They prove that the algebraic $f_!$ and the topological $f_!$ agree in the case of proper morphisms of (not necessarily smooth) quasi-projective varieties, reducing in several steps to the map from a projective space to a point, see theorem 4.1. -In other words, we have a commutative square -$$\begin{array}{ccc} -K^{alg}(X) & \to & K^{alg}(Y) \\ -\downarrow&&\downarrow& \\ -K^{top}(X) & \to & K^{top}(Y) -\end{array}$$ - -REPLY [4 votes]: Well, let me say something really straightforward. First, the map from $K^{alg}(X)$ to -$K^{top}(X)$ is defined even if $X$ is not necessarily projective. -Second, $f_!$ is defined only if $f$ is proper, and in that case I think it always agrees -with the topological push-forward.<|endoftext|> -TITLE: Accumulation of algebraic subvarieties: Near one subvariety there are many others (?), 3 -QUESTION [18 upvotes]: Part 3 of this series of questions. In the meantime, I realized that there is some very simple question that was left open in Accumulation of algebraic subvarieties: Near one subvariety there are many others (?) . -Let's work again over the field $\mathbb{C}$ of complex numbers, and let $X\subset \mathbb{P}^n$ be a projective variety. Let $\tilde{X}\subset \mathbb{P}^n$ be any small open neighborhood of $X$, in the complex topology. - -Question: Assume that $X$ is locally a (set-theoretic) complete intersection. Does there exist a smooth projective subvariety of $\mathbb{P}^n$ contained in $\tilde{X}$ of dimension equal to the dimension of $X$? - -Any ideas are welcome, also for the similar question over characteristic $p$ fields, i.e. working over a field like the completed algebraic closure of $\mathbb{F}_p((t))$, and taking a small neighborhood in the $t$-adic topology. Changing to such fields might have the advantage that bend-and-break type techniques become available, but I do not see how to make use of them. - -REPLY [10 votes]: The answer to the question for general $X$ is no: There are local obstructions coming from bad singularities. -More precisely, one can show the following: Assume that $X$ is irreducible and that there is a closed subset $A\subset X$ of codimension at least $2$, such that $X\setminus A$ is not connected. Then there are small $\epsilon$-neighborhoods $\tilde{X}$ of $X$ in $\mathbb{P}^n$ which do not contain any smooth projective subvarieties of the same dimension. -Let me sketch the argument. Let $\tilde{A}$ be a sufficiently small closed $\epsilon$-neighborhood of $A$. Choose $\tilde{X}$ such that $\tilde{X}\setminus \tilde{A}$ is still not connected. Assume that there is some smooth projective $X^\prime\subset \tilde{X}$ of the same dimension. We see that $X^\prime\setminus \tilde{A}$ is not connected. Using the sequence -$$ -H^{2\dim X - 1}(\tilde{A}\cap X^\prime)\rightarrow H^{2\dim X}_c(X^\prime\setminus \tilde{A})\rightarrow H^{2\dim X}(X^\prime)\ , -$$ -we see that $\tilde{A}\cap X^\prime$ has cohomology in degree $2\dim X -1$. If $A$ was a point, then $\tilde{A}$ would be a closed ball, and one would have a bound saying that the cohomology of $\tilde{A}\cap X^\prime$ is in degrees $\leq \dim X^\prime$. In general, one has a similar bound $\dim X^\prime + \dim A$. Now if $A$ has codimension at least $2$, one gets a contradiction. -Working slightly more carefully, one sees that the obstruction really is local in nature, and that a necessary condition is that $X$ is locally connected in codimension $1$. -On the other hand, it is clear that there are no local obstructions if $X$ is locally a set-theoretic complete intersection (as one can just deform local equations to get a local smooth deformation). Hence this argument reproves a result of Hartshorne: Set-theoretic complete intersections are locally connected in codimension $1$. -But what happens if there are no local obstructions, for example if $X$ is locally a complete intersection? I have adapted the original question accordingly.<|endoftext|> -TITLE: Can nonabelian groups be detected "locally"? -QUESTION [19 upvotes]: Suppose $m,n\geq 2$ are two integers. Is it true that for every sufficiently large nonabelian group $G$, one can find a set $A\subset G$, with $|A|=n$, so that $|A^m| >\binom{n+m-1}{m}$? -(Edit) Let's also add the condition $m\le n$ since the answer below provides a counter-example for large enough $m$. In general it would be interesting to know the range of $(m,n)$ for which the statement holds.(/Edit) -Here $A^k=\lbrace a_1a_2\cdots a_k| a_1,a_2,\dots,a_k\in A\rbrace$ is a product set. It is obvious that in every abelian group one has $|A^m| \le\binom{n+m-1}{m}$, for every $A$. -I don't have an application in mind, I was trying the case $m=2$ and I think I have a proof (still haven't checked all the steps, but it's not particularly enlightening since it splits into many cases). I'm wondering if this is true in general and if there is a slick proof, or if there is a counter-example. - -REPLY [5 votes]: Here is a collection of what I have so far thanks to the answers by Guntram and S. Carnahan. Let's denote by $P(n,m)$ the property that $|A^m| \le\binom{n+m-1}{m}$ for all subsets $|A|=n$. -We have that the only nonabelian $P(2,2)$ groups are of the form $Q_8\times G$ where $G$ is an elementary abelian 2-group, and that $P(3,2)$ groups have to be abelian by Freiman's paper "On two- and three-element subsets of groups". -In "A characterization of abelian groups", Brailovsky proves that large enough $P(n,2)$ are abelian by showing that $P(n,2)\implies P(n',2)$ for all $n\geq n'\geq 2$, so that the result follows from the previous paragraph. -In "Small squaring and cubing properties for finite groups", Berkovich, Freiman and Praeger prove that the only nonabelian group with $P(2,3)$ is $S_3$. -On the other hand there are nonabelian groups with $P(n,m)$ whenever $\binom{n+m-1}{m}\geq 2^{2n+1}$ as in Guntram's answer.<|endoftext|> -TITLE: Does there exist a complex Lie group G such that ... -QUESTION [8 upvotes]: ... every Riemann surface of genus $1$ appears as a complex one-parameter subgroup of $G$? - -REPLY [16 votes]: No. In a connected complex Lie group all compact complex subgroups must be in the center, that is, in the kernel of the adjoint representation, because in a complex general linear group there is no nontrivial compact connected complex subgroup. And in a connected abelian complex Lie group there can be only countably many compact one-parameter complex subgroups. -Edit: So in fact in a complex Lie group there are only countably many compact connected one-parameter complex subgroups (not just up to isomorphism). -Edit: Just to summarize what I have learned from this exercise: Let CCC="compact connected complex". -(1) (I already knew this.) A CCC subgroup of $GL_n(\mathbb C)$ must be trivial. Compact implies contained in a conjugate of $U_n$; complex then implies $0$-dimensional; connected then implies trivial. (Alternatively, a holomorphic map from a CCC manifold to $\mathbb C$ (therefore to $GL_n(\mathbb C)$) must be constant.) -(2) (IAKT) A CCC group is always abelian. This follows from (1) using the adjoint representation. Therefore it must be a compact complex torus group, the quotient of a complex vector space by a full lattice (i.e. a subgroup generated by an $\mathbb R$-basis). -(3) (This had never occurred to me before, but it's obvious by the same argument.) In a connected complex Lie group a CCC subgroup is always in the center. Thus a complex Lie group $G$ has a maximal CCC subgroup in the strongest sense -- one which contains all others -- namely the (unique) maximal compact subgroup of the connected component of the center of the connected component of $G$. -(4) Therefore there cannot be a nonconstant family of CCC subgroups of a complex Lie group. (3) reduces this statement to the case where the ambient group is abelian, and that case is clear. -I don't know what you mean by infinite-dimensional Lie group, exactly, but it seems to me that the arguments above can be adapted to show that in the group of all complex automorphisms of a connected complex manifold there is a CCC containing all others (again contained in the center of the component) so that again there are no families.<|endoftext|> -TITLE: 2-completeness analog of completeness theorem -QUESTION [6 upvotes]: It's not hard to see that a category is finitely complete if it has finite products and equalizers. In short, this is because one can write all limits as iterations of these two "operations". -I wonder if there is a 2-version of this. In particular, - -Does a category have all finite 2-limits if it has all 2-equalizers and 2-products? - -My instinct is no, and that we will need another(or several more) limits to build all 2-limits. -Of course the question can be generalized to n-limits, and I'm curious about that also. - -REPLY [7 votes]: Your suspicion is correct: in general, a V-category has all weighted V-limits if it has all conical V-limits and is cotensored over V (see Kelly's Basic Concepts of Enriched Category Theory, section 3.10). For V = Cat (and this is true for bicategories too), cotensors can be constructed from conical limits and cotensors with the arrow category 2, although I don't know the original reference for that. 'Finite' limits are a bit more complicated in the enriched case, but see Street's 'Limits indexed by category-valued 2-functors', JPAA 1972, for the 2-case.<|endoftext|> -TITLE: Can we collapse $\omega_1$ to $\omega$ without adding a dominating real? -QUESTION [13 upvotes]: (Disclaimer: This question was also asked at MSE (https://math.stackexchange.com/questions/71020/can-we-collapse-omega-1-without-adding-a-dominating-real). I'm posting it here because, when I asked it, I was torn between my sense that it was appropriate for MO and my suspicion that this question is much, much easier than I'm making it; and it seems to be attracting no attention at MSE. I'm not a set theorist, so it's hard for me to judge the difficulty of the questions I'm asking; please let me know if this question is too elementary for MO, and I'll delete it.) -The question is precisely that of the title: is there a notion of forcing $\mathbb{P}$ which collapses $\omega_1$ to $\omega$ but does not add a real which dominates every real in the ground model? (Here "real" means "element of $\omega^\omega$.) -It seems like the answer should be "no," and I've attempted to prove this myself. The problem is that the easiest way to do so would be to define a dominating function in terms of an arbitrary surjection $f: \omega\twoheadrightarrow\omega_1;$ however, there seems to be no clear way to do this. My first thought was to look at the set $S_f=\lbrace n: \forall m < n, f(m) < f(n)\rbrace$. This is certainly a real, but there is no reason it should be dominating, let alone not present in the original model already; in fact, we can alter the usual collapsing poset to demand that $S_f$ be precisely the evens, or precisely the powers of 17, or in fact any infinite co-infinite subset of $\omega$. -On the other hand, looking at the poset of finite increasing functions from $\omega$ to $\omega_1$ (which builds a countable sequence cofinal in $\omega_1$, hence collapsing $\omega_1$ to $\omega$), it's unclear how this would add a dominating real; so perhaps the answer to my question is yes. - -REPLY [4 votes]: Hi Noah, I'm adding an answer because this wouldn't fit in a comment. Unless I'm making a silly mistake, the situation is different with respect to adding an escaping real: -Let $P$ be the collection of finite partial functions from $\omega$ to $\omega_1$ as usual, and let $\dot f$ be a $P$-name for the generic surjection from $\omega$ onto $\omega_1$. -Given $g:\omega\rightarrow\omega$ in the ground model, and $n<\omega$, consder the set $D(g, n)$ -of conditions $p$ such that for some $m>n$, - -$m$ is in the range of $p$, -the domain of $p$ is an initial segment of $\omega$, and -the least $k$ for which $p(k)=m$ is greater than $g(m)$. - -This set is dense in $P$ for each $g$ and $n$, and so the real $h$ in the extension defined by setting $h(m)$ equal to the least $k$ such that $\dot f(k)=m$ is not bounded by a ground model real. (So essentially, we are "inverting" the surjection on the initial segment $\omega$ of its range) -Edit: Even simpler, if we define a real $h$ in the extension by setting $h(n)=\dot f(n)$ if $\dot f(n)<\omega$, and $h(n)=0$ otherwise, then $h$ is Cohen over the ground model, hence unbounded.<|endoftext|> -TITLE: Tutte polynomials of appropriate Cayley graphs -QUESTION [6 upvotes]: I was quite intrigued by Tutte polynomials in a recent talk I had been to. It was introduced as a polynomial associated to a undirected finite graph. For a graph $G=(V,E)$ we form the polynomial -$T_G(x,y)=\sum_{A\subseteq E} (x-1)^{k(A)-k(E)}(y-1)^{k(A)+|A|-|V|}$ -where $k(A)$ is the number of components of the graph $(G,A)$. -I have heard and read about the significance of this polynomial. However, I'm completely new to combinatorics and graph theory and have no prior knowledge of what is or isn't known about these things. I have two questions which should be easy enough for the experts. -(1) Let us start with a finite group $G$, chose a presentation and consider its Cayley graph $\Gamma$. Now consider the Tutte polynomial $T_\Gamma$ of $\Gamma$. We suppose that $G$ is the fundamental group of some reasonable space $M$ (it always is $\pi_1$ of some space but what I mean here is the space is a manifold or very broadly an object where distances and volumes make sense). -Question Does the Tutte polynomial $T_\Gamma$ encode any information about the topological invariants of $M$? -(2) In my own naive way I was trying to see if anything similar (like the formula at the outset) works for infinite graphs. The existence of such a thing, at least, looks very interesting as Tutte polynomials generally have a lot of significance. If such a thing doesn't exist, one may begin to study properties of the Tutte polynomials associated to finite subgraphs of an infinite graph and analyze the asymptotics of the these properties on the poset of finite subgraphs. -Question Is there a known way to define Tutte polynomials for infinite graphs? If so, how is defined and what is known? -Question If the above answer is yes then what does $T_\Gamma$ encode about the growth of $G$ when $G$ is the fundamental group of a manifold? Note that by the $\check{S}varc$-$Milnor$ lemma this growth is asymptotic to the growth (in the sense of geometry) of the universal cover of $M$. - -REMARK : I edited the first question as the growth of a finite group is not terribly interesting. The original first question should make more sense now, as presented, as the last question. - -REPLY [3 votes]: This does not really answer your questions, but has a somewhat similar flavor, so you might find it interesting anyway - if you have a sequence of finite graphs $G_{n}$ converging (in a suitable sense) to a potentially infinite graph $G$, then it is possible to say something about convergence of roots of many graph polynomials (like chromatic polynomial or Tutte polynomial) to a suitable limiting measure. See for example the preprint "Benjamini--Schramm continuity of root moments of graph polynomials" by Csikvári and Frenkel.<|endoftext|> -TITLE: Reference for nef coherent sheaves? -QUESTION [6 upvotes]: The definition and basic properties of nef locally free sheaves appear for instance in the second volume of Lazarsfeld's book "Positivity in Algebraic Geometry" (beginning of chapter 6). -However, I am in a situation where some of the sheaves I deal with are not locally free, but only coherent; so I would like to know whether there is a well-behaved notion of nefness for coherent sheaves. The only mention of this that I found is at the end of section 1 of Kodaira Dimension of Subvarieties by Peternell-Schneider-Sommese, but it is just the definition with no references and no discussion of basic properties. -So my question is: is there a reference that gives an analogue of Theorem 6.2.12 in Lazarsfeld for nef coherent sheaves? (The results I'm mostly interested in are: (a) quotient of nef is nef; (b) pullback of nef is nef; and (c) extension of nef by nef is nef.) - -REPLY [5 votes]: Definition A coherent sheaf $\mathcal{F}$ on an algebraic variety $X$ is nef if the following condition holds: For every irreducible curve $C\subset X$, the line bundle $O(1)$ is nef on $\mathbb{P}(\mathcal{F}|_C)$. -I haven't seen this definition of a nef coherent sheaf either, but I think most of the properties you mention just follow formally from the properties of ordinary nefness. Here is a proof for a) and b): -a) A quotient of a nef sheaf is nef. Let $C$ be as in the definition. If $F\to E\to 0$ is a surjection, this restricts to a surjection $F|_C\to E|_C\to 0$ and hence gives an embedding $\mathbb{P}(\mathcal{F}|_C)\hookrightarrow \mathbb{P}(\mathcal{F}|_C)$ such that $O(1)$ on $\mathbb{P}(\mathcal{F}|_C)$ restricts to $O(1)$ on $\mathbb{P}(\mathcal{E}|_C)$. Since the restriction of a nef line bundle is nef, $E$ is also nef. -b) A pullback of nef is nef. Similarily, if $f:X\to Y$ is finite, then $f$ restricts to a finite map $f:C'=f^{-1}C\to C$ and hence there is a finite map $F:\mathbb{P}(f^*\mathcal{F}|_C)\to \mathbb{P}(\mathcal{F}|_C)$ such that $O(1)=F^*O(1)$. Now the claim just follows from the corresponding statement for line bundles.<|endoftext|> -TITLE: Is there a reasonable definition of the height of a transcendental number -QUESTION [5 upvotes]: For an algebraic number $\alpha$ one can define its "height" in many ways. Informally, you could use its minimal polynomial over $\mathbf{Q}$ and consider the maximum of the heights of its coefficients. Or consider all the valuations of $\alpha$, etc. In this context, the height is supposed to be some kind of measure of complexity. -Question. Is there a reasonable definition of the "height" of a transcendental number. -I'm not sure what such a height would mean though in this context. -If there isn't any reasonable definition, is there any reasonable explanation for why this isn't possible? - -REPLY [4 votes]: Here is one potentially reasonable explanation for why such an invariant shouldn't exist. One property of height that can be useful is that the height of an algebraic number is invariant under all automorphisms of all rings that contain that number. For any algebraically independent pair of transcendental complex numbers, one may choose a ring-theoretic automorphism of $\mathbb{C}$ that exchanges them. If we want the same sort of invariance as in the algebraic setting, then all heights of transcendental numbers must be equal. -I think the existence of a single correct answer to your question would require a specific application. - -REPLY [2 votes]: It does'nt quite answer your question, but maybe there is a resaonable definition of the height of a period. The ring of period is a countable over-ring of the field of algebraic numbers, sharing (at least conjecturally) many properties with them. This ring contains most of interesting transcendental numbers.<|endoftext|> -TITLE: Proofs of parity results via the Handshaking lemma -QUESTION [18 upvotes]: I particularly like the following strategy to prove that the number of some combinatorial objects is even: to construct a graph, in which they correspond to vertices of odd degree (=valency). -Let me mention three results of such nature: - -If the graph $G$ has even number of vertices and all of them have even degree, then it has even number of spanning trees. It may be proved by using the matrix-tree theorem, though not immediately. I mean the following elementary argument: consider the new graph, in which spanning trees of $G$ serve as vertices, and two trees $T$, $T'$ are joined iff they differ by one edge (i.e. all edges of $T$ except one edge are also edges of $T'$). Then all degrees in new graph are odd and we are done. - -(This argument is due to Andrew Thomason). Given vertices $x$, $y$ in a finite graph $G$, in which all vertices except $x$, $y$ have odd degree (parity of degrees of $x$ and $y$ does not matter). Then the number of Hamiltonian paths from $x$ to $y$ is even. For proving this, consider all Hamiltonian paths starting in $y$ in the graph $G-x$. They are the vertices of a new graph. Join two such paths if they differ by one edge (i. e., all edges of the first path except one are also edges of the second). Then the degree of a path $yz\dots v$ in new graph is one less than the degree of $v$ in $G-x$. So, it is odd iff $v$ is joined to $x$, and odd vertices of our new graph correspond to Hamiltonian paths from $x$ to $y$ in $G$. - -(This is due to László Lovász). Let $G=(V,E)$ be a finite graph. Call a subset $A\subset V$ dominating if any vertex $v\in V\setminus A$ has at least one neighbor in $A$. Then any graph has an odd number of dominating sets. We prove that the number of non-dominating non-empty subsets of $V$ is even. They are vertices of a new graph, and two of them $B_1,B_2$ are joined iff there are no edges between $B_1$ and $B_2$ in $G$. All degrees are odd (degree of $B$ equals $2^k-1$, where $k$ is the number of vertices in $V\subset B$ not joined with $B$), and we are done again. - - -I have a general question: what are other examples of such flavor? Of course, proving of parity by partitioning onto pairs formally is a particular case (corresponding to the graph with degrees 1), and by partitioning onto even subsets too, but I rather ask about more involved graphs used in the proof:) -And one specific question: may Redei's result that any tournament have odd number of Hamiltonian paths be proved by such techniques? - -REPLY [7 votes]: There is a nice paper by Kathie Cameron and Jack Edmonds, Some graphic uses of an even number of odd nodes, with several examples of the use of the handshaking lemma to prove various graph-theoretic facts. -Gjergi Zaimi already mentioned the relevance of the complexity classes PPA and PPAD. In addition to Papadimitriou's paper, this paper by Beame et al. is a useful reference. -I seem to remember that the Chevalley-Warning theorem can be proven by this method, but I can't reconstruct the argument, so maybe I'm confused.<|endoftext|> -TITLE: Name this periodic tiling -QUESTION [10 upvotes]: I've been working on a problem I'm working on in ergodic theory (finding Alpern lemmas for measure-preserving $\mathbb R^d$ actions) and have found some neat tilings, that I presume were previously known. -They are periodic tilings of $\mathbb R^d$ by a single prototile consisting of any box with a smaller box removed from a corner. The two-dimensional case is illustrated in the figure below: - -There exist versions of this in arbitrary dimensions, irrespective of the size of the boxes. Furthermore, there appear to be many essentially different tilings using the same prototile. - -Does anybody recognize the $d$-dimensional version, or have a name for it? - -Thanks - -REPLY [5 votes]: Kolountzakis worked with some tilings of this sort in his paper "Lattice tilings by cubes: whole, notched and extended", Electr. J. Combinatorics 5 (1998), 1, R14.<|endoftext|> -TITLE: Can we always find a curve which doesn't have semi-stable reduction -QUESTION [9 upvotes]: Let $K$ be a number field and let $g$ be a positive integer. Does there exist a smooth projective geometrically connected curve $X/K$ of genus $g$ such that $X$ does not have semi-stable reduction over $K$? -I can write down curves over certain number fields without semi-stable reduction, but I can't do it for a general number field. - -REPLY [15 votes]: Yes. Take $f\in O_K$ a uniformizing element of some prime $\mathfrak p$. Consider the hyperelliptic curve defined by the equation -$$y^2=x^{2g+1}+f.$$ -Then this curve doesn't have semi-stable reduction at $\mathfrak p$. In fact, this equation defines a proper regular model of the curve over the localization $O_{K, \mathfrak p}$ and this model is minimal because its closed fiber is irreducible (defined by $y^2=x^{2g+1}$), and it is not semi-stable. If the curve had semi-stable reduction, the minimal regular model would also be semi-stable. Further, this curve is far from being semi-stable because its Jacobian has purely additive reduction at $\mathfrak p$. -Add a more elementary explanation on why the curve doesn't have semi-stable reduction at $\mathfrak p$. Suppose for simplicity that $\mathfrak p$ is prime to $2(2g+1)$. As saw in the comments, the curve has potentially good reduction above $\mathfrak p$. So if it had already semi-stable reduction over $K$, then it would already have good reduction over $K$. This implies (as for elliptic curves) that after a suitable homographic transformation on $x$, we will get a new polynomial with discriminant invertible in $O_{K, \mathfrak p}$. But such transformation changes the valuation of the discriminant by a multiple of $2(2g+1)$ while the initial discrimiant has valuation $2g$. Impossible.<|endoftext|> -TITLE: Normal distribution with positive SEMI-definite covariance matrix -QUESTION [6 upvotes]: In [1] is noted, that a covariance matrix is "positive- semi definite and symmetric". I wonder if it is possible to a multivariate normal distribution with a covariance matrix that is only positive semi-definite but not positive definite? -If yes, how can the density be evaluated, since it involves the inverse and the 1/determinant of the covariance matrix. -[1] http://en.wikipedia.org/wiki/Covariance_matrix - -REPLY [6 votes]: Happens all the time in ANOVA and regression problems. Here's a really simple one: Suppose -$$ -Y_i = \alpha_0 x_i + \alpha_1 + \varepsilon_i -$$ -where $\varepsilon_i \sim N(0,\sigma^2)$ and the $\varepsilon$s are independent. -Now let $\hat{\alpha}_0$ and $\hat{\alpha}_1$ be the least-squares estimates. Let $\hat{Y}_i = \hat{\alpha}_0 x_i + \hat\alpha_1$ be the fitted values. Let $\hat\varepsilon_i= Y_i-\hat Y_i$ be the residuals. -Now notice the difference between the errors $\varepsilon_i$ and the residuals $\hat\varepsilon_i$. The former are independent, so the probability that their sum is $0$ is $0$. The latter necessarily sum to $0$ and also satisfy the identity $\sum_i x_i\hat\varepsilon_i=0$. So they're not independent. Thus there are $n-2$ degrees of freedom for error. (They're also not identically distributed, by the way. That's one thing that complicates the studentization of residuals.) -The vector of errors is distributed as $N(0,\sigma^2 I_n)$ where $I_n$ is the $n\times n$ identity matrix and the $0$ is an $n\times 1$ column vector. The vector of residuals has the same expected-value vector, but its variance is a singular matrix of rank $n-2$. It's $\sigma^2$ times an orthogonal projection matrix onto the orthogonal complement of the space spanned by $(1,\ldots,1)^T$ and $(x_1,\ldots,x_n)^T$. -Multivariate normal distributions with singular variances arise in ways like this incessantly in statistics. - -More simply, suppose $A$ is any non-negative-definite symmetric real matrix. The finite-dimensional spectral theorem says $A$ has a non-negative-definite square root $A^{1/2}$. Let $Z = (Z_1,\ldots,Z_n)^T$ be a vector of i.i.d. standard normals. Then the variance (or "covariance matrix", if you like) of $A^{1/2}Z$ is $A$. -So every non-negative-definite symmetric real matrix is realizable as the variance of some vector-valued random variable. - -BTW, shall we be clear about the definition? A random vector has a multivariate normal distribution if its dot-product with every constant vector has a univariate normal distribution. (I mention this in part because the question about densities made me wonder if some people think densities are essential to the concept. They're not part of any of the standard definitions.)<|endoftext|> -TITLE: Macdonald polynomials and Macdonald positivity -QUESTION [7 upvotes]: I would like to have some order in my head about different version of Macdonald polynomials -and positivity statements about them. I understand the following: -1) There is a definition of Macdonald polynomials for any root system. These can be defined, for example as $W$-invariant polynomials on the torus $T$ of a semi-simple group $G$, which are orthogonal polynomials with respect to Macdonald scalar product and normalized -in such a way that -$$ -P_{\lambda}(q,t,x)=e^{\lambda}+\text{lower order terms} -$$ -where $\lambda$ is a dominant weight and $x\in T$. -2) In type $A$ there is a notion of transformed Macdonald polynomials, which -were extensively studied by Haiman. Haiman denotes them by $\tilde{H}_{\lambda}$ -(here $\lambda$ is a partition, which can be thought of as a domonant weight of -$GL(n)$); he proved the Macdonald positivity conjecture, which says that -${\tilde H}_{\lambda}(q,t,x)$ -is a linear combination of Schur functions in $x$ whose coefficients are polynomials in $q$ and $t$ with non-negative integral coefficients. -The definition of ${\widetilde H}_{\lambda}(q,t,x)$ appears for example on page 4 of http://math.berkeley.edu/~mhaiman/ftp/nfact/polygraph-jams.pdf -My questions are these: -a) What is the relation between $P_{\lambda}$ and ${\tilde H}_{\lambda}$? -It is not clear to me from the definition. -b) Are there positivity statements for $P_{\lambda}$ itself? Or is there a version of the -positivity conjecture for any root system? - -REPLY [6 votes]: As far as I know there is no positivity statement for $P_\lambda$ as a linear combination of characters (which are the analog of Schur functions in general type). There is a very general statement that $P_\lambda$ expands positively as a sum of monomial symmetric functions; this is proved in a root-system uniform way by Ram and Yip in -http://arxiv.org/abs/0803.1146 -The proof they give is for the "constant parameters" situation, but the technique works in general. There is some inefficiency in their formula (in type A it can be "compressed" to a formula of Haglund, Haiman and Loehr in http://arxiv.org/abs/math/0409538) which is the subject of current work. -Presumably character-positivity of $P_\lambda$ actually fails already in type A, and in general type there is no notion of plethystic substitution to save us!<|endoftext|> -TITLE: Supercuspidals : clarification -QUESTION [5 upvotes]: A supercuspidal representation is an admissible representation with trivial Jacquet module for any parabolic.This implies that the matrix coeff. are compactly supported mod center. -Cartier (in Corivalis) defines super cuspidal rep as admissible rep. with compactly supported matrix coeff. mod center. -I am wondering if this (Cartier's) definition is equivalent to triviality of Jacquet module? If yes, is there a reference for such a proof? -Thanks in advance. - -REPLY [6 votes]: The answer is yes (and this is a classical result). A good reference with complete self-contained proofs is Bill Casselman's book (http://www.math.ubc.ca/~cass/research/pdf/p-adic-book.pdf). cf Theorem 5.3.1<|endoftext|> -TITLE: Which Diophantine equations can be solved using continued fractions? -QUESTION [18 upvotes]: Pell equations can be solved using continued fractions. I have heard that some elliptic curves can be "solved" using continued fractions. Is this true? -Which Diophantine equations other than Pell equations can be solved for rational or integer points using continued fractions? If there are others, what are some good references? -Edit: -Professor Elkies has given an excellent response as to the role of continued fractions in solving general Diophantine equations including elliptic curves. What are some other methods to solve the Diophantine equations $$X^2 - \Delta Y^2 = 4 Z^3$$ and $$18 x y + x^2 y^2 - 4 x^3 - 4 y^3 - 27 = D z^2 ?$$ - -REPLY [2 votes]: I was about to mention H J S Smith's algorithm for finding integer solutions to $x^2 + y^2 = p$ for $p \equiv 1$ mod 4; but this is referred to in a related thread at Applications of finite continued fractions -(Apologies if that thread is easily found from this one; but I wouldn't have noticed it without doing a Google search, and perhaps some other readers are equally inexperienced in StackOverflow ways or unobservant!) -Also, what about higher-dimensional continued fractions, expressed as matrix recurrence relations? I seem to recall that these can be used to find rational solutions of equations involving some kinds of cubic forms.<|endoftext|> -TITLE: Set-theoretical multiverse and foundations -QUESTION [10 upvotes]: I just had a look to the article The set theoretical multiverse by (mo user) J.D.Hamkins. Not being a logician and not knowing forcing techniques, I couldn't fully appreciate the mathematical ideas, but I was fascinated by the possible philosophical perspective of being compelled (by mathematical practice of forcing in set theory) to consider a whole multiverse of sets as a natural "landscape" for set theory, without committing to any specific choice. -[There's a kind of abstract in the introduction of a n-category café blog post (which I haven't completely read yet) by David Corfield] -In the article it is also stated that it's possible to mimick the study of the "full multiverse" within ZFC. This is actually done in A natural model of the multiverse axioms: "we shall internalize the study of multiverses to set theory by treating them as mathematical objects within ZFC [...]" -Personally, I have more affinity for the formalistic viewpoint than for some version of platonism (multiversed or not). So, my first question (somehow dually to this one) is: - -Is it conceivable that the "set theoretic multiverse principles" (which at the moment are, properly, ZFC sentences - see Hamkins and Gitman-Hamkins) could fit into a formal "multiverse theory" which is carried out in its own, i.e. not within a metatheory like ZFC, hence capturing the full-blown multiverse? Could such a theory be taken as the foundation (at least in some ZFC-flavoured sense) of mathematics? - -(I will probably open one or more followup "philosophical" questions about multiverses, when I'll clarify to myself what to ask) - -REPLY [10 votes]: (It happens that I will be giving a talk this week on this topic at the Exploring the Frontiers of Incompleteness series at Harvard, which is focussing on the question of determinism in set theory. The materials for all the talks are posted on the EFI web page, which also includes videos and discussion forums.) -Set theory currently provides at least a robust informal treatment of the multiverse. What I mean by this is that it is part of standard set theoretic practice and understanding to point out when a set-theoretic assertion $\varphi$ can have different truth values in different set-theoretic universes. That is, set theorists are already paying attention to the multiverse, when they discuss such issues as independence, forcing absoluteness, indestructibility and so on, which all concern the issue of how set-theoretic assertions can vary in the mutliverse. -Another more formal treatment of the multiverse idea is provided by what I have described as the toy model formalization, where one considers a multiverse from the perspective of a much larger set-theoretic background. For example, in ZFC one may consider the multiverse of all countable models of set theory, or just of a portion of them. This is the approach that Gitman and I took in our consistency proof that you mentioned, A natural model of the multiverse axioms. One can also view the typical countable-transitive-model approach to forcing as an instance of the toy model formalization. -Perhaps the toy model formalization has some affinity with the idea of the axiom of universes in category theory, and such a formalization would lead to a tower of multiverse concepts, one existing at the level of each Grothendieck universe. So it seems clear that one may set up a formal theory of multiverses using ZFC as a background theory, supplemented with various universe axioms, and leading to a tower of multiverses. -But my opinion is that the toy model method of formalization is ultimately unsatisfying, since it leads one to adopt axioms and principles about the toy models, rather than about the background set theory, which is the intended goal. -Although many of the questions that we have about the full multiverse happen to be first-order expressible in the language of set theory, nevertheless many of the questions that we are interested in happen to be formalizable in first-order ZFC. For example, many of the concepts of set-theoretic geology are formalizable in first-order ZFC, such as the open question whether every two ground models of the universe have a deeper ground inside them. Similarly, the questions surrounding the modal logic of forcing are also first-order expressible. This is a happy phenomenon, when we are able to use our current formalization to express and understand the multiverse issues. But the troubling matter underlying your question is that some of our questions are indeed pushing up against our formalism, which may be inadequate to the task.<|endoftext|> -TITLE: How to correctly generate uniformly distibuted random elements from SO(n)? -QUESTION [8 upvotes]: I already found some way to produce such matrices from SO(n) with a method called subgroup algorithm but I would like some advice on the method I used. Nowhere I could really find any paper relating to SO(n) directly but rather O(n). -To make it simple, I generate a $(k-1) \times (k-1)$ rotation matrix $\Gamma_{k-1}$, create the matrix -$\left( \begin{array}{ccc} 1 & \cdots & 0 \\\ \vdots & \Gamma_{n-1} & \\\ 0 & & \end{array} \right)$ -then I find a rotation $R$ that transforms the first column into a vector $v$ generated randomly as a point on the unit $k$-sphere. Then generate a new $k \times k$ rotation matrix $\Gamma_k$ : -$\Gamma_{k} = R\left( \begin{array}{ccc} 1 & \cdots & 0 \\\ \vdots & \Gamma_{k-1} & \\\ 0 & & \end{array} \right)$ -Induction is doing all the work. In practice, either you start by a $1 \times 1$ matrix with single element equal to 1, or you create a first trivial $2 \times 2$ rotation matrix with a angle taken uniformly from $[0,2\pi)$ which seems more direct. I skipped the details on how to generate the matrix $R$ and the vector $v$ as it seemed useless at first. To sum up $v$ is the normalization of a vector $v'$ which elements $v'_i \sim N(0,1)$ and $R$ is obtained by a double Householder reflection. -In practice this seems like a correct incarnation of the subgroup algorithm where the group is SO(n) but I wanted to check with more mathematically inclined than me if it didn't exist a more efficient way! As I said, I found some papers explaining variations or improvements on the original method from Stewart$-$which the subgroup algorithm is a generalization$-$in papers from Anderson or Genz. All of them focused on orthogonal matrices and I would have liked to transpose these improvements to special orthogonal matrices. -See: - -"The Efficient generation of Random Orthogonal Matrices with an Application to Condition Estimators" Stewart -"Generation of Random Orthogonal Matrices" Anderson -"Methods for Generating Random Orthogonal Matrices" Genz -"The Subgroup Algorithm for Generating Uniform Random Variables" Diaconis - -EDIT: by the way, I might use this algorithm with n as large as 1000 and beyond -So the real question is (and I'm not really sure how to formulate that correctly as I have a serious lack of knowledge in that area), is there a mapping from O(n) to SO(n) that will preserve the uniformity? - -REPLY [3 votes]: The best algorithm is the one due to G.W. Stewart -Stewart, G. W., The efficient generation of random orthogonal matrices with an application to condition estimators. (With mircofiche section), SIAM J. Numer. Anal. 17, 403-409 (1980). ZBL0443.65027. -Algorithm: Let $M$ be $n\times n$ with i.i.d $N(0, 1)$ entries. Let $M=QR$ be the QR factorization ($Q$ orthogonal, $R$ upper triangular). Then $Q$ is uniformly distributed on $O(n)$ (project to $SO(n)$ if you like).<|endoftext|> -TITLE: About the category of chain complexes and Grothendieck categories. -QUESTION [5 upvotes]: Given an abelian category $\mathcal{A}$ the category of chain complexes over $\mathcal{A}$ is again an abelian category. If $\mathcal{A}$ is a Grothendieck category then the category of chain complexes over $\mathcal{A}$ is a Grothendieck category? In praticular, for a ring $R$ with unitary and the category of its left unitary modules, does this hold? -Feel free to retag. - -REPLY [6 votes]: Yes, this is stated e.g. on page 3 of Hovey: Model category structures on chain complexes of sheaves.<|endoftext|> -TITLE: determinant of the table of characters -QUESTION [18 upvotes]: I am certain that the answer to this question exists somewhere. It might be a classical exercise. -Let $G$ be a finite group. Its table of characters is a square matrix, whose rows are indexed by the conjugacy classes and the columns are indexed by the irreducible characters. It is well defined, up to the order of rows and columns. In particular, its determinant if well-defined up to the sign. Let us define $\Delta$ to be the square of this determinant (this is well-defined). Because the characters form a basis of the space of class functions, we know that $\Delta\ne0$. When $G={\mathbb Z}/n{\mathbb Z}$, $\Delta=n^n$. - -Is there a close formula for $\Delta$ for a general group? Is it always an integer? - -REPLY [9 votes]: I found the following answer after posting it: -$$\Delta=\epsilon\prod_c\frac{|G|}{|c|},\qquad\epsilon=(-1)^m,$$ -where the product is taken over the conjugacy class. And $m$ is the number of pairs of complex conjugate irreducible characters. -Proof. On the one hand, the complex conjugate of the table is itself, up to $m$ transpositions of rows. This is because the conjugate of an IC is an IC. Therefore -$$\overline{\det(TC)}=\epsilon\det(TC)$$ -($TC$ stands for ``table of characters''.) -Hence $\det(TC)$ is real if $m$ is even, pure imaginary if $m$ is odd. hence $\Delta$ is real and its sign is $\epsilon$. -Now the characters form a unitary basis. Because a unitary matrix has a unit determinant, we may compute $|\Delta|$ by taking any unitary basis. Take $\phi_c(g)$ to be $0$ if $g\not\in c$ and $|G|^{1/2}/|c|^{1/2}$ if $g\in c$. In particular $|\Delta|$ is an integer because $$\frac{|G|}{|c|}=|{\mathcal Z}(a)|,\qquad a\in c.$$ -Another Proof: Let $D$ be the diagonal matrix whose diagonal entries are the cardinals of the congugacy classes. We may assume that the first rows of $TC$ are the real characters and the $2m$ last ones are the pairs of complex conjugate characters. Then the $(i,j)$-entry of $M:=(TC)D(TC)^T$ is $|G|\langle\overline{\chi_i},\chi_j\rangle$. From the orthogonality relations, we see that $M={\rm diag}(1,\ldots,1,J,\ldots,J)$ where -$$J=\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}.$$ -The number of blocks $J$ is precisely $m$. Now take the determinant; we obtain $\Delta\det D=(-1)^m|G|^r$ where $r\times r$ is the size of $TC$. Hence the formula.<|endoftext|> -TITLE: Means of Promoting Mathematics in Young Countries! -QUESTION [28 upvotes]: We all know mathematics is life, this question is for Mankind. It's mathoverflow here when some parts of the world we have mathunderflow! I think we can do something through ideas. A similar question "Good ways to engage in mathematics outreach" has been featured but this is a different question all together. -In this question, am interested in understanding what the so called super countries in mathematics done right utilizing the little resources they have in the promotion(development) of mathematics. How can very young country (mathematics development is wanting, the research output is low, some fundamental courses are not even taught at the first place due to lack of resource persons) move on? What are some of the ideas which have helped countries with a small economy grow? Are there mathematicians who have been involved in development of mathematics in developing/ less developed countries which are experiencing an upward trend? How did you do it? - -REPLY [19 votes]: Dear Ongaro Nyang' -Although Israel is no longer so young, it was young, (even -younger than most other countries,) not so long ago, and -it is a small and rather isolated place, with some difficulties. -So some lessons from Israeli mathematics, especially in its early days -may be relevant. - -A) Immigration - -1. Immigration -Israeli mathematics was initially based on and had benefited all the -times from immigration of mathematicians to Israel. The ability to -absorb immigration, in general, and to attract and absorb immigrating scientists, in particular, -is crucial.(Keeping the relations with mathematicians who immigrated from -Israel is also an important issue.) - -B) Financial matters. - -Investing resources is, of course, crucial. There was a large public investment -in universities in Israel's first years even when the country itself -was in rather bad economic shape. Overall, theoretical academic -subjects like mathematics are "cheaper." Keeping the right balances -regarding policies for spending the money is very important. -Let me mention two items. -2. Sabbatical/Travel money -1.1 Good Sabbatical opportunities: Israeli mathematicians (and scientists -in general) had relatively good sabbatical terms which make it possible -for them to get (from the Israeli institution) a reasonable European/US -salary while in sabbatical abroad. This was especially effective when the -Israeli salaries were very low compared to salaries abroad. The academic -system is built on 15% or so of the faculty being on sabbatical at -any time. (Often people spend additional time abroad on leave.) -2.2 In addition, Israeli scientist (with academic university positions) and graduate students (who works as T.A.'s) have funds for short-term -travels: (Fixed amounts depending on the academic rank per year) This -enable participation in conferences and joint research. -Both the sabbatical and travel money apply to all people with academic -positions (Sabbatical only to lecturer/asst professor position and above) -and are essentially automatic. (Minimal amount of bureaucracy, no -committees to judge qualification and to decide on amounts, no requirement -to be an invited speaker in a conference, etc. etc..) -3. Salaries -Keeping the right balances when it comes to salaries is also important. -Low salary gives incentives to leave but very high salaries (compared to -average salaries in the country) are morally problematic and may -give incentive for corrupting the hiring and promoting system. -The salary system in Israel is based on essentially equal salary for equal -academic rank and (overall) there are no substantial salary awards for -academic excellence beside the academic rank. (There is some (modest) -awards for people getting external grants and larger but still rather small -for people serving in administrative positions.) -I think that not having overly differential salaries and not -being "in the game" of offers and counteroffers is actually beneficial. - -C) Activities - -4. Activities for national math society -There are, since early times, -regular annual meeting of the Israeli Mathematics Union and some -other local activities. -5. National mathematical journal -There was a substantial effort, again from early times, -to create and maintain an Israeli journal of -mathematics. (In the 40th there was even a professional research level -journal in Hebrew for a few years.) -6. Conferences and visitors. -There was, again from early times, some resources devoted to conferences -and visitors. Carefully administered and with attention to -the added value for the local people this can be very fruitful. -Arranging visits of top people in mathematics for lecture series and visits -can also be useful. It seems that it is a good policy (when the country is not -rich) not to over-pay for such visitors (among other reasons, because this set standards which push the cost of visitors in general too high.) -Of course, warm hospitality is priceless. - -D) Content - -7. Maintaining a sense of tradition. -Basing activities on areas with long tradition of success which are -identified with the country's mathematical strength can be a successful -and well excepted by the whole mathematical community. -8. Self-breeding can work -The success of Israeli mathematical departments was largely based on -successful self-breeding, namely absorbing as faculty member people who -graduated at the department. -9. Self-confidence, Tolerance for "sporadic"(or non-main stream) areas -(and tolerance in general) -This seems to me a strength of Israeli mathematics and looks (to me) a -good attitude especially for a peripheral and somewhat isolated place. -Tolerance is important especially since mathematical quality is rather -high dimensional (some dimensions being -importance/depth/visibility/applicability/usefulness.) -(There is also -complete tolerance and essentially indifference in the context of -mathematical life towards matters of politics, attitude towards religion, -etc, issues that Israel is very torn apart about.) -10. Patience, realistic goals, unrealistic dreams -Building a good mathematical activity takes time, and there are -ups and downs as well. - -E) Outreach - -11. Issues concerning high school mathematics -There is some efforts to promote interest in mathematics among gifted high -school students: special mathematical journal (in Hebrew), some "clubs" -and "summer camps" math Olympiads etc. I think this had some factor in -promoting math among young people. Usually, what it takes is some -mathematician in academics which is devoted to this issue and some (not -large) budget. Giving an incentive for such an activity and such a -mathematician may be a good idea. Popular Math books and text books in -Hebrew had substantial influence. A. Frankel (the set theorist) -wrote wonderful five-volume Hebrew books introducing -mathematics (It was called "An Introduction to Mathematics") in the 40s/50s. -(More recently, the Hebrew edition of Singh's -book on Wiles proof of FLT increased popularity of mathematics.) - -F) Relations with other areas - -12. Relation with CS, physics and other disciplines -In Israel CS department were largely built out of math departments (not -electrical engineering) and there are still strong academic ties between -these communities. Connection with CS seems valuable. Of course, relations with physics are very important (not so strong in Israel). Relations between math and economics seems strong in Israel. -(In Jerusalem there is an interdisciplinary "center for -rationality" which involves people from math/economics and some from -statistics/psychology/philosophy/law/biology.) -In summary, when it comes to mathematical life in a small somewhat isolated and -at times a bit troubled place, it seems that it is valuable to make the -right balances in the local mathematical community between competition and -solidarity, to practice a lot of patience and tolerance, to be open to new people and new directions, and to be careful about incentives. -Like in Brazil the efforts of few pivotals mathematicians was very crucial. -As usual, luck is useful too. Good luck. - -REPLY [6 votes]: As it was said in an other answer it seems like a good idea to send students abroad to study at strong places and have them come back, or if they don't, keep contact, have them give courses in their home country etc. -A scientifically excellent place helping with precisely such a policy is the ICTP in Trieste, see the "Education and Training" and the "Research "sections of their website (they are called institute of theoretical physics, but they have a strong mathematics section). Concretely they provide training programmes for undergrads to prepare them for a PhD - see here - and they also offer short and longer term stays for working scientists from developing countries, as well as institution partnerships, see here.<|endoftext|> -TITLE: Long exact sequence for Cech cohomology? -QUESTION [11 upvotes]: Good evening, -I have two questions concerning Cech cohomology of presheaves. -(1) Let $X$ be a topological space and $0\to\mathcal{F}\to\mathcal{G} \to \mathcal{H}\to 0$ a short exact sequence of sheaves of abelian groups on $X.$ Does there exist a long exact sequence for the Cech cohomology of these sheaves as the case of sheaf cohomology? -(2) Let $\mathcal{F}$ be a presheaf of abelian groups on $X.$ Do we have the equality $\check{H}^p(X,\mathcal{F}) = H^p(X,\mathcal{F}^+)$ where the group on the left is Cech cohomology and the one on the right is the sheaf cohomology, and $\mathcal{F}^+$ is the associated sheaf of the presheaf $\mathcal{F}.$ For example, if $\mathcal{F}$ generates a zero sheaf, is $\check{H}^p(X,\mathcal{F}) = 0$ true for $p>0$ ? -Does anyone know something about these? -Thanks in advance. -EDIT : the same question as (2) but for the Cech cohomology : do we have the equality $\check{H}^p(X,\mathcal{F}) = \check{H}^p(X,\mathcal{F}^+)$? - -REPLY [7 votes]: In a paracompact space, the answer to (1) is yes, and is done in Godement's book Topologie algébrique et théorie de faisceaux. Your last question in (2) is Theorem 5.10.2 there, and (2) is Theorem 5.10.1, and the question in youe edit is the Corollary to theorem 5.10.2. (You can replace the hypothesis on the space by some technology—«cohomolgy with supports in a paracompatifying family»...) -In other words, it is probably a good idea to read chapter 5 in that book! :D<|endoftext|> -TITLE: A question about Second-Order ZFC and the Continuum Hypothesis -QUESTION [7 upvotes]: Some logicians-such as G. Kreisel-have stated that the Continuum Hypothesis is decided in -ZFC2 ("Second-Order ZFC") although we do not know which way it is decided. This is rather -confusing, since it is not usually made clear just what the collection of axioms (both -logical and non-logical) of ZFC2-as a formalized theory-is to include. ZFC2 is presumably -formalized in the Classical Second-Order Predicate Calculus which is not recursively -axiomatizable. Is (at least) the following weaker alternative to Kreisel"s statement correct? "If T is any consistent and recursively axiomatizable sub-theory of ZFC2, then -neither the Continuum Hypothesis nor its Negation is provable in T." - -REPLY [7 votes]: Think of it this way: Let $V$ be a model of $ZFC_2$. Then I claim CH holds in $V$ if and only if $CH$ is actually true (note that in order for second-order logic to make sense, we have to make a commitment to an underlying "real" universe of sets). The proof of this is as follows. First, $\omega^V$ has order type $\omega$: clearly it has a subset of order type $\omega$, and by the second-order version of the powerset axiom, $P^V(\omega^V)=P(\omega^V)$, so if $\omega^V$ had the wrong order type $V$ would "see" the error. A fortiriori, we can deduce that $\omega^V$ is countable. -By similar reasoning, $P^V(P^V(\omega^V))=P(P(\omega^V))$. Now CH is false if and only if $P(P(\omega^V))$ contains three infinite sets $X, Y, Z$ no two of which have the same cardinalities (left-to-right is trivial; right-to-left follows from the countability of $\omega^V$). -Suppose $CH$ is false; let $X, Y, Z$ be as above. Since $P(P(\omega^V)=P^V(P^V(\omega^V))$, we have $X, Y, Z\in V$; by the axiom of extensionality, $V$ sees that the cardinalities of $X$, $Y$, and $Z$ are different, and by the second-order powerset axiom $V$ sees that $X$, $Y$, and $Z$ are infinite. So $CH\implies (ZFC_2\models \neg CH)$. -Suppose now that $CH$ is true. Let $X, Y, Z\in P(P(\omega^V))$; again, we have $X, Y, Z\in V$. Since $CH$ holds, by the second-order powerset axiom plus separation we can find a bijection $f$ between two of $X, Y, Z$, so $CH$ holds in $V$. So $\neg CH\implies (ZFC_2\models CH)$. -This shows that $ZFC_2\models CH$ or $ZFC_2\models \neg CH$. The point is that the full power of second-order logic allows $V$ to "ask" certain set-theoretic questions of the "real" underlying universe of sets; these questions include ``Is CH true?" Similarly, it seems to me that they include all questions of the form "Does $V_\alpha\models \phi$ hold?" where $\alpha$ is a computable ordinal and $\phi$ is $\Sigma_1$ over $V_\alpha$ ($\Sigma_1$ is somewhat arbitrary; higher quantifier depth can (I believe) be achieved by passing to larger computable $\alpha$). -I'd imagine that in fact this phenomenon extends much further than what I've outlined, and that a staggeringly large class of sentences of set theory are known to be decided in $ZFC_2$, even if we don't know which way they are decided. - -I just realized that I didn't answer your actual question. -As Andreas says above, your statement is not correct: both $ZFC+CH$ and $ZFC+\neg CH$ are recursively axiomatizable, and consistent (assuming $ZFC$ is), and one of them is a subtheory of $ZFC_2$ (although we can't tell which). You could try to add some effectiveness criterion to your statement - something along the lines of, "There is no recursively axiomatizable consistent theory $T$ which decides CH and that is provably a subtheory of $ZFC_2$" - but it's unclear to me how to do this in a way that results in a non-trivial, but also not false, statement. The moral is that second-order logic is really nasty. For instance, it wouldn't even make sense to ask for a derivation of CH (or $\neg CH$) from $ZFC_2$, since there's no meaningful proof system for second-order logic. To understand how ridiculously awful this is, there are proof systems for some infinitary logics that are very useful in model theory and proof theory - Lopez-Escobar developed one that Barwise used (altered? my history is a little vague on this point), but I don't know a good reference - and logics that can express concepts like "is uncountable" or can quantify over automorphisms of certain kinds of structures are even compact. Basically, second-order logic is totally unusable (although, as always, there are exceptions). -Also, it occurs to me that we don't even need all of $ZFC_2$ to decide CH. Look at the natural second-order version of the first-order theory which is commonly called, annoyingly enough, "second-order arithmetic" (so I guess its second-order counterpart should be called "second-order analysis"). This will be enough to decide CH, since the arguments above will all go through.<|endoftext|> -TITLE: Weighted area of a Voronoi cell -QUESTION [16 upvotes]: Let $X = \{ x_1,\dots,x_n\} $ denote a set of $n$ points in the unit square $S = [0,1]\times[0,1]$, and let $w = \{w_1,\dots,w_n\}$ denote a set of weights corresponding to the $n$ points in $X$. Define the "power diagram" of $X$ in $S$ to be a partition of $S$ into at most $n$ pieces $V_i$, where -$V_i = \{x\in S: \|x - x_i\|^2 + w_i \leq \|x - x_j\|^2 + w_j \forall j \neq i \}$ -i.e. a "weighted Voronoi diagram". Now let's consider varying the weight $w_1$ while fixing the other weights; specifically, consider the function -$f(w_1) = w_1\cdot \text{Area}(V_1)$ -Clearly as $w_1 \rightarrow 0$ we have $f(w_1) \rightarrow 0$ and as $w_1 \rightarrow \infty$ we have $f(w_1) \rightarrow 0$ as well. My question: is $f(w_1)$ unimodal? Convex? Is the answer different if I only have $n=2$ points? What if I define my cells slightly differently, such as -$V_i = \{x\in S: \|x - x_i\| + w_i \leq \|x - x_j\| + w_j \forall j \neq i \}$ ? - -REPLY [11 votes]: Let me answer at least some of your questions. I will only talk about your first definition of the cells, since these are somewhat nicer, as Igor Rivin pointed out. -You consider the function $f(w_1)=w_1\cdot\text{Area}(w_1)$ and asked whether it is convex. I assume you mean "Is the set $\{(x,f(x)):f(x)\geq 0\}\subset\mathbb{R}^2$ convex?", or in other words: "Is the function $f$ concave on the interval where it is positive?" -I will first show that the answer to that question is "no, not in general". -Let $p\in \mathbb{R}$ be the number where the region associated to the first weight in the diagram with $w_1=p$ is a single point and let $z\in \mathbb{R}$ be the biggest number such that $\text{Area}(z)=1$. So in general $$\text{Area}(w_1)=\begin{cases}1&w_1\leq z\\\text{interesting}&z\leq w_1\leq p\\0&p\leq w_1\end{cases}$$ -Choose the first point as $(\frac{1}{2},\frac{1}{2})$ and four more arranged in a square around it: $(\frac{1}{4},\frac{1}{2}),(\frac{1}{2},\frac{1}{4}),(\frac{3}{4},\frac{1}{2}),(\frac{1}{2},\frac{3}{4})$ and choose all weights equal to some number $c\in\mathbb{R}$. In this case -$$\text{Area}(w_1)=\begin{cases}1&w_1\leq z\\(w_1-p)^2&z\leq w_1\leq p\\0&p\leq w_1\end{cases}$$ -Here $z$ and $p$ depend on $c$, but we choose $c$ such that $0\leq z -TITLE: Solving a quadratic matrix equation -QUESTION [15 upvotes]: This might be a well-known problem but I am having trouble to find this. For square matrices $X, A, B,$ how to obtain the general solution for $X$, for the quadratic matrix equation $X A X^{T} = B$ ? What are the existence and uniqueness conditions for such solution? - -REPLY [2 votes]: I read the reference of Igor, but I have nothing shot about our equation ! -There is a case that seems less difficult. If $A,B$ are symmetric complex, then the problem is essentially equivalent to solve $2$ equations of the form $ZZ^T=D$, that is, the case $A=I$ or $B=I$. -equation 1: $YY^T=A$ , equation 2: $ZZ^T=B$ , equation 3: $XY=Z$. If $A,B$ are invertible, then $Y,Z$ exist and $X=ZY^{-1}$.In these conditions, can we find, at least in theory, all solutions ? -For Halder. The equation $XAX=B$ is easy to solve (at least when $A$ is invertible) because it can be rewritten $(XA)^2=BA$. -EDIT: Let $A,B\in\mathcal{M}_n(\mathbb{C})$. According to a result by Turnbull, Aitken (cited in the Igor's reference) our equation admits a solution iff $A,B$ are congruent iff there are invertible $P,Q$ s.t. $PAQ=B,PA^TQ=B^T$. Then, for generic $A,B$, the equation $XAX^T=B$ has no solutions. Thus, the "good" equation is $XAX^T=CAC^T$. Using my computer, I randomly choose real matrices $A,C$ ; I find that the algebraic set of complex solutions has dimension $1$ when $n=2$, $1$ when $n=3$, $2$ when $n=4$, $2$ when $n=5$.<|endoftext|> -TITLE: How much universality is there for contact processes? -QUESTION [16 upvotes]: A couple of weeks ago I had to pick my daughter up from her nursery because of suspected chicken pox. It turned out to be a false alarm, but while I was waiting at the doctor's surgery to establish that, it occurred to me to wonder how it was known that the gestation period before spots appear is five days, and that somebody with the disease is infectious for up to nine days after getting it. Presumably the answer is boring: that there are enough situations where you can pinpoint when infection must have taken place to make it possible to determine the gestation and infection periods fairly accurately. -But suppose that were not the case, and instead you wanted to deduce the infection period from the global behaviour of the spread of the disease. A rough version of my question is whether this is possible. -Here's a more precise version (but this is just one model and I don't mind answers that apply to different but similar models). Let's define a contact process as follows. At any one time, each point in $\mathbb{Z}^2$ is either diseased or healthy. If you get the disease at time $t$, then there is some probability distribution $\mu$ on $[t,\infty]$ for your four immediate neighbours: their probability of catching the disease from you during the time interval $[a,b]$ is $\mu([a,b])$. I'm thinking of $\mu(\{\infty\})$ as the probability that your neighbour doesn't catch the disease from you. And let's say that if two of your neighbours are infected, then your chances of catching the disease from one of them is independent of your chances of catching it from the other. Finally, let's assume that $\mu$ is the same for everybody (apart from the translation by $t$ to take account of when a point becomes infected). -A simple example of a distribution $\mu$ would be half the uniform distribution on $[t,t+1]$ plus half a point mass at $\infty$. That would represent the situation where if you get the disease at time $t$ then your neighbour's chance of getting the disease from you is 1/2 and if your neighbour does get the disease from you then the time of infection is uniformly distributed over $[t,t+1]$. -With this set-up, my question is this: how much can you tell about the probability distribution $\mu$ from the global spread of the disease? That's still not a completely precise question, and I think I may lack the expertise to make it completely precise, but it's the usual picture that applies to models of this kind, where you look at everything from a great distance so that you can't see the small-scale structure (so, for example, I can't just empirically test lots of pairs of neighbours to build up a picture of the probability distribution). -A different way of asking the question, which explains the title, is this. It is a well-known and fascinating phenomenon that many probabilistic models like this have global behaviour that is very insensitive to the details of their local behaviour. So I would expect, for example, that all compactly supported probability distributions for which $\mu(\infty)$ is the same would have similar global behaviour: perhaps the only parameter that mattered would be the expected time to become infected, or something like that, which would govern how quickly the disease spread. (It's not obvious to me that that is the right parameter, by the way.) But perhaps I'm wrong about this. It might, for instance, be that if the time of infection is sharply concentrated in two places, then the disease spreads in two "waves", one faster and one slower. And if that's the case, then perhaps you can work out virtually the entire distribution from the global behaviour. -I'm asking this question out of idle curiosity and nothing more. I'd just be interested to know what is known (or at least believed to be true). - -REPLY [6 votes]: Hi Gowers, -In this paper: Sidoravicius, V. and Kesten, H . A shape theorem for the spread of an infection. Annals of Mathematics, v. 167, p. 1-63, 2008, -the authors consider a different model, but they were interested in the shape problem like you. One time before publish thi paper Vladas told me that about a model, similar to what you describe and also that there is a believe about universality for those models, perhaps they discuss this on this reference. -Here is a -link for the arxiv version of this paper.<|endoftext|> -TITLE: Fejer's theorem and convergence of Fourier series in measure -QUESTION [6 upvotes]: Fejer's theorem says that for any continuous function $f \colon S^1 \to \mathbb C$ with Fourier coefficients $a=(a_n)_{n \in \mathbb Z}$ the sequence -$$\sigma_n(a) := \frac1n \sum_{k=1}^n \sum_{l=-k}^k a_l \exp(2\pi i \cdot l\phi)$$ -convergences uniformly to $f$. Moreover, by the Riemann-Lebesgue Lemma, the sequence $a=(a_n)_{n \in \mathbb Z}$ is in $c_0(\mathbb Z)$. - -Question: Let $a=(a_n)_{n \in \mathbb Z}$ be in $c_0(\mathbb Z)$. - Does $\sigma_n(a)$ converge in measure to some measurable function on $S^1$? - -More generally, is there any summing procedure (or in fact any assignment whatsoever), which leads to a linear map $\Phi \colon c_0(\mathbb Z) \to M(S^1)$, which extends Fourier summation on finitely supported functions (and preferably also Fejer summation for Fourier series of continuous functions). Here, $M(S^1)$ denotes the space of measurable functions on $S^1$ (up to measure zero) with the usual measure topology given by the metric -$$d(f,g) := \inf\lbrace\varepsilon \mid \mu(\lbrace x \mid |f(x)-g(x)|\geq \varepsilon \rbrace \leq \varepsilon \rbrace.$$ - -REPLY [3 votes]: There is no continuous linear operator from $c_0$ to $M(S^1)$ that maps the unit vector basis to the characters. In fact, any continuous linear operator from $c_0$ to $M(S^1)$ maps the unit vector basis to a sequence which converges to zero at a good rate. To see this, note that by Maurey-Nikishin, the operator factors through $L_p$ for all $0 -TITLE: Is there an analog of Kodaira vanishing for singular varieties -QUESTION [12 upvotes]: I would like to know what kind of analogs of Kodaira vanishing theorem are valid for singular varieties. For example, is the following true: let $X$ be a projective Gorenstein variety and let $\omega_X$ be its canonical bundle. Is it true that $H^i(L\otimes \omega_X)=0$ -for $i>0$ for an ample line bundle $L$? - -REPLY [14 votes]: Indeed, as JC Ottem points out, Kodaira holds for log canonical (even semi-log canonical singularities). There's also a way to quickly deduce that Kodaira vanishing holds for Du Bois singularities (either from the Ambro-Fujino machinary or mimicking arguments of Kollar, let me know if you want details, perhaps I should put it on mathoverflow since it's not written down anywhere). -However, I should probably point out that it's totally trivial to see that Kodaira vanishing holds for rational singularities. Here's the proof: -Let $\pi : Y \to X$ be a resolution. Note $R \pi_* O_Y \cong O_X$ and so $R \pi O_Y(-\pi^* L) \cong O_X(-L)$ for any line bundle $L$. Fix $L$ to be ample. By a spectral sequence/composition of derived functors argument: -$$H^i(X, O_X(-L)) = H^i(Y, O_Y(-\pi^* L)).$$ -But $\pi^* L$ is nef and big and the vanishing of the right hand side is just Kawamata-Viehweg vanishing and Serre duality.<|endoftext|> -TITLE: Software for Borel-Weil-Bott in positive characteristic? -QUESTION [9 upvotes]: I am interested in calculating cohomology of line bundles on flag varieties $G/B$ in positive characteristic. But I really just have a bunch of scattered examples. Does there exist some kind of software that will calculate this for me? For the most part I don't care about the representation structure of the cohomology modules, I just want to know dimensions. Also, I know there are various results on when this cohomology is just like the char. 0 situation, but they won't always apply to my examples. So I don't need general results, just an algorithm. -I just know how to do one example using Macaulay2: irreducible homogeneous bundles on projective space (which are pushforwards of aforementioned line bundles). But I am also interested in things like type D, and homogeneous bundles on Grassmannians. - -REPLY [6 votes]: This is a sort of negative-leaning answer to the question about existence of software for your purpose. There is quite a bit of history to the problem in prime characteristic, going back to isolated examples found in the 1970s by Mumford and his 1975 Ph.D. student W.L. Griffith Jr. (Cohomology of Flag Varieties in Characteristic p) which showed that the classical ideas could break down. The rank 2 example $G_2$ lends itself to picture drawing and has been looked at in considerable detail. See the recent updated preprint by Andersen and Kaneda here. -Andersen's clever sheaf cohomology techniques (exploiting the Frobenius map) combined with my more speculative predictions tend to imply that the results depend heavily on Kazhdan-Lusztig theory for the affine Weyl group (of Langlands dual type). Moreover, the non-vanishing of cohomology seems to involve the actual module structure, so dimensions appear only as a byproduct of the study of generic module filtrations crossing Weyl chamber walls. The algebraic group of type $G_2$ already indicates how systematic but complicated the results will be in general, so any computational approach must take this case into account. (The results for $A_2$ and $B_2$ worked out by Andersen following his 1977 MIT thesis On Schubert Varieties in G/B and Bott's Theorem are also subtle, but con't compete with the complexity of $G_2$ whose alcove geometry is richer.) -ADDED: The problem arose in the setting of algebraic geometry, as seen in the thesis work mentioned above. Seshadri wrote up his own version of the $SL_3$ case treated by Larry Griffith, in a typescript Cohomology of line bundles on $SL_3/B$ (Tata Institute, September 28, 1976). I learned about the problem from him the following spring at IAS and formulated my own tentative interpretation in a conference paper that summer. Andersen recovered Griffith's results in a general setting in his 1979 Inventiones paper here. In particular, an extra non-vanishing $H^1$ has a unique simple submodule of specified highest weight. But pinning down the dimension or formal character of this module takes more work, done first by Jantzen (before Kazhdan-Lusztig theory). There may be shortcuts in small cases, but a general algorithmic approach to the flag variety of $SL_3$ gets complicated.<|endoftext|> -TITLE: Splitting principle in equivariant cohomology -QUESTION [6 upvotes]: The following is a weaker version of what is called splitting principle in - Appendix C, page 12, see also for a lighter version Brions Eq cohom and eq intersection theory, page 6: -Let $G$ be a compact (complex) connected Lie group with torus $T\subset G$ and $N$ its normalizer, $W=N/T$ its Weyl group. Let $X$ be any $G$-variety. -Then, there is an isomorphism -$ H_G^* (X) \cong (H_T^*(X))^W $ -of graded algebras. For $X=pt$ this is known to be Chevalley's restriction theorem. -My question is, can one drop the assumption on $G$ and $T$ to be compact? I came across an article VarVas, page 12, where that has been claimed (with G=Gl_n, X quasi-projective) to be a standard result and no reference is provided. - -REPLY [5 votes]: The embedding of the unitary group $U_n$ into $GL_n(\mathbb C)$ is a homotopy equivalence; this is easily seen to imply that $H^*_{U_n}(X)$ is isomorphic to $H^*_{GL_n}(X)$. So the result for compact groups implies that for $GL_n$. -The same idea works for any reductive complex algebraic group $G$, since the embedding of a maximal compact subgroup into $G$ is a homotopy equivalence.<|endoftext|> -TITLE: Singular vectors in Verma modules. -QUESTION [6 upvotes]: Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra, and $M(\Lambda)$ the Verma module with dominant integral highest weight $\Lambda$. The singular vectors in $M(\Lambda)$ are well known (they are something like $f_i^{\Lambda(h_i^\vee)+1} v_\Lambda$). A similar fact is true for the affine Lie algebra $\widehat{\mathfrak{g}}$, with integrable highest weight. -I know of a paper by Feigin, Fuchs and Malikov (1986) where an explicit formula for singular vectors is given for Verma modules over $\widehat{\mathfrak{sl}_2}$ with more general highest weight. The answer uses a notation involving Lie algebra elements raised to complex powers! -My question is about how much is known for more general Lie algebras, and where can I find information. -For instance, have formulas for singular vectors in $M(\Lambda)$ been given for every highest weight $\Lambda$ over any affine Lie algebra $\widehat{\mathfrak{g}}$? Or has little been done since Feigin, Fuchs and Malikov. -Also, I wonder if the answer to my question exists already but in some different (perhaps more abstract) language? -Thanks, -Jethro - -REPLY [2 votes]: This was supposed to be a comment but grew too much. Some progress using the lines of work of the article you cite is given in [1] where the author gives explicit formulas (he gives an algorithm), he uses rational powers of nilpotent elements and he sketches that his algorithm works in the affine case as well. Also Feigin and Semikhatov had extended the results of $\hat{sl}_2$ to $N=2$ and $sl(2|1)$, this is not the article I had in mind, but the one I can find now [2]. Malikov himself worked out quantum group versions of their formulas (1991). -As for your actual question, I do not know of any explicit formulas but if you're the Jethro I think you are the right person to ask would be Kac. -[1] de Graaf, Willem A. -Constructing homomorphisms between Verma modules. -J. Lie Theory 15 (2005), no. 2, 415–428. -[2] Theoret. and Math. Phys. 112 (1997), no. 2, 949–987 (1998)<|endoftext|> -TITLE: Convex curves with many inscribed triangles maximizing perimeter -QUESTION [10 upvotes]: A classical nice result of Euclidean geometry states that the triangles maximizing the perimeter among all inscribed triangles of a given ellipse constitue a one-parameter family. Precisely, for each point on the ellipse there is exactly one such a maximizing triangle with vertex on that point, which can also be viewed as a period-3 bouncing trajectory. -I am wondering is this property is characteristic of the ellipses: - -Let $C$ be a compact convex subset of - the plane such that for any point - $P\in\partial C$ there is a triangle - maximizing the perimeter among all - inscribed triangles, with $P$ as a - vertex. Is it true that $\partial C$ is - necessarily an ellipse? - -REPLY [2 votes]: Please see a relevant paper -Yu. Baryshnikov, V. Zharnitsky, Sub-Riemannian geometry and periodic orbits in classical billiards, Math. Res. Lett., 13 (2006), 587–598. -As far as I know, the following question is open: are there curves, rather than ellipses, that possess circles of p- and q-periodic billiard trajectories for different p and q.<|endoftext|> -TITLE: Double Referencing in arXiv -QUESTION [15 upvotes]: I am writing two separate paper that are closely related. When I try to submit to arXiv, is it possible for each paper to refer to the other paper with an arXiv link, rather then putting a newer version of one of them just to replace the arXiv link in the reference? Since I heard from others that too many frequent update to a new paper is not preferable. -So, what is the usual practice in case such situation happens? - -REPLY [18 votes]: I asked the arxiv administrators about this last year, and received the following reply. - -Dear Scott, -Thank you for your feedback. At this time we have no plans to change -the way the submission ID and the final arXiv ID are created. We -developed the new submission system to be more flexible for our users -who need an area to work on their submission and conduct any updates for -formatting/layout/typos without being constrained on time. Hence, we -can not issue an arXiv ID until the submission has been announced. -We are looking into ways to make a submission available to all the -co-authors but can not provide a time frame when this feature will be -added. We have received feedback that validates the need to be able to -share a submission before it is announced and we are working on a solution. --- -arXiv admin -On 07/07/2010 11:38 AM, Scott Morrison wrote: - -Keyword: "submit/0016922" -Dear arxiv admins, -I'm concerned about two ways in which the new submission system is - inferior to the old one (mostly, though, it's great!) -First, we used to be told what the arxiv identifier would be. Now - we're just given a temporary identifier, like submit/0016922. This is - a problem for me, as I always used to include the arxiv identifier in - my paper (usually as a footnote, e.g. "This paper is available at - http://arxiv.org/... and at ..."). It will actually be a - real problem, presently: my coauthors and I have 3 related papers that - we want to put on the arxiv simultaneously. We'd like to be able to - refer from one paper to the others, but it seems now there's no way to - do this without submitting a second version the next day, because we - can't discover what the identifiers will be. -Second, it's somewhat annoying that there's no way for a coauthor to - see the submitted paper before it is announced. In the past, we would - always have someone other than the submitter carefully check the - produced PDF, abstract, etc., but it's now much harder to do this. -I'm not sure if it's possible for you to remedy either of these - problems, but I would encourage you to try! -scott<|endoftext|> -TITLE: Virtual algebraic calculation within proofs -QUESTION [7 upvotes]: It seems to me that the undergraduates I teach have particular difficulty with proofs that involve reasoning about algebraic calculations that arise only theoretically. Since I have in mind doing some pedagogical/expository writing around this issue, I want to assemble for myself as many characteristic examples as I can. -I'll give two examples now, and more on demand if it turns out I haven't made myself clear. -1) The proof of the transcendentality of $\pi$ such as one finds in Baker's Transcendental Number Theory, involves a product of many factors, each corresponding to hypothetical conjugate of $\pi$, and leading to a sum having size exponential in the hypothetical degree of $\pi$, all this to set up an application of the fundamental theorem on symmetric polynomials. -2) The proof of the Bruck-Ryser theorem on finite projective planes involves using geometry to create a hypothetical identity and then doing repeated substitutions to reduce the number of variables. -Challenges to clearly illustrating the mechanism within these two proofs include counterfactual hypotheses (for contradiction) and exponential complexity. Many of my students simply have very little feeling for what the distributive law says beyond the multiplication of two binomials, so I have a particular interest on proofs that hinge on analyzing the result of products with many factors each of which has many summands. -Thanks in advance for all contributions. Please give citations for paradigmatic proofs rather then mentioning theorems only, by name or statement. Please, if in doubt, post, but I'm particular interested in important results within the reach of advanced undergraduates and/or beginning graduate students. - -REPLY [5 votes]: I'm going to define the characteristic polynomial to my students next monday, and I'm going to tell them about its degree and three special coefficients. The constant coefficient is trivial by evaluation at zero. For both others and the degree, the reasoning is an example of what you ask, if I got your question right. -I will write the determinant on the board (written with parenthesis because I don't remember how to make the straight lines...): -$$\begin{pmatrix}a_{1,1}-X & a_{1,2} & \dots \\\ -a_{2,1} & a_{2,2}-X & \dots \\\ -\dots & \dots & \dots -\end{pmatrix}$$ -and consider the development along the first column : the first term is $a_{1,1}-X$ times a cofactor which looks the same as the original determinant, and all others will have lost two terms of the form $\text{something}-X$ so will at least two lower in degree. That means if I were to really develop the determinant, it would end up looking like $(a_{1,1}-X)\dots(a_{n,n}-X)+\text{at most degree}(n-2)$, so the characteristic polynomial is degree $n$, its dominant coefficient is $(-1)^n$, and the coefficient just behind will be $(-1)^{n-1}\rm{Tr}(A)$.<|endoftext|> -TITLE: $V$, $W$ are varieties. Does $V\times \mathbf{P}^1=W\times \mathbf{P}^1$ imply $V=W$? -QUESTION [39 upvotes]: If $\mathbf{P}^1$ is replaced by the affine line $\mathbf{A}^1$, this becomes the cancellation problem, and we have a pair of famous Danielewski surfaces ($xy=1-z^2$ and $x^2y=1-z^2$) as a counterexample (though I'm still seeking how to prove that..). I suppose in my case this counterexample might no longer work. -Also one may replace $\mathbf{P}^1$ by $\mathbf{P}^n$ or other fixed varieties, or ask again after imposing some conditions on $V$ and $W$ (for example dimensions) if there are counterexamples for my question. And more wildly I may ask for what kind of family $X_n$, we will have the result that $V\times X_n=W\times X_n$ implies $V=W$. Any result of these kind of variations of the problem is also welcomed. - -REPLY [78 votes]: This problem was studied by Fujita in his paper "Cancellation problem of complete varieties", Inventiones Mathematicae 64 (1981). -He showed that the obstruction to cancellation is caused by the Picard schemes, proving the following remarkable result (see Corollary 7 in the cited paper): - -Let $M$, $V$ and $W$ be compact complex manifolds such that $M \times V \cong M \times W$. Assume that $M$ is projective and that $\textrm{Alb}(M)=0$ or $\textrm{Alb}(V)=0$. Then $V \cong W$. - -In particular, cancellation problem has a positive answer for $M=\mathbf{P}^n$. -The condition on the Albanese variety is a necessary one; in fact, it is known that cancellation is not always true for abelian varieties.<|endoftext|> -TITLE: Constructing a hypersurface with given outer normals -QUESTION [10 upvotes]: Pick a point on each of the positive half-axes in $\mathbb{R}^n$. Put a (unit-norm?) vector at each of the n points. -(a) Is there a hypersurface in the orthant $\mathbb{R}^n_+$ going through these n points with the vectors as outer normals? (The answer ought to be yes, since I'm asking whether there exists a closed=exact 1-form on $\mathbb{R}^n_+$ with given values at a few points.) -(b) Is there a hypersurface with the above properties, and also constrained to have the outer normals at all points lie in some half-space? (I know the half-space in question will contain the n vectors specified above, and also the vector (1,1,...,1).) Equivalently, I'm asking that some other vector points inwards along all of the hypersurface. -(c) I'd quite like the above surface together with the coordinate hyperplanes to enclose a compact set. Is this possible? automatic? -(d) Is there a constructive way to build a hypersurface with properties (a)-(c)? Ideally in the vein of writing down some equations for the surface, or for its normal vector field, in terms of the original points and vectors. (I'd like to build one of those for each orthant and then be able to stitch them together.) -(Motivation: I'm trying to show convergence of a stochastic system by constructing (the level sets of) a Lyapunov function for it. I have quite a lot of freedom for what the function looks like on the interior of each orthant, but it's progressively more and more constrained on all the higher codimension subspaces, until at each half-axis I have a very limited range of possibilities. I'd like to know if these possibilities can be patched together in a sensible way.) - -REPLY [10 votes]: This is speculation, not a precise answer, -but I wonder if perhaps Minkowski's theorem on the existence -of a polytope with prescribed face normals and areas might help? This theorem is described in detail in, for example, Alexandrov's book, Convex Polyhedra, Chapter 7, p.311ff. -The connection to your problem is this. If you specify $k$ unit normals ${\bf n}_i$, you can imagine those -determining the orientation of $k$ faces of a convex polytope in $\mathbb{R}^d$. -Minkowski's theorem says that, if in addition, you specify face areas $F_i$ such that -$\sum_{i=1}^{k} F_i {\bf n}_i = 0$, then there exists a polytope realizing those normals and areas. -If I understand your situation correctly, you have some of the ${\bf n}_i$ prespecified. -You need to add one more ${\bf n}_i$ (one more to reach $d+1$ in total), -as well as choose areas $F_i$, in order to zero that sum. -The convexity might(?) then yield the halfspace condition you desire. -This is all discrete, of course, but it should not be difficult to pass from a polytope to a smooth hypersurface. -Example. $d=2$ (my $d$ is your $n$). You specify ${\bf n}_1=(1,0)$ and ${\bf n}_2=(-1,1)$, red below, on "the positive half-axes in $\mathbb{R}^2$" (dashed), and then throw -in ${\bf n}_3=(0,-1)$ (green) and appropriate areas, -which by Minkowski determine a triangle, which you then -approximate by a smooth curve (hypersurface), which has all outer normals in a halfspace. - -                - -For general $d$, $d+1$ vectors ${\bf n}_i$ will lead to a simplex. -Added. There is a chance that the John ellipsoid in this simplex could -serve as a starting point on your other question, -"Is there an ellipsoid with given outer normals?"<|endoftext|> -TITLE: When is a submanifold of $\mathbf R^n$ given by global equations? -QUESTION [40 upvotes]: Let $M \subset \mathbf R^n$ be a (smooth) submanifold of dimension $d$. Under which conditions does there exist global equations defining $M$? By global equations I mean : does there exist a smooth function $f: \mathbf R^n \to \mathbf R^{n-d}$, submersive at each point of $M$ and such that $M=f^{-1}(0)$. -Of course there are two necessary conditions: -1) $M$ must be a closed subset of $\mathbf R ^n$. -2) The normal bundle of $M$ in $\mathbf R^n$ must be trivial. -At first, I would have guessed that these conditions are sufficient, but I can't prove it. -I have partial answers, however. -1)The first natural thing to do is to take a tubular neighbourhood $U$ of $M$ in $\mathbf R^n$. The indentification $U \simeq M \times \mathbf R^{n-d}$ allows to define a function $f : U \to \mathbf R^{n-d}$ which has the required properties. But it is not clear to me whether $f$ can be extended to the whole $\mathbf R^n$. -2) There is a way to give an answer if we change a bit the problem: the Pontryagin-Thom construction gives a function $f: \mathbf R^n \to \widehat{\mathbf R^{n-d}} \simeq \mathrm S^{n-d}$ by sending all the points outside a tubular neighbourhood at infinity. -This maybe means that this is the good formulation of the problem, but I am still curious about the original one. -3) If $M$ has codimension $1$, then the function $f$ defined on a tubular neighbourhood $U$ of $M$ as in 1) can actually be extended to $\mathbf R ^n$ by a constant function (using the fact that the complement of $M$ has two connected components). - -REPLY [9 votes]: An example of a compact $16$-dimensional submanifold of $\mathbb{R}^{30}$ with trivial normal bundle that is not defined by such global equations may be found in [Akbulut-King, Submanifolds and homology of nonsingular real algebraic varieties, Am J Math 107 45 (1985), Theorem 5.8].<|endoftext|> -TITLE: Brauer group of a field of power series in two variables. -QUESTION [6 upvotes]: Let $k$ be the field $F_2((X,Y))$, where $F_2$ is the field with two elements and -$X$ and $Y$ are two indeterminates. Can we describe the Brauer group of $k$, or at least its $2$-torsion? -(My motivation is as one could expect: I have an irreducible representation of a group of dimension 2 over the separable closure of $k$, whose trace is in k, and I am trying to determine whether or not it is realizable over $k$) - -REPLY [6 votes]: You can get a fairly good picture of the elements of order $2$ of the Brauer -group in the following way. There is no reason to fixate on characteristic $2$ -so I assume that we are dealing with $K:=\mathbb F_p((X,Y))$ and in fact the only -reason to stick to the prime field is notational convenience as the Frobenius -map on $\mathbb F_p$ is the identity so the relative Frobenius is equal to the -absolute one which means that I won't have to distinguish between $Z$ and -$Z^{(p)}$. Technically it may be that proper references would require $Z$ to be -of finite type over the base but everything I say will be clearly true also for -$Z=\text{Spec}\mathbb F_p((X,Y))$. (On the other hand the only thing I will need -of $Z$ except for a smoothness/regularity assumption is that it is affine and -with trivial Picard group). -We define the sheaf (on the small étale site of $Z$) $\nu$ by the exact sequence -$$ -0\rightarrow\mathcal O_Z^\ast\xrightarrow{p}O_Z^\ast\rightarrow\nu\rightarrow0. -$$ -We have a map $\text{dlog}\colon\mathcal O_Z^\ast\rightarrow\Omega^1_Z$, where -$\text{dlog}(f):=df/f$ and it factors to give an injection $\nu\subseteq -\Omega^1_Z$. More precisely, it lands in the subsheaf $Z^1$ of closed forms and we -have an exact sequence (again on the small étale site): -$$ -0\rightarrow\nu\rightarrow Z^1\xrightarrow{C-\iota}\Omega^1_Z\rightarrow0, -$$ -where $C\colon Z^1/B^1\rightarrow\Omega^1_Z$, is the Cartier isomorphism ($B^1$ -being the exact $1$-forms) and $\iota\colon Z^1\subseteq \Omega^1_Z$ is the -inclusion. -Now, the first sequence (and the fact that $\text{Pic}(Z)=0$) gives that -$H^1(Z,\nu)$ is the kernel of multiplication by $p$ on the Brauer group. The -fact that $Z$ is affine gives that $H^1(Z,Z^1)=0$ and hence the second sequence -gives that $H^1(Z,\nu)$ is the cokernel of $C-\iota\colon H^0(Z,Z^1)\rightarrow -H^0(Z,\Omega^1_Z)$. This cokernel can be made very explicit (and to make it very -explicit we temporaritly assume $p=2$): -$H^0(Z,Z^1)$ is a module over $K$, where scalar multiplication is given by the -square map $f\cdot\omega=f^2\omega$, and has a basis given by $dX$, $dX/X$, -$d(XY)$, $dY$ and $dY/Y$. We have that $C$ is $0$ on $dX$, $d(XY)$ and $dY$ and -$C(dX/X)=dX/X$ and $C(dY/Y)=dY/Y$. Furthermore, $C$ is linear in the sense that -$C(f^2\omega)=fC(\omega)$. This implies that the relations in the cokernel are -given by $f^2dX=0$, $f^2dY=0$, -$f^2XY(dX/X+dY/Y)=0$, $(f^2-f)dX/X=0$ and $(f^2-f)dY/Y=0$ (where $dX$, $dY$, -$dX/X$, $dY/Y$, $XdY$ and $YdX$ is a $K$-basis for $H^0(Z,\Omega^1_Z)$). This -allows for a fairly transparent normal form for elements in $H^1(Z,\nu)$. -If one wants a direct description of the central simple algebra associated to an -element $\omega\in H^0(Z,\Omega^1_Z)$ one can apply the What Else Can It Be-principle -(a very useful though somewhat dangerous principle, in this case it is probably -OK). Recall that we have the algebra $\mathcal D$ of differential operators of order $ -TITLE: Are cubic four-folds containing a quartic scroll pfaffians? -QUESTION [6 upvotes]: Let $X\subset \mathbb{P}^5$ a smooth pfaffian smooth cubic fourfold hypersurface. It is easy to see that such a hypersurface must contain a quartic scroll surface. I wonder about the inverse question. If a cubic fourfold $X$ contains a quartic scroll, is it a pfaffian? - -REPLY [3 votes]: Part (a) of Proposition 9.2 in Beauville's "Determinantal Hypersurfaces" paper (Michigan Mathematical Journal 48, 2000) says that a cubic fourfold is linear Pfaffian precisely when it contains a quintic del Pezzo surface. One path to settling your question is to determine whether every cubic fourfold $X$ containing a quartic scroll $Q$ also contains a 2-plane $P$ for which $Q \cup P$ is a degeneration of a quintic del Pezzo in $X.$<|endoftext|> -TITLE: What problem do the adeles solve? -QUESTION [14 upvotes]: While browsing through some papers, I came across some literature discussing the Arthur-Selberg trace formula. At a conceptual level I think I understand what it is doing, but when I get down to the technical details I start to get a bit lost. Part of the problem is that James Arthur's papers are all written using adelic Lie groups, instead of working over an arbitrary field or fixing something less complicated (like R or C). Clearly, he must have had a good technical reason for doing this, but for the life of me I can't see why. -Is there any real intuition behind the adeles? Outside of algebraic number theory, why would anyone ever want to use them? What do they "do" that R doesn't? (in other words, could I safely do a mental find replace of $\mathbb{A}_{\mathbb{Q}}$ with $\mathbb{R}$ and still get basically the gist of what is going here?) - -REPLY [22 votes]: When working on a rational problem (over $\mathbb{Q}$), you can't do much analysis - so you lack quite a few tools. -The obvious solution is then to pass to the completion, where you'll be able to do analysis ; so most people go to $\mathbb{R}$. But that isn't that natural : $\mathbb{Q}$ has several completions in fact, and choosing the absolute value to measure distances wasn't the only choice. You could have gone to the $p$-adic numbers too, and use their special properties to gain further insight on your initial problem. -So somehow you gain the idea that perhaps solving your initial problem (which is called global) might involve looking into the various problems obtained by pushing it into the various available completions (which are called local). -That means you gain huge means of study, but now there are two prices to pay : you have the question whether some kind of Hasse principle applies, which is "How equivalent is it to solve the problem locally everywhere and globally?", and you have the problem that you need to work in all of those. -That last problem is where adeles come into the picture. Working adelically means you're effectively working simultaneously in all places -- and you mostly put them on an equal footing. You're still able to go into a single local place if needs arise, but you get an object which puts the pieces together. -There is another nice thing about adeles : if your initial problem wasn't just over $\mathbb{Q}$, but other any other kind of global field (a number field or a function field), then again you'll have a notion of adeles, and most tools will work the same. In fact, getting insight on a problem for a type of global fields and pushing the idea in the adeles is a good way to know what to look for in the other type of global fields. There are problems which are thus solved for some of them, and conjectures for others, precisely for this reason. -So I think what I wrote makes clear why wanting to replace adeles by just $\mathbb{R}$, while it will give some understanding of things (in a single place!), will also pretty much destroy the whole symmetry of the matter, and hence lose much of it.<|endoftext|> -TITLE: Does the set of happy numbers have a limiting density? -QUESTION [20 upvotes]: A positive integer $n$ is said to be happy if the sequence -$$n, s(n), s(s(n)), s(s(s(n))), \ldots$$ -eventually reaches 1, where $s(n)$ denotes the sum of the squared digits of $n$. -For example, 7 is happy because the orbit of 7 under this mapping reaches 1. -$$7 \to 49 \to 97 \to 130 \to 10 \to 1$$ -But 4 is not happy, because the orbit of 4 is an infinite loop that does not contain 1. -$$4 \to 16 \to 37 \to 58 \to 89 \to 145 \to 42 \to 20 \to 4 \to \ldots$$ -I have tabulated the happy numbers up to $10^{10000}$, and it appears that they have a limiting density, although the rate of convergence is slow. Is it known if the happy numbers do in fact have a limiting density? In other words, does $\lim_{n\to\infty} h(n)/n$ exist, where $h(n)$ denotes the number of happy numbers less than $n$? - -REPLY [8 votes]: I started working on this question after it was posted to MathOverflow and found bounds similar to those found by Justin Gilmer: upper asymptotic density of the happy numbers 0.1962 or greater, lower asymptotic density no more than 0.1217. However, I was also able to prove that the upper asymptotic density of the happy numbers was no more than 0.38; Gilmer mentioned in his paper that the question of whether the upper asymptotic density was less than 1 was still open. -A writeup of the result is at http://djm.cc/dmoews/happy.zip. The method used to find an upper bound on the upper asymptotic density was to start with a random number with decimal expansion $??\dots{}??\hbox{\#}\hbox{\#}\dots{}\hbox{\#}\hbox{\#}$, where the digits # are independent and uniformly distributed, and the digits ? are arbitrarily distributed and may depend on each other, but are independent of the #s. Then if there are $n$ #s, asymptotic normality implies that after applying $s$, we get a mixture of translates of a distribution which is approximately -normal, with mean $28.5n$ and standard deviation proportional -to $\sqrt{n}$. If $10^{n'}/\sqrt{n}$ is sufficiently small, each translate -of this normal distribution will have its last $n'$ digits approximately -uniformly distributed, so we get a random number which can be approximated by the same form of decimal expansion we started with, $??\dots{}??\hbox{\#}\hbox{\#}\dots{}\hbox{\#}\hbox{\#}$, where now there are $n'$ digits #. Repeating this eventually brings us to numbers small enough to fit on a computer. -The method used to find the bounds similar to Gilmer's was to start with a random number of the form $dd\dots{}dd??\dots{}??\hbox{\#}\hbox{\#}\dots{}\hbox{\#}\hbox{\#}$, where the ?s and #s are as before, the $d$s are fixed digits, and there are the same number of $d$s and #s, but very few ?s. Then if the parameters are appropriately chosen, we can show that after applying $s$, we again get a random number which can be approximated by the same form of decimal expansion, $dd\dots{}dd??\dots{}??\hbox{\#}\hbox{\#}\dots{}\hbox{\#}\hbox{\#}$, and repeat this step until the number is small.<|endoftext|> -TITLE: Consequences of the Langlands program -QUESTION [11 upvotes]: In the one-dimensional case the Langlands program is equivalent to the class field theory and the two-dimensional case implies the Taniyama-Shimura conjecture. -I would like to know: are there any other important consequences of the Langlands program? - -REPLY [26 votes]: There are many, many consequences of the general Langlands program (which I'll interpret to mean both functoriality for automorphic forms and reciprocity between Galois representations and automorphic forms). Some of these are: - -The Selberg $1/4$ conjecture. -The Ramanujan conjecture for cuspforms on $GL_n$ over arbitrary number fields. -Modularity of elliptic curves over arbitrary number fields. (Indeed, Langlands reciprocity -is essentially the statement that all Galois representations coming from geometry are attached to automorphic forms.) -Analogues of Sato--Tate for Frobenius eigenvalues on the $\ell$-adic cohomology of arbitrary varieties over number fields.<|endoftext|> -TITLE: Can taking the projective closure of an affine variety increase the degrees of its ideal generators? -QUESTION [10 upvotes]: Say we have some equations $f_1(x)=0, \ldots f_k(x)=0$ defining a variety $X$ in ${\mathbb C}^n$ (not necessarily a minimal number of generators, and not necessarily of minimal degree), and suppose we want to know the ideal $\bar{I}$ of its closure $\overline{X}$ in ${\mathbb P}^n$. A naive question: -If all the original generators $f_i$ were of degree at most $d$, can we generate $\bar{I}$ using polynomials of degree at most $d$? If not, what degree bound $d'$ can we give for a minimal generating set of $\bar{I}$? - -Motivation, in short: Ideal saturation is an extremely slow operation, but solving a linear system to identify vanishing forms up to a certain degree is polynomial time, which would give a more practical way to compute $\bar{I}$ in high dimensions. -(Sorry, I'm new to computational algebra, and maybe people already know a fast way to compute projective closure...) -Motivation, in long: To compute $\bar{I}$, it's not enough to simply homogenize the generators with respect to a new variable $z$. Denote such homogenizations by $\hat{f}$, their ideal by $\hat{I}$, and their variety by $\hat{X}$. The problem is that $\hat{X}$ might contain irreducible components on the hyperplane at infinity $(z=0)$, so we need to saturate it, $\bar{I}=(\hat{I}:z^\infty)$, to eliminate these components and get the ideal of $\bar{X}$. But saturation is extremely slow! (I think it's at least doubly exponential time in the number of variables.) So I'd like to find $\bar{I}$ instead by solving a linear system on the vector space of homogeneous polynomials of degree at most $d$ to see which ones vanish, and declare that they cut out the variety $\bar{X}$, yay! But do they? If searching up to degree $d$ is not enough, how far do I have to go? -A simple example: Say $f_1 = x$ and $f_2 = y-x^2$ in ${\mathbb C}^2$, so $X$ is just the origin $(0,0)$. Suppose we first homogenize the generators $f_i$. The parabola $\hat{f}_2=yz-x^2=0$ is tangent to the line at infinity at $(x:y:z)=(0:1:0)$, which also lies on $\hat{f}_1=0$. So together they cut out $\hat{X} = \{ (0:0:1) \cup (0:1:0)\}$, and we don't get $\bar{X}=(0:0:1)$ until we saturate $\hat{I}=\langle x,yz-x^2\rangle$ by $z$ to eliminate the component at infinity. -In this example, we already know the answer is the origin, and saturation is very fast for such a small system anyway. But for many variables, we don't already know what the variety looks like, and the saturation might not finish in a human lifetime, so it would be nice to solve a linear system instead. -Thanks for any help on this question! - -REPLY [8 votes]: Consider $R = \Bbbk[x_1, \ldots, x_5]$ and $I = (x_1x_2^2-x_3^2, x_1x_4^2-x_5^2)$. When we homogenize w.r.t $z$, we get $(x_1x_2^2-zx_3^2, x_1x_4^2-zx_5^2)$ whose saturation w.r.t $z$ contains the binomial $x_3^2x_4^2-x_2^2x_5^2$ of degree $4$. The original generators are of degree at most $3$. -Look at Section 15.10 of Eisenbud's commutative algebra book; one should homogenize a Groebner basis of $I$. Therefore, bound for $d'$ in terms of $d$ might turn out to be quite large. (This is just my feeling!) -PS: in the `Simple Example', do you mean that $\hat{f_2} = x^2-yz$? As stated the ideal $\hat{I}$ is already saturated w.r.t. $z$ and does not have a component at $[0:1:0]$.<|endoftext|> -TITLE: Finding the action of the symplectic group on the Siegel-half plane -QUESTION [6 upvotes]: Let $S$ be the Siegel-half plane of dimension $n$, i.e. the set of complex $n \times n$ matrices $Z$ which are symmetric and whose imaginary part is positive-definite. In dimension 1 we can identify $S$ with the Poincaré half-plane. -In dimension 1 there is an action of the symplectic group $Sp_2(\mathbb R)$ on $S$, given by fractional linear transformations. One can extend this action to higher dimensions; we write an element $M$ of the symplectic group $Sp_{2n}(\mathbb R)$ as a block matrix with blocks $A$, $B$, $C$, $D$, and then set -$$ M \cdot Z = (AZ + B)(CZ + D)^{-1}. $$ -Now, in dimension 1 one sees that this action is induced by the action of the general linear group on the projective line and the embedding $GL_2(\mathbb R) \hookrightarrow GL_2(\mathbb C)$. However, this does not seem to be true in higher dimensions, as $GL_{2n}(\mathbb C)$ acts on the projective space of dimension $2n$, and not on the space of dimension $n$. -The definition of the action of $Sp_{2n}(\mathbb R)$ on $S$ in higher dimensions thus seems very ad-hoc, which motivates: -Question: How can one find this action naturally in higher dimensions? -[Edit:] Question 2: David's answer is very nice, but I'd like to iterate the question before accepting it. Instead of the Siegel half-plane above, consider the space $U$ of complex $(1,1)$-forms whose imaginary part is positive-definite. In a basis we can identify this with the set of $n \times n$ matrices $Z$ whose antihermitian part is positive-definite. The action of the symplectic group now makes sense on this space as well. -In a basis the Siegel half-plane is a closed subspace of $U$ if $n > 1$. It is easiest to define the action of the symplectic group in a basis, but we can define it without reference to a basis by choosing a hermitian inner product on our vector space. Now, can we find the action of the symplectic group on $U$ in the same fashion as on the Siegel half-plane? - -REPLY [5 votes]: A nice way to think about the action of $SP_{2g}(\mathbb{R})$ on $S$ is that $S$ is an $SP_{2g}(\mathbb{R})$-invariant open set in the Lagrangian grassmannian. -Let $V$ be a $2g$-dimensional complex vector space equipped with a symplectic form. Let $L$ be the space of isotropic $g$-dimensional subspaces of $V$. Such a subspace is called a Lagrangian. Clearly, $L$ is a complex manifold equipped with a holomorphic action of $SP_{2g}(\mathbb{C})$. -Choose a splitting of $V$ as $U \oplus U^{\vee}$, where $U$ and $U^{\vee}$ are complementary isotropic subspaces of $V$ and the symplectic pairing between $U$ and $U^{\vee}$ makes $U$ and $U^{\vee}$ into dual spaces. Let $\Omega \subset L$ be the open set consisting of those Lagrangians which are the graph of a map $\phi: U \to U^{\vee}$. The condition that the graph of $\phi$ be Lagrangian is equivalent to the condition that $\phi$ be self adjoint, so $\Omega$ is identified with the space of symmetric $g \times g$ complex matrices. The action of $Sp_{2g}(\mathbb{C})$ on $\Omega$ isn't fully defined, because $\Omega$ isn't $Sp_{2g}$-invariant, but when it is defined you can check that it is given by the formula $Z \mapsto (AZ+B)(CZ+D)^{-1}$. -Choose a real $2g$-dimensional symplectic vector space $V_0$ and identify $V$ with $V_0 \otimes_{\mathbb{R}} \mathbb{C}$. Let $\Omega' \subset L$ be the open set of those Lagrangian's $X$ such that $X \cap V_0 = \{ 0 \}$. Clearly, $\Omega'$ has a holmorphic action of $Sp_{2g}(\mathbb{R})$. If we identify $\Omega$ with $g \times g$ symmetric complex matrices, then $\Omega'$ is those matrices $Z$ for which $\det \Im(Z) \neq 0$. -Obviously, one connected component of $\Omega \cap \Omega'$ is $S$: Symmetric matrices for which $\Im(Z)$ is positive definite. And it turns out that this is actually a connected component of $\Omega'$, so it inherits an action of $Sp_{2g}(\mathbb{R})$. This is the action you asked about.<|endoftext|> -TITLE: What is known about the connection of positive energy representations of loop groups and modular forms -QUESTION [14 upvotes]: At the end of Section 14.1 in Pressley, Segal "Loop Groups" there is the remark that -the partition function is a modular function in the sense that the Dedekind $\eta$ function is a modular form. I am interested about the following remark: - -It follows from the Kac character formula that the characters of all positive energy - representations of loop groups are constructed from modular functions in an appropriate - sense. No explanation of this phenomenon seems to be known at present. - -So my question is basically what is the present knowledge about this "phenomenon"? - -REPLY [15 votes]: That the characters of representations have modular properties is the content of Kac-Peterson [1]. As for an "explanation" of this, there are several but I'm afraid all of the ones I know involve in one way or the other physics. In his seminal paper [2] Zhu proves modularity for Vertex Operator Algebras with certain conditions. This explains in particular the statement in your question (at the affine Lie algebra level). However, Zhu came to his result by studying the correlation functions of these algebras on the torus, so I'll try to expand minimally on the topic although you'll find a beautiful exposition in Frenkel-Ben-Zvi's book [3]. -The idea is as follows, to each vertex operator algebra $V$ (the one associated to an affine Kac-Moody Lie algebra in the case of your question) and an algebraic curve $X$ you can associate a vector space of coinvariants of $V$, $C(X,V)$. As the curve $X$ moves in the moduli space of curves of a given genus, the vector spaces $C(X,V)$ arrange to form a vector bundle on this space (it is generally a quasi-coherent sheaf, but under certain finiteness condition which holds in your case it is a bundle). This can be generalized by looking at pointed curves and modules of $V$ over the marked points of $X$. In the particular case that you look at the moduli space of elliptic curves, the sections of this bundle are the characters of the representations (where $q = e^{2 \pi i \tau}$ is the modular parameter of the corresponding curve). This is a "reason" for those characters to have modular properties, however, the fact that we expected already these spaces to form a bundle on $M_g$ comes from CFTs considerations. -[1] Kac, Victor G.(1-MIT); Peterson, Dale H.(1-MI) Infinite-dimensional Lie algebras, theta functions and modular forms. Adv. in Math. 53 (1984), no. 2, 125–264. -[2] Zhu, Yongchang(HK-HKST) Modular invariance of characters of vertex operator algebras. J. Amer. Math. Soc. 9 (1996), no. 1, 237–302. -[3] Frenkel, Edward; Ben-Zvi, David Vertex algebras and algebraic curves. (English summary) Mathematical Surveys and Monographs, 88. American Mathematical Society, Providence, RI, 2001. xii+348 pp. ISBN: 0-8218-2894-0<|endoftext|> -TITLE: Fifth powers modulo a prime -QUESTION [21 upvotes]: This is related to Victor Protsak's approach to this question. - -Suppose that $p\gt 11$ is a prime of - the form $5n+1$. Can we prove that - $1^5,2^5,\dots,n^5$ cannot be - pairwise different modulo $p$? - -I ran a quick computer search, and this is indeed the case for $p\le 5\times 42806+1=214031$. In fact, $|\{ i^k\pmod p: 1\le i\le n\}|/n$ stays rather close to 0.672... -It is not hard to answer the same question with 3 in the place of 5: There are no primes of the form $3n+1$ with $1^3,2^3,\dots,n^3$ distinct modulo $p$. Quickly, $-3$ is a quadratic residue of any $p$ of the form $3n+1$. One easily checks that (modulo $p$) there must be a $y$ such that $y^2=-3$ and either $x=y-1\ne 1$ or $x=(y-1)/2\ne 2$ is in the interval $[1,n]$. But then $(x+1)^2=-3$ so $x^3=8$, or $(2x+1)^2=-3$ so $x^3=1$. -If instead of 3 we use a number of the form $4k$, then there are only finitely many primes $p=4nk+1$ for which $1^{4k},\dots,n^{4k}$ are distinct modulo $p$ (but there may be such $p$; for example, if $4k=84$, then we can take $n=5$). This is because there are $x,y$ with $1\le x\lt y$ and $x^2+y^2=p$, so $x^{4k}\equiv y^{4k}\pmod p$, and if $p$ is slightly larger than $(4k)^2$, then $y\le n$. -(Of course one can ask the same question with any $k$ in the place of $5$, and I suspect that as long as $k>2$, the answer is always that there are only finitely many values of $n$ for which the powers are distinct. But I also suspect that this is going to be significantly harder than for $k=5$. I would be delighted for suggestions or approaches towards this more general case.) - -REPLY [14 votes]: [Edited to describe triple and higher-order coincidences for prime $k$, recovering the observed $0.672$ proportion for $k=5$] -Darij's pretty argument, extended by GH, nicely answers the question for $k$-th powers modulo a large prime $p \equiv 1 \bmod k$ for each fixed $k>2$. Yet more can be said: that approach yields the existence of one coincidence $a^k \equiv b^k$ with $0 < a < b < p/k\phantom.$; but in fact the number of coincidences is asymptotically proportional to $p$: the count is $C_k \phantom. p + O_k(p^{1-\epsilon(k)})$, where $C_k = (k-1)/(2k^2)$ or $(k-2)/(2k^2)$ according as $k$ is odd or even, and $\epsilon(k) = 1/\varphi(k) \geq 1/(k-1)$. -Extending the analysis to triple and higher-order coincidences also yields the asymptotic proportion of $k$-th powers that arise in $\lbrace a^k \phantom. \bmod p : a < p/k \rbrace$. For example, when $k$ is an odd prime, the proportion of $k$-th powers that do not have a $k$-th root in $(0,p/k)$ is asymptotic to $((k-1)^k+1)/k^k$; for $k=5$ that's $41/125$, so the proportion with such a $k$-th root is $84/125$, which matches A.Caicedo's observed $0.672$ exactly. It also gives $1 - \frac{8+1}{27} = 2/3$ for $k=3$, matching the proportion of cubes reported by Greg Martin in comments below; as $k \rightarrow \infty$ the proportion of $k$-th powers with small $k$-th roots approaches $1 - (1/e)$. -Here's how to estimate the number of pairs. Begin with the observation that $a^k = b^k$ iff $b \equiv ma \bmod p$ where $m$ is one of the $k-1$ solutions of $m^k \equiv 1 \bmod p$ other than $m=1$. If $k$ is even, we exclude also $m=-1$, which is impossible with $0 -TITLE: Lie groups acting transitively (and isometrically) on anti de Sitter spaces -QUESTION [7 upvotes]: I hope this question is not deemed too localised. -Recall that anti de Sitter space is the lorentzian analogue of hyperbolic space; that is, a simply-connected lorentzian manifold of constant negative sectional curvature. More precisely, the $n$-dimensional anti de Sitter space $\mathrm{AdS}_n$ is the universal cover of the one-sheeted hyperboloid in $\mathbb{R}^{n+1}$ cut out by the following quadric: -$$ -x_1^2 + x_2^2 + \dots + x_{n-1}^2 - x_n^2 - x_{n+1}^2 = - R^2 -$$ -where $R>0$ is the radius of curvature. The ambient metric is flat with signature $(n-1,2)$ and the induced metric on the hyperboloid has constant negative sectional curvature proportional to $1/R$. -The group $\mathrm{SO}(n-1,2)$ acts transitively and isometric on the hyperboloid; indeed, the hyperboloid is diffeomorphic to $\mathrm{SO}(n-1,2)/\mathrm{SO}(n-1,1)$. -This is analogous to the better known fact that the round $n$-sphere is diffeomorphic to $\mathrm{SO}(n+1)/\mathrm{SO}(n)$. However, it happens that for some spheres, a proper subgroup $G \subset \mathrm{SO}(n+1)$ already acts transitively on the sphere and in some cases such $G$ is (locally) isomorphic to $\mathrm{SO}(p)$ for some $p\lt n+1$. A case in point is the round 7-sphere, on which $\mathrm{Spin}(6)$ acts transitively with stabiliser subgroup $\mathrm{SU}(3)$. -I am interested in knowing whether something like that can happen for anti de Sitter space. In particular, I would like to know whether $\mathrm{SO}(n-1,2)$ (or some other group with isomorphic Lie algebra) can act transitively on $\mathrm{AdS}_p$ for some $p\gt n$. We already know it acts on $\mathrm{AdS}_n$ and it cannot act (effectively) on $\mathrm{AdS}_p$ for $p\lt n$ by dimension. As in the case of the spheres, this can perhaps only happen for low values of $n$ and in fact, to be perfectly honest, I am presently only interested in $n=4$. -Hence let me ask the following -Question -Can $\mathrm{SO}(n-1,2)$ act locally transitively (and isometrically!) on $\mathrm{AdS}_p$ for some $p\gt n$? In particular, can this happen for $n=4$? -Thank you in advance. -Epilogue -I should have mentioned that I have checked by explicit calculation that $\mathrm{SO}(3,2)$ cannot act locally transitively and isometrically on $\mathrm{AdS}_5$, but before trying my hand at $\mathrm{AdS}_6$ or higher, I thought of asking here first. - -REPLY [2 votes]: In my paper -Isometry groups of Lobachevskian spaces, similarity transformation groups of Euclidean spaces and Lorentzian holonomy groups. Rend. Circ. Mat. Palermo (2) Suppl. No. 79 (2006), 87–97. -I classify transitive group of isometries of the real Hyperbolic space. -These groups are $SO(n,1)$ and subgroups of -$$Sim(n-1)=(\mathbb{R}_+\times SO(n-1))\times \mathbb{R}^{n-1}$$ -that preserve an isotropic line in the Minkowski space and do not preserve any non-degenerate subspace of the Minkowski space (weakly-irreducible subgroups), in addition they have non-trivial projection on $\mathbb{R}_+$. -The connected group of isometries of AdS_n is $SO(n-1,2)$, hence, if a connected $G$ acts isometrically, it is contained in $SO(n-1,2)$. From the transitivity it follows that the representation is weakly irreducible. Some of weakly irreducible subgroups of $SO(n-1,2)$ can be found in -Ikemakhen, A. Sur l'holonomie des variétés pseudo-riemanniennes de signature (2,2+n). (French) [Holonomy of pseudo-Riemannian manifolds with signature (2,2+n)] Publ. Mat. 43 (1999), no. 1, 55–84. -I think that it is possible to classify all transitively acting subgroups.<|endoftext|> -TITLE: Automorphisms of Riemann surface and mapping class -QUESTION [8 upvotes]: For a higher genus Riemann surface $\Sigma$, is it true that every nontrivial (holomorphic) automorphism is of nontrivial mapping class, i.e., not isotopic to the identity? - -REPLY [14 votes]: If a surface-diffeomorphism $h$ acts trivially on rational cohomology, the Lefschetz number of $h$ is equal to the Euler characteristic of the surface $\Sigma$. By the Lefschetz fixed-point formula, this number equals the intersection number, in $\Sigma \times \Sigma$, of the graph of the automorphism with the diagonal. -In the case that $h$ is a non-trivial holomorphic automorphism, the intersections are isolated and the intersection multiplicities positive. This can occur only when the Euler characteristic is non-negative. - -REPLY [12 votes]: A much more general fact is true: any isometry of any closed negatively curved Riemannian manifold is not homotopic to the identity. There are many proofs of this; one (perhaps not the most natural) is as follows. Hartman proved that if two harmonic maps $f_0,f_1\colon M\to N$ are homotopic, where $M$ is compact and $N$ is nonpositively curved, then $f_0$ and $f_1$ are the boundary of an isometrically embedded product $f\colon M\times [0,1]\to N$. If $M=N$ is negatively curved, this is impossible (such a product region would have sectional curvature 0 in certain directions), so we conclude that no nontrivial isometry is homotopic to the identity. -To apply this to your example, note that by uniformization the universal cover of the Riemann surface $C$ is the unit disk $\Delta$. Thus your curve $C$ is the quotient $\Delta/\Gamma$ by some group $\Gamma$ of biholomorphic automorphisms of $\Delta$, namely Aut($\Delta$)=PSL$_2(\mathbb{R})$. Now note that the action of PSL$_2(\mathbb{R})$ on $\Delta$ preserves the Poincaré metric (of constant curvature $-1$), which thus descends to a metric of constant negative curvature on $C$. The resulting metric is preserved by any automorphism of $C$ as a Riemann surface, so in particular your original map is an isometry in this metric.<|endoftext|> -TITLE: Is $SU(\infty)$ amenable? -QUESTION [8 upvotes]: We can write the finitary special unitary group $SU(\infty)$ as the direct limit -$\varinjlim SU(n)$ of ordinary special unitary groups. These groups $SU(n)$ are compact, thus amenable. In other words each of them has an invariant mean (in this case, from Haar measure). Is $SU(\infty)$ amenable? Of course one then asks the same question for the full unitary group $U(\infty)$, the special orthogonal group $SO(\infty)$ and the unitary symplectic (quaternion unitary) group $Sp(\infty)$. - -REPLY [13 votes]: The answer is that $G=SU(\infty)$ (with the direct limit topology of the usual Hilbert-Schmidt topologies) is extremely amenable. This means (by definition) that every continuous action of $G$ on a compact set has a fixed point. -This was proved as an application of the isoperimetric inequality by Gromov and Milman -M. Gromov and V.D. Milman, A topological application of the isoperimetric inequality, Amer. J. Math. 105 (1983), 843–854. -Since $SU(\infty)$ is not locally compact, various characterizations of amenability have to be adapted. One way to define amenability for such groups is to say that there exists a $G$-invariant mean on the algebra of bounded uniformly continuous real-valued functions on $G$. Extreme amenability then gives even the existence of a $G$-invariant character on this algebra. This is somewhat unintuitive, since obviously no compact group can admit such a character on the algebra of continuous functions on it. -Extreme amenability is a concept which is related to phenomena of measure concentration. Gromov and Milman proved (as an application of lower bounds on the Ricci curvature) that for every sequence of measurable subsets $A_n \subset SU(n)$ with $\liminf_{n \to \infty} \mu_n(A_n) \neq 0$, one has $$\lim_{n \to \infty} \mu_n(A_{n,\epsilon}) = 1.$$ Here, $\mu_n$ denotes the normalized Haar measure and $A_{n,\varepsilon}$ the $\varepsilon$-neighborhood of $A_n$ in the unnormalized Hilbert-Schmidt metric. This concentration phenomenon can be used to prove extreme amenability of $SU(\infty)$.<|endoftext|> -TITLE: Examples of Tate cohomology rings -QUESTION [11 upvotes]: If $G$ is a finite group with periodic cohomology then the Tate cohomology ring can be easily computed to be the localization $\hat{H}^\ast(G,\mathbb{Z}) = H^\ast(G,\mathbb{Z})_{(z)}$ where $z$ is a unit of minimal positive degree. Examples are - -Cyclic group: -$\hat{H}^\ast(C_n,\mathbb{Z}) = \mathbb{Z}[z,z^{-1}]/(nz), |z|=2$ -Quaternion group: -$\hat{H}^\ast(Q_{2^n},\mathbb{Z}) = \mathbb{Z}[z,z^{-1},a,b]/(2^nz, 2a,2b), |z| =4, |a| = |b| =2$ -Binary icosahedral group: -$\hat{H}^\ast(I,\mathbb{Z}) = \mathbb{Z}[z,z^{-1}]/(120z), |z|=4$ - -The same formular holds in mod-$p$ cohomology, if the cohomology of $G$ is $p$-periodic. -Question: Are there computations of integral or mod-$p$ Tate cohomology rings of finite groups with non-periodic cohomology in the literature ? - -REPLY [7 votes]: Bellezza's thesis is here -http://www.abdn.ac.uk/~mth192/html/archive/bellezza.html<|endoftext|> -TITLE: Rhombus tilings with more than three directions -QUESTION [20 upvotes]: The point of this question is to construct a list of references on the following subject: Fix vectors $v_1$, $v_2$, ..., $v_g$ in $\mathbb{R}^2$, all lying in a half plane in that cyclic order. I am interested in tilings by parallelograms of the form $\mathrm{Hull}(z, z+v_i, z+v_i+v_j, z+v_j)$. It is traditional in this subject to take all the $v_i$ of the same length, so that the tiles are rhombi. In my personal opinion, this convention is unmotivated, but I will obey convention and speak of rhombus tilings in the rest of this post. -The most obvious region to tile is a centrally symmetric $2g$-gon, with edges of the form $a_1 v_1$, $a_2 v_2$, ..., $a_g v_g$, $-a_1 v_1$, ..., $- a_g v_g$ for some positive integers $a_i$. One might also study tilings of the whole plane, perhaps with some periodicity conditions. I'm also open to hearing about results on regions of other shapes. -What I do want to do is rule out results which are special to the case $g=3$. Rhombus tilings -of a centrally symmetric hexagon are in bijection with plane partitions, and hence all the tools of the plane partition literature are available. They are also related to perfect matchings of the hexagonal grid, so they can be studied by Kastelyn's method and related tools. They are also in bijection with certain families of non-crossing paths, so they can be studied by the Gessel-Viennot method. There is an embarrassment of tools available for the $g=3$ case. I want to limit this discussion only to tools that work for higher $g$. -Below the line is what I know so far: - -$\bullet$ The two most obvious special cases are (1) tilings of an octagon where $(a_1, a_2, a_3, a_4) = (k,k,k,k)$ and (2) tilings of a $2g$-gon with $a_1=a_2=\cdots=a_g=1$. These are counted by Sloane A093937 and A006245 respectively. There is no closed formula in either case. -$\bullet$ Such rhombus tilings are in bijection with reduced words in $S_{a_1+a_2+\cdots+a_g}$ modulo commutation. In particular, the case where $a_1=a_2=\cdots=a_g=1$ corresponds to reduced words for the long word in $S_g$, modulo commutation. This has been rediscovered several times, but the canonical reference seems to be Rhombic tilings of polygons and classes of reduced words in Coxeter groups, by Serge Elnitsky. -$\bullet$ Drawing the paths dual to the rhombi gives us an arrangement of pseudo-lines, sometimes called deBruijn paths. -$\bullet$ Pseudo-line arrangements can be described by oriented matroids. Tracing through this language, the $a_1=a_2=\cdots=a_g=1$ case corresponds to rank $3$ oriented matroids $M$ on the set $\{ 1,2,\ldots, g, \infty \}$ such that the underlying matroid is uniform and $M/\infty$ is the uniform rank $2$ oriented matroid where $\{1,2,\ldots,g \}$ appear in that cyclic order. Call that rank $2$ matroid $U(g)$. Dualizing, we are looking for extensions of the rank $g-2$ oriented matroid $U(g)^{\ast}$ whose underlying matroid is uniform. One can generalize this description to the general case: The matroid $U(g)$ becomes the oriented matroid of rank $2$ made up of $g$ parallelism classes, of sizes $a_1$, ..., $a_g$, in that cyclic order. -Extensions of oriented matroids were studied by Sturmfels and Ziegler. In particular, they proved that a certain poset associated to this extension problem is homotopic to a sphere. -$\bullet$ Let $A$ and $B$ be two subsets of $\{ 1,2,\ldots, g \}$. We say that $A$ and $B$ are strongly separated if either every element of $A \setminus B$ is less than every element of $B \setminus A$, or is every element of $B \setminus A$ is less than every element of $A \setminus B$. A collection $\mathcal{C}$ of subsets of $\{ 1,2,\ldots, g \}$ is called strongly separated if every two elements in the collection are strongly separated. A collection is called maximally strongly separated if it is strongly separated and is not contained in any larger strongly separated collection. -Maximal strongly separated sets are in bijection with rhombus tilings of the $(1,1,\ldots,1)$-gon by a result of Leclerc and Zelevinsky. Although it doesn't appear to be in the literature, the generalization to other centrally symmetric polygons is easy enough: One uses the multiset with $a_i$ copies of $i$. Here if $A$ and $B$ are multisets which contain $u$ and $v$ copies of $i$, then $A \setminus B$ is the multiset which contains $\max(u-v, 0)$ copies of $i$. -$\bullet$ Any two rhombus tilings of the same $2g$-gon are linked by a sequence of flips, where a flip takes a single hexagon and retiles it. This is the same as the fact that any two reduced words are (up to commutation) joined by a sequence of braid moves. -$\bullet$ Take the flip graph referred to in the previous bullet point and direct it so that, in the language of braid moves, an edge points from $\cdots s_i s_{i+1} s_i \cdots$ to $\cdots s_{i+1} s_i s_{i+1} \cdots$. The resulting graph is acyclic, and is the Hasse diagram of the poset obtained by taking the transitive closure. -This poset is graded and has unique minimum and maximum elements. When $a_1=a_2=\cdots=a_g=1$, it is called the higher Bruhat order $B(g,2)$. It is a lattice for small $g$, but $B(6,2)$ is not a lattice. -$\bullet$ Let $1 \leq i < j < k \leq n$ and let $\gamma_i$, $\gamma_j$ and $\gamma_k$ be pseudolines in directions $(i,j,k)$. Then they either intersect each other in the order $(i,j)$, $(i,k)$, $(j,k)$, or the reverse of this order. A triple of such pseudo-lines is called a triangle. The triangle is ordinary if the intersections occur in the first manner, and inverted if the occur in the second. -For $\gamma$ and $\delta$ in $B(g,2)$, we have $\gamma \leq \delta$ if and only if every triangle which is inverted in $\gamma$ is also inverted in $\delta$. Presumably, this is also true when some of the sides of the polygon are longer than $1$, but I don't know a reference. -For $\gamma$ and $\delta$ two pseudo-line arrangements, the flip distance from $\gamma$ to $\delta$ is bounded below by the number of triangles which are inverted in one of $\gamma$ and $\delta$ but not the other. This lower bound is exact for $g=3$ or $4$, for $g=5$ if one of the $a_i$ is $1$, and for $(2,2,2,2,2)$, but not for any other case. -$\bullet$ The Bohne-Dress theorem: Let $C$ be the cube $\{ (x_1, \ldots, x_g) : 0 \leq x_i \leq a_i \} \subset \mathbb{R}^g$. Map $C$ to $\mathbb{R}^2$ by the linear map $\pi$ sending the $i$-th basis vector to $v_i$. So the $2g$-gon is the image of $\pi$. Then rhombus tilings of $\pi(C)$ are in bijection with continuous sections $\sigma: \pi(C) \to C$ whose image lies in the $2$-skeleton of the obvious cubical complex. -$\bullet$ Miscellaneous: some interesting lemmas about rhombus tilings are proved in my paper with Andre Henriques and in Shapiro-Shapiro-Vainshtein. - -Some things I particularly don't know: Do we know the approximate number of these tilings? Can we sample uniformly from this set? Has anyone worked on periodic rhombus tilings? (Surely someone has, but I haven't found them.) What do random tilings of a large octagon look like? (Even if nothing is proven, someone presumably has a guess.) Is there a way of comparing elements of $B(g,2)$ in time less than order of $g^3$? - -REPLY [7 votes]: This post is almost ten years old, but I find it interesting and I would like to add a few pointers from my PhD, itself about 20 years old. -First, in the paper Integer Partitions, Tilings of 2D-gons and Lattices - we explored the relations between rhombus tilings and integer partitions, and obtained a decomposition into disjoint distributive lattices of the flip graph of any 2D-gon tilings. In addition, the quotient of this set by these lattices has itself the same structure. These are direct consequences of the fact that they are isomorphic to generalized integer partitions. -Here is a simple example for the flip graph of an elemenraty octogon: -                          -Second, in the paper Graph encoding of 2D-gon tilings, we used the tile adjacency graph to process rhombus tilings and sample them, like the random decagon tiling below (the full resolution vector file is available on demand):<|endoftext|> -TITLE: Alexander duality theorem for CW-complexes and stable homotopy theory -QUESTION [5 upvotes]: In Adams, J.F. Infinite Loop Spaces Princ. Univ. Press. page 9 he states Alexander duality theorem -Theorem:[Alexander Duality] $$ H^r(X,G)=H_{n-r+1}(S^n-X,G)$$ -for finite CW-complexes with a "nice embedding". That is to say, $S^n-X$ has a CW-complex $Y$ as deformation retract and $X$ is a deformation retract of $S^n-Y$. -My first question is the following: Do you know if every CW-complex admits a nice embedding into some $S^n$? if so, could you give a reference? -Thanks in advance. - -REPLY [5 votes]: Let $X$ be a finite CW-complex. Then there is an embedding $X\to\mathbb{R}^n$ for some $n$. To show this it suffices to consider the case when $X$ is obtained from a CW-complex $Y\subset B^m\subset \mathbb{R}^m$ by attaching a cell: $X=Y\cup_f B^k$ where $B^m$ is the unit ball in $\mathbb{R}^m$ and $f:S^{k-1}\to Y$ is the attaching map. -Let us show that there is an embedding $i:B^m\sqcup B^k\to \mathbb{R}^n$ for some $n$ so that, given $x',x''\in B^m$ and $y',y''\in B^k$, the segments $[i(x'),i(y')]$ and $[i(x''),i(y'')]$ do not intersect unless $x=x''$ or $y'=y''$. Indeed, let's assume both balls live in some $\mathbb{R}^r$ and take $\mathbb{R}^n$ to be the space of polynomials $\mathbb{R}^r\to\mathbb{R}$ of sufficiently high degree (4 will do). Then for any 4 points in $\mathbb{R}^r$ the conditions that an element of $V$ are zero at those points are linearly independent. So for any $x\in B^m\sqcup B^k$ take $i(x)\in V^*$ to be the evaluation function $p\mapsto p(x)$. -Now set $X'$ to be the union of $i(Y), i(B^k)$ and all segments $[i(x),i(f(x))]$ for $x\in S^{k-1}$. there is a natural bijective continuous map $X\to X'$, which is a homeomorphism since $X$ is compact and $\mathbb{R}^n$ is Hausdorff. -Remarks: - -From this construction it is clear that one can construct an embedding of $X$ that is smooth on the interior of each cell. So by slightly modifying the proof of Whitney's embedding theorem we see that $X$ can be embedded in $\mathbb{R}^{2\dim X+1}$. However, the strong Whitney theorem does not hold. E.g. there are 2-polyhedra that can't be embedded in $\mathbb{R}^4$, see e.g. theorem 3 in this paper http://www.fmf.uni-lj.si/~repovs/clanki/2001/OnReSk01.pdf -If $X$ is not assumed finite then I would guess that a necessary and sufficient condition for $X$ to be properly embeddable in some $\mathbb{R}^n$ is that $X$ should be locally finite and should have at most countably many cells.<|endoftext|> -TITLE: bounding the trace of a matrix product by the operator norms; generalized Hölder inequality? -QUESTION [9 upvotes]: Let $A_i$ with $i=1,\dots,N$ and $p$ be real $M\times M$ matrices. -Further, let $p$ be positive definite, i.e., $p\succ 0$, with $Tr(p)=1$. Let $0< a_i<1$ and $\sum_{i=1}^N a_i = 1$. -Claim: -$Tr( A_1 p^{a_1} A_2 p^{a_2} \dots A_N p^{a_N} ) \leq \|A_1\| \|A_2\|\dots \|A_N\|$ -Here, \|X\| denotes the operator norm of $X$ (= largest singular value). -Can you prove this claim (at least for symmetric matrices A_i)? It is trivial for N=1: $Tr(A p)=\sum_i A_{i,i} p_i\leq \|A\|\sum_i p_i = \|A\|$, where I used the representation of $A$ in the eigenbasis of $p$. -The claim even seems to hold if one uses N different matrices $p_i\succ 0$, $Tr(p_i)=1$ on the left-hand side of the claim. -thanks a lot, Tom - -REPLY [7 votes]: I include some information about Hölder's inequality just for completion of details for Mikael's nice answer. -The Schatten-$p$ norm of a matrix $X$ is defined as -$$\|X\|_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$ -where $\sigma_i(X)$ denotes the $i$-th singular value of $X$. -From this definition, we see that $\|X\|_p = \|\sigma(X)\|_p$, where $\sigma(X)$ is the vector of singular values. -From von Neumann's trace inequality we know that -$$|\mbox{trace}(XY)| \le \langle \sigma(X), \sigma(Y)\rangle,$$ -while from the Hölder inequality for vectors, we have -$$\langle \sigma(X), \sigma(Y)\rangle \le \|\sigma(X)\|_p\|\sigma(Y)\|_q,$$ -where $1/p + 1/q = 1$. -On combining the above two, we immediately get alleged Schatten-p norm Hölder inequality.<|endoftext|> -TITLE: Five-lemma for the end of long exact sequences of homotopy groups -QUESTION [13 upvotes]: Consider the commutative diagram below with exact rows (from the long exact sequence of homotopy groups) and $f_1,f_2,f_4,f_5$ bijective ($f_1,f_2$ homomorphisms). Does it follow that $f_3$ is also bijective? -\begin{align} -\matrix{ -\pi_1(A)&\to&\pi_1(X)&\to&\pi_1(X,A)&\to&\pi_0(A)&\to&\pi_0(X) -\cr \downarrow f_1&&\downarrow f_2&&\downarrow f_3&&\downarrow f_4&&\downarrow f_5 -\cr\pi_1(B)&\to&\pi_1(Y)&\to&\pi_1(Y,B)&\to&\pi_0(B)&\to&\pi_0(Y)} -\end{align} -The usual proof of the five-lemma by diagram chasing makes use of the fact that the consituents are groups and all maps involved are homomorphisms. Since there is no group structure for the six sets on the right ($\pi_0$ and relative $\pi_1$), it does not apply. -However, with arguments about connected components and loops/paths, it is easily shown in an analogous diagram-chasing fashion that $f_3$ has to be surjective. The tricky part seems to be the injectivity (if it works at all): In analogy to the usual five-lemma, one can show that $f_3[x]=0\Rightarrow[x]=0$, but of course this does not immediately give injectivity due to $f_3$ only being a map between sets. -Is there a way to show injectivity? If not, is there a counterexample where this adapted five-lemma fails? -Thanks! - -REPLY [9 votes]: The abstract case is handled in -R. Brown, ``Fibrations of groupoids'', J. Algebra 15 (1970) 103-132. -where a 5-lemma type result is Theorem 4.9. I find it easier to handle the abstract case rather than the topological example. You get injectivity on Ker( $\pi_1(X, A) \to \pi_0 A)$ and an Example 4.10 (for groupoids) shows you do not always get injectivity. -There are other applications of the exact sequences of a fibration of groupoids, see -Heath, P.R. and Kamps, K.H. "On exact orbit sequences", Illinois J. Math. 26~(1) (1982) 149--154. -One aim of all editions (from 1968) of the book Topology and Groupoids is to work in the algebraic context of groupoids where appropriate, to see which results are really on groupoids.<|endoftext|> -TITLE: residually finite-by-$\mathbb{Z}$ groups are residually finite -QUESTION [10 upvotes]: I believe I read somewhere that residually finite-by-$\mathbb{Z}$ groups are residually finite. That is, if $N$ is residually finite with $G/N\cong \mathbb{Z}$ then $G$ is residually finite. -However, I cannot remember where I read this, and nor can I find another place which says it. I was therefore wondering if someone could confirm whether this is true or not, and if it is give either a proof or a reference for this result? (If not, a counter-example would not go amiss!) -Note that I definitely know it is true if $N$ is f.g. free (this can be found in a paper of G. Baumslag, "Finitely generated cyclic extensions of free groups are residually finite" (Bull. Amer. Math. Soc., 5, 87-94, 1971)). - -REPLY [9 votes]: I think what Yves meant in his comment to Andreas' answer is the result of Mal'cev [A. I. Mal'cev, On homomorphisms onto finite groups, Ivanov. Gos. Ped. Inst. Ucen. Zap., 18 (1958), pp. 49-60, in Russian] stating that any split extension of a finitely generated residually finite group by a residually finite group is residually finite. This is indeed an easy exercise. In particular, this implies that every (f.g. residually finite) - by - Z group is residually finite as every extension by a free group (e.g., by Z) splits. -If either of the conditions "finitely generated" or "split" is relaxed, the result does not hold. The counterexample with infinitely generated kernel is already provided by Yves. For non-split extensions there are even examples of the form $$1\to \mathbb Z/2\mathbb Z \to G\to Q\to 1,$$ where $Q$ is (finitely generated) residually finite while $G$ is not. In particular, being residually finite is not a quasi-isometry invariant. -Such examples can be found among solvable (moreover, center-by-metabelian) groups or even among groups of intermediate growth (see [A. Erschler, Not residually finite groups of intermediate growth, commensurability and non-geometricity, Journal of Algebra, 272 (2004), 154-172].<|endoftext|> -TITLE: Radical of $F_p[SL(2,p)]$ -QUESTION [8 upvotes]: Let $G=SL(2,p)$. Does anyone know what is the radical of the group algebra $F_p[G]$? -Does there exists any book/paper where it is calculated? -By radical here I mean maximal ideal I of $F_p[G]$ such that $I^n=0$ - -REPLY [8 votes]: The answer depends a lot on what kind of description of the radical you ask for. This family of groups of Lie type has been well-studied from the viewpoint of modular representation theory in the defining characteristic (with reference also to the ambient algebraic groups). Even the somewhat degenerate case $p=2$ fits well enough into the general pattern for odd primes. -It's easy to work out explicitly the $p$ irreducible modular representations, for instance using the characteristic 0 model of spaces of homogeneous polynomials in two variables having degree $ -TITLE: Is there something like a Heyting Ring? -QUESTION [7 upvotes]: I would like to know whether a Heyting algebra gives rise to ring in a similar way that a Boolean algebra gives rise to a Boolean ring. In a Boolean algebra $(B,\lor,\land,\lnot,0,1)$ I can define multiplication and addition as follows: -$$a * b = a \land b$$ -$$a + b = (a \land \lnot b) \lor (b \land \lnot a)$$ -And I get a Boolean ring $(B,+,*,0,1)$. From the Boolean ring we can reconstruct the Boolean algebra again via the following definitions (right?): -$$a \land b = a * b$$ -$$a \lor b = (a + 1) * (b + 1) + 1$$ -In a Heyting algebra $(H,\rightarrow,\land,0,1)$ we do only have a pseudo complement. Can we nevertheless define a ring, that in turn would allow us to reconstruct the algebra? -Best Regards - -REPLY [2 votes]: Though not directly analogous to your example of a boolean ring, a 2010 paper in Algebra Universalis approaches the problem of classifying linearly ordered Heyting Algebras via the concept of a Godel Ring. The idea is that the semiring of ideals of a suitable ring should form a Heyting Algebra. -For example, the semiring of ideals of every Von Neumann Regular Ring form a Heyting Algebra; furthermore, if the ring in question is commutative, this Heyting Algebra is in fact a Boolean Algebra.<|endoftext|> -TITLE: which integers take the form $x^2 + xy + y^2$? -QUESTION [7 upvotes]: I guess one way of putting it, when does the series $\sum_{x,y \in \mathbb{Z}} q^{x^2+xy+y^2}$ have nonzero coefficients? -The analogous answer for $\sum_{x,y \in \mathbb{Z}} q^{x^2+y^2}$ is that $q^n$ appears when $\mathrm{ord}_p(n)$ be even for all primes. -Is there a closed form for either of these with quadratic characters or theta functions or something?n - -REPLY [5 votes]: One can start showing the following: -The integers $n$ which are of the form $x^2+xy+y^2$, for two relatively prime integers $x,y$ are precisely those positive integers occurring as divisors of $m^2+m+1$, for some integer $m$. -In other words, the polynomial $f(x)=x^2+x+1$ has the property that the positive divisors of the integers it represents are precisely those integers that can be properly represented by its homogenization $F(x,y)=x^2+xy+y^2$ ("properly" here refers to the condition $(x,y)=1$). -The proof uses the fact that the imaginary quadratic order $\mathbf{Z}[x]/f(x)$ has class number one, as anticipated by Elkies. It goes as follows: -Let $n$ be a positive divisor of $m^2+m+1$, for some integer $m$. -Consider the quadratic form in $x,y$ given by: -$Q(x,y)=\frac{m^2+m+1}{n}x^2-(2m+1)xy+ny^2$; -it has integer coefficients, positive definite, and has discriminant equal to $-3$ (in particular it is primitive). -Since $h(-3)=1$, there is only one reduced, positive definite quadratic form of discriminant $-3$. This is $E(x,y)=x^2+xy+y^2$. Therefore $Q$ and $E$ are properly equivalent (that is there is a determinant-one change of variables taking one into the other), and since $Q$ certainly properly represents $n$, so does $E$. -The converse is similar and uses the fact that if a quadratic form $Q$ properly represents an integer $n$, then $Q$ is properly equivalent to a form of the type $nx^2+bxy+cy^2$ (this is lemma 2.3 of Cox's wonderful book "Primes of the form x^2+ny^2"). -Once you understand the positive integers $n$ that are properly represented by $E$, then you can get them all, after scaling by squares. -The original problem is then reduced to understanding those integers $n$ for which $x^2+x+1$ has a zero in $\mathbf{Z}/(n)$. Using the Chinese Remainder Theorem, this can be reduced to the case where $n$ is a prime power $p^s$. Then for $p\neq 3$ Hensel's Lemma tells you that your equation has a solution mod $p^s$ if and only if it has a solution mod $p$. With quadratic reciprocity you can conclude that any prime divisor of $n$ has to be congruent to $1$ mod $3$. I am at the moment missing how you can solve the equation $x^2+x+1=0$ in $\mathbf{Z}/3^s$, but I think that a version of Hensel's Lemma applies. I'll think about it. -[EDIT: The equation $x^2+x+1=0$ has a solution in $\mathbf{Z}/(3^s)$, with $s\geq 1$, if and only if $s=1$. (Therefore $x^2+xy+y^2$ represents properly only $3$ and $1$ as powers of $3$.) One can see this by checking that there is no solution for $s=2$, and therefore for $s>2$. More generally, if $p>2$ then $x^{p-1}+x^{p-2}+\ldots+x+1=0$ has no solutions in $\mathbf{Z}/(p^s)$ with $s>1$. The $p$--adic valuation of an integer of the form $x^{p-1}+x^{p-2}+\ldots+x+1=(x^p-1)/(x-1)$ is either zero or one.] -(Suggested reading. Cox's book quoted above and Serre's paper: $\Delta=b^2-4ac$)<|endoftext|> -TITLE: Koethe conjecture -QUESTION [7 upvotes]: First of all, has the Koethe conjecture been solved ?! Does a proof of the following statement really imply the proof of the conjecture ? - -The sum of two right nil ideals in any ring is nil. - -Are there other equivalent statements that imply the conjecture ? -Thanks, SB - -REPLY [3 votes]: Not solved. Many special cases though. The paper cited above looks nice, and another short survey is given in Lam's First Course in Noncommutative rings, around page 164. There are a bunch of equivalent statements there, but maybe the above paper covers them all.<|endoftext|> -TITLE: Decomposition of a complete graph into maximal matching subgraphs -QUESTION [5 upvotes]: Is there a general way to decompose a complete graph $K_n$ into an union of maximal matching subgraphs such that no two subgraphs share an edge? -For example, consider $K_4$ with vertices $V=${1,2,3,4}. It can be decomposed into the union: -$K_4=$ {{1,2},{3,4}} $\cup$ {{1,3},{2,4}} $\cup$ {{1,4},{2,3}} -In this example, none of the 3 subgraphs share an edge. For $n$ odd, I could easily find a general decomposition of $K_n$ into $n$ "independent" maximal matching subgraphs: -$K_n= \bigcup_\sigma$ { {$\sigma(1),\sigma(n-1)$}, {$\sigma(2),\sigma(n-2)$}, ... } -where the union is over all cyclic permutations $\sigma$ of the vertex set $V=${1,2,...,$n$}. For $n$ even, however, I can't find a general solution for this decomposition problem, which would involve the union of $n-1$ subgraphs. I can solve it for particular examples, like $K_4$, $K_6$ and $K_8$, but not for the general $n$ even case. - -REPLY [10 votes]: Let the $2n$ vertices be the $2n-1$ vertices of a regular polygon and its center. Join the center to one of the other vertices by a line segment $S$, then join the remaining $2n-2$ vertices in pairs by line segments perpendicular to $S$. That's one maximal matching subgraph. Now rotate the vertices around the center through $2\pi k/n$, $k=1,2,\dots,$, while keeping the line segments where they are. That will give you the rest of the maximal matching subgraphs which together decompose $K_{2n}$.<|endoftext|> -TITLE: Is this Ramsey-type problem an open problem? -QUESTION [9 upvotes]: A blog claims that the following Ramsey-type (or van der Waerden type) problem is open: -If the natural numbers are colored with finitely many colors, must there exist x and y (not both 2) such that x+y and xy are the same color? -Is it correct that this is an open problem, and can anybody help me track down a reference? - -REPLY [17 votes]: Joel Moreira has now proved that we always get monochromatic sets of the shape $\{x, xy, x+y\}$. The main result is much more general and uses a connection with topological dynamics, but he includes a beautifully simple direct proof of this special case, assuming only an easy to check consequence of van der Waerden's theorem.<|endoftext|> -TITLE: Markov Property: determined by just the law or also the realization? -QUESTION [6 upvotes]: When one says that a stochastic process is Markovian, is this a property solely of the law of the process, or does the realization of the process also come in to play? I am asking even for the simplest examples, such as a process indexed by $\mathbb{N}$. Most abstract definitions are about being Markov with respect to some filtration, which indicates that it has to do with the realization also. -My guess is that by knowing only the law one can determine if there exists a version of the process that is Markovian, but that given a realization one cannot determine if it is Markovian solely by checking its law. If the latter case is true does anyone know of simple examples? -Edit: when I say ``realization'' I mean a collection of random variables with the given law. I do not mean the value of the random variables at a given point. So the question could be rephrased as: "Can one construct a collection of random variables that has the law of a Markov process, but such that the collection itself does not form a Markov process?" - -REPLY [4 votes]: I've posted a solution on my webpage.<|endoftext|> -TITLE: Do we still need model categories? -QUESTION [76 upvotes]: One modern POV on model categories is that they are presentations of $(\infty, 1)$-categories (namely, given a model category, you obtain an $\infty$-category by localizing at the category of weak equivalences; better, if you have a simplicial model category, is to take the homotopy coherent nerve of the fibrant-cofibrant objects). -What other functions, then, do model categories serve today? I understand that getting the theory of $\infty$-categories off the ground (as in HTT, for instance) requires a significant use of a plethora of model structures. However, if we assume that there exists a good model of $(\infty, 1)$-categories that satisfies the properties we want (e.g. that mapping sets are replaced with mapping spaces, limits and colimits are determined by homotopy limits of spaces), how are model categories useful? -I suppose one example would be computing hom-spaces: a simplicial model category gives you a nice way of finding the mapping space between two objects in the associated localization. However, in practice one only considers cofibrant or fibrant objects in the $\infty$-category in the first place, as in Lurie's construction of the derived $\infty$-category (basically, one considers the category of projective complexes -- for the bounded-above case, anyway -- and makes that into a simplicial category, and then takes the homotopy coherent nerve). -One example where having a model structure seems to buy something is the theorem that $E_\infty$ ring spectra can be modeled by 1-categorical commutative algebras in an appropriate monoidal model category of spectra (in DAG 2 there is a general result to this effect), and that you can straighten things out to avoid coherence homotopies. I don't really know anything about $E_\infty$-ring spectra, but I'm not sure how helpful this is when one has a good theory of monoidal objects in $\infty$-categories. - -REPLY [23 votes]: I confess to being confused by all this $(\infty,1)$ category business, and the way $\Pi X$ is used as another name for the singular simplicial set of $X$. This is related to Peter's question on computations. -I thought one reason for moving from loops or paths to fundamental groups or fundamental groupoids, i.e. taking homotopy classes, was that one could do specific computations in groups, and also groupoids. So I began in the 1960s to look for higher dimensional versions of these groupoid methods, again with the aim, or hope, of higher dimensional nonabelian calculations. Of course we were well aware of all the laws on paths, or singular simplices, up to homotopies, e.g. Kan extension conditions, but it seemed difficult to get computational information directly at the path space or singular complex level. -What was surprising, and took a long time to realise, was that we could do these higher dimensional strict groupoid methods, using certain homotopy classes, for certain structured spaces, particularly filtered spaces (11 years), and later $n$-cubes of spaces (Loday) (17 years). In the filtered space work, the insights of model theory have also proved very useful - I think they have not yet been used in the $n$-cube situation. Grothendieck was amazed when I told him in 1985 (6?) that $n$-fold groupoids model homotopy $n$-types (Loday's theorem). Since we can use this idea for specific nonabelian colimit calculations in homotopy theory with the aid of a Higher van Kampen type theorem, I am happy as an old man to rest with the use of strict multiple groupoids of various kinds suitable for the problem at hand. Of course in the proofs, the relations between the weak (spaces of maps) situation and the strict one is crucial. -I see these ideas as another contribution to the tool kit of algebraic topology, and some younger people are using them. -It seems a useful, but not obligatory, test of a theory to ask if it can in some cases produce some numbers not previously available. -February 25, 2015 Here is a link to a talk in Galway, December, 2014, giving further background to this answer, particularly in relation to the history of algebraic topology. -November 8, 2015: Here is a June, 2015, presentation on -A philosophy of modelling and computing homotopy types, -which gives the background to using, for specific computations, some algebraic models of homotopy types in the form of strict higher groupoids. It is the strictness which leads to precise colimit computations in homotopy theory, generalising those well understood for fundamental groups. One construction these ideas have led to is a nonabelian tensor product of groups, whose current bibliography has 138 items, mainly by group theorists. -Jan 21, 2017 Parts of the above mentioned presentations have been expanded into this paper; it also discusses the increasing use of cubical sets with connections, which have been little used so far in theories such as quasi categories, but which in some areas have advantages over simplicial sets.<|endoftext|> -TITLE: L'Intermédiaire des mathématiciens -QUESTION [11 upvotes]: Does anybody know if the journal L'Intermédiaire des mathématiciens is on the web anywhere? Failing that, does anybody know of a library that has all or even some of the issues? -Thanks for any help. -Cheers, Scott - -REPLY [8 votes]: Volume 13-14 is available here, and accessible also from outside the US: -https://archive.org/details/lintermdiairede01lemogoog -EDIT: Oops, I just saw that this information was already present in a comment by Thomas Riepe. Meanwhile, I found that Volumes 1-14 are available here, and accessible also from outside the US: -http://www.bookprep.com/book/mdp.39015065219977<|endoftext|> -TITLE: When is $SL(n,R) \rightarrow SL(n,R/q)$ surjective? -QUESTION [30 upvotes]: Let $R$ be a commutative ring with unit and let $q$ be an ideal of $R$. There is thus a natural map $SL(n,R) \rightarrow SL(n,R/q)$ for all $n$. This map is surjective if $SL(n,R/q)$ is generated by elementary matrices, but I very much doubt that it is surjective in general (though I don't know any examples). -My questions are as follows. - -Can someone give me an example of a ring $R$ and an ideal $q$ of $R$ such that the map $SL(n,R) \rightarrow SL(n,R/q)$ is not surjective for any $n$? I'd like the examples to be as nice as possible. For instance, it would be great to have an example where $R$ is Noetherian and has finite Krull dimension. -What conditions can I put on $R$ and $q$ to assure that this map is surjective, at least for large $n$? - -REPLY [9 votes]: The following example relates to the second part of the question while echoing back the example $R = \mathbb{Z}[X, Y]$ and $q = (X^2 + Y^2 - 1)$ discussed above: -Let $k$ be a finite field and let $q$ be any ideal of $R = k[X, Y]$, then the natural map $SL_n(R) \rightarrow SL_n(R/q)$ is surjective for all $n \ge 2$. -This is Theorem 1.7.(2) of "On the groups $SL_2(\mathbb{Z}[x])$ and $SL_2(k[x,y])$" by F. Grunewald, J. Mennicke and L. Vaserstein, 1994 (MR1276133). -In the spirit of the requirements of the first part of the question, I would ask whether $2$ is the smallest Krull dimension we can get for a ring $R$ generated by finitely many elements as a $\mathbb{Z}$-algebra and for which surjectivity of the reduction of matrix coefficients modulo $q$ fails for some ideal $q \subset R$. -The case of $\mathbb{Z}[X]$ is somehow settled by Theorem 1.7.(1) of the same paper: for $R = \mathbb{Z}[X]$, the image of $SL_n(R)$ in $SL_n(R/q)$ is of finite index for every $n \ge 2$. In some sense, it is optimal since F. Grunewald et al. have a recipe to build quotients of $\mathbb{Z}[X]$ with non-trivial $SK_1$ (see Proposition 1.9 of the same paper and this MO post) whereas $SK_1(\mathbb{Z}[X]) = 1$. -As for the general part of the question, the group $SK_0(q)$ (see Definition II.2.6 and Exercise III.2.1 of C. Weible's K-book) is the natural obstruction to the surjectivity of coefficients reduction modulo $q$. You may argue that's kind of tautological though. -Addendum I: -T. Goodwillie's example originates from Example 13.5 of "Introduction to algebraic $K$-theory" by J. Milnor, 1971 (MR0349811). -To some extent, it is also discussed in "Serre's problem on projective modules" by T. Y. Lam, 2006 (MR2235330), see in particular Proposition I.8.12 and Remark I.8.14. -Addendum II: -It follows from Corollary 8.3 of "Stable range in commutative rings" (1967) by D. Estes and J. Ohm that the natural map $SL_n(R) \rightarrow SL_n(R/q)$ is surjective for every ideal $q$ and every $n \ge 2$ if the stable rank of $R$ is at most $2$. (This settles my question above on the minimal Krull dimension of a counter-example since $\text{sr}(R) \le \dim_{Krull}(R) + 1$ by Bass stable range theorem). Theorem 8.8 of the same article gives a surjectivity criterion for $R$ a ring of univariate polynomials over a principal ideal ring.<|endoftext|> -TITLE: Finitely presented infinite group with no element of infinite order? -QUESTION [22 upvotes]: Is there an example of a finitely presented infinite group in which every element has finite order? Or, is it known that every finitely presented infinite group has an element of infinite order? -I asked this question of math.stackexchange, thinking it might be trivial (for finitely generated groups there are numerous counterexamples...), but it seems the question is wide open. So more specifically, could someone point me to partial results, or give a good reason why this problem won't be solved any time soon? - -REPLY [36 votes]: This is just an expanded version of Igor's comment. Indeed this is an open problem. The common opinion (I believe) is that such groups do exist, but the best result in this direction so far is the Olshanskii-Sapir group, which is finitely presented and (infinite torsion)-by-cyclic. -There is a general idea, commonly attributed to Rips, which shows that such groups should exist. The idea is the following. Take the free Burnside group $B(m,n)$ on $m\ge 2$ generators and of exponent $n>>1$, so that $B(m,n)$ is infinite. Embed it into a finitely presented group $G$ by using a version of the Higman embedding theorem. Let $G=\langle x_1, \ldots , x_n\rangle$. Then take the quotient group $Q$ of $G$ by the relations -(*) $x_1=b_1, \ldots , x_n=b_n$, where $b_1, \ldots , b_n\in B(m,n)$. -Clearly $Q$ is torsion being a quotient of $B(m,n)$ and is finitely presented. And it is believable that if the embedding of $B(m,n)$ in $G$ is "reasonably hyperbolic" and elements $b_1, \ldots , b_n$ are chosen randomly, then $Q$ is infinite with probability close to $1$. This idea was essentially implemented by Olshanskii and Sapir, but they managed to impose all relations of type (*) but one. (This is in fact a very rough interpretation of what they did, so Mark may correct me here.) There are even very particular choices of $b_1, \ldots , b_n$ (e.g., long aperiodic words with small cancellation), for which the group $Q$ should be infinite. So there are even particular finite presentations which should represent infinite torsion groups, but nobody knows how to prove that.<|endoftext|> -TITLE: Relationship between Erdos and Falconer distance problems -QUESTION [6 upvotes]: Given a set $E \subset \mathbb{R}^d$, define the distance set of $E$ -$$ -\Delta(E) = \lbrace|x-y| : x,y \in E \rbrace, -$$ -where $|\cdot |$ is the usual Euclidean distance. -$\bullet$ The Erdos-distance conjecture roughly asserts that if a set $E \subset \mathbb{R}^d$ has finite cardinality $|E| = n$, then the distance set has size at least $|\Delta(E)| \ge n^{\frac{2}{d}+ \underline{o}(1)}$ $\quad$ (where $\underline{o}(1) \to 0$ as $n \to \infty$). -The case $d = 2$ was recently solved by Guth-Katz using an ingenious adaptation of Dvir's polynomial method along with the so-called algebraic method. -$\bullet$ The Falconer-distance conjecture is a continuous analogue of the Erdos-distance problem. It asserts that if the $E \subset \mathbb{R}^d$ has Hausdorff dimension $\dim_H(E) > \frac{d}{2}$ then the corresponding distance set $\Delta(E)$ has positive Lebesgue measure. -What is the relationship between these two results? It seems that the $\frac{2}{d}$ and the $\frac{d}{2}$ are related, but how, precisely? Is it possible to turn the Guth-Katz result into a result concerning the Falconer-distance problem? - -REPLY [11 votes]: Generally speaking, continuous incidence geometry problems are considered strictly harder than their discrete analogues. For instance, if (for simplicity) one replaces Hausdorff dimension with Minkowski dimension (and glosses over the distinction between upper and lower Minkowski dimension), then the Falconer problem is roughly equivalent to the following Erdos-like assertion: - -If $E \subset {\bf R}^d$ has bounded diameter and $\delta$-entropy n (i.e. it requires exactly $n$ balls of radius $\delta$ to cover $E$) for some $\delta>0$, then $\Delta(E)$ has $\delta$-entropy $\gg n^{\frac{2}{d}-o(1)}$. - -(EDIT: Strictly speaking, the above assertion is in fact false; see this paper of mine with Nets Katz for a simple counter-example. As discussed in that paper, the "correct" way to discretise the Falconer problem is in fact rather delicate.) -Informally, this is a relaxation of the Erdos problem in which one now considers two numbers to be equivalent if they differ by $O(\delta)$. This makes it significantly harder to apply algebraic methods (which do not react well to this sort of "fuzzy" arithmetic, especially when dividing by small non-zero denominators), although there have been some partial successes in extending the algebraic method to this setting, such as Guth's proof of the endpoint multilinear Kakeya conjecture. -The last part of this (now somewhat dated) survey of Wolff gives a detailed comparison of the Szemeredi-Trotter theorem and a variant of the Kakeya problem relating to sets of Furstenberg type, which is quite analogous to the relationship between the Erdos and Falconer conjectures.<|endoftext|> -TITLE: Insolvable number fields ramified only at one (small) prime -QUESTION [11 upvotes]: In his first Eilenberg Lecture at Columbia, Benedict Gross says that only recently have we been able to give examples of finite galoisian extensions $K$ of ${\bf Q}$ which are ramified only at $2$ (respectively $3$) and for which the group Gal$(K|{\bf Q})$ is not solvable. He seemed to suggest that this was made possible by recent progress in the Langlands Programme. -Question. Which such number fields have been discovered recently, and which bits of the Langlands Programme are needed to construct them ? - -REPLY [9 votes]: The construction of such number fields is not limited to the Langlands program. For instance, to what extent are there number fields with few ramified primes and Galois group S_n? You won't be able to cook these out of automorphic forms. In this connection there is a remarkable recent example of David Roberts: a number field whose Galois group is the symmetric group on 15875 letters, and whose discriminant is $-2^{130729}5^{63437}$!<|endoftext|> -TITLE: Clean Proofs of Properties of Projective Space -QUESTION [18 upvotes]: How far can one go in proving facts about projective space using just its universal property? -Can one prove Serre's theorem on generation by global sections, calculate cohomology, classify all invertible line bundles on projective space? -I don't like many proofs of some basic technical facts very aesthetic because one has to consider homogeneous prime ideals, homogeneous localizations, etc. Do there exist nice clean conceptual proofs which avoid the above unpleasantries? -If you include references in your answer it would be very helpful, thanks. - -REPLY [5 votes]: I agree with Anton that it would be too much to hope for to get serious results (e.g. cohomology of line bundles) from the "nice" universal property of projective space, but one can indeed prove that there are no non-constant regular functions on $\mathbb{P}^n$ using only the universal property. -Namely, it suffices to check that $\mathbb{P}^n$ is proper and connected. For properness, one may use the valuative criterion. Namely, let $R$ be a valuation ring and $K$ its fraction field. Then a map from $\operatorname{Spec}(K)\to \mathbb{P}^n$ is a surjection $K^{n+1}\to K$; then the image of $R^n\hookrightarrow K^n\to K$, where the inclusion is the obvious one, is isomorphic to $R$. In particular, the map $K^{n+1}\to K$ lifts to a surjective map $R^{n+1}\to R$, which is a map $\operatorname{Spec}(R)\to \mathbb{P}^n$ fullfilling the valuative criterion (its uniqueness up to automorphisms of the given diagram is also clear). -As for connectedness, it suffices to check that $\mathbb{P}^n$ is path-connected in the following sense--for any two geometric points, represented by surjections $x_0: \bar k^{n+1}\to\bar k, x_1: \bar k^{n+1}\to\bar k$, there is a ``path" connecting them; namely a map $f: \mathbb{A}^1\to \mathbb{P}^n$ with $f(0)=x_0, f(1)=x_1$. This is a surjection $\bar k[t]^{n+1}\to \bar k[t]$ such that reducing mod $(t), (t-1)$ gives the desired maps. Translating, we must choose polynomials $f_1, ..., f_{n+1}$ such that $f_i(0)=x_0(e_i), f_i(1)=x_1(e_i)$, where $e_i$ are the standard basis of $\bar k^{n+1}$, and where all the $f_i$ do not vanish simultaneously. -But one can do this by Lagrange interpolation; choose any $f_1$ with the desired values at $0,1$, then any $f_2$ with the desired values at $0, 1$ and not vanishing at the other zeros of $f_1$, then any $f_3$ analogously, and so on.<|endoftext|> -TITLE: Has anyone seen this graph? -QUESTION [8 upvotes]: I recently constructed the graph shown below in the process of investigating some problems regarding line graphs and homomorphisms, and then happened to see it on wikipedia. I was wondering if anyone knew of this graph coming up anywhere else. - -If it helps, I was partly inspired by this paper which shows that any graph with maximum degree 3 and circular chromatic index 4 must contain $K_4$ with one edge subdivided as a subgraph. Note that the graph in the link above is three copies of $K_4$ with one edge subdivided plus another vertex which is adjacent to the vertices that are subdividing the edges. -Thanks - -REPLY [3 votes]: Any connected trivalent graph realises a Schreier coset graph of a subgroup of -the modular group. This yields a transitive permutation group of degree 16 x 3 -since we expand each node into an oriented triangle. In the original graph, -switching the ends of edges & rotating the oriented triangles provides -generators of order 2 and order 3. Recall PSL(2,Z) = free product C2 * C3.<|endoftext|> -TITLE: CM abelian varieties and potential good reduction -QUESTION [5 upvotes]: Let $F$ be a number field and $A$ an abelian variety over $F$. It is known that if $A$ has complex multiplication, then it has potentially good reduction everywhere, namely there exists a finite extension $L$ of $F$ such that $A_L$ has good reduction over every prime of $L$. -And what about the inverse: if $A$ is known to be of potential good reduction everywhere, how far is it from having complex multiplication? -As the reduction behavior is determined by the Galois representations of the decompositon groups, one can reformulate the problem as follows: let $A$ be an abelian variety over $F$, $p$ a fixed rational prime, $V$ the p-adic Tate module of $A$; and for $\lambda$ primes of $F$, $\rho_\lambda$ is the $p$-adic representation on $V$ of the decomposition group $G_\lambda$ at $\lambda$. If $\rho_\lambda$ is potentially unramified for $\lambda$ not dividing $p$, and potentially cristalline for $\lambda$ dividing $p$, do we know that the global Galois representation $\rho$ on $V$ is potentially abelian, i.e. when shifting to some open subgroup, the image of $\rho$ is contained in a torus of $GL_{\mathbb{Q}_p}(V)$? What do we know about the Fontaine-Mazur conjecture in this case? -Thanks! - -REPLY [4 votes]: Potential good reduction everywhere is quite far from having complex multiplication. -For elliptic curves, the condition is equivalent to the $j$-invariant being an algebraic integer. For $F = \mathbb{Q}$ there are only finitely many $j$-invariants of CM elliptc curves (corresponding to the imaginary quadratic fields of class number $1$) but any element of $\mathbb{Z}$ is of course the $j$-invariant of an elliptic curve over $\mathbb{Q}$.<|endoftext|> -TITLE: l-adic vs complex Perverse Sheaves -QUESTION [8 upvotes]: Let $X$ be a scheme of finite type over $Spec(\mathbb{C})$. Let $X_{an}$ denote the associated complex analytic space. After fixing an isomorphism $\overline{\mathbb{Q}}_l\cong \mathbb{C}$, by $\S$6.1.2. of BBD (Asterisque 100) we know that there is fully faithfull functor -$$\mathcal{F}: D_c^b(X, \overline{\mathbb{Q}}_l) \rightarrow D_c^b(X_{an}, \mathbb{C}),$$ -where the left hand side is the usual "derived category" of $l$-adic sheaves and the the right hand side is the usual derived category of complex sheaves with constructible (for an algebraic stratification) cohomology. As is well known this functor is not essentially surjective. However, on both sides we have two natural subcategories -$$Perv_{l}(X) \subset D_c^b(X, \overline{\mathbb{Q}}_l)$$ -and -$$Perv_{\mathbb{C}}(X) \subset D_c^b(X_{an}, \mathbb{C}) $$ -of l-adic Perverse sheaves and complex Perverse sheaves respectively. My question is this: -Does $\mathcal{F}$ induce an equivalence of categories between $Perv_{l}(X)$ and $Perv_{\mathbb{C}}(X)$? - -REPLY [9 votes]: No. -If $L$ is a local system on a smooth variety $X$ of dimension $d$ then $L[d]$ is perverse. As suggested by BBD, if we take $L$ to be a rank $n$ local system of $\overline{\mathbb{Q}}_l$ vector spaces on $X_{an}$ then this is in the essential image of $\mathcal{F}$ iff the corresponding representation of $\pi_1(X_{an})$ preserves a lattice (by which I mean a rank $n$ $\mathcal{O}$-submodule, where $\mathcal{O}$ is the ring of integers in some finite extension of $\mathbb{Q}_p$). The reason for this is that the algebraic fundamental group is compact and any compact subgroup of $GL_n(\overline{\mathbb{Q}}_l)$ is contained in $GL_n(\mathcal{O})$, for some $\mathcal{O}$ as above. -This condition need not always hold: for example, one can consider representations of $\pi_1(X_{an})$ so that the eigenvalues of some element are algebraic but not $l$-adic integers.<|endoftext|> -TITLE: "Coarse" Arctic Circle Theorem -QUESTION [8 upvotes]: Uniformly random dimer tilings of aztec diamond region will have an "artic" circle appear in the middle. - - .. - .... - ...... -........ -........ - ...... - .... - .. -What happens if we make the regions a little bit coarser, do we get the same limit shape? - - .. - .. - ...... - ...... - .......... - .......... - ...... - ...... - .. - .. -I apologize for the lousy aspect ratio. In general, does the limit shape depend only on the coutour or does it depend "microscopically" on the boundary conditions as well? - -REPLY [16 votes]: To expand slightly on Joseph's answer and Noam's comment, you can describe a domino or lozenge tiling by its "height function." In the case of lozenge tilings, shown in Joseph's answer, the height function is visible, since the picture can naturally be interpreted as showing a 3d surface. (You have to make one choice, since there's an ambiguity regarding what is pointing into the screen and what is pointing out.) For domino tilings, it is not nearly as visually apparent, but height functions are explained in a beautiful paper by Thurston (Conway's tiling groups, Amer. Math. Monthly 97 (1990), 757–773). -The microscopic shape of a region influences the height function, since it determines the wire-frame boundary the height function must interpolate. For example, for Aztec diamonds the boundary is as steep as possible, while in the coarser example it stays flat. -It turns out that in the continuum limit, the shape of a typical tiling is entirely determined by the shape of the boundary, and it is determined by a variational principle: among all spanning surfaces that satisfy a certain Lipschitz constraint (that must hold for all tilings), the typical surface is singled out by maximizing the entropy, which is given by integrating local contributions that can be computed explicitly in terms of the partial derivatives of the height function. See A variational principle for domino tilings for more details. Almost all tilings will cluster around the typical tiling, so this completely determines the large-scale behavior. -The two examples from the question can be completely analyzed in this framework. In particular, the second one has the highest possible entropy at every point and therefore has nothing like an Arctic circle. See Figure 4 in Local statistics for random domino tilings of the Aztec diamond (reproduced below) for a picture of a random tiling. - -Kenyon, Okounkov, and Sheffield (Dimers and Amoebae) substantially generalized this framework, and Kenyon and Okounkov (Limit shapes and the complex burgers equation) found a beautiful description of the variational problem for lozenge tilings in terms of the complex Burgers equation. This leads to explicit descriptions of cases like the cardioid in Joseph's answer.<|endoftext|> -TITLE: Constructing a unitary matrix -QUESTION [8 upvotes]: Setting: -Given a set of $n\times n$ matrices $A_i$, I would like to find a linear combination of these matrices $Q = \sum_i A_i x_i$ with $x_i$ a set of complex numbers, such that $Q$ is unitary: $Q^{\dagger} Q = 1$. This problem is equivalent to solving a system of quadratic equations over real numbers. As far as I understand, there is no general and efficient way to solve such systems of equations, and black-box algorithms, such as Gröbner basis, struggle with systems of around 10 variables. -Question: -Does the particular structure of this system of equations make it easier than a generic one, and can this be utilized in order to speed up the calculation. -Motivation: -This problem arises in many contexts. For example, it is related to search of symmetry of a quantum Hamiltonian. Hamiltonian $H$, a finite Hermitian matrix has a symmetry if it commutes with some unitary matrix $U$ other than identity. It is very easy to construct the linear space to which $U$ should belong, it is given by the kernel of the system of linear equations $H A - A H = 0$. However the next step requires verifying whether this linear space contains unitary matrices other than identity. -Secondary question: -Given that it seems unlikely that there is an easy answer to the main question, I would also like to ask whether there are known classes of systems of quadratic equations that are quickly solvable numerically. - -REPLY [2 votes]: Every Hermitian matrix (in fact, every normal matrix) commutes with infinitely many unitary matrices: -Lemma: The square matrices A and B commute if they can be simultaneously diagonalized. -Proof: Let A=Q D inv(Q) and B=Q E inv(Q), where D and E are diagonal. Then AB = Q D E inv(Q) = Q E D inv(Q) = B A, since diagonal matrices commute. -Corollary: If H is diagonalized by the unitary matrix Q, then U = Q D Q' is unitary for any diagonal matrix D whose entries lie on the unit circle, and U commutes with H. -Thus, once you have the eigenvectors of your (discretized) Hamiltonian, you can easily form an infinite number of unitary matrices that commute with it. -Is there a constraint on the symmetrices that I'm missing?<|endoftext|> -TITLE: Learning Arakelov geometry -QUESTION [14 upvotes]: I have a complex analytic background (Griffiths and Harris, Huybrechts, Demailley etc). Also, I understand some PDE. I want to learn Arakelov geometry (atleast till the point I can "apply" computations of Bott-Chern forms and Analytic torsion to producing theorems of interest in Arakelov geometry). I know almost nothing of schemes or of number theory. I don't how much of these is needed to learn this stuff. I'd be grateful if any good references/suggestions are pointed out. - -REPLY [2 votes]: I second the suggestion of the book "Lectures on Arakelov Geometry" by Soulé, Abramovich, Burnol and Kramer. -There is this nice text by Demailly which motivates the treatment of intersection theory on the infinite fibers and probably suits you with your background. -With this in mind the analytic part of the above book should be ok to read.<|endoftext|> -TITLE: Intuition for coends -QUESTION [45 upvotes]: Let $D$ be a co-complete category and $C$ be a small category. For a functor $F:C^{op}\times C \to D$ one defines the co-end -$$ -\int^{c\in C} F(c,c) -$$ -as the co-equalizer of -$$ -\coprod_{c\to c'}F(c,c'){\longrightarrow\atop\longrightarrow}\coprod_{c\in C}F(c,c). -$$ -It is the indexed co-limit $\mbox{colim}_W F$ where the weight is the functor $W:C^{op}\times C \to Set$ given by $Hom(-,-)$. -I have two strongly related questions regarding this definition. First, what's the intuition behind this construction? Can I think of it as a kind of ''fattened'' colimit? Second, why is the integral sign used for this? Can ordinary integration be related to this construction? - -REPLY [54 votes]: I prefer a somewhat different view of ends and coends, with the intuition stemming more from classical linear algebra and functional analysis. So for me an end is really an integral in a categorical sense. -Let me explain it in more detail. The first fact about ends that show me their power and is basic in enriched category theory is the natural transformation lemma. Basically, -$$Nat(F\cdot; G\cdot) = \int_c Hom_C(Fc; Gc)$$ -This equations shows that a global natural transformation is as if "summed up" from the local, "differential" transformations on each object. That's exactly what a natural transformation is: a coherent family of morphisms. In this way an end allows us to pass from the local to the global picture, just like a real integral. Let's consider a global section functor for sheaves: -$$ \Gamma(X; \mathcal{F}) = Nat(\Bbb{Z}, \mathcal{F}) = \int_U Hom_{Ab}(\Bbb{Z}; \mathcal{F}(U)) = \int_{U\in Ouv^{op}} \mathcal{F}(U)$$ -Compare it with measure integration, where you have a (non-negative) measure defined for all measurable subsets of $X$ and you can, in principle, define the measure of $X$ analysing its subsets. At least for not-too-bad measure spaces you can find the measure of $X$ as the supremum of the subsets' measures. This can be also viewed as an end, if you consider a functor $M: Ouv \to \Bbb{R}_{+}$, where $\Bbb{R}_{+}$ is a poset category with objects $[0;\infty]$, $f:a\to b \iff a \leqslant b$. However, I don't understand at the moment how can nontrivial general integrals be treated in this conext. -Even more enlightening is the composition of distributors. A good account of distributors is in J. Benabou's article "Distributors at work". Informally, it is like a "generalized functor", the most important property being the existence of right adjoint for any functor considered as a distributor. Kan extensions also emerge miraculously. The name itself hints of this connection. A distributor is to a functor what a distribution is to a function. Formally, a (Set-valued) distibutor $F:A \nrightarrow B$ from category $A$ to category $B$ is a functor $$\hat F: A\times B^{op} \to Set$$ -A composition of distributors can be defined via Kan extensions along the Yoneda embedding, or much more neatly as a coend -$$G\circ F (a;c) = \int^b \hat G(b; c) \times \hat F(a;b)$$ -This clearly reminds of matrix composition law. A simple example of (identity) distributor is the hom-functor in a V-category: -$$[a;c] = \int^b [b; c] \otimes [a;b]$$ -Clearly we just integrate out the dummy variable and the inner hom is just a change-of-coordinates Jacobian! -A special case of these identities is the Yoneda lemma, which I will write as a left Kan extension: -$$F(a) = \int^c [c;a] \otimes F(c) $$ -and the Kan extension itself: -$$\mathrm{Lan}_K(F)(a) = \int^c [K(c);a] \otimes F(c) $$ -Clearly it's just a change of integration variables! -Another enlightening example comes from the theory of metric spaces, considered by F.W.Lawvere in "Metric spaces, generalized logic, and closed categories". A metric space is considered as a category enriched over $\Bbb{R}_+$, defined above. The objects are points, hom from a to b is the distance from a to b (the metric need not be symmetric). In this case for $\Bbb{R}_+$-valued functor $F$ it's limit is clearly it's supremum. A limit is an end -$$\mathrm{Lim} F = \int_d F(d)$$ -where $F$ is considered as a bifunctor constant in its first variable. So -$$\sup_{x\in X} F(x) = \int_{x\in X} F(x)$$ -A person familiar with tropical geometry and idempotent analysis will instantly recognise this formula as a tropical integral! Simple as it is, it is another exact shot for categorical integration. A Kan extension of $\phi: X \to \Bbb{R}$ along $f:X\to Y$ is -$$\mathrm{Ran}_{f} \phi (y) = \sup_x [ \phi(x) - \lambda Y(y, f(x)) ] $$ -If not for the nonlinear hom nature, it would be immediately recognisable as a tropical Fourier transform, aka Legendre transform. -Even the common integral itself can be considered as a kind of transfinite tensor product, but the construction is somewhat clumsy and eventually reduces to common measure, so it's not of much use practically but fits nicely in the categorical integration picture. As of yet I do not know any neat way to incorporate common integrals into the categorical framework, like tropical integrals do. -Sorry for a big post, but I just couldn't resist sharing these examples.<|endoftext|> -TITLE: Formulae of standard set theory interpreted in suitable segments of L -QUESTION [5 upvotes]: In a letter dated 16 October Georg Kreisel asked me the question that follows. He agreed to my idea of posting the question here. -I’d be grateful if you could recommend an exposition of formulae of standard set theory interpreted in suitable segments $L_\alpha$ of $L$. (a) It is commonly said that each set of $L$ is definable by such a formula with symbols for specific ordinals, usually without specifying suitable $L_\alpha$ where the formula is interpreted. (Of course, if a set is defined at all, there is a bound for the $\alpha$ beyond which the definition is stable.) (b) Are there definitions without any additional symbols interpreted in $L$ which define (in $V$) uncountable, let alone, strongly inaccessible ordinals at all? - -REPLY [4 votes]: I'm not sure I have the intended sense of the question, as there seem to be several ways to interpret it, -so let me give several replies. Please clarify if I've misunderstood. - -Perhaps you intend to ask: is there a formula -$\varphi(x)$ in the language of set theory such that -$L_\alpha\models\varphi(\kappa)$ if and only if $\kappa$ is -an uncountable cardinal in $V$? And is there a formula -$\psi(x)$ such that $L_\alpha\models\psi(\kappa)$ if and -only if $\kappa$ is inaccessible in $V$? - -In this case, the answer is no. One way to see that we -should expect this is that by forcing we may make any set -countable (and hence also non-inacessible), but forcing -does not change the satisfaction relation of formulas in -$L$ or in $L_\alpha$, and so we may change the truth of the -right-hand-side of these proposed equivalences by moving to -a forcing extension, without changing the left-hand-side. -Thus, they cannot always be equivalent. But another direct -argument is that whenever $L_\alpha\models\varphi(\kappa)$, -then by the downward Lowenheim-Skolem theorem, there is in -$L$ a countable elementary substructure $X\prec L_\alpha$ -containing $\kappa$, and the Mostowski collapse of $X$ is -isomorphic by the condensation principle to some $L_\beta$, -which will satisfy $\varphi(\kappa_0)$ for the image of -$\kappa$ under the collapse. Thus, -$L_\beta\models\varphi(\kappa_0)$, but $\kappa_0$ is -countable in $L$, violating the desired property. - -Perhaps you intend to ask: is there a formula -$\varphi(x)$ such that $L\models\varphi(\kappa)$ if and -only if $\kappa$ is uncountable in $V$ (and another formula -for inaccessibility). - -In this case, the answer is that it depends on $V$. On the -one hand, if $V=L$, then there are such a formula, because -the property, $\kappa$ is uncountable,'' is expressible -in the first-order language of set theory, as the assertion -that there is no surjective function from $\omega$ to -$\kappa$. Similarly the property of being inaccessible is -expressible. The point is that it is consistent with ZFC -that the concepts of uncountable and inaccessible are in -agreement between $L$ and $V$. -But meanwhile, it is also consistent that there are no such -formulas. For example, one quick way to see this is that if -$0^\sharp$ exists, then all the Silver indiscernible -ordinals have the same first-order properties in $L$, and -so from the perspective of $L$, the cardinal $\aleph_1^V$ -satisfies the same formulas as many countable ordinals. -But one needn't make the $0^\sharp$ assumption, and one can -do it equiconsistently with ZFC. The reason is that it is -equiconsistent with ZFC that there is a cardinal $\delta$ -with $L_\delta\prec L$, expressed as a scheme in the -language with a constant for $\delta$. In such a model, we -may move to the forcing extension $L[G]$ collapsing -$\delta$ to $\omega$. In $L[G]$, all the uncountable -ordinals are above $\delta$, but by our $L_\delta\prec L$ -hypothesis, for any ordinal above $\delta$ with a certain -property in $L$, there will be ordinals below $\delta$ with -that same property. But since these will all be countable -in $L[G]$, it violates the desired feature. -François pointed out in the comments below that you may have intended the question: - -Is it consistent that every ordinal that is definable in $L$ is countable in $V$? - -The answer here is yes, since in the model $L[G]$ above, where $L_\delta\prec L$ and $G$ collapses $\delta$ to $\omega$, we have that every definable object of $L$ is in $L_\delta$, and these are all countable in $L[G]$.<|endoftext|> -TITLE: Three-dimensional simple Lie algebras over the rationals -QUESTION [13 upvotes]: I ran into this question on math.stackexchange.com "How many 3 dimensional simple Lie algebras are there over the rationals?" -The question has been sitting idle for a long time. I thought it was interesting and would like to know myself. -If working over $\mathbb{C}$, we know all 3-dimensional simple Lie algebras are isomorphic to $\mathfrak{sl}_2(\mathbb{C})$. When we move down to the reals, there are 2 non-isomorphic real forms. -The same argument that works over $\mathbb{C}$ works over $\bar{\mathbb{Q}}$ (the field of algebraic numbers). So over $\bar{\mathbb{Q}}$ the only 3-dimensional simple Lie algebra is $\mathfrak{sl}_2(\bar{\mathbb{Q}})$. Running through a similar argument as that which classifies 3-dim simples over $\mathbb{R}$, I get a whole mess of possibilities over $\mathbb{Q}$ (due to the lack of square roots) but have no idea which forms are non-isomorphic or even how to go about proving they are non-isomorphic. -Anybody know what the answer is? Have a good reference? -Thanks! - -REPLY [10 votes]: Here is my own favorite way of understanding this problem: Let $\bigl(L,[,]\bigr)$ be a $3$-dimensional Lie algebra over a field $K$ (assumed of characteristic $0$ for my comfort, though this probably is overkill). The bracket $[,]:L\times L \to L$, which is bilinear and skew-symmetric, can be regarded as an element $\beta \in \text{Hom}\bigl(\Lambda^2L,L\bigr)\simeq L\otimes \Lambda^2 (L^\ast)$. Since the dimension of $L$ is $3$, we have that $\Lambda^2(L^\ast)$ is naturally isomorphic to $L\otimes \Lambda^3(L^\ast)$. It follows that we can regard $\beta$ as an element of $L\otimes L\otimes \Lambda^3(L^\ast)$. Since there is a canonical decomposition $L\otimes L\simeq \Lambda^2(L)\oplus S^2(L)$, there is a canonical decomposition $\beta = \lambda + \sigma$ with $\lambda\in \Lambda^2(L)\otimes \Lambda^3(L^\ast)\simeq L^\ast$ and $\sigma\in S^2(L)\otimes \Lambda^3(L^\ast)$. -Now, there is a canonical bilinear pairing -$$ -\langle,\rangle: L^\ast\times\bigl(S^2(L)\otimes \Lambda^3(L^\ast)\bigr)\longrightarrow -L\otimes \Lambda^3(L^\ast)\simeq \Lambda^2(L^\ast), -$$ -and the Jacobi identity is easily seen to be just $\langle\lambda,\sigma\rangle = 0$. -If $\lambda\not=0$, then, because $\lambda$ must be fixed by any automorphism of $L$ (in particular, the inner automorphisms), it follows that $N = \text{ker}(\lambda)\subset L$ is an ideal of codimension $1$ in $L$, so $L$ is not simple. One could follow this thread further to classify the Lie algebras in this case, but let me pass on to the simple case. -Thus, assume that $\lambda = 0$, so that $\beta = \sigma \in S^2(L)\otimes \Lambda^3(L^\ast)$. Now, there is a natural (i.e., $\text{GL}(L)$-equivariant) cubic polynomial mapping -$$ -\text{det}: S^2(L)\otimes \Lambda^3(L^\ast) \longrightarrow \Lambda^3(L^\ast) -$$ -(again, this exploits the assumption that $L$ has dimension $3$). It is easy to see that, if $\text{det}(\sigma) = 0$, then $\sigma$ lies in $S^2(N)\otimes \Lambda^3(L^\ast)$ for some nontrivial proper subspace $N\subset L$ and that this $N$ will be an ideal in $L$, so, again, $L$ is not simple. -Thus, assume that $\text{det}(\sigma)\in\Lambda^3(L^\ast)$ is not zero. In particular, $L$ is endowed with a canonical volume form, $\text{det}(\sigma)$. Since $\text{GL}(L)$ acts transitively on the nonzero elements of $\Lambda^3(L^\ast)$, it follows that classifying the $3$-dimensional Lie algebra structures on $L$ up to the action of $\text{GL}(L)$ with $\lambda=0$ and $\text{det}(\sigma)\not=0$ is equivalent to fixing a (nonzero) volume form $\Upsilon\in \Lambda^3(L^\ast)$ on $L$ and classifying, up to the action of $\text{SL}(L)$, the elements $s\in S^2(L)$ that satisfy $\text{det}(s\otimes \Upsilon) = \Upsilon$. -It is not hard to see that for such $s$, the corresponding Lie algebra is indeed simple, so this reduces the classification of the $3$-dimensional simple Lie algebras over $K$ to the classification of the unimodular quadratic forms in $3$ variables over $K$ up to the action of $\text{SL}(3,K)$, a classical problem.<|endoftext|> -TITLE: Reference request: representation of type G2 Lie algebras. -QUESTION [6 upvotes]: Let $\mathfrak{g}$ be an Lie algebra of type G2. Are there some combinatorial ways to describe a basis of a $\mathfrak{g}$-module? For classical types, there is a method used tableaux. Thank you very much. - -REPLY [2 votes]: Another direction, that works (only?) for groups of rank at most 2, is Kuperberg's "spider" model: http://arxiv.org/abs/q-alg/9712003 . The web diagrams he constructs generalize tableaux, in the sense that the $A_1$ and $A_2$ versions are in bijection with tableaux (at least of rectangular shape). This is not quite the same as constructing representations via Schur functors, but it has a combinatorial appeal.<|endoftext|> -TITLE: Kontsevich Integral without associators? -QUESTION [6 upvotes]: Recall the fact that the representations of a quantum group form a braided tensor category, and this corresponds to the fact that $U_q(\mathfrak g)$ is a quasi-triangular Hopf algebra. The braiding operations can be constructed via the KZ equations for representations of $\mathfrak g$. The first thing that the KZ equations give is a quasi-triangular quasi-Hopf algebra, i.e. we have an associator present. It turns out that this is equivalent to $U_q(\mathfrak g)$ in some suitable sense, but one needs to somehow absorb the associator into the rest of the structure in order to get a genuine quasi-triangular Hopf algebra (and this is not straightforward to do). -The Kontsevich Integral for tangles can be calculated in same the formal way one calculates the monodromy of the KZ equations (of course, since this is just the definition of the Kontsevich integral). Thus to calculate the value of the Kontsevich Integral for some parenthesized tangle, it suffices to multiply the values of together the invariant for elementary tangles and changes of parenthization (this being the use of the associator) corresponding to the given input tangle. These can be given more or less explicitly, as the (chord diagram)-valued "R-matrix" and Drinfeld associators respectively. - -Is there a (chord diagram)-valued R-matrix (hopefully somewhat explicit) which does not require an associator? (i.e. one that genuinely satisfies the relations satisfied by the elementary tangles in the category of unparenthesized tangles)? - -One would certainly need to redefine the values of the invariant of min/max elementary tangles as well. - -REPLY [10 votes]: As you point out, the relation between associators and the quasi-triangular structure of $U_q(\mathfrak g)$ (and the related tangles invariants) exists "only" at the Lie algebraic level, not (not yet) at the universal one. Roughly speaking, this is because the twisting which absorb the associator is not $\mathfrak g$-invariant, ie modifies the coalgebra structure, which a priori doesn't make sense at the level of chord diagrams. -So far I know, there is no combinatorial construction of a universal finite type invariant which can avoid associators. But of course it's something people are looking for. -But it turns out that the theory of quantum $R$-matrices is more related to the theory of virtual knotted objects (see this answer of Greg Kuperberg). This is more or less "Bar Natan's dream" that a universal finite type invariant for virtual knotted objects should corresponds somehow to Etingof--Kazhdan quantization of Lie bialgebras. -There is also a baby version of this, which is Alekseev-Enriquez-Torrossian solution of the Kashiwara Vergne conjecture based on associators. It turns out that they constructs a kind of universal twist which can "kill" the associator, and a kind of universal solution of the quantum Yang-Baxter equation, living in a bigger algebra than the algebra of horizontal chords diagrams. According again to Bar Natan, this corresponds more or less to a universal finite type invariant for "welded knots". See: http://www.math.toronto.edu/drorbn/papers/WKO/ -You may also find this paper interesting : Towards a Diagrammatic Analogue of the Reshetikhin-Turaev Link Invariants -Edit: Some details about the relation between the Alekseev-(Enriquez)-Torossian construction and Vassiliev invariants. -They start from the Lie algebra $\mathfrak{tder}_n$ of "tangential derivations" of the free Lie algebra $\mathfrak{f}_n$ on $n$ generators, that is the Lie algebra of endomorphism sending each generator $x_i$ to $[x_i,a_i]$ for some $a_i \in \mathfrak f_n$. -Let $r^{i,j}$ be the element mapping $x_i$ to $[x_i,x_j]$, and $x_k$ to 0 for $k\neq i$. Then it leads to a solution of the "classical Yang-Baxter equation whose second leg lives in a commutative subalgebra", i.e.: -$$[r^{1,3},r^{2,3}]=0$$ -and -$$[r^{1,2},r^{1,3}+r^{2,3}]=0$$ -as does, for example, the $r$-matrice of the Drinfeld double of a cocommutative Lie bialgebra. They also prove that $R=\exp(r)$ ($r:=r^{1,2}$) satisfies the Quantum Yang Baxter equation. Therefore, one get this way a representation of the (pure) braid group in $\exp(\mathfrak{tder}_n)\rtimes S_n$. The $r^{i,j}$ can be thought of as "arrow diagrams" and are related to the welded braid group. -On the other hand, exactly like in the usual theory of the Yang-Baxter equation, we have that $t^{i,j}=r^{i,j}+r^{j,i}$ satisfies the infinitesimal braid relations, also called 4t relation for horizontal chord diagrams. We thus get an injective morphism from the algebra of Horizontal chord diagram into $U(\mathfrak{tder}_n)$. Therefore we can put an associator and get another representation of the braid group in $\exp(\mathfrak{tder}_n)\rtimes S_n$ which is precisely the "Kontsevich integral for braids". -Then the main result of [AT] is the following identity: let $\Phi$ be (the image in $\exp(\mathfrak{tder}_3)$ of) an associator, then there exists some $F \in \exp(\mathfrak{tder}_2)$ such that -$$F^{2,3}F^{1,23}=\Phi F^{1,2} F^{12,3}$$ -where the indices correspond to some maps $\mathfrak{tder}_2 \rightarrow \mathfrak{tder}_3$ modelled on the coproduct of an envelopping algebra. They also show that: -$$R=F e^{t/2} (F^{2,1})^{-1}$$ -Hence $F$ is a universal version of a Drinfeld twist for Quasi-Hopf algebra, which is able to "kill" the associator, and according to the theory it implies that the corresponding representations of the braid group are the same. -Edit 2: This was actually the situation for braids, let me add a few word about knots. It turns out that while usual braids embeds into welded one, this is far from being true at the level of knots. So the twist also intertwines between the restriction of the invariant for welded knots to usual knots, and the image of the Kontsevich integral in the space of arrow diagrams, but the resulting invariant is much weaker. -Welded braids can be identified with the group of basis conjugating automorphisms of a free group, and the map it gets from usual braids is nothing but the Artin representation. This representation is faithfull, but its extension to string links (and in particular to long knots) has a huge kernel. At the level of diagrams, the algebra corresponding to knots is free commutative with two generators in degree one, and one generator in each degree greater than one. Hence it has a rather simple structure, while its analog for usual knots is very complicated. -In fact, one of the main claim of Bar Natan's paper is that the universal invariant for welded knots is roughly the Alexander polynomial. -However, let me mention that usual knots does embed into virtual one, and that the above story is an important step towards something similar in the virtual case.<|endoftext|> -TITLE: Arctic regions in higher dimensional zonotopes -QUESTION [14 upvotes]: Same way as the two dimensional tilings by rhombi come from minimal surfaces in a $D$ dimensional cubical lattice as mentioned in this answer, one can consider higher dimensional zonotopes tiled by rhombic polytopes. I've seen these usually denoted as $D\to d$ tilings, or $d$-dimensional, codimension $D-d$ tilings. We can call a region in such a tiling effectively $D'$ dimensional, if it looks locally like a $D'\to d$ tiling. -In "An n-dimensional generalization of the rhombus tiling" by J. Linde, C. Moore, and M.G. Nordahl, it is conjectured that the octahedron inscribed inside a rhombic dodecahedron plays the same role for $4\to 3$ tilings as the arctic circle does for $3\to 2$ tilings. Outside this octahedron, the tiling is most likely frozen, i.e. locally periodic and with vanishing entropy, while inside the octahedron we have the only region which is effectively $4$ dimensional. However it is also conjectured that the entropy inside the arctic octahedron approaches a uniform distribution, unlike the 2-dimensional case where the entropy peaks at the center of the arctic ellipse. -What is the status of these conjectures? Is it expected that for all zonotope tilings ($d>2$) the arctic regions are polyhedral? Is the entropy expected to be constant inside the arctic region? Is there a conceptual reason why the behaviour changes so drastically from planar tilings to higher dimensions? - -REPLY [12 votes]: Basically nothing is known beyond what you've said, and it doesn't look like that's going to change anytime soon. -There is at least one piece of relevant numerical work which I know of: J. Linde, the first author in the paper you mentioned, posted some graphs on his website which suggest that the height function isn't quite flat inside the octahedron. This means that the entropy isn't likely to be constant in there, either. -http://www.joakimlinde.se/projects/rhombusTilings/ -There are some heuristic reasons why we should expect flat facets on the arctic regions for these problems when d>3. For instance, similar problems have been studied with torus boundary conditions and a slightly different lattice, and the height function there is extremely flat. See http://arxiv.org/abs/1005.4636. -Other than that, the only other work I know about these higher dimensional tilings is numerical work on solid partitions, counted by volume. MacMahon made some guesses about how many solid partitions there are of a given volume, which Knuth proved to be incorrect in the sixties. It seems that MacMahon might have guessed an asymptotically correct answer, however, according to the results of this project. It uses massively parallel computation to count solid partitions by volume: -http://boltzmann.wikidot.com/solid-partitions -and their project has generated a paper, -http://arxiv.org/pdf/1105.6231v3 -which I just found and haven't read properly yet. There's also some approximate counting work due to Mustonen and Rajesh, which reaches the same conclusion; it's cited in the above paper.<|endoftext|> -TITLE: Do finite groups acting on a ball have a fixed point? -QUESTION [38 upvotes]: Suppose that $G$ is a finite group, acting via homeomorphisms on $B^n$, the closed $n$-dimensional ball. Does $G$ have a fixed point? - -A fixed point for $G$ is a point $p \in B^n$ where for all $g \in G$ we have $g\cdot p = p$. -Notice that the answer is "yes" if $G$ is cyclic, by the Brouwer fixed point theorem. Notice that the answer is "not necessarily" if $G$ is infinite. If it helps, in my application I have that the action is piecewise linear. -First I thought this was obvious, then I googled around, then I read about Smith theory, and now I'm posting here. - -REPLY [50 votes]: The answer is no. -A fixed point free action of the finite group $A_5$ on a $n$-cell was constructed by Floyd and Richardson in their paper An action of a finite group on an n-cell without stationary points, Bull. Amer. Math. Soc. Volume 65, Number 2 (1959), 73-76. -For some non-existence results, you can see the paper by Parris Finite groups without fixed-point-free actions on a disk, Michigan Math. J. Volume 20, Issue 4 (1974), 349-351. - -REPLY [20 votes]: Bob Oliver classified the finite groups that act without a global fixed point on some sufficiently high-dimensional disk. The conditions are somewhat complicated to state. But for finite abelian groups the conclusion is that such a group acts without fixed points on some disk if and only if it has three or more non-cyclic Sylow subgroups. Here's a link to the original announcement of his result. - -REPLY [6 votes]: The answer is "yes" (it has a fixed point) if the action is affine, i.e. if it satisfies for all $g \in G, x,y \in B^n$ and all $0 \leq t \leq 1$: $$g(tx+(1-t)y)=tgx+(1-t)gy$$. -In that case one can construct a fixed point by taking an $x \in B^n$ and averaging over its $G$-orbit: $$p:=\frac{1}{|G|}\Sigma_{g \in G}\ gx$$ -By convexity of $B^n$ the point $p$ is again in $B^n$ and by the affineness of the action $p$ is indeed afixed point. Linear actions are of course affine, now with your piecewise linear action you have to see whether you can find an orbit which falls into a linear piece, for example. -The groups which allow the above kind of argument are called "amenable groups", as I just learned on monday...<|endoftext|> -TITLE: Can the supremum of continuous functions be discontinuous on a set of positive measure? -QUESTION [7 upvotes]: Given a sequence of continuous functions $f_n(x)$, all defined on a compact set $D$ and assuming $f_n(x)$ is uniformly bounded. Let $f(x) = sup_n f_n(x)$. -It is clear that $f(x)$ is not necessarily continuous. For example, $f_n(x) = 1-x^n, D=[0,1]$. But my questions is can $f(x)$ be discontinuous on a set with positive measure? In the example I give above, $f(x)$ is discontinuous at only $x=1$. -Thanks! - -REPLY [5 votes]: Given a closed set $E$, define the distance $d(x,E)$ from $x$ to $E$ in the usual way. Let $K_n$ denote the set of $x$ so that $d(x,E)\ge 1/n$. Observe that the set $K_n$ is closed and disjoint from $E$. -Urysohn's Lemma now says that there is a continuous function $f_n:\mathbb{R}\rightarrow[0,1]$ which is 0 on $K_n$ and 1 on $E$. The infimum over $f_n$ is then simply the characteristic function of $E$. To translate this to a supremum, simply observe that $\sup (-f_n)=-\inf(f_n)$ -Now you merely need to concern yourself with producing a closed set $E$ whose boundary has positive measure. This can be done using a Cantor-type construction, as mentioned in the comments.<|endoftext|> -TITLE: P-adic C* algebras -QUESTION [18 upvotes]: I understand that there is a definition of p-adic Banach algebras and that a significant amount of functional analysis can be developed in the non-archimedean setting. Is there a p-adic version of C*-algebras? If so, is there an analogue of the GNS construction? - -REPLY [27 votes]: There exists a complete theory of non-Archimedean commutative Banach algebras. In particular, there are conditions under which an algebra is isomorphic to the algebra of continuous functions. For the commutative case, they can be seen as the counterparts of the $C^*$ condition. Note that there is no natural involution in the p-adic case. See -V. Berkovich, Spectral theory and analytic geometry over non-Archimedean fields, AMS, 1990 (the above conditions are in Corollary 9.2.7); -A. Escassut, Ultrametric Banach algebras, World Scientific, 2003. -For the noncommutative case, very little is known.<|endoftext|> -TITLE: A sum involving derivatives of Vandermonde -QUESTION [12 upvotes]: Consider the standard Vandermonde $V(x_1, \ldots, x_n) = \prod_{i < j} (x_i - x_j)$. -I am intersted in the calculation of the following expression for fixed $k$: -$$\sum_i (x_i)^k (d/dx_i)^k V(x_1 , \ldots , x_n).$$ -My guess is that it equals $c \cdot V(x_1, \ldots, x_n)$ -where $c$ is an expression depending on $k$ and $n$ but not on the $x_i$'s. Is it true ? If yes, what is this constant $c$? -I think, if it is true, then it is pretty well-known. -Would you be so kind to provide with the answer and/or proof and/or references? -What can be the context people study it? Symmetric functions? Quantum Calogero-Moser? - -REPLY [6 votes]: The Vandermonde Matrix is $V_{ij}=x^{i-1}_j$. Notice that only the $k-th$ column depends on the $k-th$ variable so we can use the Laplace expansion for the determinant of $V_{ij}$ over such column -\begin{eqnarray} -\sum_kx^n_k\frac{\partial^n}{\partial x_k^n}\underset{1 \leq i,j \leq N}{\det}(V_{ij}) & = & \sum_kx^n_k\frac{\partial^n}{\partial x_k^n}\sum_p x_k^{p-1}C_{pk} \\ - & = & \sum_kx^n_k\sum_p \frac{(p-1)!}{(p-1-n)!} x_k^{p-1-n}C_{pk} \\ - & = & \sum_p \frac{(p-1)!}{(p-1-n)!}\sum_k x_k^{p-1}C_{pk} \\ - & = & \sum_p \frac{(p-1)!}{(p-1-n)!}\underset{1 \leq i,j \leq N}{\det}(V_{ij}) -\end{eqnarray} -were $C_{ij}$ are the cofactors of $V_{ij}$. -After applying the operator, the sum over $k$ yield back a determinant as an expansion over the $p-th$ row.<|endoftext|> -TITLE: Unitary representations of the ax+b group: an accessible presentation -QUESTION [8 upvotes]: The "ax+b group" is the group of affine transformations of $\mathbb R$. It is a locally compact non unimodular group. -Its space of irreducible, continuous unitary representations has been described by Gelfand and Neumark in this 1947 paper. -I am not very familiar with this subject, but in my understanding there are now several proofs to this result. My question is: where can I find a "modern" and accessible presentation of this result? - -REPLY [17 votes]: I know of two clean approaches to classifying the unitary irreps of the $ax+b$ group. The first is to write the group as a semidirect product $\mathbb R \ltimes \mathbb R_{>0}$. There is a theory (due chiefly to Mackey) that deals with the representations of semidirect products. In this case, the semidirect is fairly simple, so the description of the unitary irreps you get from "the Mackey machine" is simple too. -The second approach is to realize that the $ax+b$ group is exponential solvable. This means that Kirillov's orbit method applies. Thus the unitary irreps lie in correspondence with coadjoint orbits of the group on its Lie algebra. As a vector space, the Lie algebra of the $ax+b$ group is $\mathbb R^2$, and the coadjoint orbits turn out to be the half-planes $\{y>0\}$ and $\{y<0\}$ and the singletons $\{(x,0)\}$, so the picture isn't very complicated. -A good, modern reference for both approaches is Folland's A Course in Abstract Harmonic Analysis (CRC Press, 1995), sections 6.7 and 7.6, respectively. Folland also describes the Plancherel formula in section 7.6. I should mention though that Folland doesn't prove absolutely everything when it comes to the orbit method, but he does give references.<|endoftext|> -TITLE: Proability of a word being fulfilled in the symmetric group -QUESTION [5 upvotes]: Let $S_n$ be the symmetric group on $n$ letters, let $g\in S_n$ be a fixed element of order, say, roughly $n^2$ (e.g. two large disjoint cycles of coprime length), let $w(a,b)$ be a reduced word in letters $a$, $b$, $a^{-1}$, $b^{-1}$. - -Question: What is the number of those $h\in S_n$ such that $w(g,h)=1$? - -I'm interested in all kinds of information (e.g. asymptotic behaviour when $n\to \infty$, restrictions on $w$, what if order of $g$ is significantly different,...). - -REPLY [2 votes]: If you search mathscinet for "word maps" you will be enlightened. See, in particular, -MR2453603 (2010g:20022) -Larsen, Michael(1-IN); Shalev, Aner(IL-HEBR-E) -Characters of symmetric groups: sharp bounds and applications. (English summary) -Invent. Math. 174 (2008), no. 3, 645–687. -20C30 (20E45 20P05 60B15)<|endoftext|> -TITLE: Is the square of a curve minus its diagonal affine? -QUESTION [8 upvotes]: Let $X$ be a smooth irreducible projective algebraic curve of genus $g\geq 1$ and $S=X^2$ the surface one obtains as the cartesian product of $X$ with itself. Let $\Delta$ be the diagonal in $S$, that is to say, a copy of $X$ embedded diagonally in $X^2$. -Is the quasi-projective surface $U=S\setminus \Delta$ affine? -One criterion for showing affinness is showing that $\Delta$ is an ample divisor in $S$. But the self-intersection of $\Delta$ is $\Delta^2=2-2g<0$ and therefore $\Delta$ can not be ample. Does this already imply that $U$ is not affine? I guess not. -Besides, Serre's criterion provides a necessary and sufficient condition for $U$ to be affine: this is the case if and only if $H^i(U,\mathcal F)=0$ for all $i>0$ and all coherent sheaves $\mathcal F$ on $U$. But I don't know how to check this in this example. - -REPLY [7 votes]: This answer is inspired by Rita's second paragraph. -It turns out that this $U$ contains projective curves for all $g\geq 1$ and hence cannot be affine: -Embed $X$ into its Jacobian $X\hookrightarrow J$ and let $Z\subseteq J$ be the image of $X\times X$ via the map $J\times J\to J$, $(x,y)\mapsto x-y$. This contracts the diagonal of $X\times X$. If $g>1$, then $Z$ is singular, but still projective. Let $H\subset Z$ be a general hyperplane cut of $Z$ inside $J$ and $C\subset X\times X$ the preimage of $H$. By choice, $H$ avoids the image of $\Delta$ and hence the projective curve $C$ is contained in $U$.<|endoftext|> -TITLE: Maximal tetrahedra inscribed in ellipsoid -QUESTION [8 upvotes]: Pietro Majer quoted the theorem of Michel Chasles in his MO question, -"Convex curves with many inscribed triangles maximizing perimeter," -which states that the triangles of maximum perimeter inscribed in an ellipse form a one-parameter family, all "billiard triangles." -(This theorem is a special case of Theorem 17.6.6 in Berger's - -Geometry II, -p.243, which establishes the result for convex $n$-gons.) -My question is whether or not any generalization to higher dimensions is known, and in particular, -whether there is an analogous theorem for tetrahedra inscribed in ellipsoids in $\mathbb{R}^3$. -Presumably the generalization would state that there is a two-parameter family of -maximum surface-area tetrahedra inscribed in an ellipsoid. -I have not found such a theorem, and would be interested to learn whether or not it (or -some variant) is known. - -                  - - -Update. Here is Henry Cohn's example, just showing the $z=0$ ellipse slice -that contains his degenerate maximal tetrahedra: - -REPLY [6 votes]: I haven't proved anything, just done some numerical experiments, but I do not think there is always a two-parameter family of maximum-surface-area tetrahedra incribed in an ellipsoid (although you do get a two-parameter family of regular tetrahedra inscribed in a sphere). For the ellipsoid defined by $x^2+2y^2+3z^2=1$, the maximum surface area appears to be $2\sqrt{2}$, and it is achieved by the one-parameter family of degenerate tetrahedra with vertices $\pm (\alpha,\sqrt{(1-\alpha^2)/2},0)$ and $\pm (-\sqrt{1-\alpha^2},\alpha/\sqrt{2},0)$ for $-1 \le \alpha \le 1$. These are the only maximum-surface-area tetrahedra my program found for this ellipsoid.<|endoftext|> -TITLE: Classical analogue of the Stone-von Neumann Theorem? -QUESTION [9 upvotes]: Let $U_s$, $V_t$ be a pair of continuous $n$-parameter groups ($n < \infty$) of unitary operators on a complex Hilbert space $\mathcal{H}$. The Stone-von Neumann Theorem establishes that any such pair forming an irreducible representation of the Weyl relations, -$U_sV_t = e^{is\cdot t}V_tU_s$ -is unitarily equivalent to the Schrödinger representation, and hence that all such representations are unitarily equivalent. (Note: the Weyl relations in this context are equivalent to the canonical commutation relations (CCRs) $[Q,P]\psi=i\psi$ for all $\psi$ in the common dense domain of $Q$ and $P$, where $Q$ and $P$ are the generators of $V$ and $U$.) -Question: Is there a known analogue of this result in the context of classical Hamiltonian mechanics? -I don't know of a classical analogue of the Weyl relations. But there is a classical analogue of the CCRs, which is the Poisson bracket $\{q,p\}=1$. So, here's how I imagine a classical analogue of the Stone-von Neumann theorem might look (just a rough attempt, really!). -Let $\mathcal{M}$ be a smooth $2n$-dimensional manifold and $\omega$ a symplectic form on $\mathcal{M}$. Let $\xi = (q,p)$ be any global coordinate system on $\mathcal{M}$, and let $Q:\mathcal{M}\rightarrow\mathbb{R}$ and $P:\mathcal{M}\rightarrow\mathbb{R}$ be the projections onto $q$ and $p$, respectively. Then (conjecture): all such pairs ($Q$, $P)$ satisfying, -$\{Q,P\}=1$ -where $\{\cdot,\cdot\}$ is the Poisson bracket associated with $(\mathcal{M}, \omega)$, are related by a single canonical transformation. -Does this seem like a reasonable way to formulate the classical analogue? Is the status of this conjecture obvious? Your thoughts are appreciated! - -REPLY [6 votes]: Expanding on Chervov's comment: the Jacobian conjecture for two -variables conjectures that if a polynomial map $(x,y) \to (X,Y)$ has for its Jacobian -$\partial(X,Y)/\partial(x,y)$ a nonzero constant, then this polynomial map has a polynomial -inverse. (The chain rule, plus the -fact that over ${\mathbb C}$ polynomials have zeros, yields the truth of the converse: -if the map has polynomial inverse, then its Jacobian is a nonzero constant.) -If $(x,y) = (q,p), (X,Y) = (Q,P)$ then $\{Q, P \}$ is the Jacobian of this transformation. So: an affirmative -answer for the Jacobian conjecture would precisely yield a 'yes' to your 'reasonable formulation' in the polynomial category when $n =1$. -Going back to Stone-von-Neumann and the Weyl relations you wrote down -suggests adding the hypothesis that the flows of the Hamiltonian vector fields -for both $Q$ and $P$ are complete. With that addition, you have a chance -of a 'yes' answer when $n=2$, perhaps in the smooth category. -Note: your hypothesis imply that your manifold is ${\mathbb R}^{2n}$.<|endoftext|> -TITLE: What is the etymology of zero-sharp? -QUESTION [9 upvotes]: I have wondered for a while what gave rise to the notation $0^\sharp$. According to wikipedia this is due to Solovay in 1967, but (perhaps unsurprisingly) there's no discussion of why that notation was chosen. In contrast, Silver mentions in a paper that he chose the symbol $\Sigma$ to represent the same thing. Nonetheless, I couldn't turn anything up with a bit of searching. -So, why did Solovay choose zero-sharp as the name for this object? (The zero part at least makes sense in the general sense of $a^\sharp$ where we consider $L[a]$ instead of $L$) Especially given that there are plenty of ways of indicating something "a little bit more/different" that are quite normal in mathematical notation. - -REPLY [9 votes]: I've heard that the symbol was originally $O^\#$, with a capital letter O, not zero. This set was viewed as a higher-level analog of Kleene's O (a universal $\Pi^1_1$ set).<|endoftext|> -TITLE: Am I allowed to do non-rigorous numerical analysis? -QUESTION [11 upvotes]: I have a paper where I am trying to show that the growth of a certain function is exponential of the order $a^n$. I would like to compute $a$, at least approximately. The base $a$ satisfies a very complicated formula. I do considerable amount of completely rigorous analysis to prove that this formula works, and this is really the interesting part of the paper. -This formula cannot be evaluated exactly, but there is a complicated method using numerical analysis techniques (numerical integration, root-finding, numerical optimization, and so forth) to compute it approximately. -I can use an off-the-shelf numerical package to compute $a$, seemingly to high precision. However, to get a rigorous bound, I would have to go through the entire recipe for computing $a$, showing that all the relevant functions are sufficiently smooth, all of the relevant local minima are in fact the global minima, the functions have the right concavity, the number of digits of accuracy is sufficiently large in each step, and so on. This would be extremely difficult, tedious, and frankly unenlightening --- if you look at a graph of the function it is clear that it has the right smoothness, and I don't want to waste a huge amount of space proving it has these properties. -Is it OK if I just numerically solve for $a$ without proving that my numerical solution is sufficiently accurate? Does it matter that this paper is in the subject of computer science, not pure math? Is there any better alternative? -Thanks for any advice - -REPLY [19 votes]: When I review papers with such assertions, here is what I look for: - -A clear description of the problem, and any known features of the quantity one is interested in (unique root, local minimizer, etc); -A clear description of the method used; -Information on the stopping/ error criteria used. -This latter is rather important - one may stop an algorithm when the successive approximations are 'close' in some norm, or when some residual measure is smaller than some threshold (presuming one's not exceeded a specified total number of iterations.) - -With this information, and a sufficiently modest claim "the computed quantity 'a' appears to provide a good approximation to the desired result'', this reviewer would be happy.<|endoftext|> -TITLE: Feasibility of linear equations with few variables mod k -QUESTION [5 upvotes]: Say I want to verify the feasibility of -$$Ax \equiv b \text{ (mod } k)$$ -where $A \in \mathbb{Z}^{m \times n}, b \in \mathbb{Z}^{m}$ and $k \in \mathbb{N}$. Is there a fast way to verify if there is such an $x$ when $n$ and $k$ are bounded by a constant. Even the case of $n=2$ is of interest to me. - -REPLY [3 votes]: There are two simple approaches. -1. If you prefer to do all of your algebra modulo k, -the best way to do this is to compute the Smith Normal form of the matrix A (or more generally, a related matrix). You can do this by multiplication of unimodular matrices on the left and right of A (i.e. elementary row and column operations respectively) to compute greatest common divisors. - -First, compute the greatest common divisor of the first column, together with k, in the (1,1) coefficient (which will change the value of the first row in general). -Then, compute the greatest common divisor of the first row, together with k, again producing the result in the (1,1) coefficient (which will change the value of the first column in general). - -... And so forth, until the (1,1) coefficient divides all coefficients in the first row and also the first column; then clear the first row and first column. Repeat this for the second row/column, etc. until you have only coefficients on the diagonal; and all the while, keep track of the products of unimodular matrices which you accumulate on the left and right. Doing so yields unimodular matrices U,V such that S = VAU has only elements on the diagonal; and the system Ax ≡ b (mod k) is then equivalent to the system Sz ≡ Vb (mod k), where we perform the substitution z = U−1x. This system is solvable if and only if each row individually is solvable; and for any solution z, we may obtain a solution x = Uz of the original system. -2. An equivalent approach which is not restricted to artihmetic mod k, which is likely to be faster in practise, is to reduce to Diophantine equations. To wit: a linear congruence of the form -$$ \mathbf a_j \cdot \mathbf x \equiv b_j \pmod{k}$$ -should instead be interpreted as the equivalent Diophantine equation -$$ \mathbf a_j \cdot \mathbf x + k d_j = b_j \ ,$$ -introducing a new auxiliary variable dj  for the multiple of k which is neglected in the modular arithmetic. One can express an entire system Ax ≡ b (mod k) of such equations — involving the same modulus; but different moduli for each equation can be easily accomodated — by the system of integer equations -$$ \Bigl[\ \ A \ \ \Big|\ \ k I \ \ \Bigr] (\mathbf x \oplus \mathbf d) = \mathbf b \ .$$ -In reference to Felix's comment above, you then compute a Hermite Normal Form (or any upper-triangular matrix, really) using unimodular transformations. If C = [ A | kI ], then one looks for the unimodular transformation U such that C' = CU is in Hermite normal form (in fact, any upper-triangular form will do); this may be done incrementally by column reduction to clear out rows, starting with the last one. One may then solve the system of equations -$$ C' \mathbf z = \mathbf b $$ -over the reals, performing the substitution z = U -1 (x ⊕ d). The final several coefficients of z will be fixed, and will be integer-valued if there are any solutions at all to the original system of equations; and the first several will be free parameters, albeit ones which should range only over ℤ. Then, we may solve for x by computing (x ⊕ d) = Uz. -[This answer is completely revised, replacing an answer which I deleted after realizing that it was wrong; I subsequently added a reference to the Smith Normal Form above.]<|endoftext|> -TITLE: A Desirable Extension of the Nerve Theorem -QUESTION [9 upvotes]: ##Backgroud -The Nerve Theorem (see nLab;) asserts that given a finite collection $\cal K$ of compact sets with the property that all non empty intersections of sets in the family are homotopically trivial, then the union $X$ of the sets in the family is homotopically equivalent to the nerve $N({\cal K})$. Here the nerve of the family is the abstract simplicial complex that records the intersection pattern so if ${\cal K}=\{K_1,K_2\dots,K_n\}$ then -$N({\cal K})=\{S \subset \{1,2,\dots,n\}: \cap\{K_i : i \in S\} \ne \emptyset$. -The nerve theorem is attributed to Borsuk and Leray (and others). There are various variations. For example if you require that all non empty intersections are homologically trivial then the homology groups of the nerve are the same as those of $X$. -##The desired extension - -Suppose that all non empty -intersections of sets in $\cal K$ are -homotopically equivalent to a single -space $Z$. Under what conditions can -you conclude that $X$ is homologically -the same as the product of Z and the -nerve. - -This is usually not the case, so the question is to find conditions (as simple as possible) that will imply it. -Perhaps the kind of condition you would need is about sets in K intersecting "nicely". For example, we can define two sets A and B as "nicely intersecting" if the usual long exact Meyer-Vietoris sequence splits into short exact sequences. So you can require that every intersection of some sets in $\cal K$ intersect "nicely" every intersection of some other sets in $\cal K$. I don't have reason to believe that this is the definition required for the conclusion but it just illustrates what type of conditions are possible. -A possible application is indicated in this question. -Edit: Oscar Randal-Williams' answer shows that my suggestion for "nice" intersection is not sufficient for the desired conclusion. Perhaps the most interesting question in view of Oscar's result is this. - -Suppose that you have a family $\cal -> K$ of compact subsets of $R^d$ -so that all non empty intersections of members of $\cal K$ -are homotopically equivalent to -to the same space $Z$ and every -intersection of some sets in $\cal K$ -intersect "nicely" every intersection -of some other sets in $\cal K$, where -"nice intersection" means that the -Meyer-Vietoris long exact sequence -splits. Is it true that the nerve of -$\cal K$ has vanishing $i$th homology -groups for $i>d$? - -I think that a positive answer would suffice for the Helly-type application. - -REPLY [12 votes]: The condition you suggest is insufficient. For example, let $F \to E \overset{\pi}\to B$ be a fibre bundle, with compact fibre say, (or more generally a local quasi-fibration), and $\mathcal{K}_B$ be a cover of $B$ satisfying the conditions of the Nerve theorem (i.e. all intersections are empty or contractible). Then -$$\mathcal{K} := \{ \pi^{-1}(K) \subset E \vert K \in \mathcal{K}_B \}$$ -is a cover of $E$ satisfying your Mayer--Vietoris condition, and it is fairly clear that -$$N(\mathcal{K}) = N(\mathcal{K}_B) \simeq B,$$ -and $E$ will not typically be equivalent to $N(\mathcal{K}) \times F$ if the original bundle is not trivial. -So, to answer a different question: given a cover $\mathcal{K}$ of a space $X$ where all non-empty intersections are equivalent to $Z$, I think there will typically be a homotopy fibre seqeuence -$$Z \to X \to N(\mathcal{K}).$$ -Whether or not this is trivial is then another problem. By the example above, every local quasi-fibration arises is this way, so there is unlikely to be an especially simple description of when it is possible. -One condition that will give what you want is the following: if you can find a map $f : X \to Z$ such that for each non-empty intersection $K_\alpha = \cap_{i \in \alpha} K_i$ the map $f\vert_{K_\alpha}: K_\alpha \to Z$ is a homotopy equivalence, then $X \simeq Z \times N(\mathcal{K})$. This is probably much too strong for most applications through.<|endoftext|> -TITLE: For what range of edge probability does the following property hold for random graphs? -QUESTION [10 upvotes]: Let $G(n,p)$ denote the Erdős–Rényi model of random graph. For a given function $p = p(n)$ we say that $G \in G(n,p)$ asymptotically almost surely has property $\mathcal{P}$ if -$$\mbox{Pr}[G \mbox{ has property } \mathcal{P}] \to 1 $$ as $n \to \infty$. -The property $\mathcal{P}$ I am interested in is the following: -For every vertex $v$ there exist vertices $x, y$ such that $N(x) \cap N(y) = v$. (Here $N(x)$ denote the set of neighbors of $x$.) -Note that this is not a monotone graph property. For random graphs it is fairly clear that $\mathcal{P}$ does not hold once $$p \ge \left( \frac{2 \log n + C \log -\log n }{n} \right)^{1/2} ,$$ -for some large enough constant $C>0$ for example, because at that point, every pair $x,y$ has large neighborhood intersection $N(x) \cap N(y)$. -On the other hand, the property also does not hold for small $p$. In particular if -$$p \le \frac{\log n - \omega}{n}$$ where $\omega \to \infty$, then $G(n,p)$ has isolated vertices $v$. -My guess is that $\mathcal{P}$ a.a.s. holds for most of the way between the thresholds for the monotone properties "minimum-degree-$2$", at about $$p = \frac{\log n + \log \log n }{n}, $$ and "for -every pair $x, y$, $N(x) \cap N(y) \neq 0$" given above. -I am more interested in the upper threshold. So the question I would most like to hear the answer to is: - -What is the largest function $p = -> p(n)$ such that $G \in -> G(n,p)$ a.a.s. has property - $\mathcal{P}$? - -I am also interested in how sharp this upper threshold is, and in particular whether the threshold is sharp in the sense of Friedgut and Kalai. -Finally, we could call this property $\mathcal{P}_2$ since it is about intersecting pairs of neighborhoods, and say that a graph has property $\mathcal{P}_k$ if for every $v$ there exist $x_1, x_2, \dots, x_k$ such that $$\bigcap_i N(x_i) = v,$$ -and I'm also interested in this more general setting. - -REPLY [7 votes]: For every $v$, we have $|N(v)|\approx np$ (I assume $p$ is not very small so the fluctuations are small enough to ignore). You want two of these such that $N(x)\cap N(y)=\{v\}$, or in other words, if you now forget about $v$ itself, you want these two sets to be disjoint. -For any specific $x$ and $y$ the probability of this is roughly $(1-p^2)^n \approx e^{-np^2}$. If we take $p=\sqrt{\alpha \log n / n}$ then we get $n^{-\alpha}$. There are $np/2$ disjoint pairs of vertices in $N(v)$ and the choices are independent (well, almost, I'll leave it for you to sort it out). Hence, the probability of not seeing such a pair is bounded by -$$(1-n^{-\alpha})^{np/2}\approx e^{-n^{1-\alpha}p/2}=e^{-n^{\frac12-\alpha}\sqrt{\alpha \log n}}$$. -Taking $\alpha<\frac12$ we get that the failure probability is less than polynomially small, and then union bound over all $v$ is enough. -If I had to bet, I'd say the right $\alpha$ is 1, but I don't have to bet. -What is the motivation for the question, if I may ask?<|endoftext|> -TITLE: A curious generalization of Helly's theorem -QUESTION [11 upvotes]: Here is a curious conjectural extension of Helly's theorem. -It may follow (if true) from a useful theorem of the kind asked in this MO question: - -Conjecture: Let ${\cal F}=P_1,P_2,\dots,P_m$ be a family all - whose members are disjoint union of - two convex sets in $R^d$. Suppose also - that -(1) $m \ge d+2$ -(2) Every intersection of $i$ - members of $\cal F$, $i < m$ is also - the disjoint union of two NONEMPTY compact convex sets. -Then the intersections of all members - of $\cal F$ is not empty. - -Remark: Micha A. Perles showed (in the 70s) that even when $d=2$ you cannot replace "two" by "48". - -REPLY [2 votes]: This answer shows that you cannot strengthen the conditions of the questions and demand condition (2) only for $m-1$ sets. -Answer for new version IF we only consider the intersection of m-1 sets (as construction fails otherwise, as pointed out by Gil in the comment): -Now it is not hard to prove that the statement is true for d=1 but there is a counterexample for d=2. Take a circle and divide its perimeter into six equal parts, A, B, .., F such that e.g. A, B and C are on the top. Now take a point somewhere high, P, and another somewhere low, Q. Our four sets will be the following: conv(A,P) $\cup$ conv(D,Q), conv(B,P) $\cup$ conv(E,Q), conv(C,P) $\cup$ conv(F,Q) and finally the last set is the disc. Now the intersection of the first three will be around P and Q, while the intersection of the fourth with any two other sets will be around two disjoint arcs of the perimeter.<|endoftext|> -TITLE: Binomial coefficient in Andrews' partition book -QUESTION [9 upvotes]: First of all, i think MathOverflow is a very great community to discuss math, either basic or advanced, and i'm glad to participate here. It's my first post, so i'm sorry if i did anything wrong, and hope that people help me to make this post better. I also hope this post helps somebody with similar problem... -I've studied partition theory for my undergraduate math monography, and the Simon Newcomb's problem is a topic of the essay. The masterpiece of George Andrews, "The Theory of Partitions" is my main guide; I reached a "binomial identity" that i cannot proof. In fact, Andrews proves it, but i cannot understand the proof clearly, and more important, his demonstration uses Gaussian Polynomials, a topic that my work doesn't cover. -That's the identity: -$$ -\sum_{\substack{i+j=s \atop i\geq 0, j \geq 0}}\binom{A-n+j}{j}\binom{n-j}{i} = \binom{A+1}{s}, -$$ -for positive integer $A$. -I'm asking here an "elementary" proof that doesn't use Gaussian Polynomials. -Sorry to bother you all with so basic question (comparing with advanced stuff that is posted here), and thanks in advance. -P.S. Sorry for my "not so good" english... - -REPLY [2 votes]: Your identity is the same as $f_s:=\sum_{j=0}^s\binom{A-n+j}j\binom{n-j}{s-j}\binom{A+1}s^{-1}=1$ for all $s\geq0$. Define -$$F(s,j):=\binom{A-n+j}j\binom{n-j}{s-j}\binom{A+1}s^{-1} \qquad \text{and} \qquad -G(s,j)=-\frac{j(n+1-j)F(s,j)}{(s+1-j)(A+1-s)}.$$ -Check routinely (say dividing both sides by $F(s,j)$ and simplifying) that -$$F(s+1,j)-F(s,j)=G(s,j+1)-G(s,j). \tag1$$ -Summing both sides over all integers $j$ and telescoping on the RHS gives -$$f_{s+1}-f_s=\sum_{j\in\mathbb{Z}}G(s,j+1)-\sum_{j\in\mathbb{Z}}G(s,j)=0.$$ -Note: we have used the convention $\binom{a}b=0$ of $b -TITLE: Fluctuations of Liouville function -QUESTION [7 upvotes]: One of my friend (who is working in mathematics) was asking the following question. Let us take Liouville λ(n) function. -et S={ λ(1), λ(2), λ(3), ..... } . Then every finite length (say l) subsequence of S occurs infinitely many times. In other words every finite block of $\pm$ signs occurs in this sequence infinitely often i.e. for any given numbers $\epsilon_i = \pm 1 ( 1 \le i \le l)$, there are infinitely many integers $n>= 1$, such that $f(n+i)=\epsilon_i (1 \le i \le l )$. -I can prove it for $l=1,2. l=1$ is the trivial case. -For $l=2$, the case for ${+1,-1}$ or ${-1,+1}$ follows trivially from $l=1$ case. -For ${+1,+1}$ or ${-1,-1}$ case: -Let us take a odd positive number n. Let us also choose f(n)=+1. This -is trivially accomplished by taking n as a product of even number of -primes. if n+1 or n-1 has f(n)=+1 then we get a pair with {+1,+1}. -Else, f(n+1)=f(n-1)= -1. Hence f(n+1/2)=f(n-1/2)=1 and this gives us a -pair {+1,+1}. This proves the case for {+1,+1}. -Similarly it may be proved for {-1,-1}. -What is known about $k \ge 3$. Is it something already proved ? - -REPLY [13 votes]: Hildebrand (On consecutive values of the Liouville function, Enseign. Math. (2) 32 (1986), 219–226) proved the conjecture for $l=3$, i.e. all 8 combinations $\pm 1,\pm 1,\pm 1$ occur infinitely often in the Liouville sequence. Christian Elsholtz proved very recently that all 16 combinations $\pm 1,\pm 1,\pm 1,\pm 1$ occur along arithmetic progressions infinitely often. I believe this is the state of the art. -Added. My response above is almost 7 years old. For the current state of the art, I recommend watching Tao's recent lecture at the Building Bridges II. conference.<|endoftext|> -TITLE: Is it consistent relative to ZF that $\frak c = \aleph_\omega$? -QUESTION [13 upvotes]: In ZFC we know that the continuum cannot have cofinality $\omega$. -However, in the Feferman-Levy model we have that $\frak c=\aleph_1$, and that $\operatorname{cf}(\omega_1)=\omega$. In fact in the Feferman-Levy model, $\aleph_\omega^L=\aleph_1^V$. -Is it consistent with ZF that $\frak c=\aleph_\omega$? Does that mean that the only restriction in ZF on the cardinality of the continuum is $\aleph_0<\frak c$? - -REPLY [13 votes]: The answer is no. The continuum cannot be $\aleph_\omega$, and this can be proved in ZF, that is, without using the axiom of choice. To see -this, suppose towards contradiction that $P(\omega)$ is equinumerous with -$\aleph_\omega$. Since $P(\omega)$ is equinumerous -with $P(\omega)^\omega$, and this does not require AC, it follows -that there is a bijection $f:\aleph_\omega\cong -(\aleph_\omega)^\omega$. Let $g(n)$ be the first ordinal -$\alpha\lt\aleph_\omega$ that is not among $f(\beta)(n)$ -for any $\beta\lt\aleph_n$. Since there are fewer than -$\aleph_\omega$ many such $\beta$, it follows that there -are fewer than $\aleph_\omega$ many such $f(\beta)(n)$, and -so such an $\alpha$ exists. Thus, $g:\omega\to -\aleph_\omega$. But notice that for any particular -$\alpha\lt\aleph_\omega$, we have $\alpha\lt\aleph_n$ for -some $n$ and consequently $g(n)\neq f(\alpha)(n)$, and thus -$g\neq f(\alpha)$. Thus, $f$ was not surjective to -$(\aleph_\omega)^\omega$ after all, a contradiction. -This is just a standard proof of Konig's theorem (that -$\aleph_\omega^\omega\gt\aleph_\omega$), and the point is -that it doesn't use AC.<|endoftext|> -TITLE: the example of ccc but not separable -QUESTION [7 upvotes]: I am interested in the relation between the property of countable chain condition (ccc) and the property of separable. Could someone recommend some papers or books about this to me? thanks in advance. - -REPLY [3 votes]: If you're interested in the relationship between the ccc and separability, you should read Stevo Todorcevic's survey "Chain condition methods in topology". -http://www.sciencedirect.com/science/article/pii/S0166864198001126 -It's a very convincing pamphlet on the power of chain conditions, clarifying that relationship with such beautiful theorems as: -1) (Todorcevic) Let X be compact Hausdorff. If every subspace of $X^2$ has the ccc then $X$ is separable. -2) (Rosenthal) A compact Hausdorff space $X$ is ccc if and only if every weakly compact subspace of $C(X)$ is separable. -Also, there's a whole book dedicated to chain conditions. It's called "Chain conditions in topology", by Wistar Comfort and Stelios Negropontis (Cambridge Tracts in Mathematics #79, Cambridge University Press).<|endoftext|> -TITLE: Is the class of elementary integrals "small" ? -QUESTION [8 upvotes]: This I read in a paper: -"The class of integrals that are elementary is very -small compared with nonelementary integrals." -What is the precise meaning of this sentence? E.g., does that mean that the former class of functions is meagre (in a suitable functional space) while the latter is not ? Is there a reference for such a subject ? -Thank you - -REPLY [7 votes]: It is small in the same sense that the set of polynomials solvable by radicals is small. The canonical reference on the subject is probably the late, lamented Manuel Bronstein's book: -Symbolic Integration I: Transcendental Functions (Algorithms and Computation in Mathematics) (v. 1) [Hardcover] -Otherwise, look up "differential algebra" or "Risch Algorithm".<|endoftext|> -TITLE: Basis for the space of Harmonic homogeneous polynomial in N variables. -QUESTION [14 upvotes]: Hello, -Does someone know an explicit basis of the space of harmonic homogeneous polynomial in N variables. -When $N=3$, if I'm not mistaking Legendre polynomial allow to write an explicit basis. -Is there a known explicit basis when $N > 3$ ? -Thanks for your answers, and reference in case you know one. - -REPLY [17 votes]: Let $K$ denote the Kelvin transform, and let $|\alpha|:=\sum_{j=1}^n\alpha_j$ denote the weight of the multi-index $\alpha\in\mathbb{N}^n$. Then, an explicit base for the space of homogeneous harmonic polynomials in $n$ variables and degreee $m$, $\mathcal H^m:=\mathcal H^m[x_1,\dots ,x_n]$, is -$$\big\{ K\big(\partial^\alpha\|x\|^{2-n}\big) : |\alpha|=m, \alpha_n\leq 1 \big\}.$$ -Indeed, an easy combinatorial computation shows it has the right cardinality $$\operatorname{card}\{\alpha\in\mathbb{N}^n : |\alpha|=m, \alpha_n\leq 1 \}= -{n+m-1\choose n-1} - {n+m-3\choose n-1} = \operatorname{dim}\mathcal H^m,$$ -(the latter dimension being already known from linear algebra considerations on the operator $\Delta$). On the other hand, one can verify that it spans the same linear space as the analogous set without the constraint on $\alpha_n$: -$$\{ K\big(\partial^\alpha\|x\|^{2-n}\big): |\alpha|=m\},$$ -which is the whole space $\mathcal H^m$. So it is a base. -A nice (and free) reference for these classical facts is Harmonic Function Theory, by S.Axler, P.Bourdon, and W.Ramey; see thm 5.25.<|endoftext|> -TITLE: Infimums of exponential sums involving primes -QUESTION [9 upvotes]: Hi, I don't know if this question is appropriate for Math Overflow but I was wondering if there is anything known about the following: Let -$$ -S(\alpha) = \sum_{n \leq x}\Lambda(n)e(n\alpha). -$$ -Then asymptotically, how small can -$$ -\inf_{\alpha}\left|S(\alpha)\right| -$$ -be relative to $x$? Also, for each $x$, if $\alpha_x$ is a value at which the sum takes the infimum, then how are the $\alpha_x$ distributed in $(0,1)$ as $x \rightarrow \infty$? -Ok, thanks. - -REPLY [14 votes]: Timothy, -This is likely to be a pretty difficult question I think. For a random sequence of $\pm 1$s in place of the von Mangoldt function $\Lambda(n)$ the answer is a little surprising: the infimum is basically $1/\sqrt{x}$, a result of Konyagin and Schlag. This is available here: -www.math.uchicago.edu/~schlag/papers/POLTRAN.pdf -I say surprising because most people, if they were given 10 seconds to guess the answer, would probably go for $\sqrt{x}$ (I certainly would have). -I'm not sure there's any real reason to suppose that the answer for the deterministic function $\Lambda(n)$ will be much different, except perhaps in logarithmic factors. -I think you have precisely no chance of saying anything useful about the $\alpha_x$, but maybe someone will prove me wrong! I would be surprised if they were not close to equidistributed, though there may be some repulsion effects away from rationals with small denominator (where $S(\alpha)$ will be large). -EDIT: Thinking about it some more, it's not obvious to me even how one would show that $S(1/x) \neq 0$, though maybe this does follow from some kind of lower bound for linear forms in $\log p$. My point is that if there is deviation from the behaviour for a random sequence I would expect that one would see it near $\alpha = 0$ (and near other rationals with small denominator).<|endoftext|> -TITLE: Etalé space construction for presheaves on a Grothendieck site -QUESTION [11 upvotes]: As it is described for example in [Mac Lane-Moerdijk, Sheaves in Geometry and Logic, II.6.], one can construct the sheafification functor very lucidly by associating to a presheaf a certain bundle (cf. espace etale) and then taking its sheaf of sections. -This construction is outlined in the reference for presheaves $\mathcal{O}(X)^{op}\to Sets$ where $\mathcal{O}(X)$ is the category of opens for a topological space $X$. -Does this method work for presheaves $C^{op}\to Sets$ on a general small Grothendieck site, too, and is this written down somewhere? -(Let's assume also that the associated topos has enough points.) - -REPLY [9 votes]: It depends on what you mean by an étalé "space". As long as $C$ has a small set of topological generators (i.e. as long as $Sh(C,J)$ isn't too large to be a topos), there always exists a certain version of an étalé space: If $F$ is a sheaf on $(C,J),$ the slice topos $Sh(C,J)/F$ has a canonical étale projection $$\pi_F:Sh(C,J)/F \to Sh(C,J).$$ This map is a local homeomorphism of topoi. This topos with this local homeomorphism is the étalé space of $F.$ Indeed, we may make this construction for each object $c \in C,$ call it $U(c):=Sh(C,J)/y(c),$ where $y(c)$ is the (possibly sheafified) Yoneda embedded object. Then, "sections of $\pi_F$ over $U(c) \to Sh(C,J)$" are in bijection with elements of $F(c).$ If the Grothendieck site $(C,J)$ happened to be the canonical site of a topological space, then each slice $Sh(C,J)/F$ is equivalent to sheaves on the étalé space of that sheaf, and the projection corresponds to the usual one. In particular, $U(c) \to Sh(C,J)$ corresponds to the inclusion of an open subset. So, this is reduces to the usual construction for spaces. Another example is, if $(C,J)$ were the small étale site of some scheme $S$, then each $Sh(C,J)/F$ is the small étale site of some algebraic space (with no seperation conditions) étale over $S,$ with $\pi_F$ corresponding to the étale map from this algebraic space to $S.$<|endoftext|> -TITLE: Image of complex conjugation by modular representations in characteristic 2 -QUESTION [8 upvotes]: The question I am going to ask looks well-known, and I even may have heard things about it (but since I used to be deaf to anything in characteristic 2, whatever I heard has never been recorded in my mind) so it may be seen as a request for -references. I have not been able to find a reference myself. -Let $f$ be a cuspidal eigenform of weight $k \geq 2$ for some congruence subgroup of $Sl_2(\mathbb{Z})$. Then as is well-known, for every prime $p$, there exists a unique, absolutely irreducible, Galois representation -$\rho : G_{\mathbb Q} \rightarrow Gl_2(K)$ where $K$ is a suitable finite extension of $\mathbb Q_p$, odd (that is such that $\rho(c)$ is conjugate to the diagonal matrix $(1,-1)$), and satisfying the Eichler-Shimura relations. -I am interested in the case $p=2$. Let $A$ be the ring of integers of $K$, $m$ is maximal ideal, and $k=A/m$ the residue field, of characteristic $2$. I want to reduce $\rho$ mod $m$. As is still well-known, there are several -way to do that, one for each choice of a stable $A$-lattice $\Lambda$ in $K^2$: one defines the representation $\bar \rho_\Lambda$ over $k$ as the action of $G_{\mathbb Q}$ on -$\Lambda/m \Lambda$. The various $\bar \rho_\Lambda$ have all the same semi-simplification. - -Now my question: What is the conjugacy class of $\bar \rho_\Lambda(c)$ in $GL_2(k)$? - -The characteristic polynomial of $\bar \rho_\Lambda(c)$ is $X^2-1 = (X-1)^2$ in $k$, -so either this matrix is the identity, or it is conjugate to the unipotent matrix -$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. I'd like to know when (that is for which $f$, $\Lambda$) we are in the first case, and when we are in the second case. -Note: the fact that $\rho(c)$ is conjugate to the diagonal matrix $(1,-1)$ does not trivially implies that -$\bar \rho_\Lambda(c)$ is necessarily the diagonal matrix $(1,1)$, because the -two eigen-lines of $\rho(c)$ may not be in good position w.r.t the lattice $\Lambda$, -that is the sum of their intersections with $\Lambda$ may be a proper sub-lattice of $\Lambda$. For example if $\rho(c)$ is the anti-diagonal matrix in the canonical basis of $K^2$, and $\Lambda = A \oplus A$ is a stable lattice, then $\bar \rho_\Lambda(c)$ -is clearly not the identity. - -REPLY [5 votes]: Joel -- it's difficult to work out what you're asking. Of course both possibilities can occur, as Wanax said. Furthermore both possibilities can occur even for the same modular form. For example, if you consider the 2-adic representation attached to Ramanujan $\Delta$ then there is a lattice for which $\overline{\rho}(c)=1$ and another lattice for which it is not 1 (in fact for any quadratic extension of the rationals with discriminant a power of two there is some lattice which gives rise to the reducible non-semisimple representation with kernel corresponding to this field). And already in Serre's 1987 paper on his conjecture he observes that for any $S_3$ extension of $\mathbf{Q}$, totally real or not, it will be modular, and for $A_5$ extensions, totally real or not, computationally they seem to work too (thanks to Mestre). -If you gave me a form in practice, I think that I would do a global calculation to try and locate the kernel of the mod 2 representation (you know where it's ramified, you know an upper bound for the degree and you know lots about how primes split so you can use tables to find the kernel), and then ask if it's totally real or not. Certainly I don't know a local method (which uses only information at 2 and infinity) and I'm not sure it would be reasonable to expect one...<|endoftext|> -TITLE: Symmetric basis of harmonic homogeneous polynomials -QUESTION [5 upvotes]: Recently, a question about the beautiful theory of harmonic polynomials made me aware there is something -I've wanted to know for a long time. -As is well known, for any number of variables $n$ and any degree $m$, there is a permutation invariant family of polynomials that generates linearly the space $\mathcal{H}_n^m$ of all (real) harmonic homogeneous polynomials in $n$ variables of degree $m$: -$$\mathcal{F}_ n^{\, m}:=\big\{ K\big(\partial^\alpha\|x\|^{2-n}\big)\, :\, \alpha\in\mathbb{N}^n\, ,\, \alpha_1+\dots+\alpha_n=m\, \big\}\, ,$$ -where $K$ denotes the Kelvin transform, $Ku(x):=\|x\|^{2-n}u(\|x\|^{-2}x)$. Here, by a permutation invariant set of polynomials I simply mean, if $P(x_1,\dots,x_n)$ is in it, then $P(x_{\sigma(1)},\dots,x_{\sigma(n)})$ also is in it, for any permutation $\sigma \in \mathfrak{S}_n$. The family $\mathcal{F}_ n^{\\, m}$ is not linearly independent; however, taking only the elements corresponding to a suitable subset of multi-indices, namely, imposing e.g. $\alpha_n\le 1$, one actually finds a basis. What makes me a bit unhappy is that this constraint breaks the symmetry. -So I wonder: - -Is there an explicit permutation invariant subset of $\mathcal{F}_ n^{\\, m}$ - which is a basis of $\mathcal{H}_n^m$ ? If not always, for what $n$ and $m$? And, more generally, how to make an explicit permutation invariant basis ? - -Actually, I do not see obstructions to the existence of such a basis, but I would also not be completely sure about how to prove its existence. Before starting and trying to answer by myself, I'd like to pose the question here. My apologies if it turns out to be trivial! - -REPLY [5 votes]: Already the desired result is false for $n = 3, m = 2$, but for simpler reasons than I suggested in the comments. In this case the polynomials you give are $x^2 - 2y^2 + z^2, xy$ and their permutations. The sum of the permutations of $x^2 - 2y^2 + z^2$ is zero, so none of its permutations can be part of a permutation-invariant basis, and the span of the permutations of $xy$ don't span the entire space.<|endoftext|> -TITLE: Is there an intuitive reason for Zariski's main theorem? -QUESTION [68 upvotes]: Zariski's main theorem has many guises, and so I will give you the freedom to pick the one that you find to be most intuitive. For the sake of completeness, I will put here one version: -Zariski's main theorem: Let $f:X\rightarrow Y$ be a quasi-finite separated birational map of varieties, where $Y$ is normal. Then $f$ is an open immersion. -There is no reason to pick this particular formulation. In fact, every formulation seems to me like a technical lemma rather than a theorem with geometrically intuitive content. -Question -Is there a formulation of Zariski's main theorem that has an intuitive/pictorial ``reason'' for it? Or is Zariski's main theorem in its core a technical result with no geometric reason? - -REPLY [20 votes]: Reading these excellent answers made me realize I did not understand the ZMT in an intuitive way, and helped me very much get at least more intuition for it than I had. I want to try to summarize what I have learned from thinking about the other answers here, in a way that is not too technical and hopefully intuitive. I do not consider the proofs, but only the intuition contained in the statements. -The first step is necessarily a little technical since it involves the definition of normality. -1) If X is a normal affine variety, i.e. if its affine ring has no non trivial module - finite extension within its quotient field, then it follows from this definition, that X accepts no non trivial finite birational morphism. Thus X is normal if and only if every finite birational morphism Y→X is an isomorphism. -2) ZMT then implies that every such normal variety is unibranch. I.e. if X has more than one branch at any point, then X must accept a non trivial finite birational map Y→X. The geometric intuition here is that it should be possible, by a finite morphism, to separate the branches of any variety that is not unibranch. This may have been suggested to Zariski by examples such as the projection cited by Francesco. Thus ZMT establishes that the only way for a variety to possess more than one branch is for it to be the target of a map similar to a “projection”. (Mumford's nice topological and power series statements are just alternate ways to say "unibranch".) -3) As Sandor pointed out, then the connectedness theorem follows naturally for a normal variety X, since if Y→X were finite, birational, and some fiber f^(-1)(p) were disconnected, then X should have at least two branches at p, hence X should not be normal. -4) The next natural piece of intuition is Grothendieck’s theorem that all quasi-finite morphisms can be completed to finite ones, as Matt and Akhil observe. Thus the only way to get a quasi - finite morphism is to restrict a finite morphism to an open set, a very intuitive geometric statement. In particular, since there are by definition no non - trivial finite birational morphisms to X, there cannot be any non trivial quasi- finite birational ones either; i.e. every such morphism Y→X is an open immersion. -5) As a consequence, a birational morphism Y→X which is not trivial, i.e. not an open immersion, cannot be quasi – finite, hence must have a positive dimensional fiber f^(-1)(p). -So for me at least, these various statements have all become geometrically natural, thanks to contemplating the other answers, admittedly in a naïve way.<|endoftext|> -TITLE: Basic properties of Nisnevich cohomology; $l'$-topology? -QUESTION [7 upvotes]: I would like to know more about Nisnevich cohomology (especially, on its properties that could be easily formulated). In particular, I would like to know which of the following statements are true, and what restrictions are needed for them. - -The higher Nisnevich cohomology of a smooth (is this necessary?) variety $V$ (over a perfect field $k$) with coefficients in a constant sheaf is zero. - -The results of Suslin and Voevodsky reduce this statement to its Zariski counterpart; yet they do not tell much about the case of a non-smooth $V$ (especially, if the characteristic of $k$ is positive, so that one cannot apply cdh-descent). Does there exist an easier reasoning? - -Assertion 1 seems to imply: for an open embedding $j$ of smooth varieties, the higher direct images $R^ij_*$ of a constant sheaf are zero (for $i>0$). Again, is smoothness necessary here? -Total direct images from derived categories of etale sheaves to those of Nisnevich ones commutes with inverse images with respect to embeddings. It seems easy to check this is stalks; yet I wonder which restrictions are required for this result. - -Upd. It seems that assertion 1 fails already when $V$ is an irreducible nodal cubic. Hence assertion 2 is wrong for the embedding of a smooth open subvariety of $V$ into $V$. -It seems that over characteristic zero field one can fix this by considering cdh-topology instead of the Nisnevich one (and h-topology instead of the etale one; this does not seem to affect the cohomology of smooth varieties with coefficients in constant sheaves). Does anybody know whether a similar scheme could be applied in positive characteristic $p$ to $l$-torsion constant sheaves ($l$ is a prime distinct from $p$) if one considers Gabber's $l'$-topology? - -REPLY [2 votes]: For smooth schemes you can use (the same argument as for) a Gersten resolution to show that the cohomology of a constant sheaf agrees with the cohomology of the generic point, and -this vanishies as fields have no higher Nisnevich cohomology. -I believe you could do the $l$-part in characteristic $p$ with Gabbers $l$-topology. As far as I know, Chisinki and Kelly have some results in this direction.<|endoftext|> -TITLE: Spectral sequence of symmetric or exterior algebras? -QUESTION [5 upvotes]: This question is inspired by Hartshorne's exercise II.5.7 (c-d): the problem reads: -Let $0\rightarrow \mathcal{F}'\rightarrow\mathcal{F}\rightarrow\mathcal{F}''\rightarrow0$ be a short exact sequence of locally free sheaves. Then for any $r$ there is a finite filtration $F$ of $S^r(\mathcal{F})$ (the sheaf of symmetric algebras over $\mathcal{O}_X$) such that $F^pS^r(\mathcal{F})/F^{p+1}S^r(\mathcal{F})$ is isomorphic to $S^p(\mathcal{F}')\otimes S^{r-p}(\mathcal{F}'')$ (and similarly for exterior algebras). -So basically, I had this feeling that a spectral sequence might be lurking around somewhere; my thinking is this: if we can write down a spectral sequence of a graded complex that converges to $S^r(\mathcal{F})$, then we would naturally be guaranteed a grading of this object which agrees with the $E_\infty$-page. So is there some sort of way to write down a complex so that we heuristically have something like $E_2^{p,q}\Rightarrow S^{p+q}(\mathcal{F})$ where if we go through and read off the sum of the diagonals of the $E_\infty$-page, we get something like $S^{p}(\mathcal{F}')\otimes S^{r-p}(\mathcal{F}'')$? - -REPLY [8 votes]: Take the symmetric power of the complex $F'' \to F'[1]$. However, this is a very exotic point of view on the question. -ADDITION. A complex is a chain of morphisms with zero compositions. You can consider complexes in the derived category as well. In some sense those can be thought of as bicomplexes in the original abelian category. Now if there is a two-term complex $A \to B$ you can consider its symmetric power: -$$ -S^n A \to S^{n-1}A\otimes B \to S^{n-2}A \otimes \Lambda^2B \to \dots \to \Lambda^n B. -$$ -If you apply this to the complex $F'' \to F'[1]$ and note that $\Lambda^k(F'[1]) = S^kF'[k]$, you will get a complex -$$ -S^nF'' \to S^{n-1}F''\otimes F'[1] \to S^{n-2}F'' \otimes S^2F'[2] \to \dots \to S^nF'[n]. -$$ -Since the original complex was quasiisomorphic to $F$, this one is quasiisomorphic to $S^nF$. Now the stupid filtration of this complex gives the required filtration of $S^nF$.<|endoftext|> -TITLE: References for bad reduction of Jacobians of modular curves? -QUESTION [6 upvotes]: Hi, -Where can I learn about the reduction of the Jacobians of modular curves -such as X_0(N) and X_1(N) at primes p dividing N? -Thanks! - -REPLY [17 votes]: The standard places that one learns this are (or at least, used to be): Mazur's Eisenstein ideal paper (which treats the case of $X_0(N)$ for $N$ prime in great detail), -Ribet's papers (his Herbrand criterion paper, his Warsaw ICM talk, his Inventiones 100 paper, -and several others as well), Gross's Tameness criterion paper, and the papers of Mazur--Wiles. Most of these references (Mazur--Wiles is one exception) -tend to focus primarily or even exclusively on the case when $p$ exactly divides $N$. -One reason for this is that, in the case when $p$ exactly divides $N$, the Jacobian -is semi-abelian when reduced mod $p$ (i.e. an extension of an abelian variety by a torus) -(at least after making a base-change, in the $\Gamma_1$ case). -The situation is worse when $p^2$ divides $N$. - -There is another way to think about this question, which is not typically how people learned it when I was a student, but might be a better way to proceed nowadays. Namely, -$J_0(N)$ or $J_1(N)$ is (up to isogeny) a product of abelian varieties $A_f$ attached -to Hecke eigenforms $A_f$. Now the reduction of $A_f$ is (more-or-less) determined -by the amount of ramification at $p$ in the $\ell$-adic Tate module of $A_f$, -which is to say, in the ramification at $p$ of the $\ell$-adic Galois representation $\rho_f$ -attached to $f$. (Here $\ell$ can be any prime distinct from $p$.) -Now the restriction to a decomposition group at $p$ of $\rho_f$, -and so, in particular, the ramification at $p$ in $\rho_f$, is determined by the the local factor at $p$ of the automorphic representation generated by $f$. (This is the celebrated -local-global compatibility theorem of Langlands--Deligne--Carayol.) This local factor -encodes information such as the power of $p$ dividing the level of $f$ and the eigenvalue -of $U_p$ acting on $f$ (but in general it contains more information than just these -facts). -A typical concrete consequence of this analysis is that if $p$ is prime to $N$, -and if we write $J_1(N;p)$ to denote the Jacobian of the modular curve of level -$\Gamma_1(N)\cap \Gamma_0(p)$, then $J_1(Np)/J_1(N;p)$ has good reduction over $\mathbb Q_p(\mu_p)$. - -Of course, the two methods aren't really different --- the arguments with arithmetic geometry and moduli of elliptic curves that appear in the first stream of literature mentioned are also used to make the deductions about local-global compatibility in Carayol's paper. But the automorphic viewpoint of Langlands--Deligne--Carayol is more powerful, -and also more efficient, in some sense: it collects all the geometry in one argument, -and then phrases the conclusions in the very efficient "local-global compatibility" language. Unfortunately, the literature in this stream is a little more austere than -in the first stream, and the first stream also benefits from the fantastic expositional skills of Ribet and Gross especially. - -In conclusion, I might begin with Ribet's Inventiones 100 paper and Gross's Tameness criterion paper, and after this, try to understand the statements of Carayol's local-global compatibility results (and see how to recover the results about Jacobians in Ribet and Gross's papers from Carayol's more general statements). - -One more thing: Serre and Tate's paper On the good reduction of abelian varieties -is a terrific read, and is good for learning how the Galois action on its Tate modules -influences the reduction type of the abelian variety.<|endoftext|> -TITLE: Cohomology groups of homogeneous spaces -QUESTION [13 upvotes]: Is there a general method to calculate the cohomology groups of homogeneous spaces ($G/H$), such as $\frac{U(4)}{U(2)\times U(2)}$, $\frac{U(5)}{U(2)\times U(3)}$, $U(4)/U(2)$, etc. If yes, could you give the references for the method? Can this be achieved by computer algebra? Thanks. -In general textbooks, there are only examples for special cases, such as $S^{n-1}=SO(n)/SO(n-1)$, and the method cannot be generalized for all homogeneous spaces. - -REPLY [5 votes]: I think that the best way to compute this cohomology is the following. The homogeneous spaces -you are looking at are all of the form $X=G/M$ where $G$ is compact and $M$ is a connected subgroup of $G$ Let $T_M$ be a maximal torus of $M$, embedded into a maximal -torus $T_G$ of $G$ and let $\mathfrak t_M$ $\mathfrak t_G$ be the corresponding (complexified) Lie algebras. Let us first look at the equivariant cohomology $H^*_G(X)$ (say, with $\mathbb C$-coefficients). It is obvious that -it is the same as $H^*_M(pt)$ (here $pt$ denote "the point") which is known -to be $Sym(\mathfrak t_M^*)^{W_M}$; here $Sym$ means "symmetric algebra", and $W_M$ means the -Weyl group of $M$. By abstract nonsense it is clear that -$H^*(X)$ -is just $H^*_G(X)\underset{H^*_G(pt)}\otimes {\mathbb C}$, -where $\otimes$ in principle means -"derived tensor product". If $M$ and $G$ have the same rank, then - you can show that -$H^*_G(X)$ is always free over $H^*_G(pt)=Sym(\mathfrak t^*)^{W_G}$ (here I denote -$\mathfrak t=\mathfrak t_M=\mathfrak t_G$) hence -you finally get -that $H^*(X)$ -is equal to $Sym(\mathfrak t^*)^{W_M}\underset{Sym(\mathfrak t^*)^{W_G}}\otimes{\mathbb C}$. -In the general case, you need to compute the above derived tensor product, which -in every specific case is usually easy to do.<|endoftext|> -TITLE: "name" for the ground model -QUESTION [5 upvotes]: I am a beginner of forcing, often I read from some articles something like "$p \Vdash \dot{G}$ is $P$-generic over $\check{M}$" (where $M$ is a countable transitive model, for instance). -Q1. I learnt from Jech's book a definition of "$p \Vdash \dot{x} \in \check{M}$", but I don't know how to translate "$p \Vdash \dot{G}$ is $P$-generic over $\check{M}$" into a formal version using this. -Q2. I also learnt from Kanamori's book that M is a definable proper class of $M[G]$ whenever $G$ is generic over $M$, why definable (i.e. of the form $\lbrace x \in M[G]: \phi(x) \rbrace$)? - -REPLY [7 votes]: In the Boolean-value approach to forcing, one may introduce a new predicate symbol $\check M$ into the forcing language, and then define that $[[\tau\in\check M]]=\bigvee_{x\in M}[[\tau=\check x]]$. That is, the Boolean value that $\tau$ is in $\check M$ is precisely the extent to which it is equal to something in the ground model. This definition obeys the equality axioms, and so one may expand the forcing language to include this new predicate for the ground model. -The forcing relation $p\Vdash\varphi$ just means $p\leq[[\varphi]]$, identifying the condition $p$ with its copy in the Boolean algebra, and therefore $p\Vdash\tau\in\check M$ just means that $p\leq[[\tau\in\check M]]$ as I defined it above. One may avoid the use of Boolean values via the equivalent formulation: - -$p\Vdash\tau\in\check M$ if and only if the set of $q$ for which there is $x\in M$ with $q\Vdash\tau=\check x$ is dense below $p$. - -It follows that whenever $G$ is $M$-generic for $P$, then the collection of $\text{val}(\tau,G)$ for which $p\Vdash\tau\in\check M$ is precisely the same as $\text{val}(\check x,G)=x$ for $x\in M$. That is, the interpretation of $\check M$ by a generic filter $G$ is precisely $M$, which is as it is desired. -For your second question, my belief is that the definability of the ground model $M$ in its set forcing extensions $M[G]$ was a recent theorem due to Richard Laver, in his paper Certain very large cardinals are not created in small forcing extensions, using a proof that is due to me, and was also observed independently by W. Hugh Woodin. This theorem is the starting point of the subject of Set-theoretic geology, which I introduced with Jonas Reitz and Gunter Fuchs, as well as J. Reitz's work on The Ground axiom. The point of this theorem is that one needn't actually introduce the predicate $\check M$, as the class that it picks out is a definable class (using parameters) in the language of set theory. To which theorem of Kanamori's book do you refer? -If you don't have parameters, then it is not true generally that the ground model $M$ is definable in the forcing extension $M[G]$. One way to see this is to add two mutually generic Cohen reals $c$ and $d$, and consider the extension $M[c][d]$, which is a forcing extension of $M$, of $M[c]$ and of $M[d]$. If $M[c]$ were a parameter-free definable class of $M[c][d]$, then the set of reals of $M[c]$ would be a parameter-free definable set in $M[c][d]$, and hence in the HOD of $M[c][d]$. But the forcing is ordinal-definable and homogeneous, and so the HOD of $M[c][d]$ must be contained in $M$, a contradiction. Thus, $M[c]$ is not a parameter-free definable class in $M[c][d]$. But it is definable with parameters, and it is also (trivially) definable in the language where you have added a predicate for the ground model. Perhaps it is this latter situation to which you refer. - -REPLY [3 votes]: Just to dot an i and/or cross a t in Joel's answer, since you asked about formalizing "$p$ forces "$\dot G$ is generic over $\check M$": Notice that, once you've introduced $\check M$ as a predicate symbol in the forcing language, as in Joel's answer, you can use it to write the usual definition of genericity as a sentence in the forcing language. The crucial clause (the only one involving $\check M$) is $$(\forall D\subseteq \check P)[(D \text{ dense in }\check P)\land (D\in\check M)\implies (D\cap\dot G\neq\emptyset)].$$ -Let me also point out something implicit in Joel's choice of expository method: Foundational things about forcing, like this, tend to be more natural and easier to understand if you think in terms of Boolean-valued models. If necessary, you can then translate into the terminology of forcing. The value of forcing (as opposed to Boolean-valued models) is not in foundational matters but rather in explicit constructions; it's usually (though not always) easier to describe a notion of forcing than to describe its associated compete Boolean algebra. - -REPLY [3 votes]: In most cases, however, you can get rid of the problem of referring to $\check{M}$. Andreas' formula above, for example, can be written as -$$ -(\forall D)((D\in \check{X})\Longrightarrow (D\cap \dot{G}\neq\emptyset)). -$$ -where $X$ is the set of dense subsets of $P$ (as defined in $M$). -Another standard trick to ensure that $M$ is a definable class in $M[G]$ is to force over $L$ -(whenever one can, this clearly rules out the use of some large cardinals).<|endoftext|> -TITLE: Maximize sum of largest eigenvalues -QUESTION [5 upvotes]: Consider the following optimization problem: -$\max_{\lambda_j(X)}\sum_{j=1}^n d_j\lambda_j(X)$ subject to $v_j^TXv_j \leq 1, X \geq 0$. -$d_j$ are such that $d_1 \geq d_2 \geq \ldots \geq d_k > 0$, $\lambda_j(X)$ is the $j$th largest eigenvalue of the positive semidefinite matrix $X$ of dimension $n\times n$. $v_j$ are vectors with elements belonging to $\{-1,0,1\}$. $T$ denotes transpose. All variables are real-valued. -Are there any theoretical results about the optimal matrix $X$ for this problem? -We know that the objective function is a convex function on the elements of $X$, so this is about maximizing a convex function over the convex set that is defined by intersecting the positive semidefinite cone with some hyperplanes. I have noticed that the optimal solution is either at the vertices of the polyhedron defined by the linear inequalites (which is then full rank), or at lower rank matrices obtained by intersecting some of the planes with the surface of the semidefinite cone (the intersection is such that these low rank matrices are uniquely defined from the hyperplanes). So basically, the low rank solution is obtained by intersecting a line, obtained from the hyperplanes, and the cone. -Grateful for any hints or references. - -REPLY [3 votes]: If I understand the question correctly, the answer is that no, optima need not occur at a unique point where some of the hyperplanes defined by tightness of the linear inequalities meet the boundary of the positive semidefinite cone. -For example, let $v_i$ be the $i^{\text{th}}$ unit vector, so the linear constraints merely say that the diagonal entries of $X$ are each at most $1$. You can check that there are many positive semidefinite matrices with all diagonal entries equal to $1$. -Since the diagonal entries are each at most $1$, the trace of $X$ is at most $n$. Since $X$ is positive semidefinite its eigenvalues are nonnegative. Thus for any $k$, the sum of the $k$ largest eigenvalues of $X$ is at most $n$. This bound is achieved simultaneously for all $k$ by the all ones matrix, so this matrix is optimal for any $k$. The matrix whose $i,j$ entry is $(-1)^{i+j}$ also achieves this bound, giving another optimal solution. -Asking that the matrices $v_jv_j^T$ span the space of symmetric matrices does not rule out this counterexample. It suffices to add some more such matrices with right hand side constants $c_j$ very large, so the corresponding linear constraints can never be tight (without violating the other constraints).<|endoftext|> -TITLE: Embeddings of finite classical groups -QUESTION [5 upvotes]: Let $q$ denote a prime power and $\text{GL}_n(q)$ and $\text{U}_n(q^2)$ the general linear and unitary group, respectively. Then $\text{U}_n(q^2)$ is naturally a subgroup of $\text{GL}_{n}(q^2)$, so one kind of groups can be embedded into the other. Let $C(g)$ be the conjugacy class of an element $g$ in its respective group. Then we can define the length of $g$ to be $\ell(g):=\frac{\log|C(g)|}{\log|G|}$. The above embedding only changes the length by a constant factor. -Note that the length function induces a biinvariant metric by $d(g,h):=\ell(gh^{-1}$. -My question is if there are any functions $f,g:\mathbb{N}\rightarrow \mathbb{N}$ such that we can always find an embedding of $\text{GL}_n(q)$ into $\text{U}_{f(n)}(q^{g(n)})$, which doesn't change the above length function to much. It would be nicest if the distortion of length would only be a constant factor. -My feeling is that this is not possible, but I somehow fail to find an explanation. -A further question is what happens if we exchange the unitary groups for orthogonal or symplectic groups. - -REPLY [7 votes]: I guess by $U_n(q^2)$ you mean the general unitary group in which the field of representation has order $q^2$? That is often denoted by ${\rm GU}_n(q)$, but I will use your notation. -In general ${\rm GL}_n(q^2)$ embeds into $U_{2n}(q^2)$, by acting on a totally isotropic space of dimension $n$, and it does not embed in $U_m(q^k)$ for any $k$ with $m < 2n$. -Similarly ${\rm GL}_n(q)$ embeds into ${\rm Sp}_{2n}(q)$ and into ${\rm GO}^+_{2n}(q)$. For ${\rm GO}^-_{2n}(q)$, the largest totally isotropic spaces have dimension $n-1$, so to embed ${\rm GL}_n(q)$, we need to go up to ${\rm GO}^-_{2n+2}(q)$. -I am not exactly sure what you looking for with your length function. For a large proportion of elements $g \in {\rm GL}_n(q)$ the centralizer of $g$ in the larger group will be only twice as large as the centralizer in the smaller group, so $\ell(g)$ will not change much. But elements of ${\rm GL}_n(q)$ with a large fixed point space will have a much larger centralizer in the large group, so there will be distortion of $\ell(g)$.<|endoftext|> -TITLE: Is $\mathbb{C}^2$ homeomorphic to $\mathbb{C}^2 - (0,0)$ with the Zariski topology? -QUESTION [28 upvotes]: A fellow grad student asked me this, I have been playing for a while but have not come up with anything. Note that $\mathbb{C}$ is homeomorphic to $\mathbb{C} - \{0\}$ in the Zariski topology - just take any bijection and the closed sets (finite sets) will biject as well. Concocting a similar thing for the plane is harder though. -I think I can show that the rational plane and the rational plane minus the origin are homeomorphic by enumerating the irreducible curves and using a back and forth argument, but I have not written it all up formally to see if I am missing something yet. -I know the question isn't natural from the point of view of algebraic geometry, because one of the objects isn't even a variety. I think it is still interesting just to see how weird the zariski topology really is. - -REPLY [15 votes]: This question was analyzed by Roger Wiegand in the context of algebraic surfaces over fields $k$ which are algebraic closures of finite fields. -[1] R. Wiegand, Homeomorphisms of affine surfaces over a finite field, J. London Math. Soc. (2) 18 (1978), no. 1, 28–32. -and in the followup paper: -[2] R. Wiegand, W. Krauter, Projective surfaces over a finite field. Proc. Amer. Math. Soc. 83 (1981), no. 2, 233–237. -In [1] he proves that the affine plane $A^2_k$ (over $k$) is Zariski-homeomorphic to any open nonempty subset of $A^2_k$ (Corollary 7). In particular, $A^2_k$ is homeomorphic to $A^2_k -\{(0,0)\}$. In particular, Greg's argument (as written) cannot be complete (as zero characteristic is not used anywhere in his sketch). -After proving the corollary Wiegand asks if the same holds for $k={\mathbb R}$ and ${\mathbb C}$. -In [2] it is proven (amon other things) that every proper nonempty open subset of $P^2_k$ is homeomorphic to either $P^2_k - \{point\}$, or to the affine plane $A^2_k$.<|endoftext|> -TITLE: Groups with a rational generating function for the word problem -QUESTION [15 upvotes]: This question comes more from curiosity than a specific research problem. Let G be a group and S a finite symmetric generating set. By the WP(G,S) I mean the set of all words in the free monoid on S mapping to the identity of G. -A classical result of Anissimov says that WP(G,S) is a regular language iff G is finite. Regular languages have rational generating functions, so I am asking: - -Dooes rationality of the generating function of WP(G,S) imply G is finite? - -I believe, but didn't check, that rationality doesn't depend on the choice do S. -I guess that my motivation is when is the generating function for probability of return to the origin at step n of a random walk rational? - -REPLY [8 votes]: I hope I'm also not misinterpreting the question, but it seems to me that the answer is yes. In fact the property of having a rational "walk generating function" characterizes finite graphs not only among Cayley graphs as in your question but also among the larger class of regular quasitransitive connected graphs (quasitransitive here means that the automorphism group acts with finitely many orbits). This is theorem 3.10 in "Counting Paths in Graphs" by L. Bartholdi, published in Enseign. Math. 45 (1999) 83-131. It is mentioned in the paper that the analogous question for arbitrary connected regular graphs is open.<|endoftext|> -TITLE: Name for an Isomorphism in a Monoidal Category that Satisfies the Braid Relation -QUESTION [6 upvotes]: Let $({\cal C},\otimes)$ be a monoidal category, $X$ an object in ${\cal C}$, and $\Psi:X \otimes X \to X \otimes X$ an isomorphism such that $\Psi$ satisfies the braid relation: -$$ -(\Psi \otimes \text{id}) \circ (\text{id} \otimes \Psi) \circ (\Psi \otimes \text{id}) = (\text{id} \otimes \Psi) \circ (\Psi \otimes \text{id}) \circ (\text{id} \otimes \Psi). -$$ -What would one call such an isomorphism? The most obvious suggestion is to call it a braiding for $X$. Might this be taken to imply that $\Psi$ comes from a braiding for the category (which I do not want to assume)? - -REPLY [12 votes]: Assuming S. Carnahan's surmise in his comment is correct, I believe the correct term for this is "Yang-Baxter" operator in a monoidal category (or, you could call an object $X$ equipped with such an automorphism $R: X \otimes X \to X \otimes X$ a Yang-Baxter object). This terminology is given in the seminal paper on the subject, Braided Tensor Categories by Joyal and Street (Adv. Math. 102, pp. 20-78, 1993). -In particular, as observed by Joyal and Street, the braid category can be characterized as the free (i.e., initial in a 2-categorical sense) monoidal category equipped with a Yang-Baxter object. -Edit: Another reference for this terminology: - -André Joyal and Ross Street, Tortile Yang-Baxter operators in a tensor category, J. Pure Appl. Alg. 71 (1991), 43-51.<|endoftext|> -TITLE: What is the mod l monodromy of a generic trigonal curve? -QUESTION [6 upvotes]: For a hyperelliptic curve H, the mod 2 monodromy is smaller than $GSp_{2g}(F_2)$ -- since the two torsion of the Jacobian H is generated by differences of Weierstrass point, the monodromy of a generic curve is the symmetric group. -The degree 3 map $C \to \mathbb{P}^1$ associated to a trigonal curve may only have ramification of type (2,1), and thus may not give any "easy to see" 3 torsion points on the Jacobian. So it is reasonable to ask if a generic such curve has maximal mod l monodromy for every l. - -REPLY [5 votes]: The paper https://arxiv.org/abs/1403.7399 shows that the monodromy of the moduli stack of trigonal curves of genus g over $\mathbb C$ is as big as possible (equal to $Sp_{2g}(\mathbb Z)$ in the topological setting, or equal to $Sp_{2g}(\widehat{\mathbb Z})$ in the arithmetic setting, by the comparison theorem). An alternate, (and to my taste, geometrically enlightening,) explanation that the mod-2 monodromy is the full symplectic group is also sketched in Anand Patel's thesis -https://www2.bc.edu/anand-p-patel/Research/ThesisTheGeometryofHurwitzSpace.pdf -in Proposition 1.11 and the paragraphs following its proof. -To explain this implication in more detail, the main theorems on the first two pages of the above cited paper shows that the lowest Maroni strata, (which is dense in the locus of trigonal curves) has monodromy equal to the full symplectic group. Since the moduli stack of trigonal curves of a given Maroni invariant is smooth, the generic point also has maximal monodromy, equal to the full symplectic group. Technically speaking, the above cited paper only applies when the genus is at least 5, but the same results holds in genus 3 and 4 because trigonal curves are dense in $M_g$ for genus $3$ and $4$, and $M_g$ is smooth with monodromy equal to the full symplectic group by a classical result (see for example 5.12 of Deligne and Mumford's "The irreducibility of the moduli stack of curves of given genus"). -Since in your question you ask about $GSp_{2g}(\mathbb F_2)$, you may be asking the question not only over an algebraically closed field, but also over a number field. In the case of a number field $K$ with $K \cap \mathbb Q^{cyc} = \mathbb Q$, we can use the geometric statement above to deduce that the monodromy group of the trigonal locus is all of $GSp_{2g}(\widehat{\mathbb Z})$. To see this, let $\chi:Gal(\overline K/K) \rightarrow \widehat{\mathbb Z}^\times$ denote the cyclotomic character, let $H_C$ denote the monodromy of a family of curves C, and let T denotes the moduli stack of trigonal curves. Then, the monodromy $H_T$ for $T$ over $K$ fits in an exact sequence -\begin{align*} -0 \rightarrow (H_T)_{\overline K} \rightarrow H_T \xrightarrow{\operatorname{mult}} \chi(K) \rightarrow 0, -\end{align*} -Since $(H_T)_{\mathbb C} = Sp_{2g}(\widehat{\mathbb Z}),$ and includes into $(H_T)_{\overline K}$, it follows that -$(H_T)_{\overline K} = Sp_{2g}(\widehat{\mathbb Z})$. Therefore, since $\chi(K) = \widehat{\mathbb Z}^\times$ when $K \cap \mathbb Q^{cyc} = \mathbb Q$, we obtain $H_T$ is a subgroup of $GSp_{2g}(\widehat{\mathbb Z})$ which contains the symplectic group and has surjective mult map, so $H_T = GSp_{2g}(\widehat{\mathbb Z}).$ In general, the monodromy over $K$ will be the preimage of $\chi(K)$ under the mult map. -It is also interesting to ask this question not only in the case of the generic trigonal curve (i.e., the generic point of the moduli space), but also in the case of a general trigonal curve over a number field. For simplicitly, let's work over $\mathbb Q$, although analogously to the previous paragraph, we can easily generalize this to any number field. While I do not think the monodromy will be maximal for a general trigonal curve, it does hold for a density-$1$ subset of the $\mathbb Q$ points of the moduli stack, with density ordered by height. That is, a density-$1$ subset of trigonal curves over $\mathbb Q$ have monodromy equal to $GSp_{2g}(\widehat{\mathbb Z})$ (as is shown in the statement and proof of Corollary 1.3 in http://arxiv.org/pdf/1608.05371v1.pdf). -Thanks to David Zureick-Brown and Jordan Ellenberg for help with answering this question.<|endoftext|> -TITLE: Varieties of groups with infinite relatively free group of rank 2 finite, infinite in rank 3 -QUESTION [7 upvotes]: Does there exist a variety of groups $\mathfrak{V}$ such that the relatively $\mathfrak{V}$-free group of rank 2 is finite, but the relatively $\mathfrak{V}$-group of rank 3 is infinite? -(In other varieties of algebras this can occur; for example, in the variety of all lattices, the free lattice of rank 2 is finite, but the free lattice of rank 3 is infinite.) - -REPLY [8 votes]: There are varieties of semigroups like that. Take any finite inherently non-finitely based semigroup $S$, say, the 6-element Brandt monoid $B_2^1$, and the variety $M$ given by all identities of $S$ depending on at most 2 variables. Then all 2-generator semigroups in $M$ are in $var S$, so are finite, but three-generated semigroups in $M$ will be infinite. Warning: the latter is not quite in the literature, it is only proved that some finitely generated semigroup in $M$ is infinite, it may have a bigger rank (in my paper from 1987, Problems of Burnside type and the finite basis property in varieties of semigroups. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 51 (1987), no. 2, 319--340, 447; translation in -Math. USSR-Izv. 30 (1988), no. 2, 295–314, it is $2^2=4$, I think). If you really want rank 3, you should take a slightly bigger monoid than $B_2^1$ or modify my construction. -For groups the answer is unknown. The variety, if exists, is necessarily of finite exponent, hence a solution on the bounded Burnside problem is required. Zelmanov proved that for every prime $p$ there exists a number $k$ such that if all groups of exponent $p$ with $k=k(p)$ generators are finite, then all finitely generated groups of exponent $p$ are finite. For non-prime exponent even that if unknown. Zelmanov's function $k(p)$ grows exponentially.<|endoftext|> -TITLE: Quaternionic-Kahler metrics whose universal covers have only discrete isometry groups? -QUESTION [5 upvotes]: I am interested in quaternionic-Kahler metrics that are "as inhomogeneous as possible." -Every complete quaternionic-Kahler manifold $X$ I can remember hearing of is a discrete quotient of some $Y$, such that $Isom(Y)$ contains a nontrivial connected Lie group. Are there any known examples of complete quaternionic-Kahler $X$ that don't arise in this way? -(By "quaternionic-Kahler manifold" I mean one with holonomy contained in $Sp(n)Sp(1)$ but not in $Sp(n)$ -- in other words, I want to exclude the hyperkahler case.) - -REPLY [6 votes]: I'm not familiar with the nonpositive case (negative since you exclude hyperkahler case) but there are no such examples known for postive quaternion Kahler manifolds (i.e. those with positive scalar curvature). They are all conjectured to be symmetric spaces (conjecture of LeBrun and Salamon) and this conjecture has been verified in dimensions 4 (Hitchin) and and 8 (Poon and Salamon). Certainly, only symmetric examples are known such as $\mathbb HP^n, Gr_2(\mathbb C^{n+2})$, $\widetilde {Gr_4}(\mathbb R^{n+4})$ (the -Grassmanian of oriented real $4$-planes) and a few exceptional spaces such as -$G_2/SO(4)$. -I looked around for what's known in negative case and in this paper LeBrun constructs an infinite dimensional family of negative quaternion Kahler metrics on $\mathbb R^{4n}$. I suspect most of these have no symmetries. But apparently, only locally symmetric compact examples are known though.<|endoftext|> -TITLE: When are roots of power series algebraic? -QUESTION [18 upvotes]: Let $K$ be a field and consider a power series $f(T) \in K[[T]]$. Under what conditions (on $K$ and/or on $f$) can we conclude that if $\alpha$ is a root of $f(T)$ then $\alpha$ is in fact algebraic over $K$? - -This question is inspired by the following: In this paper of R. Pollack, in the proof of his Lemma 3.2, the author states that if $K$ is a finite extension of $\mathbb{Q}_p$ and $f(T)$ converges on the open unit disc of $\mathbb{C}_p$ and has finitely many roots, then each of these roots is algebraic over $K$. -I haven't yet come up with a proof of this statement (and any proofs offered would be appreciated), and perhaps having a proof in hand would help me to answer my question on my own, but I'm now curious about other situations in which this can occur. - -REPLY [13 votes]: Here's another proof of the fact that any root of a (non-zero) $p$-adic power series (convergent on the open unit disc) is algebraic. The proof is a little silly in that it uses Tate's theorem on Galois invariants of ${\mathbb C}_p$ (which is quite deep) instead of the Weierstrass preparation theorem (which is fairly elementary). But here it is in any case. -The key fact I need is the following which I'll prove after explaining how it completes the proof. - -Claim: Any transcendental element of ${\mathbb C}_p$ has uncountably - many conjugates (under $\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p))$. - -Accepting this claim for the moment, let $\alpha$ be a transcendental zero of $f(x)$, our convergent power series. Then all conjugates of $\alpha$ are also zeroes of $f(x)$ in the open unit disc. Thus, by the above claim, $f(x)$ has uncountably many zeroes in the open unit disc of ${\mathbb C}_p$. However, any uncountable set in a separable metric space (e.g. ${\mathbb C}_p$) has an accumulation point. But then the zeroes of $f(x)$ have an accumulation point, necessarily in the open unit disc since that is closed in ${\mathbb C}_p$, which forces $f(x)$ to be identically zero. -Returning now to the claim, let $G=\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p)$ and let $H$ denote the subgroup of $G$ which stabilizes $\alpha$. Since the conjugates of $\alpha$ are in one-to-one correspondence with $G/H$, we will show that $H$ has uncountable index. -First note that $G$ acts continuously on ${\mathbb C}_p$ where we give ${\mathbb C}_p$ the $p$-adic topology (not the discrete topology). Thus $H$ is a closed subgroup and hence of the form $\text{Gal}(\overline{\mathbb Q}_p/M)$ for some algebraic extension$M/{\mathbb Q}_p$. -Assume $G/H$ is finite, and thus that $M$ is a finite extension of ${\mathbb Q}_p$. But then by Tate's theorem, $\alpha$ must be in $M$ as it is fixed by $\text{Gal}(\overline{\mathbb Q}_p/M)$. This is impossible as $\alpha$ is transcendental. -Thus, $G/H$ is infinite, and we must now show that it is uncountable. We use the following (standard) fact: - -Fact: Any infinite compact Hausdorff space with no isolated points is uncountable. - -Since $G$ is compact, $G/H$ is compact. The coset space $G/H$ is Hausdorff since $H$ is closed. To see that $G/H$ has no isolated points, note that if it has one isolated point, then all of its points are isolated as $G$ acts transitively (by left multiplication) on $G/H$ by homeomorphisms. But then $G/H$ is discrete which is impossible as it is infinite and compact. -Thus, $G/H$ is uncountable, and $\alpha$ has uncountably many conjugates.<|endoftext|> -TITLE: Does Godel's incompleteness theorem admit a converse? -QUESTION [50 upvotes]: Let me set up a strawman: -One might entertain the following criticism of Godel's incompleteness theorem: -why did we ever expect completeness for the theory of PA or ZF in the first place? -Sure, one can devise complete theories semantically (taking all the statements that hold -in some fixed model), but one has usually discovered something special (e.g. elimination of -quantifiers) when a naturally framed theory just turns out complete. -Now perhaps one could defend Godel's theorem as follows: -By Godel, the theory of the standard natural numbers has no recursive axiomization, but equally remarkably PA has no recursive non-standard models (Tennenbaum's theorem). That means that the incompleteness of arithmetic has a deeper character than, say, the incompleteness of group theory -- there exhibiting groups with distinct first-order properties easily suffices. -My question: -Does there exist any sort of converse to Godel's incompleteness theorem. A converse might say that when one has incompleteness and also some reasonable side condition (I'm suggesting but not committed to "there exists only one recursive model"), then there must exist some self-reference mechanism causing the incompleteness? Or stronger perhaps, the theory must offer an interpretation of some sufficiently strong theory of arithmetic? - -REPLY [2 votes]: I will attempt to answer the first question, which I think is more philosophy than math. Consider the integers. We have strong intuitions about the integers being a very definite set of things, which we all have access to. Furthermore, if we accept a naive second order axiomization, the integers are the only thing this axiomization describes. This should survive formalization: after all we only want to describe this one model that Peano arithmetic seems to describe perfectly. -Obviously what I said above is only intuition, and is in fact hopeless. But I am willing to use second order theories to save arithmetical consistency. Of course, this has a price: describing models is hard, logic on second order theories loses a lot of nice features, but if you are willing to sacrifice an entire branch of mathematics in your worldview, you can somewhat avoid having to consider weird models of the integers.<|endoftext|> -TITLE: Irreversible chess -QUESTION [21 upvotes]: Suppose we play a chess-variant, where any finite number of pieces are allowed, and the board is as large as we wish, but only two kings in total. And there is no 50 move-rule, no castling and no captures and no pawn-moves. -Then is it possible to have a a pair of positions A and B, such that we can go from configuration A to B by legal moves, but not from B to A? -No stalemates or checkmates allowed, the game must be extendable atleast two moves in both direction from both A and B. -Two configurations are different also if the pieces are in the same places but it is a different player to move. - -REPLY [7 votes]: [Edited to correct the final position (which didn't work before: 2 Bh1+! Bg2 3 Rf1 Bb1 etc.), and — while I'm at it — to remove a few superfluous pawns] -Here's yet another mechanism, which also allows arbitrarily long chains of irreversibility. The new construction is suggested by what may be an unintenional feature/bug in the original question: - -[...] the game must be extendable at least two moves in both direction from both A and B. - -The number seems arbitrary (why two moves rather than three or 23?), and I'm guessing that the intention was to guarantee that one can continue indefinitely forwards or backwards from both positions but give an easily checked criterion. However, it's possible for a position to be extendable $n$ moves and no further for any given $n$, as long as we're allowed an arbitrarily large board and supply of pieces (as the question proposal did explicitly allow). For example: let Position A have White rook a5, bishops c2,e2,g2, and pawns b3,d3,f3,h3 vs. Black bishops b1,d1,f1,h1, plus any immobile setup that traps both kings above the 3rd row: - -(source: janko.at) -and play 1 Ra1 to get Position B. Not only can't we get back from B to A, but after a few moves (but more than 2) we must reach stalemate: 1...Ba2 2 Bb1 (or 2 Rb1 stalemate, or 2 Rc1 Bc2 stalemate) 2...Bc2 3 Bd1 Be2 4 Bf1 and now 4...Bg2 is stalemate. -For a long irreversibility chain not ended by stalemate, remove the White and Black bishops on g2 and h1, put the White king on h1, and add Black pawns on h2 and g3: - -(source: janko.at) -Now after 1 Ra1 play can continue indefinitely, but only with 1...Ba2 2 Bb1 Bc2 3 Bd1 Be2 - -(source: janko.at) -and now 4 Kg2 Bf1+ 5 Kh1 Be2 etc. This extends naturally to wider boards, with a correspondingly longer sequence between 1 Ra1 and the onset of forced repetition.<|endoftext|> -TITLE: Derived Equivalence of Sheaves and Homotopy -QUESTION [5 upvotes]: This question loosely elaborates on an earlier question. It is pretty silly, but I'd like to hear some authoritative answers. -Recall that if $f:S^{\bullet}\to T^{\bullet}$ is a quasi-isomorphism of sheaves over $X$, which is, say, a manifold, then for every open set $U\subset X$, we have an induced isomorphism $R\Gamma(U,S^{\bullet})\to R\Gamma(U,T^{\bullet})$, so $H^i(U,S^{\bullet})\cong H^i(U,T^{\bullet})$ and in particular $H^i(X,S^{\bullet})\cong H^i(X,T^{\bullet})$. -To what extent is the converse true? At the coarsest level, when does a canonical isomorphism $R\Gamma(X,S^{\bullet})\to R\Gamma(X,T^{\bullet})$ reflect an underlying derived equivalence? -For a counterexample to the coarsest case, I believe the following serves: Consider a space $X$. Consider the constant sheaf on $k_X$. Let $f:X\to x_0$ be the retraction to a point $x_0\in X$. By standard theorems, we know that $H^i(X,Rf_*k_X)\cong H^i(X,k_X)$, but evaluating $Rf_*k_X(U)$ on any open subset $U$ missing $x_0$ assigns zero, as the fiber is empty. So in general these sheaves are not derived equivalent. What if $X$ deformation retracts to $x_0$? Is $k_X$ and $Rf_*k_X$ derived equivalent then? What if the homotopy doesn't have some Vietoris-Begle type behavior? See Kashiwara Schapira 2.7.8. - -REPLY [8 votes]: I think you are asking: when is the functor $R\Gamma$ conservative (in the derived sense - i.e. if $R\Gamma (f)$ is a quasi-isomorphism then $f$ is a quasi-isomorphism). This is equivalent to $R\Gamma$ having no kernel - i.e. if $R\Gamma (F) \cong 0$, then F $\cong 0$ (by taking cones). -If you restrict to the triangulated subcategory generated by the constant sheaf $k_X$, I claim that $R\Gamma$ is conservative. This category can be thought of as something like derived unipotent local systems on $X$. The Barr-Beck theorem says that it is equivalent to dg modules over the algebra $C^\ast (X)$ of cochains on $X$ (so it is still interesting even if $X$ is simply connected). -Proof of claim: Let $f : X \to pt$ be the projection, and $C$ a (complex of) sheaves on $X$ such that $R\Gamma (C) = Hom _X(k_X , C) \cong 0$. Then if $T$ is any element of the triangulated subcategory generated by $k_X$, $Hom_X (T,C) \cong 0$. In particular, if $R\Gamma (T) \cong 0$, then $Hom _X(T,T) \cong 0$, so $T\cong 0$. -I think that this should be the largest triangulated subcategory of sheaves on $X$ on which global sections are conservative, but I am not sure how to prove this in general. It would follow from the Barr-Beck theorem if we knew that $R\Gamma$ preserved geometric realizations of simplicial objects, for example. -This situation is reminiscent of quasi-coherent sheaves on an affine scheme - this category is generated by $\mathcal O_X = f^\ast k$, and is equivalent to modules over $\Gamma (\mathcal O_X)$. -About your question: if $X$ deformation retracts to a point, then are $k_X$ and $Rf_\ast k_X$ quasi-isomorphic? (where by $Rf_\ast k_X$ I assume you mean the corresponding skyscraper sheaf on $X$). Doesn't your argument just above work (provided $X$ is not a point)?<|endoftext|> -TITLE: What is the "strongest" non-local property of a ring/module that is true of all localizations at maximal ideals? -QUESTION [5 upvotes]: Given a commutative ring $A$ we say that a property P is local if - -$A$ has P $\leftrightarrow$ $A_{p}$ has P for all prime ideals $p$ of $A$ - -It is usually the case that this requirement is equivalent to $A_{m}$ having P for all maximal ideals $m$ of $A$. I was wondering which (if any) are the strongest/most interesting local properties $P$ of a commutative ring that do not satisfy the second equivalence. Similarly, I would like to know the strongest/most interesting non-local properties P that are true at all localizations at $p$. -That is to say, what are the most interesting properties P of $A$ such that: - -(1) $A_{p}$ has P for all prime ideals $p$ of $A$ but P is NOT local - -or - -(2) P is local BUT it is NOT true that if $A_m$ has P for all maximal ideals $m$ of $A$ then $A$ has P. - -EDIT: After comments and answers received have edited (and expanded) the question. Hope it is clear and unambiguous now. - -REPLY [3 votes]: If I understand correctly (1), the following properties are not local - -being a PID (including fields): take a non principal Dedekind domain; -being Noetherian (consider an infinite product of $\mathbb F_2$); -being Artinian (same example as above).<|endoftext|> -TITLE: A model of CH +$\lnot \diamondsuit$ -QUESTION [9 upvotes]: All of the models of CH which I know of also satisfy $\diamondsuit$. What is the easiest way to produce a model of CH wherein $\diamondsuit$ is false? - -REPLY [10 votes]: "The easiest way" to produce a model of CH in which $\diamondsuit$ is false is to start with a model of GCH and then do a countable support iteration of length $\omega_2$ killing off a potential $\diamondsuit$ sequence at each stage. -The forcing for doing this is straightforward: supposing $\langle A_\alpha:\alpha<\omega_1\rangle$ is a sequence with $A_\alpha\subseteq \alpha$, we force the existence of an uncountable $X\subseteq\omega_1$ such that $X\cap\alpha\neq A_\alpha$ for all $\alpha$. This is done by viewing conditions as telling us initial segments of $X$. -Iterating the above forcing $\omega_2$ times while using bookkeeping to make sure we kill of any potential $\diamondsuit$ sequence actually works. -BUT -There's a whole lot of work involved in proving that the iteration doesn't add new reals. This is where one must wrestle with the Shelah machinery of $\mathbb{D}$-completeness and $<\omega_1$-properness<|endoftext|> -TITLE: Equilibrium configurations of ions on n-Dim balls. -QUESTION [9 upvotes]: Given an n-dimensional electrically neutral, solid metal ball (a point for n=0; a rod, n=1; a disc, n=2; a solid ball, n=3; ...), place N=(n+1)! identical ions on the ball. As one of my favorite physics professors used to say, forget the mathematics, intuitively I expect the ions to equilibrate to the vertices of an n-D permutohedron inscribed in the n-ball (e.g., a hexagon inscribed in the circumference of a disc with 6 ions and a truncated octahedron for n=3 ). Similarly with N= [2(n+1)]! /[(n+1)! (n+2)!] (the Catalan numbers), I expect to see an associahedron (a Stasheff polytope, e.g., a pentagon for the disc with 5 ions). -Is my intuition correct? Has anyone seen this worked out mathematically, as an extremum problem? An electrostatics simulation for n=3 would be interesting also. (See OEIS A019538 for refined f-vectors of permutohedra / permutahedra and A133437 for associahedra.) -After reviewing the very interesting, proposed answers, I think I need to clarify and qualify my question: -I initially was thinking of a real physical situation, a conductive metal ball in 3 dimensions or disc in 2 with Coulomb's inverse-square law applying. Then, if I remember my physics correctly, free charges must always reside on the suface with the potential vanishing in the interior of the ball and with the gradient of the potential at the surface being normal to the surface in the equilibrium state. The number of ions for the 3-D ball is constrained as stated above to either 4!=24 for the 3-D permutahedron or 8!/(4! 5!) = 14 for the 3-D associahedron. Given either number, how will the ions distribute themselves on the surface? Whether one should or can sensibly modify the repulsion for higher dimensions n and whether analogous surface (n-1) physics and equilibrium topology would occur are the next logical considerations; however, the number of "ions" would still be restricted to the number of vertices of either the n-D permutohedron or associahedron. - -REPLY [10 votes]: It turns out to pretty hard to guess the answers to these kinds of extremal problems. The permutohedron is almost certainly not the answer in $\mathbb{R}^n$ for $n \ge 3$, for any reasonable potential function (for example, an inverse power law). Specifically, the truncated octahedron is not good at minimizing things, because of the square and hexagonal facets, and it seems that you can always do better with the same number of particles, although I haven't carefully proved this for all potential functions. In $\mathbb{R}^4$ the situation is even worse, since the 120 vertices of a regular 600-cell always lead to lower energy than those of a permutohedron. I'm confident that the permutohedron will be suboptimal also for $n \ge 5$, and that the associahedron behaves similarly. -More generally, it's far from clear when one should expect symmetry in ground states of physical systems like this (see Link for an expository article on this issue). For small systems you often get symmetry, but in large systems defects can actually lower energy in surprising ways. -It's natural to guess that regular polytopes should lead to energy minima, and the ones with simplicial facets actually do. (By contrast, the cube does not, because one can lower energy by rotating one facet relative to the opposite one, to get a square antiprism.) These polytopes actually minimize a very wide class of energies, including all inverse power laws; specifically, they are optimal whenever the potential function is a completely monotonic function of squared Euclidean distance. (Recall that a completely monotonic function is a nonnegative $C^\infty$ function whose derivatives alternate in sign.) Abhinav Kumar and I called this phenomenon universal optimality. -In three dimensions there is no universal optimum larger than an icosahedron, but in higher dimensions there are more examples, such as the $E_8$ root system or the minimal vectors in the Leech lattice. Another interesting example is the Schläfli configuration of 27 points on a sphere in six dimensions, which correspond to the 27 lines on a cubic surface. See Link for details and proofs. -Universal optimality seems to be quite rare, and in most cases the optimal configuration will depend on the potential function being used. See Link for some large-scale numerical experiments, as well as references to many other papers in the literature.<|endoftext|> -TITLE: More on "Transalgebraic Theories" (a 19th century yoga)? -QUESTION [26 upvotes]: Among the talks at occasion of the Galois Bicentennial, one is about "Transalgebraic Theories". Unfortunately I found only this article describing that fascinating idea as " an extremely powerful 'philosophical' principle that some -mathematicians of the XIXth century seem to be well aware of. In general terms we -would say that analytically unsound manipulations provide correct answers when -they have an underlying transalgebraic background." Do you know more? -Edit: This text tells a few words more (e.g. "This philosophy can be linked to Kronecker’s ”Judgendtraum” and Hilbert’s twelfth problem, which seems to have remained largely misunderstood.") and refers to a manuscript "Transalgebraic Number Theory". Has someone a copy? - -REPLY [8 votes]: For the most part "transalgebraic theory" seems an umbrella term for anything that relates (classic, differential, motivic, categorical) Galois theory with periods, trascendence results, special values, regularization... -Here is a compilation of papers which might be of interest. -Yves André - -Galois theory, motives and transcendental numbers -Ambiguity theory, old and new -Idées galoisiennes - -Michel Waldschmidt - -Transcendence of Periods: The State of the Art -Much more on trascendence on his page. - -Alexandru Buium - -Transcendental numbers as solutions to arithmetic differential equations -Arithmetic differential equations on GLn, III: Galois groups - -Alain Connes - -Renormalisation et ambiguïté galoisienne - -Ricardo Pérez-Marco - -The "Transalgebraic Number Theory" manuscript still doesn't seem to be avaible online. The author is clearly active (papers on arXiv), so perhaps someone can ask him directly. -The (only?) relevant online paper was alredy mentioned in the question, Notes on the Riemann Hypothesis.<|endoftext|> -TITLE: Lie Algebra of Group Scheme -QUESTION [10 upvotes]: Given a group scheme $X$ over $S$, where $S$ is an arbitrary locally noetherian scheme, then how does one define the Lie algebra of $X$? And how does it behave with respect to base change? -Is there any good reference for the theory of group schemes apart from Demazure/Gabriel's book about Algebraic Groups? -All of the treatings I have encountered only care about affine schemes, often over a base field. Where can I find a more general exposition? - -REPLY [8 votes]: To elaborate on a comment by ulrich: SGA3 Exp. 2, section 4 treats Lie algebras of arbitrary group-valued functors over an arbitrary scheme (no locally noetherian hypothesis). I'm not sure what results you want with respect to base change, but most will follow straightforwardly from some combination of Definition 1.1 and Proposition 3.4.<|endoftext|> -TITLE: A k-1 edge connected k regular graph is matching covered -QUESTION [5 upvotes]: As the title says, let $k \geq 2$ be a positive integer and let $G$ be a $(k-1)$-edge-connected $k$-regular graph with an even number of vertices. Then, for every edge $e$ of the graph there is a perfect matching of $G$ containing $e$. -First, I was wondering if this is new and if there are approaches different from mine (and for that reason I won't mention my idea at this point) -Some background: The particular case k=3 is a theorem of Petersen; I had it as a graph theory exam problem last year; my solution led to this generalization. - -REPLY [3 votes]: Ugh I just lost my post but the short version is that on top of Igor's answer, it is easy to prove this using Edmonds' characterization of the perfect matching polytope, which implies putting weight 1/k on every edge will give you a vector in the polytope. From this fact the matching-coveredness is straightforward. -EDIT: -Edmonds proved that a vector (i.e. an edge-weighting $w(e)$) is in the perfect matching polytope (i.e. the convex hull of incidence vectors of perfect matchings) if and only if the following hold: -1) Every edge has weight in $[0,1]$. -2) Every set $S$ of vertices with odd size has $\sum_{e\in\delta(S)} w(e) \geq 1$, where $\delta(S)$ is the set of edges with exactly one endpoint in $S$. -3) Every vertex $v$ satisfies $\sum_{e\in\delta(\{v\})} w(e) = 1$. -It is an easy exercise show that these conditions are necessary, but as Edmonds proved, they are also sufficient. This implies immediately that if $G$ is a $k-1$-edge-connected graph that is $k$-regular, the vector with every edge getting weight $1/k$ is in the perfect matching polytope of $G$ (in other words, $G$ is fractionally $k$-edge-colourable). Since the weight vector is nonzero everywhere, every edge must be contained in at least one perfect matching. (Again in other words, since only perfect matchings can be used to fractionally $k$-edge-colour a $k$-regular graph, every edge must be in a perfect matching.)<|endoftext|> -TITLE: Low degree polynomial approximation for the entropy function -QUESTION [6 upvotes]: Let $X$ be a discrete random variable with possible values -$\{x_1,\ldots,x_n\}$, and let $p$ denote the probability mass function of -$X$. In addition, denote $p_i=p(x_i)$. -The entropy of $X$ is defined as follows -$$H(X)=H(p_1,\ldots,p_n)=-\sum_{i=1}^n p_i\log p_i$$ -I'm looking for a low degree (up to $\log n$) polynomial $P(p_1,\ldots,p_n)$ which provides as good as possible approximation for the entropy of the distribution. - -REPLY [8 votes]: The canonical choice is the Renyi entropy: -$H_\alpha=\frac{1}{1-\alpha}\log P_\alpha$, with $P_{\alpha}(p_1,...,p_n)=\sum_{i=1}^{n}p_i^{\alpha}$ -your entropy (the Shannon entropy) is the limit $\alpha\rightarrow 1$ of $H_\alpha$ -this choice of approximation is useful because it has many meaningful applications, in a variety of contexts. -http://en.wikipedia.org/wiki/Renyi_entropy -For quantitative bounds on the rate of convergence of Renyi entropy towards Shannon entropy see -N. Harvey, J. Nelson, K. Onak, Streaming algorithms for estimating entropy, IEEE ITW '08 proceedings, online at -http://www.math.uwaterloo.ca/~harvey/Publications/StreamingEntropy/ITW.pdf<|endoftext|> -TITLE: Two questions on rational homotopy theory -QUESTION [28 upvotes]: I'm trying to read Quillen's paper "Rational homotopy theory" and am a little confused about the construction. As I understand, he associates a dg-Lie algebra over $\mathbb{Q}$ to every 1-reduced simplicial set via a somewhat long series of Quillen equivalences. But the construction that I had heard before makes spaces (rationally) Quillen equivalent to commutative dgas over $\mathbb{Q}$ via the polynomial de Rham functor. Is there a simple reason why dg-Lie algebras and commutative DGAs should be Quillen equivalent? -I believe this should be Koszul duality, but I don't really understand that right now. If someone has a (preferably lowbrow) explanation for this phenomenon (even in this specific case), I'd be interested. -In addition, I would be very interested in a "high concept" explanation of why Quillen's construction works. It seems that the crux of the proof is the Quillen equivalence between simplicial groups (localized at $\mathbb{Q}$, I guess) and complete simplicial Hopf algebras. I've been struggling with why this should proof should work, since I was not familiar with the work of Curtis on lower central series filtrations referred to there. - -REPLY [35 votes]: I'm not sure if this will still be helpful, but here is my understanding of the Quillen model. Everything correct that I write below, I learned from John Francis. (Probably in the same lecture that Theo mentioned in his comment above.) Any mistakes are not his fault---more likely an error in my understanding. -Before we begin: Quillen v Sullivan. -As others have mentioned, Quillen gets you a DG Lie algebra, where as the Sullivan model will get you a commutative DG algebra. As you write, the passage from one to the other is (almost) Koszul duality. Really, a Lie algebra will get you a co-commutative coalgebra by Koszul duality, and a commutative algebra will get you a coLie algebra. You can bridge the world of coalgebras and algebras when you have some finiteness conditions--for instance, if the rational homotopy groups are finite-dimensional in each degree. Then you can simply take linear duals to get from coalgebras to algebras. -A way to find Lie algebras. -So where do (DG) Lie algebras come from? There is a natural place that one finds Lie algebras, before knowing about the Quillen model: Lie algebras arise as the tangent space (at the identity) of a Lie group $G$. -Now, if you're an algebraist, you might claim another origin of Lie algebras: If you have any kind of Hopf algebra, you can look at the primitives of the Hopf algebra. These always form a Lie algebra. -(Recall that a Hopf algebra has a coproduct $\Delta: H \to H \otimes H$, and a primitive of $H$ is defined to be an element $x$ such that $\Delta(x) = 1 \otimes x + x \otimes 1.$) -One link between the algebraist's fountain of Lie algebras, and the geometer's, is that many Hopf algebras arise as functions on finite groups. If you are well-versed in algebra, one natural place to find Lie algebras, then, would be to take a finite group, take functions on that group, then take primitives. -A cooler link arises when a geometer looks at distributions near the identity of $G$ (which are dual to 'functions on $G$') rather than functions themselves. This isn't so obviously the right thing to look at in the finite groups example, but if you believe that functions on a Lie group $G$ are like de Rham forms on $G$, then you'd believe that something like 'the duals to functions on $G$' (which are closer to vector fields) would somehow safeguard the Lie algebra structure. The point being, you should expect to find Lie structures to arise from things that look like 'duals to functions on a group'. So one should take 'distributions' to be the Hopf algebra in question, and look at its primitives to find the Lie algebra of 'vector fields.' -A (fantastical) summary of the Quillen model. -Let us assume for a moment that your space $X$ happens to equal $BG$ for some Lie group, and you want to make a Lie algebra out of it. Then, by the above, what you could do is take $\Omega X = \Omega B G = G$, then look at the primitives of the Hopf algebra known as `distributions on $\Omega X$'. -Now, instead of considering just Lie groups, let's believe in a fantasy world (later made reality) in which all the heuristics I outlined for a Lie group $G$ will also work for a based loop space $\Omega Y$. A loop space is `like a group' because it has a space of multiplications, all invertible (up to homotopy). Moreover, any space $X$ is the $B$ (classifying space) of a loop space--namely, $X \cong B \Omega X$. So this will give us a way to associate a Lie algebra to any space, if you believe in the fantasy. -Blindly following the analogy, `functions on $\Omega X$' is like cochains on $\Omega X$, and the dual to this (i.e., distributions) is now chains on $\Omega X$. That is, $C_\bullet \Omega X$ should have the structure of what looks like a Hopf algebra. And its primitives should be the Lie algebra you're looking for. -What Quillen Does. -So if that's the story, what else is there? Of course, there is the fantasy, which I have to explain. Loop spaces are most definitely not Lie groups. Their products have $A_\infty$ structure, and correspondigly, we should be talking about things like homotopy Hopf algebras, not Hopf algebras on the nose. What Quillen does is not to take care of all the coherence issues, but to change the models of the objects he's working with. -For instance, one can get an actual simplicial group out of a space $X$ by Kan's construction $G$. This is a model for the loop space $\Omega X$, and this is what Quillen looks at instead of looking only at $\Omega X$, which is too flimsy. From this, taking group algebras over $\mathbb{Q}$ and completing (these are the simplicial chains, i.e., distributions), he obtains completed simplicial Hopf algebras. Again, instead of trying to make my fantasy precise in a world where one has to deal with higher algebraic structures (homotopy up to homotopy, et cetera) he uses this nice simplicial model. To complete the story, he takes level-wise primitives, obtaining DG Lie algebras. -Edit: This is from Tom's comment below. To recover a $k$-connected group or a $k$-connected Lie algebra from the associated $k$-connected complete Hopf algebra, you need $k \geq 0$. And $k$-connected groups correspond to $k+1$-connected spaces. This is why you need simply connected spaces in the equivalence. -I'm not sure I gave any 'high concept' as to 'why Quillen's construction works', but this is at least a road map I can remember.<|endoftext|> -TITLE: Does left-invertible imply invertible in full group C*-algebras (discrete case)? -QUESTION [24 upvotes]: The following question/problem has been bugging me on and off for some time now: so I thought it might be worth broaching here on MO, as a case of "ask the experts". -Let $G$ be a discrete group. Kaplansky observed that since the group von Neumann algebra $VN(G)$ is a finite von Neumann algebra, each left-invertible element in $VN(G)$ is invertible. A proof is outlined in - -M.S. Montgomery, Left and right inverses in group algebras, Bull. AMS 75 (1969) - -(Well, she actually states a weaker result, but inspection shows that her argument extends to give what we claim. See also my remarks on this previous MO answer.) -The basic idea is to exploit the faithful trace $T\mapsto \langle T\delta_e,\delta_e\rangle$ and how it behaves on idempotents: for if $ab=I$, then $ba$ is an idempotent. -In particular, each left-invertible element of the reduced group $C^*$-algebra is invertible. -Question. What can we say for the full group $C^*$-algebra? Is every left-invertible element in $C^*(G)$ automatically invertible? -Some basic observations: - -The case where $G$ is the free group on two generators follows from a result of M-D Choi [no relation] who showed that $C^*({\mathbb F}_2)$ embeds into a direct product of matrix algebras. -More generally, if $C^*(G)$ has a faithful trace then one can use the same argument as for the reduced $C^*$-algebra to get a positive answer. -If $C^*(G)$ has no non-trivial projections then $ab=I$ implies $ba=I$. (I think this was known to be true for $G={\mathbb F}_2$ but I've forgotten the reference at present.) -There are examples of $G$ where $C^*(G)$ has no faithful trace; these can be found in work of Bekka and Louvet, and come from exploiting Property (T). - - -Bekka, M. B.(F-METZ-MM); Louvet, N.(CH-NCH) - Some properties of $C^*$-algebras associated to discrete linear groups. $C^*-algebras (Münster, 1999), 1–22, Springer, Berlin, 2000. - -REPLY [8 votes]: There is an alternative argument for the free group; not using that free groups are residually finite-dimensional. -Let $\pi$ be a faithful representation of $C^{\ast}(F)$ on a Hilbert space $H$. Then, as $U(H)$ is connected, $\pi$ can be deformed to the trivial representation in the point-norm topology, i.e. there exists a family of unitary representations $\pi_t$ for $t \in [0,1]$, such that $t \mapsto \pi_t(a)$ is norm-continuous for each $a \in C^{\ast}(F)$, $\pi_0=\pi$ and $\pi_1(g)=1_H$ for all$ g \in F$. -Now, if $ab=1$ in $C^{\ast}(F)$, then $\pi_t(ba)$ is a continuous path of projections ending at $1_H$. Hence, $\pi_0(ba)=1_H$ and $ba=1$, as $\pi$ was faithful. -EDIT: The same argument works if the $C^{\ast}$-algebra embeds into some contractible algebra (i.e. homotopy equivalent to $\mathbb C$). However, even though many reasonable toplogical spaces are quotients of contractible topological spaces, only few reasonable $C^{\ast}$-algebras have this property. There is a close relationship with the concept of quasi-diagonality, which appeared in the work of Voiculescu.<|endoftext|> -TITLE: Arithmetic Progressions of Squares -QUESTION [24 upvotes]: Fermat may or may not have known that there are 3-term arithmetic progressions of squares (like $1^2, 5^2, 7^2$, and that there are no 4-term APs. Murky history aside, Keith Conrad has two pleasant expositions (here and here) providing a modern treatment of this from an algebraic viewpoint. -A natural combinatorial follow-up is: how large can a subset of $\{1^2,2^2,\dots,n^2\}$ be and still not have 3-term APs? In this paper, I showed that there are subsets of size $$\gg n c^{-\sqrt{\log\log n}},$$ where $c=2^{\sqrt{8}}$, but I don't know of an upper bound. - -Is there a subset of the squares with positive relative density that is free of 3-term APs? - -REPLY [23 votes]: Probably not, but a proof is hopeless. Ruzsa and Gyarmati have a preprint in which they construct such a subset of size something like $N/\log \log N$. -Even the colouring version (that is, finite colour the squares, does one of the classes contain a 3-term progression) is open. A very closely-related question (Schur's theorem in the squares) is explicitly asked as Question 11 in this paper by Bergelson: -http://www.math.iupui.edu/~mmisiure/open/VB1.pdf -It is possible to show that a positive density subset of the squares contains a solution to $\frac{1}{4}(x_1 + x_2 + x_3 + x_4) = x_5$ by adapting the technique of arXiv:math/0302311. I'd have to admit this is slightly more than a back of an envelope calculation :-)<|endoftext|> -TITLE: What is the origin of unit vector notation? (i,j,k) -QUESTION [6 upvotes]: What is the origin of this notation? Who coined them and for what purpose? - -REPLY [5 votes]: Maybe it originates from Hamilton's quaternions $\mathbb{H}$, which has a basis $1,i,j,k$ as a real vector space, and the multiplications there, namely, $i\cdot j=k, j\cdot k=i, k\cdot i=j$ correspond exactly to the wedge product in $\mathbb{R}^3$. So $\mathbb{R}^3$ can be viewed as the imanginary part of $\mathbb{H}$. -Anyway, this is just my understanding or my guess.<|endoftext|> -TITLE: modular form Fourier coefficients and associated automorphic representation -QUESTION [11 upvotes]: Hi, -Let $f$ be a cuspidal modular form of some weight and level $N$. Then it determines -an irreducible automorphic representation $\pi = \bigotimes'\pi_p$ of $GL_2(\mathbf Q)$. -Let $f = \sum_i a_i q^i$ be its fourier expansion. Then it is known that if $p\nmid N$, -then $a_p$ determines $\pi_p$ (it is an unramified principal series). Is it true that -$a_p$ determines $\pi_p$ in general? And if so, how? -Thanks! - -REPLY [11 votes]: Jared Weinstein and I wrote a paper on how to compute $\pi_p$: see here. -As Olivier says, $a_p$ will often be zero, and in fact if the central character is trivial (or has conductor coprime to $p$) this is always the case when $p^2$ divides the level of $f$. One can get a bit futher by twisting: you can always twist a newform by Dirichlet characters, and Atkin and Li have shown that $\pi_p$ is principal series or Steinberg at $p$ if and only if there is some Dirichlet character $\chi$ such that the twist of $f$ by $\chi$ is a newform with nonzero Fourier coefficient at $p$ (or an oldform attached to such a newform). -So that leaves the supercuspidal cases, and here Hecke theory won't help you at all: no matter how you twist your form, the Hecke eigenvalues are all zero. One can actually show (the "local converse theorem") that $\pi_p$ is uniquely determined by the Atkin-Lehner pseudo-eigenvalues of all of the twists of $f$; but it is not so easy to calculate these, or to explicitly identify $\pi_p$ from a list of them once you've done so. In my paper with Jared you can find details of a different approach, using Bushnell and Kutzko's theory of "types", which seems to work quite well. -These algorithms are implemented in recent versions of Magma (and should be in Sage fairly shortly as well, once someone gets around to reviewing my code).<|endoftext|> -TITLE: Arithmetic progressions in power sequences -QUESTION [14 upvotes]: In connection with this MO post (and without any applications / motivation whatsoever), here is an apparently difficult - but nice - problem. -For a non-zero real number $s$, consider the infinite sequence - $$ P_s := \{ 1^s, 2^s, 3^s, \ldots \} . $$ -What is the number of terms, say $l(s)$, of the longest arithmetic -progression contained in this sequence? -For instance, since there exist three-term arithmetic progressions in -squares, but no such four-term progressions, we have $l(2)=3$. -It is easy to see that if $s$ is the reciprocal of a positive integer, then -$P_s$ contains an infinite arithmetic progression; hence we can write -something like $P(1/q)=\infty$. It can be shown that this is actually the -only case where $P_s$ contains an infinite progression. (This is certainly -non-trivial, but not that difficult either - in fact, in a different form -this was once posed as a problem on a Moscow State University math -competition). -If now $s$ is the reciprocal of a negative integer, then $P_s$ contains an -arithmetic progression of any preassigned length; this is a simple exercise. -Are there any other values of $s$ for which $l(s)$ is infinite? - -Conjecture. For any real $s\ne 0$ which is not the reciprocal of an - integer, the quantity $l(s)$ is finite; that is, there exists an integer - $L>1$ (depending on $s$) such that $P_s$ does not contain $L$-term arithmetic - progressions. - -Three-term progressions are not rare; say, for any integer $1\sqrt{ac}$ there exists $s>0$ such that $\{a^s,b^s,c^s\}$ is an -arithmetic progression. However, I don't have any single example of a -four-term progression in a power sequence (save for the case where the -exponent is a reciprocal of an integer). - -Is it true that $l(s)\le 3$ for any $s\ne 0$ which is not the reciprocal of an - integer? - -Indeed, excepting the cases mentioned above and their immediate -modifications, I do not know of any $s$ such that $P_s$ contains two distinct -three-term progressions. - -Is it true that if $s\ne p/q$ with integer $q\ge 1$ and - $p\in\{\pm1,\pm2\}$, then $P_s$ contains at most one three-term arithmetic - progression? - -$$ $$ -As a PS: I was once told that using relatively recent (post-Faltings) results -in algebraic number theory, one can determine $l(s)$ for $s$ rational. Can -anybody with the appropriate background confirm this? - -REPLY [7 votes]: In the case $s=p/q$ with $|p| \geq 3$, there is no $3$-term AP in $P_s$. The proof is by reducing to the case $s=p$, as follows. -Let $A=a^p$, $B=b^p$, $C=c^p$ such that $A^{1/q}+C^{1/q}=2B^{1/q} \quad (*) \quad$ and $(A,B,C)=1$. Let $K=\mathbf{Q}(\zeta_q)$ be the $q$-th cyclotomic field and $L=K(A^{1/q},B^{1/q},C^{1/q})$. Then $L/K$ is a finite abelian extension of exponent dividing $q$ and by Kummer theory, such extensions are in natural bijection with the finite subgroups of $K^{\times}/(K^{\times})^q$. The extension $L/K$ corresponds to the subgroup generated by the classes $\overline{A},\overline{B}, \overline{C}$ of $A,B,C$ in $K^{\times}/(K^{\times})^q$. In view of the following lemma, it suffices to prove $L=K$. -Lemma 1. If $n^p$ is a $q$-th power in $K$, then $n$ is a $q$-th power in $\mathbf{Z}$. -Proof. Assume $n^p=\alpha^q$ with $\alpha \in K$, then taking the norm we get $n^{p(q-1)}=N_{K/\mathbf{Q}}(\alpha)^q$. Since $\alpha$ is an algebraic integer, we get that $n^{p(q-1)}$ is a $q$-th power in $\mathbf{Z}$, and since $p(q-1)$ and $q$ are coprime, we get the result. -Lemma 2. The integer $B$ is relatively prime to $A$ and to $C$. -Proof. By symmetry, it suffices to prove $(A,B)=1$. Let $\ell$ be a prime number dividing $A$ and $B$. Then $\ell^{1/q}$ divides $A^{1/q}$ and $B^{1/q}$ in the ring $\overline{\mathbf{Z}}$ of all algebraic integers. By $(*)$ it follows that $\ell^{1/q} | C^{1/q}$. Thus $C/\ell \in \mathbf{Q} \cap \overline{\mathbf{Z}} = \mathbf{Z}$ which contradicts $(A,B,C)=1$. This proves Lemma 2. -By equation $(*)$, we have $K(B^{1/q}) \subset K(A^{1/q},C^{1/q})$ which reads $\overline{B} \in \langle \overline{A},\overline{C} \rangle$ in $K^{\times}/(K^{\times})^q$. We can thus write $B \equiv A^{\alpha} C^{\gamma} \pmod{(K^{\times})^q}$ for some $\alpha,\gamma \geq 0$. By a reasoning similar to Lemma 1, we deduce that $B/(A^{\alpha} C^{\gamma})$ is a $q$-th power in $\mathbf{Q}$ but since this fraction is in lowest terms (Lemma 2), we get that $B$ is a $q$-th power in $\mathbf{Z}$. -Now let $\sigma$ be an aribtrary element in $\mathrm{Gal}(L/K)$. We have $\sigma(A^{1/q}) = \zeta \cdot A^{1/q}$ and $\sigma(C^{1/q})=\zeta' \cdot C^{1/q}$ for some $q$-th roots of unity $\zeta$ and $\zeta'$. Considering the real parts of both sides of $\sigma(*)$, we see that necessarily $\zeta=\zeta'=1$. This shows that $L=K$ as requested.<|endoftext|> -TITLE: Log resolutions on surfaces and 3-folds in characteristic p -QUESTION [5 upvotes]: If $X$ is a normal projective variety and $D$ a divisor in it, we say that $\pi\colon (\widetilde X,\widetilde D)\rightarrow (X,D)$ is a log resolution if $\widetilde X$ is a resolution of $X$, the strict transfor $\widetilde D$ of $D$ is non-singular and $\widetilde D \cup \{ \text{exceptional divisors} \}$ have simple normal crossings. -I know little about techniques for resolution of singularities and as far as I am aware, for varieties over algebraically closed fields, the problem of finding resolutions of singularities is open. -However, I was wondering if the following is solved, by whom and if someone can provide me with a 'black-box' reference: -Question: Given a projective variety $X$ over a field of characteristic $p$ and a divisor $D$ on $X$, is there a log resolution of the pair $(X,D)$ in the cases where $X$ is a non-singular variety of small dimension (1,2,3) and/or in the case the $p\neq 2,3,5\ldots$? What if the variety is a product of a projective variety and the affine line? -Of course partial answers are appreciated. However the purpose of this is just to use it in a birational proof for something else, so by no means I intend to prove it myself or get any close to it. - -REPLY [5 votes]: What you want does follow from Cutkosky's paper cited by Donu Arapura (and presumably already from results of Abhyankar, but I have not checked). One just has to combine his Theorems 1.1 and 1.2. More precisely, one can resolve singularities of $X$ using Theorem 1.1 -to get $\pi_1: X_1 \to X$ with $X_1$ smooth and then apply Theorem 1.2 to the pair $(V,S) = (X_1,D_1)$, where $D_1$ is the union of $\pi_1^{-1}(D)$ and the exceptional divisors. -This requires that the characteristic be $>5$, but if you use the result of Cossart and Piltant instead of Cutkosky's Theorem 1.1 you can get a log resolution in any characteristic (since Cutkosky's Theorem 1.2 has no characteristic restriction).<|endoftext|> -TITLE: Is there a "Cartan product" of Harish-Chandra modules? -QUESTION [5 upvotes]: If $\lambda,\mu$ are two dominant weights for a Lie group $G$, then -there is a canonical (up to scale, perhaps) -surjection $V_\lambda \otimes V_\mu \to V_{\lambda+\mu}$ -of finite-dimensional representations, -which I have occasionally heard called the "Cartan projection". -To each irreducible (or standard) Harish-Chandra module for -$({\mathfrak g},K)$, one can associate a $K$-orbit on $G/B$. -For finite-dimensional representations, this orbit is the open $K$-orbit. - -I want to know what analogues exist of the Cartan projection out of $V\otimes W$ if - $V,W$ are two Harish-Chandra modules with the same associated $K$-orbit (other than - the open orbit case above). - -The answer may be something like "every H-C quotient of $V\otimes W$ has -the wrong Gel$'$fand-Kirillov dimension for it to again -have that associated $K$-orbit," in which case I'd appreciate references -that make that most clear. (I will be sad, but not overly surprised, -if that is the case.) -EDIT: For consistency of notation, let's take $G,K$ to be the complexifications of $G_0,K_0$, where $K_0$ is a maximal compact in $G_0$. - -REPLY [4 votes]: If $G$ is simply connected semisimple and admits holomorphic discrete series, then -the tensor product of the holomorphic discrete series of two weights decomposes as a direct sum of other holomorphic discrete series, each appearing with finitely mulitplicity. (In short, holomorphic discrete series comport themselves in general just as in the case of $SL_2(\mathbb R)$.) -I forget the reference now; sorry! [Added: Actually, section 7 of the Kobyashi paper linked to by BR above gives the result, I think.] -I think that for other Harish-Chandra modules, tensor products have infinite multiplicities (or in more analytic terms, involve direct integrals rather than direct sums). Even the generic discrete series for $Sp_4(\mathbb R)$ should demonstrate this behaviour, if I remember correctly. (And tensoring a holomorphic discrete series by an anti-holomorphic discrete series would also give infinite multiplicities.)<|endoftext|> -TITLE: Riemann surfaces that are not of finite type -QUESTION [5 upvotes]: I am interested in studying Riemann surfaces that are not of finite type. By a non-finite type Riemann surface, I mean a Riemann surface that is not conformally equivalent to any Riemann sub-surface of a compact Riemann surface. I have the following questions about such surfaces:- -[1]. Is there any classification theory for such surfaces like in the compact case? -[2]. Amongst such surfaces, what are the hyperbolic ones? -[3]. Under what additional conditions is something like the Hurwitz's automorphisms theorem true, if at all, for such surfaces? -Most of the books on Riemann surfaces I have looked through do not seem to treat such surfaces. Any reference to any book/paper that deals with such surfaces would be very helpful. - -REPLY [5 votes]: Classification. You do not specify what classification (what is your equivalence relation?) -Topological classification is due to Kerekjarto. A reference is given in the answer of Richard Kent. Complete conformal classification is hopeless. -There are many different notions of hyperbolicity. They all coincide in the simply connected case. - -a) Universal cover is the disc. (All Riemann surfaces, except tori, sphere, plane and cylinder -are hyperbolic in this sense). -b) There is no Green function. -c) There is no positive harmonic function -d) There is no bounded analytic function -e) There is no analytic function with finite Dirichlet integral -f) And so on. -In 1950-s there was large area of research called Classification of Riemann surfaces. -The main subject of this research was establishing the relations between b)-f) and -other similar properties, and finding criteria for concrete surfaces to satisfy b)-e). -You can find a lot of results of this sort in the books of Tsuji, -Potential theory in modern function theory, Nevanlinna, Uniformisation, -and Ahlfors and Sario book on Riemann -surfaces. - -Hurwitz automorphism theorem is not true. There are many open surfaces with a rich -group of automorphisms. - -On your last remark. The subject is out of fasion, so modern books do not treat it. -There are just too many Riemann surfaces to have an interesting classification of -all of them. So the research is concentrated on various special classes that have -some applications. For example, on hyperelliptic surfaces (those obtained as a 2-sheeted -ramified covering of the plane); Such surfaces occur in mathematical physics. -Another interesting class which is studied is Abelian coverings of compact surfaces.<|endoftext|> -TITLE: When does mathscinet review a paper? -QUESTION [9 upvotes]: What are some reasons for mathscinet listing a article, but then stating that there will be ``no review of this item''? In particular does it at all imply that the paper has little or no merit from a mathematical viewpoint? - -REPLY [6 votes]: You can see the "official line" here: http://www.ams.org/publications/math-reviews/mr-edit In particular, - -Elementary articles or books, or articles that have not been refereed are ordinarily not listed. - -In my experience, (and in addition to Thierry's and Andreas's list) this includes: - -Books reviews -More problematic is the case of certain authors and journals who are, shall we say, excessively "prolific"-- sometimes one sees papers which are clearly "new mathematics" (or least, claim to be) but which do not get a review. For obvious reasons I won't give an example, but it's not hard to find using the MathSciNet search tool. - -You ask "In particular does it at all imply that the paper has little or no merit from a mathematical viewpoint?". Well, certainly not, in some sense. A book review, or an elementary survey article, might well contain interesting mathematics, and might well be useful to read (which also one can clearly understand why they wouldn't get a review-- a "review of a review" would be quite silly). -As to my case (2.) above-- yes, perhaps this is Math Reviews (or an editor, or someone who was sent the paper to review) making that judgement. This does seem to be a grey area, as it's not covered by the "Editorial Statement" I linked to (except maybe in the word "elementary"). It would be interesting to get more information about this... - -REPLY [4 votes]: Sometimes publications (especially in proceedings volumes) are preliminary versions of papers whose final version will appear elsewhere. In such cases (or cases suspected to be such), Mathematical Reviews will wait for the final version. This is why papers in proceedings volumes that are final versions will often have a footnote saying something like "This paper is in final form and no version of it will appear elsewhere." This amounts to "Hey, Math Reviews! Review this." - -REPLY [3 votes]: There are many things that the Mathematical Reviews does not review. Off the top of my head, - -Non-mathematical items that may appear in mathematical publications; -Many proceedings are not refereed, or if they are, the individual articles are not always refereed. - -As to why MathSciNet would maintain a list of these items without reviewing them, I'm assuming that it's for the sake of completeness.<|endoftext|> -TITLE: Where do the real analytic Eisenstein series live? -QUESTION [17 upvotes]: In obtaining the spectral decomposition of $L^2(\Gamma \backslash G)$ where $G=SL_2(\mathbb{R})$, and $\Gamma$ is an arithmetic subgroup (I am satisfied with $\Gamma = SL (2,\mathbb{Z})$) we have a basis of eigenfunctions of the hyperbolic Laplacian, and orthogonal to that we have the space spanned by the incomplete Eisenstein series, -$$ -E(z,\psi) = \sum_{\Gamma_\infty \backslash \Gamma} \psi (\Im(\gamma z)) = \frac{1}{2\pi i}\int_{(\sigma)} E(z,s)\tilde{\psi}(s)\mathrm{d}s -$$ -where $\psi \in C_c^\infty(\mathbb{R}^+)$, $\tilde{\psi}$ is its Mellin transform, and $E(z,s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} \Im(\gamma(z))^s$ is the usual Eisenstein series. -My question is, where does $E(z,s)$ itself live with respect to the vector space $V = L^2(\Gamma \backslash G)$ which can be considered as the vector space of the right regular representation of $G$, and what is this parameter $s$? -A similar question of course goes for $\mathbb{R}$, where does $e^{2\pi i x}$ live with respect to $(L^2(\mathbb{R}), \rho)$? -I would appreciate a representation theoretic flavored answer, that is why I mentioned representations, but any other answer would also be an addition to my understanding of this. -In general, is there an associated space to $(V,\pi)$, an automorphic representation, such that the elements of the vector space are of moderate or rapid growth, instead of decay. - -REPLY [2 votes]: I can give an additional point of view, which is coming from the theory of parabolic induction. Parabolic induction plays a prominent role in representation theory and gives you a better intuition for the higher rank situation. This point of view is often better stressed in the adelic theory than in the classical picture. -Let $G =PSL_2(\mathbb{R})$ with standard parabolic $B$ and $\Gamma$ a cofinite lattice. -My intuition is that te analytic Eisenstein series on $\Gamma \backslash G$ are vectors of the induced representation: -$$ Ind_{\Gamma N}^G 1,$$ -but there is one major issue with this, namely that $\Gamma N$ is not a group. -Rigorously seen they are given as $P$ series, i.e. -$$ E: C_c^\infty(N \backslash G) \rightarrow L^2(\Gamma \backslash G),$$ -by defining the $B$ series -$$E(f) (g)= \sum\limits_{B \cap \Gamma \backslash \Gamma} f(\gamma g).$$ -The image of $E$ generates a dense subspace of the orthocomponent to the cuspidal forms. -Now, we one notices that $C_c^\infty(N \backslash G)$ is a dense subspace of $Ind_N^G 1$. Induction by steps gives a decomposition -$$ Ind_N^G 1 \cong Ind_B^G Ind_N^B 1.$$ -Now for $Ind_N^B 1 \cong L^2(B/N) = L^2(M)$, where $M$ are the diagonal matrices. Pontryagin duality gives you a direct integral decomposition of $L^2(M)$, and we have as a result -$$ Ind_N^G 1 \cong \int\limits_{\Re s = 0}^{\oplus} Ind_B^G | \cdotp |^s. $$ -Certainly one hopes that $E$ extends to $Ind_B^G | \cdotp |^s$, but convergence only happens $\Re s >1/2$, and the operators has to be defined by analytic continuation to make sense on $\Re s = 0$. -Perhaps it useful to give at least one definition here: functions $f \in Ind_B^G | \cdotp |^s$ are defined as $f(bg) = |b_{1,1} / b_{2,2}|^{s+1/2} f(g)$ for $b \in B$ with $f|_K\in L^2(K)$ for $K= PSO(2)$.<|endoftext|> -TITLE: presentations of Sp(n, Z) and cocycles. -QUESTION [7 upvotes]: Question: where can I find explicit presentations of the group $Sp(n, \mathbb Z)$, for small $n$? - -It is known that $Sp(n, \mathbb Z)$ admits a $2$-cocycle $h$ with values in $\mathbb Z/2\mathbb Z$ which I'd like to view in the following way. If we fix a presentation with relations forming a set $R$, then $h$ is a function $R \to \mathbb Z/2\mathbb Z$ (which fulfils some condition). - -Question: where can I find an explicit description of this cocycle, in the form of what values does it take on elements of some presentation? - -REPLY [8 votes]: There is a simple and explicit presentation of $SP(2n,\mathbb{Z})$ in Theorem 9.2.13 of Hahn-O’Meara's book "The classical groups and K-theory". In some sense, this presentation is related to the ideas of Klingen mentioned by Rivin; however, just following Klingen's method leads to an enormous presentation (something on the order of 5 families of generators and 67 families of relations), as is shown in the paper -MR0280606 (43 #6325) -Birman, Joan S. -On Siegel's modular group. -Math. Ann. 191 1971 59–68. -Another simple presentation can be found in the paper -MR1152494 (93b:11054) -Lu, Ning(1-RICE) -A simple presentation of the Siegel modular groups. -Linear Algebra Appl. 166 (1992), 185–194. -The basic idea is that there exist nice presentations for the mapping class group, which surjects onto the symplectic group.<|endoftext|> -TITLE: Continuous cohomology of semi-simple Lie group. -QUESTION [5 upvotes]: Let $G$ be a real connected semi-simple Lie group. Let $M$ be a finite dimensional representation of it. Are there general criteria when the continuous cohomology groups $H_{cont}^q(G,M)$ vanish? -A situation of particular interest for me is $G=SO^+(n-1,1)$, namely the connected Lorentz group, and $M$ is the standard representation of it. Is it true that the first continuous cohomology $H^1_{cont}(G,M)=0$ ? - -REPLY [6 votes]: As pointed out by Konrad, this follows from the generalisation of van Est's theorem to the continuous case (due to Hochschild and Mostow); namely, that -$$ -H_c^m(G,M) \cong H^m(\mathfrak{g},\mathfrak{k};M) -$$ -where $\mathfrak{g}$ is the Lie algebra of $G$ and $\mathfrak{k}$ is the Lie algebra of the maximal compact subgroup of $G$. -For the case in question, $m=1$, $\mathfrak{g} = \mathfrak{so}(n-1,1)$ and $\mathfrak{k}=\mathfrak{so}(n-1)$. I will take $n>2$. -According to Chevalley and Eilenberg, the cohomology $H^m(\mathfrak{g},\mathfrak{k};M)$ is computed from 'horizontal' 'equivariant' cochains in $C^m(\mathfrak{g},M)$, where 'horizontal' means that the cochain vanishes whenever any of its entries belongs to $\mathfrak{k}$ and 'equivariant' means with respect to the action of $\mathfrak{k}$. -Now for the algebras in question, $\mathfrak{g}$ breaks up as $\mathfrak{k} \oplus V$ under the action of $\mathfrak{k}$, where $V$ is the fundamental vector representation of $\mathfrak{k}$, whereas $M = V \oplus \mathbb{R}$, with $\mathbb{R}$ the trivial one-dimensional representation. -Since -$$ -C^0(\mathfrak{g},\mathfrak{k};M) = M^{\mathfrak{k}} -$$ -it follows that -$$ -\dim C^0(\mathfrak{g},\mathfrak{k};M) = 1~. -$$ -The differential $\delta: C^0 \to C^1$ is injective, since if $T \in M^{\mathfrak{k}}$ ($T$ is 'timelike' hence the notation) -$$ -\delta T(X) = X \cdot T -$$ -which does not vanish identically. -On the other hand, -$$ -C^1(\mathfrak{g},\mathfrak{k};M) = \text{Hom}(V,M)^{\mathfrak{k}} -$$ -is again one-dimensional, hence $C^1 = \delta C^0$ and hence $H^1 =0$.<|endoftext|> -TITLE: Homotopy groups of spheres in a $(\infty, 1)$-topos -QUESTION [28 upvotes]: Let $H$ be an $(\infty,1)$-topos (seen as a generalization of the homotopy category of spaces). -You can define the suspension of an object $X$ as the (homotopy) pushout of $*\leftarrow X \to *$, hence you can define inductively the spheres $\mathbb{S}^n$ (the sphere of dimension $-1$ is the initial object of $H$ and the sphere of dimension $n+1$ is the suspension of the sphere of dimension $n$). -You can also define the loop spaces of a pointed object as the (homotopy) pullback of $*\to X \leftarrow *$. It will be itself pointed (because there is an obvious commutative diagram with a $1$ instead of $\Omega{}X$, so there is (I think) an arrow between this $1$ and $\Omega{}X$). -Then, given two integers $n, k$, you can define $\pi_k(\mathbb{S}^n)$ as the set of connected components (global elements up to homotopy) of the $k$-fold loop space of the $n$-sphere (I don’t know if this definition is one of the two described in the nlab) -Is there a natural group structure on $\pi_k(\mathbb{S}^n)$? -Is there something known about these groups in general? -For example, - -Are they completely known for some $H$? -Is it always true that $\pi_k(\mathbb{S}^n)$ is trivial for $k -TITLE: Infinitesimally Desarguian curve families -QUESTION [9 upvotes]: The $2$-dimensional family of solution curves $y = u(x, \xi, \eta) \approx \eta + \xi x + \mathcal O(x^2)$ near $y = 0$ of a differential equation -$y'' = \Phi(x, y, y')$ -has been called infinitesimally Desarguian if -\begin{equation*} - \Phi(x, y, p) = \mathcal O(|y|^3 + |p|^3) \quad \textrm{as $(y, p) \to (0, 0)$} . -\end{equation*} -The reason for this terminology is that such curve families in some sense are close to the family of straight lines. The curve family in the $\xi \eta$-plane that is defined by $y = u(x, \xi, \eta)$ while $x$ and $y$ are considered as parameters is called the dual curve family. -It has been stated without proof for instance in Gelfand, Gindikin, and Shapiro, A local problem of integral geometry in a space of curves, Functional Anal. Appl. 13 (1979), p. 88 and p. 99 (doi: 10.1007/BF01077241, mathnet) and in the book by Arnold, Geometrical methods in the theory of ordinary differential equations, ch. 1, § 6 G Exercise (!), that a curve family is infinitesimally Desarguian if and only if its dual family is defined by a differential equation of the form -$\eta'' = \Psi(\xi, \eta, \eta') $, -where $\Psi(\xi, \eta, p)$ is a polynomial in $p$ of degree at most $3$ with coefficients that are smooth functions of $(\xi, \eta)$; that class is known to be preserved by arbitrary smooth diffeomorphisms in the $\xi \eta$ plane, see the cited book by Arnold. -I would very much like to know a reference for the proof of this fact. - -REPLY [2 votes]: A proof can be extracted from Cartan's "On manifolds with projective connections." (Translated by D. H. Delphenich). -In section VII Cartan defines a projective connection in the projectivised tangent bundle ("manifold of elements") with connection one-form: -$$ -\omega = \begin{pmatrix} -\omega_0^0&\omega^1&\omega^2\\ -\omega_1^0&\omega_1^1&\omega_1^2\\ -\omega_2^0&\omega_2^1&\omega_2^2\\ -\end{pmatrix} -$$ -After normalising the connection he arrives at the curvature two-form -$$ -\Omega = \begin{pmatrix} -\gamma&0&0\\ -\Omega_1^0&\gamma&0\\ -\Omega_2^0&\Omega_2^1&\gamma\\ -\end{pmatrix}, -$$ -where -$$ -\begin{aligned} -\Omega_2^1 &= a\omega^2\wedge\omega_1^2,\qquad a=-\frac{1}{6}\frac{\partial^4f}{\partial y'^4}\\ -\Omega_1^0&=b\omega^1\wedge\omega^2,\qquad b=\text{complicated function}\\ -\Omega_2^0&=\text{complicated two-form}\\ -\gamma&=\Omega_0^0=\Omega_1^1=\Omega_2^2=\text{complicated two-form}.\\ -\end{aligned} -$$ -In section 23 Cartan then states the dual connection with connection one-form -$$ -\varpi = \begin{pmatrix} -\varpi_0^0&\varpi^1&\varpi^2\\ -\varpi_1^0&\varpi_1^1&\varpi_1^2\\ -\varpi_2^0&\varpi_2^1&\varpi_2^2\\ -\end{pmatrix}=\begin{pmatrix} -\omega_2^2&\omega_2^1&\omega^2\\ -\omega_2^1&\omega_1^1&\omega^1\\ -\omega_2^0&\omega_1^0&\omega_0^0\\ -\end{pmatrix} -$$ -which is simply the primal connection transposed with respect to the antidiagonal and the same goes for the the curvature two-form: -$$ -\Pi = \begin{pmatrix} -\gamma&0&0\\ -\Pi_1^0&\gamma&0\\ -\Pi_2^0&\Pi_2^1&\gamma\\ -\end{pmatrix} = \begin{pmatrix} -\gamma&0&0\\ -\Omega_2^1&\gamma&0\\ -\Omega_2^0&\Omega_0^1&\gamma\\ -\end{pmatrix}, -$$ -so that we have -$$ -\begin{aligned} -\Pi_1^0=\Omega_2^1\\ -\Pi_2^1=\Omega_1^0 -\end{aligned} -$$ -We have that $\Omega_2^1=0$ is the condition that the right-hand side of the differential equation is a polynomial in $y'$ with degree at most 3 and the condition $\Omega_1^0=0$ is then "infinitesimal desargueness". In the dual connection these two conditions are swapped. -It might be argued that "infinitesimal nondesargueness" should really mean that any of $\Omega_2^1$ or $\Omega_1^0$ is non-vanishing since they are both obstructions to the construction of homogenous coordinates which then gives a projective structure on the $x,y$-plane with the solution curves as lines. The availability of homogenous coordinates is equivalent to the Desarguesian property of a projective plane. -Here's a couple of blog posts discussing these things: - -The geometry of y''+y=0 -The geometry of a second order differential equation, part 1 -The geometry of a second order differential equation, part 2<|endoftext|> -TITLE: Cone of movable curves -QUESTION [9 upvotes]: Let $X$ be a smooth complex projective variety of dimension $n$. -Under the duality between $N_1(X)$ and $N^1(X)$ we know that closure of cone of effective curves $\overline{NE}(X)$ is dual to closure of ample cone $\overline{Amp}(X)$. -It was proved in 2004 that the closure of cone of effective divisors $\overline{Eff}(X)$ is dual to the closure of cone of movable curves $\overline{Mov}(X)$. A movable curve by definition is a curve class $C \in N_1(X)$ such that $C=\pi_*(H_1 H_2 \cdots H_{n-1})$, where $\pi: X' \rightarrow X$ is a birational morphism and $H_i$'s are ample classes on $X'$. -My question: Let $Q(X)$ be the cone obtained by curve classes $H_1 H_2 \cdots H_{n-1}$ where $H_i$ are ample divisors on $X$ itself. Is it true/false that $\overline{Q}(X)=\overline{Mov}(X)$? i.e. as long as I am interested only in the closure of these cones; do I really miss some curve class if I only restrict my self to intersection of ample classes on $X$ itself. -Can any body give an example where $\overline{Q}(X) \neq \overline{Mov}(X)$? -Meanwhile, I am only interested in $n=3$ case. - -REPLY [3 votes]: The definition you give for movable is actually the definition for strongly movable. A movable curve on X is one that lives in a family that covers a dense open subset of X, or the closure of the convex span of this concept. That the definitions are equivalent is a strong result. -I am mentioning this because I needed the other definition, and Google brought me here. -Cheers.<|endoftext|> -TITLE: What items MUST appear on a mathematician's CV? -QUESTION [74 upvotes]: Writing a CV makes me paranoid that I'm failing to abide by unwritten rules. Of course CVs are flexible to capture the diversity of accomplishments someone might have. But there must be plenty of things a hiring committee absolutely expects. So I'm interested in anything that must be on a CV — whose omission would raise a red flag — of a mathematician looking for an academic job. -"Obvious" answers are welcome. Even things which sound obvious like "your name." What is obvious to someone who has read and evaluated lots of CVs is different from someone preparing one for the first time. In your answer, please also be fairly specific about scope: have you served on hiring committees? for what types of positions? in the US or Europe or? -(Note: As suggested in the comments, it is very good to ask people "in the know" directly for such career advice. One reason for asking this question on MO is to have more open, less clubby answers — there is an echo chamber effect when you ask a bunch of people in the same subcommunity.) - -REPLY [56 votes]: From my perspective, the critical question isn't what must be included on your CV, but rather what mustn't, since that seems to be the more common problem (judging by the ones I see). What I'm about to describe is based on my experience at a U.S. research lab; I imagine it generalizes quite a bit beyond that, but I can't say how far, and it is certainly country-specific. -I'll discuss five rules below, with some overlap between them. Of course these rules are not absolute (except for the last one), but you certainly shouldn't break them without thinking carefully about it and deciding there's a good reason to do so. -(1) Your CV should represent you as a professional mathematician. Anything that is not relevant to your professional life should be left out. For example, you should generally not describe non-math-related summer or part-time jobs, hobbies, side interests outside of mathematics and related fields, etc. If there's something unusually interesting or impressive (you published a novel or are a chess champion) or that displays relevant skills (you write free software in your spare time), it's OK to mention it, but just briefly and not in a prominent position. -I've seen some hair-raising violations of this rule, in which applicants devoted considerable space to things that have nothing to do with working as a mathematician. Nobody is going to reject your application just because you put something weird in your CV, but it's not good for your image as a professional. -(2) Your CV shouldn't include anything unless you think the search committee might need or want to know it. For example, contact information is valuable, as is anything that can legitimately help judge your application. However, in the U.S. you should not list your age or birthdate, your marital status, information about your children, or your religion (unless you are applying to a religious institution). I realize this is common in some countries, and of course people will be understanding about that, but it comes across strangely to give people information they don't want and shouldn't be influenced by. -(3) You should try not to seem desperate to impress, particularly with awards and distinctions. Some people provide enormous lists of very minor distinctions, sometimes with no relevance to research/teaching/service (for example, a college scholarship from a local business club). Coming across as insecure can make you seem less attractive: an ambitious department wants to hire people who are marginally too good for them, not people who are trying hard to be good enough. As a rule of thumb, when you get your Ph.D. and apply for your first job, it's OK to list any substantive distinction from grad school. You can list a few undergraduate honors, but only if they are impressive (Putnam fellow or major university-wide prize, yes; random scholarship, no). You shouldn't list high school honors at all (well, just maybe an IMO medal, but be careful not to look like you consider it your proudest achievement). -(4) Be sure not to give the impression you are trying to obfuscate anything. I don't just mean you should tell the truth, but also that you should be clear and straightforward. For example, people sometimes feel bad about not having enough items to list in their publication or talk sections, and it can be tempting to reorganize the CV to try to obscure this. For example, you could replace the "publications" section with a "research" section in which you list not just publications but also talks and poster presentations, or even current/future research topics. This is a bad idea, since it can look like you are trying to make the information less accessible, and then everything on your CV will be looked at more skeptically. Instead, you want to make it easy to understand your CV and easy to see that you aren't doing anything tricky. -(5) Don't lie. Don't say a paper will appear in a journal until it has been accepted, even if you are sure it will be. Don't say a paper is submitted until it is, even if you plan to submit it by the time the committee meets. Don't call something a preprint until it is written down and ready to distribute (you can say "in preparation" before then, but many people will ignore this since it is unverifiable). Don't say you have received a fellowship or prize if you haven't. You'd think all these things go without saying, but I've seen a couple of people get caught on one of them. You really don't want to be the person who gets asked for a copy of their preprint and can't produce one.<|endoftext|> -TITLE: Effective versus movable cones of curves -QUESTION [8 upvotes]: Let $\overline{NE}(X)$ be the closure of the cone generated by the numerical classes of effective curves and $\overline{\mathrm{Mov}}(X)$ the closure of the cone of moving curves. -(Q) Is there an example of a smooth projective variety $X$ such that - -$\overline{NE}(X)$ is (finite) polyhedral, but -$\overline{\mathrm{Mov}}(X)$ is not? - -Here are some trivial observation: -1 -If $X$ is a surface then the two cones are dual to each other and hence they are either both polyhedral or not. -2 -If $X$ is a Fano variety then - -$\overline{NE}(X)$ is polyhedral by the Cone Theorem, and -$\overline{\mathrm{Mov}}(X)$ is polyhedral by a result of Barkowski if $\dim \leq4$ (see here) and by [BCHM] in general. - -REPLY [5 votes]: As J.C. indicates in the comments, an example for Q1 can be gotten from the variety considered in this paper. This isn't spelled out in the paper, so let me explain it here. -First let's change the question into its dual form. The cone of curves is dual to the nef cone, and Boucksom--Demailly--Peternell--Paun showed that the cone of moving curves is dual to the cone of pseudoeffective divisors. So we want to find an example of $X$ such that $Nef(X)$ is rational polyhedral but the cone of pseudoeffective divisors $PsEff(X)$ isn't. -I claim the variety $X$ in the linked paper is such an example. Here, $X$ is constructed by blowing up $\mathbf{P}^3$ at the base locus of a general net of quadrics. -The variety $X$ is then elliptically fibred over $\mathbf{P^2}$, with the generic fibre having an infinite abelian group (more precisely, rank 8) of sections. Call this group $MW(X)$. Translating by differences of sections gives an action of $MW(X)$ on $X$ by so-called pseudo-automorphisms (meaning birational automorphisms that are isomorphisms in codimension 1). This group action preserves effective divisors, and hence the cone $PsEff(X)$. One can calculate the action fairly explicitly, and in particular one sees the orbit of a divisor $E_i$ (the exceptional divisor of one of the blowups) in N^1(X) is infinite. Now it is easy to see that each $E_i$ spans an extremal ray of $PsEff(X)$, and hence so does any $MW$-translate of $E_i$; since there are infinitely many of these, $PsEff(X)$ has infinitely many extremal rays. -On the other hand, it's not hard to show that $Nef(X)$ is rational polyhedral. This is done more or less by brute force: enumerate some curve classes, find the dual cone to the convex hull of those classes (which is then an upper bound for $Nef(X)$), and check that it's spanned by nef classes. Details are in the paper.<|endoftext|> -TITLE: Theorems proved with AD whose proof is also known in the ZF world -QUESTION [8 upvotes]: This question arises from discussions with my professor and from Todd Eisworth comments in this question Large cardinal axioms and the perfect set property -In $L(\mathbb{R})$ we have $AD$ and it is a powerful tool to prove theorems. Almost all of the theorem proved with $AD$ come in a very natural way: we use games and determinacy as in the First/Second/Third Periodicity Theorems. However no "$ZF$+Large Cardinal" proof is known for the Periodicity Theorems. Another example is that of the Perfect Set Property: Using $AD$ all sets of reals have the Perfect Set Property, but is a proof of the statement "Assuming infinitely many Woodin cardinals with a measurable above then every set of reals has the perfect set property" known? -So my question is: which theorems proved with $AD$ also have a known proof in the $ZF$+large cardinals world? - -REPLY [2 votes]: I don't know if this is what you are looking for, but many important examples of statements that are consequences both of AD and of large cardinals are themselves phrased in terms of determinacy or large cardinals. For example, AD and "there is a measurable cardinal" both imply "every analytic set is determined" and "every real has a sharp". (In fact, these two conclusions are equivalent to one another.) There are more such phenomena at other levels of consistency strength.<|endoftext|> -TITLE: Is every solvable subgroup of $GL(n,\mathbb{Z})$ polycyclic? -QUESTION [8 upvotes]: Is every solvable subgroup of $GL(n,\mathbb{Z})$ polycyclic? -The first solvable group that is not polycyclic is $\mathbb{Z}[1/2]\rtimes \mathbb{Z}$ (where the automorphism is given by multiplication with 2) and I do not see a way of embedding it into $GL_n(\mathbb{Z})$ for some $n$. - -REPLY [17 votes]: It is a theorem of Mal'cev that all solvable subgroups of $GL(n,\mathbb Z)$ are polycyclic, and a theorem of Auslander that every polycyclic group is isomorphically embeddable in $GL(n,\mathbb Z)$, for some $n$. Auslander's theorem was later reproved by Swan purely algebraically by adapting the proof of Ado's theorem. - -A. I. Mal’cev, "On certain classes of infinite solvable groups", Mat. Sb. 28 (1951) 567–588; Amer. Math. Soc. Transl. (2) 2 (1956) 1–21 -R.G. Swan, "Representations of polycyclic groups", Proc. Amer. Math. Soc, 1967 -L. Auslander, "On a problem of Philip Hall", The Annals of Mathematics, 1967<|endoftext|> -TITLE: An identity involving an infinite integral with a sinh in the denominator -QUESTION [9 upvotes]: I recently encountered the rather appealing looking integral, which appears in the theory of random matrices : -$$\int_{-\infty}^{\infty}\prod_{j=1}^{p-1}(j^{2}+z^{2})\frac{zdz}{\mathrm{sinh}(2\pi z)} = \frac{\Gamma(p+1/2)^{2}}{2\pi p}$$ -I wondered if anyone had seen it before in some table of integrals or such like, or had an idea how to prove it? Here are some things I noticed/tried : -1) The following identity is straightforward to prove, -$$\int_{0}^{\infty}z^{k}\frac{dz}{\mathrm{sinh}(2\pi z)} = \frac{2(1-2^{-k-1})\zeta(k+1)k!}{(2\pi)^{k+1}}$$ -and has a well known closed form expression for odd k. It is proved basically by expanding the sinh in a geometric series and integrating term by term. Noting that the first integral I mentioned has all odd powers of z in the numerator, I tried to multiply out the polynomial and integrate term wise. The trouble is one obtains a quite complicated combinatorial sum (involving central factorial numbers and bernoulli numbers) which I didn't know how to evaluate. -2) Contour integration methods. The pole structure of the integrand is quite easy to write down, and I tried a few box type contours which basically involved shifting the contour up the imaginary axis, and joining at the sides (noting that the integrand decays exponentially as Re(z) goes to + or - infinity). The residues at the poles can be evaluated easily, the trouble is that the integral along the top edge of the box is not obviously related to the original one, along the real line. -3) Generating function methods. The generating function, -$$\sum_{p=0}^{\infty}\frac{\prod_{i=1}^{p}(i^{2}+z^{2})}{(p!)^{2}}y^{p}$$ -appears to be expressible as a hypergeometric function. How this could be used constructively, I'm not sure. -p.s. I know the identity is true because it appears as the consequence of solving a certain random matrix problem in two completely different ways. - -REPLY [13 votes]: This integral is equivalent to the special case $\alpha=\gamma=p$, $\phantom.\beta=\delta=1/2$ of the known contour integral -$$ -\int_{-i\infty}^{i\infty} \Gamma(\alpha+s) \phantom. \Gamma(\beta+s) \phantom. \Gamma(\gamma-s) \phantom. \Gamma(\delta-s) \phantom. ds -= 2\pi i \frac{\Gamma(\alpha+\gamma) \phantom. \Gamma(\alpha+\delta) \phantom. \Gamma(\beta+\gamma) \phantom. \Gamma(\beta+\delta)} {\Gamma(\alpha+\beta+\gamma+\delta)}, -$$ -valid for all $\alpha,\beta,\gamma,\delta$ of positive real part. This is formula 641.2 on page 655 of Gradshteyn and Ryzhik [GR], which in turn cites formula 302(32) in Erdélyi et al. [E]. [EDIT Thanks to Richard Borcherds for recognizing this integral as a result of Barnes 1908. See postscript below about references and proofs for this integral.] -Indeed, the factor $z \phantom./\sinh(2\pi z)$ of the integrand factors as -$$ -\frac{1}{2} \frac{iz}{\sin(\pi i z)} \frac1{\cos(\pi i z)} = -\frac1{2\pi^2} \Gamma(1+iz) \phantom. \Gamma(1-iz) \phantom. \Gamma\bigl(\frac12+iz\bigr) \phantom.\Gamma\bigl(\frac12-iz\bigr), -$$ -while the factor $\prod_{j=1}^{p-1} (j^2+z^2)$ is -$$ -\prod_{j=1}^{p-1} (j+iz) \phantom. \prod_{j=1}^{p-1} (j-iz) -= \frac{\Gamma(p+iz)}{\Gamma(1+iz)} \frac{\Gamma(p-iz)}{\Gamma(1-iz)}. -$$ -Hence the integral is -$$ -\frac1{2\pi^2} \int_{-\infty}^{\infty} \Gamma(p+iz) \phantom. \Gamma(p-iz) \phantom. \Gamma\bigl(\frac12+iz\bigr) \phantom.\Gamma\bigl(\frac12-iz\bigr) \phantom. dz. -$$ -Now take $z=is$ to obtain the known integral, multiplied by $1/2\pi^2 i$. -The formula for that integral, multiplied by the same factor $1/2\pi^2 i$, then yields -$$ -\frac1\pi \frac{\Gamma(2p) \phantom. \Gamma(p+(1/2))^2 \phantom. \Gamma(1)}{\Gamma(2p+1)} , -$$ -which is equivalent to the desired answer because $\Gamma(1)=1$ and $\Gamma(2p+1) = 2p \phantom. \Gamma(2p)$. -Postscript about the contour integral formula: Gary couldn't locate the Erdélyi reference; I found it, but it turns out the book just displays the formula without proof or source. Fortunately Richard Borcherds (in a comment to this answer) recognized it as a result of Barnes [B]; see Lemma 15 on pages 154-155. There's even a Wikipedia page on Barnes' work in this area, including this result which is called the "first Barnes lemma". Barnes' proof uses a hypergeometric identity of Gauss, which is natural in the context of his paper. Here's an alternative approach using Fourier convolutions. For $\mu,\nu$ of positive real part, and real $t>1$, we have -$$ -\int_{-i\infty}^{i\infty} \Gamma(\mu+s) \phantom. \Gamma(\nu-s) \phantom. t^s \phantom. ds -= 2\pi i \phantom. \Gamma(\mu+\nu) \phantom. t^\nu / (1+t)^{\mu+\nu}. -$$ -[This is basically [GR, p.657, 6.422#3], and can be proved by shifting the contour integral to the left: the poles of $\Gamma(\mu+s)$ yield the terms in the Laurent expansion of the right-hand side about $t = \infty$.] Now take $(\mu,\nu) = (\alpha,\gamma)$ and $(\delta,\beta)$ and multiply, finding that -$$ -\int_{-i\infty}^{i\infty} F(s) \phantom. t^s \phantom. ds = (2\pi i)^2 \Gamma(\alpha+\gamma)\phantom.\Gamma(\beta+\delta)\phantom.t^{\beta+\gamma} / (1+t)^{\alpha+\beta+\gamma+\delta} -$$ -where $F(\cdot)$ is the convolution on the imaginary $s$-axis of $\Gamma(\alpha+s) \phantom. \Gamma(\gamma-s)$ with $\Gamma(\delta+s) \phantom. \Gamma(\beta-s)$. -Since this integral uniquely determines $F$, it follows that -$$ -F(s) = 2\pi i \frac{\Gamma(\alpha+\gamma) \phantom. \Gamma(\beta+\delta)} {\Gamma(\alpha+\beta+\gamma+\delta)} \Gamma(\alpha+\delta+s) \phantom. \Gamma(\beta+\gamma-s). -$$ -Taking $s=0$ recovers the Barnes formula. -References -B] Barnes, E.W.: A new development of the theory of the hypergeometric functions. Proc. LMS (1908) s2-6(1): 141–177. -[E] Erdélyi, A., et al.: Table of Integral Transforms II. New York: McGraw Hill, 1954. -[GR] Gradshteyn, I.S., and Ryzhik, I.M.: Table of Integrals, Series, and Products (4th ed.). New York: Academic Press, 1980.<|endoftext|> -TITLE: Definition of enriched caterories or internal homs without using monoidal categories. -QUESTION [8 upvotes]: I know this question may seem nonsensical at first but let me exlain what i have in mind: -In enriched category theory we define categories enriched over a monoidal category $(\mathcal{V},\otimes, I)$. An enriched category then is given by a set/class of objects $\mathcal C$ and a rule assigning to every pair $X,Y$ of such objects a hom-object $[X,Y]$. Furthermore we define composition and identities using $\otimes$ and $I$, remodelling the definitions of usual category theory. -Now for the question: Can we go the other way around? -Let's stick to internal homs for the beginning: Given a category $\mathcal V$ ; can say what additional data turn a functor $$[-,-]:\mathcal{V}^{\mathrm{op}}\times\mathcal V\to \mathcal V$$ into something like an internal hom? -In the case of $[X,-]$ having a left adjoint $-\otimes X$ for every $X$, these additional data should result in $(\mathcal V,\otimes)$ becomming a closed monoidal category with internal hom isomorphic to $[-,-]$. - -REPLY [12 votes]: This is exactly the notion of a closed category. See Eilenberg and Kelly's article in the 1965 La Jolla proceedings (Springer 1966). I think they also describe categories enriched in a closed category.<|endoftext|> -TITLE: Intuitive understanding of the Stieltjes transform -QUESTION [14 upvotes]: I have been using random matrix theory in signal processing and have some trouble understanding what the Stieltjes transform does. -The gist of my work is that I have an $N\times N$ true covariance matrix of the population with $k$ eigenvalues $\lambda_i>1,\ i=\lbrace 1,\ldots,k\rbrace$, corresponding to the signals and the remaining $N-k$ eigenvalues are all $1$, corresponding to the background noise. Assuming Gaussian statistics, I'm deriving the eigenvalue density of the sample covariance matrix of a set of observations from this population, which then is used to detect the signals. -From J. W. Silverstein, “Strong convergence of the empirical dis- -tribution of eigenvalues of large dimensional random ma- -trices”, J. Multivar. Anal. 54, 175–192 (1995), I understand that the Stieltes transform of the density of the sample covariance matrix can be written as -$$m=\sum_{i=1}^{N\to\infty} \frac{1}{\lambda_i (1-c-c z m)-z}$$ -where $c$ is the ratio of number of dimensions $N$ to number of observations $M$ as $N,M\to\infty$. Taking it as a polynomial in $m$ and looking for solutions in the imaginary plane, I can retrieve my density using the inverse transform. -So far, all this is well and good and I have my numerical routines that solve this for any model I input and give me the correct density. However, my question is more fundamental, as to the role that Stieltjes transforms plays here. How does one go from the definition in wikipedia -$$S_\rho(z)=\int_I \frac{\rho(x)}{z-x}\ dx$$ -where $I$ is the interval of support of $\rho(x)$ to the equation in Silverstein's paper. I understand that Stieltes transforms deal with measures of density and if I had in fact taken the density of the population covariance matrix, i.e., $\rho(\lambda)=\sum_{i=1}^N \delta(\lambda-\lambda_i)$, its Stieltjes transform would've looked like -$$S=\frac{1}{N}\sum_{i=1}^N\frac{1}{z-\lambda_i}$$ -So I can see the similarity with the first equation above (barring a sign change that's trivial to take care of), but don't understand how the $(1-c-czm)$ term comes into the picture and how that helps map the density of the eigenvalues of the population covariance matrix to those of the sample. -I've tried reading Terry Tao's blog on the topic, and while that was enlightening, it didn't fix this conceptual gap. It seemed like the original 1965 paper by Marchenko and Pastur dealt with these from a more first principles approach than Silverstein. However, that paper is too dense to tackle, and I'm not sure how using partial differential equations like they do help solve it. Perhaps I'm missing something here because I'm not a mathematician, but I'd appreciate someone helping me fill this gap. - -REPLY [20 votes]: Firstly, the equation you attribute to Silverstein (and is sometimes known as the "self-consistent equation" for the Stieltjes transform) is not exact, but only asymptotically valid in the limit $n \to \infty$. The definition given in Wikipedia is the exact formula. (Your final formula, by the way, is missing a normalisation factor of either $1/n$ or $1/p$, depending on conventions.) -The self-consistent equation arises by viewing the Stieltjes transform as a trace -$$ S = \frac{1}{n} \hbox{tr}( A - zI )^{-1} = \frac{1}{n} \sum_{j=1}^n ((A-zI)^{-1})_{jj}.$$ -It is possible to solve for the $jj^{th}$ component $((A-zI)^{-1})_{jj}$ of the resolvent $(A-zI)$ using the method of Schur complements, in terms of an expression involving the inverse of an $n-1 \times n-1$ minor of $A$, which in turn can be approximated in terms of the eigenvalues of that minor. The eigenvalues of the minor can in turn be estimated by the eigenvalues of the original matrix by means of the Cauchy interlacing law, and the resulting expression involving these eigenvalues turns out to be a function of the original Stieltjes transform, leading to the (asymptotic) formula mentioned in Silverstein's paper (and also in my blog post you mentioned, in the case of Wigner matrices). -The Stieltjes transform can be viewed as a complexification of the spectral measure. Indeed, if one looks at the "jump" in the Stieltjes transform as one passes from the upper half plane to the lower half plane, this jump is (up to some factors of $\pi$) essentially the spectral measure. The real part of the Stieltjes transform is thus essentially the harmonic extension of spectral measure (or, if you like, a smoothed out version of the eigenvalue counting function $N_I$), and the imaginary part is its harmonic conjugate. As such, it neatly packages the spectral information in a way that can be easily manipulated by the methods of complex analysis. -Another way to view the Stieltjes transform is as a sort of moment generating function for the spectral density. This is basically the viewpoint taken in free probability (see my later lecture notes on this topic).<|endoftext|> -TITLE: Why so difficult to prove infinitely many restricted primes? -QUESTION [25 upvotes]: I wondered whether there were an infinite number of -palindromic primes written in binary (11, 101, 111, 10001, 11111, 1001001, 1101011, ...) -and quickly discovered that it is unknown -(OEIS A117697). -Indeed, even though almost all palindromes in any base are composite, -whether there are an infinite number of palindromic primes in any base is unknown -(Wolfram article). -Earlier -(in the MO question, -"Why are this operator’s primes the Sophie Germain primes?"), -I learned that -it is unknown if there are an infinite number of -Sophie Germain Primes. -In addition, is not known if there are an infinite number of -Mersenne primes, -Fibonacci primes (OEIS A005478), -Wilson primes, -Cullen primes, -not to mention prime twins, quadruplets, sextuplets, and -$k$-tuples. -No doubt this list of our ignorance could be extended. -It appears to the naïve (me) that there is no nontrivial restriction on the -primes for which we know there remain an infinite number in the restricted sequence. - -Q1. Is this superficial perception in fact true? -Q2. If so, is there any high-level reason why it is so difficult -to prove these statements? Or is each difficult for its own idiosyncratic -reason? - -I ask this out of curiosity, without expert knowledge of number theory. -Thanks for enlightening me! -Questions Answered. Thanks for the wonderfully rich and informative answers! -Essentially both questions have been answered: My superficial perception (Q1) -is not in fact accurate, as detailed in the examples provided by quid, Anthony Quas, and Joël, -augmented by comments by several. -A high-level reason (Q2) explaining the difficulty in the examples I listed was nicely encapsulated -by Frank Thorne, enriched by appended comments. Thanks! - -REPLY [23 votes]: "It appears to the naïve (me) that there is no nontrivial restriction on the primes for which we know there remain an infinite number in the restricted sequence. -Q1. Is this superficial perception in fact true?" -No, it is definitely not true. Here are a few examples: -(1) Let $a>0, b$ be two relatively prime integers. Are there infinitely many prime of the form $an+b$? Yes. -(2) Let $P(X)$ be a monic polynomial of degree $n$ with coefficients in $\mathbb{Z}$. Are there infinitely many prime $p$ such that $P(x)$ has $n$ distinct roots mod $p$? Yes. -(3) Let $X$ be a projective smooth variety over $\mathbb{Q}$, $\chi$ the Euler-Poincaré characteristic of the manifold $X(\mathbb{C})$, $n$ an integer. Are there infinitely many prime $p$ such that the -the number of points of $X(\mathbb{F}_p)$ is $\chi$ modulo $n$? Yes. Same question with $X(\mathbb{C})$ replaced by $X(\mathbb{R})$? Yes. -(4) Are there infinitely many primes $p$ that can be written $a^2+b^2$? Yes. $a^2+8b^2$ with $b$ odd? Yes... -(5) Let $a,b$ be two integers (such that $4a^3+27b^2 \neq 0$), $a_p$ the number of solutions of $y^2=x^3+ax+b$ modulo $p$, minus $p$. Are there infinitely many primes $p$ such that $a_p=0$? yes. -Are there infinitely many primes $p$ such that $a_p \neq 0$? Yes. Let $\alpha$ and $\beta$ be two reals between $-1$ and $1$, and $\alpha$ smaller than $\beta$; are there infinitely many -primes $p$ such that $\alpha < a_p/2 \sqrt{p} < \beta$? Yes. -One could multiply those examples. They all belong to algebraic number theory, -and a line of thought that has begun with Dirichlet's theorem (example 1), and has developed into the modern theory of algebraic number fields, Galois representations, automorphic forms and the Langlands program. Perhaps the most salient result is Cebotarev's density theorem, of which (1) is a very special case, (2) is a consequence, (3) also a consequence -in combination with Grothendieck's étale cohomology, (4) also a consequence. Only (5) -lies really beyond this result, due respectively to Noam Elkies, Jean-Pierre Serre, -and the long list of people responsible for the proof of Sato-Tate. -Admittedly, there are many natural and interesting -sequences of integers in which we can reasonably conjecture that there are infinitely many primes, and to which this line of thought is not supposed to apply (Mersenne's primes, -to name one).<|endoftext|> -TITLE: theta functions from fock space -QUESTION [5 upvotes]: Is it possible to get theta functions from free fermions? I'm looking for proof of identity -\[ \sum_{n = -\infty}^\infty (-1)^n q^{n^2} = \prod_{j=1}^\infty \frac{1-q^j}{1+q^j} \] -maybe as a matrix element or trace in a Fock space. - -REPLY [2 votes]: Here is another point of view inspired by conformal field theory. If rewritten as -$$\prod_{n=1}^\infty \frac{1+q^n}{(1-q^n)} \sum_{j \in \mathbb{Z}} (-1)^j q^{j^2}=1$$ -this identity (I think) follows as a consequence of a BGG-type resolution of the trivial representation for the N=1 (Ramond) superconformal algebra. -Similarly, famous Euler's pentagonal number theorem -$$\frac{\sum_{j \in \mathbb{Z}} (-1)^j q^{j(3j-1)/2}}{\prod_{n=1}^\infty (1-q^n)}=1$$ -follows from a similar resolution for the Virasoro algebra.<|endoftext|> -TITLE: In what way do the Weil Conjectures pertain to Langlands? -QUESTION [5 upvotes]: For a relative variety $X$ over a ring of integers $O_K$, we can define a zeta function. This zeta function is defined as the product of the zeta functions of the variety specialized to $O_K/\mathfrak{p}$ as $\mathfrak{p}$ runs over $Spec(O_K)$. In turn, zeta functions of varieties over finite fields are easy to define using the counting of rational points. As Grothendieck proved, these zeta functions can be expressed as a product of $L$-functions indexed by $i$, where the $i^{th}$ $L$-function is related to the $i^{th}$ (Weil) cohomology of $X_{O_K/\mathfrak{p}}$. The $i^{th}$ $L$-function of $X$ over $O_K$ is defined to be the product over $\mathfrak{p}$ of the $i^{th}$ $L$-function of $X_{O_K/\mathfrak{p}}$. -The Weil conjectures give us a lot of information about the zeta functions of varieties over finite fields, and in fact about their $L$-functions. -The Langlands program is about properties of $L$-functions of $X$ over $O_K$. -Is it possible to interpret the Weil conjectures as telling us something meaningful about the Langlands program? - -REPLY [7 votes]: If $\pi$ is a regular algebraic cuspidal automorphic representation of $GL_n/K$ with K totally real or CM, and $\pi$ satisfies a certain self-duality condition, then $\pi_v$ is tempered for all finite $v$. This monumental theorem is a vast generalization of Deligne's proof of the Ramanujan conjecture, and the proof ultimately appeals to the Weil conjectures, by proving an identity $L(s,\pi)=L(s,M(\pi))$ where $M(\pi)$ is (essentially) a submotive of the cohomology of a Shimura variety.<|endoftext|> -TITLE: Completion of local rings in the exceptional divisor of a blow-up -QUESTION [5 upvotes]: Let $X=\mathrm{Spec}(A)$ be an affine variety, $Z\subseteq X$ a closed, reduced subscheme. Let -$$\beta:Y=\mathrm{Bl}_Z(X)\to X$$ -be the blow-up of $X$ in $Z$. In other words, $Y=\mathrm{Proj}(A[IT])$ for $I:=I(Z)$. Let $E:=\beta^{-1}(Z)$ be the exceptional divisor. For a point $Q\in E$, I am now wondering how the completion $\hat{\mathcal{O}}_{Y,Q}$ looks like. I would like to understand its relation to $\hat{\mathcal{O}}_{X,\beta(Q)}$, in particular. -Edit: Some remarks and my thoughts on the matter. -If the base field $k$ is algebraically closed and both $X$ and $Y$ are nonsingular, then -$$\hat{\mathcal{O}}_{X,\beta(Q)}\cong k[[x_1,\ldots,x_d]]\cong\hat{\mathcal{O}}_{Y,Q}.$$ -It becomes tricky when they are (possibly) nonsingular. I thought it would be nice to know something about the local rings. Since $X$ and $Y$ share the same function field $K$ and $\beta$ is dominant, -$$\mathcal{O}_{X,\beta(Q)}\hookrightarrow\mathcal{O}_{Y,Q}\hookrightarrow K.$$ -We have adjoined the fractions $a/b$ for $a,b\in I_{\beta(Q)}$ and $bT\notin Q$. -Clearly, $\mathfrak{m}_{\beta(Q)}\cdot\mathcal{O}_{Y,Q}\ne\mathfrak{m}_Q$ which denies us access to the exactness property of the completion. We can write $\mathfrak{m}_{\beta(Q)}=(x_1,\ldots,x_d)$ with $x_i\in I_{\beta(Q)}$ iff $i\le r$. Then, I feel $\mathfrak{m}_Q$ should be equal to $$(z_1,\ldots,z_r,x_{r+1},\ldots,x_d)$$ with $z_i=x_i/b$ for some appropriate $b$, at least under certain conditions. However, I don't know if that is correct and if it even helps. - -REPLY [2 votes]: We can write $A = K[X_1, \ldots, X_n]/\mathfrak a$. -Suppose that $I = (a_1, \ldots, a_m)$; then -$A[It] \cong A[T_1, \ldots, T_m]/\mathfrak b$ for some ideal -$\mathfrak b$. A description for $\mathfrak b$ is given in Section 1.1 of -W. Vasconcelos, \textit{Integral Closure, Rees algebras, multiplicities and -Algorithms}, (Springer). -Let us suppose, for simplicity, that $P := \beta(Q)$ is defined by $(X_1, -\ldots, X_n)$ and that $Q$ is defined by $(X_1, \ldots, X_n, T_1, \ldots, -T_m)$. Let us write $R$ and $S$ for the local rings at $P$ and $Q$, respectively. - Since we are interested in completions, we may consider the affine -open subset of $Y$ defined by $T_1 \neq 0$. This can be thought of as the -spectrum of $A[\frac{a_2}{a_1}, \ldots, \frac{a_m}{a_1}] \cong -A[Y_2, \ldots, Y_m]/\mathfrak c$ for some ideal $\mathfrak c$. -Then $\mathfrak c$ -can be obtained from $\mathfrak b$ by `dehomogenizing' with respect to -$T_1$; see, \textit{e.g.}, Section 5.5 of \textit{Integral Closure of -Ideals, Rings, and Modules} by I. Swanson and C. Huneke, (LMS Lecture Note -Series 336). It contains the relations $a_1Y_i = a_i$ for all $1 \leq i -\leq m$, but, in general, could have more. -Hence -$\widehat R \cong K[[X_1, \ldots, X_n]]/\mathfrak a$ -and -$\widehat S \cong\widehat R[[Y_1, \ldots, Y_m]]/\mathfrak c$. -(By abuse of notation, we write $\mathfrak a$ for an ideal of $K[X_1, -\ldots, X_n]$ and the ideal of $K[[X_1, \ldots, X_n]]$ generated by it.) -Therefore, in the non-singular situation, with $A = -K[X_1, \ldots, X_n]$ and $I = (X_1, \ldots, X_n)$, we have -$\widehat R \simeq K[[X_1, \ldots, X_n]]$ and -$\widehat S \simeq K[[X_1, Y_2, \ldots, Y_n]]$. -(We use the fact that since $X_1, \ldots, X_n$ is a regular sequence in -$A$, the ideal $\mathfrak b$ is generated by their `Koszul syzygies', i.e., -by $X_iT_j - X_jT_i, 1 \leq i < j \leq n$. Hence $\mathfrak c$ is -generated by $X_1Y_j - X_j, 2 \leq j \leq n$.) -They are (abstractly) isomorphic as formal power series rings, but the -structure -morphism is not an isomorphism.<|endoftext|> -TITLE: Stable normal bundle of a manifold -QUESTION [12 upvotes]: Hi, -in bordism-theory and many bordering areas one has the following construction: Given a manifold M (say closed for the purposes of this discussion and k-dimensional), we embed it into some $\mathbb R^n$ for n large and look at the normal bundle of that embedding. This pulls back to give an n-k-dim vectorbundle over M, and we consider the homotopy class $M \rightarrow BGL(n-k) \rightarrow BGL$, where the first map is the classifying map for that bundle and the second one is induced by the obvious inclusion. -One now finds that the homotopy class of this composition does not depend on the particular embedding chosen. Since $BGL$ classifies principal-$GL$-bundles we have thus constructed an isomorphism class of such bundles and from what I gather this is what is called the stable normal bundle. -Now my question is: -Is there a sufficiently nice construction of an actual $GL$-bundle representing this isomorphism class? -There certainly seems to be none for the individual normal bundles (for they of course DO depend on the embedding for small n), but for the infinite one there just might be, right? By 'construction' I mean construction out of intrinsic data of the manifold and not one along the lines of 'embed M into $\mathbb R^\infty$ and look at the frames of the arising normal bundle'. -If a construction can be found at all then there are probably many, so there won't be a canonical one, which is why I don't really want to specify what 'nice' is supposed to mean. -Thank you for any answers - -REPLY [3 votes]: Perhaps I'm confused about what you are looking for, but haven't you already constructed an "actual" $GL$-bundle in your question? -What I mean is the following. The usual definition of $GL$ is a direct limit of $GL(m)$'s. So an element of $GL$ is just an element of $GL(m)$ for some $m$. Similarly, if $M$ is compact then a $GL$ bundle over $M$ is just a $GL(m)$ bundle over $M$, for some $m$. As you note in your question, one can construct such bundles by embedding $M$ in $\mathbb{R}^{k+m}$, taking the frame bundle of the normal bundle, and then interpreting this as a $GL$-bundle rather than a $GL(m)$-bundle. Any two such embeddings of $M$ give isomorphic $GL$-bundles. - -In response to pudin's comment below, here's a second construction. Embed $M$ into $\mathbb{R}^\infty$. Define a bundle $F$ over $M$ whose fiber at $x$ is frames of the normal bundle of the embedding at $x$ which eventually coincide with the standard framing of $\mathbb{R}^\infty$. (This is possible because the image of the embedding will lie in some $\mathbb{R}^n \subset \mathbb{R}^\infty$ if $M$ is compact.) $F$ is a principal $GL$ bundle, where in this case we take $GL$ to be invertible linear maps $\mathbb{R}^\infty \to \mathbb{R}^\infty$ which differ from the identity only on a finite subspace.<|endoftext|> -TITLE: Sum of trigonometric functions -QUESTION [16 upvotes]: Do somebody know the closed form of the following sum (m is an integer) -$$f(\beta)=\sum _{k=1}^{2m+1} \sin^{2 m+1}\left[\frac{-\beta+k \pi -}{1+2 m}\right]$$ -If instead of $n=1+2m$ we put $n=2m$, then the above sum do not depends on $\beta$ and is equal to $\frac{2 \Gamma(1/2+m)}{\sqrt{\pi} \Gamma(m)}$. -I need the maximum of $f(\beta)$, it seems that it is achieved in $0$ or in $\pi/2$ but I don't have the proof. -Indeed I need the maximum $M_{k,q}$ of this integral (at least for q=1) -$$F_k(\beta)=\int_0^{2\pi -}\left(1- -\cos\left(\frac{2\beta}{k+1}+s\right)\right)^{\frac{(k+1)q-2}{2}}\left| -\cos\frac{(1+k) (s-\pi)}{2}\right|^q ds.$$ -It is related to the sharp inequality $$|f^{(k)}(z)|\le (1-|z|^2)^{1-q-kq}M_{k,q}\|\Re f\|_{p}$$ where $f$ is an analytic function in the unit disk and the corresponding norm is the Hardy norm. For $n=1+k$ an even integer I have the proof... - -REPLY [6 votes]: Some further manipulation past my answer of a week ago yields a formula that should reduce the proof of the observed behavior to routine (if not entirely pleasant) estimates. Whereas $f$ is constant at $2/B(\frac12,\frac{n}2)$ for even $n$, in the present case of odd $n$ the maximum exceeds $2/B(\frac12,\frac{n}2)$ by a tiny amount that is very nearly -$$ -\frac{4}{\pi} \phantom. \frac1{n+2} \frac2{n+4} \frac3{n+6} \cdots \frac{n}{3n} = \frac4\pi n! \frac{n!!}{(3n)!!} -= (27+o(1))^{-n/2} -$$ -for large $n$. Here and later we use "$u!!$" only for positive odd $u$ to mean the product of all odd integers in $[1,u]$; that is, $u!! := u!/(2^v v!)$ where $u=2v+1$. -Recall the previous notations: we take $n=2m+1$ and -$$ -g(x) = f(x+\frac\pi2) = g(-x) = -g(x+n\pi), -$$ -which has a finite Fourier expansion in cosines of odd multiples of $X := x/n$, namely -$$ -f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos \phantom. tX}{\sin \frac{\pi t}{2n}} -$$ -where $t = n-2j$. Even before we use this expansion, we deduce from the original formula -$$ -f(\beta)=\sum_{k=1}^n \sin^n\frac{-\beta+k \pi}{n} -$$ -that $f(\beta)-f(\beta+\pi) = 2\phantom.\sin^n (\beta/n)$, from which it follows that $g(x)$ is maximized somewhere in $|x| \leq \pi/2$, but that changing the optimal $x$ by a small integral multiple of $\pi$ reduces $g$ by a tiny amount; this explains the near-maxima I observed at $x=\pm\pi$ for $2|m$, and indeed the further oscillations for both odd and even $m$ that I later noticed as $n$ grows further. -This also suggests that in and near the interval $|x| \leq \pi/2$ our function $g$ should be very nearly approximated for large $n$ by an even periodic function $\tilde g(x)$ of period $\pi$. We next outline the derivation of such an approximation, with $\tilde g$ having an explicit cosine-Fourier expansion -$$ -\tilde g(x) = g_0 + g_1 \cos 2x + g_2 \cos 4x + g_3 \cos 6x + \cdots -$$ -where $g_0 = 2/B(\frac12,\frac{n}2)$ and, for $l>0$, -$$ -g_l = (-1)^{m+l-1} \frac4\pi \frac{n!}{2l+1} \frac{((2l-1)n)!!}{((2l+1)n)!!} -$$ -with the double-factorial notation defined as above. Thus -$$ -\tilde g(x) = g_0 + (-1)^m \frac{4n!}\pi \left(\frac{n!!}{(3n)!!} \cos 2x - \frac13 \frac{(3n)!!}{(5n)!!} \cos 4x + \frac15 \frac{(5n)!!}{(7n)!!} \cos 6x - + \cdots \right). -$$ -For large $n$, this is maximized at $x=0$ or $x=\pm\pi/2$ according as $m$ is even or odd. Since we already know by symmetry arguments that $g'(0) = g'(\pm \pi/2) = 0$, this point or points will also be where $g$ is maximized, once it is checked that $g - \tilde g$ and its first two derivatives are even tinier there. -The key to all this is the partial-fraction expansion of the factor $1 / \sin (\pi t /2n)$ in the Fourier series of $g$, obtained by substituting $\theta = \pi t / 2n$ into -$$ -\frac1{\sin \pi\theta} = \frac1\pi \sum_{l=-\infty}^\infty \frac{(-1)^l}{\theta-l} -$$ -with the conditionally convergent sum interpreted as a principal value or Cesàro limit etc. I already noted in the previous note that the main term, for $l=0$, yields the convolution of $\cos^n (x/n)$ with a symmetrical square wave, which is thus maximized at $x=0$ and almost constant near $x=0$; we identify the constant with $2/B(\frac12,\frac{n}2)$ using the known product formula for $\int_{-\pi/2}^{\pi/2} \cos^n X \phantom. dX$. The new observation is that each of the error terms $(-1)^l/(\theta-l)$ likewise yields the convolution with a square wave of $(-1)^l \cos(2lx) \phantom. \cos^n(x/n)$. If we approximate this square wave with a constant, we get the formula for $g_l$ displayed above, via the formula for the $n$-th finite difference of a function $1/(j_0-j)$. The error in this approximation is still tiny (albeit not necessarily negative) because $\cos^n (x/n)$ is minuscule when $x$ is within $\pi/2$ of the square wave's jump at $\pm \pi n / 2$. -I've checked these approximations numerically to high precision (modern computers and gp make this easy) for $n$ as large as $100$ or so, in both of the odd congruence classes mod $4$, and it all works as expected; for example, when $n=99$ we have $f(0) - g_0 = 2.57990478176660\ldots \cdot 10^{-70}$, which almost exactly matches the main term $g_1 = (4/\pi) \phantom. 99! \phantom. 99!!/297!!$ but exceeds it by $5.9110495\ldots \cdot 10^{-102}$, which is almost exactly $g_2 = (4/\pi) \phantom. 99! \phantom. 297!!/(3 \cdot 495!!)$ but too large by $7.92129\ldots \cdot 10^{-120}$, which is almost exactly $g_3 = (4/\pi) \phantom. 99! \phantom. 495!!/(5 \cdot 693!!)$, etc.; and likewise for $n=101$ except that the maximum occurs at $\beta = \pi/2$ and is approximated by an alternating sum $g_1 - g_2 + g_3 \ldots$ (actually here this approximation is exact because $x=0$).<|endoftext|> -TITLE: Parameters in arithmetic induction axiom schemas -QUESTION [9 upvotes]: The induction schema of Peano Arithmetic is standardly given as the universal closure of $\phi(0)\land \forall x (\phi(x)\rightarrow \phi(x+1)) \rightarrow \forall x\phi(x)$. However, since the language of arithmetic has a name for every standard number, it is not obvious (to a beginner like me) why parameters are necessary in the induction schema; why not restrict to the case where $x$ is the only free variable in $\phi$? - -Does having parameters in the induction schema really make the system stronger, and, if so, how is that proven? -Are there natural theorems that can only or most easily be proven using the stronger system? -Is the weaker system of any interest? - -REPLY [13 votes]: $\def\fii{\varphi}$While the other answers resolve the question, a simple way of deriving induction for a formula $\fii(x,\vec p)$ with parameters $\vec p$ is to use parameter-free induction on the formula -$$\psi(x)=\forall\vec p\,[\fii(0,\vec p)\land\forall y\,(\fii(y,\vec p)\to\fii(y+1,\vec p))\to\fii(x,\vec p)].$$ -In fact, this derives the induction schema with parameters from the (parameter-free) induction rule, since $\psi(0)$ and $\psi(x)\to\psi(x+1)$ are provable in Q without any assumptions on $\fii$.<|endoftext|> -TITLE: Presentations of PSL(2, Z/p^n) -QUESTION [9 upvotes]: As is well known, the group $PSL(2,\mathbb Z)$ is isomorphic to the free product $C_2 \ast C_3$ of cyclic groups of order $2$ and $3$. Call the generators of the cyclic groups $S$ and $T$. - -Problem: Given a prime number $p$ and a natural number $n$, write a presentation of the quotient $PSL(2, \mathbb Z/p^n\mathbb Z)$ with the images of $S$ and $T$ as generators. - -REPLY [4 votes]: The group $PSL_2(\mathbb{Z}/p^n)$ is the automorphisms group of the $(p+1)$ regular tree of depth $n$, where at level $m$ of the tree you have the points of $\mathbb{P}(\mathbb{Z}/p^m)$. The main benefit of this view, is that you can understand the relations at each level, and then move inductively to the next one.<|endoftext|> -TITLE: Pathologies of analytic (non-algebraic) varieties. -QUESTION [9 upvotes]: Note: By an "analytic non-algebraic" surface below I mean a two dimensional compact analytic variety $X$ (over $\mathbb{C}$) which is not an algebraic variety. -A property of Nagata's example (see the end of the post for the construction) of a non-algebraic normal analytic surface $X$ is the following: -($\star$) $\quad$ There is a point $P$ on $X$ such that every (compact) algebraic curve $C$ on $X$ passes through $P$. -In a paper I am writing I also constructed (to my surprise) some examples of non-algebraic normal analytic surfaces which have this peculiar property. - Questions: Is this sort of behaviour "normal" for such surfaces? Or, more precisely, if an analytic surface does not satisfy ($\star$), is it necessarily algebraic? How about for higher dimensions? - Nagata's Construction (following Bădescu's book on surfaces): Start with a smooth plane cubic $C$ and a point $P$ on $\mathbb{P}^2$ such that $P - O$ is not a torsion point (where $O$ is any of the inflection points of $C$) on $C$. Let $X_1$ be the blow up of $\mathbb{P}^2$ at $P$, and for each $i \geq 1$, let $X_{i+1}$ be the blow up of $X_i$ at the point of intersection of the strict transform of $C$ and the exceptional divisor on $X_i$. Each blow up decreases the self-intersection number of the strict transform $C_i$ of $C$ by $1$, so that on $X_{10}$ the self-intersection number of $C_{10}$ is $-1$. $X$ is the blow down of $X_{10}$ along $C_{10}$. By some theorems of Grauer and Artin, $X$ is a normal analytic surface. - -REPLY [13 votes]: The answer is no. -A counterexample is provided by the so-called Hopf surfaces (they were actually constructed by Kodaira, see Donu Arapura's comment). -A Hopf surface of type $\alpha=(\alpha_1, \, \alpha_2)$, where $0 < |\alpha_1| \leq |\alpha_2| < 1$, is the compact complex surface $H_{\alpha}$ obtained as the quotient of $\mathbb{C}^2 \setminus (0,0)$ by the infinite cyclic group generated by the automorphism $$ (z_1, \, z_2) \to (\alpha_1 z_1, \, \alpha_2 z_2).$$ -One can prove that $H_{\alpha}$ is a compact complex surface diffeomorphic to $S^1 \times S^3$, so it admits no Kähler metric. In particular, it is not algebraic. -However, $H_{\alpha}$ does not satisfy your property $(\star)$. In fact, there is the following result: - -The Hopf surface $H_{\alpha}$ is an elliptic fibre space over $\mathbb{P}^1$ if and only if $\alpha_1^l=\alpha_2^k$ for some $l, \, k \in \mathbb{Z}$. Otherwise $H_{\alpha}$ contains exactly two compact curves, which are disjoint (they are the images of the $z_1$-axis and the $z_2$-axis). - -So $H_{\alpha}$ contains either infinitely many or exactly two disjoint elliptic curves. -For more details, see [Barth-Peters-Van de Ven, Compact Complex Surfaces, Chapter V].<|endoftext|> -TITLE: How to calculate [10^10^10^10^10^-10^10]? -QUESTION [38 upvotes]: How to find an integer part of $10^{10^{10^{10^{10^{-10^{10}}}}}}$? It looks like it is slightly above $10^{10^{10}}$. - -REPLY [81 votes]: I think the number in question is $10^{10^{10}}+10^{11}\ln^4(10)$ plus a tiny positive number. That is, it starts with a digit $1$, followed by $10^{10}-13$ zeros, then by the string $2811012357389$, then a decimal point, and then some garbage (which starts like $4407116278\dots$). -To see this let $x:=10^{-10^{10}}$, a tiny positive number, and put $c:=\ln(10)$, an important constant. We have -$$10^x=1+cx+O(x^2)$$ -$$10^{10^x}=10^{1+cx+O(x^2)}=10+10c^2x+O(x^2)$$ -$$10^{10^{10^x}}=10^{10+10c^2x+O(x^2)}=10^{10}+10^{11}c^3x+O(x^2)$$ -$$10^{10^{10^{10^x}}}=10^{10^{10}+10^{11}c^3x+O(x^2)}=10^{10^{10}}+10^{10^{10}}10^{11}c^4x+O(x^2),$$ -where $O(x^2)$ means something tiny all the way. -In the last expression we have $10^{10^{10}}10^{11}c^4x=10^{11}c^4$, which justifies my claim.<|endoftext|> -TITLE: What space does "Loop infinity of the infinite dimensional sphere" refer to? -QUESTION [6 upvotes]: I've been reading Hatcher's survey "A Short Exposition of the Madsen-Weiss Theorem". In it, he discusses the Barratt-Priddy-Quillen theorem, which says that the homology of the infinite symmetric group is the same as the homology of one component of $\Omega^{\infty} S^{\infty}$. My question : what space exactly is being referred to in this notation? It would appear that he intends to take the limit of $\Omega^n S^n$ as $n$ goes to infinity; however, I don't see a natural inclusion $\Omega^n S^n \subset \Omega^{n+1} S^{n+1}$, so I don't know how to make sense of this. Another possible interpretation would be to take the limit of $\Omega^n S^{\infty}$ as $n$ goes to infinity, but again I don't know a natural inclusion $\Omega^n S^{\infty} \subset \Omega^{n+1} S^{\infty}$. - -REPLY [15 votes]: Your first statement is correct: it is taking a colimit of $\Omega^n S^n$. The inclusion from one to the next is called the "suspension" map. The points of $\Omega^n S^n$ are basepoint-preserving functions $S^n \to S^n$, and the suspension map takes such a function $f$ and sends it to $f \wedge id: S^n \wedge S^1 \to S^n \wedge S^1$. -This is easier to describe in terms of another model. $S^n$ is homeomorphic to the space obtained by taking $[0,1]^n$ and identifying the boundary to a single point. Under this identification, you could describe a function $S^n \to S^n$ using coordinates as a function -$$(x_1,\ldots,x_n) \mapsto f(x_1,\ldots,x_n),$$ -and the suspension map sends this to the function -$$(x_1,\ldots,x_n,x_{n+1}) \mapsto (f(x_1,\ldots,x_n), x_{n+1})$$ -Of course, you have to check that this preserves the equivalence relation if $f$ does and so on. -Most models of $S^\infty$ are contractible, and hence so are $\Omega^n S^\infty$; these are definitely not what you want. The notation is not to be taken literally; at best, $S^\infty$ should denote an "infinite suspension" operator which doesn't actually produce spaces as output.<|endoftext|> -TITLE: What is due diligence when translating a paper? -QUESTION [21 upvotes]: EDIT: Part of the community has decided on a less catchy and more representative title -than "Socially acceptable plagiarism (with regards to translation)". Let's run with that for a while. GRP.10.27 END EDIT. -The catchy part of the title refers to reusing words or ideas without permission, but without -'rocking the boat'. In mathematics, this is often done by proper attribution of the -source, by naming the author/speaker, or providing other clues that the phrasing one has -just used is not original. -If this were a discussion forum, I would invite answers that lists various forms of socially -acceptable plagiarism; instead I prefer brief mention of such to the comments. This post is about the propriety and etiquette involving translations. -My situation is that I want to devote time to producing an English translation of some papers in German, and make the results available. Assume (although I may end up doing something different) that I place my efforts in a PDF and put it on some web page for public Internet access. - -Primary question: What trouble do I get into by doing that? - -Answers to this may depend on what things I did wrong, especially if I use certain words, -phrases, images, parts of images, without crediting the source or asking for permission. So let's add to the context that the papers are from before 1980, the authors are likely unresponsive, and the source journals are likely discontinued, although online -access to the sources are not freely available. I will list the original sources in the PDF, but the following question arises: - -What trouble may happen if I do not contact some representative and ask for permission to use old material? - -For material published in 1980 and afterwards, there is less of an excuse not to seek such permission or rights, and I would accept anecdotes that might lend guidance and are not obvious applications of common sense, but I am interested in the amount of effort someone in academics expends in order to reuse published material, especially in translated form. -I know of a few examples in book publishing where more effort is made in getting such -permission, however that appears more expensive than I feel the current scenario merits. This leads to: - -What is "due diligence" in producing such a translation? - Does it matter if the translation is provided gratis or for a fee? - -I don't expect to sell access to the translation; it is my intent to make it freely available to any individual researcher. If someone else wants to put it in a book and sell that -book, however, perhaps I could grant them such rights in exchange for a small monetary (or caffeine-ary) consideration. -Finally, let's assume I provide my own translation, except that for some small -sections (possibly in a different language) I use someone else's translation of the same or related source. Let's say that I am concerned especially about a fragment that -is (roughly) three paragraphs or about 200 words long. The answer to the following questions may be length dependent: if so, consider that I also have a 20 word fragment that is -of concern. - -How do I attribute this fragment? Do I use a footnote, or mention it in a preface? What is due diligence for making sure I can use this fragment? - -This question is barely suitable for MathOverflow; I ask it here because the papers and output are mathematical, and the conventions in mathematics and mathematical publishing may not -be addressed were I to ask these questions elsewhere. If they are addressed elsewhere, please provide a pointer to such material. -Also of interest, although I do not need the answers here, is if I have the same situation as above, except that I provide an interpretation (which is laced with my own perspective) rather than a translation (which attempts semantic fidelity and objectivity with respect to the -source paper). The answers may stay the same, but it feels like a different situation to me. -(A similar situation is mentioned in Mathematical etiquette: Rephrasing / restructuring a work, limited release (with attribution) acceptable? , which has some useful advice, although it does not involve natural language translation.) -Gerhard "Yes, It's About Jacobsthal's Function" Paseman, 2011.10.26 - -REPLY [2 votes]: I want to acknowledge the contributions others have made; I choose the answer format to help with emphasis. -I should have mentioned copyright issues in the question, as that is the major consideration when asking about trouble; thanks to -Michael Greinecker for pointing this out and to Henry Cohn for -expanding upon this. (This teaches me to preview questions on -meta.mathoverflow where being, shall we say, less thoughtful carries less stigma or less embarrassment.) Since this handles the primary -question, I will accept one of their answers if a) they submit an -answer with that content and b) no one else gives a more thorough answer regarding due diligence. -Much as I liked my original title, I thank darij grinberg for changing it -to reflect a key issue of the series of questions. (I do have mixed -feelings, grumble, grumble. At least I read the bit in the FAQ about -collaborative editing before posting.) Although I am still interested in hearing of other examples of socially acceptable plagiarism, I agree with darij that this question is more about due diligence with respect to translations. -I am glad for Gerald Edgar's contribution; I am hoping to see more -anecdotes like those, and I should have said so more loudly. I will -wait at least three days and then accept his answer if I do not see one I like even better. -I also thank David Speyer and KConrad for their remarks. I have not yet decided, but their comments sway me towards writing an expository article which contains my intepretation of the papers of interest, and adds some original material. That would force me to do more summaries and cutting, but that may be a good thing. -Based on the number of views and votes, and the fact that the question is still open, I thank the community for tolerating this kind of question. I hope more good answers will appear so that this can serve future readers. -Gerhard "Can't Wait For Oscar Night" Paseman, 2011.10.28<|endoftext|> -TITLE: surjective morphism of schemes or epimorphism of sheaves? -QUESTION [9 upvotes]: I have a technical question coming from reading Toen's master course on stacks. -If we view schemes as locally ringed spaces then there we could define a morphism to be surjective if it the underlying morphism of topological spaces is surjective. -I believe this is the same as saying that for a morphism $f: X \to Y$ and every $k$-point $Spec k \to Y$ (with k a field) there exists a field extension $Spec l \to Spec k$ and an $l$-point of $X$ $Spec l \to X$ such that $Spec l \to X \to Y = Spec l \to Spec k \to Y$. (I believe one can equivalently work with geometric points, i.e. only with algebraically closed fields) -So let's take the latter as a definition of surjective, as it is more natural if we work functorially (and also because it makes sense for arbitrary presheaves on Aff, the category of affine schemes) -Now, Toen uses the etale topology on affine schemes and defines schemes (and more general beasts) as particular sheaves on the etale site. His notion of surjectivity is being an epimorphism of sheaves. -My question is: do these notions coincide? What happens if we take a different topology (like fpqc)? -To be honest, the case I care about the most is when defining atlases, i.e. for $\coprod U_i \to X$, with $U_i$ affine schemes and $U_i \to \coprod_i U_i \to X$ an open immersion (or etale, or smooth, or more generally flat I guess). Nevertheless I would like to understand how much epis of sheaves capture the notion of being surjective for arbitrary morphisms. -EDIT: The most concrete description I know for an epi of sheaves $X \to Y$ is that given any affine $T \to Y$ there exists a covering (etale, or of the chosen topology) $T_i \to T$ such that there exist maps $T_i \to X$ such that $T_i \to T \to Y = T_i \to X \to Y$. - -REPLY [3 votes]: I think (what currently is) Lemma 5.9 (tag 05VM) of the Algebraic Spaces chapter of the stacks project is exactly what we want. -Let F,G fppf sheaves and let $F \to G$ be a schematic, flat, locally of finite presentation and surjective (as in: surjective on fields). Then it is an epimorphism of sheaves.<|endoftext|> -TITLE: A hypercube-related graph -QUESTION [8 upvotes]: For integer $n\ge 3$, consider the graph on the set of all even vertices of the $n$-dimensional hypercube $\{0,1\}^n$ in which two vertices are adjacent whenever they differ in exactly two coordinates. This is an $(n(n-1)/2)$-regular graph on $2^{n-1}$ vertices. Is there any standard name / notation for this graph? Is there a way to construct it from some "basic" graphs using standard graph operations (like products of graphs)? Has anybody ever studied the isoperimetric problem for this graph? -Thanks! - -REPLY [8 votes]: Conway & Sloane's "Sphere Packings, Lattices and Groups" references Coxeter's "Regular Polytopes" for the phrase "halfcube", but Coxeter only uses the notation $h\Pi_n$, saying $h$ can be taken to stand for half- or hemi-, for an arbitrary polytope $\Pi_n$ {$p, q, \ldots, w$} with even $p$ (in your case, {$4,3,3,\ldots, 3$}) This construction is section 8.6 in Coxeter. Since then, halfcube seems to have lost favour, and hemi-cube has become the name for a construction of quotienting out vertices, while the term demicube (or demihypercube if you want to be explicit about using hypercubes and not cubes) is reserved for the construction of deleting vertices of a hypercube. See Conway, Burgiel and Goodman-Strass's "Symmetries of Things." Chapter 26 covers this, where they call them hemicubes, and draw some lovely pictures. -Specific dimensional cases have different names. Your $n=3$ case is the complete $K_4$. $n=4$ is the 16-cell, also called a hexadecachoron in older books, and happens to be a cross-polytope (this does not continue in higher dimensions). By $n=5$, the polytopes begin to take shape as their own specific family and no longer have multiple names. See http://en.wikipedia.org/wiki/Demihypercube, and various dimension specific pages there. -I do not know anything about the isoperimetric problem for these graphs, but there has likely been work done on the $n \leq 4$ cases, since those graphs also show up as other constructions.<|endoftext|> -TITLE: $F_4$ flag variety -QUESTION [5 upvotes]: As flag variety or a homogeneous variety is a quotient $\Sigma=G/P$ of a reductive Lie group $G$ by one of its parabolic subgroups $P$. The subgroup $P$ fixes a flag of subspaces of standard representation $V$ of $G$. There is an embedding of projective varieties $\Sigma\subset \mathbb P V_{\lambda}$, where $V_{\lambda}$ is some highest weight representation of $G$. -For the exceptional Lie group of type $G_2$, if we consider its highest weight representation for highest weight $\omega_2$ then we have an embedding of a homogeneous variety $\Sigma\subset \mathbb P V_{\omega_2}$. Since its a subvariety of Gr(2,7), which can easily be seen to be a "flag variety", so we can some how realise this $G_2$ variety as a flag variety. -If we consider an exceptional Lie group $F_4$ and take its highest weight representation with highest weight $\omega_1$ then $V_{\omega_1}$ is 26 dimensional and we have an embedding of $F_4$ homogeneous variety $\Sigma ^{15}\subset \mathbb P V_{\omega_1}$, which is a codimension 10 embedding. My question is that how can we realise this variety as a "flag variety" or is it also a subvariety of some other standard flag variety? - -REPLY [2 votes]: As far as I remember $F_4/P_{\omega_1}$ is a hyperplane section of $E_6/P_{\omega_1}$.<|endoftext|> -TITLE: Is the polynomial de Rham functor a Quillen equivalence? -QUESTION [9 upvotes]: It is known that the rational homotopy theory of spaces (e.g. simplicial sets) is equivalent in some sense to the homotopy theory of cdgas over $\mathbb{Q}$. This has been expressed in various forms in the literature. For instance, Felix-Halperin-Thomas show that -homotopy classes of maps between simply connected rational spaces with finite-dimensional homology in each dimension can be computed via homotopy classes of cdga maps between Sullivan models for each of them. -However, there is a stronger statement that I would like to be true, have heard asserted (without proof) that it is true, but haven't been able to figure out: I would like for that statement to be true without finite-dimensional hypothesis. -Consider the following two model categories: -First, we can take simplicial sets with the usual (Kan) model structure, and then left Bousfield localize at the class of "rational homology equivalences." In other words, the model structure is such that the cofibrations are the injections, weak equivalences are maps inducing isomorphisms on $H_*(\cdot, \mathbb{Q})$, and everything else is determined. That this model structure exists follows from a combination of the small object argument and the Bousfield-Smith cardinality argument (or one can probably appeal to general facts on existence of Bousfield localizations). -Second, we can take commutative dgas (nonnegatively graded) over $\mathbb{Q}$. The model structure is obtained by transfer from a slight variant of the model structure on nonnegatively graded chain complexes. In other words, a fibration of cdgas is a surjection, and a weak equivalence is a quasi-isomorphism. The cofibrations are thus determined; there is a standard generating set (basically, what one gets by applying the free functor to generating sets for chain complexes). -Edit: As Tyler Lawson observes below, this is not actually a model structure. I am not sure at the moment what the right one is. Perhaps we should relax "surjections" to "surjections in degrees $\geq 1$." Alternatively, we could consider the model category of all cdgas (in which case the functor below is obviously not anywhere near a Quillen equivalence). -Now, we have a Quillen adjunction between the first and the opposite of the second, which sends a simplicial set $X_\bullet$ to the "polynomial de Rham algebra" $A^{PL}(X_\bullet)$, a cdga which is quasi-isomorphic to the rational cochain algebra (and which, in particular, solves the commutative cochain problem). -Q1: Is this a Quillen equivalence? (As Tyler Lawson points out, there is a simple reason why this is not the case: namely, that elements in $H^0$ could be nilpotent. Is this, however, the "only" reason why it fails? For instance, what if one restricts to cdgas that are "connected," i.e. have only $\mathbb{Q}$ in degree zero?) -I have not seen this statement in what I've read (not that much). In fact, most authors seem to prefer to work only with simply connected spaces; the advantage is that there you can localize at the rational homotopy equivalences (which are the same as the rational homology equivalences by the mod $\mathcal{C}$ theory). Granted, simply connected spaces do not form a model category. Quillen dealt with it by working with 2-reduced simplicial sets (those with only one vertex and edge). Other authors (e.g. Felix-Halperin-Thomas) don't use the framework of model categories at all, but state something essentially equivalent to this in the case when one works with simply connected spaces with finite-dimensional homology groups. -Here's what I understand of the argument, and why I don't know how to extend it. The basic claim, as before, is that if $X_\bullet $ and $Y_\bullet$ are simplicial sets, with $Y_\bullet$ an abelian space and a rational Kan complex (if I'm not mistaken, these are fibrant in the model structure thus constructed), and $A, B$ are cofibrant (e.g. Sullivan) models in cdgas for $X_\bullet, Y_\bullet$, then homotopy classes of maps $[X_\bullet, Y_\bullet]$ are the same as homotopy classes of maps $ B \to A$. By using a Postnikov tower to express $Y_\bullet$ as a homotopy inverse limit under a whole bunch of fibrations with Eilenberg-MacLane spaces as fibers, we can assume that $Y_\bullet $ is a $K(V, n)$ for $V$ a vector space over $\mathbb{Q}$. (If I understand correctly, fibrations and homotopy inverse limits of spaces obtain good Sullivan models.) If $V$ is finite-dimensional, then we have an explicit Sullivan model (just a free (graded)-commutative algebra), and so we can compute homotopy classes of maps $B \to A$; they'll be the same as maps from $V$ into the cohomology of $X_\bullet$. But maps from $X_\bullet$ into an Eilenberg-MacLane space classify cohomology classes, so we're done. -But, this doesn't seem to work when $V$ is infinite-dimensional. I don't know what a good Sullivan model for a $K(V, n)$ is anymore. The cohomology in dimension $n$ is $V^\vee$, but, say, if $n$ is even, then the cohomology in dimension $2n$ should be $(V \otimes V)^{\vee}$, not what would be nice: $V^\vee \otimes V^\vee$. So, is this statement even true? At the very least, can we get some kind of equivalence of $\infty$-categories? -Quillen himself stated the result using a Quillen equivalence, but it's actually a somewhat complex series of them. He starts with reduced 2-simplicial sets, and then goes to reduced simplicial groups via the loop group construction, then takes the completed group ring dimensionwise to get a simplicial complete Hopf algebra, and then takes primitive elements to get a simplicial Lie algebra, and then applies the lax symmetric monoidal Dold-Kan functor to get dg-Lie algebras. So maybe dg-Lie algbras (or dg-coalgebras) are better than cdgas for describing rational homotopy theory. -Still, one thing that I would like to be true, but is not proved in Quillen's paper, is a direct Quillen equivalence starting with (localized) simplicial sets and ending in some algebraic category. The problem is, the Quillen equivalences I've described go in the opposite direction. The loop group is a left adjoint, but taking primitive elements is a right adjoint. So, while there is an honest Quillen equivalence between reduced simplicial sets and simplicial complete Hopf algebras, there is not proved (in this paper, as far as I can tell) the existence of a single Quillen equivalence (not a zig-zag) which ends in a category with no reference to topology. -Q2: Is there a direct Quillen equivalence of the rational homotopy category with a nice algebraic category? -For instance, it would be interesting if there was some coalgebra version of the de Rham complex. - -REPLY [11 votes]: This is more like a comment, but the usual space allowed for comments is too small! -It seems (after Toën's work) that we cannot describe rational homotopy types in terms of algebra, but rather in terms of (derived) algebraic geometry; the fact that commutative dg algebras (i.e. affine derived schemes) are sufficient for nice simply connected spaces is really due to the lack of monodromy. This is one of the reasons why Toën developped his theory of affine homotopy types (partly following ideas which were already sketched in Grothendieck's Pursuing stacks): for any ring $k$ and any space $X$, there is the pro-unipotent completion of the $\infty$-groupoid corresponding to $X$, denoted by $(X\otimes k)^{uni}$, which is the higher analog of the pro-unipotent completion of a group. If $k$ is a field of characteristic zero, and if $X$ is $1$-connected, nilpotent and of finite type, then the stack $(X\otimes k)^{uni}$ is the spectrum (in the sense of derived algebraic geometry) of the commutative dg algebra of de Rham cohomology of $X$, so that this theory includes classical rational homotopy theory. The essential information given by the stack $(X\otimes k)^{uni}$ is essentially the $\infty$-category of $k$-linear local systems over $X$. -All this is explained in the paper -B. Toën, Champs affines, Selecta Math. (N.S.) 12 (2006), no. 1, 39-135. -(see in particular Cor. 2.4.11, Cor. 2.5.3 and Cor. 2.5.4, to see that this extends nicely classical rational homotopy theory to non-simply connected homotopy types).<|endoftext|> -TITLE: Cantor-Bernstein for notions of forcing -QUESTION [13 upvotes]: For forcing notions $\mathbb{P}$ and $\mathbb{Q}$ let us write $\mathbb{P}\triangleleft\mathbb{Q}$ if forcing with $\mathbb{Q}$ always adds a $\mathbb{P}$-generic filter over $V$. In other words, $\mathbb{P}\triangleleft\mathbb{Q}$ holds if there is a $\mathbb{Q}$-name $\tau$ such that $\tau[H]$ is a $\mathbb{P}$-generic filter over $V$ whenever $H$ is a $\mathbb{Q}$-generic filter over $V$. If $\mathbb{P}\triangleleft\mathbb{Q}$ and $\mathbb{Q}\triangleleft\mathbb{P}$ need $\mathbb{P}$ and $\mathbb{Q}$ be forcing isomorphic, ie must they produce the same generic extensions? I expect not, but haven't been able to think of a counterexample. -I believe this question can be phrased in terms of Boolean algebras; ie if two complete Boolean algebras are each isomorphic to a complete Boolean subalgebra of the other, need they be forcing isomorphic? -(Incidentally, if the generics the forcings add are done in a 'reversible' way, ie if $\tau,\sigma$ are the $\mathbb{Q},\mathbb{P}$ names for the added generics and always $\tau[\sigma[G]]=G$ and $\sigma[\tau[H]]=H$ then the forcing notions will be equivalent. This is mentioned in the beginning of Shelah's text Proper Forcing). - -REPLY [6 votes]: I realize I'm a bit late to the party, but I wanted to add that counterexamples can be easily constructed out of almost any type of forcing. I'll give a particular example, but the method is flexible. All we need are three forcings, $\mathbb{R}_1$, $\mathbb{R}_2$ and $\mathbb{R}_3$, pairwise incomparable by the $\lhd$ relation described in the question. For concreteness, let's take $\mathbb{R}_1 = Add(\omega,1)$, $\mathbb{R}_2 = Add(\omega_1,1)$ and $\mathbb{R}_3 = Add(\omega_2,1)$. -Let $\mathbb{P}$ be the lottery sum of $\mathbb{R}_1$ and $\mathbb{R}_1 \times \mathbb{R}_2$ (so $\mathbb{P}$ first chooses generically whether to force with just $\mathbb{R}_1$ or with the product $\mathbb{R}_1 \times \mathbb{R}_2$, and then forces accordingly). Similarly, let $\mathbb{Q}$ be the lottery sum of $\mathbb{R}_1$ and $\mathbb{R}_1 \times \mathbb{R}_3$. -$\mathbb{P}$ and $\mathbb{Q}$ are clearly not forcing equivalent, since the first has the option of adding a generic for $\mathbb{R}_2$, which the second cannot (and vice versa, adding a generic to $\mathbb{R}_3$). However, forcing with one will always add a generic for the other, since either one will add a generic for $\mathbb{R}_1$, which is one option in the lottery in each case. -This does not answer the second question, about Boolean algebras, and I think it demonstrates that the order relation on Boolean algebras given by "embedding into a subalgebra" is really distinct from the $\lhd$ order -- here we have two forcings satisfying $\mathbb{P} \lhd \mathbb{Q}$ and $\mathbb{Q} \lhd \mathbb{P}$, but neither Boolean algebra embeds into a subalgebra of the other. -This suggests a natural additional condition, namely that $H$ give the entire extension $V[G]$. That is, in every extension $V[G]$ by $\mathbb{P}$ there is $H$ generic for $\mathbb{Q}$ such that $V[G]=V[H]$. I believe this guarantees forcing equivalence in the case where $\mathbb{P} \lhd \mathbb{Q}$ and $\mathbb{Q} \lhd \mathbb{P}$, since every extension by $\mathbb{P}$ equals and extension by $\mathbb{Q}$ and vice versa.<|endoftext|> -TITLE: Stirling number identity via homology? -QUESTION [29 upvotes]: This is a question about the well-known formula involving both types of Stirling numbers: -$\sum_{k=1}^{\infty}(-1)^{k}S(n,k)c(k,m)=0$, -where $S(n,k)$ is the number of partitions of an $n$-element set into $k$ blocks, and $c(k,m)$ is the number of permutations of a $k$-element set having $m$ cycles. I assume that $n \neq m$. The question is whether there is a known proof of the formula of the following nature: -(1) there is a complex of vector spaces with ranks $S(n,k)\cdot c(k,m)$, -(2) the complex is exact. I would be interested in a general answer or, if it makes the situation simpler, just an answer that applies in case $m=1$. -The context is this: I am trying to work out, in algebraic geometry, an effective method of computing the Severi degrees for plane curves of specified genus and number of nodes. The method depends on constructing an analogous homological argument for which the Stirling number argument (if it exists) would provide a prototype. - -REPLY [5 votes]: This is an example in some notes that I worked on. It's a bit involved, and I don't know how to simplify the approach further, so let me just offer a sketch (in particular, I want to try to ignore a bunch of boundary cases which result in awkward constructions that I don't know how to explain well at the moment), which at first glance, looks similar to the approaches sketched above. If a more detailed version is useful, please feel free to contact me by email. -Consider the category $C$ of ordered nonempty finite sets and where $Hom_C(S,T)$ is the set of surjections $f \colon T \to S$ such that $\min f^{-1}(i) < \min f^{-1}(j)$ whenever $im$, you will get the desired complex because $\dim_k P_i([n]) = S(n,i)$. -How do we construct such a projective resolution? It is enough to calculate the Ext groups between the different $k_m$ since $Hom(P_n,k_m)$ is $1$-dimensional if and only if $n=m$ and is $0$ otherwise. To do this calculation, we can reduce to calculating the homology of certain simplicial complexes: given $n0$) are chains of morphisms -$[n] \to S_1 \to \cdots \to S_i \to [m]$ -and a simplex contains another if it can be obtained by composing morphisms (this is related to the nerve construction of a category except I fix the endpoints). I want to consider the augmented simplicial cochain complex of this, but with a weird caveat: when $i=0$, I want to allow "multiple" empty simplices, which are indexed by the set $Hom_C([n],[m])$. (This is defined this weird way just to incorporate boundary cases correctly.) Anyway, let $H^i$ be the cohomology of this weird complex with coefficients in $k$. Then the claim is that -$Ext^i(k_n, k_m) = H^{i-2}$ for $i>0$. -This follows from modifying an argument of Cibils in -http://dx.doi.org/10.1016/0022-4049(89)90058-3 -(the relevant result is Proposition 2.1). -Anyway, the final punchline: these nerve-like things I've introduced are order complexes of truncations of the partition lattice. Specifically, we look at set partitions of the set $[m]$ which have at least $n$ parts and less than $m$ parts. It can be shown that this poset (with a min and max adjoined) is EL-shellable and hence its order complex is homotopy equivalent to a wedge of spheres of maximal dimension (in this case $m - n - 2$ -- let me ignore discussing the case $m-n$ being small). One can calculate the number of spheres from general poset topology tools -- but instead we could appeal to the fact this wedge of spheres fact implies that a complex like I mentioned exists (i.e., a linear resolution and modulo calculating the ranks) and that the ranks have to be the cycle numbers $c$ in order for the Euler characteristic to work out.<|endoftext|> -TITLE: How do you present a non-existence theorem? -QUESTION [10 upvotes]: (This question might be too vague, feel free to edit or vote for closing.) -In math there are usually lots of non-existence theorems. When someone presents such a theorem, one natural response is "why shall I even care", or "why should such a thing be impressive". -The problem is, in the case of a non-existence theorem, usually all examples are trivial. If you tell some undergrad non-constant bounded entire functions don't exist, he/she will probably reply with a shrug. Similar thing happen to me and my friends when we talk about some fancier theorems (or when I see a paper stating such a non-existence theorem). I feel like it's really hard to convince people (or convince myself) "a priori this thing could exist, however by this awesome theorem it doesn't." -I think people would be impressed if all hypothesis look innocent, like the one in Liouville's theorem on entire functions, or maybe the fact that people have seen the existence of differentiable bounded non-constant functions helps. In general it is not necessary that all hypothesis look friendly. How would one figure out whether a non-existence theorem is a good one or is true just because one of the hypothesis is insanely strong? - -REPLY [14 votes]: One important class of "non-existence theorems", which includes Liouville's theorem as a model example, are the various rigidity theorems throughout mathematics that tend to have the general flavour of "any object in class X which is weakly regular, is automatically strongly regular and/or unexpectedly algebraic". Or, in non-existence form, "There does not exist any object of class X which is weakly regular, but not strongly regular". These results tend to be very powerful in applications (because one can then verify a strong regularity property simply by establishing a much weaker, and thus presumably easier to check, regularity property) and are also often important psychological bridges between otherwise disjoint categories (e.g. between the topological and smooth categories). It also often allows one to think of strongly regular objects in X as being "isolated" in some sense from the irregular objects, since there are no transitional objects of intermediate regularity. -Rigidity results can often be motivated as an unexpected converse: one can state the stronger and weaker notions of regularity, and show that the stronger implies the weaker with so much room to spare that it would seem ridiculous that the converse claim can be established - and yet it is true. -An elementary example of a rigidity theorem is the high school geometry theorem that any two triangles with the same side lengths, must in fact have the same angles as well and are thus congruent; or in non-existence form, one cannot find a pair of triangles with the same side-lengths but differing angles. Thus, triangles are rigid in a way in which quadrilaterals, for instance, are not. -Deeper examples of rigidity theorems include - -Liouville type theorems: solutions to a certain PDE which are "uniformly bounded" or "precompact" in some sense are necessarily constant (or trivial, or a soliton...). Such theorems have become a central part of the modern theory of critical dispersive PDE, see e.g. this survey of Killip and Visan. Similar themes also come up in Perelman's proof of the Poincare conjecture, when he finds that the asymptotics of Ricci singularities are in some sense governed by the very special solutions known as gradient shrinking solitons. Somewhat related here are the various elliptic regularity theorems, such as the theorem that every weakly harmonic function (or distribution) is strongly harmonic. -Hilbert's fifth problem: C^0 Lie groups are necessarily C^infty Lie groups (and in particular come with the rich algebraic structure of a Lie algebra, which is certainly not obvious at all if one only begins with C^0 regularity). This is actually part of a large constellation of related theorems, such as Cartan's theorem that any closed subgroup of a Lie group is again a Lie group, or Gromov's theorem that any group of polynomial growth is virtually nilpotent, or the Peter-Weyl theorem which asserts (among other things) that every compact group is the inverse limit of linear Lie groups. -Eliashberg-Gromov symplectic rigidity: roughly speaking, the C^0 limit of symplectomorphisms is again a symplectomorphism, which is surprising as one would naively imagine one would need something like C^2 control instead. I'm not an expert on this topic, but I understand that this is a fundamental theorem in symplectic topology. A related result which also has some rigidity flavour to it (and is certainly a non-existence theorem) is Gromov's symplectic non-squeezing theorem that one cannot sympletically map a large ball into a thin cylinder. (Hmm, Gromov's name is coming up a lot...) -Dynamical rigidity: For certain very special types of dynamics (such as homogeneous dynamics coming from unipotently generated groups), the only minimal closed invariant sets or ergodic measures are those that come from algebraic constructions, such as algebraic subgroups; Ratner's deep theorems on this subject are model examples of this phenomenon. These theorems can be incredibly powerful for establishing equidistribution or density of orbits in homogeneous spaces, for instance the Oppenheim conjecture can be solved as a quick corollary of these theorems (though this was not quite the historical chain of events, as Margulis' proof of this conjecture preceded Ratner's work by a few years). Another example of dynamical rigidity is superrigidity, that in some cases the action of a continuous group can be controlled by the subaction of a discrete lattice; this in turn is related to hyperbolic geometry rigidity theorems such as Mostow rigidity, though this is far from my own area of expertise. Kazhdan's property (T), which asserts that approximately invariant vectors in actions of certain groups must be close to genuinely invariant ones, is another related property of groups that certainly pulls in the direction of rigidity.<|endoftext|> -TITLE: Are plethories a theory of basis-free polynomials? -QUESTION [5 upvotes]: This question is a follow-up to a question about the theory of polynomials. -It should be quite clear by now that matrix theory and linear algebra are quite different topics. As the various answers to that question clearly state, the difference is largely because of basis choice. But we know how to do a lot of abstract linear algebra, without ever having to choose a basis. -But when one switches to polynomials, the situation changes: even very general definitions of polynomials (like that in Lang's Algebra, for example) still use monomials in the standard basis. Why that basis, and not the rising factorial basis, or falling factorial, or Chebyshev basis, or any other? And the most important question: why, in fact, choose a basis at all? -Which leads to the question in the title: Are plethories a theory of basis-free polynomials? -I think the motivation should be clear: if going 'basis free' was so successful for linear algebra, shouldn't we also expect similar success with polynomials? Unfortunately, I have not been able to find any work which I could readily understand as being about developing a theory of basis-free polynomials. -EDIT: After reading the current comments and answers, I am starting to think that perhaps what I am really seeking is a theory for basis-independent polynomials, or even basis-generic rather than leaping right away to basis-free. There are many applications of polynomials where bases other than the monomial one greatly simplify both reasoning and computations (which is what I am eventually after), but the tools for doing this seem to be hard to find. - -REPLY [5 votes]: Some naive remarks. It seems to me that the simplest reason to choose the standard basis is because it exhibits the universal property of a polynomial ring. -One way to exhibit a partially basis-free definition of a polynomial ring is to define it as the symmetric algebra on a vector space. This retains the grading but is otherwise basis-free. If one wants to ignore the grading then I am not sure what options are available and depending on the application it may not be natural to ignore the grading.<|endoftext|> -TITLE: Primes of the form $x^2+ny^2$ and congruences. -QUESTION [22 upvotes]: The answer of following classical problem is surely known, but I can't find a reference - -For which positive integer $n$ is the set $S_n$ of primes of the form $x^2+n y^2$ ($x$, $y$ integers) determined by congruences? - -A set of prime $S$ is said determined by congruences if there is a positive integer $m$ -and a set $A \subset (\mathbb{Z}/m\mathbb{Z})^\ast$ such that a prime $p$ not dividing $m$ is in $S$ if and only if $p$ modulo $m$ is in $A$. There is a natural place to look for this question: the book by Cox "primes number of the form $x^2+ny^2"$". Unfortunately I don't have it, my library doesn't have it, and I can't find it on the internet, except for some preview at Amazon and Google. From the table of content and the preview it seems that the book does not contain the answer to my question (otherwise I wouldn't ask) but it is still possible -that the answer be hidden precisely in one of the sporadic pages that amazon doesn't -want me to see. -From that book one knows that a prime $p$ is in $S_n$ if and only if it splits in the -ring class field $L_n$ of the order $\mathbb{Z}[\sqrt{-n}]$ in the quadratic imaginary field $K_n:=\mathbb{Z}[\sqrt{-n}]$. Therefore the question becomes: is $L_n$ abelian over $\mathbb{Q}$? Now $H_n:=Gal(L_n/K_n)$ is the ring class group of $\mathbb{Z}[\sqrt{-n}]$, hence abelian, and $Gal(L_n/\mathbb{Q})$ is a semi-direct extension of $\mathbb{Z}/2\mathbb{Z}$ by $H_n$, the action of the non-trivial element of $\mathbb{Z}/2\mathbb{Z}$ on $H_n$ being $x \mapsto x^{-1}$. Hence, if I am not mistaken (am I?), the question is equivalent to - -For which $n$ is the ring class group $H_n$ killed by $2$? - -Thanks for any clue or reference. I am especially interested in the case $n=32$. - -REPLY [23 votes]: Even though some of the main ideas have been pointed out above, I will try to add (in all modesty) a complete answer. -The following outline comes from my master thesis and further develops some of the ideas in David Cox's book Primes of the form $x^2+ny^2$. - -$\color{red}{\text{PART 1: Preliminary results on the genus field.}}$ - -Lemma: Given a number field $K$, there is a one-to-one correspondence between unramified Abelian extensions $M$ of $K$ and subgroups $H$ of the ideal class group $C(\mathcal{O}_K)$. Furthermore, if the extension $K \subset M$ corresponds to the subgroup $H \subset C(\mathcal{O}_K)$, then the Artin map induces an isomorphism - $$C(\mathcal{O}_K)/H \xrightarrow{\sim} \mathrm{Gal}(M/K)$$ - -Proof. See Corollary 5.21 in Cox for the details. - -Definition: The genus field $M$ is the unramified Abelian extension of $K$ corresponding to the subgroup $C(\mathcal{O}_K)^2 \subset C(\mathcal{O}_K)$. - -Possibly there are other definitions for the genus field, but it can be shown that the above definition implies the following characterization: - -Theorem: Let $K$ be an imaginary quadratic field of discriminant $d_K$. Let $\mu$ be the number of primes dividing $d_k$ and let $p_1, \ldots, p_r$ be the odd primes dividing $d_K$. (So that $\mu=r$ or $r+1$ according to whether $d_K \equiv 0 \textrm{ or } 1 \bmod{4}$). Set $p_i^*=(-1)^{(p_i-1)/2}p_i$. Then: - -The genus field of $K$ is the maximal unramified extension of $K$ which is an Abelian extension of $\mathbb{Q}$. -The genus field $M$ of $K$ is $K(\sqrt{p_1^*}, \ldots, \sqrt{p_r^*})$ and $[M:\mathbb{Q}]=2^{\mu}$. - - -Proof. See Theorem 6.1 in Cox for the details. - -$\color{red}{\text{PART 2: When is the Genus field equal to the HCF?}}$ - -Lemma 1: The Hilbert class field $L$ of $K=\mathbb{Q}(\sqrt{-n})$ is Abelian and Galois over $\mathbb{Q}$ (besides being Galois over $K$) if and only if the genus field equals the Hilbert class field - -Proof. It can be shown that $[G,G]=C(\mathcal{O}_K)^2$ with $G=\mathrm{Gal}(L/\mathbb{Q})$ (see Theorem 6.1 in Cox). So if $\mathrm{Gal}(L/\mathbb{Q})$ is Abelian then $C(\mathcal{O}_K)^2=1$ and via Galois theory of unramified Abelian extensions this gives: -$$C(\mathcal{O}_K)/1 \xrightarrow{\sim} \mathrm{Gal}(M/K)$$ -But the left hand side corresponds to $\mathrm{Gal}(L/K)$ via Artin Reciprocity, so that -$$[L:K]=|\mathrm{Gal}(L/K)| = |\mathrm{Gal}(M/K)|=[M:K]$$ -Since we already know that $M \subset L$, we actually have $M = L$. -Before we continue, we recall the following result: - -Lemma 2: Let $K$ be an imaginary quadratic field and let $L$ be Hilbert class field of $K$. Then $L$ is a Galois extension of $\mathbb{Q}$, and its Galois group can be written as a semidirect product - $$\mathrm{Gal}(L/\mathbb{Q}) \cong \mathrm{Gal}(L/K) \rtimes (\mathbb{Z}/2\mathbb{Z})$$ - where the nontrivial element of $\mathbb{Z}/2\mathbb{Z}$ acts on $\mathrm{Gal}(L/K)$ by sending $\sigma$ to its inverse element $\sigma^{-1}$. - -Proof. See p122 in Cox for the details. - -Lemma 3: Let $H$ and $K$ be two finite groups. Then $H\rtimes_\theta K$ is abelian if and only if $H$ and $K$ are abelian and $\theta$ is the trivial homomorphism. - -Proof. Sufficiency is obvious. For necessity, if $\theta$ is nontrivial, then some $k\in K$ maps to some $\operatorname{id}\ne \theta_k\in \operatorname{Aut}(H)$, which means that $\theta_k(h)\ne h$ for some $h\in H$. Thus $hk=k\theta_k(h)\ne kh$, so we have two elements that do not commute. - -Theorem 1: The Hilbert class field $L$ of $K=\mathbb{Q}(\sqrt{-n})$ is Abelian if and only if $CL(K) \cong (\mathbb{Z}/2\mathbb{Z})^r$ for some $r \geq 0$. - -Proof. From the two previous lemmas it follows that $\mathrm{Gal}(L/\mathbb{Q})$ is abelian if $\mathrm{Gal}(L/K)$ is abelian, and $\sigma$ acts trivially. The last condition is only possible if for every element of $\sigma \in \mathrm{Gal}(L/K)$ we have $\sigma=\sigma^{-1}$. But this means that $\sigma^2=1$ for every $\sigma$, such that $\mathrm{Gal}(L/K)$ must be an elementary abelian $2$-group (i.e. of the form $\mathbb{Z}/2\mathbb{Z}^r$ for some $r \geq 0$). Under these assumption, the second requirement, that $\mathrm{Gal}(L/K)$ must be abelian, is automatically satisfied. Because of Artin reciprocity we now have that -$$CL(K) \cong \mathrm{Gal}(L/K) \cong (\mathbb{Z}/2\mathbb{Z})^r$$ -and we are done. - -Definition: Let $K=\mathbb{Q}(\sqrt{-n})$ be an imaginary quadratic field. Then we call the positive integer $n$ a convenient number if the class group of $K$ is an abelian $2$-group, i.e. $CL(K)=(\mathbb{Z}/2\mathbb{Z})^r$ for some $r \geq 0$. - -But which numbers $n$ are convenient numbers? Euler discovered that the following 65 idoneal numbers are convenient numbers. - -Weinberger proved in 1973 that at most one other convenient number can exist, and that if the generalized Riemann hypothesis holds, then the list is complete. -Combining Lemma 1 and Theorem 1, we have: - -Theorem 2: The genus field equals the Hilbert Class field of $K=\mathbb{Q}(\sqrt{-n})$ if and only if the class group of $K$ is an abelian $2$-group if and only if $n$ is one of Euler's 65 idoneal numbers or Weinberger's (conjectured, unique) 66th number. - - -$\color{red}{\text{PART 3: Primes of the form $x^2+ny^2$ and congruence relations.}}$ -Another curious fact of Euler's idoneal numbers is that they are very closely related to situations in which congruences can be used to express the necessary and sufficient conditions for a prime to be of the form $p=x^2+ny^2$. More specifically, - -Theorem 3: Let $n$ be a convenient number. Then the necessary and sufficient conditions for a prime to be expressed as $p=x^2+ny^2$ can be stated by congruence relations. - -Before we can prove this claim we need the following lemma - -Lemma 4: Let $n$ be one of Euler's 65 convenient numbers such that the genus field of $K=\mathbb{Q}(\sqrt{-n})$ equals the Hilbert class field. Since the genus field is given by - $$M=K(\sqrt{p_1^*}, \sqrt{p_2^*}, \ldots, \sqrt{p_n^*})$$ - the HCF can be written as a composite of quadratic fields - $$L=K_0 K_1 K_2 \cdots K_n$$ - with $K_i=\mathbb{Q}(\sqrt{p_i^*}$) for $i \geq 1$, and $K_0=K$. Moreover, a prime $p$ splits completely in $L$ if and only if it splits completely in all of the $K_i$. - -and the following major result - -Theorem 4: A prime can be represented as $p=x^2+ny^2$ if and only if the prime $p$ splits completely in the HCF of $\mathbb{Q}(\sqrt{-n})$. - -Proof. See Theorem 5.26 in Cox. -We can now start proving Theorem 3: -Proof. If $n$ is one of Euler's 65 convenient numbers, then Lemma 4 argues that the HCF can be written as the composite field of extensions $\mathbb{Q}(\sqrt{p_i^*})$ over $\mathbb{Q}$ and that a prime splits completely over the composite field if and only if it splits completely over all the fields $\mathbb{Q}(\sqrt{p_i^*})$. So, Theorem 4 can now be reread as follows: a prime can be expressed in the form $p=x^2+ny^2$ iff $p$ splits completely in the HCF iff $p$ splits completely over all the $K_i$. Using Dedekind's theorem on the splitting behavior in quadratic extensions, this just means that $\left(d_{K_i}/p\right)=1$ for all $i$. Each condition gives us a congruence relation. Taking the different congruences together, we can conclude that the necessary and sufficient conditions can be expressed by congruence relations. - -Corollary: Let $n$ be a convenient number. Let $q_i$, for $i=1, \ldots, r$ be the odd divisors of $n$ and set $q^* = (-1)^{(q-1)/2}$. Then - $$p=x^2+ny^2 \Leftrightarrow \left( \frac{-n}{p} \right) = 1 \textrm{ and } \left( \frac{q_1^*}{p} \right) = \left( \frac{q_2^*}{p} \right) = \cdots = \left( \frac{q_r^*}{p} \right) = 1$$ - -Example. The above theory gives us a very easy way to find congruence relations when $n$ is a convenient number: just take any number Euler's list and find the divisors, .e.g. -\begin{eqnarray*} -p=x^2+93y^2 &\Leftrightarrow& \left( \frac{-93}{p} \right) = \left( \frac{-3}{p} \right) = \left( \frac{-31}{p} \right) = 1 \\ -&\Leftrightarrow& p \equiv 1, 25, 49, 97, 109, 121, 133, 157, \\ -&& \ \ \ \ \ \ \ 169, 193, 205, 253, 289, 349, 361 \mod{372} -\end{eqnarray*}<|endoftext|> -TITLE: T-bundles and the Borel-Weil-Bott theorem -QUESTION [8 upvotes]: Hi, -Let $G$ be a reductive, connected group, $T$ a maximal torus, and $B$ a Borel subgroup containing $T$ with unipotent radical $U$. Then it turns out that the functions on the algebraic variety $G/U$ give a representation of $G$ where each irreducible representation appears exactly once. Geometrically, $G/U$ is a $B/U = T$-bundle over the flag manifold $G/B$, -and I think one can deduce Borel-Weil-Bott by studying this $T$-bundle. -This much was explained to me some time ago, and now I would like to understand this circle of ideas better, but I can't find it anywhere... Any explanations/details/references/etc. would be appreciated! -Thanks! - -REPLY [11 votes]: I'm very skeptical about the possibility of getting the full Borel–Weil–Bott theorem just by studying $G/U \to G/B$. Probably the closest thing I can think of is Bott's original proof of his theorem, which involves studying certain $\mathbb P^1$-bundles $G/B \to G/P$. On the other hand, you can prove the Borel–Weil theorem by studying the function space $\mathcal{O}(G/U)$, but even here you need to know a little more than just that this space contains every irrep of $G$ exactly once. More specifically, you want to know how each irrep shows up. Let me sketch the argument. To be safe, I assume we're working over $\mathbb C$, but what follows probably works over any algebraically closed field of characteristic zero. -To start off, note that $G$ acts on $\mathcal{O}(G)$ by left and right translation. Viewing $\mathcal{O}(G)$ under the latter action, we can think of -$$ \mathcal{O}(G/U) = \{ f \in \mathcal{O}(G) \colon f(gu) = f(g) \text{ for all } g \in G, u \in U \} $$ -as the space $\mathcal{O}(G)^U$ of $U$-invariants. Now recall that there's a $G\times G$-equivariant decomposition -$$ \mathcal{O}(G) = \bigoplus V \otimes V^\ast \qquad -\text{[an algebraic Peter–Weyl theorem]}$$ -where the sum runs over the irreps of $G$, and $G$ acts on $V$ by left translation and on $V^\ast$ by right translation. Therefore we find that -$$ \mathcal{O}(G/U) = \mathcal{O}(G)^U = \bigoplus V \otimes (V^\ast)^U. $$ -Let's assume that $U$ is built up using negative roots, so that $(V^\ast)^U$ is the lowest weight space of $V^\ast$, and in particular is one-dimensional. This shows that every irrep of $G$ appears in $\mathcal{O}(G/U)$ exactly once. But that's not all: using the right $G$-action, we can "capture" the irrep of highest weight $\lambda$. Indeed, as a $T$-module, $(V^\ast)^U = \mathbb C_\mu$, where $\mu$ is the lowest weight of $V^\ast$, or said differently, $-\mu$ is the highest weight of $V$. So, using the fact that $\text{Hom}_T(\mathbb C_\lambda, \mathbb C_\mu) = \delta_{\lambda\mu} \mathbb C_\lambda$, we see that the irrep of $G$ of highest weight $\lambda$ can be gotten as -$$ \text{Hom}_T(\mathbb C_{-\lambda}, \mathcal{O}(G/U)) = \bigoplus V \otimes \text{Hom}_T(\mathbb C_{-\lambda}, (V^\ast)^U). $$ -We can re-write the left side of the above as -$$\begin{align} -(\mathbb C_\lambda \otimes \mathcal{O}(G/U))^T &= \{ f \in \mathcal{O}(G) \colon f(gtu) = \lambda(t)^{-1} f(g) \text{ for all } g \in G, t \in T, u \in U \} \\ -&= \{ f \in \mathcal{O}(G) \colon f(gb) = \lambda(b)^{-1} f(g) \text{ for all } g \in G, b \in B \}, -\end{align}$$ -which of course we can think of as the space of global sections of the line bundle $L_\lambda = G \times_\lambda \mathbb C$ over $G/B$. This proves the first part of the Borel–Weil theorem, namely that if $\lambda$ is dominant then $H^0(G/B,L_\lambda)$ is the irrep of highest weight $\lambda$. The other part, that $H^0(G/B,L_\lambda)=0$ if $\lambda$ is not dominant also follows easily. Indeed, all of the above works just as well for such $\lambda$, except in this case we have $\text{Hom}_T(\mathbb C_{-\lambda}, (V^\ast)^U)=0$ for all irreps $V$.<|endoftext|> -TITLE: regular singularities and logarithmic singularities -QUESTION [5 upvotes]: What is the difference between regular singularities and logarithmic singularities? Could someone give me a reference where the distinction is clearly explained? I apologize in advance if this question is very trivial (and inappropriate for the level of MO), but I am a bit confused because I feel, in some papers, people treat these both singularities as being the same object, but I think it should exist an important difference between them otherwise I can not see what is the point in introducing the same object with two different names. - -REPLY [5 votes]: My guess is that the OP is thinking of singularities of integrable connections rather than singularities of varieties. If so logarithmic singularities usually refer to solutions of $y'=1/x$ whose (many-valued) solution is exactly $\log x$. It may possibly also refer to higher order equations whose solutions are powers of $\log x$. In any case $y'=\alpha/x y$ with solution $x^\alpha$ also has regular singularities. In one dimension the general connection with regular singularities is locally a combination of these two cases. In higher dimension there are even more complicated connections with regular singularities (even when the divisor of singularities of the connection only has normal crossing singularities). However, there is the notion of log-connections and by a result (of Deligne I believe) when the divisor of singularities has normal crossings any connection with regular singularities has an extension across the divisor which is a log-connection. - -REPLY [3 votes]: If you mean the difference between notions such as terminal and log terminal or canonical and log canonical, then the difference is a shift in requirements. -The word "log" might be a little misleading, but one possible interpretation for it that it refers to logarithmic differentials as opposed to differentials. -Here is an example: -$X$ has canonical singularities if for a resolution of singularities $\pi:Y\to X$, -$$\pi^*K_X\leq K_Y.$$ -In other words, the pull-back of top regular differential forms from $X$ remain regular on $Y$. -On the other hand $X$ has log canonical singularities if for a resolution of singularities $\pi:Y\to X$, -$$\pi^*K_X\leq K_Y + E,$$ -where $E$ is the reduced exceptional divisor. -In other words, the pull-back of top regular differential forms from $X$ remain regular with only logarithmic poles on $Y$. -Does this answer your question or do you mean something entirely different?<|endoftext|> -TITLE: Evidence for integer factorization is in $P$ -QUESTION [24 upvotes]: Peter Sarnak believes that integer factorization is in $P$. It is a well-known open problem in TCS to identify the real complexity class of integer factorization. Take a look at this link for Peter Sarnak's lectures where he mentions that he does not believe factoring is not in $P$. -What evidence is there that integer factorization is in $P$ other than the fact that polynomial factorization is in $P$? - -REPLY [42 votes]: I don't think there is any compelling evidence that integer factorization can be done in polynomial time. It's true that polynomial factoring can be, but lots of things are much easier for polynomials than for integers, and I see no reason to believe these rings must always have the same computational complexity. (Strangely, if you do believe that, it means the shortest lattice vector problem should also be efficiently solvable, but it doesn't seem to tell you anything about discrete logarithms. This puzzles me, since the parallels between factoring and discrete logs are also strong.) It's also true that primality testing can be done in polynomial time, but that is a fundamentally different problem: when a modulus is prime, it has enormous consequences for modular arithmetic, and it is not difficult to test for primality by looking for those consequences, but the actual methods shed no light on factoring. -On the other hand, there is also no compelling evidence that factoring can't be done in polynomial time (see http://research.microsoft.com/~cohn/Thoughts/factoring.html for a little more detail about this). -At our current level of knowledge, I view the complexity of factoring as a matter of opinion, speculation, and wishful thinking, not principled arguments. By contrast, there are exceedingly good reasons why the Riemann hypothesis should be true, and good reasons why P shouldn't equal NP. I'm certainly open to arguments that fall far short of rigorous proof; I've just never heard a convincing one about the complexity of factoring. -My interpretation of Sarnak's belief isn't that he sees some good reasons other people don't appreciate. Rather, it just feels plausible to him, and he's perhaps a little annoyed that lots of people firmly believe the opposite for no good reason, so he makes a point of stating a strong opinion.<|endoftext|> -TITLE: Hilbert's 3rd problem,number theory, motives, cyclic homology,... -QUESTION [11 upvotes]: This talk by Jinhyun Park connects a lot of interesting themes, making me curious to read more about that. Do you know where? - -REPLY [25 votes]: This circle of topics is certainly one of my favourite surprising connections in mathematics. I will try to outline what little I understand of the big picture. Apologies for the length. -Hilbert's 3rd problem and Dehn complexes: As is well-known, Hilbert's 3rd problem asked for examples of tetrahedra of equal volume which are not scissors congruent, and Dehn solved it using the invariant now named after him. However, this only gave rise to further questions: what about higher-dimensional euclidean spaces, and what about the other classical geometries? By now, I guess the proper objects to consider are the Sah algebra and Dehn complexes I will outline. -To study scissors congruences, it is best to consider the scissors congruence groups for all dimensions at the same time. Join of simplices induces a product, dimension a grading. In the spherical case, the Dehn invariants provide a coproduct, and the duality of spherical simplices gives an involution, making the total spherical scissors congruences into a graded Hopf algebra. This object is called Sah algebra. -The coproduct statement does not work in the other geometries since the dihedral-angle part of the Dehn invariant always introduces a spherical geometry component. However, in the other geometries, we still get comodule structures over the spherical scissors congruences. It is also possible (and interesting) to consider the graded cobar complexes for these modules over the coalgebra of spherical scissors congruences. These complexes were introduced by Goncharov who called them Dehn complexes (in the JAMS paper discussed below). -The Hopf algebra and comodule picture is discussed in the following book (not easy to get but very much worth reading): - -C.-H. Sah. Hilbert's third problem: scissors congruence. Research notes in mathematics 33, Pitman, 1969. - -Another book concerning scissors congruences (also very much worth reading but does not discuss the Hopf algebra view) is: - -J.-L. Dupont. Scissors congruences, group homology and characteristic classes. Nankai tracts in mathematics 1, World Scientific, 2001. - -A very general form of Hilbert's 3rd problem can then be formulated for all the classical geometries: are Dehn invariant and volume sufficient to completely characterize the scissors congruence class of a polytope? This is only known for euclidean space in dimensions $\leq 4$, for hyperbolic and spherical not even in dimension $3$. Even more general, one can wonder about the exact computation of the cohomology of the Dehn complexes. -Goncharov's conjectures then relate the above Dehn complexes to K-theoretic stuff. In the spherical and hyperbolic cases, Goncharov conjectures that the Dehn complexes compute the ($+1$ and $-1$-eigenspaces of complex conjugation on the) weight-graded pieces of algebraic K-theory of $\mathbb{C}$. This is formulated and discussed in - -A. Goncharov. Volumes of hyperbolic manifolds and mixed Tate motives. J. Amer. Math. Soc. 12 (1999), 569-618. - -Maybe it would also make sense to expect an explicit quasi-isomorphism between Bloch's cycle complexes (computing motivic cohomology) and the Dehn complexes. -I am not sure if a formulation like this exists in the literature, but probably the above conjecture says something like ``the Sah algebra should be isomorphic to the motivic Galois group of mixed Tate motives over $\mathbb{C}$'' (only that we do not have the latter). -Anyway, the conjecture is known in weight 2 due to the work of Dupont, Sah, Bloch, Suslin (this goes under the name of motivic weight $2$ complex, dilogarithm complex etc). The trilogarithm case has been worked out by Goncharov, see e.g. - -A. Goncharov. Geometry of configurations, polylogarithms and motivic cohomology. Adv. Math. 144 (1995), 197-318. - -Regulators, volumes and number theory -Goncharov's conjecture translates the regulator on K-theory to the volume of hyperbolic simplices (with Dehn invariant $0$). From this point of view, Goncharov's conjecture is a far-reaching generalization of the fact that volumes of hyperbolic 3-manifolds can be expressed in terms of dilogarithms. Under Goncharov's conjecture, the generalized version of Hilbert's 3rd problem for hyperbolic spaces and spheres translates into an injectivity conjecture for K-theoretic regulators due to Ramakrishnan. Currently, it is not even known if $K_3(\mathbb{C})$ (which is relevant for scissors congruences in $\mathbb{H}^3$) is bigger than $K_3(\overline{\mathbb{Q}})$ (which we understand in terms of Borel regulators). This is the number theory appearing in the title (and might be related to other conjectures on the description of the period algebra which I know nothing about). -Probably the consequences relating regulators and volumes actually motivated the conjecture in the first place. Certainly the relation between K-theory and polylogarithms is part of the motivation for Bloch's definition of mixed Tate motives as modules over a suitable Lie algebra. -Euclidean geometry and cyclic homology: So far, we have talked about the relation between the noneuclidean geometries and algebraic K-theory. Now, infinitesimally, hyperbolic space looks like euclidean space. As a consequence, euclidean Dehn complexes are the ``tangent spaces'' to hyperbolic Dehn complexes. Again, Goncharov has a precise conjecture how the euclidean Dehn complex should be related to mixed Tate motives over the dual numbers $\mathbb{C}[\epsilon]/(\epsilon^2)$, cf. - -A. Goncharov. Euclidean scissors congruence groups and mixed Tate motives over dual numbers. Math. Res. Lett. 11 (2004), 771-784. - -This is how we get a relation to cyclic homology, because cyclic homology is related to the tangent space of algebraic K-theory. This way, the euclidean part of Hilbert's 3rd problem involves talking about additive Chow groups (about which you can learn by reading e.g. the papers of Park, but also Cathelineau, Bloch, Esnault,...). -You see the appearance of groups related to cyclic homology by reformulating the classical Dehn-Sydler theorem as an exact sequence: -$$ -0\to \mathbb{R}\to\operatorname{SC}(\mathbb{E}^3)\stackrel{D}{\longrightarrow}\mathbb{R}\otimes S^1\to\Omega^1_{\mathbb{R}/\mathbb{Q}}\to 0, -$$ -where the map $D$ is the Dehn invariant from 3-dimensional scissors congruences (mapping a three-simplex to the sum of edge lengths tensor dihedral angles). The kernel of the Dehn invariant is detected by the volume map to $\mathbb{R}$, and the cokernel of the Dehn invariant is a group of Kähler differentials. -Further reading: check the works of Sah, Dupont, Cathelineau, Goncharov,... There is too much literature to mention in this already oversized answer.<|endoftext|> -TITLE: About the surface area vs. volume of polytopes -QUESTION [8 upvotes]: Given a convex body $K\in\mathbb{R}^n$, represented by a set of linear inequalities (intersection of halfspaces), I am interested in understanding how much of its volume can be close to its perimeter (under certain restrictions). -More formally, given a parameter $k$, and a partition of $\mathbb{R}^n$ into boxes of side $k$, I would like to know how large can the ratio $|P|/|I|$ be, where $P$ is the set of such boxes which intersect the perimeter, and $I$ is the set of boxes fully contained in $K$ (such a bound would be important for determining the accuracy of integration, for example). -I believe that as in 2 and 3 dimensions, the smallest ratio would be achieved by a ball (sphere), and that the worst ratio would be achieved for polytopes whose volume approaches 0 (by having a width smaller than k in one dimension, for example, which gives $|I|=0$). -However - are there some reasonable limits (for example, containing the 1/n unit sphere, or just having a volume > 0, and a poly(n) representation length of the linear inequalities) that can determine an upper bound on this ratio? -Many thanks, -Guy - -REPLY [2 votes]: A few updates to the question: - -Thanks for the correction about the minimality of a square (I was thinking of surface area/volume ratio, which is similar but different). -$k$ is not a fixed parameter, but can vary with a given $K$. However, I cannot just let $k\rightarrow 0$, since I am thinking about an algorithm which is limited in space (and running time), and would like to understand how small does $k$ need to be in order to guarantee that the above ratio $|P|/|I|$ is not too bad (for example, that it is $\leq 1$). - -So - to rephrase the above question: Let's assume that $K$ is a polytope in $\mathbb{R}^n$, and that it has a representation using poly(n) bits, as a set of linear inequalities. - We also assume that it has volume greater than 0 (easy to verify). If it is helpful, we may even assume that it contains the $\frac{1}{n}$ sphere in $\mathbb{R}^n$. -How small should the parameter $k$ (determining how fine we integrate) be, to determine that $|P|/|I|\leq 1$ ? -Thanks again, -Guy<|endoftext|> -TITLE: Calculating a curvature tensor by polarization -QUESTION [8 upvotes]: I'm reading some articles by Siu and Nannicini on the Weil-Petersson metric associated to families of compact Kahler-Einstein manifolds. In each article Siu and Nannicini construct a Weil-Petersson metric $h$, show that it is Kahler, and obtain results on the holomorphic sectional curvature by heroic calculations. -From what I can piece together, the curvature estimates go as follows: 1) show that the holomorphic sectional curvature, given by $R_{jjjj}$, is negative, 2) use a polarization trick to calculate the general tensor $R_{jklm}$, 3) then a miracle occurs, 4) so the holomorphic sectional curvature $R_{jjkk}$ is negative. -I'm trying to fill in the gaps in my understanding between the first and fourth steps, and I'm stuck on the second one. For Kahler manifolds the holomorphic sectional curvature determines the entire curvature tensor (see for example Lemma 7.19 of Zheng's "Complex differential geometry"), so I'm perfectly willing to believe that knowing $R_{jjjj}$ lets us calculate $R_{jklm}$. The problem is that I don't know how to do it. -This is a purely algebraic calculation, so I imagine it's written down somewhere, but the only results I've found are of the same type as in Zheng's book, i.e. they show that these calculations are theoretically possible but don't say how to do them. -Question: Is there a reference where this calculation is done explicitly? -[Edit] As Deane pointed out in the comments, one needs to know $R(X,\bar X,X,\bar X)$ for all holomorphic tangent fields $X$ to know the curvature tensor, just knowing $R_{j\bar jj\bar j}$ doesn't cut it. This makes two phrases from Siu's and Nannicini's papers a bit mysterious: -Siu: "We now polarize the expression for $R_{i\bar i i\bar i}$ to get the expression for $R_{i \bar j k \bar l}$." in "Curvature of the Weil-Petersson metric in the moduli space of compact Kahler-Einstein manifolds of negative first Chern class" (beginning of paragraph 5.4). -Nannicini: "The complete expression for the Riemann tensor can now be obtained by polarization of $R_{i\bar ii\bar i}$" in "Weil-Petersson metric in the moduli space of compact polarized Kahler-Einstein manifolds of zero first Chern class" (the page before Theorem 1). -Revised question: What exactly are Siu and Nannicini doing, if not applying the lemma on the holomorphic sectional curvature? - -REPLY [10 votes]: The explicit polarization formula is the following, taken from this paper of Bishop and Goldberg. -Working with real tangent vectors (instead of $(1,0)$ vectors, but it's easy to switch from one point of view to the other) the holomorphic sectional curvature of a unit vector $X$ is $Q(X)=R(X,JX,JX,X)$, and a general sectional curvature $R(X,Y,Y,X)$ can be expressed as -$$R(X,Y,Y,X)=\frac{1}{32}(3Q(X+JY)+3Q(X-JY)-Q(X+Y)-Q(X-Y)-4Q(X)-4Q(Y) ).$$ -If you want to switch back to vectors of type $(1,0)$, you can call $V=\frac{1}{\sqrt{2}}(X-iJX)$ -and then you have that $Q(X)=2 R(V,\bar{V},V,\bar{V})$.<|endoftext|> -TITLE: minimum space dimension to place n-points knowing pairwise distances -QUESTION [5 upvotes]: Hi everyone, -Let P be a set of n points. -Assuming I know the pairwise distances for each pair of points. -What would be the minimum dimension of the space in which I could place those n points with respect to the different pairwise distances. -The idea would be to set a first point at random coordinates in a multi-dimensional space. -Then, add the other n-1 points so that the pairwise distances are respected. -Sorry, it's maybe a trivial question for mathematicians but I'm still wondering if the relation: "number of points-> minimum dimension of space" does exist. -Thank you for your comments. - -REPLY [4 votes]: Obviously, for some special distances, you can embed the points in a fewer-dimensional space. If this is important, one way to count the correct dimension would be the rank of a certain matrix. -$(b-a,c-a)=(d(a,b)^2+d(a,c)^2-d(b,c)^2)/2$ -If you fix a point $a$ and place the other values of this dot product into an $n-1$ by $n-1$ symmetric matrix, the rank of that matrix will be the dimension of the space you can embed the points in.<|endoftext|> -TITLE: Eckmann-Hilton for $A_{\infty}$-spaces? -QUESTION [8 upvotes]: Suppose I have a grouplike $A_{\infty}$-space $G$ that carries an additional structure as a topological group (which does not coincide with the $H$-space structure of the $A_{\infty}$-part). Denote the $H$-space multiplication by $*$ and the group multiplication by $\cdot$ and suppose that -$$ -(a*b) \cdot (c *d) = (a \cdot c) * (b \cdot d) -$$ -also known as the Eckmann-Hilton condition holds (equality, no "up to homotopy" here). Moreover, the identity of the group is a homotopy unit for the $H$-space structure. - -Is this enough to deduce that $BG$ - coincides with the space $B(G_{*})$, - i.e. that the classifying space of the - topological group is (up to homotopy - of course) the delooping of the - $A_{\infty}$-space? Do I get homotopy - commutativity as well? - -REPLY [15 votes]: EDIT: Here is a counterexample to the stated question. -I'm going to start with a topological group $G$ which is a product of Eilenberg-Mac Lane spaces. Specifically, I'm going to choose $G \simeq K(\mathbb{Z},2) \times K(\mathbb{Z},7)$. The Dold-Thom theorem allows me to actually construct this as an abelian topological group: I take a based CW-complex with $H_2 = \mathbb{Z}$, $H_7 = \mathbb{Z}$, and other (reduced) homology groups zero (e.g. $S^2 \vee S^7$). Then I take the free topological abelian group on it, and this has the homotopy groups I want. - -The classifying space of $G$ using this multiplication is a $K(\mathbb{Z},3) \times K(\mathbb{Z},8)$. -$G$ has a unique H-space structure. Because Eilenberg-Mac Lane spaces represent cohomology, $[G \times G, G]$ is $H^2(G \times G) \times H^7(G \times G)$. A calculation with the Kunneth formula tells you that the restrictions to each factor makes the map -$$[G \times G, G] \to [G,G] \times [G,G]$$ -into an isomorphism. Thus, up to homotopy there is a unique binary multiplication that's homotopy unital; the group multiplication on $G$ is a representative. -As a result, if I construct any other $A_\infty$-structure on $G$, the binary multiplication is homotopic to the multiplication of $G$. Since $A_\infty$-structures are insensitive to such a change by homotopy (yes, this is handwaving) I could equally well assume that the $A_\infty$-structure starts with the multiplication of $G$. Because $G$ is abelian, this satisfies the interchange law. -Since $A_\infty$-structures pass along homotopy equivalences, it suffices for me to find a space $Y$ and a homotopy equivalence $G \to \Omega Y$ where $Y$ is not $K(\mathbb{Z},3) \times K(\mathbb{Z},8)$. -There is a map $f: K(\mathbb{Z},3) \to K(\mathbb{Z},9)$ classifying the cohomology operation $x \mapsto x^3$ (which is nontrivial, and 2-torsion). I'm going to let $Y$ be the homotopy fiber of $f$. Then $\Omega Y$ is the homotopy fiber of a map $\Omega f: K(\mathbb{Z},2) \to K(\mathbb{Z},8)$ which is 2-torsion, and hence $\Omega f$ is trivial. Therefore $\Omega Y$ is homotopy equivalent to the homotopy fiber of the trivial map, and is a product homotopy equivalent to $G$. -The map $f$ is the first k-invariant in the Postnikov tower of $Y$, and is nontrivial, so $Y$ is not a product of Eilenberg-Mac Lane spaces. - -An exercise for the reader who is not in bed right now, for whatever reason, is to trace through why several simpler choices of cohomology operation (e.g. $Sq^2: H^2 \to H^4$) don't make this argument work. - -Below follows my original post. I assumed that the interchange law held for the entire $A_\infty$-structure, meaning that for any of the $n$-ary operations $m_n$ we would have -$$ -m_n(x_1 \ast y_1, \ldots, x_n \ast y_n) = m_n(x_1, \ldots, x_n) \ast m_n(y_1, \ldots, y_n). -$$ -However, the question only says that the H-space structure itself satisfies the interchange law, and not the higher coherences. -The Eckmann-Hilton condition implies that you can take classifying spaces twice, and the order of taking classifying spaces doesn't matter. You can write $BBG$ for this iterated classifying space. Both $B(G)$ and $B(G_*)$ are grouplike, and so the group completion theorem implies that the natural maps $B(G) \to \Omega BBG$ and $B(G_*) \to \Omega BBG$ are weak equivalences. Thus, under the standard types of point-set assumptions on $G$, both classifying spaces are weakly equivalent. -In addition, this implies that, because $G$ is grouplike, it is weakly equivalent to the 2-fold loop space $\Omega^2 BBG$ with either H-space structure, and hence both multiplications are homotopy commutative. -This is related to work of Brun, Fiedorowicz, and Vogt in "On the multiplicative structure of topological Hochschild homology." (preprint version here) They show that the "tensor product" of the associative operad with the little $n$-cubes operad is weakly equivalent to the little $(n+1)$-cubes operad. This means that having an associative multiplication and an $A_\infty$-multiplication is, up to homotopy equivalence, basically the same as having an $E_2$-multiplication.<|endoftext|> -TITLE: Random real forcing -QUESTION [5 upvotes]: Let $\kappa$ be an infinite cardinal and let $B$ be the random forcing for adding $\kappa$-many random reals. -Question: What are the elements of $B$. More precisely given a condition $p \in B$, what are the domain and the range of $p$, if there are any?. What does it mean "a coordinate of $p$"?. - -REPLY [4 votes]: There are different ways of representing $B$. One possibility is to let $B$ consist of all closed subsets of $2^\kappa$ of positive measure; another possibility is to allow all positive Borel sets. Two conditions are equivalent if their symmetric difference has measure zero. -Each basic clopen set has a finite set of coordinates. By ccc, each open set is almost the same as some open set that uses only countably many coordinates. So every closed/Borel set $p$ is equivalent to a set that uses only countably many coordinates. I guess that these are the "coordinates of $p$".<|endoftext|> -TITLE: What element of $\pi_2(S^2)$ do rational functions represent? -QUESTION [7 upvotes]: Take a rational function of a single complex variable. View it as a continuous function from the Riemann sphere to itself. Is there a nice way to compute which element of $\pi_2(S^2)$ this corresponds to? - -REPLY [16 votes]: It's the cardinality of the preimage of a generic point, because generically the local degree of a complex analytic function is always +1. If the rational function is $a(x)/b(x)$, then the number of solutions in $x$ to $a(x)/b(x) = y$ for generic $y$ is $\max(\deg a,\deg b)$.<|endoftext|> -TITLE: Exponential (or other) families of distributions on manifolds. -QUESTION [7 upvotes]: The exponential family is a general parametrized class of probability distributions on $R^n$ that has many nice properties (ML estimation among them) and includes most of the "standard" distributions one encounters (Gaussian, multinomial, exponential, $\chi^2$ etc). -Are there similarly well-defined parametrized families of distributions for manifold-valued random variables ? Specifically, if you have a general Riemannian manifold ? Or asked another way, is there an equivalent notion of an "exponential family" for a Riemannian manifold ? - -REPLY [4 votes]: There are several subtle aspects to these questions that I am sure have emerged for you since you first posted this question. The primary one is of course that probability spaces are assumed to have algebraic structures so that moments and other elements for statistical investigation of a random variable can be found -- or at least defined -- through direct analytic methods. -For example, the kth moment of a random variable with values in $(0,1)$ and probability distribution $f(x)$ is $\int_0^1 x^k f(x) dx.$ Note that you need to be able to multiply the values of the random variables to themselves and a real-valued function ($f$). Even multivariate statistics (which lacks multiplication) utilizes the vector-algebraic structure of $R^N$ to calculate moments etc. -All of this is lacking on a general manifold. Embedding the manifold into Euclidean space will not help, since you would typically get expected values, etc. living outside of the manifold, although it would be rather entertaining to tell someone that the expected location of the crash site of a satellite with decaying orbit is the center of the Earth. -Anyway, that's a long-winded introduction to point out that you should check out the following papers: -Pennec, X. "Probabilities and statistics on Riemannian manifolds: a geometric approach" -http://hal.inria.fr/inria-00071490/ -Pennec, X. "Probabilities and statistics on Riemannian manifolds: Basic tools for geometric measurements." IEEE Workshop on Nonlinear Signal and Image Processing. Vol. 4. 1999. -Bochner in his text ${\it Harmonic\ Analysis\ and\ the\ Theory\ of\ Probability}$ describes stable distributions for general probability spaces, which is picked up for hyperbolic spaces in Getoor, R. K. "Infinitely divisible probabilities on the hyperbolic plane." Pacific Journal of Mathematics 11.4 (1961): 1287-1308.<|endoftext|> -TITLE: Generalising Gelfand's spectral theory -QUESTION [7 upvotes]: This is primarily a request for references and advices. - -Question (edited on 10/29/2011). What's known about comprehensive - generalisations of Gelfand's spectral - theory for unital [associative] normed - algebras [over the real or complex field] (*)? - -Here, a generalisation should be meant as a framework, say, with the following distinctive features (among the others): - -It should be founded on somehow different bases than the classical theory - especially to the extent that the notion itself of spectrum isn't any longer defined in terms of, and cannot be reduced to, the existence of any inverse in some unital algebra. -It should recover (at least basic) notions and results from the classical theory for unital Banach algebras in some appropriate incarnation (more details on this point are given below), for which the "generalised spectrum" does reproduce the classical one. -It should be unsensitive to completeness [under suitable mild hypotheses] in any setting where completeness is a well-defined notion (**), so yielding as a particular outcome that an element in a unital normed algebra, $\mathfrak{A}$, shares the same spectrum as its image in the Banach completion of $\mathfrak{A}$. - -(*) If useful to know, my absolute reference here is the (let me say) wonderful book by Charles E. Rickart: General Theory of Banach Algebras (Academic Press, 1970). -(**) At least in principle, the kind of generalisation that I've in mind is tailored on the properties of topological vector spaces, though I've worked it out only in the restricted case of normed spaces. - -Historical background. -As acknowledged by Jean Dieudonné in his History of Functional Analysis, the notion of spectrum (along with the foundation of modern spectral theory) was first introduced by David Hilbert in a series of articles inspired by Fredholm's celebrated work on integral equations (the word spectrum seems to have been lent by Hilbert from an 1897 article by Wilhelm Wirtinger) in the effort of lifting properties and notions from matrix theory to the broader (and more abstract) framework of linear operators. Especially, this led Hilbert to the discovery of complete inner product spaces (what we call, today, Hilbert spaces just in his honour). In 1906, Hilbert himself extended his previous analysis and discovered the continuous spectrum (already present but not fully recognised in earlier work by George William Hill in connection with his own study of periodic Sturm-Liouville equations). -A few years later, Frigyes Riesz introduced the concept of an algebra of operators in a series of articles culminating in a 1913 book, where Riesz studied, among the other things, the algebra of bounded operators on the separable Hilbert space. In 1916 Riesz himself created the theory of what we call nowadays compact operators. Riesz's spectral theorem was the basis for the definitive discovery of the spectral theorem of self-adjoint (and more generally normal) operators, which was simultaneously accomplished by Marshall Stone and John von Neumann in 1929-1932. -The year 1932 is another important date in this story, as it saw the publication of the very first monography on operator theory, by Stefan Banach. The systematic work of Banach gave new impulse to the development of the field and almost surely influenced the later work of von Neumann on the theory of operator algebras (developed, partly with Francis Joseph Murray, in a series of articles starting from 1935). Then came the seminal work of Israil Gelfand (partly in collaboration with Georgi E. Shilov and Mark Naimark), who introduced Banach algebras (under the naming of normed rings) and elaborated the corresponding notion of spectrum starting with a 1941 article in Matematicheskii Sbornik. -Now, it is undoubtable that Gelfand's work has deeply influenced the subsequent developments of spectral theory (and, accordingly, functional analysis). Yet, as far as I can understand in my own small way, something is still missing. I mean, something which may still be done, on the one hand, to clean up some inherent "defects" (or better fragilities) of the classical theory and, on the other, to make it more abstract and, then, portable to different contexts. -Naïve stuff. -As I learned from an anonymous user on MO (here), the term spectrum, in operator theory as well as in the context of normed algebras, is seemingly derived from the Latin verb spècere ("to see"), from which the root spec- of the Latin word spectrum ("something that appears, that manifests itself, vision"). Furthermore, the suffix -trum in spec-trum may come from the Latin verb instruo (like in the English word "instrument", which follows in turn the Latin noun instrumentum). So, the classical (or, herein, Gelfand) spectrum may be really considered, even from an etymological perspective, as a tool to inspect (or get improved knowledge of) some properties. I like to think of it as a sort of magnifying glass; we can move it through the algebra, zoom in and out on its elements, and get local information about them and/or global information about the whole structure. -Now, taking in mind (some parts of) another thread on this board about "wrong" definitions in mathematics, we are likely to agree that the worth of a notion is also measured by its sharpness (let me be vague on this point for the moment). And the classical notion of spectrum is, in fact, so successful because it is sharp in an appropriate sense, to the extent that it reveals deep underlying connections, say, between the algebraic and topological structures of a complicated object such as a Banach algebra (which is definitely magic, at least in my view). On another hand, what struck my curiosity is the consideration that the same conclusion doesn't hold (not at least with the same consistency) if Banach algebras are replaced by arbitrary (i.e. possibly incomplete) normed algebras, where the spectrum of a given element, $\mathfrak{a}$, can be scattered through the whole complex plane (in the complete case, as it is well-known, the spectrum is bounded by the norm of $a$, and indeed compact). So the question is: Why does this happen? And my answer is: essentially because the classical notion of spectrum is too algebraic, though completeness can actually conceal its true nature and make us even forgetful of it, or at least convinced that it must not be really so algebraic (despite of its own definition!) if it can dialogue so well with the topological structure. Yes, any normed algebra can be isometrically embedded (as a dense subalgebra) into a Banach one, but I don't think this makes a difference in what I'm trying to say, and it does not seriously explain anything. Clearly enough, the problem stems from the general failure in the convergence of the Neumann series $\sum_{n=0}^\infty (k^{-1}\mathfrak{a})^n$ for $k$ an arbitrary scalar with modulus greater than the norm of $\mathfrak{a}$. And why this? Because the convergence of such a Neumann series follows from the cauchyness of its partial sums, which is not a sufficient condition to convergence as far as the algebra is incomplete. According to my humble opinion, this is something like a "bug" in the classical vision, but above all an opportunity for getting a better understanding of some facts. -Motivations. -In the end, my motivation for this long post is that I've seemingly developed (the basics of) something resembling a spectral theory for linear (possibly unbounded) operators between different normed spaces. To me, this stuff looks like a sharpening of the classical theory in that it removes some of its "defects" (including the one addressed above); and also as an abstraction since, on the one hand, it puts standard notions from the operator setting (such as the ones of eigenvalue, continuous spectrum, and approximate spectrum) on a somehow different ground (so possibly foreshadowing further generalisations) and, on the other, it recovers familiar results (such as the closeness, the boundedness, and the compactness of the spectrum as well as the fact that all the points in the boundary are approximate eigenvalues) as a special case (while revealing some (unexpected?) dependencies). -Then, I'd really like to know what has been already done in these respects before putting everything in an appropriate form, submitting the results to any reasonable journal, and being answered, possibly after several months, that I've just reinvented the wheel. It would be really frustrating... Yes, of course, I've already asked here around (in Paris), but I've got nothing more concrete than contrasting (i.e. negative and positive) feelings. Also, I was suggested to contact a few people, and I've done it with one of them some weeks ago (sending him something like a ten page summary after checking his availability by an earlier email), but I've got no reply so far and indeed he seems to have disappeared... Then, I resolved to come here and consult the "oracle of MO" (as I enjoy calling this astounding place). :-) -Thank you in advance for any help. - -REPLY [3 votes]: There exists an approach by Berezansky and Brasche to define a kind of generalized selfadjointness and develop some eigenfunction expansions for operators between different Hilbert spaces. This is done within the theory of rigged Hilbert spaces. Their motivation was apparently different from yours - they wanted to cover the case of Schroedinger operators with distribution potentials. Nevertheless there can be some connection. See -Yu.M. Berezansky and J. Brasche, Generalized selfadjoint operators and their singular perturbations. Methods Funct. Anal. Topol. 8, No. 4, 1-14 (2002); -Yu.M. Berezansky, J. Brasche, and L. P. Nizhnik, On generalized selfadjoint operators on scales of Hilbert spaces, Methods Funct. Anal. Topol. 17, No. 3, 193-196 (2011). -The second paper is available online: http://www.imath.kiev.ua/~mfat/<|endoftext|> -TITLE: Quadratic extension of quadratic extension -QUESTION [6 upvotes]: Let $K_1$ be a perfect field. Let $K_2/K_1$ and $K_3/K_2$ be quadratic extensions. Let $K_4/K_3$ be the Galois closure of $K_3$ over $K_1$. Is it true that either $K_3 = K_4$ or $K_4/K_3$ is quadratic such that the Galois group of $K_4$ over $K_1$ is isomorphic to $\mathcal{D}_4$, the dihedral group with 8 elements? - -REPLY [10 votes]: Yes. Let $a_1$ be a generator of $K_3$ over $K_1$ and let $a_2$, $a_3$, and $a_4$ be the other three roots of the polynomial of $a_1$. Then say that $a_2$ is the other root of the polynomial of $a_1$ over $K_2$. Then The Galois group of $K_4/K_1$ acts on the partitioned set $\{\{a_1,a_2\},\{a_3,a_4\}\}$, and is therefore a transitive subgroup of $D_4$, either $D_4$ itself or $C_4$ or $C_2 \times C_2$. -I think that the following is a more general fact that can be proven in the same way. If $K_2/K_1$ is separable and its Galois closure has Galois group $G$, and if $K_3/K_2$ is separable and its Galois closure has Galois group $H$, then the Galois group of the Galois closure of $K_3/K_1$ is a subgroup of a wreath product of $G$ and $H$. If you take a longer chain of extensions you get an iterated wreath product.<|endoftext|> -TITLE: Question about 0-dimensional Polish spaces -QUESTION [6 upvotes]: Hello everybody, -I'm stuck with proving (or disproving) the following statement. -Statement: -For every $0$-dimensional Polish space $(X,\mathcal{T}\ )$, and a countable basis of clopen sets $\mathcal{B}$ for $\mathcal{T}$, every open set is the disjoint union of clopen sets in $\mathcal{B}$. -Every open set is the union $O=\cup_{n}B^{0}_{n}$ of the basic clopen sets contained in it (say ordered with a given numbering of $\mathcal{B}$). The idea is to make it a disjoint union by considering, iteratively, -$O= B^{0}_{1} \cup O^{1}$ -where $O^{1} = O\setminus B^{0}_{1}$, which is open. Then again we have -$O^{1}=\cup_{n}B^{1}_{n}$. -So consider $O^{2}= O^{1}\setminus B^{1}_{1}$ -etcetera. The resulting union -$\cup_{m} B^{m}_{1}$ -is open. However, I'm stuck in proving that, in general, a point $x\in O$ ends up necessarily in some $B^{k}_{1}$, for $k\in \mathbb{N}$, i.e., I can't prove that -$O= \cup_{m}B^{m}_{1}$. -Googling around I found this interesting paper [1]. The author says that it is a known fact (unfortunately he doesn't give a reference) that for every $0$-dimensional Polish space, the Borel sets are generated from the clopens by closing under countable disjoint unions and complements. This does not solve my problem, but still, I would be interested in reading a proof. Could you point me to some relevant literature? -Thanks in advance, -[1] Abhijit Dasgupta. Constructing $\Delta^{0}_{3}$ using topologically restrictive countable disjoint unions. - -REPLY [2 votes]: If you have such a union of $O = \cup_n B_n$, consider the sets $B'_n = B_n \setminus \cup_{i=0}^{n-1} B_i$, where $B'_0 = B_0$. Show these sets are disjoint, clopen, and have the same union as the original sets, as for every $x \in O$ there is a first index $n(x)$ such that $x \in B_{n(x)}$.<|endoftext|> -TITLE: Repeatedly indexing into an $\infty$-sequence of integers -QUESTION [6 upvotes]: Suppose one has in hand an infinite sequence $s$ of distinct natural numbers, -for example, -$$s=s_1=(1, 3, 5, 7, 9, 11, 13, 15, 17, 19,\ldots) \;.$$ -So this sequence can be considered an injection -$f: \mathbb{N} \mapsto \mathbb{N}$. -Now replace $s_1$ with $s_2$ by indexing in $s_1$ using $s_1$: -$$s_2=(1, 5, 9, 13, 17, 21, 25, 29, 33, 37,\ldots) \;.$$ -So we take the 1st, 3rd, 5th, ... elements of $s_1$ to form $s_2$. -To construct $s_3$, index into $s_2$ using $s_2$: -take the 1st, 5th, 9th, ... elements of $s_2$, i.e., -$$s_3=(1, 17, 33, 49, 65, 81, 97, 113, 129, 145,\ldots) \;.$$ -Note that, e.g., the 2nd element of $s_3$ is not $f^3(2) = 9$, -but rather $f^2( f^2( 2)) = f^4(2) = 17$. -Iterating once more we reach -$$s_4=(1, 257, 513, 769, 1025, 1281, 1537, 1793, 2049, 2305,\ldots)\;. $$ -Here, e.g., the 2nd element of $s_4$ is $f^8(2)=256 \cdot 2 - 255 = 257$. -Several questions: -Q1. -For which starting sequences $s$ does this process lead to a fixed -sequence, $s_k = s_{k+1}$? Certainly it does if $s$ represents -the identity: $s_1=(1,2,3,4,5,\ldots)$. Are there any other -fixed sequences? -Q2. -For which starting sequences $s$ does this process lead to a cycle -among the sequences, $s_k = s_{k+m}$, $m>1$? And can the length of such a cycle be predicted from -the structure of the starting sequence? -Q3. What is the expected behavior under iteration of a "typical" (random?) starting injection $s$, -under any reasonable sense of "typical"? -I feel certain this has all been studied before, and I am just not phrasing it in an easily -recognizable manner. I would appreciate pointers—Thanks! - -REPLY [3 votes]: Q1. If the sequence is strictly increasing, there will be neither nontrivial fixed sequences nor cycles. Each sequence will increase faster than the next. -In general, the equation is obviously f(f(n))=f(n). To construct a sequence satisfying that equation, divide the natural numbers into a some number of equivalence classes, then map each class to one element of itself. -Q2. To solve the equation $f^{2^k}(n)=f(n)$, first divide the natural numbers into equivalence classes, then group the equivalence classes into cycles of order dividing $2^k-1$, then map each class to an element of the next class in its cycle. -Q3. In general, some registers of the sequence will stay fixed, some will stay in cycles, and some will go off to infinity. The proportions of each depend on the exact random distribution used to construct the sequence. If each number of the sequence is independent, the paths of the different registers will be relatively independent, interfering only if they intersect. -I think it probably makes more sense to study $f^1(n),f^2(n),f^3(n),\dots$ of which your sequences just form the power-of-$2$ component. - -REPLY [2 votes]: Suppose a sequence $s_k = (a_1, a_2, \ldots)$ has $s_k = s_{k+1}$. Then $a_i = a_{a_i}$, and by injectivity of $f$ (and hence iterations of $f$), we have $a_i = i$. So there is just one fixed sequence. -Here are a family of sequences where $s_1 = s_{2n + 1}$. Only the first $2^n + 1$ terms are permuted, and the rest of the sequence is a copy of the identity. If any number $a_i = i$ at some point during the iteration of sequences, then it will remain stationary forever, so we need only consider the first $2^n + 1$ terms. -$$ -s_1 = (2^n + 1, 1, 2, 3, \ldots, 2^n) $$ $$ -s_2 = (2^n, 2^n + 1, 1, \ldots, 2^n - 1) $$ $$ -s_3 = (2^n - 2, 2^n - 1, 2^n, 2^n + 1, 1, \ldots, 2^n - 3) $$ $$ -s_4 = (2^n - 6, 2^n - 5, 2^n - 4 , 2^n - 3 , 2^n - 2, 2^n - 1, 2^n, 2^n + 1, 1, \ldots, 2^n - 7)$$ $$ -\cdots $$ $$ -s_{n+1} = (2, 3, \ldots, 2^n + 1, 1) $$ $$ -s_{n+2} = (3, \ldots, 2^n + 1, 1, 2) $$ $$ -s_{n+3} = (5, \ldots, 2^n + 1, 1, 2, 3, 4) $$ $$ -\cdots $$ $$ -s_{2n+1} = (2^n + 1, 1, 2, 3, \ldots, 2^n) $$ -Doing the same thing beginning with $2^n$ instead of $2^n + 1$ leads to the identity. Doing it with $2^n + k$ where $0 < k < 2^n$ leads to a cycle with some term $s = (2^n + 1, 2^n + 2, \ldots, 2^n + k, 1, 2, \ldots, 2^n)$. -If there is some simple bounding condition as follows, then iterations will cycle locally, but may not globally. -Suppose there is a sequence of natural numbers $0 = m_0 < m_1 < m_2 < \cdots$ such that for every $j$, $a_i \leq m_j$ for all $i \leq m_j$. Then the first $m_1$ terms will remain in the first $m_1$ places, the next $m_2 - m_1$ terms will remain in the next $m_2 - m_1$ places, etc. Since there are only $(m_{j+1} - m_j)!$ permutations of each of these sets, they must all eventually be cyclic individually. But we can choose these cycles to have no common multiple, e.g. take $m_{i+1} - m_i = 2^i + 1$, and repeat the construction above.<|endoftext|> -TITLE: Number of distinct values taken by x^x^...^x with parentheses inserted in all possible ways -QUESTION [19 upvotes]: For what positive x's the number of distinct values taken by x^x^...^x with parentheses inserted in all possible ways is not represented by the sequence A000081? Is it exactly the set of positive algebraic numbers? Is it a superset of positive algebraic numbers? Is it countable? Is $2^{\sqrt 2}$ or $\log_2 3$ in the set? - -REPLY [15 votes]: The answer to the second question is "no". Consider the unique solution $x > 0$ to the equation $x^x = 3$. By the Gelfond-Schneider theorem, this number is transcendental. But we have -$$((x^x)^x)^x = x^{x^3} = x^{(x^{(x^x)})}$$ -so that two of the parenthesizations coincide. So evidently this set contains transcendental numbers. Lots of other solutions can be similarly generated (e.g., solve $x^{(x^x)} = 4$).<|endoftext|> -TITLE: Some questions from the paper "Forcing the failure of CH by adding a real" by Shelah and Woodin -QUESTION [5 upvotes]: 1-Let $P=Add(\omega_1, \kappa)$, and let $D$ be a dense open subset of $P$. Then there is a dense subset $S$ of $D$ such that for every $f \in S$ and any $g \in P$, if $domf=domg$ and the set $\{ \beta: f(\beta) \neq g(\beta) \}$ is finite then $g \in D$. -Why is this true? -2-Assume $0^{\sharp}$ exists, and let $\lambda \geq \aleph_2^{L[0^{\sharp}]}.$ In $L$ let $P$ be the Easton support product of the forcing notions $Add(\delta^{++}, 1)$ where $\delta < \lambda$ is a limit cardinal in $L$, and for $p \in Add(\delta^{++}, 1)$ we require $domp$ be a subset of $(\delta, \delta^{++})$. Let $F_{0}$ from the union of the intervals $(\delta, (\delta^{++})^{L})$ where $\delta$ is a limit cardinal in $L$ into $2$ be the resulting generic function. Let $I$ be the class of silver indiscernibles and for $\delta \in I$ let $\langle \alpha_{n}^{\delta}: n < \omega \rangle \in L[0^{\sharp}]$ be a cofinal sequence through $(\delta^{++})^{L}.$ Also let $\langle r_{\alpha}:\alpha < \lambda \rangle$ be a sequence of Cohen reals generic over $L[0^{\sharp}]$. Define $F_{0}^{*}$ by the same domain as $F_{0}$ by: -$F_{0}^{*}(\beta)=r_{\delta}(n)$ , if $\beta=\alpha_{n}^{\delta}, \delta \in I$ -$F_{0}^{*}(\beta)=F_{0}(\beta)$ otherwise -Show that $F_{0}^{*}$ is $P-$generic over $L$. - -REPLY [8 votes]: Let me answer question 1. Conditions in $P$ are partial -functions $p$ from $\omega_1\times\kappa\to 2$, with -countable domain, ordered by inclusion. For any condition -$p$, since $\text{dom}(p)$ is countable, there are only countably -many finite modifications of $p$ on this domain. For each -such finite modification $p^\ast$, there is a stronger -condition $q^\ast\leq p^\ast$ with $q^\ast\in D$, and hence -a corresponding $q\leq p$ such that the same modification -to $q\mapsto q^\ast$ places it into $D$. So we may build a -descending sequence of conditions $p_0\geq p_1\geq -p_2\geq\cdots$ such that at each step, moving from $p_n$ to -$p_{n+1}$, we handle one finite modification of $p_n$ to -$p_n^\ast$, and extend $p_n$ to $p_{n+1}$ such that the -corresponding finite modification $p_{n+1}^\ast$ is in $D$. -Since the domains are countable, we can arrange by suitable -bookkeeping to handle all the finite modifications that -arise (details: for example, we could handle the $k$-th -modification to $p_r$ at stage $\langle k,r\rangle$, using -a pairing function). By the countable closure of the -forcing, the union $p_\omega=\bigcup_n p_n$ is a condition, -and this condition has the property that for every finite -modification $p_\omega^\ast$ of it, there is some $p_n$ -supporting this modification, and we already arranged that -$p_{n+1}^\ast\in D$, and since $D$ is open this means -$p_\omega^\ast\in D$ as well. Thus, all finite -modifications of $p_\omega$ land into $D$. So the set $S$ -of all such $p_\omega$ is a dense subset of $P$ with the -desired property. -For question 2, perhaps you can provide more details. I don't have the paper here.<|endoftext|> -TITLE: Why "adding" a single extender cannot give an L-like inner model for say, a strong cardinal? -QUESTION [11 upvotes]: The constructible universe $L$ is too thin for large cardinals greater than measurable. To build $L$-like inner models for large cardinal, it is natural to think about "adding" the evidences into the model. For example, $L[U]$ is an inner model for measurable cardinal. However, the situation becomes much more complicated beyond measurable. One reason is that $L[U]$ can contain only one measurable cardinal not two, which I knew. And I was told somewhere that -if $\kappa$ is a some strong cardinal witnessed by an/some elementary embedding(s) $j$, and $E$ is an extender generated from $j$, Then $L[E]=L[U]$ where $U=E_{\{\kappa\}}$. -Is the statement true? If true, how to prove it? If not, how to argue that it is necessary to develop much more complicated technique to build inner model for larger cardinals? - -REPLY [6 votes]: A fairly complete answer to this question appears in Woodin's "In search of Ultimate $L$" at the beginning of the section on Martin-Steel extender sequences. Woodin omits the proofs, so I'll fill in some of the details. -If $E$ is a short extender, then $L[E] = L[U]$ where $U$ is the normal measure of $E$. To see this, let $j : V\to M$ be the ultrapower of $V$ by $E$. Kunen's analysis shows that $j\restriction V_{\kappa+1}\cap L[U]$ coincides with the unique iterated ultrapower embedding $i : L[U]\to L[j_E(U)]$. (Every element of $V_{\kappa+1}\cap L[U]$ is $\Sigma_2$-definable in $L[U]$ from parameters in $\kappa\cup \Gamma$ where $\Gamma$ is the class of common fixed points of $i$ and $j$; since $i$ and $j$ are elementary and agree on these parameters, they agree on $V_{\kappa+1}\cap L[U]$.) Since $i$ is definable over $L[U]$, $E\cap L[U] = \{(a,X)\in L[U] : X\subseteq [\kappa]^{<\omega}, a\in i(X)\}$ is in $L[U]$. Since $L[E]$ is the minimum proper class model $N$ of ZFC such that $E\cap N \in N$, $L[E]\subseteq L[U]$. The reverse inclusion is obvious. -If $E$ is a long extender, then $L[E]$ can be larger than $L[U]$. For example, if $U$ and $W$ are measures on distinct measurable cardinals and $E$ is the extender of $j_W\circ j_U$, then $L[E] = L[U,W]$ has two measurable cardinals. Woodin's Theorem 4.20 asserts that if $E$ is a nice enough long extender (that is not "too long"), then $L[E]$ will not contain an inner model with a Woodin cardinal. This is probably pretty hard. Woodin also claims that dropping the niceness condition, consistently, one can code arbitrary sets into long extenders, so these do not have an inner model theory. There are still some open questions here, but these results together argue that inner model theory needs more than one extender.<|endoftext|> -TITLE: Vector bundles on graphs -QUESTION [6 upvotes]: Vector bundles over manifolds have fundamental importance in differential geometry, algebraic topology etc. Are there any applications of this concept (or some variation of it) for graphs (finite or infinite)? -The only place I have seen something like this is in a paper on spanning forests by Kenyon, where the application seems somewhat specialized. - -REPLY [3 votes]: Another cool paper is -http://arxiv.org/abs/1107.5588 -Where they talk about line bundles on graphs (and define very nice new class of integrable systems).<|endoftext|> -TITLE: Is there a "knot theory" for graphs? -QUESTION [31 upvotes]: I think knot theory has been studied for quite a while (like a century or so), so I'm just wondering whether there is a "knot theory" for graphs, i.e. the study of (topological properties of) embeddings of graphs into R^3 or S^3. -If yes, can anyone show me any reference? -If the answer is basically no, then why? Is it just too hard, uninteresting, or can it be essentially reduced to the study of knots (and links)? - -REPLY [5 votes]: It is somewhat surprising (to me) that what to me seems the simplest nontrivial example of theorems exactly fitting the question in the OP has not yet been mentioned in this thread: the embeddings of the Möbius ladders, which are finite simple undirected graphs, into $\mathbb{R}^3$. -This is an answer to - -whether there is a "knot theory" for graphs, i.e. the study of (topological properties of) embeddings of graphs into R^3 or S^3. - If yes, can anyone show me any reference? - -at least in the sense that in the very interesting article Erica Flapan: The Symmetries of the Möbius Ladder. Math. Ann. 283, 271-283 (1989), which I think, could be given a fruitful revival from a point of view of constructive mathematics (e.g., how much of Flapan's proofs/theorems can be done constructively?), the following was done, inter alia (there is more in Flapan's paper): - -in Section 1 of loc. cit. a proof is given that there exists a graph $G$ and an automorphism $\alpha$ of $G$ as an abstract graph, such that there does not exist any embedding of (the geometric realization $\lvert G\rvert$ of ) $G$ into $\mathbb{R}^3$ such that $\alpha$ could be realized by an element of the group of leaving-$\lvert G\rvert$-invariant-as-a-point-set diffeomorphisms of $\mathbb{R}^3$. Note that in this rendition I rendered the author's "group of homeomorphisms of $G$ up to isotopy" into what to me evidently seems equivalent and clearer "automorphism of $G$ as an abstract graph". The example graph used by the author is $G:=$complete graph with six vertices. -in Section 2 of loc. cit., first a proof in classical logic is given that for any embedding $\eta\colon M_r\to S^3$ of the $r$-rung Möbius ladder $M_r$ into the $3$-sphere $\mathbb{S}^3$, any orientation-reversing diffeomorphism $\varphi\colon S^3\to S^3$ has the property that if it fixes1 $\mathrm{im}(\eta)$ as a pointset, then $\varphi$ does not fix at least $r-2$ of the $r$ rungs of $M_r$. This is, partly, an interesting counterfactural (since the author later gives a proof that for an odd number of rungs, such $\varphi$ are impossible): roughly, if there is a symmetry of the Euclidean-space-embedded Möbius ladder, then it must needs jumble almost all the ladder's rungs. This implication is then put to use to give a proof that for an odd number of $r$ungs, such a $\varphi$ is impossible. Roughly: you cannot reverse the orientation of an embedded odd-rung-number Möbius ladder graph by a Euclidean symmetry. The author then gives an example that for even number of rungs, such isometries do exist. I perceived this to be a result which is very relevant to the OP; it in particular is an interaction between a combinatorial property of the abstract graph (number of rungs) and a concept studied in knot theory: slightly vaguely, one could say: Flapan gave a proof that for rung-numbers $\geq 4$, each odd-rung Möbius ladder is non-amphichiral, while each even-rung Möbius ladder is amphichiral. The smallest example of the latter ladders is the embedding represented by the illustration - - -on p. 272 of loc. cit., when thought of to have precisely four rungs, about which loc. cit. says that it is amphichiral, while if the picture is thought of to represent a Möbius ladder with five rungs, then loc. cit. says that there does not exist any orientation-reversing self-diffeomorphismof $\mathbb{R}^3$ which would fix it as a point set. -(The latter is a precise and usual statement of the matter. A more vague alternative statement one often encountered is 'It is not equivalent to its mirror-image.' wherein 'mirror-image' is either left undefined, or is defined via orientation-reversing self-homeomorphism of $\mathbb{R}^3$, whereupon the alternative statement becomes at least not simpler.) -The amphichirality of the even-rung Möbius ladders is easy to see, the substance of Flapan's results is non-amphichirality of the odd-rung Möbius ladders with at least 5 rungs. -The author on p. 272 of loc. cit. writes - -The property of topological achirality for graphs is analogous to the property of amphicheirality for knots. - - -in Section 3 of loc. cit., the focus shifts from (non-)existence of orientation-reversing Euclidean symmetries of embedded Möbius ladders, to properties of orientation-preserving such symmetries. The emphasis is on a difference between embeddings into $\mathbb{R}^3$ and embeddings into $S^3$. - -This difference is illustrated by the author using the following illustration on p. 278 of loc. cit. - -which is a an example of what the OP is asking for: this meant to represent a non-knot embedded into $S^3$, namely the abstract undirected simple graph $M_3$, the three-rung Möbius ladder. -Furthermore, one should mention that there is a very recent preprint - -Erica Flapan, Thomas Mattman, Blake Mellor, Ramin Naimi, Ryo Nikkuni: Recent Developments in Spatial Graph Theory. arXiv:1602.08122v2 [math.GT] - -which is relevant to the OP. Therein, more results on embedded Möbius ladders are summarized, in particular the new paper (published after the OP) - -E. Flapan and E. Davie Lawrence, Topological symmetry groups of Möbius ladders, J. Knot Theory Ramifications, vol. 23, no. 14, (2014) - -Remarks. -1 Note that in loc. cit. there is a slightly stronger hypothesis than mere fixing the image of the embedding in its entirety. This hypothesis is essential in the special case $r=3$, yet can be left out for all $r\geq 4$, as loc. cit. itself points out on p. 272. -2 The rung-number $r=3$ is an exceptional case. The $3$-rung Möbius ladder is isomorphic to the complete graph $K^{3,3}$, and its geometric realization in $\mathbb{R}^3$ happens to be amphichiral: it is evident that the geometric graph $G$ represented by the illustration - -in loc. cit. is isotopic to its 'mirror-image'.<|endoftext|> -TITLE: Does every finite nontrivial group have two distinct irreducible representations over the complex numbers of equal degree? -QUESTION [18 upvotes]: Is it true that for any finite nontrivial group G, there exist two inequivalent irreducible representations of G over the complex numbers that have the same degree. -If so, is there an easy proof? If not, what is the smallest counterexample? -Note: Any counterexample group must be perfect, because if the abelianization is nontrivial, we get multiple irreducible representations of degree one. [EDIT: Further, as Colin Reid notes in the comment, a minimal counterexample must be a simple (non-abelian) group]. This whittles down our search considerably. The general expressions for the degrees of irreducible representations for the families of simple groups that I've checked suggests that there is plenty of repetition of degrees in these cases. - -REPLY [6 votes]: It has already been noted in earlier answers and comments that a finite non-trivial group of minimal order subject to having all complex irreducible characters of different degrees would have to be non-Abelian simple with all its characters rational-valued. -From the point of view of historical perspective, it is worth pointing out G. Seitz and W. Feit proved (using the classification of finite simple groups) that there only finitely many (in fact 5) possible non-Abelian composition factors, other than alternating groups, of a finite group with all characters rational valued. By a result of James and Kerber, for example, no alternating group has all its characters rational-valued. In fact, JG Thompson and I have given a precise description of the field generated by the character values of $A_{n}$ for every $n.$ -Hence, using these facts, the question asked here reduces to checking the character tables of 5 non-Abelian simple groups.<|endoftext|> -TITLE: Infinite index subgroup of polycyclic group has strictly lower Hirsch length? -QUESTION [7 upvotes]: This question has been frustrating me for a while, because I'm pretty sure the solution is simple and I'm just not seeing it. I have found the claim (e.g., in Segal's Polycyclic Groups) that if $G$ is polycyclic and $H\leq G$, then $h(G) = h(H)$ if and only if $[G:H]<\infty$. The "if" direction is trivial, but I'm having trouble with the "only if" direction. (In Segal, this is left as an exercise.) If $H$ is subnormal in $G$, then the claim is easy. But I don't think a subgroup of a polycyclic group is necessarily subnormal, and I haven't managed to prove the claim if $H$ is not subnormal. Can anybody point me in the right direction? - -REPLY [10 votes]: Consider the series $A_0 < A_1 < ... < A_n=G$ with cyclic factors. Suppose $G/A_{n-1}$ is infinite (if it is finite, replace $G$ by $A_{n-1}$). So $h(G)=h(A_{n-1})+1$. Let $H$ be your subgroup of $G$. Consider its intersection $H'=H\cap A_{n-1}$. Then $H'$ is normal in $H$ and $H/H'$ is cyclic. Assume $h(H)=h(G)$. Then $h(H')=h(A_{n-1})$. By induction, $H'$ is of finite index in $A_{n-1}$. Also note that $HA_{n-1}$ is of finite index in $G$ (since every non-trivial subgroup of a cyclic group is of finite index and $H$ is not inside $A_{n-1}$). We can as well assume that $HA_{n-1}=G$ (we can always replace $G$ by its finite index subgroup containing $H$). But the representatives of the cosets of $H'=H\cap A_{n-1}$ in $A_{n-1}$ are representatives of cosets of $H$ in $HA_{n-1}$, so $H$ has finite index in $G$.<|endoftext|> -TITLE: F.p. groups where all elements of the same order are conjugate -QUESTION [5 upvotes]: The question I want to ask is related to the Boone-Higman conjecture (see -Embedding in f.p. simple groups for the details). -We discussed recently with Ievgen Bondarenko this conjecture and he noticed that it would follow if the answer to the following question was "yes": -\textbf{Question: Does every f.p. group with soluble word problem embed in a f.p. group with all elements of the same order being conjugate?} -Does anybody know if that is "no" actually? Or, at least, is it "no" if we take out the condition of solubility of WP? -Could anybody suggest some references with examples of f.p. groups with elements of the same order being conjugate. Do there exist such groups where occur all possible orders of elements? - -REPLY [7 votes]: I think the answer to the main question is negative. That is, there exists a finitely presented group $G$ with decidable WP that does not embed into any f.p. group where all elements of the same order are conjugate. -Here is the construction. Let us start with a recursively presented group $S$ such that the WP in $S$ is decidable while the order problem (that is, given an element, find its order) is not. Such groups exist and, in fact, are not hard to construct. Then we take the direct product of $S$ and cyclic groups $\mathbb Z$ and $\mathbb Z/n\mathbb Z$ for all $n$. The resulting group, call it $R$, is recursively presented, has decidable WP, and has the following property: -(*) There is no algorithm that allows to decide for any given $x,y\in R$ whether $|x|=|y|$. -Indeed if such an algorithm existed, given $x\in S$ we could run countably many copies of this algorithm taking the generators of the cyclic subgroups $\mathbb Z$ and $\mathbb Z/n\mathbb Z$ as $y$'s. All these algorithms can be run simultaneously using the standard diagonal argument and one of them always stops, which gives the solution to the order problem in $S$. -Further let us embed $R$ into a finitely presented group $G$ with decidable WP. This is always possible since the Higman embedding preserves decidability of the WP. Suppose now that the group $G$ is embedded into a f.p. group $P$ where all elements of the same order are conjugate. We will derive a contradiction by showing that $P$ can be used to decide whether two elements of $R$ have the same order, thus violating (*). -Given two elements $x,y\in R$, we run two algorithms. -Algorithm A. Using the finite presentation of $P$, this algorithm produces the list of all elements of $P$ conjugate to $x$. If at some point we see $y$ in this list, the algorithm stops, otherwise it runs forever. -Algorithm B. This algorithm starts by solving the WP in $R$ for the elements $x,y, x^2, y^2, x^3, y^3 \ldots$. If all these elements are nontrivial, the algorithm does not stop. Otherwise let $x^n$ (or $y^n$) be the first trivial element in the list. We will only consider the case of $x^n$, the case of $y^n$ is symmetric. Then $|x|=n$. We can then use decidability of the WP in $R$ to effectively answer the question of whether $|y|=n$. Then the algorithm stops and returns the answer. -Observe now that at least one of the algorithms always stops. Indeed if $|x|=|y|$, then $x$ and $y$ are conjugate in $P$ and hence Algorithm A stops. Otherwise at least one of the orders $|x|$, $|y|$ is finite. Hence Algorithm B will stop and tell us if $|x|=|y|$. Thus we can decide whether $|x|=|y|$, which contradicts (*).<|endoftext|> -TITLE: What are the smooth manifolds in the topos of sheaves on a smooth manifold? -QUESTION [13 upvotes]: The category of internal locales in the Grothendieck topos of sheaves on a locale X -is equivalent to the slice category over X. -In other words, internal locales over X are precisely morphisms of locales of the form Y→X. -Is the same statement true for smooth manifolds instead of locales -and submersions instead of continuous maps? -More precisely, consider the site of smooth manifolds equipped with its standard -Grothendieck topology of surjective submersions. -Is the category of internal smooth manifolds in the Grothendieck topos of sheaves -on a smooth manifold M equivalent to the category whose objects are surjective -submersions of the form N→M and morphisms are commuting triangles? -Definitions tend to branch when we internalize them, -so an answer to this question should include the correct definition -of a smooth manifold in an arbitrary Grothendieck topos. -According to nLab, internal dualizable modules over the locale of real numbers -in the category of sheaves on a smooth manifold M -are precisely finite-dimensional vector bundles over M. -This can be seen as a further motivation for the above question. - -REPLY [4 votes]: See the book (a kind of culmination point of the theory: "synthetic differential geometry") -MR1083355 Moerdijk, Ieke; Reyes, Gonzalo E. Models for smooth infinitesimal analysis. Springer-Verlag, New York, 1991.<|endoftext|> -TITLE: Bijective function on a dense set -QUESTION [7 upvotes]: Suppose X is a complete metric space, and $f:X↦X$ a continuous surjective function. Let D be a dense set. Suppose $f:D↦D$ is injective and $f^{-1}(D)=D$. -Is $f$ injective ? -Is there a family of metric spaces where you can conclude $f$ is injective? - -REPLY [10 votes]: Sorry but I could not resist: -     -The map is defined on $[0,1]\times[0,1]$, and can be written as -$$ - f(x,y) = \begin{cases} - (x, (2-x)y) & \text{ if }y\leq 1/2, \\ - (x, xy+1-x) & \text{ if }y> 1/2. - \end{cases} -$$ -Just take $D$ to be the set of $(x,y)$ with rational coordinates in $(0,1)\times(0,1)$, then $f$ is bijective on $D$ (because it can be easily inverted), but it is clearly not injective on $[0,1]\times[0,1]$.<|endoftext|> -TITLE: Length of the last edge when visiting points by nearest neighbor order -QUESTION [16 upvotes]: Take $n$ points uniformly in $[0,1] \times [0,1]$. Then pick uniformly $X_0$ one of these points as your starting point. Then let $X_1$ be the nearest neighbor of $X_0$, let $X_2$ be the nearest neighbor (not yet visited) of $X_1$ and so on. What can be said of the asymptotic of $X_n-X_{n-1}$ the length of the last crossed edge? What about the length of the longest crossed edge? I stumbled upon these kind of models in the area of environmental statistics where one tries to find clusters in a geographical dataset but I am not sure if this question is interesting. -Edit : Many more questions could be asked about this model : If one makes the model dynamic by adding the points in $[0,1]^2$ sequentially, most of the time the addition of an extra point changes the path only locally, but from time to time it will have a big impact. Does the path converges locally? (I guess not). How often do you see catastrophic modifications of the paths? -Finally, is there a way to find an interesting local spectrum in this object by renormalizing it and looking at the sizes of the edge ? It is probably related to the local dimension of the counting measure of the uniform PPP on [0,1]^2. But as pointed out in the answer below the first step would be to obtain the asymptotic of $L_n$ the total length. - -REPLY [6 votes]: This is closely related to a nice open problem of David Aldous, from the list of open problems on his web site, some version of which in fact has quite a long history in the combinatorial optimization community. At the above link Aldous has references to existing knowledge about the problem. The state of the art is that the sum of all edge lengths is $O(n^{1/2})$ in expectation.<|endoftext|> -TITLE: Constructing the surreal numbers as iterated Hahn series -QUESTION [13 upvotes]: A theorem due to N. Alling (Foundations of Analysis over Surreal Number Fields, §6.55) states that the surreal numbers are isomorphic, as an ordered and valued field, to the field of Hahn series with real coefficients and value group the surreal numbers themselves. There is also a restricted version, which I'll refer to in order to avoid the (IMHO uninteresting) foundational difficulties in dealing with classes: if $\kappa$ is a regular uncountable cardinal, the set $\mathrm{No}_\kappa$ of surreal numbers with birth date $<\kappa$ is isomorphic to the field of Hahn series of length $<\kappa$ with real coefficients and exponents in $\mathrm{No}_\kappa$ itself (in the indeterminate $\frac{1}{\omega}$). -Upon reading this, I thought to myself, “well, this is nice, this means the surreal numbers can be given a construction as iterated Hahn series, something along the lines of: start with the reals, take the Hahn series over that, then take the Hahn series over that (as value group), repeat transfinitely, and voilà, surreal numbers”. Unfortunately, it seems I was being a bit naïve there. -Let us define $F_0 = \mathbb{R}$ and inductively $F_{\alpha+1}$ to be the field of Hahn series of length $<\kappa$ with real coefficients and exponents in $F_\alpha$ (the indeterminate being written $\frac{1}{\omega}$); and for $\delta$ a limit let $F_\delta = \bigcup_{\alpha<\delta} F_\alpha$ with the obvious embeddings. Then if I am not mistaken, $F_\kappa$ is indeed isomorphic to Hahn series of length $<\kappa$ over itself, it is indeed an $\eta_\xi$ field for $\kappa=\omega_\xi$, of cardinality $2^{<\kappa}$, just like $\mathrm{No}_\kappa$, and it is quite conceivable (I didn't check) that the two are isomorphic (as ordered—and valued—fields over $F_0 = \mathbb{R}$). But this can't possibly respect the map $x \mapsto \omega^x$ because in $\mathrm{No}_\kappa$ the latter has plenty of fixed points whereas in $F_\kappa$ it has none. So this construction is “wrong” in that it doesn't explain surreal numbers properly. -Thus, my question is: is there some variant of this construction that will succeed in constructing $\mathrm{No}_\kappa$, including its map $x \mapsto \omega^x$? Perhaps the answer depends on what is done at limit ordinal steps, but I'm rather confused so I wish someone could clear up the confusion. - -REPLY [2 votes]: I'm a novice in this area so I don't think my reply should be taken too seriously, but I recently read Alling's Foundations of Analysis over Surreal Number Fields and have been corresponding with Philip Ehrlich about this topic. I've seen Ehrlich's proof of the isomorphism between Conway Names and the Hahn Series, which I find particularly interesting. Ehrlich also, this year, published the document Conway names, the simplicity hierarchy -and the surreal number tree Journal of Logic & Analysis 3:1 (2011) 1–26. -What I want to draw your attention to is Conway's original 1976 On numbers and games. In the section where he introduces us to these Conway names he includes a caveat. Every surreal number can be expressed in the form of a Conway name, but the use of the Conway name to generate surreal numbers is problematic because it involves feedback loops between surreal numbers of the same generation so is not amenable to use for the generation of the surreals by induction. This differs from the definition of the surreal numbers using a Dedekind-like cut which does generate all the surreal numbers by induction. -To see the problem. Consider the two expressions $\exp(\omega)$ and $\sin(\omega)$. Neither of these can be expressed by Conway names without feedback. The expression $\exp(\omega)$ as a Conway name appears as $\omega^{\omega/\ln(\omega)}$ only because $\omega/\ln(\omega)$ is a surreal number. On the other hand, on the surreals the expression $\sin(x)=\{x-x^3/3!,\ldots|x,x-x^3/3!+x^5/5!,\ldots\}$ evaluates to $\sin(\omega)=0$. As a Conway name $\sin(\omega)$ appears as $\omega^{\ln(\sin(\omega))/\ln(\omega)}$. You see the feedback problem? -The isomorphism to Hahn series involved an interesting change of the sign of the power. So $\exp(\omega)$ on the surreals translates to $1/\exp(\omega)$ on Hahn series.<|endoftext|> -TITLE: Limits in an $(\infty,1)$-category -QUESTION [11 upvotes]: In ordinary category theory, the notion of limit in a category $C$ is usually formulated with a category (of indices) $J$ and a functor $F:J\to C$ (a diagram in $C$), and a limit of this diagram is something satisfying some universal property. -In the context of quasi-categories (I looked in the nLab and in Higher Topos Theory), the category $J$ is replaced by a simplicial set $K$ and the functor $F$ is replaced by a map of simplicial sets $K\to C$ (a quasi-category is a particular simplicial set). But this definition is specific of quasi-categories, if you take another model for $(\infty,1)$-categories, it does not make sense to talk about a map between a simplicial set and an $(\infty,1)$-category. -Is there a definition of limits for $(\infty,1)$-categories independent of the model? -In particular, can we replace $K$ by an $(\infty,1)$-category and $F$ with an $(\infty,1)$-functor? If we can, why does everybody take a simplicial set instead (for quasi-categories)? - -REPLY [3 votes]: One notion which is sometimes convenient, and can be considered to "underlie" all notions of $(\infty,1)$-category, is a derivator. A prederivator is just a 2-functor $Cat^{op} \to Cat$ (modulo size questions). And any $(\infty,1)$-category, in any model, will have an underlying "representable" prederivator defined by $X\mapsto Ho(C^X)$, where $C^X$ is the "functor $(\infty,1)$-category" in whatever model you've chosen and $Ho$ denotes the homotopy category of an $(\infty,1)$-category. -A derivator is a prederivator satisfying axioms saying that (among other things) all limits and colimits exist. But it is also possible to characterize when a particular limit or colimit exists in a prederivator; see for instance here. One expects, though I don't know to what extent it has been proven, that a limit in any model for $(\infty,1)$-categories will give rise to a limit in the underlying prederivator. I don't know about the converse, but for many applications of limits and colimits, the structure of a derivator is sufficient.<|endoftext|> -TITLE: Incarnations of a theorem of Eilenberg -QUESTION [11 upvotes]: Let $R$ be any ring, let $\text{Mod}_R$ be the category of right $R$-modules and let $\text{Ab}$ be the category of abelian groups. There is a classical theorem of Eilenberg (I think) which says that for any right exact functor $F:\text{Mod}_R \to \text{Ab}$ which preserves direct sums, there exists a left module structure on $F(R)$ making $F$ naturally isomorphic to the functor $- \otimes_R F(R)$. -Does anyone know any nice "incarnations" of this theorem? By this I mean "nice", simple and concrete right exact functors from $\text{Mod}_R$ to $\text{Ab}$ (for some $R$) preserving direct sums, for which it is not immediately clear that they are given by a tensor product (= isomorphic to a tensor functor) with a left $R$-module, but for which this left $R$-module can still be constructed in a concrete way. - -REPLY [4 votes]: There is a paper by Mark Hovey which discusses incarnations of the Eilenberg-Watts theorem in homotopy theory. First he reviews the algebraic version and provides an equivalent formulation wherein the theorem says that the category of abelian groups is left self-contained (note that Eilenberg also proved Ab is right self-contained, though it seems Watts did not prove this). Hovey then finds the correct definition of this concept in homotopy theory and proves a very general theorem about when a model category has this property. His theorem holds for topological spaces, simplicial sets, chain complexes, and all the models of spectra. Note, there is a more recent version of this paper on Hovey's website, but it does not compare the new results to the classical ones as fully. -On page 2 of Hovey's paper above is an open problem which you might be interested in. It asks for conditions such that a closed symmetric monoidal category is self-contained (rather than homotopically self-contained as Hovey's paper answers).<|endoftext|> -TITLE: Can any smooth hyperelliptic curve be embedded in a quadric surface? -QUESTION [8 upvotes]: Let's consider algebraic curves over a fixed algebraically closed field $K$. -It's well known, that every smooth elliptic curve (genus $g = 1$) can be embedded in a quadric surface in $\mathbb{P}^3$. This fact follows simply from the Riemann–Roch theorem. -More generally, for smooth hyperelliptic curves of higher genus ($g \ge 2$) it's known that such curves can be embedded in a quadric in weighted projective space $\mathbb{P}(1,1,g)$, see, for example, work of D. Eisenbud. (So, in the case $g=2$ we have embedding in $\mathbb{P}^4$). - -But, can any smooth hyperelliptic curve $H$ be embedded in a quadric surface in $\mathbb{P}^3$? - -It's natural question, because, by the definition we have mophfism $\phi:H \to \mathbb{P}^1$ of degree 2. -I think, this problem is connected with topics "Families of hyperelliptic curves and double covers of quadric surface" and Quotient Surface of A Hyperelliptic Involution, but I don't get it. (I'm interested in the case of any algebraicaly closed field and in the case of finite field). - -REPLY [11 votes]: An explicit realization of degree $2$ and degree $g+1$ maps that separate points can be provided. Suppose the equation of a hyperelliptic curve is $$C:y^2=f(x)$$ with $\deg(f)=2g+2$. "Complete the square" by writing $$f(x)=r(x)^2+q(x)$$ with $\deg(r)=g+1$ and $\deg(q)\leq g$. Then, the maps $$A:(x,y) \mapsto x$$ $$B:(x,y)\mapsto y-r(x)$$ are degree $2$ and degree $g+1$ respectively. The map $B$ is degree $g+1$ because if we assume that $B(x,y)=c$, then we get the equation $c(c+2r(x))=q(x)$, which generically has $g+1$ solutions. Furthermore $(A,B)$ is clearly injective. So, the image $$(A,B):C\mapsto C'\subset\mathbb{P}^1\times\mathbb{P}^1$$ is a degree $(2,g+1)$ curve that $C$ normalizes and the logic of Jack's answer applies.<|endoftext|> -TITLE: What is the current status of the Kaplansky zero-divisor conjecture for group rings? -QUESTION [59 upvotes]: Let $K$ be a field and $G$ a group. The so called zero-divisor conjecture for group rings asserts that the group ring $K[G]$ is a domain if and only if $G$ is a torsion-free group. -A couple of good resources for this problem that gives some historical overview are: - -Passman, Donald S. The algebraic structure of group rings. Pure and Applied Mathematics. Wiley-Interscience [John Wiley & Sons], New York-London-Sydney, 1977. -Passman, Donald S. Group rings, crossed products and Galois theory. CBMS Regional Conference Series in Mathematics, 64. Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI, 1986. - -The conjecture has been proven affirmative, when $G$ belongs to special classes of groups. I tried to write down some of the history: - -Ordered groups (A.I. Malcev 1948 and B.H. Neumann 1949) -Supersolvable groups (E. Formanek 1973) -Polycyclic-by-finite groups (K.A. Brown 1976, D.R. Farkas & R.L. Snider 1976) -Unique product groups (J.M. Cohen, 1974) - -Here are my questions: - -Was Irving Kaplansky the first one to state this conjecture? Can someone provide me with a reference to a paper or book that claims this? -Since the publications of Passman's expository note (above) in 1986, has there been any major developments on the problem? Are there any new classes of groups that will yield a positive answer to the conjecture? Can someone help me to extend my list above? - -The zero-divisor conjecture (let's denote it by "(Z)") is related to the following two conjectures: -(I): If $G$ is torsion-free, then $K[G]$ has no non-trivial idempotents. -(U): If $G$ is torsion-free, then $K[G]$ has no non-trivial units. -Now, if $G$ is torsion-free, then one can show that: -(U) $\Rightarrow$ (Z) $\Rightarrow$ (I). -Has there been any developments, since 1986, to any partial answers on conjecture (U)? Passman claims that "this is not even known for supersolvable groups". Is this still the case? -I want to point out that this post is related to another old MO-post. - -REPLY [26 votes]: Apologies for the self-promotion, but there is now a counterexample to the unit conjecture (U) with $K=\mathbb{F}_2$ and virtually abelian $G = \langle a, b \,|\, (a^2)^b=a^{-2}, (b^2)^a=b^{-2} \rangle$ (as mentioned in David Craven's answer) in arXiv:2102.11818 -.<|endoftext|> -TITLE: What's the easiest example of a morphism of topoi that is not from that of a site? -QUESTION [14 upvotes]: A topos is defined to be a category that's equivalent to the category of sheaves on a site. Morphisms between topoi is defined by a pair of adjoint functors that behave like pull-back/push-forward of sheaves. But I was told one of the cool thing about topos is that sometimes there are morphisms of topos that are not from morphisms of a site. When people talk about this they mention the word "crystalline"... -But is there a toy example I can play around with? What's the easiest example of this? - -REPLY [4 votes]: As for you word "crystalline" it comes from the Crystalline Topos. In the book "Notes on crystalline cohomology" by Berthelot and Ogus, they show that a morphism $X\to X'$ of schemes (over a fixed base $S$) induces a morphism of the associated crystalline topoi althugh there is no morphism of the corresponding sites. This is discussed in Section 5, page 5.1.<|endoftext|> -TITLE: How do the rings of level $N$ quasi-modular forms related to the rings of modular forms? -QUESTION [7 upvotes]: It is well known that the graded algebra $\mathcal{M}(1)$ of Modular forms for $\Gamma = PSL_2(\mathbb{Z})$ is the polynomial algebra -$$ -\mathcal{M}(1) = \mathbb{C}[E_4, E_6] -$$ -where $E_4$ and $E_6$ are the Eisenstein series of weights 4 and 6, respectively. It is also true that, while $E_2 = -\frac{1}{24} + \sum_{k=1}^\infty \sigma(k) q^k$, the Eisenstein series of weight 2 is not modular, it is quasi-modular, and satisfies a similar transformation law. Moreover, the graded algebra $\mathcal{QM}(1)$ of quasi-modular forms is given by -$$ -\mathcal{QM}(1) = \mathcal{M}(1)[E_2] = \mathbb{C}[E_2, E_4, E_6]. -$$ -What can be said about quasi-modular forms for congruence subgroups of $\Gamma$? Is it still the case that, if we denote by $\mathcal{M}(N)$ and $\mathcal{QM}(N)$ the algebras of modular (resp. quasi-modular) forms for, say $\Gamma(N)$ (or perhaps $\Gamma_0(N)$, etc), that we can write -$$ -\mathcal{QM}(N) = \mathcal{M}(N)[E_2]? -$$ -If not, is there some way of determining generators for $\mathcal{QM}(N)$ over $\mathcal{M}(N)$, say for even small values of $N$ ($N = 2, 4$ are of interest to me)? -If this is not known, are there at least dimension formulae? - -REPLY [10 votes]: Dear Simon: -Your assertion is right. Let $\Gamma$ be a subgroup of finite index in $SL_2(Z)$. Then any quasi-modular form for $\Gamma$ can be written uniquely as a polynomial in $E_2$ with coefficients which are modular forms for $\Gamma$. This is proved in a paper by Kaneko and Zagier, A generalized Jacobi theta function and quasi-modular forms in The Moduli Space of Curves, Progress in Math., Vol. 129, Birkhauser, 1995, pp. 165-172. Hope this helps. -Ram Murty<|endoftext|> -TITLE: When does every point in a polytope lie along a chord between its edges? -QUESTION [10 upvotes]: Consider the 3-simplex, or tetrahedron, in 3-space. Regardless of the positions of the vertices, every point in the simplex lies on a chord between two non-adjacent edges of the simplex. Or, equivalently, every interior point lies along a straight line segment which intersects two non-adjacent edges. -When is this property true of other convex (or non-convex) polyhedra? How does this property extend to the general $N$-simplex? - -REPLY [3 votes]: I think that the proposed solution is slightly incomplete since the original question asks for nonadjacent edges of a 3-polytope. It can be easily fixed by studying the intersection of the graphs G and -G. -Regarding possible n-dimensional versions of the problem, the following result can be proved. Let P be an n-dimensional polytope and k and m positive integers such that k + m = n + 1. For any point x in P there are two faces, F and G, of P such that dim F \le k - 1, -dim G \le m - 1, and x is in conv(F U G).<|endoftext|> -TITLE: Groups whose centralisers are finite -QUESTION [6 upvotes]: Let $G$ be an infinite group such that the centraliser of any non central element is finite (and bounded). -Is there any structure theorem known about $G$ ? -Such a group seems to be at the other extreme of an FC-group (whose centralisers all have finite index). I can add the following requirements alltogether if need be : -1- $G$ has finitely many conjugacy classes. -2- $G$ has a trivial centre. -3- $G$ has no involution. -4- $G$ is simple. - -REPLY [4 votes]: In the opposite direction, B. Hartley and M. Kuzucuoglu proved that in an infinite locally finite simple group the centralizer of every element is infinite. See Theorem A2 of the following paper: -B. Hartley - M. Kuzucuoglu: “Centralizers of Elements in Locally Finite Simple -Groups”, Proc. London Math. Soc. 62 (1991), 301-324.<|endoftext|> -TITLE: Irreducible mod-p representation of a semidirect product with trivial p-core -QUESTION [6 upvotes]: Consider the group $G=H\rtimes{}C$, where $H$ has order prime with $p$ and $C$ is cyclic of order $p^k$, and $C\rightarrow{}\mathrm{Aut}(H)$ is faithful (or equivalently $G$ has trivial $p$-core). Assume $V$ to be an irreducible (not just indecomposable) faithful representation of $G$ over $\mathbb{F}_p$. -I need a reference for the following (supposedly well known?) fact: -Proposition: -as $\mathbb{F}_p[C]$-module $V$ is free (that is isomorphic to $\mathbb{F}_p[C]^n$ for some n). -This can be proved going to the algebraic closure $\overline{\mathbb{F}_p}$, and decomposing $V$ into irreducible $H$-representations. Then it's possible to see (I omit the details) that $C$ should act with orbits of cardinality $p$ on the irreducible $H$-factors of $V$, and hence that $V$ is induced from an irreducible $H$-representation. -In particular I would like to have an easy way to say that every extension of $G$ by $V$ is split, this is an easy consequence of the proposition above and Schur-Zassenhaus theorem. And I would be surprised if this was not well-known as well. - -REPLY [5 votes]: Here is an easy non-cohomological proof of the splitting question. The set-up is this. We have a group $\Gamma$ having a minimal normal subgroup $V$, where $V$ is a $p$-group. Also, $\Gamma/V = G$, and $G$ has a normal $p'$-subgroup $H$. In the original question, $H$ was complemented by a cyclic $p$-group in $G$, but we do not need to assume that. Also, in the original question, $V$ was a faithful $G$-module, but we need a much weaker assumption: that $H$ acts nontrivially on $V$. We want to show that $\Gamma$ splits over $V$. -Write $H = K/V$ and by Schur-Zassenhaus, let $X$ be a complement for $V$ in $K$. Let $N = N_\Gamma(X)$. We argue that $N$ is the desired complement for $V$ in $\Gamma$. First, $KN = \Gamma$ by a Frattini argument, using the fact that all complements of $V$ in $K$ are conjugate to $X$ in $K$. It follows that -$\Gamma = VN$. Now $V \cap N$ is normal in $G$ since $V$ is abelian. Since $V$ is minimal normal in $\Gamma$, we have either $V \cap N = 1$, as wanted, or -$V \subseteq N$. The latter would imply that $X$ centralizes $V$, and this is not the case since $H$ acts nontrivially.<|endoftext|> -TITLE: Two rectangular parallelepiped -QUESTION [8 upvotes]: Prove that if we have two rectangular parallelepiped (cuboids) such that one of them is placed inside the other then the sum of the three lengths of the inner parallelepiped is at most the sum of the three lengths of the exterior parallelepiped. -In 2 dimensions the problem is trivial. -Does this hold in higher dimensions? -There is a way to prove it in higher Euclidean dimensions? - -REPLY [10 votes]: This version works for all parallelepipeds, not only rectangular ones: -If you replace each parallelepiped by all points that have distance at most $\varepsilon$ to a point in the parallelepiped, you can still place the smaller inside the bigger one. In particular, the smaller object has a smaller volume. -We divide up the extended parallepipeds by extending the planes corresponding to the six faces. This gives the volume of the original solid in the center, parallelepipeds of height $\varepsilon$ on top of each face, partial cylinders (with slanted parallel ends) of radius $\varepsilon$ and length the corresponding edge, and partial spheres of radius $\varepsilon$ around each vertex. -The partial spheres add up to exactly one whole sphere simply by translation. The partial cylinders corresponding to parallel edges add up to one whole cylinder by translation. -Now let $\varepsilon$ tend to infinity (yes, really). The term with $\varepsilon^3$ comes from the sphere of radius $\varepsilon$ and does not depend on the parallelepiped at all, . The coefficient of $\varepsilon^2$ comes from the cylinders and is clearly the sum of the edges times some constant. -So, for the inequality to be valid the sum of the edges of the smaller box must be smaller than the sum of the edges of the larger box. -I don't see any problem with the generalization to higher dimension.<|endoftext|> -TITLE: Did Grothendieck introduce vertical arrows that denote morphisms? -QUESTION [16 upvotes]: It is usual in algebraic geometry to represent morphisms by vertical arrows pointing downwards, like that : -$$\begin{matrix} X \\\\ \downarrow \\\\ S \end{matrix}$$ -I suppose this stemmed from Grothendieck's amazingly original idea that a morphism of schemes should always be considered as some sort of fibre bundle, even in cases apparently very distant from the bundles considered in topology. -Many geometers have since adopted these vertical arrows, which they find suggestive and psychologically helpful. -My question is simply whether anybody had drawn maps vertically before Grothendieck a) in topology b) in algebraic geometry. -While on the subject I can't resist telling an anecdote I heard, according to which in some seminar led by Grothendieck, a joker ( Serre?) always drew the vertical morphism above on the blackboard just before Grothendieck arrived. So an auxiliary question might be: c) is this true? - -REPLY [9 votes]: In Hasse's school of number theory, it was quite common to represent an extension of fields by writing the bigger field above the smaller one and drawing a line segment between them, without an arrowhead. This is one possible source of Grothendieck's notation. - -REPLY [2 votes]: Vertical arrows appear everywhere in the book by Cartan and Eilenberg "Homological Algebra" (1956). -For instance, at page 5 one finds the statement of the $5$-lemma, with the usual commutative diagram, and a page 6 there is the definition of projective module, again with a diagram containing a vertical arrow. -I do not know where vertical arrows originated from. My guess is that they arise naturally in Topology when one tries to prove some kind of lifting theorem.<|endoftext|> -TITLE: How similar are discrete stable RVs to their continuous analogues? -QUESTION [7 upvotes]: The generalized central limit theorem of Gnedenko-Levy describes the asymptotic behavior of a sum of IIDRVs which may not have finite mean or variance. Only a small class of limit laws can be realized, and Cauchy distributions are a notable example. -Steutel and van Harn introduced a notion of stability for RVs defined on $\mathbb{N}$: similarly to the "traditional" case (where characteristic functions are extensively utilized), this is done in terms of generating functions. -In my cursory survey, I have found oblique remarks to the effect that discrete stable distributions are very similar to their continuous analogues. I have, however, been unable to find a usable reference. -In my work I have recently come across a rather surprising bit of behavior. I have some RVs on $\mathbb{N}$ that (after a bit of reasonable kernel density estimation) are profoundly good fits to (discrete analogues of) Cauchy distributions (in fact, up to single-parameter scaling, the same distribution--which is remarkable because their peaks are not at the origin and rescale to each other). This is also pretty weird because--again--these RVs are defined on $\mathbb{N}$, not $\mathbb{Z}$. If the fits weren't so good, I'd toss it off as a coincidence, but I am (I think naturally) curious about the possibility that this is no accident. If the discrete stable distributions are close in a suitable sense to continuous analogues, I would have the seed of an explanation. So: - -How similar are discrete and - continuous stable distributions, and - in particular is there something like - a Cauchy distribution centered away - from 0 that can occur as a discrete - stable distribution supported on - $\mathbb{N}$? - -REPLY [6 votes]: I realise you asked this question over two years ago, but: -The one-sided continuous-stable distributions can be directly linked to the discrete-stable distributions through a formulation called the 'Poisson transform', where the mean of a Poisson distribution is modulated by another distribution (one example of this is where the other distribution is a Gamma distribution - the resultant distribution is negative binomial). -In the case when the modulating distribution is one-sided continuous-stable of power-law index nu, the resultant distribution is discrete-stable with index nu. -I talk about this in depth in my thesis ("Continuous and discrete properties of stochastic processes") which also has an in-depth literature review you might find useful and, for a tl;dr version, in the paper "Continuous and discrete stable processes" (on Academia.edu or DOI link to APS.org).<|endoftext|> -TITLE: Fake projective spaces -QUESTION [6 upvotes]: I'm looking for examples of "fake" projective spaces. -Question: Are there smooth manifolds other than $\mathbb{C}\mathbb{P}^n$ whose cohomology ring is the truncated polynomial ring $\mathbb{K}[h]/h^{n+1}$ with $h$ of degree 2? -I know that there is a classification of so-called fake projective planes in the world of algebraic geometry and results restricting the existence of higher-dimensional examples. -However, I'm interested more in the topological question so don't need these manifolds to arise as varieties, but I do want to have the same ring structure. I don't mind about what coefficient ring we work over. -Has anyone come across examples of such things? -Edit Sorry, I should clarify: I guess I'm really looking for something with the same cohomology ring as $\mathbb{C}\mathbb{P}^n$ but, say, nontrivial fundamental group. Thanks for those answers about exotic smooth structures though. - -REPLY [5 votes]: Quote from the first page of the article Uniqueness of the complex structure on Kähler manifolds of certain homotopy types by Libgober and Wood (Journal of Differential Geometry 32, 1990, no. 1, 139-154): -"On the other hand it is known that for every $n>2$ the homotopy type of $\mathbb{CP}^n$ supports infinitely many inequivalent differentiable structures distinguished by their Pontryagin classes (see Montgomery and Yang [25] or Wall [30] for $n=3$ and Hsiang [15] for $n>3$)."<|endoftext|> -TITLE: Does integrating with respect to a finitely additive measure respect addition? -QUESTION [6 upvotes]: Let $X$ be a set and $\mathcal{A} \subseteq P(X)$ a $\sigma$-algebra. Assume $\nu : \mathcal{A} \to [0,\infty]$ is a finitely additive measure. If $f : X \to [0,\infty]$ is a measurable function, we can define $$ \int_{X}f\,d\nu$$ in the standard way. If $f,g :X \to [0,\infty]$ are simple measurable functions then it is easy to prove that $$\int f + g\,d\nu = \int f\,d\nu + \int g\,d\nu. $$ However, if $f$ and $g$ are just measurable functions, then it is only obvious that $$ \int f\,d\nu + \int g\,d\nu \leq \int f + g\, d\nu. $$ - -Question : Does integration with respect to a finitely additive measure respect addition? - -Note, that if $\nu$ is countably additive, then the standard way to prove that integration respects addition is to appeal to the monotone convergence theorem. - -REPLY [5 votes]: As mentioned in the question, the inequality -$$ -\begin{align}\int(f+g)\, d\nu\ge\int f\,d\nu+\int g\,d\nu&&{\rm(1)}\end{align} -$$ -follows easily from the definition of the integral $\int f\,d\nu$ (as the supremum of the integrals of nonnegative simple functions bounded by $f$). So, I'll just show the reverse inequality which will establish additivity of the integral. -Choose any nonnegative simple function $h\le f+g$. We need to show that $\int f\,d\nu+\int g\,d\nu\ge\int h\,d\nu$. However, we can write $h=\sum_{k=1}^nc_k1_{A_k}$ for $c_k\in(0,\infty)$ and pairwise disjoint $A_k\in\mathcal{A}$. As long as it can be shown that $\int_{A_k}f\,d\nu+\int_{A_k}g\,d\nu\ge c_k\nu(A_k)$, then the required inequality will follow by summing over $k$ and applying (1). So, replacing $f,g$ by $1_{A_k}f,1_{A_k}g$ respectively (for a fixed $k$), we reduce to the case where $n=1$. Dividing through by $c_k$ reduces to $c_k=1$. -So, we have reduced to the situation with $f+g\ge1_A$ and just need to show that $\int f\,d\nu+\int g\,d\nu\ge\nu(A)$. Without loss of generality (capping $f,g$ by 1 if necessary), we further reduce to the case with $0\le f,g\le1$. Then, for each positive integer $N$, consider the simple functions -$$ -\begin{align} -f_N&=\sum_{j=0}^{\lfloor N\rfloor}1_{f^{-1}((j/N,(j+1)/N])}\frac jN\le f,\\\\ -g_N&=\sum_{j=0}^{\lfloor N\rfloor}1_{g^{-1}((j/N,(j+1)/N])}\frac jN\le g. -\end{align} -$$ -We have $f_N+g_N\ge(1-\frac2N)1_A$. So, using additivity for simple functions -$$ -\int f\,d\nu+\int g\,d\nu\ge\int f_N\,d\nu+\int g_N\,d\nu\ge\left(1-\frac2N\right)\nu(A). -$$ -Letting $N$ increase to infinity gives the required inequality.<|endoftext|> -TITLE: Deriving the Hilbert spaces for Chern-Simons TQFTs with complex gauge group -QUESTION [7 upvotes]: One method for finding the Hilbert spaces corresponding to surfaces in Chern-Simons TQFT is by geometrically quantizing the phase space, which is just the character variety of the surface. I know that this has been done by Anderson and others in the holomorphic polarization and by Jeffrey and Weitsman in the real polarization when the gauge group is $SU(2)$. -Is there a derivation of the state spaces for Chern-Simons TQFTs with gauge group $SL(n,\mathbb{C})$ that uses geometric quantization? Equivalently, is there a geometric quantization of the $SL(n,\mathbb{C})$ character varieties of surfaces? Or at least for $n=2$? - -REPLY [5 votes]: This was done in a paper by E. Witten: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1104202513 -The paper was written before the arxiv came to be, so unfortunately it is not on the arxiv. -Another paper by Witten may be relevant here: http://arxiv.org/PS_cache/arxiv/pdf/1001/1001.2933v4.pdf but I haven't studied it in detail.<|endoftext|> -TITLE: When is $\mathbb{G}_m(R)$ enough to determine $R$? -QUESTION [5 upvotes]: Say I have a ring, $R$, with 1 which I consider my universe, and I know its group of units $G=\mathbb{G}_m(R)$. Then given a subgroup, $H\le G$, can I determine if there is there a subring $S_H$ such that $\mathbb{G}_m(S)=H$? If so, is $S_H$ unique with this group of units? If so, is there in fact a--canonical in the sense above--1-1 correspondence between subgroups of $G$ and subrings of $R$ with 1? Preliminary attempts at a solution don't indicate any problems with the truth of the statement, but naturally one should be skeptical of limited data especially in a subject with so many intricacies as groups and rings. -The motivating example is $\overline{\mathbb{Q}}/\mathbb{Q}$, due to some interesting number theory that could come out of such a correspondence. -In the case of fields the question is supposed to collapse into the question "Can I add 0 to a subgroup of the group of units of some big field and get a subfield without doing anything else?" -There is no possibility for general rings, but are there assumptions on $R$ or $G$ which can ensure existence or uniqueness? And it is also fine to induce assumptions on what kind of $S$ we are allowed to have as well, fields instead of just rings for example. - -REPLY [3 votes]: It is also worth to recall that if $F$ is a field then a free algebra $F[X]$ has $F^\times$ as group of units for any set $X$... -Anyway, there is an extensive literature devoted to study rings with a fixed group if units. A sample of this is the paper: -I. Stewart: Finite rings with a specified group of units, Math. Z. 126 (1972), 51-58.<|endoftext|> -TITLE: Fourier transform of a real-valued function. -QUESTION [6 upvotes]: My chemist roommate asked me the following question. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a real-valued function and $F$ its Fourier transform. Suppose we know the modulus function $|F| : \mathbb{R} \rightarrow \mathbb{R}$. What can we deduce about $f$, can we determine it completely? -Feel free to assume any regularity conditions on $f$. - -REPLY [12 votes]: To try to determine a function from the absolute value of its Fourier transform is actually the famous "hidden phase problem". In X-ray crystollography one measures the absolute value of the Fourier transform of a function that describes where the atoms in the molecule are located. However, using clever tricks and some a priori knowledge of the unknown functions (for instance the fact that it is non-negative) one has been able to handle this problem in practice. The Nobel prize in chemistry 1985 was awarded for progress on this problem.<|endoftext|> -TITLE: Can the Riemann hypothesis be undecidable? -QUESTION [53 upvotes]: The question is contained in the title; I mean the standard axioms ZFC. The wiki link: Riemann hypothesis. There are finite algorithms allowing one to decide if there are non-trivial zeroes of the $\zeta$-function in the domains whose union exhausts the whole strip $0<\Re z<1$, but this does not seem to be the obstacle for undecidability. Are there other arguments? - -REPLY [14 votes]: Without worrying about reductions, if RH is false, it is provably false: suppose $\rho\in\mathbb{C}$ is a zero in the critical strip but off the critical line (say, $\Re(\rho)>\frac12$). Then for a little rectangle $R$ with (say) rational corners, containing $\rho$ in its interior but not containing the pole at $s=1$ or intersecting the critical line we'd have -$$\frac{1}{2\pi i}\oint_{\partial R} \frac{\zeta'}{\zeta}(s)\mathrm{d}s \geq 1$$ -by the argument principle. -But we can approximate $\zeta(s),\zeta'(s)$ to arbitrary precisition by a finite computation, and similarly we can approximate the integral to arbitrary precision by a numerical computation. In other words, there is a finite computation which provably approximates the integral above to within $\frac{1}{2}$. Then the non-vanishing of the approximation proves the falsity of RH. -Similarly, RH is equivalent to estimates on the prime-counting function, for example to $$|\psi(x)-(x)|<\sqrt{x}\log^2 x$$ -where $\psi(x) = \sum_{p^r -TITLE: Why does a group action on a scheme induce a group action on cohomology? -QUESTION [13 upvotes]: This is probably totally obvious but I have no clue how this is done: Say you have an endomorphism $f:X \rightarrow X$ of schemes. Why (if true, perhaps some additional assumptions are necessary!) do you get for a Zariski/étale/l-adic sheaf $\mathcal{F}$ on $X$ an induced endomorphism on the corresponding cohomology? How is this constructed? Are there conditions, when the induced morphism is an isomorphism (I'm having a Frobenius in mind)? -Perhaps the above is too general, so my real question is: Why does a group/monoid action on the Deligne-Lusztig variety induce a group/monoid action on the l-adic cohomology (with compact support) of this variety? In every book I looked at this is just mentioned but not explained. - -REPLY [17 votes]: If a (say constant) group $G$ acts on a scheme $X$, you may want to consider the notion of a $G$-sheaf : a sheaf $\mathcal F$ endowed with isomorphisms $\lambda_g: g^* \mathcal F\simeq \mathcal F$, for $g\in G$ satisfying the usual cocycle conditions. Then by functoriality of cohomology for $g:X\to X$ you get an isomorphism $H^i(X, \mathcal F) \to H^i(X, g^*\mathcal F)$ that you can compose with the morphism induced by $\lambda_g$, that is $H^i(X, g^* \mathcal F)\simeq H^i(X,\mathcal F)$. Thus you get for each $g\in G$ an automorphism of $H^i(X, \mathcal F)$ and it is easy to check that this gives an action of $G$ on this cohomology group. -A probably better way to see this is to use functoriality of $G$-sheaves : the global section functor goes from $G$-sheaves of abelian groups to abelian groups endowed with an action of $G$. Since the abelian category of $G$-sheaves has enough injectives (a classical fact, you can find it in Grothendieck's famous Tohoku paper) you can derive it. You get cohomology groups naturally endowed with an action of $G$. Once you apply the forgetful functor, you recover the usual cohomology groups (this is easy to see directly, or you can use Grothendieck's theorem on derivation of a composition of functors, the only point is that the forgetful functor is acyclic). -There is a natural generalization to action of non constant groups, and also to action of monoids. -I think you can apply this in your situation, since the $l$-adic sheaf defining $l$-adic cohomology is naturally endowed with the structure of a $G$-sheaf (the sheaf $\mathbb Z/l^k\mathbb Z$, as any constant sheaf, has a canonical structure of $G$-sheaf).<|endoftext|> -TITLE: Is the category of rings co-well-powered? -QUESTION [7 upvotes]: Can anybody explain me if the category of (associative) rings is co-well-powered? (This is the MacLane definition, in Russian literature this is called "locally small from the right side".) I mean, it is well-powered, of course, since for any ring A one can easily find a skeleton in the category Mono(A) (of all monomorphisms with values in A) and this will be a set (the set of all subrings in A). But is it true, that it is co-well-powered, i.e. for any ring A there exists a skeleton in the category Epi(A) (of all epimorphisms from A into other rings) and is this skeleton again a set? - -REPLY [7 votes]: Sergio Buschi already gave a general answer, and here is a pedestrian way of thinking about it. First, apparently if $e : R \to S$ is epi, then the cardinality $|S|$ of $S$ cannot exceed the cardinality $|R|$ of $R$. This is not obvious, as $e$ need not be surjective. I am no expert on rings, so I simply Googled the fact and found this. For each cardinality below $|R|$, there are can be only set-many non-isomorphic rings. So we have a set of sets of candidates, which is again a set. That's co-well-poweredness.<|endoftext|> -TITLE: associated sheaf functor doesn't preserve arbitrary products -QUESTION [5 upvotes]: I need example that associated sheaf functor doesn't preserve arbitrary products. I think that one can provide an example for sheaves over topological space. Thanks for your help. - -REPLY [8 votes]: An example where sheafification does not preserve arbitrary products is where we take sheaves over a (sober) space $X$ that is not locally connected, for example the space of irrationals or Cantor space. -Recall that a Grothendieck topos $E$ is locally connected if the (essentially unique) geometric morphism $\Gamma = f_\ast: E \to Set$ has a left adjoint $f^\ast$ that in turn has a left adjoint. More generally, a geometric morphism $f_\ast: E \to F$ between toposes is an essential geometric morphism if its left adjoint $f^\ast$ has a left adjoint. -We have the following facts: - -A presheaf topos $Set^{C^{op}}$ is locally connected. Here the left adjoint to the global sections functor $\Gamma: Set^{C^{op}} \to Set$ is the diagonal functor $\Delta: Set \to Set^{C^{op}}$, which of course has a left adjoint. -A geometric morphism $f_\ast: E \to F$ is essential if and only if $f^\ast$ preserves arbitrary products. (Of course $f^\ast$ is already left exact and so preserves equalizers, if it preserves small products, then it preserves small limits. Using the fact that Grothendieck toposes are cototal, this is enough to ensure that $f^\ast$ has a left adjoint.) -Then in particular for a small site $(C, J)$, the inclusion functor $i: Sh(C, J) \to Set^{C^{op}}$ is an essential geometric morphism if and only if sheafification $a: Set^{C^{op}} \to Sh(C, J)$ preserves small products. Thus, putting the last two facts together, the composite geometric morphism -$$Sh(C, J) \stackrel{i}{\to} Set^{C^{op}} \stackrel{\Gamma}{\to} Set$$ -is essential, i.e., $Sh(C, J)$ is locally connected, if $a$ preserves products. -In the case where the site is $(\text{Open}(X), J)$ where $J$ is the canonical Grothendieck topology given by covering families, $Sh(X) = Sh(C, J)$ is locally connected if and only if $X$ is a locally connected space. Thus $a: Set^{\text{Open}(X)^{op}} \to Sh(X)$ preserves small products only if $X$ is locally connected. - -REPLY [8 votes]: Let $X$ be a space. An abelian group (or a set, if you prefer) $G$ determines a constant presheaf, call it $C_G$, and the associated sheaf is the sheaf of locally constant maps into $G$. Given a family $G_i$ with product $G$, the product of the presheaves $C_{G_i}$ is the presheaf $C_G$, but the product of the associated sheaves is not in general the sheaf of locally constant maps to $G$. (If $X$ is not locally connected then one can easily have a map to $G$ which projects to a locally constant map to $G_i$ for each $i$ but which is not locally constant itself. ) -This is presumably a down-to-earth special case of what Todd Trimble is saying.<|endoftext|> -TITLE: Can $L^p(\mathbb{R})$ and $ L^q(\mathbb{R})$ be isomorphic? -QUESTION [39 upvotes]: Let $p,q \in (1,\infty)$ with $p\neq q$. Are the Banach spaces $L^p(\mathbb{R})$, $L^q(\mathbb{R})$ isomorphic? - -REPLY [41 votes]: The proof Fabian alludes to in the book reference Mark gave is a modern one using the notions of cotype and type. One way to prove that a Banach space $X$ is not isomorphic to a Banach space $Y$ is to exhibit a property that is preserved under isomorphisms that $X$ has but $Y$ does not. Type and cotype are examples of such properties. The (best) type and cotype of $L_p$ are calculated in many books. I suggest you look at Theorem 6.2.14 in the book of Albiac and Kalton. From the statement you see that if $p\not=q$, then $L_p$ and $L_q$ either have different (best) type or different (best) cotype. -Type and cotype depend only on the collection of finite dimensional subspaces of a space (we call such a property a local property). So you cannot use either to prove, e.g., that $L_p$ is not isomorphic to $\ell_p$ when $p\not= 2$. One way of proving this is to show that $\ell_2$ embeds isomorphically into $L_p$ but not into $\ell_p$ when $p\not=2$. These facts you can also find in Albiac-Kalton. You can also use infinite dimensional techniques to prove that $L_p$ and $L_q$ are not isomorphic when $p\not=q$. Banach knew this result through infinite dimensional considerations--the concepts of type and cotype came on the scene only 40 years ago. -You will also find in Albiac-Kalton a discussion of when $L_p$ or $\ell_p$ embeds isomorphically into $L_q$. That is more complicated and in fact Banach did not know everything. He called the question the problem of the linear dimension of $L_p$ spaces, IIRC.<|endoftext|> -TITLE: Does a notion of convex graph make sense? -QUESTION [9 upvotes]: Let $X=(V,E)$ be a finite connected graph. I would be interested in some notion of convexity. - -General question: Is there a notion of convexity for finite connected graphs? How does it look like? - -As pointed out below by David Eppstein, there is a standard notion for a subsets of a graph to be convex. I am actually interested in something different: a notion of convexity for the graph itself. -I want to share some thoughts, hoping that someone is interested. Taking inspiration from the unit ball in $\mathbb R^2$ and also from the properties that I would need, I am tempted to require the following properties: -Let $\mathcal C$ be the set of paths inside $X$. I want to axiomatize a convex structure, saying that some of these paths are lines. So, a convex structure on $X$ should be a (possibly proper) subset $\Gamma$ of $\mathcal C$ such that -First Property. For all $x,y\in V$, $x\neq y$, the set of $\gamma\in\Gamma$ passing through $x,y$ is non-empty and closed under intersection, I will denote by $[x,y]$ the intersection of them. (Notice that I am not supposing that $[x,y]=[y,x]$ as a set of points). -Before stating the other properties, I need to define what are the $\Gamma$-extremal points -Definition: $x_0\in V$ is called $\Gamma$-internal if for all $x\in V$, $x\neq x_0$, there is $y\in V$, with $y\neq x_0$, such that $[x,x_0]\subseteq[x,y]$. A vertex is called $\Gamma$-extremal if it is not $\Gamma$-internal. -Now, let $Extr(V)$ be the set of extremal vertexes. I can state the remaining properties. Next property states that I can prolonge uniquely the line until hitting the boundary. -Second Property. For all $x,y\in V$, $x\neq y$, there exists a unique $l(x,y)\in Extr(V)$ such that $[x,y]\subseteq[x,l(x,y)]$ -Now, I want some version of continuity, for the points obtained prolonging line till hitting the boundary. -Third Property. If $x_1\sim x$ and $y_1\sim y$, then $l(x_1,y_1)$ and $l(x,y)$ are either adjacent or coincide, where $\sim$ stands for the usual adjacency relation. -At this point, one can says Well, take $l(x,y)$ to be constant!. But I don't want this triviality. -Fourth property. The set $Extr(V)$ has to be connected as a subgraph of $V$; $V$ has to be contractible and $Extr(V)$ has to be non-contractible (contractibility is defined in Def. 17 in http://arxiv.org/abs/1111.0268. Intuitively, keep in mind the following example: the square $[0,n]^2$ is contractible; the boundary of this square, for $n\geq3$, is not contractible, since there is a hole.). -The point is that I am not able to find any example of such graphs! :) I can imagine that some huge discretization of the ball might play the game, but I am not quite sure. - -More specific question: Does there exist some non trivial examples of such graphs? - -Thanks in advance, -Valerio - -REPLY [2 votes]: (i) If $[x,y]$ can be empty, then taking an $n\times n$ square in the square grid and the vertical and horizontal paths as the set of paths $\mathcal C$, all properties are satisfied except the second property for pairs $x,y$ on the boundary; to get all properties satisfied, instead of a square take the vertices of the grid in the lozenge $|x|+|y|\le n$. -(ii) If you relax the third property by allowing that $l(x_1,y_1)$ and $l(x_2,y_2)$ either are adjacent or coincide'', then the 5-cycle with the set of shortest paths as $\mathcal C$ seems to satisfy all conditions. -(iii) Condition (i) needs to be written in a more precise way: as I understand, $[x,y]$ is the (vertex-set) union of the portions between $x$ and $y$ of all paths of $\mathcal C$ passing via $x$ and $y$ (and not their intersection). -(iv) Bibliography remarks: on a related topic (but not for graphs), see the paper R. Dhandapani, J. E. Goodman, A. Holmsen, R. Pollack, S. Smorodinsky, Convexity in Topological Affine Planes. Discrete & Computational Geometry 38(2): 243-257 (2007). About abstract convexity, see the book Theory of convex structures by M. Van de Vel.<|endoftext|> -TITLE: Elliptic curve group law, Sum of intersection points -QUESTION [5 upvotes]: If a plane curve of degree n intersects an elliptic curve in 3n points, then do those points always sum to zero when added using the group law on the points of an elliptic curve ? - -REPLY [8 votes]: I assume that elliptic curve means a cubic curve $E$ with some base point $O$. -Then the answer to your question is yes if and only if the base point O for the group law is a flex point of $E$. (In many texts it is assumed that the base point of $E$ is a flex line, but to define the group law this is completely unnecessary, see Fulton's book on plane curves for the geometric definition of the group law in this case.) Fulton's geometric definition of the group law on a cubic is equivalent with -$P_1,..P_{3n}$ add up to zero if and only if $P_1+..P_{3n}-3n O$ is zero in $Pic^0(E)$, i.e., is a divisor of a function on $E$. -Assume first that $O$ is a flex point. Let g be polynomial defining the flex line, -let $f$ be a polynomial defining $C$ and let $P_1,...P_{3n}$ be the $3n$ intersection points of $C$ and $E$. -As David commented above we have $div(f/g^n)=P_1+..P_{3n}-3nO$, hence the points add up to zero. -Suppose now that O is not a flex. Let $R$ be the third intersection point of $E$ with the tangent line of $E$ at $O$. Since $O$ is not a flex we have that $R\neq O$. To check whether these points add up to zero we have to calculate $O-O+O-O+R-O$ in $Pic^0(C)$. It is easy to see that this divisor is nonzero in $Pic^0(C)$.<|endoftext|> -TITLE: cohomology of complete intersections -QUESTION [10 upvotes]: Let $X\subseteq\mathbb{P}^n(\mathbf{C})$ be a complete intersection (smooth if you want). -Q: Is there a good reference which gives (and proves in enough details) an explicit description of the graded ring $H^*(X,\mathbb{Z})$ (or $H^*(X,\mathbb{Q})$)? - -REPLY [9 votes]: One can compute the Betti numbers of a smooth complete intersection $X$ of multidegree $d=(d_1,\ldots,d_r)$ by induction on $r$. The case $r=0$ corresponds to $\mathbb{P}^n(\mathbb{C})$. -A smooth complete intersection of $X$ multidegree $(d_1,\ldots, d_r)$ is a section of the positive line bundle $\mathcal{O}(d_1)$ over a smooth complete intersection of $Y$ multidegree $(d_2,\ldots, d_r)$. By the Lefschetz hyperplane theorem we have $H^i(X,\mathbb{Q})\cong H^i(Y,\mathbb{Q})$ for $i\leq \dim Y-2=\dim X -1$. Using the induction hypothesis and the Poincar\'e duality we can compute all $H^i(X,\mathbb{Q})$ apart from $H^{\dim X}(X,\mathbb{Q})$. -But we can also compute the Euler characteristic of $X$ as follows. The normal bundle of $X$ is $\mathcal{O}(d_1)\oplus\cdots\oplus\mathcal{O}(d_r)$, which extends to $\mathbb{P}^n$. The Chern class of $T\mathbb{P}^n$ is $(1+H)^{n+1}$ where $H$ is the class of the hyperplane section. So the Chern class of $TX$ is $$\frac{1+H)^{n+1}}{\Pi(1+d_i H)}$$ restricted to $X$. So the Euler characteristic of $X$ is $d_1\cdots d_r$ times the coefficient of $H^{\dim X}$ in the above expression. Now we can compute the rank of $H^{\dim X}(X,\mathbb{Q})$. -Moreover, from the above and the universal coefficients formula it follows that $H^*(X,\mathbb{Z})$ is torsion free. Indeed, we may assume that $H_{<\dim X}(X,\mathbb{Z})$ and $H^{<\dim X}(X,\mathbb{Z})$ are torsion free by induction hypothesis and hence so are $H_{>\dim X}(X,\mathbb{Z})$ and $H_{>\dim X}(X,\mathbb{Z})$ by the Poincar\'e duality; moreover $H^{\dim X}(X,\mathbb{Z})=Hom (H^{\dim X}(X,\mathbb{Z}),\mathbb{Z})$ is torsion free and so is $H_{\dim X}(X,\mathbb{Z})$ by the Poincar\'e duality again.<|endoftext|> -TITLE: Why is BG infinite dimensional for G finite ? -QUESTION [24 upvotes]: If $G \neq \lbrace 1 \rbrace$ is a finite group with classifying space $BG$ -then there are infinitely many i such that $H^i(BG,\mathbb{Z}) \neq 0$. This -can be found, for example, there: -Non-vanishing of group cohomology in sufficiently high degree -As a consequence, the CW-complex $BG$ (unique up to homotopy) can not be of finite dimension. -Question: Are there alternative proofs for this observation. In particular, I would be interested in knowing if there is a purely topological proof without homological algebra. - -REPLY [15 votes]: A proof based on fixed point theory: If $BG=EG/G$ is finite-dimensional, then $EG$ is as well and has the homology of a point. Choose a non-trivial Sylow subgroup $P$ of $G$. By a well-known theorem of P.A. Smith, the fixed point set $EG^P$ is non-empty, contradicting the free action of $G$ (and hence $P$) on $EG$. Consequently $BG$ must be of infinite dimension.<|endoftext|> -TITLE: Motivating Algebra and Analysis for Average Undergraduates -QUESTION [16 upvotes]: I work at a small liberal arts college, where many of our mathematics majors will not attend graduate school in mathematics. My hope in asking the following question is to gather innovative ideas for motivating an average student for the development of theory usually found in first courses in algebra and analysis. - -Question: What are the most productive ideas for motivating algebra and analysis for average undergraduate math majors not destined for graduate school? - -I'm interested in hearing about motivation of the subjects in general, and bits of theory in particular. -On the first front, around half way through a typical undergraduate linear algebra course (around when abstract vector spaces appear) there is an abrupt change in the focus of the material from highly computational exercises meant to produce what the students think of as an "answer", to proofs of basic theorems. (Of course good theory facilitates computation, but without experience students probably can't even appreciate the questions driving the theory!) What are some ideas for smoothing this transition? -On the second front, how do we motivate individual concepts? Ideally it would be nice if there were collections of questions that made the need for theory absolutely clear for students, e.g. the very nice problem found in Herstein's Topics in Algebra: - -Let $G$ be a finite group whose order is not divisible by 3. Suppose that $(ab)^{3}=a^{3}b^{3}$ for all $a,b \in G$ prove that $G$ is abelian. - -Students can compute until they are blue in the face with this one, but are basically forced to consider the properties of homomorphisms to solve the problem. I'd like to be pointed to questions like these. -Addendum: MO may be an ideal place to quickly synthesize "problem hikes" through theory, as many of us have encountered favorite problems that illuminate important ideas. Please feel free to submit your favorite problems as answers to this question. It would be great if this site could generate undergraduate problem books in (or at least problem hikes through) algebra and analysis. - -REPLY [2 votes]: My experience is that people learn better if they have an intuitive understanding for what purpose they are learning all that stuff. This is why I would start with symmetry groups of crytsals. This ensures for instance also interdisciplinarity and something people can understand. M. Artin's book "Algebra" is really good there. -Normally, I would have advised you to tell them about the fascinating applications in differential geometry, such as the geometry of symplectic spaces. But, as you are dealing with "average undergraduates" as you named them I would use the above. -Remark: Although I would like to oppose to the nomenclature: Average undergrad at a samll college doesn't mean they shouldn't be motivated. Perhaps some of them will be fascinated by your course so intensely that they plan to go on to grad school.<|endoftext|> -TITLE: Geometric interpretation of Simpson's correspondence -QUESTION [10 upvotes]: What is the exact geometric meaning of the Simpson's correspondence between Higgs bundles and local systems ? I know that it should have a rich geometric content but don't know an explicit geometric interpretation which reveals the significance f this correspondence. - -REPLY [11 votes]: The nonabelian Hodge theorem, i.e. Simpson's correspondence, for smooth projective -varieties refines a number of earlier results by several authors (Narasimhan-Seshadri, Donaldson...) where things can be understood more explicitly. For example, a unitary local system $L$ gives rise to polystable vector bundle $E= L\otimes \mathcal{O}_X$ with zero Higgs field. In general, the correspondence is highly transcendental, and I think it would -be fair say that the geometric meaning is a very deep mystery. -We can see this even in rank one. On the local system side, the moduli space is -the character variety -$$M_{Betti}=Hom(\pi_1(X), \mathbb{C}^*);$$ -on the Higgs bundle side -the moduli space is the cotangent bundle -$$M_{Dol}=T^*Pic(X) = Pic(X)\times H^0(X,\Omega_X^1).$$ -As algebraic varieties or even as complex manifolds these are very different, $M_{Betti}$ is Stein and the other is not. Yet they correspond as sets or topological spaces -because they can both be identified using polar coordinates/Hodge theory with -$$Hom(\pi_1(X),U(1))\times Hom(\pi_1(X),\mathbb{R})$$ -I could go on, but perhaps I've made my point that the algebro-geometric meaning of the correspondence is by no means clear. -But I should add the significance should not be underestimated. For example, Simpson's -work shows that among all representations of the fundamental group of a variety, the -ones having Hodge theoretic (e.g. geometric) origin hold special status in this framework. -In particular, he showed that any representation can be deformed to such a representation, -which I think was totally unexpected.<|endoftext|> -TITLE: Combinatorial Problem -QUESTION [7 upvotes]: Any ideas or references on how to approach this problem? Every element in a set has a parameter $p_i \geq 0$ and $c_i \geq 0$. The objective is to find a subset which maximizes -$\prod_{i \in S} p_i + \sum_{i \in S} c_i$ -I tried to prove complexity results but was unsuccessful. My guess is that it is NP-Hard, but I cannot find a reduction from any known problem. -A greedy heuristic which sequentially adds elements which give the highest increase works fine, but can give arbitrarily bad results. - -REPLY [14 votes]: It is actually an easy one. Suppose that you were allowed to take any part of each item adding $tc_i$ and multiplying by $p_i^t$ for any $t\in[0,1]$. First, you should take everything that has $p_i\ge 1$ (because it never hurts). Then you have to maximize $C\prod_j e^{-b_j t_j}+\sum_j t_j c_j$ where $b_j=-\log p_j>0$. Now the points $(\sum t_j b_j, \sum t_j c_j)$ form a convex polygon and you know its boundary (half of it comes from arranging $c_j/b_j$ in the decreasing order and the other half from arranging them in the increasing order). Moreover, the functional $(x,y)\mapsto Ce^{-x}+y$ is convex, so its maximum is attained at some vertex on the boundary (which correspond to some subsets) and you are done in $n\log n$ time most of which is spent on sorting. -Now, if the task were to minimize that quantity, that would be worse than the knapsack packing problem.<|endoftext|> -TITLE: Why is the cardinality of the codomain of a ring epimorphism at most the cardinality of the domain? -QUESTION [8 upvotes]: According to this page and thence linked text, if $e : R \to S$ is an epimorphism of rings, then the cardinality of $S$ cannot exceed the cardinality of $R$. This is a non-trivial observation because epimorphisms of rings need not be surjective. Is there a "layman's" explanation of this fact, one that does not require me to learn French? - -REPLY [2 votes]: An explanation of a layman to a layman. -Let $T={\rm Im}\ e$. Then embedding $T\to S$ is again an epimorphism. With respect to $T$ the ring $S$ behaves like the ring (not necessarily the field!) of fractions (compare $\mathbb{Z}$ and $\mathbb{Q}$), so it has the same cardinality. It is enough? :-) I think it is possible to give a rigorous proof by so-called "zigzag-theorem" from Theory of semigroups.<|endoftext|> -TITLE: What are the irreducible modular representations of $SU(n,p)$? -QUESTION [6 upvotes]: To fix notation, by $SU(n,p)$, I mean the subgroup of $SL_n(\mathbb F_{p^2})$ consisting of matrices $A$ which satisfy $\overline A^t A = 1$, where $\overline A$ is the matrix given by raising all the entries of $A$ to the power of $p$. -So my question is what are the irreducible algebraic representations of this group defined over $\overline {\mathbb{F_p}}$? In particular, are they the same as the representations of $SL(n,p)$? Mucking around with the MeatAxe in GAP, I've checked this is true for a few small primes and $n\leqslant 3$. -Is it true in general? - -REPLY [7 votes]: Yes, it's a theorem of Steinberg from his fundamental 1963 Nagoya Math. J. paper, in the pre-Meataxe era. This is treated in Chapter 2 (especially 2.11) of my LMS Lecture Note Series No. 326 -Modular Representations of Finite Groups of Lie Type along with full references and discussion. The main point here is that the irreducible modular representations (in the defining characteristic) come from the ambient algebraic group (special linear group) by restriction to either of the two finite groups. -Of course, Steinberg's theorem is not obvious. -By the way, the ordinary characters of the two groups are related in an interesting way as well, though of course the degrees can't match precisely because the group orders are somewhat different.<|endoftext|> -TITLE: Infinite exponential representation of real numbers -QUESTION [39 upvotes]: I was thinking about infinite exponential representation of real numbers (like $2=e^{e^{-e^{-e^{e^{-e^{e^{e^{-e^{-e^{-e^{-e^{-e^{e^{-e^{e^{e^{-e^{e^{\cdot^{\cdot^{\cdot}}}}}}}}}}}}}}}}}}}}}$. -The sequence of signs before exponents can be obtained by repeated application of $\ln|x|$ to $2$ and taking a sign of each result. It seems that this gives an almost 1-1 correspondence between $\mathbb{R}$ and the set of infinite sequences of signs (or positive and negative 1's) , except that $0$ has two representations $\pm e^{-e^{e^{e^{\cdot^{\cdot^{\cdot}}}}}}$ (this is also true for every number that has representation as a finite exponent tower ending with $0$) and sequences $\pm e^{e^{e^{\cdot^{\cdot^{\cdot}}}}}$ diverge to $\pm \infty$. -Has this representation been studied? Does any algebraic number (except $0$, $1$ and $-1$) have an eventually periodic representation? What we can say about frequency of each sign in representation of a particular number (say, $2$)? (first several hundreds of elements suggest that $-1$ appears two times more often than $1$). Are there arbitrarily long runs of the same sign? - -REPLY [10 votes]: First, even though I think this is a fun question, -it's not really research mathematics and I'm not sure it belongs on mathoverflow. -(You know that some really smart people answer questions on math.stackexchange, right?) -As was noted in Robert's answer, one is investigating the sequence $x_{n+1} = | \log(x_n)|$, which makes -sense for all $x_0$ outside some countable set. Moreover, the periodic points -will (surely) be transcendental, and it will be impossible to prove this fact -(except for periods of length $1$). -So what should one expect for the number of $+$ and $-$ signs? -Here is a heuristic description of what happens, with the caveat that I have made no attempt to be rigorous (to do so, I would compare this with the probability theory of the -Gauss–Kuzmin distribution). -First, choose $x_0$ in $[0,\infty]$ according to some probability measure -$f_0(x)$, with cumulative probability distribution $F_0(x)$. -Let $f_n(x)$ be the distribution of $x_n$. -By definition, $f_n(x)$ must satisfy the following equation: -$$\int^{b}_{a} f_n(x) dx = \int_{e^a}^{e^b} f_{n-1}(x) dx + \int_{e^{-b}}^{e^{-a}} f_{n-1}(x) dx$$ -for all $b \ge a \ge 0$. -If $F_n(x) = \int^{x}_{0}f_n(t)dt$ is the cumulative distribution function of $x_n$, then this equation becomes: -$$F_n(b) - F_n(a) = F_{n-1}(e^b) - F_{n-1}(e^a) + F_{n-1}(e^{-a}) - F_{n-1}(e^{-b}).$$ -Letting $a = 0$, one obtains: -$$F_n(z) = F_{n-1}(e^z) - F_{n-1}(e^{-z}).$$ -Given some basic assumption on $F_0(z)$, the sequence -of functions $F_n(z)$ converges to the unique increasing function $F(z)$ -such that -$$F(z) = F(e^z) - F(e^{-z}).$$ -The independence of $F(z)$ on $F_0(z)$ implies that this function -should describe the cumulative distribution function of $x_n$ for $n$ sufficiently large -for almost all initial values $x_0$. -By choosing random functions $F_0(z)$, one can estimate that -$$F(1) \simeq 0.6518\ldots$$ -Since the sign in the exponential is determined by whether $x_n > 1$ or not, -this implies that the ratio of $-$ signs to $+$ signs (for almost all initial values, which presumably includes $x_0 = 2$) - is roughly $1.872$ to $1$. This seems to confirm what you observed experimentally. -Moreover, since the function $F(z)$ is strictly increasing, -it follows by Kolmogorov's zero-one law that, for almost all initial values $x_0$, that there are arbitrarily long -runs of $+$ signs, $-$ signs, etc. - - Edit: To make this completely rigorous, one can -make $\mathrm{R}^{+}$ a compact measure space given by the measure -specified by $\mu([a,b]) = F(b) - F(a)$. -The function $T:=|\log(x)|$ (modified so that $T(1) = 1$) is then measure preserving and (as is relatively easy to check) ergodic. -The claims then follow from the Birkhoff Ergodic theorem, for suitable choices of -test function $f$ (like the step function which is zero for $x < 1$ and one for $x > 1$). -(BTW, I may have added a few more decimal digits above than was really justified.)<|endoftext|> -TITLE: Homotopy equivalence between the Grassmannian Gr_{n,m} and Gr_n \times Gr_m. -QUESTION [6 upvotes]: The following assertion appears in a paper I am reading, and I can't seem to verify it. -Let $\text{Gr}_{n,m}$ denote the set of pairs $(V,W)$ where $V$ and $W$ are as follows. - -$V$ is an $n$-dimensional subspace of $\mathbb{C}^{\infty}$. -$W$ is an $m$-dimensional subspace of $\mathbb{C}^{\infty}$. -$V$ and $W$ are orthogonal. - -The space $\text{Gr}_{n,m}$ has an obvious topology. If $\text{Gr}_n$ and $\text{Gr}_m$ are the usual Grassmannians of $n$ and $m$ planes in $\mathbb{C}^{\infty}$, then there is an obvious map $\psi : \text{Gr}_{n,m} \rightarrow \text{Gr}_n \times \text{Gr}_m$. -The map $\psi$ is almost a homeomorphism, but not quite because of condition 3 above. The paper claims that $\psi$ is a homotopy equivalence. -Thanks for any help! - -REPLY [10 votes]: The forgetful map $Gr_{n,m} \to Gr_n$ that drops $W$ is a fiber bundle (exercise), and the map $Gr_{n,m} \to Gr_n \times Gr_m$ is a map of fiber bundles. It's an equivalence on the (connected) base space, so it suffices to check that the map of fibers is an equivalence. -The fibers over $V$ are, respectively: $m$-dimensional subspaces in $V^\perp \subset \mathbb{C}^\infty$, and $m$-dimensional subspaces in $\mathbb{C}^\infty$. -The inclusion of one infinite-dimensional complex vector space in another induces a homotopy equivalence of Grassmannians; you could construct an explicit homotopy equivalence by choosing an appropriate basis, or you could argue that the associated map of Stiefel manifolds is a homotopy equivalence (both are contractible, so this is easy) and so it passes to an equivalence after taking the quotient by the general linear group. - -REPLY [5 votes]: Look at the canonical principal $U(n)\times U(m)$ bundle over $Gr_{n,m}$ given by pairs of orthonormal frames $(v_1,\ldots, v_n), (w_1,\ldots w_m)$. Its total space is the set of all orthonormal $n+m$-frames in $\mathbb C^\infty$. It's contractible (that's well-known) and hence $Gr_{n,m}$ is a homotopy $B_{U(n)\times U(m)}$ which is clearly homotopy equivalent to $B_{U(n)}\times B_{U(m)}=Gr_n\times Gr_m$. To see that the natural map $Gr_{n,m}\to Gr_n\times Gr_m$ is the one inducing an equivalence notice that it's obviously covered by a map of principal bundles and hence the result follows by 5-lemma since total spaces are contractible and the fibers are the same.<|endoftext|> -TITLE: Margulis-Ruelle inequality for piecewise continuous interval maps -QUESTION [5 upvotes]: The Margulis-Ruelle inequality states that measure-theoretic entropy is controlled by Lyapunov exponents; more precisely, if $f$ is a $C^{1+\alpha}$ diffeomorphism on a $d$-dimensional manifold $M$ and $\mu$ is a Borel $f$-invariant ergodic probability measure with Lyapunov exponents $\lambda_1, \dots, \lambda_d$, then $h_\mu(f) \leq \sum_{\lambda_i>0} \lambda_i$. -This also holds for non-invertible $C^1$ interval maps: we have $h_\mu(f) \leq \max(0,\lambda(\mu))$, where $\lambda(\mu) = \int \log|f'(x)|\\,d\mu(x)$. (See, for example, Proposition 4.1 of [Ledrappier, Some properties of absolutely continuous invariant measures on an interval, Ergodic Theory Dynam. Systems 1 (1981), 77-93].) -Question: Has this result been proved for piecewise $C^1$ interval maps? I would be surprised if it is not true in this setting, but neither I nor anyone I've asked has been able to produce a reference to a proof in the case where $f$ is a piecewise monotonic interval map without any assumption of Markov structure. - -REPLY [2 votes]: You can find a complete proof of a slightly more general result in my paper "A. Barrio Blaya and V. Jimenez Lopez, On the relations between positive Lyapunov exponents, positive entropy, and sensitivity for interval maps, Discrete Contin. Dyn. Syst. 32 (2012), 433-466", see Theorem 7.1. You can download the paper from here.<|endoftext|> -TITLE: Example of a topos that violates countable choice -QUESTION [6 upvotes]: At this nLab page we have the line - -In contrast, any topos that violates countable choice, of which there are plenty, must also violate internal COSHEP. - -It doesn't give an example, and neither does the page on countable choice. So, what are these all-so-common examples? - -REPLY [7 votes]: If you're looking for a purely topos-theoretic model, I think you don't need to go through set theory (even though the end result may end up being basically equivalent). Look at the topos of continuous actions of the pro-completion of the integers, which is to say, the category of sets equipped with an automorphism all of whose orbits are finite. Here we have an N-indexed family of objects (one orbit of each cardinality) which are all inhabited, but whose product is empty -- hence the NNO is not internally projective. -Have you read P. Freyd's paper "The Axiom of Choice"?<|endoftext|> -TITLE: Two kinds of equivalence: conjugate vs. isomorphic objects -QUESTION [7 upvotes]: Conjugate vertices in a graph1 or conjugate elements of a group2 are equivalent (indistinguishable, essentially the same) in one specific structural sense. -Isomorphic objects in a category are equivalent in another specific structural sense. -In both cases we don't look inside the objects but declare them equivalent from the outside. -In both cases equivalence has to do with isomorphism: with structure preserving maps between the structure the conjugate elements live in and itself (automorphisms) resp. with iso arrows between the elements themselves. -Define two objects $A,B$ in a category to be conjugate when there is an isomorphism endofunctor $F$ with $F(A) = B$. - -Question 1: Is it true that any two - isomorphic objects are conjugate - (since there is an isomorphism - endofunctor that permutes them)? - -The reverse is most certainly false: There are categories with conjugate objects that are not isomorphic. E.g. the graphs -     (source) -in the category of graphs over two fixed vertices (with graph homomorphisms as morphisms) are two such objects (#9 and #6 in the diagram below). Note that there is no morphism at all between these two graphs. - -Question 2: Might it be the case - that whenever two objects are conjugate-but-not-isomorphic there is no - morphism between them? Or is this true only in special categories and/or special cases? -Question 3: How "normal" is it that a category contains - conjugate-but-not-isomorphic objects? - -Most of all I'd like to know how to think about this bewildering pair of equivalences in general terms. -Appendix -Here is the complete category of graphs over two fixed vertices and an arrow whenever there is a graph homomorphism. Compositions and identities are omitted. -     (source) -The numbers are derived from the adjacency matrices: 0 = 00|00, 1 = 10|00, ..., 15 = 11|11. -The numbers of the two graphs above are 10|01 = 9 and 01|10 = 6. -Footnotes -1 $x,y$ are conjugate iff there is a $g \in \text{Aut}(G)$ with $g(x) = y $. -2 $x,y$ are conjugate iff there is a $g \in G$ with $gx= yg$. - -REPLY [4 votes]: Question 2 is false for some finite categories. There are two objects $A$ and $B$ that both have an idempotent endomorphism to themselves. There is a morphism from $A$ to $B$ and one from $B$ to $A$, their composition is the idempotent endomorphism. (The best realization I have for this is two finite simple groups, neither being a subgroup of the other, without their automorphisms or with the same automorphism group.) -However, Question 2 is true for finite categories where all endomorphisms are invertible. -Suppose there is an arrow $A\to B$ and an isomorphism $F(A)=B$. $F$ must have some finite order $n$. Consider the arrows $F(A\to B),F(A\to B), F^2(A\to B),...,F^{n-1}(A\to B)$. These arrows form a cycle, therefore they all have inverses.<|endoftext|> -TITLE: Is the Gelfand-Graev character isomorphic to a cohomology group for some sheaf on a Deligne-Lusztig variety? -QUESTION [11 upvotes]: Deligne-Lusztig theory -is awesome. You take a maximal torus $T$, you take a character $\theta$, construct a variety $X_T$$^*$, take etale cohomology, get a virtual character $R_T^\theta$, maybe it's reducible, so you try to decompose it. -Gelfand-Graev character -is awesome. You take a maximal unipotent subgroup in some maximal split Borel subgroup, take a generic character, induce the character to the whole group, and you get many interesting subrepresentations. -My question - -Is the Gelfand-Graev character equal to the character of the cohomology of some sheaf on some nice variety, similar to a Deligne-Lusztig character? - -Why is this interesting? -Say you have an $R_T^\theta$ that is reducible. Before trying to find explicitly all constituents, let's try to decompose it first into constituents of the Gelfand-Graev character (generic), and the rest (not generic). If $R_T^\theta$ has exactly two subrepresentations, one generic and one not generic, then we needn't look further. -What am I looking for? -The best thing would be if there was a sheaf $F_{GG}$ on $X_T$ with cohomology, in the $\ell(w)$-degree, with character equal to the Gelfand-Graev character. Then we might have a sequece of sheaves -$$0\rightarrow F' \rightarrow F_\theta \rightarrow F_{GG} \rightarrow F^{''} \rightarrow 0$$ -and we might get that the cohomologies of $F'$ and $F^{''}$ will break our $R_T^\theta$ into two parts. -So, in essence, what I'm looking for, is a geometric way to break a Deligne-Lusztig character into its genereic and non-generic parts. -This might not be possible, at least not in the way I described, which is very naive and wishful. The sentence before last should be regarded as the real question. -(*) Non-standard notation, I know. Fix some maximal $F$-stable torus, let $w$ be the Weyl element that twists the torus to desired $T$, and let $X_T=X(w)$, where $X(w)$ is the standard notation Deligne-Lusztig variety. - -REPLY [3 votes]: This is not precisely what you're looking for, but Bonnafé and Rouquier have made in -"Coxeter orbits and modular representations." Nagoya Math. J., 183 :1–34, 2006, -the conjecture that the DL-restriction of a GG-module is a (suitably shifted) GG-module. -In particular it is concentrated in one (explicit) degree. This is even stated for integral coefficients and they give a geometric proof for T=Coxeter torus. Olivier Dudas has proved this statement for any torus in "Deligne-Lusztig restriction of a Gelfand-Graev character", -Annales scientifiques de l'ENS (42), 2009, pp 653-674. He also explains a couple of consequences of the statement, such as the fact that generic constituents of the cohomology of $X(w)$ only occurs in degree $l(w)$, etc. -As for the possibility of realizing the full GG module in the cohomology of a single DL variety, I am as skeptical as Jim Humphreys.<|endoftext|> -TITLE: Geometric examples of the Serre intersection formula -QUESTION [16 upvotes]: The Serre intersection formula, as an alternating sum of contributions from Tor-groups, is something that combines a lot of ingredients that I'm interested in, but I've never really felt that I have a "grip" on it. One of the reasons for this is that, despite making attempts on a couple of occasions, I never seem to have been able to concoct an example where I get contributions from higher Tor-terms that's geometrically satisfying. -This is, unfortunately, a vague description of what I'm looking for. When you describe intersection multiplicity, it's relatively easy to give examples of double and triple intersection of planar curves, give a proto-definition in terms of degree of tangency, show how this is captured by their scheme-theoretic intersection, and talk about what happens when the curves are moved slightly. This is the kind of thing I miss having. -I should also clarify that I'm looking for something slightly more complicated than intersecting a planar curve with a point. -Are there examples of the Serre intersection formula in this vein? - -REPLY [10 votes]: Consider a flat morphism $f:X\to Y$ of smooth connected varieties. For instance let $X=Y\times F$ with $X,Y,F$ all smooth. Further let $Z\subset X$ be a generically reduced subvariety such that $f|_Z:Z\to Y$ is a finite morphism, which is not flat. For any $y\in Y$ let $X_y\subset X$ denote the fiber of the original $f$. -As any two points on $Y$ are equivalent, the intersection product $Z\cdot_X X_y$. is independent of $y\in Y$. -Since $Y$ is smooth and $Z$ is generically reduced, most fibers of $f|_Z$ are smooth, so the intersection product $Z\cdot_XX_{y}$ for a general $y\in Y$ is given by the length of $\mathscr O_{X_{y}}$, in other words, there is no $\mathrm{Tor}$ contribution there. -Now take a point $y\in Y$ such that $f_Z$ is not flat in any neighbourhood of $y$ in $Y$. -It follows that the length of $\mathscr O_{X_y}$ (which is also the Hilbert polynomial of the fiber) has to be different from the value that we get from the smooth fibers. Therefore there has to be $\mathrm{Tor}$ contribution there. -Perhaps this example shows why it is $\mathrm{Tor}$ that comes in: $\mathscr O_{Z_y}$ is not flat on $Z$, so it does not give the right length and thus has to be corrected. It is arguably intuitive that the alternating sum of the length of the $\mathrm{Tor}$'s will be constant as $y$ runs through the (closed) points of $Y$. -I don't know if you find this geometrically satisfying. I would paraphrase what Hailong said and say that anything involving $\mathrm{Tor}$ should not be geometrically obvious, on the other hand $\mathrm{Tor}$ measures the failure of exactness of the tensor product so the failure of flatness is an obvious condition to look at. Then again, all of this just reiterates the fact that flatness is a difficult notion to truly comprehend (and I don't claim I do). - -The examples/comments given by JC and Hailong fit into this example: - -Since $Y$ is smooth and $f|_Z$ is finite, it is flat if and only if $Z$ is Cohen-Macaulay, which gives us Hailong's comment, and -a concrete example is given by $X=\mathbb A^4$, $Y=\mathbb A^2$, $f$ the obvious projection, and $Z$ the union of two planes meeting in a single point both of which project onto $Y$ isomorphically, which gives us JC's example.<|endoftext|> -TITLE: SL(2,C) Chern-Simons theory in genus 1 -QUESTION [5 upvotes]: In http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1104202513, Witten claims (p. 54) that to quantize the moduli space of flat $SL(2,\mathbb{C})$ connections on a torus, one can simply quantize the cotangent bundle of a real torus and take the part invariant by the Weyl group $W$. -I agree that $T^*(T\times T)/W$ is the moduli space of flat connections on the torus. However, it seems to me that the correct symplectic form should be the following: if we parametrize $T\times T$ with coordinates $(r_1, \theta _1, r_2, \theta _2)$ (quotienting the $\theta$ coordinates appropriately) then the symplectic form $\omega$ should be induced by $dr_1\wedge dr_2+d\theta _1 \wedge d\theta _2$. This restricts to the correct symplectic form on the real torus $T\times T$ obtained by taking the $r_1 = r_2 = 1$ subspace (which appears to be the $SU(2)$ character variety with its correct symplectic structure). But this is not the symplectic form induced by the cotangent structure, which would be $\omega'=d\theta _1 \wedge dr_1 + d\theta _2 \wedge dr_2$. On the other hand $\omega'$ vanishes on the real torus corresponding to $SU(2)$ representations, and does not seem to be the form induced by the character variety, which comes equipped with a natural symplectic form. -This makes a substantial difference in quantization because $\omega'$ is exact and hence we can take the trivial line bundle for prequantization, and then in a real polarization obtain a Hilbert space $L^2(T\times T)$ (which is what Witten claims to be the quantization of the moduli space). However, if we take $\omega$ as the symplectic form, $\omega$ is non-zero in cohomology, and we will end up with a more complicated quantization. - -REPLY [7 votes]: Let me call your $\omega$ as $\omega_I$. The symplectic form you get from the Chern-Simons action is $k\omega_I+s\omega_K$, where $\omega_K$ is one of the Kähler forms on the Hitchin space, which, in particular, is exact. If you choose a real polarization as Witten does, the Hilbert space is $\Gamma(Bun_GX,Det^{\otimes k})$, where $Det$ is the determinant bundle whose first Chern class $[\omega_I]$. One should note that the polarization is not the naive vertical polarization on $T^*Bun_GX$ since the fibers are not Lagrangian for $k\neq 0$. -Narasimhan-Seshadri identifies $Bun_GX$ with the character variety for the compact group, which in genus 1 is $T\times T\ /W$. So, the Hilbert space is $\Gamma(T\times T\ /W, Det^{\otimes k})$, precisely what Witten claims after eq. (5.11).<|endoftext|> -TITLE: Eigenvectors and partitions of graphs -QUESTION [8 upvotes]: Let G be an undirected graph with the node set $V$ and the Laplacian matrix $L$. Let $N(v)$ denote the neighbors of a node $v$ and $|N(v)|$ its degree. Then a partition $\pi=(V_1, V_2, \ldots, V_k)$ is almost equitable if it holds that $\forall i \ne j\in\{1,\ldots,k\}$ $\forall v, u\in V_i$ $|N(v)\cap V_j|=|N(u)\cap V_j|$, i. e. that the number of neighbors of a node $v$ in $V_i$ in a different component $V_j$ does not depend on the choice of $v$. -It is known that if $\pi$ is an almost equitable partition of $L$, then (after reordering the vertices appropriately) $L$ has an eigenvector of the form $x=(x_1,x_2,\ldots,x_k)^T$, where $x_i\in {\mathbb{R}}^{|V_i|}$ and $x_i=c_i\mathbf{1}$, $c_i\in \mathbb{R}$. -I had previously asked whether the converse was also true. The converse is not true. For example, the graph with the Laplacian $\begin{pmatrix} 2 & -1 & -1 & 0 & 0 & 0\\\ -1 & 3 & -1 & -1 & 0 & 0\\\ -1 & -1 & 3 & -1 & 0 & 0\\\ 0 & -1 & -1 & 4 & -1 & -1\\\ 0 & 0 & 0 & -1 & 1 & 0\\ 0 & 0 & 0 & -1 & 0 & 1\end{pmatrix}$ has an eigenvector $x=(c_1, c_2, c_2, c_2, c_3, c_3)^T$, which does not correspond with an almost equitable partition. However, if you consider the graph structure, you see that each node in a component $V_i$ has either the same amount of edges to a component $V_j$ or no edges to $V_j$. Just from looking at examples, it seems to always be the case if an eigenvector with repeated entries is present in the graph. However, so far, I have not been able to prove this. -Are there any results about this in the literature? Alas I am not very familiar with algebraic graph theory, but all the literature I could find on Laplacian eigenvectors either did not relate the eigenvector to the graph structure (apart from the Perron ev), or only studied graphs that allowed almost equitable partitions. -Are there necessary and sufficient conditions on the graph structure, such that the graph Laplacian does not have eigenvectors of the form $x$ (apart from $\mathbf{1}$)? - -My original question has led to a different one. I have edited my original question, hoping that this is the right way to proceed? - -REPLY [2 votes]: final thoughts Let us start with a vector (or just specify the distinct entries) and then try to build a graph. With only 2 distinct entries the partition does need to be almost equitable. I think that it is easy to create high irregularity if some entries are 0. Here is an example cobbled up ad hoc (goals three distinct entries, none being zero. every vertex has neighbors with the other two labels, every label showing two types of vertices): Suppose that the entries are $1$,$2$,$-3$ (how many times each to be determined later). We can ignore edges between vertices with the same label. Suppose $u,v$ are two vertices labeled $1$ , that $u$ has $a$ and $b$ neighbors labeled $2$ and $-3$. Then the eigenvalue is $1(a+b)-2a+3b=-a+4b$ I'll pick $a=1$ and $b=5$ giving eigenvalue $19.$ Then $-c+4d=19$ where $c$ and $d$ are the number of edges from $v$ to vertices labelled $2$ and $-3$ I'll say that some have $c=5$ and $d=6$. Similar calculations and choices lead to this possible situation - -$[p;1 \rightarrow 2^1,-3^5]$ meaning $p$ vertices labeled 1 each having 1 neighbor labeled 2 and 5 labeled -3 -$[q;1 \rightarrow 2^5,-3^6]$ -$[r;2 \rightarrow 1^3,-3^7]$ -$[s,2 \rightarrow 1^8,-3^6]$- -$[t,-3 \rightarrow 1^{13},2^1]$ -$[u,-3 \rightarrow 1^8,2^5]$ meaning $u$ vertices labelled -3 with 8 and 5 neighbors labeled 1 and 2 respectively - -Now we have 6 equations in the (positive integer) variables which will need to be satisfied. Counting in two ways edges with ends labelled 1 and -3 gives $5p+6q=13t+8u$ -If my calculations are correct then one possible solution is $[p,q,r,s,t,u]=[21m,4m,3m,4m,5m,8m]$ I do not think that there would be any problem (other than tedium) in assigning edges to make that work (say with m=10 or 20). At any rate, the idea is clear. -later thoughts You rephrase your question as (essentially) "what conditions on a graph are necessary for it to have at least one positive eigenvalue (for the Laplacian) with at least one eigenvector with at least one repeated entry?" I don't think that is exactly what you intended but for that question I don't expect a satisfying answer. However many conditions are sufficient. -In your example of $[a,a,-a,-a,0,0,0,0]$ the corresponding graph does have the equitable partition with 6 singleton cells along with $\lbrace6,8\rbrace$ and that partition does support the eigenvector. Every vertex is adjacent to both or neither of $6,8$ so one also has the eigenvector $[0,0,0,0,0,1,0,-1]$ for $3$ (as $6$ and $8$ are not connected and have degree $3$) This means that every eigenvector for other eigenvalues does have equal entries in positions 6 and 8 (as they will be orthogonal to that one). -It turns out that $\lbrace 1,4\rbrace,\lbrace2,3,6,8\rbrace,\lbrace5\rbrace,\lbrace7\rbrace$ is almost equitable. This means that it supports $4$ eigenvalues with eigenvectors. Leaving out the edges inside each part gives a bipartite graph with those same eigenvalues and eigenvectors and a nice structure. -Any time you have an eigenvalue of multiplicity greater than 1 you can arrange for it to have an eigenvector with some repeated entry by taking the right linear combination of two independent eigenvectors. Your graph has $3$ and $\frac{7 \pm \sqrt{17}}{2}$ as eigenvalues of multiplicity $2$ along with $0$ and $4$ of multiplicity $1$. -Given your comments, your question seems to be better put as: What conditions on a partition of the vertices is necessary for it to support an eigenvector for (a positive eigenvalue of) the Laplacian? Perhaps you wish to add the condition that the eigenvector takes distinct values on distinct cells. There i don't know but I still don't expect much. -earlier I am having trouble editing my previous answer so here are later thoughts. -With regard to the Laplacian Matrix: - -For any graph the all $1$'s vector is an eigenvector of the Laplacian matrix although the partition with one class is only equitable if the graph is regular. -Consider a graph and an eigenvector $v$ of its Laplacian matrix which has a healthy number of repeated values (perhaps due to an underlying equitable partition) now freely add and delete edges joining vertices with the same value. I claim that for the new graph the vector $v$ will still be an eigenvector for the same eigenvalue although the partition need not be especially nicely behaved. -If one has an equitable partition into $k$ parts then not only does it have an eigenvector of the form you mention but it has $k$ such eigenvectors (with multiplicity). However the same argument as above shows that you can change the graph by arbitrary changes of edges within each part and have the same eigenvectors for the same eigenvalues. -Consider a graph with $n$ vertices which has an eigenvalue with multiplicity $m>1$ ( say that $W$ is the space of eigenvectors for that value) then for any $p0$. Partition the vertices into subsets (preferably, many small ones) each with sum of corresponding entries 0. Now assign to each subset a new vertex joined to all its members and getting the weight 0. The new vector is an eigenvaue for the new matrix for eigenvalue $\lambda+1.$ -If an eigenvector has one or more $0$'s then one can hang off of them any structure of further edges and vertices which all get a value of zero and this gives an eigenvector for the new graph. - -With regard to the incidence matrix: I think that my argument in the previous answer does show that a partition into $k$ parts is equitable if and only if it supports $k$ linearly independent eigenvectors.<|endoftext|> -TITLE: convexity of images of space-filling curves -QUESTION [19 upvotes]: Suppose $f:[0,1]\to[0,1]^2$ is continuous and for each $t\in[0,1]$, the area of $\lbrace f(s) : 0\le s\le t \rbrace$ is $t$. For what sets of values of $t\in[0,1]$ can $\lbrace f(s) : 0\le s\le t \rbrace$ be convex? All $t$? Only countably many $t$? If so, which countable sets? Topologically discrete ones? Dense ones? - -REPLY [12 votes]: As I said in the comment above, I think that the set of points $t\in[0,1]$ where a square-filling curve with strictly increasing area defines a convex $f([0,t])$, is a nowhere dense closed set containing $0$ and $1$, and conversely, any such set can be obtained this way. While I'm not sure about how to show that such a set is always nowhere dense, the other direction seems easier to pursue, and allows a nice construction (I'll try to include a picture too). Precisely: - -For any closed nowhere dense subset -$C$ of $I:=[0,1]$ containing $0$ and -$1$ there exists a square-filling -curve $f:I\to I\times I$ with the -property that $f([0,t])$ has area $t$ -for all $t\in I$, and $f([0,t])$ is -convex exactly for all $t\in C$. - -For convenience, I'll describe the construction with a slight variation in the parametrization, requiring that the curve satisfies, for all $t\in C$, $f(t)=(0,t)$ (thus, at any time $t\in C$ it touches the right vertical edge of the square, at heigh $t$). The area will be strictly increasing, for instance with $\operatorname{Area}\big(f([0,t])\big)=\phi(t):=3t^2-2t^3$ for all $t\in I$ (any other homeomorphism $\phi$ of $[0,1]$ in itself such that $\phi(t)=o(t)$ and $\phi(1-t)=o(t)$ as $t\to0$ works as well). Of course, if one started with $C\:':=\phi(C)$, then one finds a curve $f\circ \phi^{-1}$ parametrized in "arc-area", as initially stated. -To start the construction we first need to fix the subsets $f([0,t])$, for all $t\in C$. To this end, note that there exists a nested family of closed, convex subsets of the square, $\{A_t\}_{t\in C}$, such that $A_0:=\{(1,0)\}$, $A_1:=I^2$, $\operatorname{Area}(A_t)=\phi(t)$ for all $t\in C$, and $\operatorname{diam}(A_ s \setminus A_r)=o(1)$ as $|s-r| \to0 $, (uniformly for $r$ and $s$ in $C$). -Instead of entering the details of the construction of these $A_t$, let's just say that they can be realized e.g. as sub-graphs of a family of concave functions $\alpha_t:I\to ]-\infty,1]$: -$$A_t:=\{(x,y)\in I^2\, :\, \alpha_t(x)\ge y\}$$ -where $\alpha _ s\leq\alpha _t$ for $s\leq t$ and -$\int_0^1\alpha^{+} _ t(x)dx=\phi(t)\, .$ -The graphs of these functions appear as a forest of binary trees leaning their branches towards the right vertical edge, with (possibly uncountable) leaves exactly at the set $\{1\}\times C$. They disconnect the square into a countable family of open regions, one for each component $J$ of $I\setminus C\, .$ -The curve $f:I\to I^2$ is defined to be $f(t)=(0,t)$ as said, for all $t\in C$. On any open interval $J:=]r,s[$ which is a component of $I\setminus C$, define $f_{|J}$ to be a Peano-like curve filling the set ${ A_s\setminus A_r }$ up to its closure, with end-points $f(s)=s$ and $f(r)=r$ as said, and parametrized in such a way that $\operatorname{Area}(f[s,t])=\phi(t)-\phi(s)$ for all $t\in J\, .$ -This defines a curve $f:I\to I^2$ with the stated properties. Note that the continuity is ensured by the requirement that $\operatorname{diam}(A_t \setminus A_s)=o(1)$ as $|t-s| \to0 $, (uniformly). Since we want the sets $f([0,t])$ to be convex at exactly the points $t\in C$, a small care is needed in order to avoid creating new convex sets $f[0,t]$ for $t\in I\setminus C$, but a small thoughts shows that this is not a problem (for instance the original Peano curve does have this property).<|endoftext|> -TITLE: Software for Computing Baker-Campbell-Hausdorff -QUESTION [13 upvotes]: Does anyone have a recommendation for software which can efficiently calculate the Baker-Campbell-Hausdorff series in classical Lie algebras? -Right now, I have a problem which boils down to understanding Baker-Campbell-Hausdorff with respect to a basis in su(2), and this seems like the kind of thing Sage or Mathematica should be able to handle. However, I haven't had to use computer algebra packages for Lie theory before, so I would love to be pointed in the right direction. -Many thanks, -Jesse - -REPLY [3 votes]: There is a quite comprehensive package for Lie algebras in Maple. It is developed by Ian Anderson (from Utah State not Jethro Tull).<|endoftext|> -TITLE: Constants for Rolle's Theorem applied to polynomials -QUESTION [19 upvotes]: Rolle's Theorem states that $f(1/2)=f(-1/2)+f'(x)$ has a root in the open real -interval $(-1/2,1/2)$ if $f$ is continuous and differentiable. How large can the absolute value of such a root -$\xi$ -be if $f$ is a polynomial of degree $d$? (We choose $\xi$ to be the real root of minimal norm in the case of several solutions.) -A few comments: -$\xi$ can be arbitrarily close to the boundary if there is no condition on the degree. -$\xi$ is always $0$ in degree $2$. -I have examples (using random polynomials) with $\xi=.28867...$ for $d=3$ -(the correct value is perhaps $1/(2\sqrt{3})$ given for example with $f(x)=x^3$), -and $\xi=.324...$ for $d=5$. -Considering for $f$ a complex polynomial, the equation $f(1/2)=f(-1/2)+f'(x)$ -seems always to have a complex solution of norm at most 1/2. Is this true? -If yes, does the real constant working for real polynomials of degree $d$ -also work for complex polynomials of degree $d$? (I have no counterexample.) -A positive answer for complex polynomials would imply the statement for entire -functions (where the smallest root could perhaps be on the boundary of the -complex disc of radius $1/2$). - -REPLY [16 votes]: For real polynomials $\xi$ can be $1/2 - O(1/d^2)$ but no closer to $1/2$ than that. The value $1/\sqrt{12} = 0.288675\ldots$ is correct for $d=3$, and also for $d=4$; for larger $d=2n-1$ or $d=2n$, the supremum occurs at half the largest root of the $n$-th Legendre polynomial $P_n$, and is not attained because at the extremal points $f$ also has points of inflection satisfying the Rolle condition, as at $x=0$ for the quintic polynomial $f(x) = x^3-4x^5$ which shows the supremum $\sqrt{3/20} = .387298\ldots$. (The reason $n=2$ is different is that $P_2(x) = (3x^2-1)/2$, and indeed $1/\sqrt{12}$ is half the largest root $1\sqrt{3}$, but there are no inflection points because $P_2$ has no zeros in the interior of $[-1/\sqrt{3},+1/\sqrt{3}]$. For $n=1$ the Legendre polynomial vanishes only at zero, which we know must be a Rolle point when $f$ is quadratic.) -Here's a Wolfram Alpha plot for $d=11$, for which $\xi = .4662\ldots$: -    (image source) -Given $n$, let $p(x)=P_n(2x)$, so that $p$ is a degree-$n$ orthogonal polynomial on $[-1/2,+1/2]$; denote its roots by $x_i$ in increasing order, $x_1 < x_2 < x_3 < \cdots < x_n$, with each $x_i = -x_{n+1-i}$ by symmetry. In particular $x_1 = -x_n$. For any polynomial $f$ of degree $d \leq 2n$, let $g(x) = f'(x) - cx$ where $c = f(1/2) - f(-1/2)$. I claim $g$ cannot be positive on $[x_1,x_n]$. Indeed $\int_{-1/2}^{1/2} \phantom. g(x) \phantom. dx = 0$, but by Gaussian quadrature $\int_{-1/2}^{1/2} \phantom. g(x) \phantom. dx$ is a positive linear combination of the $g(x_i)$ — contradiction. On the other hand, $g$ can be nonnegative but not identically zero on $[x_1,x_n]$, which happens iff -$$ -g(x) = (ax+b) \phantom. p(x)^2 / (x_n^2-x^2) -$$ -for some $b>0$ and $a \in [-b/x_n,b/x_n]$. (Note that $a=0$ deals with $d=2n-1$.) Indeed the same Gaussian-quadrature formula shows that $g$ must vanish on each $x_i$, with double roots except possibly at $x_1$ and $x_n$; this gives $g(x) = (ax+b) \phantom. p(x)^2 / (x_n^2-x^2)$, and then the inequalities on $a,b$ follow from nonnegativity on $[x_1,x_n]$. Conversely, if $g$ is of that form then -it is indeed nonnegative on $[x_1,x_n]$, and -$$ -\int_{-1/2}^{1/2} \phantom. g(x) \phantom. dx = -\int_{-1/2}^{1/2} \phantom. p(x) \frac{(ax+b)\phantom. p(x)}{x_n^2-x^2} \phantom. dx = 0 -$$ -because the integral is the inner product on $[-1/2,+1/2]$ of $p$ with a polynomial of degree less than $n$. -We can then take $f_0(x)$ to be an indefinite integral of $g(x)$, which satisfies $f_0(-1/2)=f_0(+1/2)$ and is strictly increasing on $[x_1,x_n]$. For each $\epsilon>0$ we can then perturb $f_0$ to get a polynomial $f$ of the same degree, still taking the same values at $-1/2$ and $+1/2$, but with a strictly positive derivative on $[x_1+\epsilon,\phantom, x_n-\epsilon]$. For instance we may take $f(x) = f_0(\theta x) + \delta\cdot x$ with $\theta$ slightly larger than $1$ and $\delta>0$ chosen so that $f(-1/2)=f(+1/2)$. This completes the proof.<|endoftext|> -TITLE: What does Mellin inversion "really mean"? -QUESTION [45 upvotes]: Given a function $f: \mathbb{R}^+ \rightarrow \mathbb{C}$ satisfying suitable conditions (exponential decay at infinity, continuous, and bounded variation) is good enough, its Mellin transform is defined by the function -$$M(f)(s) = \int_0^{\infty} f(y) y^s \frac{dy}{y},$$ -and $f(y)$ can be recovered by the Mellin inversion formula: -$$f(y) = \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} y^{-s} M(f)(s) ds.$$ -This is a change of variable from the Fourier inversion formula, or the Laplace inversion formula, and can be proved in the same way. This is used all the time in analytic number theory (as well as many other subjects, I understand) -- for example, if $f(y)$ is the characteristic function of $[0, 1]$ then its Mellin transform is $1/s$, and one recovers the fact (Perron's formula) that -$$\frac{1}{2\pi i} \int_{2 - i \infty}^{2 + i \infty} n^{-s} \frac{ds}{s}$$ -is equal to 1 if $0 < n < 1$, and is 0 if $n > 1$. (Note that there are technical issues which I am glossing over; one integrates over any vertical line with $\sigma > 0$, and the integral is equal to $1/2$ if $n = 1$.) -I use these formulas frequently, but... I find myself having to look them up repeatedly, and I'd like to understand them more intuitively. Perron's formula can be proved using Cauchy's residue formula (shift the contour to $- \infty$ or $+ \infty$ depending on whether $n > 1$), but this proof doesn't prove the general Mellin inversion formula. -My question is: - -What do the Mellin transform and the inversion formula mean? Morally, why are they true? - -For example, why is the Mellin transform an integral over the positive reals, while the inverse transform is an integral over the complex plane? -I found some resources -- Wikipedia; this MO question is closely related, and the first video in particular is nice; and a proof is outlined in Iwaniec and Kowalski -- but I feel that there should be a more intuitive explanation than any I have come up with so far. - -REPLY [9 votes]: Have a look at Zagier's appendix: -http://people.mpim-bonn.mpg.de/zagier/files/tex/MellinTransform/fulltext.pdf -It provides a nice description of the Mellin transform when $f(x)$ is sufficiently smooth at $x=0$, and of rapid decay at infinity. -For example, assume $f(x) = \sum_0^\infty a_n x^n$, in some neighbourhood of the origin, and decays rapidly as $x \to \infty$, then its Mellin transform has meromorphic continuation to all of $\mathbb{C}$ with simple poles of residue $a_n$ at $s=-n$, $n=0,1,2,3,\ldots$. This is nicely explained in Zagier's appendix. -So, rate of decay issues aside, shifting the inverse Mellin transform to the left, i.e. letting $\sigma \to -\infty$, picks up the residues of the integrand at s=-n, i.e. $a_n x^n$, i.e. recovers the Taylor expansion about $x=0$ of $f(x)$. -Of course, it only applies to a limited class of functions $f$, but, in many practical examples, this reasoning gives one explanation of why the Mellin inversion formula is true, without resorting to Fourier inversion.<|endoftext|> -TITLE: Is the Hausdorff metric on sub-$\sigma$-fields separable? -QUESTION [5 upvotes]: Let $(X,\mu,\mathcal{F})$ be a probability space. The paper Equiconvergence of Martingales by Edward Boylan introduced a pseudometric on sub-$\sigma$-fields (sub-$\sigma$-algebras) of $\mathcal{F}$ as follows: -$\rho(\mathcal{G},\mathcal{H}) - := \sup_{A\in \mathcal{G}} \inf_{B\in \mathcal{H}} \mu(A \triangle B) + \sup_{B\in \mathcal{H}} \inf_{A\in \mathcal{G}} \mu(A \triangle B)$ -where $A \triangle B$ is symmetric difference. -It seems to be called the Hausdorff pseudometric on $\sigma$-fields in later papers. (Does anyone know why?) Further, if we only consider a $\mu$-complete $\sigma$-fields then $\rho$ is a metric. Also, the paper shows $\rho$ is complete. - -Is this metric $\rho$ - separable---assuming, say, $X=[0,1]$ - and $\mu$ is the Lebesgue measure? - -My guess is that it is not, but I cannot off-hand come up with a witnessing set to show this. Considering the paper is 40 years old, I imagine this might be well-known. And if it is not separable, then my follow up question is this? - -Is there a known separable, complete - metric on the space of - $\mu$-complete sub-$\sigma$-fields? - -For reference, I found the following list $20of$20sigma-fields%22%7Csort:date/sci.math/iz249jCUvEU/rkKOei1NV5YJ" rel="nofollow noreferrer">online, compiled by Dave L. Renfro, of papers dealing with metrics on $\sigma$-fields (listed in Chronological order). I quickly looked though these papers and didn't find what I was looking for, but maybe I missed something. - -Edward S. Boylan, "Equiconvergence of martingales", -Annals of Mathematical Statistics 42 -(1971), 552-559. [MR 44 #7603; Zbl 218.60049] -Jacques Neveu, "Note on the tightness of the metric on the -set of complete sub sigma-algebras of a probability space", -Annals of Mathematical Statistics 43 (1972), 1369-1371. -[MR 48 #5133; Zbl 241.60036] -Hirokichi Kudo, "A note on the strong convergence of -sigma-algebras", Annals of Probability 2 (1974), 76-83. -[MR 51 #6900; Zbl 275.60007] -Lothar Rogge, "Uniform inequalities for conditional -expectations", Annals of Probability 2 (1974), 486-489. -[MR 50 #14858; Zbl 285.28010] -Louis H. Blake, "Some further results concerning -equiconvergence of martingales", Revue Roumaine de -Mathématiques Pures et Appliquées 28 (1983), 927-932. -[MR 86i:60130; Zbl 524.60029] -Hari G. Mukerjee, "Almost sure equiconvergence of -conditional expectations", Annals of Probability 12 -(1984), 733-741. [MR 86c:28012; Zbl 557.28001] -Beth Allen, "Convergence of sigma-fields and applications -to mathematical economics", pp. 161-174 in Gerald Hammer -and Diethard Pallaschke (editors), SELECTED TOPICS IN -OPERATIONS RESEARCH AND MATHEMATICAL ECONOMICS (Proceedings, -Karlsruhe, West Germany, 22-25 August 1983), Lecture Notes -in Economics and Mathematical Systems #226, Springer-Verlag, 1984. -[MR 86f:90029; Zbl 547.28001] -Dieter Landers and Lothar Rogge, "An inequality for the -Hausdorff-metric of sigma-fields", Annals of Probability -14 (1986), 724-730. [MR 87h:60006; Zbl 597.60003] -Abdallah M. Al-Rashed, "On countable unions of sigma -algebras", Journal of Karachi Mathematical Association -8 (1986), 57-63. [MR 88f:28001; Zbl 639.28001] -Maxwell B. Stinchcombe, "A further note on Bayesian -information topologies", Journal of Mathematical Economics -22 (1993), 189-193. [MR 93k:60011; Zbl 773.90016] -Timothy Van Zandt, "The Hausdorff metric of sigma-fields -and the value of information", Annals of Probability 21 -(1993), 161-167. [MR 94d:62012; Zbl 777.62007] -Xikui Wang, "Completeness of the set of sub-sigma-algebras", -International Journal of Mathematics and Mathematical -Sciences 16 (1993), 511-514. [MR 94f:28002; Zbl 782.28001] -Zvi Artstein, "Compact convergence of sigma-fields and -relaxed conditional expectation", Probability Theory and -Related Fields [= Zeitschrift für Wahrscheinlichkeits- -theorie] 120 (2001), 369-394. [MR 2002g:28003; Zbl 992.28001] - -REPLY [5 votes]: Take a sequence $A_n$ of independent sets of measure $1/2$. Given two different subsets $B$ and $C$ of natural numbers, suppose WLOG that there is an $n$ in $B\sim C$. Now $\mu(A_n\Delta A) = 1/2$ for all sets $A$ which are independent of $A_n$, so the distance from the sigma algebra generated by $(A_n)_{n\in B}$ to the sigma algebra generated by -$(A_n)_{n\in C}$ is at least $1/2$. This shows that the density character of your space is at least the continuum.<|endoftext|> -TITLE: Galois theory and algorithms -QUESTION [8 upvotes]: Steven Weintraub's book {\em A Guide to Advanced Linear Algebra} includes the following remark: -"Of course, there is no algorithm for factoring polynomials, as we know from Galois theory." -I can't make sense of this. I feel confident that Galois theory doesn't speak to the question of algorithms, and confident that there do exist algorithms for factoring integer polynomials over the integers (after Kronecker), and strategies for computing in the field of algebraic numbers that make tautological the question of factoring polynomials irreducible over the rationals. -Have I missed some way to salvage this remark? - -REPLY [2 votes]: Hermite ca 1858 in Comptes Rendus, solved the general 5-ic using the j-function -and the modular polynomial Phi_5(j(z), j(5z)).<|endoftext|> -TITLE: Patching together homeomorphisms: how badly can it fail? -QUESTION [8 upvotes]: Suppose we have a set $X$ with $X=U \cup V$. If we pick a permutation $f$ of $U$ and a permutation $g$ of $V$ which agree on the intersection $U \cap V$, we can coalesce them into one big endo-map $F$ of $X$. In general, of course, $F$ is no longer a permutation - we can only say that it is onto. A counter-example is fun to think of - for example, if $X$ is finite, then $F$ is always a permutation. -Now let $X, U$ and $V$ be open subsets of $\Bbb{R}^n$, and suppose $f$ and $g$ are smooth homeomorphisms. I have been unable to come up with a single example for which the above patching fails "badly", say for which the set of points $x$ having more than one pre-image had positive measure. -I'd love to know if anyone knows anything related to this problem, or could provide a counter-example. Thank you! - -REPLY [3 votes]: The set of points having more than one pre-image is open (so it always has positive measure if it is non-empty). It follows that $U \cap V$ must be disconnected and in fact it has to have infinitely many connected components. However there are examples in $\mathbb{R}^2$ where both $U$ and $V$ are connected: -For a negative integer $n$ we define the real intervals $$I_n=(2n-\frac{1}{3},2n+\frac{1}{3}) \mbox{ ; } J_n=(2n+\frac{2}{3},2n+\frac{4}{3}).$$ For a non-negative integer $n$ we let $$I_n=J_n=(n-\frac{1}{3},n+\frac{1}{3}).$$ -Let $f$ and $g$ be two increasing (smooth if you wish) homeomorphisms from $\mathbb{R}$ onto $\mathbb{R}$ such that $f[I_n]=I_{n+1}$ and $g[J_n]=J_{n+1}$ for all $n \in \mathbb{Z}$. Let $I=\bigcup_{n\in \mathbb{Z}}I_n$ and $J=\bigcup_{n\in \mathbb{Z}}J_n$. -Now in $\mathbb{R}^2$ we define the open sets $$U=I \times (-1, \infty) \cup \mathbb{R} \times (2,\infty)$$ $$V=J \times (-\infty, 1) \cup \mathbb{R} \times (-\infty ,-2).$$ -Finally we let $F=f \times id$ and $G=g \times id$ be the autohomeomorphisms of $U$ and $V$ respectively. The set of points with two pre-images is precisely $I_0 \times (-1,1)$. -Similar examples exist in $\mathbb{R}^n$ for $n\geq3$, but of course in $\mathbb{R}$ we cannot get $U,V$ to be connected.<|endoftext|> -TITLE: Tags for theorems, definitions, examples, etc. in monographs -QUESTION [9 upvotes]: I write many telegraphic reviews of monographs and textbooks. -Occasionally I see certain books do something I think should become -an industry standard, namely tagging nearly every theorem, lemma, definition, -example, exercise, etc. with some short mnemonic nickname, so readers need -not depend solely on a numbering scheme for cross-references, etc. -Examples of books that employ this laudable practice (and just happen to come to mind) -include - Epstein's Word Processing in Groups and - Thurston's Three-Dimensional Geometry and Topology. -My question: what author and/or publisher initiated this practice? -(I'd like to cite the original source as a comparison when I feel the -visual design of a book makes it difficult for browsing.) - -REPLY [6 votes]: First a comment. I recently wrote an intro level text on group representation theory with such tags. One of the reviewers found the tags annoying and said it seemed like I was sharing my latex labels with the world. I had to convince Springer to let me keep them. So it may be a matter of taste. -To answer your question partially I doubt any publisher started this. I think it is an author thing. Notice both books you mentioned have Thurston and Silvio Levy involved. My advisor tended to name theorems so that one could have something to refer to them by. I think many of his students picked this up. -I doubt it originated with one person.<|endoftext|> -TITLE: What is a complex inner product space "really"? -QUESTION [19 upvotes]: This is an extended re-post of a question that I have asked on MSE not a long time ago. But anyway, it seems more appropriate for MO. -To begin with, in a real inner product space we have a geometric intuition for the inner product. In the finite dimensional case, the inner product of two vectors is the product of their lengths (norms) times the cosine of the angle between them. Reverse engineering this suggests that the purely algebraic properties of an abstract inner product give it the properties of the "length of the projection" (scaled somehow). In particular, we can think of two vectors with zero inner product as orthogonal geometrically (and even call them that way) and bring all the geometric notions related to orthogonality to the abstract, possibley infinite dimensional case (with the appropriate care and restrictions of course). My question is, what is the geometric or otherwise intuition behind the abstract notion of a complex inner product space? -Here are a few thoughts: -1) This is a good mathematical structure to model some physical phenomena. An (if not the) example is quantum mechanics. This is an interesting line of thought. One problem is that I don't know enough quantum mechanics to follow it more deeply. If it is possible to explain in an elementary as possible way, What does it mean that a state of a particle is an element of a complex Hilbert space I would very much want to hear about it. I would also like to hear about other, hopefully more elementary, phenomena modeled by complex inner spaces. In particular, ia there any such phenomenon modeled by a finite-dimensional complex inner product space? -2) This is a good mathematical tool for other mathematical theories. Perhaps, unitary representations of groups or of other algebraic structures. Again, I don't have enough background in representation theory myself. explicit examples of such utility are welcome. -3) it is a good tool to investigate real structures by complexification and exploit of the good properies of the complex field (such as being algebraically closed). There are a lot of such examples in linear algebra, such as the classification of orthogonal maps, but I haven't seen such examples in the context of inner product spaces. -I would like to stress that saying that this is somehow algebraically natural analogue of real inner product spaces and that it has a lot of nice properties is somehow not enough in my opinion. Also not very satisfying is saying that it has application in such and such very advanced theories without elaboration. -Thanks! - -REPLY [10 votes]: This is a purely algebraic answer, so maybe not what you're looking for. -What would an inner product on a left $A$-module $M$ over a noncommutative algebra $A$ be? It should give a map $M \to M^*$ where $M^* = Hom_A(M,A)$, but that's a right $A$-module. We can teach it to be a left $A$-module again, if we have a chosen anti-automorphism $i:A\to A$. Eventually we'd like the natural function $M \to M^{**}$ to be $A$-linear, probably, so we should also ask that $i$ be an involution. -If $A = \mathbb C$ then there are two (continuous) such $i$: the identity, and the complex conjugate. I claim that the latter is the more natural one, or at least that this isn't crazy. If you think of $\mathbb C = \left\{\begin{pmatrix} a&b\\ -b&a \end{pmatrix} : a,b\in \mathbb R\right\}$, then (1) it's weird that $\mathbb C$ is commutative, since matrices shouldn't commute, and (2) $i$ is transpose. -If you go up to $A = \mathbb H$ then you no longer have this weird choice of $i = $ identity available. But transpose still works (thinking of $A$ as certain $4\times 4$ real matrices). -Anyway, if you now unwind the $A$-linearity of $M \to M^*$, where $A=\mathbb C$ and $i(z) = \overline{z}$, you get sesquilinearity.<|endoftext|> -TITLE: Failure of the GCH -QUESTION [9 upvotes]: What is the (currently known) consistency strength of global failure of the GCH? -I do not have access to the exact statement of the original Foreman-Woodin result. My searches seem to indicate that they used an assumption at the region of a supercompact, although I have seen comments stating that the result has been improved to require something in the region of a hypermeasurable. Is this correct? What this exact upper bound? -Thanks a lot. - -REPLY [9 votes]: By work of Gitik-Mitchell a $(\kappa+2)$-strong cardinal $\kappa$ is required, and by work of Merimovich a $(\kappa+3)$-strong cardinal $\kappa$ (in fact a cardinal $\kappa$ with $o(\kappa)=\kappa^{++}+\kappa^{+}$) is enough. Gitik and Merimovich have a project to get the total failure of $GCH$ from optimal hypotheses. It's yet incomplete. If I remember it correctly, it says something like this: -Theorem. The following are equiconsistent: -1-For any $\alpha$, there are stationary many cardinals $\kappa$ with $o(\kappa)=\kappa^{++}+\alpha,$ -2-GCH fails everywhere, -3-$\forall \lambda, 2^{\lambda}=\lambda^{++}.$<|endoftext|> -TITLE: Polynomials with prescribed points to match prescribed bounds -QUESTION [6 upvotes]: Consider real polynomials on the interval $I=[-1,1]$. It is easy -to see that the smallest degree for a non-negative polynomial -with given zeros $x_1,\dots,x_s\in I^\circ$ is $n=2s$ (e.g. -$P(x) = \prod_{i=1}^s (x-x_i)^2$ works). - -My question is: -What is the smallest degree for a polynomial such that it is -bounded by $\pm 1$ on $I$ and attains the value $1$ on a set -$x_1^+,\dots,x_s^+$ and the value $-1$ on a set -$x_1^-,\dots,x_r^-$? - -Background: I know that the fact about nonnegative polynomials -with presribed zeros can be generalized to "generalized -polynomials" built from Tchebycheff-systems (due to a theorem by -Krein). I would love to see a similar theorem on bounded -generalized polynomials which attain the bounds at prescribed points. -Edit: In this question I leanerd from the answer of Gjergji Zaimi that there are bounds on the degree of increasing interpolating polyomials. How does the bounds change for monotone interpolation are described above? Are there (algebraical or numerical) methods to calculate the polynomial? -It seems to me that monotone interpolating polynomials are not treated in the current literature and are not subject of current research. Is that right, and if so is there a special reason for that? - -REPLY [4 votes]: Let $D$ be the minimum distance between $x$'s, merging all the $x$'s into one list of length $N$. -Let $k$ be an integer $\ge \max(16\log(8/D^2),10N)\ /\ D^2$. -Then a polynomial of degree of $6(k+1)(N-1)$ suffices. -Proof: -Let $p(x) = \frac{1}{2}(3 q(x) - q^3(x))$, where $q(x) = \sum_i \pm \Pi_{j \neq i} r_{ij}(x)$ and -$$r_{ij}(x) = \left(1-\frac{(x-x_i)^2}{4}\right)^k \frac{(x-x_j)^2}{(x_i-x_j)^2}$$ -Then $p$ is clearly of the specified degree, and has the specified values at the $x_i$'s. -The key is to show that $p$ is bounded by $\pm1$. -Small terms: When $|x-x_i| > D/2$, $r_{ij}(x) \le (1-D^2/16)^k (4/D^2) < -%(1-D^2/16)^{\large(16/D^2)\log(8/D^2)}(4/D^2) < -1/2$, so $\Pi_{j \neq i} r_{ij}(x) < 2^{-(N-1)}$. -Large terms: When $|x-x_i| < D/2$, -$$r_{ij}(x) -\le -\left(1-\frac{(x-x_i)^2}{4}\right)^k \left(1 - \frac{x-x_i}{x_j-x_i}\right)^2 -\le -\left(1-\frac{(x-x_i)^2}{4}\right)^k \left(1 + \frac{2|x-x_i|}{kD}\right)^k -$$ -$$ -\le -\left(1+ \frac{2|x-x_i|}{kD}-\frac{(x-x_i)^2}{4}\right)^k -\le -\left(1+\frac{4}{k^2D^2}\right)^k -\le -e^{\large 4/D^2k} -\le -e^{\large 4/10N} -\le -(3/2)^{\large 1/N}. -$$ -So $\Pi_{j \neq i} r_{ij}(x) < 3/2$ -Since each $x$ is within $D/2$ of at most one $x_i$, $q(x)$ is the sum of at most one -large term bounded by 3/2, and by $N-1$ small terms bounded by $2^{-(N-1)}$. So $|q|\le2$, -and $|p|\le1$.<|endoftext|> -TITLE: Jacobian criterion for smoothness of schemes -QUESTION [11 upvotes]: An affine scheme $X = Spec(A)$ is said to be smooth if for any closed embedding -$X\subset\mathbf A^n$, of ideal $I$, it is true that, locally on $x\in X$, the ideal $I$ -can be generated by a sequence $f_{r+1},\dots,f_n$ such that their Jacobian has maximal rank. -My question is: - -Will the Jacobian of ANY set of $n-r$ generators of $I$ be of maximal rank? - -REPLY [11 votes]: Yes, the rank of the Jacobian matrix doesn't depend on the set of generators of $I$. The Jacobian matrix at $x$ represents the subspace generated by the differentials at $x$ of all $f\in I$. -Note that the rank of the Jacobian matrix at $x$ is computed in the fiber where $x$ lives, it has nothing to do with the base scheme. -Some more explanations We work over a base field $k$. Let $I$ be the ideal defining $X$ in $Y:=\mathbb A^n$. Then we have a canonical exact sequence -$$ I/I^2 \to \Omega_{Y}|_X \to \Omega_X \to 0 $$ -where the first map is $\bar{f}\mapsto df\otimes 1$. Tensoring by $k(x)$ we get -$$ I/I^2 \to \Omega_{Y}\otimes k(x) \to \Omega_X\otimes k(x) \to 0.$$ -If $g_1,...,g_m$ are a system of generators of $I$, then $dg_1,...., dg_m$ generate the image of $I/I^2$ in $\Omega_{Y}\otimes k(x)\simeq k(x)^n$. Call this image $C$. Let $J_x$ be the Jacobian matrix associated to $g_1,...,g_m$ at $x$ in a system of coordinates of $Y$. Then the columns of $J_x$ correspond to the images of $dg_1,...,dg_m$ in $C\subseteq \Omega_{Y}\otimes k(x)$. Therefore the rank of $J_x$ is just the dimension of $C$ over $k(x)$, and this is independent on the choice of the system of generators $g_1,...g_m$. -By the way, these discussions show that -$$\dim_{k(x)} (\Omega_{X}\otimes k(x))=n - \mathrm{rank} J_x.$$ -As the smoothness at $x$ is equivalent to $ \dim_{k(x)} (\Omega_{X}\otimes k(x))= \dim_x X$, we see that it is also equivalent to $\mathrm{rank} J_x=n-\dim_x X$ which is the Jacobian criterion of smoothness.<|endoftext|> -TITLE: Krull dimension less or equal than transcendence degree? -QUESTION [30 upvotes]: Let $k$ be a field, and $A$ a $k$-domain, so that the fraction field of $A$ has transcendence degree $n$ over $k$. -If $A$ is finitely-generated over $k$, then $A$ has Krull dimension $n$ (Theorem A in Eisenbud). -However, if $A$ is infinitely-generated, then it is possible for the dimension of $A$ to be less than the transcendence degree of its fraction field. Take, for example, rational functions in one variable $A=k(x)$. Dimension 0, transcendence degree 1. -Is it always true that the dimension of $A$ is less than or equal to $n$? - -REPLY [7 votes]: There is already a perfect, accepted answer, but I thought I add this to point out that there is a geometric reason for the dimension to possibly drop versus the transcendence degree. -For simplicity assume that the fraction field of $A$ is a finitely generated field extension of $k$ and that $A$ is integrally closed. If we are doing geometry these assumptions seem reasonable (and there is already a complete answer in the general case). I think with a little more work one can follow a similar argument without these assumptions, but this post's goal is mainly to explain the geometry behind this phenomenon, so I am not striving for completeness especially that (as I already mentioned twice) since there is already a complete solution there is no need to do that. -So, first, $A$ contains a transcendence base for its fraction field over $k$, in other words we have -$$ -k\subseteq k[x_1,\dots,x_n]\subseteq A \subseteq \mathrm{Frac} (A) -$$ -where $\mathrm{Frac} (A)$ is a finite algebraic extension of $k(x_1,\dots,x_n)$. Now let $B$ be the integral closure of $k[x_1,\dots,x_n]$ in $\mathrm{Frac} (A)$. Since $A$ is assumed to be integrally closed, $B\subseteq A$ and since $\mathrm{Frac} (A)$ is a finite algebraic extension of $k(x_1,\dots,x_n)$, $B$ is a finitely generated $k[x_1,\dots,x_n]$-module and a finitely genrated $k$-algebra and $\mathrm{Frac} (B)=\mathrm{Frac} (A)$. -The integral $k$-algebra extension $k[x_1,\dots,x_n]\subseteq B$ corresponds to a finite surjective morphism $X\to \mathbb A^n_k$ of $k$-varieties, so for $X$ we have the quoted result: $\dim X$ equals the transcendence degree of the fraction field of $A$. -Now we have that $B\subseteq A \subseteq \mathrm{Frac} (B)=\mathrm{Frac} (A)$. In geometric situations this typically happens if $A$ is a localization of $B$. In that case we get that the Krull dimension of $A$ is at most the Krull dimension of $B$ because the dimension of $A$ is just the maximum height of the points of $\mathrm{Spec}B$ contained in $\mathrm{Spec} A$. -This is really why I wrote this whole answer: the way the Krull dimension drops versus the transcendence degree is if we localize at non-closed points: In your original example $k(x)$ corresponds to the generic point of the line $\mathbb A^1$, so its geometric dimension = transcendence degree is $1$, but its algebraic dimension is $0$ because it is a field (or more generally the local ring of a non-closed point).<|endoftext|> -TITLE: Efficient enumeration of Bruhat intervals -QUESTION [6 upvotes]: Hi everyone. -I'm currently programming some stuff for Hecke algebras. My current implementations have several bottlenecks and I'd like to improve that as much as I can so that I can use stuff like $E_8$ as input one day (without waiting until the heat death of the universe for an answer). -Given a Coxeter group W (let's start with a finite one), one of the things that I would like to improve are some iterations over intervals in the Bruhat order in W. For example a recursion for the Kazhdan-Lusztig-polynomials $P_{y,w}$ and the $\mu$-polynomials $\mu_{y,w}^s$ involves formulas like -$P_{y,w}^\ast = v_t P_{y,w}^\ast + P_{ty,tw}^\ast - \sum_{z\in W: y\leq z < tw \wedge tz < z} P_{y,z}^\ast \mu_{z,tw}^t $ -The crucial step of calculating $\mu_{y,w}^s$ involves a similar sum over the interval $(y,w)$. Some other calculations I'd like to perform in the future also have this structure. -One way (this is how I do it at the moment) of realizing this summation over intervals $[a,b]$ is to iterate through all elements of $W$ whose length is between $l(a)$ and $l(b)$ and check everytime if $a\leq z \leq b$ is satisfied. This works fine if both $a$ and $b$ are near the bottom or near the upper end of the Bruhat order, but if they are in the middle, the level sets $\lbrace z | l(z)=const\rbrace$ are much, much bigger than the interval $[y,z]$: Considering the characterization of the Bruhat order in terms of subwords something like $2^{l(w)-l(y)}$ should be an upper bound for the cardinality of this interval, but the level sets in the middle of the Bruhat order have more than $|W|/l(w_0)$ (which is more than $5,8*10^6$ in the $E_8$ case). -A solution could be to use the subword-characterization and iterate through all subwords of $w$. It is no difficult to write a program iterates that way through $[1,w]$. But this approach enumerates all $2^{l(w)}$ subwords and discards several of them to ensure that only reduced words appear and each appears only once. Now an additional check whether $y\leq z$ or not only slows thing down even further. So this also is not a very efficient way if $w$ is somewhere in the middle of the Bruhat order. -Hence my question is: -Is there an efficient way to enumerate all elements of Bruhat intervals? - -REPLY [2 votes]: There is a recursive way to do this that will probably do better for intervals in the middle than the two methods that you said you are unhappy with. Check out Section 5 of N. Reading, "The cd-index of Bruhat Interals:" -http://www.combinatorics.org/ojs/index.php/eljc/article/view/v11i1r74 -Basically, here's how it goes: Suppose the interval is $[u,w]$. We choose a reduced word for $w$ and use it to guide a recursive construction. The recursion starts from the singleton interval $[u',w']=[e,e]$ at the identity and, for every letter $s$ in the reduced word, we have a way of constructing either $[u',w's]$ or $[u's,w's]$ from $[u',w']$ by performing a simple global operation (either "doubling" the interval or doing something slightly more complicated) and then killing off certain elements. -There is a lot of computation still, but here is why it is better, for example, than checking all $2^{l(w)}$ subwords of w to find the interval $[1,w]$: In effect, you start with one element, but multiply by $2$ a total of $l(w)$ times, but at each step you get rid of elements you don't want, rather than producing $2^{l(w)}$ elements and then deleting the ones you don't want. You get a similar savings for general intervals $[u,w]$ (which can be much "fatter" than $2^{l(w)-l(u)}$ elements).<|endoftext|> -TITLE: Weil's descent criteria for covers from the critereon for varieties? -QUESTION [5 upvotes]: I have read several articles which use a version of the Weil decent criterion for covers, but the reference is always to Weil's original paper (1956 - The field of definition of a variety). I would like to know how one makes the transition. Note that when I say covers I mean a morphism between covers $(f:V\rightarrow W), (f':V'\rightarrow W)$ is a $g:V\rightarrow V'$ commuting with the covering maps. -For reference, here is the theorem from Weil's paper, which I simplified assuming Galois: -Let $k/k_0$ be a finite, Galois extension, $H=Gal(k/k_0)$. Let $V$ be a projective variety defined over $k$. Elements of $H$ have a natural action on varieties and morphisms defined over $k$ (for example by acting on the coefficients if we embed into projective space). Suppose for each $\sigma, \tau \in H$, we have an isomorphism $f_{\tau,\sigma}:\sigma(V)\rightarrow \tau(V)$. Then we have a model $V_0$ over $k_0$ for $V$ if the following are satisfied: -(i) $f_{\tau,\rho}=f_{\tau,\sigma}\circ f_{\sigma,\rho}$ for all $\sigma,\tau,\rho\in H$. -(ii) $f_{\tau \omega, \sigma \omega}=\omega(f_{\tau,\sigma})$ for all $\sigma,\tau\in H$, $\omega \in Gal(k_0^{sep}/k_0)$. -Now I suspect that to make the translation, one takes $f:V\rightarrow W$ and looks at the graph $\Gamma_f\subseteq V\times W$. Let's suppose we can satisfy the criteria above (with the $f_{\tau,\sigma}$ morphisms of covers) for $\Gamma_f$. Then we get some $\Gamma_0$ defined over $k_0$ and an isomorphism $\varphi:\Gamma_0\times k \rightarrow \Gamma_f$. There are two points I can't resolve: -(1) How do we know we have $\Gamma_0\subseteq V_0\times W_0$ for some models $V_0,W_0$ of $V,W$ (we can get the models of $V$ and $W$ over $k_0$ from the covering data). And further that it is the graph of a morphism $V_0\rightarrow W_0$. -(2) How do we know that the map $\varphi$ corresponds to a morphism of covers of $W$? -Thanks - -REPLY [2 votes]: As I understand it, question 1 is about whether closed immersions descend along Galois covers, and question 2 is about whether being an open immersion or a surjection is a property that descends along Galois covers. -You can find the answer to the first question in SGA 1 Exp 8, section 5, where the Galois cover condition is replaced by a much weaker condition of fpqc cover. -You can find the answer to the second question in EGA IV vol. 2 2.7.1, or SGA 1 Exp 8, section 4, again in somewhat more general language than what you strictly need. -There are also proofs in Vistoli's notes. I'm afraid I don't know about any proofs that use Weil's old terminology, and I couldn't think of a good reason to look for one.<|endoftext|> -TITLE: Complexity of a weirdo two-dimensional sorting problem -QUESTION [13 upvotes]: Please forgive me if this is easy for some reason. -Suppose given $S$, a set of $n^2$ points in $\mathbb{R}^2$. -I want to choose a bijective map $f$ from $S$ to the set of lattice points in $\lbrace 0,\ldots,n-1\rbrace \times \lbrace 0,\ldots,n-1\rbrace$ so as to maximize the sum, over all $p$ in $S$, of the dot product $p \cdot f(p)$. -If, instead of $\mathbb{R}^2$, I had $\mathbb{R}^1$, and I was putting $S$ in bijection with $\lbrace0,\ldots,n-1\rbrace$, then this would simply be sorting $S$ and one knows how to do that fast. -For this problem, it's not even obvious to me how to do it in a number of steps that's polynomial in $n$. -Is this easy? Is it an example of a known genre of optimization problem? - -REPLY [2 votes]: Here are some more cross-connections: The problem can be massaged into a least-squares matching problem -$$\sum_{p\in S} \|(-f(p))-p\|^2 \to \min$$ between $S$ and the grid rotated by $180^\circ$ (which in this case happens to be the same as the original grid, after translation), and hence it becomes a problem of optimal mass transport. The classical reference is: - -Franz Aurenhammer, Friedrich Hoffmann, and Boris Aronov, - Minkowski-type theorems and least-squares clustering. Algorithmica 20: - 61–76 (1998). DOI: 10.1007/PL00009187 - -There is an implementation in R: https://rdrr.io/cran/transport/man/aha.html -See also my little note -Two applications of point matching. I would also be interested to find a provably faster algorithm than the general matching procedures, which would make use of the geometric situation. (Fast practical algorithms and fast approximation algorithms are mentioned in the references of these papers and the software.)<|endoftext|> -TITLE: Do all exact $1 \to A \to A \times B \to B \to 1$ split for finite groups? -QUESTION [41 upvotes]: Let $A$, $B$ be finite groups. Is it true that all short exact sequences $1 \rightarrow A \rightarrow A \times B \rightarrow B \rightarrow 1$ split on the right? -In other words, do there exist finite groups $A$, $B$ and homomorphisms $f: A \rightarrow A \times B$, $g: A \times B \rightarrow B$ such that $1 \rightarrow A \rightarrow A \times B \rightarrow B \rightarrow 1$ is exact and there does not exist a homomorphism $h: B \rightarrow A \times B$ such that $g \circ h = \text{id}_B$? -An example when $A$, $B$ are not finite is given by $A = \prod_{i=1}^\infty \mathbb{Z}$, $B = \prod_{i=1}^\infty \mathbb{Z}/2\mathbb{Z}$, $f((n_i)) = ((2n_i),0)$, and $g((n_i),(m_i)) = (\overline{n_1}, m_1, \overline{n_2}, m_2, \ldots)$. - -REPLY [47 votes]: This is true (1). It was extended to finitely generated profinite groups here (2). Surprisingly, it is also true in the category of finitely generated modules over a Noetherian commutative ring (3). -(1) Joseph Ayoub, The direct extension theorem, J. Group Theory 9 (2006) 307-316. -(2) Goldstein, Daniel, Guralnick, Robert The direct product theorem for profinite groups. J. Group Theory 9 (2006), no. 3, 317-322. -(3) Takehiko Miyata Note on direct summands of modules. J. Math. Kyoto Univ. -Volume 7, Number 1 (1967), 65-69.<|endoftext|> -TITLE: Topology on the space of Schwartz Distributions -QUESTION [14 upvotes]: If we equip the Schwartz space $\mathcal{S}$ with its usual Fréchet space topology, then the space of continuous linear functionals $\mathcal{S}^\ast$ is known as the space of Schwartz distributions or tempered distributions. If we equip this space with the strong topology, is there anything we can say about the resulting topological vector space? Evidently, the resulting space will not be a Fréchet space, but perhaps it will have other nice properties. In particular, I am interested in the space of continous linear operators on $\mathcal{S}^\ast$. Is there anything interesting we can say about this space? -Unfortunately, a quick google search did not turn up many sources that dealt with the particulars of the topology on $\mathcal{S}^\ast$, much less the topology on the space of continuous linear operators on $\mathcal{S}^\ast$, so a point in the right direction to a reference would also be greatly appreciated. -EDIT: After thinking about this more deeply, I realize that I am interested in a specific type of operator on $\mathcal{S}^\ast$. $\mathcal{S}$ occurs naturally inside of $L^2$, so after identifying the dual of $L^2$ with itself via the Riesz Representation Theorem, we can in turn regard $\mathcal{S}$ as a subspace of $\mathcal{S}^\ast$. With this in mind, I am interested in the operators on $\mathcal{S}^\ast$ that restrict to operators on $\mathcal{S}$. -The motivation for this question comes from quantum mechanics, where I have in mind the position and momentum operators acting on $\mathcal{S}^\ast$. I am thus interested in the operator algebra they generate. Furthermore, these of course restrict to operators on $\mathcal{S}$, and so I am likewise interseted in the operator algebra of operators on $\mathcal{S}^\ast$ that restrict to operators on $\mathcal{S}$. In particular, I would like to abstractly characterize this space. -As this is the natural space for observables in quantum mechanics, there has to be at least something known about this space. . . - -REPLY [3 votes]: Just as a variation on the other answers: the Schwartz space can easily be written as a (projective) limit of Hilbert spaces $V_s$, basically requiring that both a function and its Fourier transform be in a Levi-Sobolev space. The inclusions for $s>t$ are compact (in fact, trace-class, seen by proving that they're composites of Hilbert-Schmidt) inclusions. This adds a bit to the assertion that it is Frechet, which would indeed give a proj lim of Banach spaces, but if we can have Hilbert spaces, it's even better. Then, as in the question, identifying $L^2(\mathbb R)$ with its own dual (up to complex conjugation, anyway), the dual of $V_s$ for positive $s$ is $V_{-s}$... and we can give it the strong (=Hilbert-space) topology. It is not completely formal-categorical, but easy enough, to show that the dual of the limit is the colimit of the duals, however we topologize them. With the Hilbert-space topologies we obtain (yet another presentation of) the strong topology on tempered distributions. -It is completely formal that the dual of a colimit is the limit of duals, and a virtue of this set-up, with Hilbert spaces, is that the reflexivity is trivial. -Also, operators on $V_{-\infty}=\bigcup_s V_s$ are easy to understand in this presentation. The operators that stabilize $V_{+\infty}=\bigcap_s V_s$ must stabilize each Hilbert space $V_s$, etc. -Edit: in light of the additions/edits to the question: the very tangible instance of operators as in the question on $\mathbb R$ uses the "Dirac factorization" $-\Delta+x^2=RL+1=LR-1$, with raising and lowering operators $R=i({\partial\over \partial x}-x)$ and $L=i({\partial\over \partial x}+x)$. That "Schr\"odinger/Hamiltonian" $S=-\Delta+x^2$ can be used to give a family of semi-norms $|f|^2_k = \langle S^k, f\rangle$, with respective Hilbert spaces $V_s$, and $V_{s+1}\to V_s$ is trace-class. The projective limit is a nuclear Frechet space, and exhibits the Schwartz space as such. Likewise, the colimit of the Hilbert space duals $V_{-s}$ of $V_s$'s exhibit tempered distributions as dual-of-nuclear-Frechet. -This Hilbert-space case of more general constructions, with fairly obvious generalizations, suffices for many purposes.<|endoftext|> -TITLE: Why are people interested in Cohen-Macaulay of codimension 2? -QUESTION [6 upvotes]: In deformation theory, Cohen-Macaulay in codimension 2 is the first to be considered in higher order deformation. Does Cohen-Macaulay in codim. 2 have some good property to work with? Does it somehow measure how singular a scheme is? -I am reading Hartshorne's deformation theory, and I have difficulty to "see" why Cohen-Macaulay in codim. 2 is a good object to work with. -Thank you very much - -REPLY [11 votes]: Angelo has already mentioned the Hilbert-Burch theorem in a comment. -One could present its importance this way: The ideal of a codimension $1$ subscheme in a regular affine scheme is locally free of rank $1$, so everything is pretty easy in this case. -The next case is codimension $2$. Here things can get hairy, but the Hilbert-Burch theorem says that if the subscheme is Cohen-Macaulay its deformation theory is still nice. -To answer your other questions: - -Cohen-Macaulay has good properties regardless of the codimension (see below). -Yes, it measures how singular the scheme is and it says that in some sense it is not too badly singular. Local complete intersections are Cohen-Macaulay and although Cohen-Macaulay is not necessarily local complete intersection, many things that work for l.c.i.'s work for CM as well. - -So, why is CM a good property: -A practical reason is that CM is absolutely immune to passing to or from hyperplane sections, so in general using induction is relatively robust. -A less obvious but extremely useful property CM schemes have is that their dualizing complex is supported at only one place, in other words they admit a dulaizing sheaf. This means that duality theory is much simpler for CM schemes. -There are many properties that families of CM schemes satisfy that are not true in general. Most of these actually stem from the two properties I mentioned above. -A more sophisticated version of the CM property is Serre's condition $S_n$. This requires less, in some sense it requires that the scheme is CM "up to some degree". -The property $S_2$ appears as a necessary condition for a scheme to be normal (the other is being regular in codimension $1$). Property $S_2$ is also important as it is the algebraic version of the Hartog property, namely that regular functions be determined by their codimension $1$ behaviour. For more on $S_2$ see this MO question.<|endoftext|> -TITLE: Given a small category with some colimits, can the rest of the colimits be added? -QUESTION [9 upvotes]: Let $\mathcal{A}$ be a small category with some ( maybe no) colimits. What I would like to be able to do is add the rest of the colimits in a universal way. The Yoneda lemma will not work, since this simply adds all colimits formally. That is to say that you have new colimits that are different than the old. We do have maps from the newly created colimits to the old. The question is can we localize about these morphisms making the old and new colimits isomorphic? - -REPLY [10 votes]: Yes. -More generally, let $A$ be a small category, and $D$ some set of "distinguished" colimits in $A$ — for example, you could take the set of all colimits that exist. A sheaf on $(A,D)$ is a functor $F: A^{\mathrm{op}} \to \mathrm{Set}$ such that for every colimit diagram $d\in D$, $Fd$ is a limit diagram in $\mathrm{Set}$. In particular, every representable functor $\hom(-,a)$, for $a\in A$, is a sheaf, by definition of colimit. Denote by $\operatorname{Shv}(A,D)$ the full subcategory of $\operatorname{Fun}(A^{\mathrm{op}},\mathrm{Set})$ whose objects are sheaves (thus the morphisms are all natural transformations). Then we have a full faithful embedding of categories $\gamma: A \to \operatorname{Shv}(A,D)$ sending $a\mapsto \hom(-,a)$. -It should be obvious that the category $\operatorname{Shv}(A,D)$ is complete. Indeed, given any limit diagram in $\operatorname{Shv}(A,D)$, I claim that their limit in $\operatorname{Fun}(A^{\mathrm{op}},\mathrm{Set})$ actually lies in $\operatorname{Shv}(A,D)$, and this basically turns on the fact that "limits commute". It is less obvious, but also true, that $\operatorname{Shv}(A,D)$ is cocomplete. Also, there is a "sheafification" functor $\operatorname{Fun}(A^{\mathrm{op}},\mathrm{Set}) \to \operatorname{Shv}(A,D)$ which is adjoint to the forgetful map $\operatorname{Shv}(A,D) \to \operatorname{Fun}(A^{\mathrm{op}},\mathrm{Set})$. -In fact, $\operatorname{Shv}(A,D)$ is universal in the following sense. Let $C$ be any cocomplete category, and $F : A \to C$ a functor such that for every distinguished diagram $d\in D$, $Fd$ is a colimit diagram in $C$. Then $F$ factors as $F = \tilde F \circ \gamma$, where $\gamma$ is the Yoneda embedding $A \to \operatorname{Shv}(A,D)$ from above, and $\tilde F : \operatorname{Shv}(A,D) \to C$ is a cocontinuous functor (preserves all colimits). Moreover, $\tilde F$ is unique up to unique isomorphism. -The buzzwords for this construction are that $(A,D)$ is a sketch and that $\operatorname{Shv}(A,D)$ is a presentable category. The best reference I know for this material is Jiří Adámek and Jiří Rosický, Locally presentable and accessible categories, Cambridge University Press, 1994. We review some of this material (enough for some hands-on understanding) in the second section of my paper with Alex Chirvasitu — probably you only want to look at pp 6–13 or so, but maybe it is easier to find what you're looking for there than in the big Adámek and Rosický book.<|endoftext|> -TITLE: Entire calculus and clmc algebras -QUESTION [5 upvotes]: If $\mathcal{A}$ is a complete locally convex (Hausdorff) associative unital algebra (over $\mathbb{C}$) one is interested in defining "transcendental" functions of a given algebra element $a \in \mathcal{A}$ like e.g. exponentials $\exp(a)$ by means of the power series expansion. This works fine for complete locally multiplicatively convex algebras. Recall that $\mathcal{A}$ is called lmc if there is a defining system of seminorms which are submultiplicative for the product. Equivalently, such an algebra is a (suitable) projective limit of Banach algebras. Then the (algebraic) polynomial calculus sending a polynomial $p \in \mathbb{C}[z]$ to the algebra element $p(a)$ extends by completion to an entire calculus -$$ -\mathcal{O}(\mathbb{C}) - \ni f \mapsto f(a) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} a^n \in \mathcal{A}, -$$ -which is a continuous algebra homomorphism for a given $a$. Here $\mathcal{O}(\mathbb{C})$ is equipped with its usual Fréchet topology of locally uniform convergence. Equivalently and more convenient in this context, one can use the seminorms given by $p_R(f) = \sum_{n=0}^\infty -\frac{|f^{(n)}(0)|}{n!} R^n$, from which one sees the continuity of the entire calculus on the nose. -Now there are many lc algebras which are definitely not lmc like the Weyl algebra generated by the canonical commutation relations $[Q, P] = i\hbar \mathbb{1}$ (whatever lc topology you may put on it). -My question is whether it is possible to have an entire calculus in the sense that there is a continuous algebra homomorphism extending the polynomial calculus to $\mathcal{O}(\mathbb{C})$ without having a lmc algebra but just a locally convex algebra. Can one give examples, reasonable conditions etc? - -REPLY [4 votes]: At least for commutative and completely metrizable locally convex algebras there is a theorem of Mityagin, Rolewicz, and Zelasko (Studia Math. 21 (1962), 291-306) which says that the algebra has to be locally m-comvex if all entire functions operate on it (meaning that $f(a)=\sum\limits_{n=0}^\infty\frac{ f^{(n)}(0)}{n!} a^n$ converges for all entire functions -$f$ and all elements of the algebra).<|endoftext|> -TITLE: Is the Duflo map for Lie algs. unique ? -QUESTION [8 upvotes]: The Duflo map is the map S(g) -> U(g), which known to satisfy the following properties: -1) identity on g -2) isomorphism of g-modules (and in particular vector spaces) -3) restricted to Poisson center on S(g) it is ISOMORPHISM of commutative algebras -S(g)^g to ZU(g) (the center of U(g)). -(This is highly non-trivial property). It predicts that the centers on the "classical" and "quantum" level are the same. (Kontesevich generalized to arbitrary Poisson variety). -The question: -Is it the only map satisfying such properties ? -(At least for semisimple Lie algebras) ? -(I think answer is YES, and it should be known, but I have not seen the reference). -========= -Example let g-commutative, then it is true: -Since S(g)=U(g) and since the map is required to be identity on g and it is homomorphism, -so it is identity map on the S(g). -=== -If you read this question I guess you know what S(g) and U(g) means :) -But may be nevertheless to keep good spirit of MO it is more polite to include definitions: -S(g) - means commutative algebra - symmetric algebra of g (i.e. just take some basis xi in g and consider polynomial algebra C[x1 ... xn] - this is S(g) as a vector space. Lie algebra g acts on - it simple way - it acts on xi by adjoint action, and continued further by Leibniz rule). -U(g) - means NON-commutative algebra - universal enveloping algebra of g - which is: you take basis xi and consider non-commutative polynomials C[xi] where generators satisfying the relations: [xi, xj] = C_ij^k x_k , where C_ij^k - are structure constants of g. -The Duflo map is for example discussed here: -D. Calaque, C. Rossi "Lectures on Duflo isomorphisms in Lie algebras and complex geometry" -http://people.mpim-bonn.mpg.de/crossi/LectETHbook.pdf -========= -Kontsevich mentions that in general situation (of Poisson manifolds) Grothendieck-Teichmuller groups should act on quantizations and in particular on "Duflo" type maps, but he writes that for semisimple algebras this action is trivial. So we should not have problem from this side. Moreover as far as I understand since nobody seen non-trivial example of this action it might be always like this. - -REPLY [7 votes]: Choose a map $\varphi$ satisfying these properties and make the difference $\psi=\varphi^{-1}\varphi_D$ with the Duflo map. Then $\psi$ is an automorphism of the $\mathfrak g$-module $S(\mathfrak g)$ which is the identity on $\mathfrak g$ and is multiplicative on invariants. -If you want this map to be universal (namely it should only involve universal formulae in terms of the Lie bracket) then it is very likely to be unique. -But if you want the statement to be true independantly for every single Lie algebra, I would believe the answer is "no". Namely, consider the $2$-dimensional solvable Lie algebra $\mathfrak g:={\bf k}x\oplus {\bf k}y$ with $[x,y]=y$. Now we have that $S(\mathfrak g)={\bf k}[x,y]$ and that $S(\mathfrak g)^{\mathfrak g}={\bf k}$. Therefore any non-trivial automorphism of the $\mathfrak g$-module $S^{\geq2}(\mathfrak g)$ (e.g. a non-trivial multiple of the identity) gives a counter-example. .<|endoftext|> -TITLE: The metric space associated to a measure space -QUESTION [10 upvotes]: Let $(X, \mathcal{A}, \mu)$ be a measure space such that $\mu(X) < \infty$. We say that two measurable sets $A$ and $B$ are equivalent if $\mu (A \Delta B) = 0$. The equation $$ d(A,B) = \mu (A \Delta B)$$ defines a metric on $\mathcal{A}$ modulo equivalence. - -Question: What information if any is encoded in this metric space? - -Here is a trivial example: Assume that $X$ is a finite set, $\mathcal{A} = P(X)$ and $\mu$ is the counting measure. We can give $P(X)$ the structure of an undirected graph: Join $A$ and $B$ by an edge if they differ by a point. Then the metric induced on $\mathcal{A}$ by the measure is exactly the graph metric (i.e the distance between two points is the length of the shortest path). - -REPLY [7 votes]: This metric recovers the measure space up to measure-preserving transformations. Fix a point to be $0$. You can take unions and intersections relative to that point, using only the metric. For instance, $X\cap Y$ is the point farthest from $0$ such that two triangle inequalities are exact: $d(X,P)+d(P,0)=d(X,0)$ and $d(Y,P)+d(P,0)=d(Y,0)$. Similarly for anything else you want to do with sets. So you get the entire sigma-algebra structure, modulo sets of measure $0$. -However, that structure isn't very much. Any measure-preserving transformation between two measure spaces is an isomorphism from this perspective.<|endoftext|> -TITLE: Sums of binomials with even coefficients -QUESTION [6 upvotes]: While looking for a closed form of a expression I worked myself to a formula that resembles the Vandermonde convolution, but is summed over even binomial coefficients only. -$\sum_{k=0}^n\sum_{l=0}^n{{2k+2l}\choose{2l}}{{4n-2k-2l}\choose{2n-2l}}$ -I'm at a loss as to what to do with it. I can re-write it in several ways, but the principal problem remains. Is there a known technique to attack such sums? Thanks. - -REPLY [15 votes]: We have -$$\sum_l\binom{a+l}lx^l=\frac1{(1-x)^{a+1}},$$ -hence the generating function for the even terms of the sequence is -$$\sum_l\binom{a+2l}{2l}x^{2l}=\frac12\left(\frac1{(1-x)^{a+1}}+\frac1{(1+x)^{a+1}}\right).$$ -Consequently, -\begin{multline*}\sum_l\binom{a+2l}{2l}\binom{b+2(n-l)}{2(n-l)}=\\\\ -[x^{2n}]\frac14\left(\frac1{(1-x)^{a+1}}+\frac1{(1+x)^{a+1}}\right)\left(\frac1{(1-x)^{b+1}}+\frac1{(1+x)^{b+1}}\right),\end{multline*} -where $[x^n]f$ denotes the $n$th coefficient of the power series for $f$. Plugging in the actual values for $a$ and $b$ and summing over $k$ gives -\begin{align*} -\sum_k\sum_l&\binom{2(k+l)}{2l}\binom{2(n-k)+2(n-l)}{2(n-l)}\\\\ -&=[x^{2n}]\sum_{k=0}^n\frac14\left(\frac1{(1-x)^{2k+1}}+\frac1{(1+x)^{2k+1}}\right)\left(\frac1{(1-x)^{2(n-k)+1}}+\frac1{(1+x)^{2(n-k)+1}}\right)\\\\ -&=[x^{2n}]\left[\frac{n+1}4\left(\frac1{(1-x)^{2n+2}}+\frac1{(1+x)^{2n+2}}\right)+\frac{1-x^2}{8x}\left(\frac1{(1-x)^{2n+2}}-\frac1{(1+x)^{2n+2}}\right)\right]\\\\ -&=\frac{2n^2+4n+1}{2n+1}\binom{4n}{2n}=(2n^2+4n+1)C_{2n}. -\end{align*}<|endoftext|> -TITLE: Using slides in math classroom -QUESTION [27 upvotes]: I am toying with the idea of using slides (Beamer package) in a third year math course I will teach next semester. As this would be my first attempt at this, I would like to gather ideas about the possible pros and cons of doing this. -Obviously, slides make it possible to produce and show clear graphs/pictures (which will be particularly advantageous in the course I will teach) and doing all sorts of things with them; things that are difficult to carry out on a board. On the other hand, there seems to be something about writing slowly with a chalk on a board that makes it easier to follow and understand (I have to add the disclaimer that here I am just relying on only a few reports I have from students and colleagues). -It would very much like to hear about your experiences with using slides in the classroom, possible pitfalls that you may have noticed, and ways you have found to optimize this. -I am aware that the question may be a bit like the one with Japanese chalks, once discussed here, but I think as using slides in the classroom is becoming more and more common in many other fields, it may be helpful to probe the advantages and disadvantages of using them in math classrooms as well. - -REPLY [5 votes]: Finally, I question in MO I feel qualified to answer! -I am a PhD student in Ireland doing an amount of lecturing. As a first remark, I am lucky in the sense that undergraduate maths was never especially easy for me and therefore I empathise with the average student. My second remark is that I hope for a career lecturing in the Irish Institute of Technology sector where the role in primarily teaching as opposed to the university sector where research is the primary role. Hence I am acutely interested in the skills as a mathematics teacher. -The second half of the answers here are closer to my philosophy than the first. A particular distinction must be put on the classroom environment and facilities. Regardless, my first instinct is that slides alone is sub-optimal. -The alternative to this is to produce everything on the blackboard. I did this last year in a differential calculus module (the students were maths studies --- by and large headed towards a career as "high school" mathematics teachers). The emphasis in this course is to convey to the students that although differential calculus is a relatively intuitive subject with the motivation coming from geometric concerns, as mathematicians we must also be rigorous, logical and precise in our thinking. Hence, we are not merely making a series of calculations and passing exams --- we must understand the content. When I wrote blackboard after blackboard of notes, the students did not have any chance of understanding the material. While I am a fervent believer that exercises and reflection are the best way for a student to achieve this aim, I am reminded of my undergraduate experience where certain obstacles lay in the path of me putting in this work and luckily my presence at lecture-time was sufficient for me to grasp the general theory and progress (eventually with first class grades) despite less than exemplary exam results in previous years. Put simply, ordinary students do not have the faculties to take down written notes and consider the important comments of the lecturer in real time. -However, slides do not work because mathematics is not a spectator sport (not a cliche when the average student is first interested in passing exams --- its is the goal of the educator to transcend this). It takes a superlative lecturer and a cohort of motivated and enthusiastic students to assimilate a lecture purely by ear. At least once I had a lecturer of this standard but I would vouch that were engineering, scientific or humanities students subjected to his fantastic delivery and questioning, they would simply fall asleep. It is a curse but a fact (among my students at least --- none of which are Math majors), that the average student does not have that aptitude to bask in such splendour. -My compromise, therefore is the very similar to what has been suggested above. I produce a set of notes (available soft-bound in a local printing house), with gaps which we fill in during the class (I print the notes onto an acetate sheet which I project onto a screen and can write on with a marker). All the theorems are writ-large, and everything else is teased out per a blackboard with suitable prior fillings in to both give the students a sneak preview and for the practical reasons of properly spacing out my scribblings. Does the need arise, I can put more complicated graphics in this set of notes. Today we introduced implicit differentiation and I projected this Wikipedia page list of curves onto the screen and this was but a two minute interlude. -The issue of students looking ahead was served by a motivation at the start of term (we are studying continuous and differentiable (smooth) functions. We draw a picture. We translate these geometric pictures into an algebebraic ones and never lose sight of this fact). -I have covered more content this year than last using this method, the first continuous assessment results showed a marked improvement and I am ahead of schedule despite being able to allocate a lot more time to comments and explanation of subtleties.<|endoftext|> -TITLE: numerically track spectrum curves of a parameter dependent linear operator -QUESTION [6 upvotes]: Hi, I am interested in how to numerically track spectrum curves of a parameter dependent linear operator. -Given a linear operator in square matrix form $M(t)$, where the matrix is smooth dependent of parameter $t\in[0,\infty]$. We also could know any order derivatives $M^{(k)}(t)$, whether analytic or not. -Based on some perturbation theory, we may compute the derivatives of eigenvalues and eigenvectors of $M$ in theory. The main challenge is the treatment of repeated eigenvalues(or its derivatives). The general idea in literature, is to find a canonical form of eigenvectors $\{X_i\}$, such that $X_i$ is corresponding to the repeated eigenvalues of $M^{(0)},...,M^{(k-1)}$ and distinct eigenvalue of $M^{(k)}$. Thus the derivative of eigenvalue is given as -$\lambda'_i = X^T_i M' X_i$ -However, in numerical setting, non pair of eigenvalues are exactly the same, and the situation turn to be difficult when I would like to compute and observe the evolving of eigenvalues w.r.t parameter $t$ in its domain. -For example, at $t=0$, suppose all eigenvalues are distinct. But during the $t$ evolves, the curves corresponding to distinct eigenvalue could intersect. I found it is still hard to handle the problem of numerical tracking, namely to find $n$ distinct smooth curves in $t$ domain which represent the spectrum of $M(t)$. -Here are some basic ideas: -In numerically sense if I have the eigenvalues $\lambda_{i}(t_j)$ and $\lambda_{i_q}(t_{j+1})$, $i=1,...,n$(the matrix dimension) for $t_{j+1} = t_j + \Delta t$(where $\Delta t$ should be reasonably small), computed by some standard numerical routines(such as the ones implemented in SLEPc), How could I figure out the permutation $i_p$ which indicates the correspondence between two consecutive group of eigenvalues. -Since the standard routines for eigenvalue problem only solve a static matrix $M$, the eigenvectors computed does not have the canonical form described previously. By providing all eigenpairs of $M(t_j)$, $M(t_{j+1})$ and any derivatives of $M(t)$, is it possible to establish the eigenpair correspondence(maybe a probability sense in term of $\Delta t$). -Or alternatively, by providing the eigenpairs of $M(t_j)$, is there a way to compute eigenpairs of $M(t_{j+1})$ in a sense to establish those correspondence implicitly. -The problem is only challenged when two curves intersect(enough close at some $t$ comparing to $\Delta t$), otherwise I would simply assign the correspondence by comparing eigenvalues or inner product of eigenvectors of two consecutive eigen group. -Due to my limited knowledge, I hope there are other people encountering the same problem, whether from theoretical interest or application background. All comments which may provide related informations are welcome. - -REPLY [4 votes]: I faced this problem a few years ago. In that case, I obtained a satisfactory approach along the lines of one of your suggestions. Specifically, I found that the eigenvectors changed relatively slowly with $t$. So I could associate the corresponding eigenvalue curves after an intersection with the right ones before an intersection by considering the similarity of the corresponding eigenvectors. I found that the inner-product $X^T_i Y_j$, where $X_i$ is one of the eigenvectors after the intersection and $Y_j$ is one of the eigenvectors after the intersection was a good measure of their similarity. -By the way, you might get more answers by posting this on https://scicomp.stackexchange.com/<|endoftext|> -TITLE: Examples of Non-Power Associative Algebras -QUESTION [6 upvotes]: What are the most important examples of non-power associative algebras? Why are they important, and what are their applications? - -REPLY [9 votes]: The algebras known variously as left-symmetric algebras (or right-symmetric algebras), Vinberg algebras, or pre-Lie algebras are in general not power associative. These algebras occur in at least two apparently unrelated contexts. One is the study of flat affine structures, the other is the study of the algebraic structure of renormalization. -Let $[x, y, z] = (xy)z - x(yz)$ be the associator. An algebra is left-symmetric if -\begin{align} -(LSA)\qquad [x, y, z] = [y, x, z] -\end{align} -for all $x$, $y$, and $z$ (the opposite algebra of a left-symmetric algebra is right-symmetric, meaning it satisfies the identity $[x, y, z] = [x, z, y]$). -These certainly do not have to be power associative. A simple example, taken from E. Kleinfeld's paper "Assosymmetric rings" is the following. The algebra is spanned by $x_{1}$, $x_{2}$, and $x_{3}$, with the non-zero products $x_{1}^{2} = x_{2}$ and $x_{2}x_{1} = x_{3}$. The only non-zero associator of the form $[x_{i}, x_{j}, x_{k}]$ is $[x_{1}, x_{1}, x_{1}] = x_{3}$. This shows that the algebra is left symmetric but not power associative. Here's a more interesting example. Consider the algebra with basis $x_{1}, x_{2}, x_{3}, \dots$ and product $x_{i}x_{j} = (j+1)x_{i+j}$. Then $[x_{i}, x_{j}, x_{k}] = -k(k+1)x_{i+j+k} = [x_{j}, x_{i}, x_{k}]$, so the algebra is left symmetric. Since $[x_{i}, x_{i}, x_{i}] = -i(i+1)x_{3i}$, this algebra is not power associative. The associated Lie algebra is the Lie algebra of polynomial vector fields on the line. A concrete realization is given by $x_{i} = z^{i+1}\partial_{z}$. The left symmetric multiplication is that given by the standard (flat) covariant derivative. -Left symmetric algebras were introduced by E.B. Vinberg (hence the terminology "Vinberg algebras") and used by him to give a sort of classification of homogeneous convex cones (the title of the English translation is "The Theory of convex homogeneous cones"). In a related context, a left invariant flat (torsion free) connection on a Lie group gives its Lie algebra a structure of a left symmetric algebra. (If $\nabla$ is an affine connection, then the product on vector fields defined by $xy = \nabla_{x}y$ satisfies $[x, y] = xy - yx$ if $\nabla$ is torsion-free, and satisfies (LSA) if $\nabla$ is moreover flat). -Roughly contemporaneously, M. Gerstenhaber defined a product on the space of Hochschild cochains of an associative algebra and showed that makes the cochains into what he called a (graded) pre-Lie algebra. This alternative terminology reflects that for any left symmetric algebra the commutator $[x, y] = xy - yx$ satisfies the Jacobi identity, so is a Lie bracket (the condition (LSA) means that the left regular representation of the algebra is a Lie algebra homomorphism). In the context in which Gerstenhaber was working, the algebras are graded, meaning the rule of signs has to applied to all brackets, associators, etc. -There is a more sophisticated story involving operads and rooted trees and related to the work of Connes and Kreimer on the Hopf algebra of rooted trees, but I don't know the details well enough to recount it here. The key thing is that there is a product on rooted trees which makes them into a pre-Lie algebra. See the paper of Chapoton and Livernet ("Pre-Lie algebras and the rooted trees operad") and a survey by D. Manchon ("A short survey on pre-Lie algebras") for this point of view and references. For the more classical material related to cones and affine structures, in addition to Vinberg's article, some of the standard references are D. Segal's "Complete left-symmetric algebras" and J. Helmstetter's "Radical d'une algebra symetrique gauche". There are surveys by D. Burde and P. Cartier ("Vinberg algebras").<|endoftext|> -TITLE: The hypercube: $|A {\stackrel2+} E| \ge |A|$? -QUESTION [15 upvotes]: I have a good motivation to ask the question below, but since the post is -already a little long, and the problem looks rather natural and appealing -(well, to me, at least), I'd rather go straight to the point. -Let $n\ge 3$ be an integer. If $E$ denotes the standard basis of the -vector space ${\mathbb F}_2^n$, then for any subset $A\subset{\mathbb -F}_2^n$ we have $|A+E|\ge|A|$. This trivial estimate is easy to improve -in various ways, but this is not my concern here. What I am interested in -instead is the sumset $A{\stackrel2+} E$ consisting of all those vectors -of ${\mathbb F}_2^n$ with at least two representations as $a+e$, -where $a\in A$ and $e\in E$; that is, vectors at Hamming distance $1$ -from at least two elements of $A$. How small can this sumset be? -It is not difficult to find a linear subspace $L<{\mathbb F}_2^n$ of -co-dimension ${\rm codim}\,L=\lfloor\log_2 n\rfloor+1$ such that every two -elements of $L$ are at least distance $3$ from each other. Clearly, -$L{\stackrel2+} E$ is empty, showing that if $|A|<2^n/n$, then, in general, -no lower bound for $|A{\stackrel2+} E|$ can be obtained. Let's assume, -however, that $A$ is large; what can be said in this case? A simple double -counting shows that - $$ |A{\stackrel2+} E| > \Big( 1-\frac{2^n}{n|A|} \Big) |A|. $$ -My question is: assuming that $A$ is large enough (say, $|A|=2^{n-1}$), -can this estimate be improved to $|A{\stackrel2+} E|\ge|A|$? Here is a -way to put it in a particularly simple, notation-free form: - -Suppose that half of the vertices of the $n$-dimensional hypercube are - colored, say, red. Is it true that under any such coloring, at least half - of the vertices have two (or more) red neighbors? - -Notice that, if true, the estimate $|A{\stackrel2+}E|\ge|A|$ is best -possible: equality is attained, for instance, if $A$ is the set of all -vectors of the same parity (alternatively, all vectors with the first -coordinate equal to $0$, or all vectors with the sum of all coordinates, but -the first one, equal to $0$). -Another remark is that $|A{\stackrel2+} E|\ge|A|$ holds true if $A\subseteq{\mathbb F}_2^n$ is an affine subspace with $|A|>2^n/n$. - -REPLY [8 votes]: OK, suppose that $n\ge 3$, let $A$ be a set of even vertices of cardinatily $2^n\mu\ge 2^{n-2}$ (so $\mu\ge \frac 14$), and write $B:=A{\stackrel2+}E$; that is, $B$ is the set of odd vertices with at least two neighbors in $A$. Assume that $|B|=2^n\xi$. Our aim is to show that $\xi\ge\mu$. Let us consider the action of the averaging (over neighbors) operator $T$ in $L^2$ with respect to the Haar measure. -Let $f$ be the characteristic function of $A$. Let $g$ be $f$ with the constant and the alternating components removed; thus, $g(z)=f(z)-2\mu$ if $z$ is even, and $g(z)=0$ if $z$ is odd. Then $\|g\|_2^2=\mu-2\mu^2$ and, thereby, $\|Tg\|_2^2\le (1-\frac 2n)^2(\mu-2\mu^2)$ because we removed the eigenspaces corresponding to the eigenvalues $\pm 1$ and every other eigenvalue is at most $1-\frac 2n$. -On the other hand, we know that $Tg\le \frac 1n-2\mu$ on the complement of $B$ in the set of odd vertices. To balance it to the average $0$, we should have $Tg$ at least $\frac{\frac 12-\xi}{\xi}(2\mu-\frac 1n)$ on $B$ on average and, since the quadratic average can be only larger, we get -$$ -\|Tg\|_2^2\ge \left[(\frac 12-\xi)+\xi\left(\frac{\frac 12-\xi}{\xi}\right)^2\right] -(2\mu-\frac 1n)^2=\frac 12\frac{\frac 12-\xi}{\xi}(2\mu-\frac 1n)^2 -$$ -Thus -$$ -\frac 12(\frac 12-\xi)(2\mu-\frac 1n)^2\le \xi(1-\frac 2n)^2(\mu-2\mu^2) -$$ -Now, $2\mu-\frac 1n=2\mu(1-\frac{2}{4\mu n})\ge 2\mu(1-\frac 2n)$ under our assumption $\mu\ge \frac 14$. So, we get -$$ -\frac 12(\frac 12-\xi)4\mu^2 \le \xi(\mu-2\mu^2) -$$ -or, equivalently, -$$ -(1-2\xi)\mu\le (1-2\mu)\xi -$$ -i.e., -$$ -\mu\le\xi. -$$ -I hope I haven't made a stupid mistake anywhere though I do not really like this proof: it works for $\stackrel{2}+$, but not for $\stackrel{4}+$ and you are, probably, interested in $\stackrel{K}+$ for all fixed $K$ as $n\to\infty$. Anyway, it gives the desired cutoff at $1/2$ for fixed parity and, thereby, the cutoff at $\frac 34$ in general.<|endoftext|> -TITLE: What are "good" examples of spin manifolds? -QUESTION [43 upvotes]: I'm trying to get a grasp on what it means for a manifold to be spin. My question is, roughly: - -What are some "good" (in the sense of illustrating the concept) examples of manifolds which are spin (or not spin) (and why)? - - -For comparison, I'd consider the cylinder and the mobius strip to be "good" examples of orientable (or not) bundles. - -I've read the answers to Classical geometric interpretation of spinors which are helpful, but I'd like specific examples (non-examples) to think about. - -REPLY [14 votes]: If you know about Steenrod operations, here's a very convenient characterization: - -A manifold $M$ is Spin iff its Poincare duality in $H^*(M,\mathbb Z/2)$ is - compatible with $Sq^1$ and $Sq^2$. - -Similarly, oriented manifolds are those whose Poincare duality in $H^*(M,\mathbb Z/2)$ is compatible with $Sq^1$. -The story continues: String manifolds have a Poincare duality in $H^*(M,\mathbb Z/2)$ that is compatible with $Sq^1$, $Sq^2$ and $Sq^4$ (but now, that's no longer an if and only if). My paper http://arxiv.org/abs/0810.2131 with Chris Douglas and Mike Hill describes all that in detail and provides many concrete examples.<|endoftext|> -TITLE: The Schwartz Space on a Manifold -QUESTION [13 upvotes]: I asked this question a couple of days ago on math.stackexchange, but have yet to receive a response, so I have decided to post this here. -This question is also vaguely related (both questions arose from the same thing I was working on) to this question I just asked last night. -The question is simple: on a general manifold $M$, can one generalize the space of Schwartz functions on $\mathbb{R}^n$ to a space of smooth functions $\mathcal{S}(M)$ on $M$ that obeys similar properties? I would like to be able to define the convolution of two Schwartz functions, so I guess I better require that $M$ at least be a (unimodular) Lie group. Is it then possible to define $\mathcal{S}(M)$? What about the Fourier transform? Is there a natural definition of the Fourier Transform on $\mathcal{S}(M)$? - -REPLY [8 votes]: To define a Schwartz space, you need a notion of decay at infinity, so you need a ``norm'', i.e. a distance to some origin. So the convenient framework is a complete Riemannian manifold. However, even on a Lie group, it is not enough to choose an invariant Riemannian structure to get a Schwartz space having the properties that you require (convolution algebra, good Fourier transformation...). See e.g. the subtlety in the definition of Harish-Chandra's Schwartz space on a semi-simple Lie group, where you have to throw in the $\Xi$-function. -For simply connected solvable Lie groups, the definition of the Schwartz algebra is (I believe) fairly recent: see a paper by Emilie David-Guillou: https://arxiv.org/pdf/1002.2185<|endoftext|> -TITLE: structure of the variety of normal matrices -QUESTION [6 upvotes]: I need some help analyzing the variety of normal matrices $M M^\dagger -M^\dagger M = 0$. Each entry in this equation satisfies a quadratic equation -$$\sum_j h_{ij} \overline{h_{kj}} = \sum_j \overline{h_{ij}} h_{kj} $$ -Where $i,j = 1, \dots, n$. This reminds me of the Plucker embedding of the Grassmanian as the intersection of quadratics. -If this were a single non-degenerate conic it would be similar to some kind of cone $x_1^2 + \dots + x_k^2 - x_{k+1}^2 -\dots -x_n^2=0 $. But this is an intersection of (possibly degenerate) quadrics. Not at all sure the structure of this variety. -I'm looking for general information about this variety. Is this variety reducible or singular? What kind of components does it have? What's it's dimension? - -EDIT: By the spectral theorem normal matrices are similar to diagonal matrices, $\mathrm{diag}(\lambda_1, \lambda_2, \dots, \lambda_n)$ by a unitary matrix so there is a $U(n)$ action on the variety of normal matrices fibered by the diagonal matrices themselves. Some of these orbits are degenerate (like when some $\lambda = 0$). -So maybe $\dim \{ [M, M^\dagger]=0\} = \dim U(1)^n + \dim U(n) = n^2 + n$ - -REPLY [2 votes]: If you need a reference, most of what Torsten Ekedahl explains above can be found in the paper - -Kh. Ikramov, The dimension of the variety of normal matrices, Zh. Vychisl. Mat. Mat. Fiz., 38 (1998), 5–10 - -In fact Ikramov has quite a few papers devoted to similar problems (for example for the variety of conjugate normal matrices which satisfy $XX^{\dagger}=\overline{X^{\dagger}X}$ etc.). The paper above is hard to find in electronic form, I believe, but some of its results are reproduced in S. Friedland's "Normal Matrices and the Completion Problem" (which is also available here). In particular Lemma 2.4 proves that the variety of (complex) normal matrices as a real variety is irreducible of dimension $n(n+1)$, and its set of smooth points corresponds to normal matrices with distinct eigenvalues.<|endoftext|> -TITLE: When is $\ker AB = \ker A + \ker B$? -QUESTION [21 upvotes]: Prove/ Disprove: Let $n$ be a positive integer. Let $A$, $B$ be two $n \times n$ square matrices over the complex numbers. If $AB = BA$ and $\ker A = \ker A^2$ and $\ker B = \ker B^2$ -then $\ker AB = \ker A + \ker B$. -(Recall that $\ker A$ is the set of all vectors $v$ such that $Av = 0$.) -Background: I am teaching linear algebra this semester. I did not like the standard proof of the Jordan canonical form I found in the textbooks, and thought I could prove it differently, directly from the axioms for a vector space, without using either the determinant, or the classification theorem for finite abelian groups. If the statement above is true, I believe I have a proof for the Jordan canonical form for $T$ by setting $A = (T-\lambda_1I)^{n_1}$ and $B=(T-\lambda_2I)^{n_2}$ for appropriate $n_1$ and $n_2$. -Note 1: If $A = B = \left( \begin{array}{cc} -0 & 1\\\0 & 0 \end{array} \right)$ then $AB = BA$ but $\ker AB \neq \ker A + \ker B$. -Note 2: It is easy to find $A, B$ such that $\ker A = \ker A^2$ and $\ker B = \ker B^2$ and $B$ maps a vector outside $\ker A + \ker B$ to $\ker A$, so that $\ker AB \neq \ker A + \ker B$. -Hence, both conditions are necessary. - -REPLY [7 votes]: Here's another statement of more or less the same result. The ideas in the proof are from this proof in German that Martin Brandenburg linked to in his comment. -Claim: Let $n$ be a non-negative integer. Let $A$, $B$ be two $n×n$ square matrices over the complex numbers. If $AB=BA$ and $\ker A \cap \ker B = \{0\}$ then $\ker AB=\ker A \bigoplus \ker B$. -Note: Assuming that $\ker A\cap\ker B=\{0\}$ is not a big restriction, since we can always quotient out to eventually reduce to this case. -Proof: Since $A,B$ commute, it is clear that $\ker A \bigoplus \ker B\subseteq \ker AB$. Further, $\ker AB$ is invariant under $A, B$. Since all the action is taking place within $\ker AB$, we may assume without loss of generality that $\ker AB$ is the entire space, of dimension $n$. This implies $\operatorname{im} B\subseteq \ker A$. -By the rank-nullity theorem, $\dim\ker B + \dim\operatorname{im} B = n$. Hence, $\dim\ker A + \dim\ker B\geq n$. Since these spaces intersect trivially by assumption, we are done.<|endoftext|> -TITLE: Being a subgroup: proof by character theory -QUESTION [28 upvotes]: Let me first cite a theorem due to Frobenius: - -Let $G$ be a finite group, with $H$ a proper subgroup ($H\ne (1)$ and $G$). Suppose that for every $g\not\in H$, we have $H\cap gHg^{-1}=(1)$. Then - $$N:=(1)\cup(G\setminus\bigcup_{g\in G}gHg^{-1})$$ - is a normal subgroup of $G$. - -The proof is fascinating. One never proves directly that $N$ is stable under the product and the inversion. Instead, one constructs a complex character $\chi$ over $G$, with the property that $\chi(g)=\chi(1)$ if and only if $g\in N$. This ensures (using the equality case in the triangle inequality) that the corresponding representation $\rho$ satisfies $\rho(g)=1$ if and only if $g\in N$. Hence $N=\ker \rho$ is a subgroup, a normal one! - -Does anyone know an other example where a subset $S$ of a finite group $G$ is proven to be a subgroup (perhaps a normal one) by using character theory? Is there any analogous situation when $G$ is infinite, say locally compact or compact? - -Edit: If the last argument, in the proof that $S$ is a subgroup, is that $S$ is the kernel of some character, then $S$ has to be normal. Therefore, an even more interesting question is whether there is some (family of) pairs $(G,T)$ where $T$ is a non-normal subgroup of $G$, and the fact that $T$ is a subgroup is proved by character theory. I should be happy to have an example, even if there is another, character-free, proof - -REPLY [3 votes]: Let me re-phrase my remark. -Give sufficient conditions for a character to be a permutation character.<|endoftext|> -TITLE: Article about partitions with forbidden parts/multiplicities -QUESTION [5 upvotes]: I am looking for an article, most likely from the 90s, that generalized the bijection between partitions with odd and distinct parts by explaining how a bijection between the forbidden parts could be transformed into a bijection of the partitions. -(So, in the example above, a bijection can be formed using the fact that $(2k)$ is in bijection with $(k,k)$ which form the forbidden parts on both sides.) - -REPLY [9 votes]: Possible articles you might have seen could be - -M.V. Subbarao, Partition theorems for Euler pairs, Proc. Amer. Math. Soc. 28 (1971), - no. 2, 330-336. - -Here the author characterizes all triples $(A,B,r)$ so that the number of partitions with parts in $A$ is equal to the number of partitions with parts in $B$ with no part repeated more than $r-1$ times. These are the sets satisfying $rB\subset B$ and $A=B-rB$. - -J.B. Remmel, Bijective Proofs of Some Classical Partition Identities, J. Combin. - Theory Ser. A 33 (1982), 273–286 - -This paper proves a very general statement phrased in terms of forbidden patterns. Namely given two sequences of non-empty multisets $\mathcal A=\lbrace A_i\rbrace _{i\in \omega}$ and $\mathcal B=\lbrace B_i\rbrace _{i\in \omega}$, let $|M|$ denote the sum of the elements in $M$ as a multiset. Then if -$$|\bigcup _{i\in S} A_i|=|\bigcup _{i\in S} B_i|$$ -holds for all $S\subset \omega$, the number of partitions with no $\mathcal A$ patterns is equal to the number of partitions with no $\mathcal B$ patterns. - -K.M. O’Hara, Bijections for Partition Identities, J. Combin. Theory Ser. A 49 (1988), - 13–25. - -While Remmel's approach gives a bijection based on the Garsia-Milne involution, this paper shows that the algorithm for producing bijections can be made considerably faster at least in the case of disjoint multisets. -On the other hand, it could have been a more recent article which most probably references at least one of these papers. You can find an extended list of references as well as various statements which generalize the bijection you mention, in Igor Pak's survey on partition bijections, the relevant section being section 8 (survey is also available from his website). This is also the topic of Herbert S. Wilf's notes "Lectures on Integer Partitions" (which you can find here).<|endoftext|> -TITLE: Expected norm of sum of random orthogonal matrices -QUESTION [8 upvotes]: Somehow I got wondering about the following question today: -Suppose $Q_1,\ldots,Q_n$ are random (uniformly sampled) $d \times d$ orthogonal matrices. - -What is the expected value of the quantity $\|\sum_i Q_i\|$? - -Additionally, suppose I actually generate random skew-symmetric matrices $S_1,\ldots,S_n$, and then obtain corresponding orthogonal matrices via the matrix exponential, $e^{S_i}$. - -What is the expected value of the quantity $\|\sum_i e^{S_i}\|$ - -EDIT: From the comments (and from Mikael's answer) it seems like this is a tough question. But already the case with large $d$ is quite useful. - -REPLY [8 votes]: EDIT: My answer only deals with the $d \to \infty$ regime. -This question is not too naive (or at least the answer is hard). I am almost sure that for fixed $d$ there is no exact formula. For the limit as $d \to \infty$ I think that one expects that the norm of $\sum_1^n Q_i$ almost surely converges to $2 \sqrt{n-1}$, but I don't know if a proof exists yet (my guess is that everything works the same way as for unitaries, see below). -If one replaces orthogonal by unitaries, the result is known to hold from the very recent work of Collins and Male, see part 3.2 here. Their result is more general and they compute the liming norm of any sum of products of independant random unitaries in term of free probability. In fact the proof uses a simple but clever coupling argument together with the deep work of Haagerup and Thorbjornsen A new application of Random Matrices: Ext(C*_{red}(F_2)) is not a group, where Haagerup and Thorbjornsen prove the corresponding result for gaussian hermitian matrices instead of unitaries. -The work of H-T has been generalized to random real symmetric matrices by Hanne Schulz, and this implies that the answer to the second question can be computed in finite time (assuming that you take a suitable Gaussian probability measure on the set of skew-symmetric matrices). I can try to do the exact computation if you want. -Second edit: I deleted a remark where I said that the norm of $\sum_1^N e^{S_i}$ is of order $\sqrt n$. This would be true if $E(e^{S_i})$ was $0$, which is not the case as Terry Tao points out in his comments to your question. In fact it is the norm of $\sum (e^{S_i} - E(e^{S_i}))$ which is of order $\sqrt n$.<|endoftext|> -TITLE: Is the Mendeleev table explained in quantum mechanics? -QUESTION [56 upvotes]: Does anybody know if there exists a mathematical explanation of the Mendeleev table in quantum mechanics? In some textbooks (for example in "F.A.Berezin, M.A.Shubin. The Schrödinger Equation") the authors present quantum mechanics as an axiomatic system, so one could expect that there is a deduction from the axioms to the main results of the discipline. I wonder if there is a mathematical proof of the Mendeleev table? -P.S. I hope the following will not be offensive for physicists: by a mathematical proof I mean a chain of logical implications from axioms of the theory to its theorem. This is normal in mathematics. As an example, in Griffiths' book I do not see axioms at all, therefore I can't treat the reasonings at pages 186-193 as a proof of the Mendeleev table. By the way, that is why I did not want to ask this question at a physical forum: I do not think that people there will even understand my question. However, after Bill Cook's suggestion I made an experiment - and you can look at the results here: https://physics.stackexchange.com/questions/16647/is-the-mendeleev-table-explained-in-quantum-mechanics -So I ask my colleagues-mathematicians to be tolerant. -P.P.S. After closing this topic and reopening it again I received a lot of suggestions to reformulate my question, since in its original form it might seem too vague for mathematicians. So I suppose it will be useful to add here, that by the Mendeleev table I mean (not just a picture, as one can think, but) a system of propositions about the structure of atoms. For example, as I wrote here in comments, the Mendeleev table states that the first electronic orbit (shell) can have only 2 electrons, the second - 8, the third - again 8, the fourth - 18, and so on. Another regularity is the structure of subshells, etc. So my question is whether it is proved by now that these regularities (perhaps not all but some of them) are corollaries of a system of axioms like those from the Berezin-Shubin book. Of course, this assumes that the notions like atoms, shells, etc. must be properly defined, otherwise the corresponding statements could not be formulated. I consider this as a part of my question -- if experts will explain that the reasonable definitions are not found by now, this automatically will mean that the answer is 'no'. -The following reformulation of my question was suggested by Scott Carnahan at http://mathoverflow.tqft.net/discussion/1202/should-a-mathematician-be-a-robot/#Item_0 : -"Do we have the mathematical means to give a sufficiently precise description of the chemical properties of elements from quantum-mechanical first principles, such that the Mendeleev table becomes a natural organizational scheme?" -I hope, this makes the question more clear. - -REPLY [7 votes]: I'm arriving after the war, but this is an interesting question, so I'm going to write up what I understand about it. -First of all, for a comprehensive mathematical understanding of the periodic table, you have to settle on a model. The relevant one here is quantum mechanics (for large atoms, relativistic effects start to become important, and that's a whole mess). It's entirely axiomatic, and requires no further tweaking. Then you basically have to solve an eigenvalue on a space of functions of $6N$ coordinates (ignoring spin). That gives you a "mathematical explanation" of the table, in the sense that knowledge of the solution $\psi(x_1,x_2,\dots,x_N)$ is all there is to know about the static structure of an atom. Notice that in this formulation, all electrons are tied together inside one big wavefunctions, so an "electronic state" has no meaning. Mendeleev table is not even compatible with this formulation. -Of course, solving the full eigenproblem is not possible, so all you can do is mess around with approximations. A simplistic but illuminating approximation is to completely neglect electron repulsion. Great simplification occurs, and it turns out one can speak of "electronic states". Non-trivial behaviour occurs because of the Pauli exclusion principle. This is known as the "Aufbau" principle: one builds atoms by successively adding electrons. The first electron gets itself into the lowest energy shell, then the second one gets into the same state, but with opposite spin. The third begins to fill the second shell (which has three spaces, times two because of spin), and so on. This is the basic idea behind the table, and provides a clue as to why it is organised the way it is. So this might be the theory you're looking for. It's explicitely solvable, and only requires the theory of the hydrogenoid atoms. -Of course, because of the approximations, the quantitative results are all wrong, but the organisation is still there. Except for larger elements, where the Mendeleev table is, from what I understand, an ad-hoc hack. You can improve the approximation using ideas like "screening", and this leads to the Hartree-Fock method, which still preserves the notion of shells. -Hope that helps. Then again, if you're looking for a completely logical approach to physics that'll readily explain real life, you're bound to be disappointed. Even simple theories such as the quantum mechanics of atoms are too hard to be solved exactly, which is why we have to compromise and make approximations.<|endoftext|> -TITLE: What is significant about the half-sum of positive roots? -QUESTION [57 upvotes]: I apologize for the somewhat vague question: there may be multiple answers but I think this is phrased in such a way that precise answers are possible. -Let $\mathfrak{g}$ be a semisimple Lie algebra (say over $\mathbb{C}$) and $\mathfrak{h} \subset \mathfrak{g}$ a Cartan subalgebra. All the references I have seen which study the representation theory of $\mathfrak{g}$ in detail make use of the half-sum of positive roots, which is an element of $\mathfrak{h}^\ast$: e.g. Gaitsgory's notes on the category O introduce the "dotted action" of the Weyl group on $\mathfrak{h}^\ast$, the definition of which involves this half-sum. -Is there a good general explanation of why this element of $\mathfrak{h}^\ast$ is important? The alternative, I suppose, is that it is simply convenient in various situations, but this is rather unsatisfying. - -REPLY [3 votes]: Parts of what I will explain here have been said in other answers, but not quite in this way. -One thing about the weight $\rho$ is that the places it tends to show up, it is really the weight $-\rho$ that is the important one. Note that the "dot" action is just a shift to make $-\rho$ the new "$0$". -So why would we like to have $-\rho$ be our new "$0$"? Because it turns out to be the weight that behaves like a $0$ should when we consider cohomology. -This becomes a bit clearer when we go to the closely related setting of a semisimple simply connected algebraic group $G$ with Borel subgroup $B$. Here we consider the functor $\operatorname{Ind}_B^G$ inducing from $B$-modules to $G$-modules (for full details of definitions and proofs, one can consult for example Jantzen's book). -To make things simpler, for a weight $\lambda$ we denote also by $\lambda$ the $1$-dimensional $B$-module where the maximal torus in $B$ acts as $\lambda$, and $B$ acts via the projection to this torus. -For $i\geq 0$ write $H^i(\lambda) = R^i\operatorname{Ind}_B^G(\lambda)$ for the $i$'th right derived functor of induction applied to the $B$-module $\lambda$. Note that this is the $i$'th cohomology of the flag variety $G/B$ with coefficients in the line bundle defined by $\lambda$. -Now the reason for wanting $-\rho$ to be our "$0$" starts to be clear, as $H^i(-\rho) = 0$ for all $i\geq 0$. But there are other weights satisfying this, so what makes $-\rho$ extra special? -What makes $-\rho$ so special is that the vanishing of the above cohomology groups in some sense happens "as early as possible". More precisely, if $P = P(\alpha)$ is a parabolic subgroup corresponding to a simple root $\alpha$, then already $R^i\operatorname{Ind}_B^P(-\rho) = 0$, and we can factor $\operatorname{Ind}_B^G$ as $\operatorname{Ind}_P^G\operatorname{Ind}_B^P$. -Moreover, if $\lambda\neq -\rho$ then there is some simple root $\alpha$ such that either $\operatorname{Ind}_B^{P(\alpha)}(\lambda)\neq 0$ or $R^1\operatorname{Ind}_B^{P(\alpha)}(\lambda)\neq 0$.<|endoftext|> -TITLE: Are $G$-spectra the same as modules over a "group ring spectrum"? -QUESTION [17 upvotes]: Let $G$ be a finite group (maybe this will also work when $G$ is compact, or something, but to be safe we'll let it be finite). I imagine it's quite natural to ask: is the category of $G$-spectra equivalent to the category of module spectra over some ring spectrum, probably denoted by $SG$? -For definiteness, we can take "the" category of $G$-spectra to be the symmetric-spectra model (or orthogonal if we want to try and deal with compact groups), and similarly for modules over a ring spectrum. -I imagine if such a thing existed it would have the property that $\pi_0SG$ should be the actual group ring, $SG$ should be an $A_\infty$-ring, and if $G$ is abelian it should be an $E_\infty$-ring. -It seems like such a thing should exist since the ($\infty$-)category of $G$-spectra (at least coming from, say, the orthogonal model) looks presentable and generated by the equivariant sphere... so by some theorem in Lurie it should be equivalent to an ($\infty$ -)category of modules over some ring spectrum. -A brief search online and glance at May's book on the subject revealed nothing, but I could have easily missed it. Any pointers to the literature or brief epositions of a construction of such a thing would be much appreciated! -(I would say "take the free spectrum on the objects of $G$ and mod out by some relations" and maybe this is how it's done, but I wonder how to make this precise in the "brave new" world, also such a construction probably would not generalize to the compact group case, so maybe there's some better way of doing it.) - -REPLY [17 votes]: Eric wrote a really nice response telling that your initial hope is incorrect and why. I'd just like to write some positive results that you can find. -Disclaimer: I understand little to nothing about the case of a compact Lie group. -Schwede and Shipley have a paper entitled "Stable model categories are categories of modules" from 2003. In particular, G-spectra form a stable model category to which their results apply. Schwede-Shipley show that if you pick a set of "generators", then you'll get an category $I$ enriched in spectra, with $Hom_I(i,j)$ being a spectrum together with units $\mathbb S \to Hom_I(i,i)$ and composition maps $$Hom_I(j,k) \wedge Hom_I(i,j) \to Hom_I(i,k)$$ which are unital and associative. (This is the spectrum version of a DG-category, if you like). Then there is an equivalence between $G$-spectra and enriched functors from $I$ to the category of spectra. -In $G$-spectra, we can pick generators given by the spectra $\Sigma^\infty G/H_+$, which are representing objects for the standard "fixed point" functors. So a $G$-spectrum is equivalent to the data of a collection of spectra $Y^H$ as $H$ ranges over the subgroups of $G$, together with "action maps" -$$F(\Sigma^\infty G/H_+, \Sigma^\infty G/K_+) \wedge Y^H \to Y^K$$ -that are unital and associative. -If you're feeling like it, you could instead replace several generators with $\bigvee_H G/H_+$, and establish $G$-spectra as equivalent to modules over one ring spectrum which is a big "matrix algebra" containing a bunch of commuting idempotents. It's not clear to me whether this is generally profitable. (It certainly doesn't make looking at the symmetric monoidal structure on $G$-spectra easier.) -To go further, we need to specify a little about which category of $G$-spectra you're interested in. This is often phrased in terms of a choice of universe. -At one end, you have $G$-spectra indexed on the trivial universe, which are formed by taking the category of $G$-spaces and inverting the suspension functor. There's a "coalescing" result of Elmendorf (his "Systems of fixed point sets") that essentially shows that the homotopy category of $G$-spaces is equivalent to the homotopy category of functors from the orbit category of $G$ to spaces; $G$-spectra indexed on the trivial universe satisfy a similar result. -At the other end, you have $G$-spectra indexed on a complete universe, where all the spheres based on orthogonal representations of $G$ become invertible. These are more complicated, because they're generated by more than just actions $g: Y^H \to Y^{gHg^{-1}}$ and restrictions $Y^K \to Y^H$ for $H < K$. They also have transfer maps. -If you've done any looking into $G$-spectra, you've probably encountered the notion of a Mackey functor, which is a collection of abelian groups with restriction, transfer, and conjugation maps. One compact way to phrase this is that Mackey functors are additive functors from the "Burnside category" to the category of abelian groups. $G$-spectra indexed on a complete universe satisfy a similar result: they are enriched functors from a topological Burnside category to the category of spectra. In particular, every $G$-spectrum produces a Mackey functor to the stable homotopy category. (There are several places I could insert some more or less gratuitous $\infty$-category theory here.) -I know that Clark Barwick has given several talks on this, and is likely in the process of writing it up. -Whether Mackey functors make you happy might depend on whether you're in that pleasant zone between understanding their definitions and trying to do serious homological algebra with them. While I'm writing pithy asides, it's kind of depressing that the Burnside category doesn't have entries on Wikipedia or the nLab for me to link to, and Mackey functors only have this. Many of the presentations in the literature are worth looking at. -There are several categories of $G$-spectra in between, and I don't know much about the general properties of those.<|endoftext|> -TITLE: Are there any interesting connections between game theory and engineering? -QUESTION [9 upvotes]: I am doing a senior project and it must be based off game theory, but I am having trouble finding any connections to engineering, possibly structural, or architectural, maybe even civil or mechanical. Are there any branches of engineering that use a significant amount of game theory? - -REPLY [12 votes]: A Spring 2010 course was offered by Asu Ozdaglar @MIT, entitled "Game Theory with Engineering Applications." It is included in MIT's Open CourseWare, so extensive information is available. -Here is the course description: - -Course Description. - This course is an introduction to the fundamentals of game theory and mechanism design. Motivations are drawn from engineered/networked systems (including distributed control of wireline and wireless communication networks, incentive-compatible/dynamic resource allocation, multi-agent systems, pricing and investment decisions in the Internet), and social models (including social and economic networks). The course emphasizes theoretical foundations, mathematical tools, modeling, and equilibrium notions in different environments. -            - -(The point of this image is that the stag hunt is "a game which describes a conflict between safety and social cooperation.") - -Here is a high-level syllabus of this course: - - Introduction to game theory - Strategic form games - Learning, evolution, and computation - Extensive games with perfect information - Repeated games - Games with incomplete information - Mechanism design - Mechanisms in networking - -See also the LLNL (U.S. Lawrence Livermore National Laboratory) Optimization and Game Theory project, which they describe as follows: - -Researchers apply a wide range of optimization and game theory techniques to address challenging problems in national security, energy economics, and environmental management. Our problems typically involve complex objectives in multiple dimensions (e.g., improving safety while reducing cost), and uncertainty of the outcomes of actions. Problems can require high-resolution models or national-scale infrastructure, so efficient algorithms are required that exploit problem structure and large-scale computational hardware. - -Finally, I will mention the 2006 (short) book, Game Theory for Wireless Engineers, whose abstract contains -these two sentences: - -In the early to mid-1990's, game theory was applied to networking problems including flow control, congestion control, routing and pricing of Internet services. More recently, there has been growing interest in adopting game-theoretic methods to model today's leading communications and networking issues, including power control and resource sharing in wireless and peer-to-peer networks.<|endoftext|> -TITLE: Reference needed for Lucas' Theorem for multinomial coefficients modulo a prime -QUESTION [5 upvotes]: I am looking for a reference for a proof of the following: -Let $n$ and $a,b, \ldots ,z$ be non-negative integers with $a + b -+ \ldots + z = n$, and let $p$ be a prime. Write $n = n_0 + n_1 p + \ldots + -n_m p^m$ in $p$-ary notation, similarly for $a, b, \ldots , z$. -Then, modulo $p$, the multinomial coefficient ${n \choose a,b, \ldots, z}$ is zero if there is -some "carrying" in computing the sum $a + b + \ldots + z$ (i.e., $a_i + b_i + \ldots + z_i \geq p$ for some $i$). Otherwise, it is the product of the multinomial coefficients of -the individual $p$-digits, that is, -${ n \choose a, b, \ldots, z} = {n_0 \choose a_0, b_0, \ldots, z_0}{n_1 \choose a_1, b_1, -\ldots, z_1} \ldots {n_m \choose a_m, b_m, \ldots , z_m}$ -I've found plenty of references for this theorem restricted to binomial coefficients, otherwise known as Lucas' theorem, but can't seem to find one for arbitrary multinomials. -Thanks in advance for the help. - -REPLY [4 votes]: You can prove this by induction on the maximum number of base $p$ digits, and to make the argument simpler it's better to formulate a mildly stronger theorem where the leading base $p$ "digit" is allowed to be nonnegative rather than be constrained between 0 and $p-1$: for $d \geq 0$, $t \geq 1$ and nonnegative integers $m_0,m_1,\dots,m_t$, write the $m_i$'s as -$$ -m_0 = c_0 + c_1p + \cdots + c_dp^d \text{ and } -m_j = c_{0j} + c_{1j}p + \cdots + c_{dj}p^d -$$ -where $0 \leq c_i, c_{ij} \leq p-1$ for $i < d$ and $c_d \geq 0$, $c_{dj} \geq 0$. Then -you want to show -$$ -\binom{m_0}{m_1,\dots m_t} \equiv \binom{c_0}{c_{01},\dots,c_{0t}}\cdots \binom{c_d}{c_{d1},\dots,c_{dt}} \bmod p. -$$ -Your description of this result treats separately the case when one of the multinomial coefficients on the right doesn't have a combinatorial meaning (because the numbers in the bottom have a sum exceeding the top), but the congruence is true even in that case if you define a multinomial coefficient as a polynomial in the upper parameter, in the same way you can define $\binom{a}{n}$ as $a(a-1)\cdots(a-n+1)/n!$ in order to give a meaning to $\binom{a}{n}$ even if $a$ is not a positive integer greater than or equal to $n$. -By allowing $c_d,c_{d1},\dots,c_{dt}$ to be just nonnegative, a proof by induction on $d$ immediately reduces to the case $d = 1$. Now use the multinomial theorem in ${\mathbf F}_p[X_1,\dots,X_t]$: in the mod $p$ equation -$$ -(1 + X_1 + \cdots + X_t)^{c_0+c_1p} = (1 + X_1 + \cdots + X_t)^{c_0}(1 + X_1^p + \cdots + X_t^p)^{c_1} -$$ -comparing the coefficient of $X_1^{m_1}\cdots X_t^{m_t}$ on both sides implies -$$ -\binom{m_0}{m_1,\dots m_t} \equiv \binom{c_0}{c_{01},\dots,c_{0t}}\binom{c_1}{c_{11},\dots,c_{1t}} \bmod p -$$ -and we're done. -I gave this argument when I needed the result in www.math.uconn.edu/~kconrad/articles/jacobistick.pdf. See congruence C2 at the bottom of the third page. (Appeared in Enseign. Math. 41 (1995), 141−153.)<|endoftext|> -TITLE: Asymptotic bounds for a confluent hypergeometric function $F_{1}[;1;x]$ -QUESTION [6 upvotes]: I know that for infinite series and $|z|<1$ there exists a confluent hypergeometric expression -$ -\sum_{k=0}^{\infty} \frac{z^k}{k!k!} = F_{1}[;1;z] -$ -This is not very helpful though, and I 'd like to know if it is possible to get some asymptotic expansion for this function and if there exists some general approach to bounding hypergeometric functions asymptotically. - -REPLY [2 votes]: Although not directly for $z < 1$, here is a general paper that might help with some of the techniques that one can use: -The confluent hypergeometric functions $M(a; b, z)$ and $U(a; b, z)$ for large $b$ and $z$ by J. L. López and P. J. Pagola. -Also, if not this paper, please take a look at other papers by López; from what I heard, he is an expert on asymptotics of hypergeometric functions.<|endoftext|> -TITLE: A Model-Theoretic Helly's Theorem -QUESTION [12 upvotes]: There is a combinatorial question posed to me (or rather, posed near me) by my adviser. I am having quite a lot of difficulty proving it. It goes: - -For any NIP theory $T$ (complete with infinite models as usual) and any partitioned formula $\phi(x; y)$, there are natural numbers $k$ and $N$ such that for any finite sets $A$ and $B$, if $A$ is $k$-consistent in $B$, then there is a $B_0\subset B$ with $|B_0|=N$ and so that for any $a\in A$, there is $b\in B_0$ such that $\phi(a;b)$ holds. - -Here, by $k$-consistent, we mean for any $a_1, \ldots, a_n\in A$ there is $b\in B$ such that $\bigwedge_{i=1}^n \phi(a_i,b)$ holds. Also, while it isn't stated, we can take $A$ and $B$ to be sets of tuples; I don't think this affects much of anything. -This is used in a proof that NIP theories have UDTFS, in an unpublished paper by Pierre Simon and Artem Chernikov. They claim this lemma has been proved "in the literature;" neither I nor my adviser has been able to verify this, except for a very long-winded probabilistic argument which feels non-illustrative. -So, what I'm hoping for is either a reference to where it has been proved (or even discussed), some direction on the literature hunt, or just an explanation of why it may or may not be true (or a counterexample, if it is false). Just anything, really. -Note: this fails immediately in the unstable NIP case (take $A=B$ to be an $\omega$-sequence, ordered by $\phi$, so the sets $\phi(a_i,B)$ are strictly decreasing, infinite, with empty intersection). - -REPLY [6 votes]: First, let me point out that the proof of udtfs is joint work with Artem Chernikov. -Hunter Johnson's answer is correct. The reference is Matousek's paper. I don't know of any proof not using probabilty theory. I tried a little bit to look for a model theoretic proof, but did not find any. -However, Matousek's $(p,k)$-theorem gives us more than we need, and the proof can be made much shorter as: - -we are only interested in the case $p=k$ of the $(p,k)$-theorem, -we do not need an optimal value of $k$. - -I will sketch the argument: -Let $\phi(x,y)$, $A$ and $B$ as in your question. -Pick any $\epsilon>0$. -Use the VC-theorem to get $N$ such that for any probability measure on $A$, there is an $\epsilon$-net of size $N$ (namely, a subset $A_0$ of size $N$ such that for any $b$ the measure of $\phi(A,b)$ is within $\epsilon$ of $\frac {|\phi(A_0,b)|}{N}$.) -Assume that $A$ is $N$-consistent in $B$ (using your terminolgy). It follows that for any probability measure on $A$, there is $b$ such that $\phi(A,b)$ has measure at least $1-\epsilon$. -Next, apply Farkas' lemma exactly as in the paper by Alon and Kleitman to which Matousek refers (there are some unnecessary intermediate steps there). You obtain a measure on $B$ such that every subset $\phi(a,B)$ has large measure. Then apply the VC-theorem again to extract an $\epsilon$-net $B_0 \subseteq B$ of bounded size (for some, possibly smaller $\epsilon$). Then every $\phi(a,B)$ intersects $B_0$, so we have what we want. -Compared with Matousek's proof, we can avoid the fractional Helly property and Lemma 2.2 in Alon and Kleitman.<|endoftext|> -TITLE: Critical to Ribet's method -QUESTION [13 upvotes]: At the end of the Introduction to a recent work of Joël Bellaïche to appear in Inventiones one can read a paragraph that I try to rephrase as follows (see the author's webpage, or the "Online First Articles" section in the aforementioned journal webpage -- subscription required -- for the original statements): -The currently available strategies to produce as many independent extensions in the Selmer group as predicted by the $p$-adic Bloch-Kato conjecture rely on (vast) generalizations of the ideas first appeared in Ribet's gem on the converse to Herbrand (the so-called Ribet's method). Loosely, if one wants to produce elements in ${\rm Sel}(\rho)$ one picks a suitable automorphic form $f$ whose attached Galois representation is reducible, containing $\rho$ and the trivial character as (some of the) constituents, and deforms $f$ into a $p$-adic family whose Galois representation is generically irreducible; then a variant of Ribet's lemma yields extensions as the ones desired that can be shown (when one is lucky enough to be able to rule out the possible ``parasit'' extensions) to actually land in ${\rm Sel}(\rho)$. Further, for the methods to succeed, a good knowledge of the relation between the $p$-adic $L$-functions of $f$ and $\rho$ seems necessary. -What motivates my question is the following tantalizing comment of the author in the above section: - -``and there are fundamental reasons that - the form $f$ [as above] we work with - is critical or $\theta$-critical, that - it would take us too far to explain - here'' - -(As explained in the paper, in the case of a classical elliptic cuspidal eigenform $f$, the notions of critical and $\theta$-critical are equivalent, the former meaning that there is some non-classical overconvergent modular form which is a generalized eigenvector for the eigensystem of $f$.) -Hence I would like to openly ask for some explanation on why it is so fundamental that the form $f$ is chosen to be critical or $\theta$-critical. - -REPLY [20 votes]: If we start with a Galois representation $\rho_\pi = 1 \oplus \rho \,\oplus$ other terms, with a refinement, and try -to apply the method in question, we need at three different steps to assume that the refinement is critical: - - -To construct by automorphic methods a non-trivial deformation of $\rho_\pi$ with its chosen refinement. - -To prove that the deformation constructed in 1. is generically irreducible - -And finally, to prove that the non-trivial extensions constructed from this deformation (residually reducible, generically irreducible) by some generalizations of Ribet's lemma have "good reduction" in the sense of Bloch-Kato at places dividing $p$. - - - -Of the three reasons, 3. seems to me the more fondamental. In 2. we can in certain situations do without the criticality hypothesis, -but we get much better result using it. In 1., the criticality hypothesis is needed in all cases that have been considered in the literarure so far, but I can imagine cases where it should not be so. -I now want to explain those reasons in more detail, but for this I need to define the terms I use, in particular -what is a refinement and what it means for it to be critical (which can be done precisely), -and also what is the method I am talking about (which by necessity can only be done with a certain vagueness). -So the method I am considering here, or I should rather say a family of closely related methods, is a proper (and small) subset of the huge world created by Ribet's 1976 paper, used by Chenevier and me in certain of our papers (mainly in our 2004 Annales de l'ENS and our 2009 book at Astérisque), and also by Skinner and Urban (mainly in their 2006 paper at JIMJ and their 2006 announcement at the ICM -- to delimitate the method I should say I don't consider their work on the main conjecture for elliptic curves as part of it). The nethod can be roughly described as follows: the aim is to produce some non-trivial element(s) in the Bloch-Kato Selmer group $H^1_f(\rho)$ of a $p$-adic Galois representation $\rho$ of automorphic origin and of motivic weight $-1$ (this is the most intersting case). The method involves finding, using an hypothesis on $\rho$ (typically a vanishing of its $L$-function) an automorphic representation $\pi$ for a reductive group $G$ over a number field (say $\mathbb{Q}$ to fix ideas), whose attached Galois representation $\rho_\pi$ (semi-simple by construction) is $1 \oplus \rho$ plus other factors deforming $\pi$ $p$-adically in such a way that the corresponding deformation of $\rho_\pi$ is generically irreducible, and use some generalization of Ribet's lemma to construct one or several non-trivial independent extensions of $1$ by $\rho$ which are provably in $H^1_f(\rho)$. -Let us assume to fix ideas that $\rho$ is crystalline at $p$, and $\rho_\pi$ as well (in practice the extra factors are crystalline, so -this is a consequence of the assumption on $\rho$). Let $D$ be the filtered $\phi$-module -attached the the restriction of $\rho_\pi$ to the decomposition group at $p$, and $\phi$ the crystalline Frobenius acting on it. For simplicity, extend the field of ceofficients so that all eigenvalues of $\phi$ are in it, and assume -these eigenvalues are distinct. Then a {\it refinement} of $\rho_\pi$ is an ordering of the eigenvalues of $\phi$ on $D$, or equivalently a, $\phi$-stable complete flag in $D$. So there are $d!$ refinments, if $d$ is the dimension of $\rho_\pi$. We say that a refinement is {\it non-critical} if this flag is in generic position w.r.t. the Hodge filtration on $D$. It is an exercise that for every representation there is at least one non-critical refinement (and in general, many). For a representation which is decomposable like $\rho_\pi$ there is also always one (and actually many) critical refinement. -There is also a notion of refinement on the automorphic side: Let say that $G(\mathbb{Q}_p)=GL_d(\mathbb{Q}_p)$ for simplicity, to avoid mentioning the $L$-group; for $\pi$ an automorphic form such that $\pi_p$ is unramified, an {\it automorphic refinement} is a character of the Atkin-Lehner algebra $A_p$ (for $G=Gl_2$, this is just the lagebra generated by the Atkin=Lehenr operator $U_p$) appearing in $\pi_p^{I_p}$, where $I_p$ is the Iwahori subgroup. There is an injective map from the set of automorphic refinements of $\pi$ to the set of refinements of $\rho_\pi$. It is surjective when $\pi_p$ is a full unramified principal series (that is, a represresentation induced from a Borel of an unramified character of a maximal torus which happens to be irreducible), which is the generic situation. -Now I can explain why the criticality of the refinement is needed in step 1. It turns out that if we want a Galois representation $\rho_\pi$ containing $1 \oplus \rho$ which is attached to an automorphic representatin $\pi$, then in the current state of knowledge, this representation also has to contain the cyclotomic character $\chi$. This is because we only know how to attach Galois representations to groups like unitary or symplectic groups, and the Galois representation attached to those groups are (conjugate) self-dual up tp a twist by $\chi$ -- so if they contain $1$, they must also contain $\chi$. (Actualy even if we could work for $G=GL_d$, if we wanted our $\pi$ to appear in the discrete spectrum, Arthur's conjectures would also force $\rho_\pi$ to contain $\chi$. Only for Eisenstein series for $GL_d$ would it perhabs be possible to avoid the $\chi$). Now the very presence of a $1$ and a $\chi$ in $\rho_\pi$ forces the unramified representation $\pi_p$ to be a proper subquotient of the unramified principal series where it belongs (cf. the classification of unramified principal series). -In other words, in our case, not all refinements of $\rho_\pi$ comes form an automorphic refinement of $\pi$. -A closer analysis shows that all the refinements of $\rho_\pi$ that comes from $\pi$ are critical. Since the theory of deformations of automorphic representations (a.k.a eigenvarieties) can only deform representations with a given automorphic refinement, this forces our refinement of $\rho_\pi$ to be critical in order to carry out step 1. -For step 2. it is sometimes possible to prove by a tedious case analysis using the classification of automorphic representation for $G$ when it is known, that the deformation of $\rho_\pi$ we constructed in 1. -is generically irreducible (this is what I did in my thesis, and that Skinner-Urban did in their JIMJ paper), but there is a better argument, purely Galois-theoretical, which proves that if we start with a Galois representation $\rho_\pi$ with a suitable refinement, and deform it (in a way compatible with the refinement, a notion I have not explained), then the deformation obtained -has very strong irreducibility properties, even after restriction to $G_{\mathbb{Q}_p}$. Again, the suitable condition to impose on the refinement implies it to be critical. This is the method used in my paper with Chenevier at Annales de l'ENS and in our book, as well as by Skinner-Urban in their ICM's paper. -After step 2, the details vary a lot and can be very messy, but let me explain the common idea: a generalization of Ribet's lemma ensures the existence of either a non-trivial extension of $1$ by the cychotomic character $\chi$, or a non-trivial extension of $1$ by $\rho$. We need to know that those extensions have good reduction at $p$: this will ensure that the first one do not exist, and tha the secons is really in $H^1_f(\rho)$. This is done by applying a lemma of Kisin, whose result depend of the refinement we started with. Ensuring by this method that the extension of $1$ by $\chi$ has good reduction always requires the refinement to be critical (this is also true in most case for the extension of $1$ by $\rho$, though when the Hodge-Tate weights of $\rho$ are well positionned we can avoid the use of Kisin's lemma and the criticality hypothesis for this particular point). So the criticality assumption is necessary for step 3.<|endoftext|> -TITLE: Many p,q-Sylow subgroups -QUESTION [8 upvotes]: It is a fact that the symmetric groups have as many 2-Sylow subgroups as possible. More precisely, for all $n \geq 1$, the number of 2-Sylow subgroups in $S_n$ is exactly $n!/2^{\nu_2(n!)}$, which is the index of a 2-Sylow subgroup of $S_n$. This follows from (or, depending on which direction you're coming from, proves) the fact that one (equivalently, all) 2-Sylow subgroup is self normalizing. -It isn't too hard to show that given a prime $p$, there is a family of finite groups $(G_n)$ such that $\nu_p(|G_n|) \rightarrow \infty$ and all the $p$-Sylow subgroups of $G_n$ are self-normalizing. -I want to generalize this to two primes in the obvious way, but I am encountering difficulty. The following would be a good start. - -Given distinct primes $p$, $q$, does there exist a finite group $G$ such that $pq \ \Big| \ |G|$ and all $p$-Sylow, all $q$-Sylow subgroups are self-normalizing? - -REPLY [10 votes]: The answer is no: see Corollary 1.3 in - -Robert M. Guralnick; Gunter Malle; Gabriel Navarro, Self-normalizing Sylow subgroups, Proc. Amer. Math. Soc. 132 (2004), 973-979.<|endoftext|> -TITLE: book on calabi yau manifolds -QUESTION [5 upvotes]: hi, -does anybody know a good book on calabi yau manifolds (i am a beginner) ? -thanks in advance -lois - -REPLY [6 votes]: I would also add the following book: -Dominic Joyce, Compact Manifolds with Special Holonomy -The early parts of the book include an introduction to the Riemannian geometry of Calabi-Yau manifolds. It also includes a proof of the Calabi conjecture.<|endoftext|> -TITLE: Most general context for the Morse Lemmas -QUESTION [13 upvotes]: Among the foundational results in differential topology are the Morse lemmas: - -Suppose that $f\colon\, M\to \mathbb{R}$ is a smooth function on a closed manifold M, that $f^{−1}[-\epsilon,\epsilon]$ is compact, and that there are no critical values between $-\epsilon$ and $\epsilon$. Then $f^{-1}(-\infty,-\epsilon]$ is diffeomorphic to $f^{-1}(-\infty,\epsilon]$. - -Let $f\colon\, M\to \mathbb{R}$ be a smooth function on a closed manifold M, with no critical points on $f^{-1}[-\epsilon,\epsilon]$ except k nondegenerate ones on $f^{-1}(0)$, all of index $s$. Then $f^{-1}[-\infty,\epsilon]$ is diffeomorphic to $X(f^{-1}[-\infty,-\epsilon];f_1,\ldots,f_k;s)$ (for suitable fi). -Here $X(M;f;s)$ for $f\colon\,(\partial D^s)\times D^{n-s}\to M$ is M with an s-handle attached by f. - - -In plain English, the Morse lemmas give us instructions for how to build M out of simple pieces, like a child would build a structure out of Lego blocks. The first lemma says "if f has no critical point, do nothing", while the second lemma says "if f has a critical point, glue in an appropriate handle". -One of the things that makes me feel that I don't understand Morse's lemmas as well as I would like to is that the conditions on the source and target of the Morse function f seem unnecessarily restrictive. Maybe we'd like M to be a manifold with boundary or with corners, or a stratified space, or an infinite-dimensional something-or-other? Indeed, analogues of the Morse lemmas continue to hold (but how much CAN we relax our requirements on M?). -On the target side, what about if we want the target to be something other than $\mathbb{R}$? Circle-valued Morse theory and Morse 2-functions deal with Morse functions to S1 and to R2 correspondingly, and are quite useful. -And so, in order to feel I have a bit more of a grip on the meta-mathematical conceptual framework of the Morse Lemmas, I'm very much interested in the following question: - -Let $f\colon\, M\to N$ be a smooth function, with non-degenerate critical points (whatever that means in context). What conditions do I have to impose on M and on N to obtain reasonable analogues of the Morse Lemmas? - -By reasonable analogues, I mean lemmas which explicitly relate $f^{-1}(N^{\prime})$ to $f^{-1}(N^{\prime\prime})$ up to diffeomorphism (by some sort of generalized handle attachment operation?), where $N^{\prime}\subseteq N^{\prime\prime}\subseteq N$ are some reasonable analogues in N to $(-\infty,-\epsilon]$ and $(-\infty,\epsilon]$ correspondingly. -Is there any work, or any results along these lines? Is this all well-known (and easy?), is it open, or is it known to be impossible (as in: "if the target isn't R then something goes disastrously wrong")? - -REPLY [5 votes]: The questions that you asked are addressed by the once very sexy field of catastrophe theory. The story is a bit too long to tell here. The conditions you are asking for are called stability conditions. Hassler Whitney is one of the pioneers. He gave beautiful answers for $f: M\to N$, $\dim N=\dim M=2$. (The folds and cusps were discovered by him.) See the two volume book of Arnold-Varchenko-Husein-Zade, or the Golubitsky-Guillemin book: Stable Mappings and Their Singularities<|endoftext|> -TITLE: Connes' embedding conjecture for uncountable groups -QUESTION [14 upvotes]: In this topic, I will use the word uncountable group referring to groups whose cardinality is $\leq|\mathbb R|$. -Notation: $R$ is the hyperfinite $II_1$-factor, $\omega$ is a free ultrafilter on the natural numbers, $R^\omega$ is the tracial ultrapower, $\tau$ is the unique normalized trace on $R^\omega$, $U(R^\omega)$ is the unitary group of $R^\omega$. -Definition: A group $G$ is called hyperlinear if there is a group monomorphism $\theta:G\rightarrow U(R^\omega)$ such that $\tau(\theta(g))=0$ for all $g\neq1$. - -Question 1: Does there exist an uncountable non-hyperlinear group? - -A bit of background: the same question for countable groups is known as Connes' embedding problem for groups and it's still unsolved. When I began my PhD (Nov. 2008), my former advisor Florin Radulescu told L. Paunescu and myself that he would have liked to have a better understanding of the problem in higher cardinality. In particular, he asked us to see whether the free group on uncountable many generators, here denoted by $\mathbb F_c$, and the circle group $S_1$ were hyperlinear. Maybe he expected that one of them was not, but we came up with the general result that every subrgroup of $U(R^\omega)$ is hyperlinear (this is basically proved in http://arxiv.org/abs/0911.4978); in particular $S_1$ is hyperlinear and, with an additional 10-line argument, also $\mathbb F_c$ turns out to be hyperlinear. Moreover, the result is quite general and excludes many possible example a priori, making Question 1, in my modest opinion, non-trivial and interesting. At some point, maybe also talking with someone else (but I don't remember exactly who), I got quite convinced that Question 1 has the same level of difficulty of Connes' problem and that they may be actually equivalent. In this case, I would like to find a formal way to express that. In this view, a positive answer to the following question would not completely solve the problem, but would be a nice starting point. - -Question 2: Let $G$ be an uncountable group. Do there always exist countable groups $G_1,G_2,\ldots$ and a free ultrafilter $\omega$ such that $G$ embeds into the algebraic ultraproduct of the $G_i$'s? - -Update: as shown my Simon Thomas below, the answer is positive assuming CH. On the other hand, Joel's answer shows that without CH we have some weaker result. For instance, Question 2 has affirmative answer if we allow the sequence $G_i$ to be indexed by a possibly non-countable set $I$. -Thanks in advance, -Valerio - -REPLY [9 votes]: The general situation, where CH fails, may be informed by -the Keisler-Shelah isomorphism -theorem, which -asserts that two first-order structures have isomorphic -ultrapowers if and only if they have the same first-order -theory. -In particular, for any infinite group $G$ at all, of any size, we -may take a countable elementary subgroup $H$, meaning in -particular that they have the same first-order theory, and -so there is a nonprincipal ultrafilter $U$ on an index set $I$ -such that the ultrapowers $G^I/U\cong H^I/U$ are isomorphic. Since every first-order structure maps -elementarily into its ultrapowers, this means in particular -that $G$ maps elementarily (and hence monomorphically) into -an ultrapower of $H$, a countable group. -Thus, this fully answers the version of question 2 in which -we allow the ultrafilter to live on a bigger index set: -Theorem. For every group $G$ there is a countable group $H$ and a free ultrafilter $U$ on a set, such that $G$ embeds into the ultrapower $H^I/U$. -If you want to insist that the ultrafilter -concentrate on index set $\mathbb{N}$, however, then things become more complicated. If the CH holds, then the -Keisler-Shelah theorem shows that any two groups of size at -most $2^{\aleph_0}$ and with the same theory have -isomorphic ultrapowers by an ultrafilter on $\aleph_0$, and -so the desired result is attained. In the non-CH case, -however, what we seem to get is that for any cardinal -$\lambda$, if $\beta$ is smallest such that -$\lambda^\beta\gt\lambda$, then any two groups of size -$\beta$ with the same theory have isomorphic utrapowers -using an ultrafilter on $\lambda$. Thus, they each map into -an ultrapower of the other. -The Keisler-Shelah theorem was proved first by Keisler in -the case that GCH holds, using saturation ideas as in -Simon's answer. The need for the GCH was later removed by -Shelah.<|endoftext|> -TITLE: Point sets in Euclidean space with a small number of distinct distances -QUESTION [15 upvotes]: It is well known and not hard to prove that the regular simplex in n-dimensions is the only way to place n+1 points so that the distance between distinct pairs of points is always the same. My general question is what happens when we allow more than one distance? -On the plane it is a good exercise to show that this is the complete list of diagrams with 2 distances and 4 or 5 points: - -You can go further on the plane for example see: -Harborth, H and Piepmeyer, L (1996). Three distinct distances in the plane -Geometriae Dedicata 61, No. 3, 315-327 -http://www.springerlink.com/content/u35183155g06115r/ -Shinohara, M (2008). Uniqueness of maximum planar five-distance sets. -Discrete Mathematics, 308(14), 3048-3055. -http://linkinghub.elsevier.com/retrieve/pii/S0012365X07006498 -What about higher dimensions? The cross-polytope in n-dimensions is always a 2-distance set with 2n points. Even better taking the set of mid-points of edges of the n-simplex gives a 2-distance set with n(n+1)/2 points (of course in 3d this gives the vertices of the octahedron). Are their better examples? -My motivation for this is mainly visual, the requirements that a small set of distances places on symmetry mean that these sets should give interesting forms. It should also be noted that (perhaps unsurprisingly given the elementary nature) it was also a problem that attracted Erdös, for example see: -Erdös, P (1970) On Sets of Distances of n Points -The American Mathematical Monthly 77, No. 7, pp. 738-740 -http://www.jstor.org/pss/2316209 -To finish with a precise question: What is known about n-distance sets in 3 and 4 dimensions? - -REPLY [4 votes]: I have some maximal results up to 20 points at Maximal Unit Lengths in 3D. -The maximal log(unit edges)/log(points) so far is 1.43392 with 14 points producing 44 unit edges. The points are {{0,0,0}, {3,0,3}, {3,3,0}, {3,-3,0}, {3,0,-3},{0,3,-3}, {6,0,0}, {2,-1,1}, {-1,2,1}, {-1,-1,-2}, {-1,-1,4}, {-1,-4,1},{-4,-1,1}, {2,-4,4}}/(3 sqrt(2)). The graph looks like the following:<|endoftext|> -TITLE: Complexity of detecting a convex body in $\mathbb{R}^n$? -QUESTION [6 upvotes]: Let $K_0$ be a bounded convex set in $\mathbb{R}^n$ within which lie two sets $K_1$ and $K_2$. $K_0,K_1,K_2$ have nonempty interior. Assume that, - -$K_1\cup K_2=K_0$ and $K_1\cap K_2=\emptyset$. -The boundary between $K_1$ and $K_2$ is unknown. (To avoid the trivial case, I assume that the boundary is not a hyperplane.) -Either $K_1$ or $K_2$ is a convex set, but we don't know which one is. -We have two initial points $\mathbf{x},\mathbf{y}$ on hand, where $\mathbf{x}\in K_1$ and $\mathbf{y}\in K_2$. - -Essentially, $K_0$ can be viewed as a black box. Further assume that one can query any point in $K_0$ with a membership oracle, namely a procedure that given a point $\mathbf{p}\in K_0$, reports the set contains $\mathbf{p}$. -The goal is to determine which set is convex using membership queries. -My questions: - -Can this be done with finite number of queries? -What is the complexity class of this problem? - -Thanks. - -REPLY [4 votes]: You haven't said precisely what you mean by an algorithm here, and this is actually a nontrivial issue since your question is outside the usual finitary context of computability theory, so it isn't clear what you might mean, and there are choices to be made about it. -One way to make the concept precise would be to work with $\mathbb{Q}^n$ rather than $\mathbb{R}^n$ (or some other countable dense set), that is, to finitize the problem by regarding all the relevant reals as rational. In this case, we can use a standard Turing machine concept of computability to make the question precise. For this manner of formalization, we regard the "input" as fixing the oracles for the $K_i$, and the question is whether there is a Turing machine program that can detect the answer correctly regardless of the oracle (subject to them satisfying your hypotheses). Let me assume first that the hypotheses are that exactly one of the $K_i$ are non-convex. -Theorem. With the $\mathbb{Q}^n$ formalization just described, the answer is yes, there is an algorithm. -Proof. Consider the algorithm that systematically produces all triples $(p,q,r)$ of rational points in $\mathbb{Q}^n$, such that $q$ is on the line from $p$ to $r$, and checks membership in each $K_i$. Note that any instance of non-convexity will therefore be revealed. Thus, we will at some finite time in the algorithm, discover which of the $K_i$ is not convex, and thus produce the correct answer in finitely many steps. QED -The previous algorithm relied on the fact that instances of non-convexity can be discovered in finitely many steps. -Theorem. In the $\mathbb{Q}^n$ formalization, in the context where the input sets $K_1$ and $K_2$ might both be convex (and in such a case only, we allow the boundary to be a hyperplane), then there is no computable procedure that will always work. -Proof. Suppose that there were such a procedure in the case $n=1$. Let $K_0=(0,2)$, $K_1=(0,1)$ and $K_2=[1,2)$. The algorithm will halt in finitely many steps, having made queries about reals $q_0$, $q_1$, ... $q_n$, and stating that one of the sets, lets suppose $K_1$, is convex, which is true. Consider now the modification of the sets by letting $r$ be the largest $q_i$ below $2$ (or $\frac32$, whichever is larger), and using $K_1'=(0,1)\cup (r,2)$ and $K_2'=[1,r]$. Since the answers to the queries about the $q_i$ are exactly the same for this configuration as in the original problem, the algorithm will again say that $K_1'$ is convex, but in this case, it would be incorrect. QED -Since the $K_i$ input is infinitary in nature, it isn't immediately clear what you might mean by "complexity class". However, the problem does have an inherent semi-decidable nature, since instances of non-convexity are revealed by a single instance. In the case where one side or the other must be non-convex, this semi-decidable nature becomes decidable, since we can know which is convex by discovering the other set to be non-convex. -Let us now try to consider the general case of $\mathbb{R}^n$. We don't have a standard formal notion of computability here (there are several distinct notions of computability on the reals, such as BSS machines, computable analysis, descriptive set-theoretic notions of "computability", infinite time computability). Note that the input to the algorithm is the sets $K_i$, rather than a real number. -But let's try to be flexible and allow a more generous concept of algorithm, subject to the following properties: an algorithm is a deterministic procedure that (somehow) produces points $p$ in $\mathbb{R}^n$, makes inquiries about whether they are in $K_0$, $K_1$ and $K_2$, and then uses the resulting yes/no answers to those queries to produce additional real numbers about which queries may be made. Eventually, based on the result of the queries, the algorithm is supposed to give an output by specifying whether $K_1$ or $K_2$ is convex. In particular, in this set-up, the same algorithm can be used with many different $K_0$, $K_1$ and $K_2$, and the reals produced depend only on the answers to the queries, rather than on the oracle sets themselves. We give the "input" of $K_i$ to the algorithm by attaching the black boxes, without providing any additional information about the $K_i$ sets, except that they satisfy the assumptions you identified. In this case, it doesn't matter whether one insist that exactly one or at most one $K_i$ is non-convex. -Theorem. With the $\mathbb{R}^n$ manner of formalization just described, there can be no algorithm correctly identifying a convex $K_i$. -Proof. Suppose that there were such an algorithm in the case $n=1$. Call the algorithm $e$. Consider the set $Q$ of all real numbers that will ever be produced by the algorithm $e$ for a membership query for any combination of sets $K_0$, $K_1$ and $K_2$. Our assumptions on the nature of the algorithm ensure that $Q$ is a countable set, since any real produced during the course of any computation is produced as the result of a finite sequence of yes/no answers to the previous queries, and there are only countably many such possible patterns of membership. Let $r\lt s\lt t$ be real numbers not in $Q$, and consider the two possible inquiries: - -$K_0=[r,t]$, $K_1=\{r\}\cup[s,t]$, $K_2=(r,s)$. -$K_0=[r,t]$, $K_1=[s,t)$, $K_2=[r,s)\cup\{t\}$. - -Because these two input configurations agree on $Q$, the algorithm must give the same output for both of them. But in the first case, it $K_2$ that is convex, while in the second, it is $K_1$ that is convex. So there can be no such algorithm. QED - -REPLY [3 votes]: Without more constraints on the problem, there is no deterministic algorithm that can decide which is convex with a finite number of queries. For randomized algorithms, I think that you can never get a finite expected number of queries. -Take $K_0$ to be a sphere. Run your algorithm, and let every query response be $K_1$. Now, whatever the response of the algorithm, we can construct a contradicting example with the same responses. -If the response was $K_1$ is convex, take $K_2$ to be a small sphere in $K_0$ that is disjoint from the query set. Then, $K_2$ is convex and $K_1$ is not. -If the response was $K_2$ is convex, we are going to take a slightly flattened sphere to be $K_1$ and the non-convex spherical cap to be $K_2$. Take the convex hull of the query points and choose a supporting hyperplane of any face of the convex hull. If we remove the spherical cap defined by the intersection of $K_0$ and the hyperplane and bulge the resulting flat face outwards, we get $K_1$ as convex and $K_2$ is not convex.<|endoftext|> -TITLE: What is the state of the art for the Turán number of $K_{4,4}$? -QUESTION [9 upvotes]: In Chung and Graham's "Erdős on Graphs: His legacy of unsolved problems," they discuss several open problems concerning Turán numbers for bipartite graphs. -There is a construction which gives graphs on $n$ vertices with $C n^{5/3}$ edges, with no $K_{3,3}$ subgraphs, and at the time of the writing of this book there was no better lower bounds on the Turán number of $K_{4,4}$. - -Is it known whether $$\lim_{n\to \infty} \frac{ \mbox{ex}\left(n ; K_{4,4}\right) }{n^{5/3}} = \infty ? $$ - -(Here $\mbox{ex}(n ; H)$ is the maximum number of edges for an $n$-vertex graph with no subgraphs isomorphic to $H$.) - -REPLY [3 votes]: Here is a near answer. In Turan Numbers of Bipartite Graphs and Related Ramsey-Type Questions, Alon, Krivelevich, and Sudakov prove that $ex(n; K_{s,t}) \leq O(n^{2-1/s})$. They note that this bound is tight for all values of $s \geq 2$. -That is, by previous results, if $t > (s-1)!$, then the Turan number of $K_{s,t}$ is in fact $\Theta (n^{2-1/s})$. Therefore, if we replace $K_{4,4}$ by $K_{4,7}$, then the answer to your question is yes.<|endoftext|> -TITLE: Doubly stochastic matrices as squares of entires of unitary matrices -QUESTION [8 upvotes]: Given a unitary matrix $A$ with entries $a_{ii}$, it's clear that the matrix $B$ with entries $b_{ii} = |a_{ii}|^2$ is doubly stochastic. Is the inverse of this statement true? Namely, given a doubly stochastic matrix $B'$ with entries $\beta_{ii}$, does there exist a unitary matrix with entries $\alpha_{ii}$ such that $|\alpha|^2_{ii} = \beta_{ii}$? - -REPLY [3 votes]: A brief googling yielded another interesting paper: -http://www.sciencedirect.com/science/article/pii/0024379578900228 -Topological properties of orthostochastic matrices ☆ -Tony F. Heinz - -"In this paper, it is shown that for n⩾3 the orthostochastic matrices are not everywhere dense in the set of doubly stochastic matrices, thus answering a question of L. Mirsky in his survey article on doubly stochastic matrices [2]"<|endoftext|> -TITLE: Line bundles with integrable connection on abelian varieties -QUESTION [11 upvotes]: Let $X$ be an abelian variety over an algebraically closed field $k$. -Let $L$ be a line bundle on it equipped with an integrable connection $\nabla: L \rightarrow L \otimes \Omega^1_{X/k}$. -Does it then automatically folllow that $L$ is a bundle in $Pic^0(X)$? -And how general can one make such a statement? -I mean: Does it also hold for $k$ arbitrary or at least of characteristic zero? And does it hold for abelian schemes over a base $S$, say of characteristic zero? - -REPLY [10 votes]: Yes, it is true, though an algebraic proof seems (there may be a simpler proof however) somewhat tricky. - - Such a line bundle lies in $\mathrm{Pic}^\tau(X)$. This is a general fact as a line bundle lies in $\mathrm{Pic}^\tau(X)$ if its rational Chern classes are trivial (this follows from Riemann-Roch) and the Chern classes of a line bundle with integrable connection are torsion (an algebraic proof is given in Dix exposés sur la cohomologie des schémas). This works without the assumption of $X$ being abelian. - -It is then a further fact that for abelian varieties $X$ we have that $\mathrm{Pic}^\tau(X)= -\mathrm{Pic}^0(X)$ (this I think is an Mumford's abelian varieties somewhere). - -This extends to immediately to families by checking fibre by fibre. -Addendum: Sorry forgot to say that this all requires characteristic zero. In characteristic $p$ every $p$'th power line bundle has an integrable connection (even of $p$-curvature $0$) but will in general not lie in $\mathrm{Pic}^0(X)$. -Addendum 1: Lars (in a comment) makes an interesting point about the positive characteristic situation. A module structure over the ring of differential operators (aka a stratification) implies in particular that the line bundle is a $p^n$'th power for each $n$ and as $\mathrm{Pic}(X)/\mathrm{Pic}^\tau(X)$ is a finitely generated group this implies that the line bundle lies in $\mathrm{Pic}^\tau(X)$. The same idea could be applied to the characteristic zero situation if the $p$-curvature of the reduction modulo $p$ for an infinite number of primes $p$ were zero. However that should be true only if the line bundle has finite order so it doesn't give very much. -Addendum 2: Veen is asking about the equality $\mathrm{Pic}^0(A)=\mathrm{Pic}^\tau(A)$ particularly in a family (when $A$ is abelian). The easiest way to answer all of these questions simultaneously is to assume that there is an ample line bundle $\mathcal L$ on $A$ (which is true locally on the base) and then consider the map $A\to \mathrm{Pic}(A)$ given by $a\mapsto \mathcal L_a\bigotimes\mathcal L^{-1}$. To get all equalities needed it is enough to show that the image is $\mathrm{Pic}^\tau(A)$. This is something that can be checked fibrewise and then it can be extracted from Mumford.<|endoftext|> -TITLE: What are "good" examples of string manifolds? -QUESTION [18 upvotes]: Based on this mathoverlow question, I would like to have a similar list for the case of string manifolds. An $n$-dim. Riemannian manifold $M$ is said to be string, if the classifying map of its bundle of orthonormal frames $M \to BO(n)$ lifts to a map $M \to BO(n)<8> = BString(n)$, which is the case if and only if the class $\frac{p_1}{2} \in H^4(M, \mathbb{Z})$ vanishes. There are a lot of models that yield geometric realizations of $String(n)$ either as a topological group (see Stolz-Teichner), infinite-dimensional Lie group (see Nikolaus-Sachse-Wockel) or a 2-group (see Schommer-Pries). - -What are enlightening examples of - string manifolds? What are - non-examples? When do you have a - geometric interpretation of the - obstruction class? - -So far, I am aware of the list given at the end of Douglas-Henriques-Hill. What else is out there? - -REPLY [11 votes]: Qingtao Chen and Fei Han have constructed [arxiv:0612055] tons of examples of string manifolds and non-string manifolds as complete intersections in products of complex projective spaces. -For integers $s$ and $n_q$ with $1\leq q \leq s$ consider the product -$$ -Z:=\mathbb{C}P^{n_1} \times ... \times \mathbb{C}P^{n_s} -$$ -and the projection $pr_q:Z \to \mathbb{C}P^{n_q}$ to the $q$th factor. -For further integers $t$ and $d_{pq}$ with $1 \leq p \leq t$ and $1 \leq q \leq s$, consider for $1 \leq p \leq t$ the line bundle -$$ -E_p := \bigotimes_{q=1}^s \;\; pr_q^{*}\mathcal{O}^{d_{pq}}_{q} -$$ -over $Z$, where $\mathcal{O}_q$ is the canonical line bundle over $\mathbb{C}P^{n_q}$. -Now let $V_{d_{pq}}$ be the intersection of the zero loci of smooth global sections of $E_1,...,E_t$. -$V_{d_{pq}}$ is always a smooth manifold, and the statement of Proposition 3.1 of the above paper is: - -Theorem. Let $m_q$ be the number of non-zero elements in the $q$th column of the matrix $D := (d_{pq})$. Suppose that $m_q +2 \leq n_q$. Then, $V_{d_{pq}}$ is string if and only if - $$ -D^t D = diag(n_1 + 1,...,n_s+1). -$$<|endoftext|> -TITLE: Groups surjecting onto a free group -QUESTION [14 upvotes]: Is there something resembling a characterization of which groups can map onto a non-abelian free group? Obviously they cannot have property T, and should have nontrivial abelianization, but are there some positive results? - -REPLY [13 votes]: Such groups are often called 'very large'. A group with a very large subgroup of finite index is called 'large'. Here are some miscellaneous facts: - -Baumslag and Pride showed that every group of deficiency two (ie with a presentation with two more generators than relators) is large. -One can deduce from Wise's residually finite version of the Rips construction that there is a 'large' version of the Rips construction; that is, for every fp group $Q$ there is a short exact sequence - -$1\to K\to\Gamma\to Q\to 1$ -where $K$ is 3-generated and $\Gamma$ is large. -So large (and hence very large) groups are quite common. I doubt there is any kind of characterisation, but there are some open questions that are relevant. For instance: -Question: Is there a finitely presentable group $\Gamma$ with $vb_1(\Gamma)=\infty$ which is not large? -Note that $\mathbb{Z}\wr\mathbb{Z}$ gives a non-finitely presentable counterexample. (Here $vb_1(\Gamma)$ is of course the maximum of the first Betti number over all subgroups of finite index. It is obvious that $vb_1$ of a large group is infinite.) - -And another thing... -I just remembered that there is a (not implementable) algorithm to determine whether a finitely presented group is very large. The point is that the group $G=\langle x_1,\ldots,x_m\mid r_1,\ldots,r_n\rangle$ maps onto a non-abelian free group if and only if some system of equations and inequations -$[x_p,x_q] \neq 1\wedge\bigwedge_j r_j(x_1,\ldots,x_m)=1$ -has a solution in $F_2$, for some $p\neq q$. Now, such systems of equations and inequations over a free group $F_n$ can be solved by Makanin's algorithm.<|endoftext|> -TITLE: Pullback of the canonical divisor between smooth varieties -QUESTION [16 upvotes]: I have a surjective morphism $\pi: Y \to X$ between smooth projective varieties of the same dimension over some algebraically closed field $k$. Debarre claims in his book "Higher-Dimensional Algebraic Geometry" (1.41) that there is an effective divisor $R$ such that -$$ K_Y \equiv \pi^* K_X + R$$ -if $K(X) \subset K(X)$ is a separable field extension. In particular, this is always true if $\pi$ is birational. Debarre also claims that in this case the support of $R$ is the exceptional locus. -I search for a reference for this fact because Debarre does not provide one. I'm in particular interested in the case of positive characteristic. In characteristic $0$ this is a well known fact and I would like to confirm that it is still fine in positive characteristic. - -REPLY [15 votes]: EDIT Originally I claimed a more general statement and along the incremental generalizations I reached a statement that was not true. Thanks to Carlos for pointing out this error! So, I thought it would be fair to point out where the error lied. -The main issue is that the original proof works for a finite morphism, but not if there are exceptional divisors, because then on the target the localization would not happen at a height $1$ prime. In order to preserve the original proof, here is a fix that divides the statement into two parts. -0 - -By Stein factorization it is enough to prove the statement for finite or birational morphisms. - -1 - -Thm Let $\pi:Y\to X$ be a separable projective finite morphism between normal varieties of the same dimension. Assume that $K_X$ is a $\mathbb Q$-Cartier divisor (i.e., there exists an $m\in \mathbb N_+$ such that $mK_X$ is a Cartier divisor). Then there exists an effective divisor $R\subset Y$ whose support is contained in the ramification locus of $\pi$ (that is, the complement of the largest open subset of $Y$ on which $\pi$ is smooth) such that - $$K_Y\sim \pi^*K_X + R.$$ - -Proof: -We need to prove that the divisor $\pi^*K_X-K_Y$ is linearly equivalent to an effective divisor supported on the exceptional locus . This can be done by localizing at height $1$ primes, so the question reduces to a question about regular schemes of dimension $1$. One may apply the usual proof of the Hurwitz formula: -Consider the short exact sequence $$0\to \pi^*\Omega_X\to \Omega_Y\to \Omega_{Y/X}\to 0$$ -and observe that $\Omega_{Y/X}$ is a torsion sheaf whose support is the ramification locus (this is where you need that the map is separable), which in this case is the same as the exceptional divisor (the smaller parts of the exceptional locus disappear at the localization). This is a finite set, so $\Omega_{X/Y}$ maybe considered as the structure sheaf of a finite subscheme of $Y$. Let $R\subseteq Y$ be this subscheme. Tensoring the above short exact sequence by $\Omega_Y^{-1}$ gives: -$$ 0\to \pi^*\Omega_X\otimes \Omega_Y^{-1}\to \mathscr O_Y \to \mathscr O_R \to 0,$$ -which shows that $\pi^*\Omega_X\otimes \Omega_Y^{-1}\simeq \mathscr O_Y(-R)$. This implies the needed linear equivalence. -To find the original $R$ all you need to do is to take the divisor that localizes to the $R$ we found in the $1$-dimensional case. Since we're talking about divisors the codimension $2$ ambiguity makes no difference. -2 - -Thm Let $\pi:Y\to X$ be a projective birational morphism between smooth varieties. Then there exists an effective divisor $R\subset Y$ whose support is the exceptional locus $E\subseteq Y$ of $\pi$ such that - $$K_Y\sim \pi^*K_X + R.$$ - -Proof: -Consider (again) the short exact sequence $$0\to \pi^*\Omega_X\to \Omega_Y\to \Omega_{Y/X}\to 0$$ -and notice that $\pi$ is an isomorphism on $Y\setminus E$, so similarly $\pi^*\Omega_X\to \Omega_Y$ is an isomorphism there. Now take the determinant of these locally free sheaves and conclude that $\pi^*\omega_X\subseteq\omega_Y$ and $\omega_Y/f^*\omega_X$ is supported on $E$. This implies that the divisor $K_Y-f^*K_X$ is an effective divisor supported on $E$. -We need to prove that the support of $R$ is the entire $E$. For this, first notice that in order to prove the desired statement, using what we have proved already, we may pass to another birational model that dominates $Y$. (The point is that a $\pi$-exceptional divisor will be exceptional for the combined map to $X$ but not for the map to $Y$.) Second, a theorem of Zariski says that every exceptional divisor can be reached by a sequence of blow-ups (see Kollár-Mori98, 2.45). We know how to compute the canonical divisor of a blow-up and we know that the entire exceptional locus is contained in the discrepancy divisor, so the desired statement follows. -3 -Comment The reason the second part requires smoothness is that one needs $\Omega_X$ to be locally free so when pulled back it would give the right thing. Otherwise it might pick up torsion or co-torsion. The statement is true in a little bit more general situation, if $X$ has at worst canonical singularities, but that is essentially the definition of those singularities and this statement says that smooth points are canonical, so it is a reasonable condition to use to define singularities.<|endoftext|> -TITLE: Intrinsic description of the image of $V \to V^{**}$ -QUESTION [10 upvotes]: Let $V$ be a vector space over a field $K$. Call a linear map $F : V^* \to K$ representable if there is some $v \in V$ such that $F(w) = \langle w,v \rangle$ for all $w \in V^*$. Here, $\langle w,v \rangle := w(v)$. Remark that this representing vector $v$ is then uniquely determined. -Remark the similarity to category theory: A functor $F : C^{\mathrm{op}} \to \mathrm{Set}$ is representable if there is an object $v \in C$ such that $F \cong C(-,v)$. By the Yoneda Lemma, this object is uniquely determined up to canonical isomorphism. Now there are various theorems which say when a certain functor is representable. Often we require that $F$ preserves limits and satisfies some finiteness condition. -Now back to linear algebra, is there a criterion when a linear map $V^* \to K$ is representable? In other words, is there an intrinsic description of the image of $V \to V^{**}$? If $V$ is finite-dimensional, then we get representability for free. I'm interested in the general case. I would like to mimic somehow the category theoretic conditions. -// Moosbrugger has given below the following nice characterization: Give $K$ the discrete topology, $K^V$ the product topology, and $V^* \subseteq K^V$ the subspace topology. Then a linear map $V^* \to K$ is representable iff it is continuous! However, the proof is rather trivial, except that it uses the result for finite-dimensional vector spaces. So in practise, I doubt that this will be a good criterion. Therefore I would add the following requirement to the criterion: it should be useful in practise (whatever this means) or linear algebraic / geometric: When is a hyperplane in $V^*$ cut out by a single vector $v \in V$? - -REPLY [9 votes]: Moosbrugger has given an excellent answer to the question on which I'll elaborate a bit. -A similar result holds in functional analysis: -Let $V$ be a Banach space and $V'=hom(V,\mathbb C)$its dual (i.e., the set of continuous linear functionals from $V$ to $\mathbb C$). If you equip $V'$ with the norm topology, then -its dual is typically bigger than $V$. However, if you equip it with the weak-star topology, then its dual is exactly $V$. In other words, $V\subset V''$ is the subset of weak-star continuous functionals (and the norm on $V$ is induced by that on $V''$). In that way, you can recover $V$ from $V'$ if you known its weak-star topology. -Now, for the sake of completess, let me also say what the weak-star topology on $V'$ is: it's the topology under which which a net $\{\varphi_i\}_{i\in I}$ converges to some element $\varphi$ if and only if $\{\varphi_i(v)\}_{i\in I}$ converges to $\varphi(v)$ for every $v\in V$.<|endoftext|> -TITLE: Is the stalk of the (co)limit of sheaves equal to the (co)limit of the stalks? -QUESTION [6 upvotes]: More precisely, if $\mathcal F_i$ is a system of sheaves, is it the case that -$$ -(\lim \mathcal F_i)_p = \lim ((\mathcal F_i)_p) -$$ -and similarly for colimits? I can see how to get a map -$$ -(\lim \mathcal F_i)_p \rightarrow \lim ((\mathcal F_i)_p) -$$ -by taking the stalks in the diagram for $\lim \mathcal F_i$ and then using the universal property of $(\lim \mathcal F_i)_p$, but I can't see how I should use the sheaf condition to go the other way. -I learning algebraic geometry and taking a class with Hartshorne's book, and we showed that a map of sheaves is injective/surjective iff it is injective/surjective on stalks. Since one can check injectivity/surjectivity by computing the kernal/cokernal, I am thinking that what I wrote above would be a generalization of this fact. -Looking at sheafication as a adjoint and how it interacts with limits and colimits really made me feel more comfortable with sheafication, so I'm hoping that looking the interaction between sheaves and their stalks in terms of limits and colimits will give me some intuition. - -REPLY [9 votes]: Let $F$ be a sheaf on $X$ and $p \in X$. Then $F_p$ is just the pullback $i^{-1} F$, where $i : \{p\} \to X$ is the inclusion of a point. Now $i^{-1}$ is left adjoint to $i_*$, thus cocontinuous, i.e. preserves all colimits. This shows that the canonical morphism $\mathrm{colim}_i(F_p) \to \mathrm{colim}_i(F)_p$ is an isomorphism. -Now for limits, we get a canonical morphism $(\mathrm{lim}_i F)_p \to \mathrm{lim}_i(F_p)$. This is almost never an isomorphism (neither injective nor surjective). Consider infinite products and see what happens. However, the (left) exactness of $F \mapsto F_p$ means that this functor preserves finite limits.<|endoftext|> -TITLE: Multiplicativity of Euler characteristic for non-orientable fibrations -QUESTION [14 upvotes]: Let $E\to B$ be a fibration with fiber F, and assume for simplicity that B is connected. Suppose moreover that B and F have Euler characteristics (perhaps they are manifolds). Then often, one can conclude that E has an Euler characteristic as well, and that -$$ \chi(E) = \chi(B)\cdot \chi(F). $$ -The only proof of this that I have been able to find uses a spectral sequence argument, and requires that $\pi_1(B)$ act trivially on the homology of F, so that the homology in the spectral sequence can be taken with constant coefficients. This condition is sometimes referred to as orientability of the fibration (with respect to the homology theory, normally rational homology). -Is the result known to be true any more generally than this? Is there any other known proof? Are there any examples where it is known to be false? - -REPLY [7 votes]: Since this question seems to be attracting some renewed interest, I may as well point out that a few years after I asked it, Kate Ponto and I proved a generalization of this formula by purely homotopical/categorical methods, which applies to any fibration and yields a result about the Lefschetz number and even the Reidemeister trace. The paper is here.<|endoftext|> -TITLE: The word problem in the ring of polynomials -QUESTION [11 upvotes]: This question must be well known but I cannot find it in the literature. -Question: What is the computational complexity of the word problem in a subring of the ring of polynomials in $n\ge 1$ variables ring over rationals? Is it in $P$? -Thus we fix $m$ polynomials $f_1,...,f_m$ from $\mathbb{Q}[x_1,...,x_n]$ Then for every polynomial $g(y_1,...,y_m)$ over $\mathbb{Q}$, we need to check if the polynomial $g(f_1,...,f_m)$ is 0. The size of the input is the amount of space needed to write down $g$. - Update I was thinking about the following algorithm: we need to check that $g(f_1,...,f_m)=0$. First plug in $x_i=0$ and check that the result is 0. It is easy. Then take the derivative $\partial g(f_1,...,f_m)/\partial x_i$ and plug in 0's there, etc. The number of possible derivatives is polynomial in the size of $g$, and the length of the computation seems to be polynomial in the size of $g$ also. We need to pre-compute enough products of values of $f_i$ and their derivatives at 0. Does this algorithm ran in polynomial time? Is it well-known, and if so what is a reference? - -REPLY [5 votes]: I'm going to summarize what is said in the comments and my own remarks about the update, to make a complete answer: -The algorithm suggested in the update is basically checking lower-order coefficients of the composition. Certainly if the exponents are in binary, then the algorithm doesn't work: It takes exponential time to distinguish $x^n$ from $0$. Even if the exponents are in unary, it takes exponential time to distinguish $x_1x_2\ldots x_n$ from $0$. -The paper of Ibarra and Moran places a more general problem, the word problem in the sense of circuits or straight-line programs, in co-RP. -A straight-line program is a sequence of formulas arranged like a tax form. It's a sequence of intermediate variables defined by formulas. It's equivalent to an arithmetic circuit. It's an interesting definition for groups as well as for polynomial arithmetic. It is more general than the stated question because it's easy to decompose a polynomial with listed terms into a circuit --- the only hard part is exponents in binary and you can do that by squaring up --- and composition of polynomials can be implemented in a straight-line program by the definition of a straight-line program. -The complexity class BPP means yes-no questions answered in polynomial time by a randomized algorithm which gives a probably-correct answer, with probability $p > c > 1/2$. The complexity class RP is the same thing, except that if the program answer yes, it is certainly correct; if it answers no, it is only probably correct. The probability of correctness can be amplified to very close to 1. It is a well-known conjecture that BPP = P, and obviously BPP contains RP and RP contains P. The basis of the conjecture is that good pseudo-random number generators seem to exist. However, there are strong theoretical reasons that even RP = P is a very hard conjecture. -Ibarra and Moran give a polynomial-time algorithm in RP to show that an arithmetic circuit over $\mathbb{Q}$ (say) is non-zero. (So there is an algorithm in co-RP that the circuit is zero -- this is just switching "yes" and "no".) The algorithm consists of plugging in large values at random. It is a bit easier to see what they do over $\mathbb{Z}$, and not really different because you can reduce $\mathbb{Q}$ to $\mathbb{Z}$ by decomposing fractions into numerators and denominators. They make an exponentially large box in $\mathbb{Z}^n$, and they show that if the circuit is non-zero, then it is usually non-zero in the box, using simple (but skillful) counting estimates for zero values. -Kabanetz and Impagliazzo show that some of the same strong theoretical difficulties that obstruct the conjecture that BPP = P or RP = P, also make it hard to test circuit equality in P. So you should recognize Ibarra-Moran as efficient in practice because it is in co-RP, believe that Ibarra-Moran can be converted to P using pseudo-random numbers, and be satisfied that no one can provably find any algorithm for the problem in P.<|endoftext|> -TITLE: Examples of exotic modules for the additive group -QUESTION [8 upvotes]: Let $k$ be an algebraically closed field of positive characteristic $p > 0$, and let $X$ be an intedeterminate over $k$. I am interested in the additive group scheme $\mathbb{G}_a$, that is, the affine algebraic group scheme having coordinate ring $k[X]$. We can identify $\mathbb{G}_a$ with the unipotent radical of a Borel subgroup $B$ for the group scheme $SL_2$, e.g., if we identify $B$ with the subgroup of upper triangular matrices in $SL_2$, then we can identify $\mathbb{G}_a$ with the subgroup $U$ of upper triangular unipotent matrices in $SL_2$. -I am specifically interested in rational modules for $\mathbb{G}_a$, or equivalently, $k[X]$-comodules. Given a rational $B$-module $M$, I can of course restrict $M$ to $U = \mathbb{G}_a$ and obtain a rational $\mathbb{G}_a$-module. But what I'd really like are some explicit examples of rational $\mathbb{G}_a$-modules that do not arise via restriction from a rational $B$-module, and I haven't had any luck so far finding explicit examples in the literature. - -Can anyone point me to explicit examples in the literature of (finite-dimensional) rational $\mathbb{G}_a$-modules that do not arise as the restriction of rational $B$-modules? - -Here's the (single) example I've been able to cook up from scratch for the case $p=2$. Let $V$ be the $k$-vector space with basis vectors $v_0$ and $v_1$. I define a linear map $\Delta: V \rightarrow V \otimes k[X]$ by -$\Delta(v_0) = v_0 \otimes 1 + v_1 \otimes (X+X^2)$, -and $\Delta(v_1) = v_1 \otimes 1$. Then it's straightforward to check that this defines a $k[X]$-comodule structure on $V$ (equivalently, a rational $\mathbb{G}_a$-module structure on $V$), which cannot come from the restriction of a rational $B$-module structure on $V$ because of the non-homogeneous term $X+X^2$. - -REPLY [3 votes]: At first glance it seems that your $p=2$ example can be generalized to arbitrary primes by using $p$-polynomials such as $X + X^p$. In prime characteristic there are lots of exotic embeddings of the additive group involving such polynomials, which in turn translate into morphisms of (affine) algebraic groups. See for instance the discussion in Chapter 20 of my 1975 Springer GTM book, with references to older literature including Demazure-Gabriel. On the other hand, representations of the multiplicative group are pretty much the same in all characteristics, so you would typically see the same incompatibility you note for $p=2$ when the additive group is represented using as coordinate functions some nontrivial $p$-polynomials. -I'm not sure whether concrete examples of the type you want are written down explicitly, though people studying group actions and invariant theory in prime characteristic may have dealt with the situation.<|endoftext|> -TITLE: Which small finite simple groups are not yet known to be Galois groups over Q? -QUESTION [46 upvotes]: The subject line pretty much says it all. To expand just a little bit: -1) What is the smallest simple group that is not yet known to occur as a Galois group over $\mathbb{Q}$? (Variants: not known to occur regularly over $\mathbb{Q}$; not known to occur regularly over every Hilbertian field.) -2) Is there a convenient list of the simple groups of order, say, at most $10,000$ together with information about which of them are known to be Galois groups over $\mathbb{Q}$? -3) Is there perhaps some nice website keeping track of information like this? (There should be!) -Added: The table in $\S 8.3.4$ of Serre's Topics in Galois Theory (the 1992 edition; I don't know if this matters) lists $\operatorname{SL}_2(\mathbb{F}_{16})$ -- which has order $4080$ -- as the smallest simple group which is not known to occur regularly over $\mathbb{Q}$. On the other hand this 2007 paper of Johan Bosman shows that $\operatorname{SL}_2(\mathbb{F}_{16})$ occurs as a Galois group over $\mathbb{Q}$, but mentions that the problem of realizing it regularly is still open. Thus the answer to the "regular" variant of the question seems to be $\operatorname{SL}_2(\mathbb{F}_{16})$. But to be clear, I am really looking for as much data as possible, not just "records". - -REPLY [24 votes]: I decompose the question into four parts, all of which Pete already knows a lot about, and/or are mentioned in the comments: -(1) The books by Malle-Matzat and Völklein thoroughly explain the workhorses of the field such as the rigidity method of John Thompson. -(2) Later results: Pete himself is responsible for a significant further result in the $A_1(2,p)$ case. It seems well-known that the case $A_1(2,q)$ (and $G(q)$ for other many other algebraic groups $G$) is difficult when $q$ is a prime power rather than a prime. (But that was the wisdom 10 years ago --- see the update below.) Pete already gave an interesting recent reference for the case $A_1(2,16)$. -(3) People tabulating results. It seems that the main activity is by Klüners with help from Malle. The on-line database of Klüners is organized by the permutation degree of the group, which is of course not the same as the cardinality. Also Klüners looks at all finite groups, not just finite simple ones. -(4) Sizes of finite simple groups. The Wikipedia page, list of finite simple groups, is excellent. It lists all 16 non-abelian finite simple groups of order less than 10,000. -So it could be best to correspond directly with Klüners concerning specific inverse Galois results. -If you're interested in organizing finite simple groups by cardinality --- this is a question that comes up from time to time --- it could be very useful to convert the Wikipedia page to a Python/SAGE program. -Update 2: David Madore had the same idea 8 years ago and wrote a code not in Python but in Scheme. So if you're interested in the smallest simple group not proven to be a Galois group, you only need to go down Madore's list. Note the Shih-Malle theorem that Pete summarizes in his paper: $A_1(2,p)$ is $\mathbb{Q}$-Galois when $p$ is prime and not a square mod 210. The first prime $p$ that is a square mod 210 is 311. (Pete's paper settles some other such $p$, but not that one.) Madore's table shows that the open case $A_1(2,311)$ shows up just a few entries after the only open sporadic case, $M_{23}$. So what does that leave that's smaller than $M_{23}$? - -Update 1: Some Googling found an interesting paper by Dieulefait and Wiese and a more recent paper by Bosman. It seems that a lot more is known about the $A_1(2,q)$ case in recent years using the number of theory of curves Galois representations attached to modular forms, rather than the rigidity method of Thompson. In particular, no open cases remain in the Wikipedia list of finite simple groups of order less than 10,000, other than possibly $A_1(27)$ and ${}^2A_2(9)$, which is the unitary group $\text{PSU}(3,\mathbb{F}_9)$. Actually it looks like ${}^2A_2(9)$ doesn't survive either because (like $M_{11}$) it is handled by older methods, according to Malle and Matzat. The smallest finite simple group which is not known to be a Galois group over $\mathbb{Q}$ could be an interesting trivia question for the moment.<|endoftext|> -TITLE: In what ways did Leibniz's philosophy foresee modern mathematics? -QUESTION [22 upvotes]: Leibniz was a noted polymath who was deeply interested in philosophy as well as mathematics, among other things. From my mathematical readings I have the impression that Leibniz's stature as a mathematician has grown in the last fifty years as some of his philosophically oriented mathematical ideas have connected with modern mathematicians and mathematics. That because of Leibniz's philosophical reflections, he foresaw aspects or parts of modern mathematics. Can anyone elaborate on these connections and recommend any references? -EDIT, Will Jagy. Editing mostly to bump this to the front of active. It is evident that Jacques and Sergey have good, substantial answers in mind. Please do not answer unless you have read Leibniz at length. I kind of liked philosophy in high school and college, or thought I did. Recently, I read one page of Spinoza and gave up. - -REPLY [6 votes]: To answer the title question, "In what ways did Leibniz's philosophy foresee modern mathematics?" one could mention the distinction between assignable and inassignable number that closely parallels the distinction between standard and nonstandard number in Abraham Robinson's (or Edward Nelson's) framework. Furthermore, Leibniz's notion of a generalized relation of equality closely parallels the modern notion of shadow (or standard part). Leibniz's law of continuity finds a close procedural proxy in the transfer principle of nonstandard analysis. The remainder of this answer will explain how one can make such claims without falling into the trap of presentism. -Daniel Geisler speculates that "because of Leibniz's philosophical reflections, he foresaw aspects or parts of modern mathematics" and asks: "Can anyone elaborate on these connections and recommend any references?" -Several responders mentioned the connection to Robinson's theory. On the other hand, François Brunault rightly cautioned: "The statement that someone (even Leibniz) foresaw parts of modern mathematics is potentially controversial because of its subjectivity. I think most historians of mathematics now insist on the fact that the works by earlier mathematicians should also be studied from the point of view of that time, before extrapolating possible connections." -François Brunault is correct in suggesting that there is resistance among historians of mathematics to the idea of seeing continuity between Leibniz and Robinson. Indeed, the prevalent interpretation of Leibnizian infinitesimals is a so-called syncategorematic interpretation, pursued notably by R. Arthur and many other Leibniz scholars. On this view, Leibnizian infinitesimals are merely shorthand for ordinary ("real") values, assorted with a (hidden) quantifier, viewed as a kind of a pre-Weierstrassian anticipation. These scholars rely on evidence drawn from various quotes from Leibniz where he refers to infinitesimals as "useful fictions", and explains that arguments involving infinitesimals can be paraphrased a l'ancienne using exhaustion. In this spirit, they interpret the Leibnizian "useful fictions" as LOGICAL fictions, denoting what would be described in modern terminology is a quantified formula in first-order logic. -For example, Levey writes: -"The syncategorematic analysis of the infinitely small is ... fashioned around the order of quantifiers so that only finite quantities figure as values for the variables. Thus, -(3) the difference $|a-b|$ is infinitesimal -does not assert that there is an infinitely small positive value which measures the difference between~$a$ and~$b$. Instead it reports, -($3^*$) For every finite positive value $\varepsilon$, the difference $|a-b|$ is less than $\varepsilon$. -Elaborating this sort of analysis carefully allows one to express the now-usual epsilon-delta style definitions, etc." -This comment appears in the article -Levey, S. (2008): Archimedes, Infinitesimals and the Law of Continuity: On Leibniz's Fictionalism. In Goldenbaum et al., pp.~107--134. The book is -Goldenbaum U.; Jesseph D. (Eds.): Infinitesimal Differences: Controversies between Leibniz and his Contemporaries. Berlin-New York: Walter de Gruyter, 2008, see http://www.google.co.il/books?id=tWWuQ9PHCusC&q= -I personally find it hard to believe Levey is talking about Leibniz, but there you have it. Whether or not Levey's analysis stems from a "Desire To Preserve The Orthodoxy of Epsilontics Against The Heresy of Infinitesimals", as Yemon likes to put it, is anybody's guess. -What the "syncategorematic" view tends to overlook is the presence of DUAL methodologies in Leibniz: both an Archimedean one, and one involving genuine "fictional" infinitesimals. On this view, Leibnizian infinitesimals are PURE fictions (rather than logical ones). Such a reading is akin to Robinson's formalist view, and sees continuity not merely between Leibniz's and Robinson's mathematics, but also their philosophy. This view is elaborated in a text entitled "Infinitesimals, imaginaries, ideals, and fictions" by David Sherry and myself, to appear in Studia Leibnitiana, and accessible at http://arxiv.org/abs/1304.2137 -HOPOS (Journal of the International Society for the History of Philosophy of Science) just published our rebuttal of syncategorematist theories that seek to sweep Leibnizian infinitesimals under a Weierstrassian rug.<|endoftext|> -TITLE: Algebraic curve cannot suddenly end -QUESTION [13 upvotes]: This is a literature request for (hopefully) an English version to a rigorous proof that a complex algebraic curve cannot abruptly end. -That is, if the algebraic curve enters a closed region it must also leave it. -This has a historic significance because Gauss's proof in his Phd thesis assumed this property holds. From looking around it seems that A Ostrowski rigorously proved the result around the 1920's. Is this correct? I am unable to find the title of the paper. -Is there also a proof that a real algebraic curve does not end abruptly? -I don't regard this property as obvious, but it doesn't seem to be well commented in the literature. Maybe, I'm wrong. -Thanks in advance. -Abruptly end: -Given an irreducible polynomial $p$, we define $V(p)$ to be the complex algebraic curve associated to $p$. I say that $V(p)$ does not abruptly end at $(x,y)\in\mathbb{C}^2$ with $p(x,y)=0$ if there is a disc small enough so that the boundary contains exactly two points in $V(p)$. -(This is a first attempt and maybe needs some corrections.) - -REPLY [7 votes]: Following the idea of Felipe Voloch, I try to give a simple proof based on Puiseux series expansion. Let $C$ be a real algebraic curve at the origin. Look at the Puiseux series expansion (say in terms of $x$) of $C$ near $O$. By assumption one of the branches (over $\mathbb{C}$), call it $C_1$, has the form -$$y = a_1x^{r_1} + a_2x^{r_2} + \cdots \quad\quad\quad (1)$$ -for $a_i \in \mathbb{R}$. Let $q$ be the least common multiple of the denominators of $r_i$'s. If $q$ is odd, then the branch expands to both sides of the origin and therefore $C_1$ does not end abruptly. So assume $q$ is even. Let $\zeta := e^{2\pi i/q}$. For each $j$, $1 \leq j \leq q$, the complex curve corresponding to $C_1$ has a Puiseux expansion of the form $y = \sum_i a_i \zeta^{jp_i}x^{r_i}$, where $p_i = qr_i$. In particular, taking $j =q/2$ (so that $\zeta^j = -1$), we see that the complex curve corresponding to $C_1$ has an expansion of the form -$$y = \sum_i a_i (-1)^{p_i}x^{r_i}. \quad\quad\quad (2)$$ -It follows by the minimality of assumption on $q$ that there is $i$ such that $a_i\neq 0$ and $p_i$ is odd , and consequently, $(1)$ and $(2)$ give different real curves, and it follows that $C_1$ does not end abruptly. -PS: The above arguments only show that $C_1$ has at least two end points on the boundary of a small enough disk centered at $O$. But it can not have more than two, because for all $j \not\in \lbrace q/2, q\rbrace$, $\zeta^j$ is non-real, so the corresponding parametrization does not give any real points.<|endoftext|> -TITLE: Nielsen equivalence in one-relator groups -QUESTION [9 upvotes]: Let $X=(x_1, \ldots, x_n)$ be an $n$-tuple of elements of a given group $G$. Then two $n$-tuples $X$ and $Y$ are Nielsen equivalent if there exists an automorphism of the free group on $n$-generators, $\phi\in \operatorname{Aut}(F_n)$, such that $X\phi=(x_1\phi, \ldots, x_n\phi)=Y$. Also, $X$ and $Y$ are said to lie in the same $T$-system if there exists an automorphism of $G$, $\psi\in \operatorname{Aut}(G)$, such that $X\psi$ is Nielsen equivalent to $Y$. -It is (relatively) well-known that both "Nielsen equivalence" and "lying in the same $T$-system" are equivalence relations. -I am interested in what is known about generating $n$-tuples of one-relator groups, so groups of the form $\langle x_1, \ldots, x_n; R^m \rangle$ where $m \geq 1$ and $R \neq S^i$ for any words $S\in F(X)$ and $i > 1$ (so $R$ is not a proper power of any element of the free group). -Now, I know that if $n = 2$ and $m > 1$ then there is only one Nielsen equivalence class. This is in a paper published in 1977 by S. J. Pride (The isomorphism problem for two-generator one-relator groups is solvable, Trans. Amer. Math. Soc. 227 (1977) 109-139) which uses this fact to solve the isomorphism problem for such groups. There is a recent paper published in 2005 by I. Kapovich and P. Schupp (Genericity, the Arzhantseva-Ol'shanskii method and the isomorphism problem for one-relator groups, Math. Ann. 331 (2005) 1-19) about what happens for almost all finite $n$ and $m > 1$. -However, I was wondering what happens if $m = 1$. My question is the following, - -Let $G=\langle X; R^m\rangle$, $|X|=2$, $m=1$. What can be said about the number of Nielsen equivalence classes in the $T$-system of $X$? Are there finitely many, or infinitely many? What if $|X|>2$? - -Note that it is well-known that if $m > 1$ then these groups are residually finite (whether or not the 185 page proof of this result is true or not is beside the point!) and so generating them should be easy. However, if $m = 1$ then you might end up with non-Hopfian groups, never mind non-residually finite! So, the epimorphism which is surjective but not injective in your group will (if I remember correctly) necessarily move your generating tuple out of its $T$-system. By a result of A. M. Brunner (Transitivity-systems of certain one-relator groups, Springer Lecture Notes in Math. 372 (1974) 131-140), this is certainly what happens with the Baumslag-Solitar group $BS(2, 3) = \langle a,b \, \vert \, ab^2a^{-1} = b^2 \rangle$ and the map $a\mapsto a$, $b\mapsto b^2$. Indeed, this map takes your generating pair to a new $T$-system every time! That is, the generating pairs $(a, b^{2^n})$ lie in different $T$-systems for all $n$. Clearly this means there are infinitely many Nielsen equivalence classes, and so infinitely many $T$-systems. This is why I want to know about what happens in your given $T$-system! - -REPLY [2 votes]: The solvable Baumslag-Solitar group $BS(1, n) = \langle a, b \, \vert \, aba^{-1} = b^n \rangle$ with $n \in \mathbb{Z} \setminus \{0\}$, has only one $T$-system of generating pairs and any number of Nielsen classes, including infinity, can be achieved for a suitable choice of $n$. This preprint presents of proof of my claim, see Corollary E. -The proof is elementary and can be summarized as follows. First identify $G = BS(1, n)$ with $\mathbb{Z}[\frac{1}{n}] \rtimes_n \mathbb{Z}$ where the canonical generator of $\mathbb{Z}$ acts as the multiplication by $n$ on $R = \mathbb{Z}[\frac{1}{n}]$. Then the commutator of a generating pair of $G$ is of the form $(1 - n)u$ for some unit $u$ of $R$ and every unit arises in this way. By Higman's lemma, if two generating pairs $\mathbf{g}_1$ and $\mathbf{g}_2$ of $G$ are Nielsen equivalent then the corresponding units $u_1$ and $u_2$ are related by an element of the form $(-1)^kn^l$ with $k, l \in \mathbb{Z}$. It is not difficult to show that every generating pair of $G$ can be Nielsen reduced to $(a, b')$ for some $b'$ that identifies with a unit of $R$. Now it is immediate to see that if $u_1$ and $u_2$ are related by an element of $(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$ then $\mathbf{g}_1$ and $\mathbf{g}_2$ are Nielsen equivalent. So $G$ has as many Nielsen classes of generating pairs as elements in $R^{\times}/(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$. A careful inspection of $R^{\times}$ shows that the cardinality of the latter is infinite if $n = 6$ and equal to $d \in \mathbb{N} \setminus \{0\}$ if $n = 2^d$. As the group $G$ has only one $T$-system of generating pairs by [Theorem 2.4, 4], it answers OP's question with residually finite instances. -There are some classical examples which can provide insight on OP's question. The fundamental group $\pi_1(\Sigma_g)$ of a closed surface $\Sigma_g$ of genus $g > 0$ is long known to have a single Nielsen equivalence class of generating $2g$-tuples, see [2, 7]. -The family of one-relator groups $T(m, n) = \langle a, b \, \vert \, a^m = b^n \rangle$ for $m,n \in \mathbb{Z}$ has been also extensively studied. It follows from [1] and [5] that each $T$-system of generating pairs of $T(m, n)$ contains a unique Nielsen equivalence class (see also [3] for the specific case of the trefoil group). There are infinitely many such $T$-systems if $\vert m \vert,\vert n \vert \ge 2 $ and $\vert m \vert + \vert n \vert > 4$. -But all these examples are again residually finite. -A. M. Brunner proved in [4] that for each $n \ge 0$ the generating pair $(a, b^{2^n})$ of the non-Hopfian Baumslag-Solitar group $BS(2, 3) = \langle a, b \, \vert \, ab^2a^{-1} = b^3 \rangle$ lies in a different $T$-system of $BS(2, 3)$ but he left open the question whether each $T$-system has a representative of this form. By [Theorem B, 7], every automorphism of $BS(2, 3)$ is inner except the automorphism induced by $(a, b) \mapsto (a, b^{-1})$. Therefore each $T$-system contains at most two Nielsen equivalence classes and the $T$-system of $(a, b^{2^n})$ contains a single Nielsen equivalence class. More generally, if $p, q > 1$ and neither divides the other, then $\Gamma = \operatorname{Out}(BS(p, q))$ is isomorphic to the dihedral group of order $d = 2\vert p - q \vert$ generated by the images of $(a, b) \mapsto (ab, b)$ and $(a, b) \mapsto (a, b^{-1})$. As a result every $T$-system of generating pairs contains at most $d$ Nielsen equivalence classes and the $T$-system of $(a, b)$ contains a single Nielsen equivalence class. If $p$ properly divides $q$, then $\Gamma$ is not finitely generated [6]. But also in this case, inspection the automorphism group shows that the $T$-system of $(a, b)$ contains a single Nielsen equivalence class. - -[1] "Über die gruppen $A^{a}B^{b}=1$", O. Schreier, 1923. -[2] "Über die Nielsensche Kürzungsmethode in freien Produkten mit Amalgam", H. Zieschang, 1970. -[3] "Presentations of the trefoil group", M. J. Dunwwoody, A. Pietrowsky, 1973. -[4] "Transitivity systems of some one-relator groups", A. M. Brunner, 1974. -[5] "Generators of the free product with amalgamation of two infinite cyclic groups", H. Zieschang, 1977. -[6] "Automorphisms and Hopficity of certain Baumslag-Solitar groups", D. J. Collins and F. Levin, 1983. -[7] "Tree actions of automorphism groups", N. D. Gilbert et al., 2000. -[8] "Nielsen equivalence in closed surface groups", L. Louder, 2010, arXiv:1009.0454.<|endoftext|> -TITLE: reference request: John Baez on (-1)- and (-2)-categories and properties+structure+stuff -QUESTION [5 upvotes]: I vaguely recall reading a long time ago a 50-or-so page paper, either by John Baez or linked from his page (I think the former), which among other things gave a justification for his table of n-categories and a very cute explanation of the realization that really it should start at n=-2. I believe this paper also introduced (to me) the ideas of properties, structure, and stuff. I was just thinking about how I'd love to look back at this, but among the wealth of resources available on Baez's webpage I can't seem to find what I'm looking for. Googling hasn't turned it up either. Does anyone know what paper I'm talking about? And if it doesn't talk about properties, structure, and stuff, can anyone recommend a friendly and polished exposition of these ideas? - -REPLY [8 votes]: I think you mean the paper "Lectures on n-Categories and Cohomology". You can also look at the appropriate pages at the nLab.<|endoftext|> -TITLE: Smoothness of vector fields with smooth flow -QUESTION [5 upvotes]: Suppose you have a Vector field $X$ on a smooth (complete) manifold $M$, whose flow $\phi_X^t$ is, for each time $t\in (-\varepsilon,\varepsilon)$, smooth (of class $C^k$). -Questions: -Is $X$ smooth (of class $C^{k-\textrm{something}}$)? Can I control that "something"? What about the case $M=\mathbb{R}^n$? -Thank you in advance! - -REPLY [2 votes]: Consider for instance the case of $M:=\mathbb{R}$, and any $C^k$ field $X$ on $M$, that we may identify with a function. Say, such that $0 < a \le X(s)\le b$. The field $X$ has a globally defined flow, conjugated to the translation flow $x\mapsto x+t$: -$$\phi^t(x):= h^{-1}(h(x)+t)\,,$$ -where the conjugation $h:\mathbb{R}\to\mathbb{R}$ is $$h(x):=\int_0^x\frac{1}{X(s)}ds \, .$$ -Thus if $X$ is of class $C^k$, possibly not $C^{k+1}$, $\phi $ is in any case in $C^{k+1}(\mathbb{R}\times \mathbb{R},\mathbb{R})\, .$<|endoftext|> -TITLE: Does the derived category of coherent sheaves determine the hodge theory? -QUESTION [21 upvotes]: Given two smooth algebraic varieties (proper or not), if the two derived categories of the bounded complexes of coherent sheaves over them are equivalent (if necessary we assume there is a fully faithful equivalence between the two categories) , do they have the same hodge numbers, or even the same hodge decomposition in the hochschild homologies of the two categories? -Baranovsky in arXiv:math/0206256 discussed this problem and gave a affirmative answer, but I think he only proved that the two varieties have the same odd and even part of homologies, but not the individual degrees of homologies, i.e. I don't know why they have the same betti numbers from his argument, let alone the hodge numbers. -The point I feel is that, we can define hochschild homology from a dg-category (e.g. a dg-enhancement of the derived category), but can't define hodge decomposition of the hochschild homology from only the information of the derive category. -Orlov in arXiv:math/0512620v1 discussed this problem as a consequence of his theorem on motives and gave a sufficient condition which demands the support of the corresponding Fourier-Mukai transform to be equal to the dimension of the concerned varieties. I think this condition is too strong and difficult to verify. -A related problem is, does the homological mirror symmetry conjecture imply the coincedence of hodge numbers of two mirror duals? - -REPLY [6 votes]: I don't know the state of the art. In physics if two smooth Calabi-Yau varieties, $X$ and $X^\vee$, of dimension $n$ are mirror to each other then $h^{p,q}(X)=h^{n-p,q}(X^\vee)$. In general they will not smooth but the physics suggests that we take crepant resolutions $Y\to X$ and $Y^\vee\to X^\vee$ and hence we should have $h^{p,q}(Y)=h^{n-p,q}(Y^\vee)$. But, if a crepant resolution exists it is not necessarily unique. This motivated Maxim Kontsevich to introduce motivic integration and prove that if $X_1$ and $X_2$ are two crepant resolutions of $X$ then $h^{p,q}(X_1)=h^{p,q}(X_2)$. -As for the decomposition of Hochschild cohomology holding for derived equivalences, this is not true. In a paper by Oren Ben-Bassat, Jon Block, and Tony Pantev, they show that for a torus $X$ and its dual $X^\vee$ the Fourier-Mukai transfrom induces an isomorphism $H^0(X,\wedge^2T_X)\simeq H^2(X^\vee,\mathcal{O}_{X^\vee})$. In other words, a noncommutative deformation of $X$ goes to a gerby deformation of $X^\vee$.<|endoftext|> -TITLE: Number fields with same zeta function? -QUESTION [13 upvotes]: Given a zeta function $\zeta_K$ of some number field $K$ how much information will this give us about $K$? Specifically, if two number fields have the same zeta function, what shared properties are they known to have? Is there a way to construct distinct number fields that have the same zeta function? - -REPLY [3 votes]: Coming very late to the party, here is a small complement to Alex's excellent answer. There is a recent paper of Marcolli and Cornelissen (arXiv link) which among other things discusses this question. The following two points give partial answers to the question posed here: - -If two number fields are Galois over $\mathbb{Q}$ and have the same zeta function, then they are isomorphic. -In general, one can say something similar if one is willing to consider all twists of the zeta function by Dirichlet characters. More precisely, assume that there is an isomorphism between the Pontryagin duals of the abelianized Galois groups of the two number fields. Assume also that whenever two characters are identified under this isomorphism, the corresponding twisted zeta functions agree. Then the two number fields are isomorphic. See Theorem 2 in the paper linked above for more details. - -The introduction of the paper actually gives a rather good overview over criteria which can or cannot determine whether two number fields are isomorphic.<|endoftext|> -TITLE: Presentation of the pure Artin groups -QUESTION [6 upvotes]: Let $W$ be a Coxeter group attached to a Coxeter matrix with entries $m_{ij}$ . The presentation of $W$ is given by -$$W= < T_1, \dots, T_n | T_i^2=1, T_iT_jT_i \ldots = T_jT_iT_j \ldots, i \neq j>$$ -where each side of the second equation has $m_{ij}$ terms. -Suppose $B_{W}$ is the corresponding braid group (aka Artin group) obtained by removing the relations $T_i^2=1$: -$$B_{W}=< T_1, \dots, T_n | T_iT_jT_i \cdots = T_jT_iT_j \cdots>.$$ -Then there is a canonical surjective homomorphism $B_{W} \to W$, the kernel of this homomorphism is called pure Artin group. -Does anyone know the presentation of the pure Artin group? -Thank you! - -REPLY [2 votes]: I don't know any reference where such a presentation is written down for any Artin group. But: - -You can find In this paper of Enriquez a presentation for the pure Artin group of type B: http://arxiv.org/abs/math/0408035 Proposition 1.1 (by setting $N=2$ is the formulaes) -For all the infinite families, the corresponding pure braid groups are iterated semi-direct products of free groups. (For all the families but the $D_n$ one, it follows from the fact that the corresponding hyperplane arrangements are of fiber type, hence these are even almost-direct product). It should leads quite easily to a nice presentation of these groups. - -Edit: You may also be interested by this paper of Crips and Paris: http://arxiv.org/abs/math/0210438. Recall that $W_{B_n}=(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$ and that $W_{D_n}=(\mathbb{Z}/2\mathbb{Z})^{n-1}\rtimes S_n$. It is proved in this paper that $B_{B_n}=F_n \rtimes B_n$ and that $B_{D_n}=F_{n-1}\rtimes B_n$ where $B_n$ is the braid group of type $A$ and $F_n$ is a free group, and that the canonical projection from the braid group to the Coxeter group is compatible with these decomposition (in the sense that in both cases it restricts to the obvious projections $B_n \rightarrow S_n$ and $F_k \rightarrow (\mathbb{Z}/2\mathbb{Z})^k$).<|endoftext|> -TITLE: Analog of residue for meromorphic quadratic differentials -QUESTION [7 upvotes]: Hi I had asked this already on math.stackexchange.com but got no answers. -I was wondering if there was any sort of (natural) analog of the residue of a meromorphic one form that made sense for a meromorphic quadratic differential. Ideally, one would take a differential $Q$ express it in local coordinates around a singular point and get such a residue as the factor in front of the $\frac{1}{z^2}$ term. However, taking -$$Q=\left(\frac{1}{z^3}+\frac{1}{z^2}\right) dz^2$$ -on $\mathbb{D}\backslash \lbrace 0 \rbrace$ shows that this is not well defined independent of coordinate changes. -One idea I had was the following. Let $U$ be a neighborhood of a singular point $p$ of $Q$. Let $\hat{U}^*$be the double cover of $U^*=U\backslash \lbrace p \rbrace$ and $\pi:\hat{U}^*\to U^*$ the (conformal) double cover. If I am not mistaken there is then a meromorphic one form $\omega$ on $\hat{U}^*$ so that $\pi^*Q=\omega \cdot \omega $ (i.e. in the double cover the square root of $Q$ should exist). If we define $\alpha$ to be the square of the residue of $\omega$ at the point over $p$ then $\alpha$ should be defined independent of choices and so is a (not necessarily natural) candidate. -Perhaps this is well known? (or incorrect...) - -REPLY [12 votes]: Let me point out a somewhat different answer. Rbega explicitly mentions -$$ -Q = \left(\frac1{z^3} + \frac1{z^2}\right)\ (dz)^2 -$$ -as an example of the sort of meromorphic quadratic differential that is of interest, and this is not at all covered by dalakov's answer. -Consider the more general problem of asking when there is a normal form (possibly with parameters) for differentials of the form -$$ -Q = \frac{h(z)\ (dz)^l}{z^k}\ , -$$ -where $k$ and $l$ are fixed positive integers and $h$ is holomorphic near $z=0$ with $h(0)\not=0$. -Now, when $k$ is not a multiple of $l$, there exists a local coordinate $w$ centered on $z=0$ such that -$$ -Q = \frac{h(z)\ (dz)^l}{z^k} = \frac{(dw)^l}{w^k}\ , -$$ -so all of these differentials are locally equivalent. -However, when $k=ml$ for some integer $m>0$, the story is quite different. For example, it is clear that the 'residue' -$$ -R_{l,l}\left(\frac{h(z)\ (dz)^l}{z^l}\right) = h(0) -$$ -is well-defined, independent of the choice of 0-centered local coordinate $z$. Moreover, in this case, there exists a coordinate $w$ centered on $z=0$ such that -$$ -\frac{h(z)\ (dz)^l}{z^l} = \frac{h(0)\ (dw)^l}{w^l}, -$$ -so the 'residue' $h(0)$ is the only invariant in the case $k=l$. -Now, in all the cases in which $k=ml$, there is such an invariant. For example, when $k=2l$, one has the 'nonlinear residue' -$$ -R_{l,2l}\left(\frac{h(z)\ (dz)^l}{z^{2l}}\right) = h(0)\left(\frac{h'(0)}{h(0)}\right)^l, -$$ -which is independent of choice of coordinates. One also has -$$ -R_{l,3l}\left(\frac{h(z)\ (dz)^l}{z^{3l}}\right) -= h(0)\left(l\ \frac{h''(0)}{h(0)}-(l{-}1)\ \left(\frac{h'(0)}{h(0)}\right)^2\right)^l, -$$ -and -$$ -R_{l,4l}\left(\frac{h(z)\ (dz)^l}{z^{4l}}\right) -= h(0)\left(l^2\ \frac{h'''(0)}{h(0)}-3l(l{-}1)\frac{h'(0)h''(0)}{h(0)^2} - +(2l{-}1)(l{-}1)\ \frac{h'(0)^3}{h(0)^3}\right)^l, -$$ -and so forth. -The general rule is that, in these cases, one can write -$$ -Q = \frac{h(z)\ (dz)^l}{z^{ml}} = \left(\frac{\bigl(h(z)\bigr)^{1/l}\ dz}{z^m}\right)^l -= \omega^l\ , -$$ -where $\omega$ is a meromorphic $1$-form, well defined near $z=0$ up to multiplication by an $l$-th root of unity. The $l$-th power of the usual residue of $\omega$ at $z=0$ then provides an invariant that works out to be a rational expression in the coefficients of the power series of $h$, as in the cases noted above. -Finally, when $m>1$ and $h(0)\not=0$, it can be shown that there exists a local coordinate $w$ centered on $z=0$ such that -$$ -\frac{h(z)\ (dz)^l}{z^{ml}} = \left(\frac{(1 + a\ w^{m-1})\ dw }{w^m}\right)^l\ . -$$ -In this case, the 'residue' is (up to a universal constant multiple) simply $a^l$. An alternative normal form is -$$ -\frac{h(z)\ (dz)^l}{z^{ml}} = \frac{(1 + b\ \zeta^{m-1})\ (d\zeta)^l}{\zeta^{ml}} -$$ -for some constant $b$, and the nonlinear residue in this case is (a universal constant multiple of) $b^l$.<|endoftext|> -TITLE: Large cardinal axiom: everything that happen once must happen an unbounded number of times -QUESTION [13 upvotes]: I remember reading something about a large cardinal axiom saying something like - -If some cardinal $\kappa$ has some property $P$, then there should be a proper class of cardinals with the property $P$. - -Of course this is inconsistent if you allow any property $P$, but this was used to justify for example the existence of uncountable inaccessible cardinals, because $\aleph_0$ is inaccessible, so there should be a proper class of inaccessible cardinals (because having only one inaccessible cardinal is not very homogeneous, and the ordinal hierarchy should be homogeneous). -Unfortunately I don't remember where I read that. -My questions are: Is there a precise formalisation of this axiom, and (if there is one) how high is it in the large cardinal hierarchy? - -REPLY [14 votes]: Here are two contexts in which such conclusions follow. -Strong reflection axioms. Consider the strong reflection axiom, sometimes denoted $V_\delta\prec V$, which is axiomatized in the language of set theory augmented with a constant symbol for $\delta$, axiomatized by the assertions $$\forall x\in V_\delta\ \ [\varphi(x)\iff \varphi^{V_\delta}(x)].$$ -This theory is a conservative extension of ZFC, since every finite subset of the theory can be interpreted in any given model of ZFC by the reflection theorem. Meanwhile, in this theory, if the cardinal $\delta$ (or any larger cardinal) has a property $P$ expressible in the language of set theory, then $V$ satisfies $\exists\kappa P(\kappa)$, and so $V_\delta$ also satisfies this assertion, so there is a $\kappa\lt\delta$ with $P(\kappa)$. Similarly, if the collection of cardinals $\kappa$ with $P(\kappa)$ is bounded, then $V_\delta$ would have to agree on the bound by elementarity, but it cannot agree on the bound since $\delta$ itself has the property. So the collection of cardinals with property $P$ must be unbounded in the ordinals. -A stronger formulation of the strong reflection axiom includes the assertion that $\delta$ itself is inaccessible (or more), in which case it carries some large cardinal strength. It is exactly equiconsistent with the assertion "ORD is Mahlo", meaning the scheme asserting that every definable stationary proper class contains a regular cardinal, which is weaker in consistency strength than a single Mahlo cardinal. -Another seemingly stronger formulation of the theory, but still conservative over ZFC, is due to Feferman, and this asserts that there is a closed unbounded proper class $C$ of cardinals $\delta$, all with $V_\delta\prec V$. This theory can be stated as a scheme in the language of set theory augmented by a predicate symbol for $C$. Feferman proposed it as a suitable substitute and improvement of the use of universes in category theory, because it provides a graded hierarchy of universe concepts, which moreover agree between them and with the full universe on first-order set-theoretic truth. -The maximality principle. The maximality principle is the scheme asserting that any statement $\varphi$ which is forceable in such a way that it continues to be true in all further forcing extensions, is already true. This axiom is the main focus of my article, "A simple maximality principle", JSL 62, 2003, and was introduced independently by Stavi and Vaananen. The axiom asserts in short, that anything that you could make permanently true in forcing extensions, is already permanently true. The point now is that under MP, one gets your phenomenon: -Theorem. Under MP, if there is any inaccessible cardinal, then there is a proper class of such cardinals. And the same for Mahlo cardinals and many other large cardinal concepts. -Proof. Assume MP. If there are no inaccessible cardinals above some ordinal $\theta$, then consider the forcing to collapse $\theta$ to be countable. In the resulting forcing extension $V[G]$, there will be no inaccessible cardinals at all, and there never will be such cardinals in any further extension (because any inaccessible cardinal of $V[G][H]$ would also be inaccessible in $V[G]$). Thus, in $V$ the assertion that there are no inaccessible cardinals is forceably necessary, and so by MP, it must already be true. Thus, under MP, either there are no inaccessible cardinals or there are a proper class of them. The same argument works with Mahlo cardinals or any large cardinal concept that is downwards absolute. QED<|endoftext|> -TITLE: General form of Schwarz reflection principle -QUESTION [5 upvotes]: Hello all, -It is easy to find results on reflecting holomorphic functions over circles and lines, but I am wondering what there is for reflecting over smooth curves, or analytic arcs, etc. In particular, I am interested in the conformal map f from the upper half-plane to $\{x+yi : y>1/(1+x^2)\} $ which maps $0$ to $i$ and fixes infinity (so, say, maps $i$ to $2i$). It seems to me that I should be able to extend $f$ to be analytic in a neighborhood of infinity, but I cannot find a reference. Any help will be appreciated. -Greg - -REPLY [2 votes]: For your specific question, note that the domain you describe $\mathbb{D}$, consisting of those $z = x+iy$ for which $y > 1/(1+x^2)$, when regarded as a domain in the extended complex plane, $\mathbb{CP}^1= \mathbb{C}\cup\infty$, is a disk with a smooth, real-analytic boundary in $\mathbb{CP}^1$, so the mapping you are describing does, in fact, extend analytically across the boundary, everywhere along the boundary. In particular, it extends analytically (meromorphically, actually) to a neighborhood of $z = \infty$.<|endoftext|> -TITLE: Do simplicial objects in a Topos form a model category? -QUESTION [8 upvotes]: Sometimes people say "If you don't like the word 'topos', just think the category of Sets", but I'm not sure to what extent this analogy holds. -The real question here is, do simplicial object in a topos have the structure of a model category? I'm just not sure if you really need the geometric realization functor (and topological spaces) to define the usual model category structure on simplicial sets. - -REPLY [3 votes]: As pointed out, the answer is no – because we cannot hope to have fibrant replacements in general. On the other hand, by following the programme of van Osdol [1977, Simplicial homotopy in an exact category], it is possible to quickly establish a weaker result: -Theorem. For any regular category $\mathcal{S}$ (e.g. an elementary topos), the category of internal Kan complexes in $\mathcal{S}$ is a category of fibrant objects where the fibrations are the internal Kan fibrations and the trivial fibrations are the internal trivial Kan fibrations. (By Ken Brown's lemma, this suffices to determine the weak equivalences.) -Here, "internal" refers to the internal logic of regular categories: for example, an internal trivial Kan fibration in $\mathcal{S}$ is a morphism between simplicial objects in $\mathcal{S}$ such that the matching morphisms (à la Reedy) are regular epimorphisms. Note however that we are using the "external" notion of simplicial objects. -In the special case where $\mathcal{S}$ is a sheaf topos, the internal Kan fibrations and internal trivial Kan fibrations turn out to be the same thing as Jardine's local fibrations and local trivial fibrations. (See Theorem 1.12 in [1987, Simplicial presheaves].) It follows that the weak equivalences in the sense above are the same as Jardine's local weak equivalences.<|endoftext|> -TITLE: Is it possible to check two curves on birational equivalence by some computer algebra system? -QUESTION [5 upvotes]: I have two curves, for example hyperelliptic: -\begin{align} -&y^2 = x^6 + 14x^4 + 5x^3 + 14x^2 + 18, \\\\ -&y^2 = x^6 + 14x^4 + 5x^3 + 14x^2 + 5x + 1 -\end{align} -Is it possible to check them on birational equivalence (is able one curve be birationally transformed to another?) via some computer algebra system (like GAP, Sage, Magma, Maple, Maxima or something)? -It would be great if such system be free, but it is almost OK if It isn't. - -REPLY [6 votes]: I suppose Magma's IsIsomorphic will do the job. -From the documentation - -IsIsomorphic(C, D) : Crv, Crv -> BoolElt,MapSch - Given irreducible curves C and D this function returns true is C and D are isomorphic over their common base field. If so, it also returns a scheme map giving an isomorphism between them. The curves C and D must be reduced. Currently the function requires that the curves are not both genus 0 nor both genus 1 unless the base field is finite. - -Example code in the web calculator -K:=AffineSpace(Rationals(),2); -C1A:=Curve(K,x^10-1-y^2); -C2A:=Curve(K,x^10-2^10-y^2); -C1:=ProjectiveClosure(C1A); -C2:=ProjectiveClosure(C2A); -IsIsomorphic(C1,C2); - -true Mapping from: CrvPln: C1 to CrvPln: C2 -with equations : --2*$.1 - 32*$.2 -$.3 - -Wish this is implemented in sage.<|endoftext|> -TITLE: Hard-sphere gases and the wave equation -QUESTION [5 upvotes]: I'm trying to bridge a basic gap in my own education: -Where can I find a written discussion concerning the derivation of the wave -equation (for the propagation of sound, say), assuming nothing (much?) more than -having a gas consisting of colliding hard spheres? -NB I plead total ignorance concerning whether such a derivation even -exists, so I may need help here just to ask the right question. - -REPLY [5 votes]: Even if C. Wong's answer contains good advices, I wish to add my stone to the wall. -I doubt that you can find a simple mathematic explanation that uses only hard sphere model. Point 2) in Wong's answer is fundamental: the sound speed does depend on the number $D$ of freedom degrees of the molecules, through the adiabatic constant $\gamma=c_p/c_v$. For a monoatomic gas (Hydrogen, Argon, ...) $\gamma=\frac53$. For a diatomic gas (Oxygen, Nitrogen, ..., air in first approximation) $\gamma=\frac75$. More generally $\gamma=1+\frac2D$. -The second fundamental point is that in an ideal gas, the sound speed is proportional to the square root of the temperature. The latter quantifies the standard deviation of the molecule velocities around their mean velocity. Thus it seems difficult to do anything meaningfull without digging into the Boltzman equation. Mind that when $D\neq3$ (there are rotational degrees of freedom), the Boltzmann model is quite complicated, and not that much realistic.<|endoftext|> -TITLE: CLT for the squares of unbounded non-identically independently distributed random variables -QUESTION [8 upvotes]: I have a sequence of independent but not identically-distributed random variables $X_1, X_2, \ldots, X_n$ where $X_i\sim A_i$, with each $A_i$ having a support over $\mathbb{R}$ and subject to the following properties with known parameters: - -$\mathbf{E}[X_i]=0$ (zero mean) -$\mathbf{var}[X_i]=\sigma^2_i<\infty$ (finite variance) -$\mu_3(X_i)=0$ (symmetry) -$\mu_4(X_i)<\infty$ - -I am interested in the convergence properties of the sum of squares of $X_i$: $\sum_{i=1}^n X_i^2$ -Specifically, I am interested in the properties of $A_i$ such that -$$\frac{\sum_{i=1}^n (X_i^2-\sigma^2_i)}{\sqrt{\sum_{i=1}^n (\mu_4(i)-\sigma_i^4)}}\xrightarrow{d}\mathcal{N}(0,1)$$ -That is, I am interested what additional conditions would $A_i$ have to obey in order for the normalised sum of squares to converge to a standard Gaussian distribution. -I am aware that the Lendeberg-Feller condition specifies if that most of the mass of $X^2_i$ is in the small interval around the mean, then CLT applies. This basically excludes heavy-tailed $A_i$. Unfortunately, it's unwieldy to apply. -I also know about the Lyapunov Condition: one can easily show that if $A_i$'s are Gaussian, then the normalised sum of their squares converges to a standard Gaussian. -I have three specific and one general questions: - -If $A_i$'s are platykurtic, does its square obey either Lyapunov and/or Lendeberg-Feller conditions (or any other conditions that ensure convergence to CLT)? -If $A_i$'s belong to the exponential family, what are the least restrictive conditions that one can impose on their parameters so that the sum of squares of random variables drawn from them converges to a Gaussian? (I am looking for a looser condition than parameters yielding Gaussian $A_i$'s) -If I impose constraints on higher central moments (as opposed to absolute moments) of $A_i$'s (but not absolute moments), can that ensure that the CLT applies to the sum of squares? If so, how loose can the constraints be (i.e. can I get away by saying that moments are finite)? - -General question: the above three questions relate to specific distribution properties I thought of (and that are relevant to my project). Are there other properties of $A_i$'s that I should look into? Are there classes of $A_i$ for whose sum of squares CLT always applies? -Background: I am working on a variance detection problem and need to analytically characterise the asymptotic performance of a detector. - -REPLY [4 votes]: The easiest condition would be a bound on $\sup_i \mathbb{E} X_i^6$, which would allow you to apply the Berry–Esseen theorem. More generally, if for some $0<\varepsilon < 2$ you have a uniform bound on $\mathbb{E} |X_i|^{4+\varepsilon}$, then you can apply the Lyapunov condition (which, as I noted above, follows from Lindeberg's condition by the Markov/Chebyshev inequality).<|endoftext|> -TITLE: diagonalizability of the curvature operator -QUESTION [20 upvotes]: When can the curvature operator of a Riemannian manifold (M,g) be diagonalized by a basis of the following form -'{${E_i\wedge E_j }$}' where '{${E_i}$}' is an orthonormal basis of the tangent space? If the manifold is three dimensional then it is always possible. But what about higher dimensional cases? - -REPLY [13 votes]: There are still a few interesting things to say about this question, so I thought I'd add some comments. -In one sense, the answer to the question of when a Riemannian metric has an orthonormal coframing that diagonalizes the curvature in the manner requested by the OP is an algebraic problem: As is well-known, the space of curvature tensors (when regarded as quadratic forms on $\Lambda^2$ that satisfy the first Bianchi identity) has dimension $D_n = \tfrac{1}{12}n^2(n^2{-}1)$, while the set of those diagonalizable in some orthonormal coframe is the $\mathrm{O}(n)$-orbit of a linear subspace of dimension $\tfrac{1}{2}n(n{-}1)$, so (when $n\ge 3$) it is a cone $\mathcal{R}_n$ of dimension $n(n{-}1)$. Thus, a Riemannian metric will have to satisfy a set of at least -$$ -R_n = \tfrac{1}{12}n^2(n^2{-}1) - n(n{-}1) = \tfrac{1}{12}n(n{-}1)(n{-}3)(n{+}4) -$$ -polynomial relations on its curvature in order for such a diagonalization to be possible at every point. Writing out a set of generators for these relations is not likely to be easy and probably won't be enlightening, even for $n=4$, which is when it first becomes nontrivial. Moreover, there is no guarantee that this ideal $\mathcal{I}_n$ of polynomial relations is generated by only $R_n$ polynomials or that you won't still have to impose some inequalities to make sure that the curvature is diagonalizable by an element in $\mathrm{O}(n)$ rather than by an element of $\mathrm{O}(n,\mathbb{C})$ that doesn't lie in $\mathrm{O}(n)$. However, this approach would, in theory, give the answer to the OP's literal question. -On the other hand, one might want to interpret the question as asking how one could 'generate' all of the metrics that satisfy this diagonalizability property, at least locally. This is a more interesting (and more challenging) problem. Willie and Thomas have each given examples of classes of such metrics that essentially depend on one function of $n$ variables: Willie cited the conformally flat metrics, which are locally of the form $e^u g_0$ where $g_0$ is the standard metric on $\mathbb{R}^n$, and Thomas cited the induced metrics on hypersurfaces in a space form of dimension $n{+}1$, each of which, locally, can be described as the graph of one function of $n$-variables). The interesting question is whether these are, themselves, special cases of some more general class of metrics with the desired property. Might there be a class of examples that depend on more than one arbitrary function of $n$ variables? Another interesting question is whether their examples 'reach' all the curvature tensors that satisfy the relations $\mathcal{I}_n$ and, if not, whether are there other examples that do. -This latter question is easier to answer than the former. It is easy to see, just by an algebraic count, that neither the conformally flat metrics nor those induced on hypersurfaces in space forms can actually `reach' all of the $\mathrm{O}(n)$-orbits in $\mathcal{R}_n$. (The two sets of orbits that they reach overlap, and they are distinct, proper closed subsets of $\mathcal{R}_n$.) On the other hand, examples provided by É. Cartan of nondegenerate submanifolds of dimension $n$ in $\mathbb{R}^{2n}$ that have flat normal bundle turn out to have their curvature tensors in $\mathcal{R}_n$ and, using these, one can reach an open subset of the orbits in $\mathcal{R}_n$. Now, Cartan's examples depend locally on $n^2{-}n$ arbitrary functions of two variables (not $n$ variables), and it turns out that they satisfy many more differential equations (of higher order) than just the $\mathcal{I}_n$. For example, in Cartan's examples, the diagonalizing coframing $\omega=(\omega_i)$ turns out to be integrable, i.e., $\omega_i\wedge d\omega_i = 0$ for all $i$, so that the metric itself can be diagonalized in a local coordinate chart and thus is locally of the form -$$ -g = e^{2f_1}\ {dx_1}^2 +e^{2f_2}\ {dx_2}^2 + \cdots + e^{2f_n}\ {dx_n}^2. -$$ -Meanwhile, the condition for a metric in this form to have its curvature tensor be diagonal with respect to the coframing $\omega = (\omega_i) = (e^{f_i}\ dx_i)$ and, hence, take values in $\mathcal{R}_n$ turns out to be an involutive system of second order PDE for the functions $f_i$ whose general local solution depends on $n^2{-}n$ arbitrary functions of two variables. These turn out to be slightly more general than the ones that arise as Cartan's examples, and, using solutions of this type, one can reach all of the $\mathrm{O}(n)$-orbits in $\mathcal{R}_n$. -However, the question of how to 'generate' the 'general' metric whose curvature tensor takes values in $\mathcal{R}_n$ for $n\ge 4$ seems to be a very difficult problem. It is an overdetermined system for the metric that is not involutive, and computing its first two prolongations, even in the $n=4$ case, yields a system that is extremely algebraically complicated and still not involutive. Thus, I do not know (and I believe that it is not known) whether the general local solution of this problem depends (modulo diffeomorphism) on more than one arbitrary function of $n$ variables.<|endoftext|> -TITLE: Covering $\mathbb{N}$ with prime arithmetic progressions -QUESTION [19 upvotes]: For every prime $p_i>2$ choose a $k_i\ge p_i$ , $k_i \in \mathbb{N}$ and take the arithmetic progression $A_i=k_i+np_i$ $n \ge 0$ . Is there any choice of the $k_i's$ such that $|\mathbb{N} \backslash \bigcup A_i | < \infty $ ? -ADDED Does it makes any diferrence if we omit some other prime number (not 2)? - -REPLY [3 votes]: I like Aaron Meyerowitz's efforts and think his and similar methods deserve further study. I want to post my skepticism as a counter, and hope that something will arise from the contrast. I do not consider this post as being an acceptable answer though. -The problem is essentially a shifted sieving problem. After taking the first $n$-many (finitely) primes $q_i$ with offsets $r_i$, one has an eventually periodic pattern of uncovered integers which repeats with period $Q_n = \prod_{i \leq n} q_i$, which contains $U_n = \prod_{i \leq n} (q_i - 1)$ uncovered numbers in each period, and has the first period starting somewhere near $M_n = \max_{i \leq n} r_i$. -If the $q_i$ are the primes in ascending order, we have (Mertens) that $U_n$ is -$O(Q_n/\log(q_n))$, which is (roughly) about $n$ times as many primes in the interval -$(M, M + Q_n)$ when $n$ gets large, especially when $n$ is comparable to the largest -integer $M$ allowed to be uncovered. -If the distribution of coprimes to $Q_n$ were amenable to being nicely covered by arithmetic progressions of primes less than $q_n$, I might share Aaron's confidence. -However, each later prime $q$ used is itself coprime to $Q_n$, and with small deviation will cover only about $1/q$ of what needs to be covered. I suspect that when $n$ gets -to be about $Q_{24}/2$ using Aaron's sequence $Q_i$, he will run short on primes. It -might be prudent to try more extensive simulations which leave no numbers greater than -50 uncovered. -Gerhard "Saying As I Feel It" Paseman, 2011.11.18<|endoftext|> -TITLE: l-adic Turrittin -QUESTION [7 upvotes]: What would be an l-adic analogue of the Turrittin-Levelt decomposition theorem? -Turrittin-Levelt is a structure theorem of meromorphic connections on a complex curve in the formal neighbourhood of a singularity. - -REPLY [8 votes]: Here's a naive formulation of an analogue, which is false. (This fits very well the conditional phrasing from your question, since it would be an analogue if it were true!) -Levelt-Turrittin says that after a finite extension of the disc, every de Rham local system is a direct sum of factors which are of the form a 1-dimensional de Rham local system tensor something with tame ramification (i.e., regular). -The naive $\ell$-adic guess would be that an $\ell$-adic representation of a local Galois group is, after restricting to a finite index subgroup, a sum of factors which are tensor products of 1-dimensional characters of inertia with tamely ramified characters. -However, this is false -- any such thing is a sum of 1-dimensional characters when restricted to the wild inertia subgroup. To construct something that doesn't have this property, take a finite totally ramified non-abelian extension of a your field which is wild (i.e., the Galois group is a $p$-group) and an irreducible representation of the corresponding finite Galois group with non-abelian image. -However, Google seems to think that there is a (true) analogue in the theory of $p$-adic differential equations. I don't know that subject, so I'll leave any such description to an expert in that field.<|endoftext|> -TITLE: A question about groups of intermediate growth -QUESTION [11 upvotes]: Let $G$ be a finitely generated group, $S$ a fixed symmetric generating set and $B(n)$ the ball of radius $n$ about the identity with respect to the word length induced by $S$ on $G$. -Fix $k\geq1$ and denote by $\zeta_k(G,d_S)$ the infimum over $n\geq1$ of $\frac{|B(nk+k)|}{|B(nk)|}$. -Observe that: - -$\zeta_k(G,d_S)=1$ for all $k$ (well, $k=1$ is enough) implies that $G$ has sub-exponential growth. -If $G$ has polynomial growth, then $\zeta_k(G,d_S)=1$, for all $k$ (Gromov + Pansu - by the way, is there a direct proof of this fact, without using such a big theorems?). - -What happens in the middle? More formally: - -Question: What can we say about $\zeta_k(G,d_S)$ when $G$ has intermediate growth? Is it always $1$? Is it always $>1$? Can be both? - -Update: The answer has been provided by Martin Kassabov below: the condition $\zeta_k(G)=1$ is equivalent for $G$ to have sub-exponential growth. -Thanks in advance, -Valerio - -REPLY [9 votes]: If $G$ has sub exponential growth one has that $\lim_s \sqrt[s]{B(s)}=1$ -if you assume that $\zeta_k(G) = c >1$ then by an easy induction -we have $B(nk) > K c^n$ which implies that -$$ -\limsup_s \sqrt[s]{B(s)} > \limsup_n \sqrt[nk]{B(nk)} > \limsup_n \sqrt[nk]{kc^n} = -\sqrt[k]{c} > 1 -$$ -Therefore $\zeta_k(G) \leq 1$, but it is clear that $\zeta_k(G)\geq 1$, i.e -$\zeta_k(G)=1$. -I.e. for any group of sub exponential growth $\zeta_k(G) =1$. -The same is true if you replace the infimum in the the definition of $\zeta_k(G)$ -with limsup, but the argument is more involved and uses that sub-multiplicaticity estimates.<|endoftext|> -TITLE: The complexity of the word problem in linear groups -QUESTION [8 upvotes]: This question is related to this one . Let $G$ be a finitely generated subgroup of $GL_n(K)$ for some field $K$ of characteristic 0. Then $G$ is a linear group over $\mathbb{Q}(x_1,...,x_m)$, the field of rational functions over $\mathbb{Q}$. This follows from the fact that one can assume $K$ to be a finitely generated field, which is a finite extension of the field $\mathbb{Q}(x_1,...,x_m)$, and one can get rid of the finite extension by considering matrices of bigger size. -Question. What is the maximal possible computational complexity of the word problem in such $G$? -It seems clear that the complexity always is at most co-NP (one can check that a product of matrices with entries rational functions is not equal to 1 by plugging not very large values for the variables $x_1,...,x_m$ and computing the product of matrices over $\mathbb{Q}$. Is co-NP the best we can get in general (assuming $m\ge 2$, $n\ge 3$)? - -REPLY [6 votes]: The word problem is in deterministic logspace for linear groups, so it is very fast! See Word problems solvable in logspace by Lipton and Zalcstein, J. Assoc. Comput. Mach. 24 (1977), no. 3, 522–526.<|endoftext|> -TITLE: Affine scheme on spec(A) of a ring A as the sheafification of a pre-sheave on spec(A)? -QUESTION [25 upvotes]: It is obvious that there is a parallel between the definition of structure sheaf of $\operatorname{Spec}(A)$ -versus the sheafification of a pre-sheaf. -The definition of the sheaf $\mathscr F^+$ associated to pre-sheaf $\mathscr F$ is (Hartshorne p.64): - -For any open set $U$, let $\mathscr F^+ (U)$ be the set of functions $s$ from $U$ to the union of stalks $\mathscr F_P$ of $\mathscr F$ over points $P$ of $U$ such that: - -For each $P$ in $U$, $s(P)$ is in $\mathscr F_p$. - -For each $P$ in $U$, there is a neighborhood $V$ of $P$ , contained in $U$, and an element $t$ in $\mathscr F(V)$, such that for all $Q$ in $V$, the germ $t_Q$ of $t$ at $Q$ is equal to $s(Q)$. - - - -While, in (Hartshorne p.70), the definition of the sheaf of rings $\mathscr O$ on $\operatorname{Spec}(A)$ is: - -For any open set $U$ of $\operatorname{Spec}(A)$, let $\mathscr O(U)$ be the set of functions $s$ from $U$ to the union of localizations $A_\mathscr{p}$ of $A$ at $\mathscr{p}$ such that: -For each $\mathscr{p}$ in $U$, there is a neighborhood $V$ of $\mathscr{p}$, contained in $U$, and elements $a,f$ of $A$, such that for each $\mathscr{q}$ in $V$, $f$ not in $\mathscr{q}$, and $s(\mathscr{q}) = a/f$ . - -So, is there a naturally occurring pre-sheaf on $\operatorname{Spec}(A)$ (which in general is not a sheaf) that exists for any ring $A$ such that its sheafification gives exactly the structure sheaf $\mathscr O$ of $\operatorname{Spec}(A)$? - -REPLY [45 votes]: For any open subset $U\subseteq\mathrm{Spec}(A)$ let $S_U=A\setminus\bigcup_{\mathfrak p\in U}\mathfrak p$ and $\mathscr O'(U)=A[S_U^{-1}]$. It is obviously a presheaf. -Claim: For open subsets of the form $U=\mathrm{Spec}(A_f)$ with $f\in A$ we have $\mathscr O'(U)=A_f$. (This shows that the associated sheaf of $\mathscr O'$ is indeed $\mathscr O_{\mathrm{Spec}(A)}$.) -Proof: Assume there is an $s\in S_U$ which does not divide $f^n$ for any $n$. The ideal $(s)$ does not meet the multiplicative set $S_f=\{1,f,f^2,\dots\}$, so it is contained in an ideal $\mathfrak q$ which is maximal with respect to this property, but it is well-known that such an ideal $\mathfrak q$ is prime. By construction, $s\in\mathfrak q\in U$, contradicting $s\in S_U$. -Applying the usual associated sheaf construction to $\mathscr O'$ seems to be what Hartshorne does when he defines $\mathscr O_{\mathrm{Spec}(A)}$.<|endoftext|> -TITLE: Finite index subgroups of free groups and torsion-free amenable quotients of free groups -QUESTION [13 upvotes]: Let $F$ be a finitely generated free group and let $A$ be a finite index subgroup of $F$. -Does there exist a subgroup $B\subset A$ such that $F/B$ is (elementary) amenable and torsion-free? -A group $G$ which is amenable and torsion-free has (at least conjecturally) the following nice properties: $\Bbb{Z}[G]$ has conjecturally no zero divisors (this is known if $G$ is say locally indicable or left-orderable) and $\Bbb{Z}[G]$ embeds in its Ore localization. -I have a certain application for the above question in mind, where using an Ore localization plays a role. But I am also curious since my intuition fails me on that question. -Note that if $F/A$ is solvable, then one could just take $F$ to be an iterated commutator subgroup. But if $F/A$ is non-solvable, then I don't know what to do. - -REPLY [10 votes]: If F is a free group and R is a normal subgroup of F, then I thought it was well-known that F/R' is torsion free (cannot find an explicit reference now, though this is stated just after Lemma 5 of [Farkas, Daniel R. Miscellany on Bieberbach group algebras. Pacific J. Math. 59 (1975), no. 2, 427–435]). Of course if F/R is finite, then F/R' is virtually abelian and certainly elementary amenable.<|endoftext|> -TITLE: What is an isomorphism of Banach spaces? -QUESTION [14 upvotes]: The nLab page on Banach spaces (http://ncatlab.org/nlab/show/Banach%20space) was recently criticised as being, in effect, too heavily biased to category theory (not of the Baire kind) and not enough reflecting how Banach spaces are treated "in the real world" (or the closest approximation thereof that functional analysts live in). A couple of functional analysts have stepped in and are helping out, but I thought I'd also ask here just to see if there was anything more that we were missing. -So far, we have the following notions of morphism and isomorphism: - -Morphisms are continuous (aka bounded) linear maps, isomorphisms are linear homeomorphisms (aka bi-Lipschitz linear equivalences). -Morphisms are "short maps", aka continuous linear maps of norm at most 1, isomorphisms are isometries. - -Are there any others that are in reasonably common use or do we have them all? - -REPLY [10 votes]: Mark gives a good answer. I thought to make this a comment on his answer, but I rambled on past the allowed length and so post as an answer. -It is all a matter of what maps one wants to study. As Mark noted, most analysts as well as probabilists are most interested in (1), the category Ban. The less flexible and easier to treat category Ban$_1$, given by (2), is interesting for many people and was the first to be developed to a high degree. PDE people, interested in Lipschitz mappings, naturally are care about biLipschitz equivalence: some geometers like uniform equivalence; while geometric group theorists are mostly interested in coarse equivalence. I am interested in all of these notions of equivalence and more. -No matter what category one works in, the word isomorphism means linear homeomorphism (usually into; one adds onto or surjective when called for). Other notions of being the same are called isometric, Lipschitz equivalent, uniformly equivalent, coarsely equivalent. For a time, geometric group theorists called "coarse equivalence" "uniform equivalence", but this fortunately is passing. -From a Banach space theoretic perspective, one major challenge is to determine when a weaker notion of equivalence (or embedding) implies isomorphic equivalence (or isomorphic embedding). This is interesting also for people who use Banach spaces without doing Banach space theory. Take, for example, geometric group theorists. Yu and then Kasparov and Yu proved numerous results about finitely generated groups whose Cayley graphs coarsley embed into a "nice" Banach space. For a time it was open whether every "nice" (in this case uniformly convex) Banach space embeds into a Hilbert space--were this true, they could have ignored other Banach spaces. Alas (or YES!, depending on your point of view), that is not the case. It is now a research topic of interest to a large group to determine when a Banach spaces embeds coarsely into a special Banach space $X$. For $X$ a Hilbert space (or, more generally, $L_p$ and $\ell_p$ for $p \le 2$), the answer was provided by Nirina Randrianarivony, but there are only partial results for $L_p$ when $2 -TITLE: When is a power series with coefficients in {0,1} algebraic? -QUESTION [7 upvotes]: If we have a series $F(x)=\sum a_n x^n$ where $a_n$ is in {0,1} for every integer $n$. Is $F$ algebraic over $\mathbb Q$ (set of rational numbers). If it is, under what conditions? -Thank you. - -REPLY [15 votes]: There is a classical rational-transcendental dichotomy for power series with coefficients which don't grow too fast. Here are two famous results: - -P. Fatou, "Series trigonometriques et series de Taylor", Acta Math. (1906), no. 30, 335–400. - -Fatou proves that if a power series $F(x)\in \mathbb Z[[x]]$ converges inside the unit disk, then either $F(x)\in \mathbb Q(x)$ or $F(x)$ is transcendental over $\mathbb Q(x)$. - -F. Carlson, "Uber Potenzreihen mit ganzzahligen Koeffizienten.", Math. Zeitschr. (1921), no. 9, 1–13. - -Carlson proves that if $F(x)\in \mathbb Z[[x]]$ converges inside the unit disk, then either $F(x)$ is rational or $F(x)$ admits the unit circle as a natural boundary. -This tells you that your series if it is algebraic then it must be a rational function with rational coefficients. This happens if and only if the binary number $0.a_1a_2\dots$ is eventually periodic (to see this just plug in $x= \frac{1}{2}$). - -REPLY [2 votes]: I think the answer to Asymptotics/growth for coefficients of algebraic power series will be helpful.<|endoftext|> -TITLE: Galois groups at closed points from Galois group at generic point? -QUESTION [7 upvotes]: Consider the finite map $\mathbb{A}^1_\mathbb{Q}\rightarrow \mathbb{A}^1_\mathbb{Q}$ given by $z\mapsto z^5-z$. The fiber over generic point is the field extension $\mathbb{Q}(t)[z]/(z^5-z-t)$ over $\mathbb{Q(t)}$ whose normalization has Galois group equal to the full symmetric group $ S_5 $ (can prove this by tensoring with $\mathbb{C}$ and calculating the monodromy action - but how to obtain this is unimportant for the question). What I want to know is, can we use this information about the generic point to deduce anything about the analogous Galois group at closed points whose fiber is a field extension? (I don't expect a statement at all such points, just something about some of them - see the motivation below) -If the answer is positive, is there a generalization to arbitrary finite morphisms of curves or of varieties? -I am thinking about this problem after spending some time trying to ferret out the exact difference between what is given by two different proofs of the insolubility of the quintic - the first by demonstrating that the Galois group of $z^5 - z - 1$ (or any other specific quintic over $\mathbb{Q}$) is the full $S_5$, the second by demonstrating that the "generic" quintic, i.e. the quintic with indeterminate coefficients has Galois group the full $S_5$ over the field obtained by adjoining its symmetric polynomials to $\mathbb{Q}$. In particular, I am interested to know if you can deduce that this fact must hold for some polynomial over $\mathbb{Q}$ from the fact that it holds for the "generic" polynomial (it is less interesting to me that it holds for independent complex transcendentals over $\mathbb{Q}$, even though this is enough to deduce that there is no "quintic formula" for quintics with complex coefficients). -Any answers, references, or comments on either the problem posed or the motivation given would be highly appreciated! Thanks. - -REPLY [11 votes]: What you need is the Hilbert irreducibility theorem. -This implies that the Galois group for "most" rational points (i.e. outside a thin subset) is the full symmetric group. More generally one can consider dominant generically finite Galois morphisms $f:X \to Y$ over a number field $K$, where $Y$ is a rational variety (this is important) and if the generic Galois group is $G$ then the Galois group outside a thin subset of $Y(K)$ will also be $G$. -A thin set is defined to be any subset of a finite union of sets of two types: 1) $Z(K)$ where $Z$ is a proper subvariety of $Y$ and 2) $f(X(K))$ for $f:X \to Y$ a quasifinite morphism over $K$ with $X$ irreducible and $f$ not having a rational section. -Another way of stating Hilbert's irreducibility is to say that $\mathbb{A}^n(K)$ is not thin for $K$ any number field. -A nice reference is "Lectures on the Mordell-Weil theorem" by Serre.<|endoftext|> -TITLE: Rokhlin lemma for arbitrary infinite groups. -QUESTION [15 upvotes]: Let $G$ be an at most countable discrete group acting freely on a standard probability measure space $X$ in a measure preserving way. -It is well known that if $G$ is a finite group then this action admits a fundamental domain. As pointed out by Andreas below, by Rokhlin lemma, if $G$ contains an element of infinite order we can find an $(\varepsilon, N)$-fundamentalish domain $U$, where the latter is defined as follows: -Call a set $U\subset X$ an $(\varepsilon, N)$-fundamentalish domain iff there exist $N$ elements $g_1, \ldots, g_N$ of $G$ such that the sets $g_i(U)$ are pairwise disjoint and the measure of their union is at least $1-\varepsilon$. - -Question: If $G$ is an infinite group, $N_0$ is a natural number, $\varepsilon_0$ is a positive real number, does there exist an $(\varepsilon, N)$-fundamentalish domain with $\varepsilon<\varepsilon_0$ and $N>N_0$? - -For example when the action is profinite and "transitive on each level", then clearly answer is positive: there exist $(0,N)$-fundamentalish domains for arbitrary large $N$. - -REPLY [8 votes]: If you fix $N$ and group elements $g_1$, ..., $g_N \in G$, then your question becomes closely related to tilings of groups. Specifically, in Chapter 2, section 2 of "Entropy and Isomorphism Theorems for Actions of Amenable Groups," Ornstein and Weiss prove: -Let $G$ be a countable group acting freely and measure preservingly on a standard probability space $(X, \mu)$. Fix a finite set $T \subseteq G$. If for every $\epsilon > 0$ there is a measurable set $U \subseteq X$ such that the $T$-translates of $U$ are disjoint and $\mu(T \cdot U) > 1 - \epsilon$, then $T$ tiles $G$ in the sense that there is a set of centers $C \subseteq G$ such that the sets $Tc$ ($c \in C$) partition $G$. -They also prove that if $G$ is amenable then the converse holds. So if $G$ is amenable and $T$ is a tile for $G$, then for every free probability measure preserving action of $G$ and every $\epsilon > 0$ there is a $(\epsilon, |T|)$-fundamentalish domain. Thus a natural question is: which amenable groups admit arbitrarily large finite tiles? -Weiss called a group $G$ MT (mono-tileable) if for every finite set $F \subseteq G$ there is a finite tile $T \subseteq G$ containing $F$. In "Monotileable amenable groups," Weiss proved that all solvable groups and all residually finite groups are MT. In "Elementary Amenable Groups," Chou proved that all elementary amenable groups and all free products of non-trivial groups are MT. So in particular, your question has a positive answer whenever the group $G$ is elementary amenable. A stronger tiling condition, called ccc, is studied in chapter 4 of "Groups Colorings and Bernoulli Subflows" (this paper is in preparation). Weaker properties of poly-MT and poly-ccc are studied in "Burnside's Problem, spanning trees, and tilings." To the best of my knowledge these are the only papers which study tilings of countable groups.<|endoftext|> -TITLE: Given a Levy Exponent find the jump-measure and drift -QUESTION [5 upvotes]: A Levy subordinator is an finite variation Levy process with non-negative drift and positive jumps. The Levy exponent is given by -$$\phi(\lambda) = \gamma \lambda + \int_0^\infty ( 1 - e^{-\lambda s} ) \nu(ds)$$ -where $\gamma>0$ is the drift of the subordinator and $\nu$ is the jump measure (Levy measure). If the jumps are a compound Poisson process with (net) jump intensity $\alpha$ and jump-size distribution $\mu$ then $\nu = \alpha \mu $ and the levy exponent becomes -$$\phi(\lambda) = \gamma \lambda + \alpha (1 - \widehat{\mu}( \lambda ) )$$ -where $\widehat{\mu}( \lambda ) = \int_0^\infty e^{- \lambda s} \mu(ds)$. -My Questions are as follows: - -Given a function $\phi(\lambda)$, how do I know if there is a $\nu$ and a $\gamma$ that generates it? -For a given $\phi$ is the pair that generates it $(\gamma,\nu)$ unique? -Assume the jumps are a compound Poisson process. If you are given $\phi(\lambda)$ can you find $\alpha$ and $\gamma$? Finding $\alpha$ and $\gamma$ would uniquely determine $\widehat{\mu}( \lambda )$ and allow us to reconstruct $\mu(ds)$ from the inverse Laplace transform. Then $\nu(ds) = \alpha \mu(ds)$. -More generally, given $\phi(\lambda)$, can you find $\nu$ and $\gamma$. - -The reason for these questions is that I am going to numerically construct $\phi(\lambda)$ from data. Ideally, I would like to then construct $\gamma$ and $\nu$ (or $\alpha$ for a Poisson process) as well. At this point, it isn't clear to me that I actually need $\gamma$ and $\nu$ for my calculations. It may be that $\phi(\lambda)$ is enough (this project is in its nascent stage at the moment). But, even if I don't need $\nu$ and $\gamma$ I am curious to see if I can construct them. And an existance and uniqueness result would definitely strengthen my paper. - -So, I have a partial answer to the construction of $(\gamma,\nu)$ from $\phi$. Clearly $\gamma = \lim\limits_{\lambda \to \infty} \phi(\lambda)/\lambda$. -Still looking for a construction of $\nu$ at the moment. - -REPLY [2 votes]: Once you get $\gamma$, you can calculate -$\frac{\phi'(\lambda)-\gamma}{\lambda}=\int_0^\infty e^{-\lambda s}\nu(ds)$ -and $\nu$ can be obtained by Laplace inversion.<|endoftext|> -TITLE: Why study simplicial homotopy groups? -QUESTION [19 upvotes]: The standard definition for simplicial homotopy groups only works for Kan complexes (cf. http://ncatlab.org/nlab/show/simplicial+homotopy+group). I learned that the hard way, when I tried to compute a very simple example, i.e. the homotopy group of the boundary of the standard 2-simplex. -My naive idea to actually compute simplicial homotopy groups for arbitrary simplicial sets was taking the fibrant replacement. But obviously we need a model structure for that. Then again, a weak equivalence in the usual model structure for simplicial sets is precisely a weak equivalence of the geometric realization.(cf. http://ncatlab.org/nlab/show/model+structure+on+simplicial+sets) -As I understand it so far, the only satisfactory way to talk about simplicial homotopy groups requires the notion of "classical" homotopy groups. Hence my question: why does it still make sense, to actually talk about simplicial homotopy groups in the first place? - -REPLY [6 votes]: Well, both in the book of Peter May and the paper from Curtis it is proved that the homotopy category of the category of Kan simplicial sets is equivalent to the homotopy category of CW complexes. The equivalence is given by the adjoint functors "geometric realisation" and "total singular simplicial set". This is one of many motivations to study simplicial sets as a very suitable "discrete model" for topological spaces. One very nice thing about simplicial sets is how constructive things are (often :) ). As far as I have read, simplicial sets have been first used for constructing classifying spaces for topological groups. -The fact that we can define the homotopy groups within the category of Kan simplicial sets is great, because this gives us the opportunity to study them using some combinatorial tools as well. You can check this paper -http://arxiv.org/abs/1211.3093 -Here the authors develop a polynomial time algorithm, which computes homotopy groups of simply connected simplicial complexes. For example, starting with a finite simplicial complex model of the 2-sphere, we can introduce an ordering of the vertices and generate a simplicial set. This simplicial set will never be Kan, but using Postnikov towers and a suitable model for Eilenberg-MacLane spaces, one can build up Kan simplicial sets, which approximate it homotopically and use the Postnikov stages for the computations. The result is beautiful and very constructive. -I hope this gives some motivation for the study of simplicial homotopy groups.<|endoftext|> -TITLE: Is there a spectral theory approach to non-explicit Plancherel-type theorems? -QUESTION [15 upvotes]: Teaching graduate analysis has inspired me to think about the completeness theorem for Fourier series and the more difficult Plancherel theorem for the Fourier transform on $\mathbb{R}$. There are several ways to prove that the Fourier basis is complete for $L^2(S^1)$. The approach that I find the most interesting, because it uses general tools with more general consequences, is to use apply the spectral theorem to the Laplace operator on a circle. It is not difficult to show that the Laplace operator is a self-adjoint affiliated operator, i.e., the healthy type of unbounded operator for which the spectral theorem applies. It's easy to explicitly solve for the point eigenstates of the Laplace operator. Then you can use a Fredholm argument, or ultimately the Arzela-Ascoli theorem, to show that the Laplace operator is reciprocal to a compact operator, and therefore has no continuous spectrum. The argument is to integrate by parts. Suppose that -$$\langle -\Delta \psi, \psi\rangle = \langle \vec{\nabla} \psi, \vec{\nabla \psi} \rangle \le E$$ -for some energy $E$, whether or not $\psi$ is an eigenstate and even whether or not it has unit norm. Then $\psi$ is microscopically controlled and there is only a compact space of such $\psi$ except for adding a constant. The payoff of this abstract proof is the harmonic completeness theorem for the Laplace operator on any compact manifold $M$ with or without boundary. It also works when $\psi$ is a section of a vector bundle with a connection. -My question is whether there is a nice generalization of this approach to obtain a structure theorem for the Laplace operator, or the Schrödinger equation, in non-compact cases. Suppose that $M$ is an infinite complete Riemannian manifold with some kind of controlled geometry. For instance, say that $M$ is quasiisometric to $\mathbb{R}^n$ and has pinched curvature. (Or say that $M$ is amenable and has pinched curvature.) Maybe we also have the Laplace operator plus some sort of controlled potential --- say a smooth, bounded potential with bounded derivatives. Then can you say that the spectrum of the Laplace or Schrödinger operator is completely described by controlled solutions to the PDE, which can be interpreted as "almost normalizable" states? -There is one case of this that is important but too straightforward. If $M$ is the universal cover of a torus $T$, and if its optional potential is likewise periodic, then you can use "Bloch's theorem". In other words you can solve the problem for flat line bundles on $T$, where you always just have a point spectrum, and then lift this to a mixed continuous and point spectrum upstairs. So you can derive the existence of a fancy spectrum that is not really explicit, but the non-compactness is handled using an explicit method. I think that this method yields a cute proof of the Plancherel theorem for $\mathbb{R}$ (and $\mathbb{R}^n$ of course): Parseval's theorem as described above gives you Fourier completeness for both $S^1$ and $\mathbb{Z}$, and you can splice them together using the Bloch picture to get completeness for $\mathbb{R}$. - -REPLY [2 votes]: Too big to fit well as comment: There is a seeming-technicality which is important to not overlook, the question of whether a symmetric operator is "essentially self-adjoint" or not. As I discovered only embarrasingly belatedly, this "essential self-adjointness" has a very precise meaning, namely, that the given symmetric operator has a unique self-adjoint extension, which then is necessarily given by its (graph-) closure. In many natural situations, Laplacians and such are essentially self-adjoint. But with any boundary conditions, this tends not to be the case, exactly as in the simplest Sturm-Liouville problems on finite intervals, not even getting to the Weyl-Kodaira-Titchmarsh complications. -Gerd Grubb's relatively recent book on "Distributions and operators" discusses such stuff. -The broader notion of Friedrichs' canonical self-adjoint extension of a symmetric (edit! :) semi-bounded operator is very useful here. At the same time, for symmetric operators that are not essentially self-adjoint, the case of $\Delta$ on $[a,b]$ with varying boundary conditions (to ensure symmetric-ness) shows that there is a continuum of mutually incomparable self-adjoint extensions. -Thus, on $[0,2\pi]$, the Dirichlet boundary conditions give $\sin nx/2$ for integer $n$ as orthonormal basis, while the boundary conditions that values and first derivatives match at endpoints give the "usual" Fourier series, in effect on a circle, by connecting the endpoints. -This most-trivial example already shows that the spectrum, even in the happy-simple discrete case, is different depending on boundary conditions.<|endoftext|> -TITLE: Textbook for Etale Cohomology -QUESTION [32 upvotes]: What is the best textbook (or book) for studying Etale cohomology? - -REPLY [2 votes]: Here are some extra references: - -Amazeen, Étale and Pro-Étale Fundamental Groups; -Belmans, Grothendieck Topologies and Étale Cohomology; -Bergström--Rydh, Étale Cohomology Spring 2016; -Conrad (?), Étale Cohomology; -Hajj Chehade, Sheaf Cohomology on Sites and the Leray Spectral Sequence, Chapter -3; -Klingler, Étale Cohomology and the Weil Conjectures; -Kunkel, Étale Fundamental Group: An Exposition; -Laskar, Étale Cohomology; -Puttick, Galois Groups and the Étale Fundamental Group; -Robinson, Étale Cohomology; -Sarlin, The Étale Fundamental Group, Étale Homotopy and Anabelian -Geometry; -Szamuely, Galois Groups and Fundamental Groups, Chapter 5; -The Stacks Project Authors, Étale Cohomology; -Yang, Fundamental Groups of Schemes; -Zarabara, Étale Cohomology over $\mathrm{Spec}(k)$.<|endoftext|> -TITLE: What information is contained in the Kazhdan-Lusztig polynomials? -QUESTION [21 upvotes]: The Kazhdan-Lusztig polynomials contain all kinds of representation theoretic (and other kinds of) informations. -For example the character of a simple module over a Lie algebra with Weyl group $W$ can be read of from the KL-polynomials in the following way: -$$ch(L_w)=\sum_y (-1)^{l(w)-l(y)}P_{y,w}(1) ch(M_y)$$ -here $L_w$ resp. $M_w$ denote the simple resp. Verma module of highest weight $-w(\rho) -\rho$. -What are other examples of important information encoded in KL-Polynomials? - -REPLY [3 votes]: I'm not sure whether this would be considered an answer but there is a "concrete" interpretation of the coefficients given in http://arxiv.org/abs/1212.0791 .<|endoftext|> -TITLE: P-adic representations -QUESTION [8 upvotes]: Hi, - I am reading about p-adic representations from Fontaine's book which can be found at http://staff.ustc.edu.cn/~yiouyang/research.html. On page 145 -where they prove Proposition 5.24 which is essentially the theorem -of Tate-Sen, they show $H^{n}(Gal(L/K_{\infty},C(i)^{G_L})=0$ and -the argument is essentially same as in the proof of hilbert Thm 90. But -then they are concluding that implies $H^{n}(H_K,C(i))=0$ by passing to the -limit. I am confused because I thought that you can only pass through -the limit in case of discrete modules. I think this same argument will -also show that $H^{1}(G_K,C_K)=0$ which is not true. I am sure I am missing something obvious. I will greatly appreciate any kind of clarification. - -REPLY [7 votes]: The whole thing is done with more details in Tate's original article "p-divisible groups", section 3.2. Tate proves that one can approximate a cocyle in $C_p(i)$ by cocyles with values in $Q_p^{alg}(i)$ and this is how he reduces the computation to the "discrete case". -I would suggest that it's better to prove the result by $p$-adic approximation. This way, you can basically work with cocycles with values in $O_{C_p}(i)/p^n$, also a discrete space.<|endoftext|> -TITLE: Subtlety in the definition of the Kobayashi metric -QUESTION [9 upvotes]: When defining the Kobayashi metric on a connected complex analytic space $X$, one makes the following auxiliary definition: - -A holomorphic chain from $x\in X$ to $y\in X$ is a finite sequence of holomorphic maps $f_1,\ldots ,f_n\colon\Delta\to X$ (where $\Delta$ is the unit disk in $\mathbb{C}$) together with points $z_1,\ldots ,z_n,w_1,\ldots ,w_n\in\Delta$ such that $f_1(z_1)=x$, - $f_i(z_i)=f_{i+1}(w_{i+1})$ for $1\le i< n$ and $f(w_n)=y$. -The length of a holomorphic chain is, with this notation, is $\sum_{i=1}^nd(z_i,w_i)$ (Poincaré metric on $\Delta$). -Finally the Kobayashi pseudo-distance on $X$ is obtained by setting $d(x,y)=$ infimum of lengths of all holomorphic chains from $x$ to $y$. - -This "pseudo-distance" is obviously symmetric, satisfies $d(x,x)=0$ and the triangle inequality. The space is called Kobayashi hyperbolic if $d$ is in addition non-degenerate, i.e. if $d$ is a metric. -Now one could as well begin with the following much simpler construction: - -Consider the function $\delta :X\times X\to [0,\infty ]$ with $\delta (x,y)=\inf d(z,w)$, the infimum running over all triples $(f,z,w)$ with $f:\Delta\to X$ holomorphic, $z,w\in\Delta$ and $f(z)=x$, $f(w)=y$. - -This is still symmetric and satisfies $\delta (x,x)=0$, but now it is unclear whether -a) $\delta (x,y)$ is finite, i.e. the set of triples $(f,z,w)$ is non-empty; -b) $\delta$ satisfies the triangle inequality. -Clearly, a) and b) together are equivalent to $d=\delta$, and $d$ can be obtained from $\delta$ by an easy construction. Finally, my questions: - -Under which circumstances is $d=\delta$? -Is there a simple example where $d\neq\delta$? -What is the logical relation between a) and b)? - -REPLY [4 votes]: This is actually more subtle than you might think. A classification of the spaces for which $\delta = d$ is far from known, even for domains in $\mathbb{C}^n$. -However, if $\Omega \subset \mathbb{C}^n$ is convex (or biholomorphic to a lineally convex domain), then $\delta = d$, which was shown by Lempert [Lempert, László . La métrique de Kobayashi et la représentation des domaines sur la boule. Bull. Soc. Math. France 109 (1981), no. 4, 427--474.] There are also some other examples known where $\delta = d$. -One fairly simple example where $\delta$ fails to satisfy the triangle inequality is the following. Let $$\Omega_\epsilon = \lbrace z \in \mathbb{C}^2 : |z_1| < 1, |z_2| < 1, |z_1z_2| < \epsilon \rbrace.$$ Also, let $P = (1/2, 0)$ and $Q = (0, 1/2)$. You can check that $\delta(P,0)$ and $\delta(0,Q)$ (with respect to $\Omega_\epsilon$) are independent of $\epsilon$, but $\delta(P,Q) \to \infty$ as $\epsilon \to 0$. Hence, if $\epsilon$ is sufficiently small, $\delta_\Omega$ violates the triangle inequality.<|endoftext|> -TITLE: is the presheaf of automorphisms a sheaf? -QUESTION [7 upvotes]: In Chapter III,$\S 4$ of Milne's Etale cohomology a correspondence between twisted forms and Cech cohomology cocycles is described. -Fix some Grothendieck topology, say, etale, and let $A$ be a sheaf of algebras over a scheme $X$. A sheaf of algebras $A'$ is called a twisted form of $A$ if there exists a cover $\mathscr{U}=(U_i \to X)$ such that $A \times_X U_i \cong A' \times_X U_i$. Milne then writes that to such an isomorphism one can associate a cocycle in $\check{H}^1(\mathscr{U}, \underline{Aut}(A))$ where $\underline{Aut}(A)$ is the sheaf associated to the presheaf of groups $Aut(U)=Aut_U(A \times_X U)$. -Why does one need to sheafify $Aut$? Isn't it true that $Aut$ is a sheaf already? It is clearly separated, because $A$ is separated, and I don't see why the second sheaf condition would fail, and I couldn't find a counterexample. - -REPLY [4 votes]: You are absolutely right, it is a sheaf, you can glue local automorphisms.<|endoftext|> -TITLE: Stable homotopy theory of orbifolds -QUESTION [11 upvotes]: Is there a notion of stable homotopy, spectrum, or a stable homotopy category which corresponds to orbifolds and orbispaces, in the same way that classical stable homotopy theory corresponds to ordinary manifolds and spaces? -I'm not so interested in speculations about what such a thing might look like; I can come up with my own speculations. I'm curious whether orbifold theorists have actually come up with such notions before and used them for things. - -REPLY [5 votes]: Since Mike asked this question almost 10 years ago, Schwede's work on global equivariant homotopy theory, alluded to by Tyler in the comments, has become a classic. I'm personally not aware of the connection between orbifolds and equivariant homotopy theory having been deeply explored, but maybe it has (and maybe somebody will chime in on this question to talk about it!). It appears that a seminar on the intersection of the two subjects was held at Columbia in 2019. From the syllabus, I would guess that there hadn't been much work bridging the fields to that point, because the topics appear to be pretty clearly delineable between "equivariant homotopy theory" talks and "orbifold" talks. -However, there is recent work of Juran constructing a global equivariant stable homotopy type from any orbifold. This seems like an area ripe for cross-fertilization. -Something about the way both subjects are especially fond of finite groups really screams to me that they must be related. -One would think there'd be an analogy -orbifolds : global equivariant homotopy theory :: manifolds : ordinary homotopy theory<|endoftext|> -TITLE: On a result about genus two pencils -QUESTION [5 upvotes]: I am reading the paper "Canonical models of surfaces of general type" by E. Bombieri. In the last section of this paper, there is a statement saying that surfaces with $K^2=1$ and $p_g=0$ do not have pencils of genus $2$, and there is no proof. Is there a proof of this statement? - -REPLY [5 votes]: In fact it seems that the statement is not correct. -The paper [Calabri, Ciliberto, Mendes Lopes, -Numerical Godeaux surfaces with an involution. -Trans. Amer. Math. Soc. 359 (2007), no. 4] contains the classification of numerical Godeaux surfaces (i.e., minimal surfaces of general type with $K^2=1$ and $p_g=0$) that have an automorphism of order 2. The examples described in section 6 have a pencil of curves of genus 2 (cf. Remark 6.3).<|endoftext|> -TITLE: The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$ -QUESTION [15 upvotes]: It seems that the normalizer of $H=\mathrm{GL}(n,\mathbf Z)$ in $G=\mathrm{GL}(n,\mathbf Q)$ is "almost" equal to itself, that is, -$$ -N_G(\mathrm{GL}(n,\mathbf Z))=Z(G) \cdot \mathrm{GL}(n,\mathbf Z) -$$ -where $Z(G)$ is the centre of $G$ (one may guess so by applying the description of automorphisms of groups $\mathrm{GL}(n,\mathbf Z)$ by Hua and Reiner). Is there, however, a simpler and direct proof/disproof of this fact? More generally, for which integral domains $R$ it is known that $\mathrm{GL}(n,R)$ "almost" coincides with its normalizer in the group $\mathrm{GL}(n,Q(R))$ where $Q(R)$ is the quotient field of $R?$ (The question has been earlier posted at mathunderflow). - -REPLY [4 votes]: Let me simplify a bit (if I may) a nice argument by Matthew Emerton, by omitting the part with $p$-reductions. We start as above: let $g \in\mathrm{GL}(n,\mathbf Q)$ normalize $\mathrm{GL}(n,\mathbf Z).$ Then the subgroup -$$ -zg(\mathbf Z^n) -$$ -is in $\mathbf Z^n$ for a certain $z \in \mathbf Q$ and is invariant under all elements of $\mathrm{GL}(n,\mathbf Z).$ By the description of the subgroups of free abelian groups, there is a basis $f_1,\ldots,f_n$ of $\mathbf Z^n$ and integers $m_1,\ldots,m_n$ with -$m_k | m_{k+1}$ $(k=1,\ldots,n-1)$ such that -$$ -zg(\mathbf Z^n) =\langle m_1 f_1,m_2 f_2,\ldots,m_n f_n \rangle. -$$ -Hence -$$ -(z/m_1)g(\mathbf Z^n) =\langle f_1,(m_2/m_1) f_2,\ldots,(m_n/m_1) f_n \rangle \leqslant \mathbf Z^n. -$$ -Thus the subgroup $(z/m_1)g(\mathbf Z^n)$ contains a unimodular/basis element (namely, $f_1$) and then since $(z/m_1)g(\mathbf Z^n)$ is invariant under $\mathrm{GL}(n,\mathbf Z)$ we have that -$$ -(z/m_1)g(\mathbf Z^n) =\mathbf Z^n. -$$ -Thus $(z/m_1) g \in \mathrm{GL}(n,\mathbf Z).$<|endoftext|> -TITLE: Top chern class in positive characteristic -QUESTION [8 upvotes]: Given a nonsingular, projective variety $X$ of dimension $n$ over an algebraically closed field $k$. - Over $k=\mathbb{C}$, the top chern class $c_n(T_X)$ of the tangent sheaf is the Euler characteristic of the associated complex manifold. Is there some kind of (geometric) intuition or well-known formulas for the value of $c_n(T_X)$ in the case of $p=\mathrm{char}(k)>0$? - -REPLY [12 votes]: The same thing is true in positive characteristic, the degree of $c_n$ is equal to the Euler characteristic (except if you consider de Rham cohomology where it only is the Euler characteristic mod $p$). The proof of course cannot use the standard proof in the complex case, using Hopf's theorem that says that the degree of the Euler class is the Euler characteristic and the identification of the Euler class with the top Chern class). One can instead use the Riemann-Roch theorem and the identification of the Euler characteristic with the de Rham Euler characteristic. Given the latter the rest is just to verify that the seemingly complicated expresssion of the Riemann-Roch simplifies by a calculation using the splitting principle to just $c_n$. Alternatively, if I remember correctly one can use a Lefschetz pencil and induction over the dimension. -Addendum: It's all coming back to me, a third possibility is to use that the self-intersection of the diagonal is on the one hand the Euler characteristic (as it is gives the trace of the identitity map), on the other hand that self-intersection is given by the top Chern class of the normal bundle of the diagonal which is exactly the cotangent bundle.<|endoftext|> -TITLE: History of a conjecture/problem non-inner automorphisms of order p in finite p-groups -QUESTION [9 upvotes]: The following conjecture/problem posed in The Kourovka Notebook in 1973 by Ya. G. Berkovich: -Problem 4.13. Prove that every finite non-abelian $p$-group admits an automorphism of order $p$ which is not an inner one. ($p$ as usual denotes a prime number) -I would like to know if anybody knows anything concerning history of this problem. -What I know are as follows: -W. Gaschütz has proved in 1966 that every finite $p$-group of order greater than $p$ has a non-inner automorphism of $p$-power order. -One year before Gaschütz, H. Liebeck has proved problem 4.13 for $p$-groups of class $2$ whenever $p>2$. -The problem in its own is of course interesting, but I think maybe the main motivation to propose the problem is to strengthen Gaschütz's famous result, am I right? -What I really look for is to know: does the problem come from other mathematics problems that people already noted to and they stated it? -For example, has it any relation to the classification of finite simple groups?! -I am sorry if you feel my questions are so vague! and I apologize in advance for. - -REPLY [2 votes]: One relevant recent direction is given in my 2007 paper with Geir Helleloid, The automorphism group of a finite p-group is almost always a p-group. J ALGEBRA vol. 312, (1) 294-329. -http://www.sciencedirect.com/science/article/pii/S0021869307000142 also at http://arxiv.org/abs/math/0602039 -We show that the automorphism group of a finite p-group is almost always a p-group. The asymptotics in our theorem involve fixing any two of the following parameters and letting the third go to infinity: the lower p-length, the number of generators, and p. The proof depends on a variety of topics: counting subgroups of a p-group; analyzing the lower p-series of a free group via its connection with the free Lie algebra; counting submodules of a module via Hall polynomials; and using numerical estimates on Gaussian coefficients.<|endoftext|> -TITLE: Chevalley's Theorem on Constructible Sets -QUESTION [6 upvotes]: I'm having a hard time understanding the theorem in the title, more specifically the proof of the related fact that the image of a dominant morphism contains a dense open set of it's closure. (My sources are Liu "Algebraic Geometry and Arithmetic Curves" pg. 98, Hartshorne "Algebraic Geometry" pg. 94 or Humphreys "Linear Algebraic Groups" pg. 32) -I would like to understand this theorem from a naive stand point of polynomial maps and algebraic sets (i.e. zero sets of polynomials in some affine space), however I need to work over an imperfect field. Upon attempting to understand the commutative algebra (seems to rely on only Noether normalization, localization, and integrality) behind the statement in the initial line, I am unable to see where things fail (if they do) when working over an imperfect field. -The following example confuses me. Suppose $K$ is an imperfect field of characteristic $2$, and $f: K \to K$ is given by $f(k) = k^2$. The Zariski closure of the image is then $K$ and this map is dominant. However, the image is a subgroup of the algebraic group $(K,+)$, and so if it is constructible, then it must be Zariski closed (Humphreys pg. 54). However, the imperfectness of $K$ ensures us that the image is not surjective, contradicting the fact that the Zariski closure of the image is all of $K$. -Apologies if I am missing something obvious, as I am not an algebraic geometer by trade! - -REPLY [6 votes]: As Jim Humphreys has pointed out in the comments, you have to either work over an algebraically closed field, or use scheme language. It is clear that over an algebraically closed field your example does not make any problems. So let us look at your example from the point of view of schemes. -What you are considering is the morphism $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ of $K$-schemes which is obtained as follows: recall that $\mathbb{A}^1_K=\operatorname{Spec} K[x]$, hence to give a morphism as above is the same as to give a homomorphism of $K$-algebras, and our homomorphism is given by $x\mapsto x^2$. Then, the morphism $f$ is not only dominant but surjective as a morphism of schemes. -How does one see this? $\mathbb{A}_K^1$ is the set of prime ideals in $K[x]$. There are two types of these: the prime ideal $(0)$, which defines the generic point, and the maximal ideals. Every maximal ideal is generated by a unique monic irreducible polynomial $f$. Hence if $K$ is not algebraically closed, there are strictly more points in $\mathbb{A}_K^1$ than the generic point and those corresponding to elements of $K$. By this classification of prime ideals it is now an easy exercise to deduce surjectivity of $f$. -However, as you rightly remarked, for $K$ imperfect of characteristic two, the induced map on $K$-rational points is not surjective and the image is "weird" in the sense that it is not of the form $S(K)$ for a subscheme $S\subseteq\mathbb{A}^1_K$. But your argument is wrong. The result from Humphreys that you quote again has to be understood as a statement about schemes, and only in the case of an algebraically closed base field can it be interpreted in the "naive" way as a statement about $K$-rational points. -One correct argument is this: if a subset $S\subseteq\mathbb{A}^1_K$ is constructible (i.e., a subscheme), then it is either finite or the complement of a finite subset. This is because the closed proper subsets of $\mathbb{A}_K^1$ are precisely the finite sets of closed points. But if $K$ is imperfect of characteristic two, than the subset $f(K)\subset K$ is infinite and has infinite complement. -By the way, there is a much easier example: take $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ as above, but with $K=\mathbb{R}$. Then the image of the induced map $\mathbb{R}\to\mathbb{R}$ is the set of nonnegative reals, clearly not "constructible", by the same reason. -Finally, here's the correct version of Chevalley's theorem: - -Theorem (EGA IV, 1.8.4.) Let $f:X\to Y$ be a finitely presented morphism of schemes (any morphism between varieties over a field is of this type). Then the image of any constructible subset of $X$ under $f$ is a constructible subset of $Y$. - -If $K$ is an algebraically closed field and $X$ and $Y$ are varieties over $K$ and $f$ is a morphism of $K$-varieties, then $X(K)$ can be identified with the set of closed (!) points on $X$, and you obtain the "naive" version.<|endoftext|> -TITLE: Is there an axiomatic approach of the notion of dimension ? -QUESTION [13 upvotes]: There are many notions of dimension : algebraic, topological, Hausdorff, Minkowski... (and others). -While the topological one generalize the algebraic one, the last three need not coincide for every sets. Yet it is generally acknowledged that the Hausdorff dimension has "nice enough" properties to work with (the interest of the Minkowski dimension lies mainly in the fact that it's easier to compute). -So my main question is this : is there an axiomatic approach that would tidy up this mess ? For example, is there a result of the form : if you ask these axioms then the only map from "reasonnable sets" to the set of positive real integers is the Hausdorff dimension ? (or another one ?). If so what are they ? -Are there also a clearly identified list of properties that you would ask from any notion of dimension ? I give the following as an example : - -it should coincide with the algebraic dimension for finite dimensional vector spaces -dim A $\leq$ dim B if $A \subset B$ -some sort of nice behaviour for cartesian products (at least for reasonnable sets) -some sort of nice behaviour for infinite increasing unions and/or decreasing intersections - -REPLY [4 votes]: There is a notion of "Krull dimension" valid for arbitrary topological spaces, whose definition is totally analogous to that of algebraic varieties, but using lattices of closed subsets instead of rings of functions. -As far as I know, it is the only dimension function $\dim$ defined for arbitrary topological spaces, with the following properties: - -If $Y $ is a subsapce of $X$, then $ \dim Y \leq \dim X $. -$\dim (X \times Y) \leq \dim X + \dim Y $. -It coincides with Grothendieck's combinatorial dimension on noetherian spaces, and with the standard dimensions (cover and ind) on separable metric spaces. - -This beautiful idea goes back to the 60's. You can check the following papers and the references therein. -Section 2 of: - -Sancho de Salas, J.B. and M.T.: "Dimension of dense subalgebras of $C(X)$", in Proceedings of the American Mathematical Society, 105 (1989) - -or the introduction of: - -Sancho de Salas, J.B. and M.T.: "Dimension of distributive lattices and universal spaces", in Topology and its Applications, 42 (1991), 25-36<|endoftext|> -TITLE: What are the values of the derivative of Riemann's zeta function at the known non-trivial zeros? -QUESTION [11 upvotes]: Specifically I am interested in the quotients - $$-\frac{\zeta'(\rho)}{\zeta'(1-\rho)}=2(2\pi)^{\rho-1}\Gamma(1-\rho)\sin(\pi\rho/2).$$ -Obviously they are in $\mathbb{T}$ for all known non-trivial zeros. But how often are these number $\pm 1$? I would find some tables of the derivative at the known zeros rather usefull, or even perhaps tables of the quotients above? I would be very grateful if somebody can provide me with a good reference. Thanks! - -REPLY [13 votes]: Here is an answer in a few parts. - -Ghaith Hiary has computed fairly large tables of zeros at large height, which are at -http://sage.math.washington.edu/home/hiaryg/page/index.html -I believe that each table has 10 million zeros starting at t = 10^n, for n from 12 to 28. There are tables of derivatives of Z'(t) at the zeros, but not zeta'(1/2 + it). At at zero, it is not hard to go from one to the other, though. -To compute what you want to compute, I think the formula you wrote down is not going to be computationally practical due to the exponential decay of the gamma function and the exponential growth of the sine function. (I think may already be precision issues in the picture in Stopple's answer.) If I calculate correctly (I hope I'm not embarrassing myself), then -$$ --\frac{\zeta'(1/2 + i\gamma)}{\zeta'(1/2 - i\gamma)} = e^{-2 i \theta(\gamma)} -$$ -where $\theta$ is the Riemann-Siegel theta function. This is probably a better way to compute the quotient. -To make the above picture using sage, if you have the optional package database_odlyzko_zeta installed, you can use: - - -import mpmath -mpmath.mp.prec = 200 - -L = [] -for gamma in zeta_zeros()[:100]: - z = exp(-2 * CC.0 * CC(mpmath.siegeltheta(gamma))) - L.append((z.real(), z.imag())) - -P = points(L) -P.save('plot.png', figsize=[5,5]) - - -Something similar will work if you want to parse Hiary's files. -As for whether there is any tendency for these numbers to be close to -1: I think there is not. This looks like the question of whether there is any asymptotic relation between the zeros of the zeta function and gram points, and I think that the answer is expected to be no, but I don't know if that is a theorem. (Those thoughts should not stop you from checking, though.)<|endoftext|> -TITLE: Embedding $G(2,n)$ into $G(k,n)$ -QUESTION [6 upvotes]: Let -$$M=\begin{pmatrix} -u_1 & u_2 & \ldots & u_n \\ -v_1 & v_2 & \ldots & v_n \\ -\end{pmatrix}$$ -be a $2 \times n$ matrix. Define $\nu(M)$ to be the $k \times n$ matrix -$$\nu(M) = \begin{pmatrix} -u_1^{k-1} & u_2^{k-1} & \ldots & u_n^{k-1} \\ -u_1^{k-2} v_1 & u_2^{k-2} v_2 & \ldots & u_n^{k-2} v_n \\ -\ldots \\ -v_1^{k-1} & v_2^{k-1} & \ldots & u_n^{k-1}\\ -\end{pmatrix}.$$ -This immediately descends to a map $\nu: G(2,n) \to G(k,n)$, by thinking of a matrix as its rowspan. -EDIT Actually, as Sasha points out below, it only gives a rational map. The matrix $\nu(M)$ has full rank if and only if $M$ has at least $k$ pairwise linearly independent columns. -I was talking to a bunch of physicists today who are interested in the cohomology class of the image of this map. Does anyone know of papers which look at this? -In particular, if I understood them correctly, they have been computing the intersection of $\nu(G(2,n))$ with $G(k-2, n-4)$ embedded in the more normal way. (That is to say, we embed $G(k-2, n-4)$ into $G(k,n)$ as the space of $k$ planes containing a given $2$-plane $A$, and contained in a given $(n-2)$-plane $B$, with $A \subseteq B$.) Experimentally, the number of intersection points is coming out to be the Eulerian number $\left\langle \begin{matrix} n-3 \\ k-1 \end{matrix} \right\rangle$. For example, the intersection $\nu(G(2,6)) \cap G(1,2)$ inside $G(3,6)$ gives $4$. -Has anyone seen this? - -REPLY [6 votes]: Let $V = k^n$. The map in question is the composition of the canonical map -$$ -f:G(2,V) \to G(k,S^{k-1}V) -$$ -given by the $(k-1)$-th symmetric power of the tautological bundle, and the (noncanonical) linear map -$$ -g:G(k,S^{k-1}V) \to G(k,V) -$$ -given by the choice of the basis in $V$. For both maps it is easy to compute the pullbacks of the Chern classes of the tautological bundles. Since the Chow ring is generated by those classes you can compute the pullback of any Schuber class and this determines the class of the image. -In fact, the map $g$ (as well as the initial map) is not regular but only rational. But it has a simple resolution. Let $m = \dim S^{k-1}V - (n - k)$ and $K = Ker(S^{k-1}V \to V)$. Consider the partial flag variety $F(k,m;S^{k-1}V)$ and its subscheme $Z$ consisting of all flags $U_k \subset U_m \subset S^{k-1}V$ such that $K \subset U_m$. Then the projection $Z \to G(k,S^{k-1}V)$ is birational (for generic $U_k$ we have $U_m = U_k \oplus K$) and the projection to $G(k,S^{k-1}V/K) = G(k,V)$ is regular. On the other hand, by definition $Z$ in $F(k,m;S^{k-1}V)$ is the zero locus of the vector bundle $K^*\otimes Q_m$, where $Q_m$ is the tautological quotient bundle (with fiber $S^{k-1}V/U_m$). This description allows to compute the pullback for $g$. -Addition. The precise formula in case $n = 6$, $k = 3$ for the map is the following. Let $U$ denote the tautological rank 2 subbundle on $G(2,6)$ and let $Q$ denote the tautological rank 3 quotient bundle both on $G(3,6)$ and $G(18,21)$. Consider the -product $X = G(2,6)\times G(18,21)$. Let $z = c_9(S^2U^*\otimes Q)$ on $X$. Let $p_*$ denote the map on the Chow rings induced by the projection $X \to G(2,6)$. Then the map on the Chow rings induced by the composition $g\circ f$ takes -$$ -\prod c_i(Q)^{k_i} \mapsto p_*(z\cdot c_3(Q)^{15}\cdot \prod c_i(Q)^{k_i}) -$$ -(in the LHS $Q$ is considered as a bundle on $G(3,6)$ and in the RHS as a bundle on $G(18,21)$).<|endoftext|> -TITLE: Why should the anabelian geometry conjectures be true? -QUESTION [20 upvotes]: I had probed friends of mine about Grothendieck's motivation for making the anabelian geometry conjectures, and they gave me the following explanation: -If $X$ is a hyperbolic curve over some field $K$ (think projective and of genus $\geq 2$), then, intuitively, its universal cover is the upper half plane. This means that to distinguish between any two hyperbolic curves, it suffices to distinguish between the actions on the upper-half plane that induce those two hyperbolic curves. In some vague way, this should be the same as distinguishing between their fundamental groups. -This seems a little tenuous to me. Is there a modification of the above argument that gives a moral reason for why anabelian geometry should be correct? Is there a completely different moral reason for anabelian geometry? If so, what is it? What intuitive reason should I have to believe anabelian geometry (beside the mounting evidence that it is indeed true)? - -REPLY [3 votes]: I think you should take a look at: -http://www.renyi.hu/~szamuely/heid.pdf -The section there about anabelian geometry gives several reasons why one might believe it is true.<|endoftext|> -TITLE: Minimize Energy for Charge Distributions -QUESTION [7 upvotes]: I am considering [positive] charge distributions $\rho:M\rightarrow\mathbb{R}_+$ (nonnegative reals) with unit charge $\int_M\rho=1$ for convenience. Here $M$ is a nice-enough region, say a submanifold of $\mathbb{R}^n$ (or perhaps simply a metric space ?). -The electrostatic energy is (up to constant factor) $W=\int_M\rho V$, where $V(r)=\frac{\rho(r)}{r}$ is the potential at distance $r$. We can rewrite this $W=\int_M \frac{\rho(x)\rho(x')}{|x-x'|}dx\ dx'$. -What charge distributions minimize the energy for a given region? -In dimension $n=3$, for a bounded region $M\subset\mathbb{R}^3$, charges are being placed on a conductor (which by definition is an object where charges are free to move). Charges seek minimum potential $V$, and end up moving out and lieing on the boundary $\partial M$. One way to see this is via Earnshaw's theorem, which says that there is no stable equilibrium for a collection of charges acted upon only by electrostatic forces (hence the minimum is attained on a boundary, which provides a normal force). A minimum potential (for a collection of charges) corresponds to a minimum energy $W$, since general $\rho=\frac{1}{n}\sum^n_{i=1}\delta(x-x_i)$ gives $W=\frac{1}{n}\sum_iV(x_i)$. -So in this case I know that any $\rho$ will be zero on the interior of $M$... but how does $\rho$ behave on $\partial M$? I seem to have reduced the problem to compact manifolds without boundary. A brief electric-field argument shows that the boundary is actually at an equipotential $V_0$, which gives the minimum $W$. -Is it true/obvious that $\rho=\frac{1}{4\pi}$ on $M=S^2$ ? (yes, see Henry Cohn's answer below) -But for dimension $n=1,2$ the charges do NOT all run to the boundary. It is a fact (Am. J. of Phys. 61, 1993 by R. Friedberg) that the distribution on a conducting disk is not zero on the interior. And charge on a conducting needle does not all go to the ends (Am. J. of Phys. 64, 1996 by D. Griffiths). **These are the n-dimensional objects, so that the boundary of a needle is two points and the boundary of a disk is a circle. -Sorry if my questions/thoughts are too broad... I was just randomly proposing a physics problem to myself. I will clarify / be more specific if necessary. - -REPLY [7 votes]: The behavior for continuous charge distributions amounts to classical potential theory; for discrete charges, you get this behavior in the continuum limit. -It is true that the distribution is uniform for a sphere. On other manifolds things can be more complicated. See http://www.ams.org/notices/200410/fea-saff.pdf for a very nice exposition and further references. For example, Figure 5 from that paper (included here thanks to Joseph O'Rourke) shows the limiting distribution for particles on a torus under an inverse $s$-th power law: -           -In this example, for $s \ge 2$ you get a uniform distribution, which is the default behavior when the energy for a continuous charge distribution diverges. For $s < 2$ the particles converge to the continuous distribution that minimizes energy. When $s<1$, this distribution is not even supported on the entire torus. -You don't see these phenomena for the sphere, because of its symmetry, but they are typical for less symmetric manifolds. -Incidentally, the behavior of 1 and 2 dimensions is not so strange. The charges do indeed end up on the boundary, if one uses a harmonic potential function (for example, a logarithmic potential in $\mathbb{R}^2$). The difficulty is that the Coulomb potential is not harmonic in $\mathbb{R}^1$ or $\mathbb{R}^2$. One way of thinking about it is that if you view the needle or disk as sitting inside $\mathbb{R}^3$, then the charges do all end up on the boundary, because the boundary in $\mathbb{R}^3$ is the entire set. -More generally, in $\mathbb{R}^n$, if you use an inverse $s$-th power law for the potential function, then all the charge will be on the boundary if $s \le n-2$ (because the potential function is superharmonic and therefore satisfies the minimum principle). When $s > n-2$, that does not happen.<|endoftext|> -TITLE: Joint law of the time integral of Brownian motion and its maximum -QUESTION [10 upvotes]: Suppose $W_t$ is a standard one dimensional Brownian motion. Let $M_t$ and $I_t$ be its running maximum and time integral, respectively: - -$$M_t=\max_{0\leq s\leq t}\,W_s$$ - - -$$I_t=\int\limits_0^tW_s\,\mathrm{d}s$$ - -The laws of $M_t$ and $I_t$ can be easily derived by any beginnner studying stochastic processes. However, I haven't seen anything in the literature about their joint law. Is the joint law of $M_t$ and $I_t$ known? - -REPLY [3 votes]: You can find explicit formula of the joint density of these two variables in (2.29)---(2.31) of the following articles. The authors cited the formula from other books. -Gerber, Hans U., Elias SW Shiu, and Hailiang Yang. "Valuing equity-linked death benefits and other contingent options: a discounted density approach." Insurance: Mathematics and Economics 51.1 (2012): 73-92.<|endoftext|> -TITLE: A "rewiring process" on graphs -QUESTION [8 upvotes]: I am interested in a discrete process defined as follows. We start with a given graph. At each time step we delete an edge $(i,j)$ and add two edges $e$ and $f$; the edge $e$ is incident with $i$ and a neighbour of $j$ (which is not already a neighbour of $i$), and the edge $f$ is incident with $j$ and a neighbour of $i$ (which is not already a neighbour of $j$). A stable configuration is a graph in which a step of the process gives an isomorphic graph. I would like to collect some information about the behaviour of the process and its stable configurations. - -REPLY [6 votes]: As Barry Cipra points out in his answer, the rewiring process continues as long as the graph contains some path $hijk$, such that $hj$ and $ik$ are not the edges of the graph. Equivalently, the terminal graphs are exactly the graphs which contain no induced subgraphs isomorphic to the $3$-edge path $P_4$ or to the $4$-cycle $C_4$. Such graphs have been studied before under the name trivially perfect graphs, and the Wikipedia page contains a number of equivalent characterizations of graphs in this class.<|endoftext|> -TITLE: If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too? -QUESTION [27 upvotes]: Suppose $V$ is a vector space, we say that $\mathcal B$ is a basis for $V$ if: - -Every $v\in V$ can be written as a linear combination of elements of $\mathcal B$; -If $\sum\alpha_i b_i = 0$, where $\alpha_i$ are scalars and $b_i\in\mathcal B$, then $\alpha_i=0$ for all $i$. - -Assuming the axiom of choice, every vector space has a basis. In particular, every subspace have a basis. -However assuming the axiom of choice does not hold, there are spaces without a basis. Of course that if $V$ is a vector space without a basis it may have a subspace which has a basis, e.g. a span of a single vector. -It is simple to have a vector space which has a non-$\aleph$ basis as well, since in the absence of choice there are sets whose cardinals are not $\aleph$ numbers, let $A$ be such set and consider the functions from $A$ into $\mathbb F$ with finite support. That is: -$$V=\left\lbrace f\colon A\to\mathbb F\ \colon\ |A\setminus f^{-1}(0)|<\aleph_0\right\rbrace$$ -Addition and multiplication by scalar defined pointwise make it pretty clear how this is a vector space over $\mathbb F$. Every such function can be defined as a linear combination of $\delta$ functions, that is functions which are $1$ at a single point only. -It is also pretty clear that $a\mapsto\delta_a$ is a bijection between $A$ and this basis, therefore we have a basis which is not well-orderable. -Question: $(\lnot AC)$ Suppose $V$ is a vector space, and $\mathcal B$ is a basis of $V$. Is it true that every subspace of $V$ has a basis? Or can we find a counterexample, namely a vector space spanned by a basis with a subspace which has no basis? -Does this depend on the definition of basis above? - -REPLY [17 votes]: The answer is no, I think. Here is a proof sketch. (An unclear point in a previous version has now been removed, by slightly modifying the construction of the sequence.) -Let $(S_n)_{n\in\omega}$ be a family of ``pairs of socks''; that is, each $S_n$ has 2 elements, the $S_n$ are disjoint, but there is no set which meets infinitely many $S_n$ in exactly one point. Let $S$ be the union of the $S_n$. -Let $V$ be a vector space with basis $S$ over the 3-element field. For each $v\in V$, each $s\in S$ let $c_s(v)$ be the $s$-coordinate of $v$. (In your notation: $v(s)$.) -Consider the subspace $W$ of all vectors $w$ with the following property: For all $n$, if $S_n = \{a,b\}$, then $c_a(w)+c_b(w)=0$. The set of all $n$ such that for both/any $a\in S_n$ we have $c_a(w) \neq0$ will be called the domain of $w$. Clearly, each domain is finite, and for each finite subset of $\omega$ of size $k$ there are $2^k$ vectors $w\in W$ with this domain. -[Revised version from here on.] -I will show - -From any basis $C$ of $W$ we can define a 1-1 sequence of elements of $W$. -From any 1-1 sequence of elements of $W$ we can define a 1-1 sequence of elements of $S$. -Together, this will show that there is no basis, as $S$ contains no countably infinite set. - -For each set $D$ which appears as the domain of a basis vector, let $x_D$ be the sum of all basis vectors with this domain. So $x_D \neq 0$, and for $D\neq D'$ we get $x_D\neq x_{D'}$. -From a well-order of the finite subsets of $\omega$ we thus obtain a well-ordered sequence of nonzero vectors. Since there must be infinitely many basis vectors, and only finitely many can share the same set $D$, we have obtained an infinite sequence of vectors in $W$. -We are now given an infinite sequence $(w_n)$ of distinct vectors of $W$. The union of their domains cannot be finite, so we may wlog assume that the sequence $k_n:= \max(dom(w_n))$ is strictly increasing. (Thin out, if necessary.) -Now let $a_n$ be the element of $S_{k_n}$ be such that $c_{a_n}(w_n)=1$. Then the set of those $a_n$ meets infinitely many of the $S_k$ in a singleton.<|endoftext|> -TITLE: Did Grothendieck have a plan for proving Riemann Existence algebraically? -QUESTION [47 upvotes]: A recent question reminded me of a question I've had in the back of my mind for a long time. It is said that Grothendieck wanted the center-piece of SGA1 to be a completely algebraic proof (without topology) of the following theorem: -$\pi_1^{et}(\mathbb{P}^1_{\mathbb{C}}\smallsetminus a_1,...,a_r)\cong$ the profinite completion of $\langle \alpha_1,...,\alpha_r|\alpha_1...\alpha_r=1\rangle$. -As you may know, he did not succeed. -From my experience with Grothendieck's ideas, he often has a proposed proof in mind that would take years (if it all) to be realized. Did Grothendieck have an idea of how to prove this fact algebraically? If so, what was the missing element in his proposed proof? - -REPLY [21 votes]: Nice question. I hope that someone more knowledgeable than me will step in -and bring more information, but let me share what I think. -It doesn't seem to me that Grothendieck had an idea of how to prove this fact algebraically. -It is true that Grothendieck had many ideas, and that some of them was so ambitiously genial that they would take year even to him to be realized, the most notable example being the -idea of the proof of the Weil's conjecture. But in this case, he would say that he had the idea. For example he announced his strategy for the Weil conjectures very early, at the ICM -1958, and in the introduction of the EGA. From my readings of Récoltes et Semailles, -I believe he would have considered unethical to raise a question without saying everything he -had in mind on the subject. -This being said, back on the subject of determining algebraically the étale fundamental group -of the projective line minus n points, here is what he says in SGA 1 (remarks 2.7 page 267): -"A cet égard, la situation de la droite rationnelle privée de $n$ points, et l’étude des revêtements d’icelle modérément ramifiés en ces points, est plus sympathique, puisque la considération des groupes de ramification en ces $n$ points fournit $n$ éléments du groupe fondamental à étudier, dont on montre en effet qu’ils engendrent topologiquement ce groupe fondamental, comme nous verrons ultérieurement. Mais même dans ce cas particulièrement concret, il ne semble pas exister de démonstration purement algébrique. Une telle -démonstration serait évidemment extrêmement intéressante." -So he says that one can algebraically construct $n$ elements in the $\pi^1$, and -using transcendental methods prove that those elements topologically generates the $\pi^1$. -But he doesn't seem to imply that he has the slightest idea about a purely algebraic proof. -Edit: An other argument came to my mind. The subject of the étale fundamental group of the -projective line minus n points is one of the very few question that interested Grothendieck a lot in the two part of his mathematical life (before his abrupt depart from IHES in 1970, -and after that). So if he had an idea about how to prove this fact around 1960, and not the time to follow it, would he not have come back to it later when in the nineties if he was interested in the "Dessins d'enfants" and all those anabelian things. To my knowledge he didn't. (Again this is just a guess, that would happily yield to facts or even more educated guesses)<|endoftext|> -TITLE: Given an integer polynomial, is there a small prime modulo which it has a root? -QUESTION [24 upvotes]: I am looking at a paper by Pascal Koiran on the computational complexity of certifying the solvability of integer polynomial equations in several variables. With the aid of some important theorems in algebraic geometry, Koiran reduces everything to the following univariate question: Suppose that $f \in \mathbb{Z}[x]$ is a polynomial of degree $D$, and suppose that the $\ell_1$ norm of the coefficients (or the $\ell_\infty$ norm or anything in between; they are all equivalent for this purpose) is bounded by $R$. Then it is a theorem of Lagarias, Odlyzko, and Weinberger that there is a prime -$$p = \exp(\text{poly}(\log D,\log R))$$ -modulo which $f$ has a root. The only catch is that they assume the generalized Riemann hypothesis. It could be somewhat easier to prove that there is a prime power $q$ of this size and a root in $\mathbb{F}_q$. That seems just as good in context, but in any case there is a prime $p$ that does the job. This theorem is closely related to the "effective Chebotarev density theorem" of Lagarias, Odlyzko, et al. -Koiran needs an ample supply of such primes, but my question is about just finding one. My hunch at the moment is that it is still an open problem to find a $p$ in the range given above unconditionally, in particular without GRH. What is the current status of the question? Could it be easier just to find a root than to establish full effective existential Chebotarev (rather than the density result), or are these equivalent results? Is it viewed as difficult for the same reasons that GRH is difficult, or is GRH just one possible approach? -(By the way, you can get an interesting but inadequate bound unconditionally as follows: $f(x)$ only attains a unit value at most $2D$ times, so choose some other $x$ with $|x| \le D$ and then pick a prime divisor of $f(x)$.) - -REPLY [3 votes]: I am sorry to parasite this venerable question with something which is less an answer than another question, but is the theorem of Lagarias, Odlyzko, Weinberger you mention even proved under GRH ? (you ask if it is true without GRH). -I assume the paper of Pascal Koiran you refer to is this one. One finds indeed in section 5 the theorem you state, and the proof is essentially a -reduction to Lemma 3 of this article by Adleman and Odlyzko (and not by Lagarias and Odlyzko). Now there seems to be a problem in the proof of Lemma 3, -already in the first six lines of that proof. There a prime-counting function $S(x)$ (it counts the primes $p$ up to $x$ with multiplicity $1$ if $P$ has no root mod $p$, $0$ if $P$ has one root mod $p$, $-1$ if $P$ has 2 roots, etc.) and it is said without further justification that the bound $S(x) = x^{1/2} (n \log x + \log D_P)$ (where $n$ is the degree of $P$ and $D_P$ its discriminant) -follows from Theorem 1.1 of Lagarias-Odlyzko (that is the famous effective Chebotarev under GRH). But I don't see to which Galois extension of number fields this theorem is applied. It has to be an extension of $\mathbb Q$ since the result is counting prime numbers, not prime ideals in some extension. Let $K$ denotes the field generated by a root of $P$, and $L$ its splitting field; then the theorem would give more or less what is stated if applied to $K/\mathbb Q$, but it can't be applied to that extension because in general it is not Galois. If instead -applied to $L/\mathbb Q$, which is Galois, then the degree $n$ has to be replaced -by the degree of $L$ which my be as large as $n!$, ruining the estimate. -So I am perplexed... Since there are at least two people here having thought on this kind of results, what am I missing?<|endoftext|> -TITLE: Magic square of fibered products: vague/unclear? -QUESTION [9 upvotes]: I'm working through Ravi Vakil's notes on algebraic geometry, and I'm at exercise 2.3.R, which states that, if there are maps of objects $X_1\rightarrow Y,X_2\rightarrow Y,Y\rightarrow Z$, then -$\begin{matrix}X_1\times_YX_2\rightarrow&X_1\times_ZX_2\\\\\downarrow&\downarrow\\\\Y\rightarrow&Y\times_ZY\end{matrix}$ -is a fibered/Cartesian square, assuming all relevant fibered products exist. I believe I have written a full proof, but a thing or two bothered me. In writing down the fibered square for $Y\times_ZY$, I guess I assumed that the maps to both $Y$ terms were the same. But I'm sure there are other ways of building this fibered product. Also, a similar situation in seeing what the map along the bottom of the magic square is; I wrote it down assuming that $Y$ mapped to itself via the identity and then applying the universal property, but again, I'm sure that's not the only way. -The map on the right side of the square... Is there some sort of natural map there? I mean, I guess you could look at the maps $X_1\times_ZX_2\rightarrow X_1\rightarrow Y$ (and similarly for the other factor) and again appealing to the universal property... But there are, again, other ways to do this (I think..). -Am I wrong? Or could this really be a lot more precise? - -REPLY [25 votes]: This cartesian square is important for establishing some basic results about base change in algebraic geometry (although, of course, it holds in every category). The maps are constructed as follows: $X_1 \times_Y X_2 \to X_1 \times_Z X_2$ corresponds to a pair of maps $X_1 \times_Y X_2 \to X_i$ whose composition to $Z$ is the same; well just take the projections from the fiber product and remark that since their composition to $Y$ is the same, the same is true for $Z$. The map $X_1 \times_Z X_2 \to Y \times_Z Y$ is induced by the two maps $X_i \to Z$. The map $Y \to Y \times_Z Y$ is the diagonal map, which is defined to correspond to the identity of $Y$ in both factors. Finally, the morphism $X_1 \times_Y X_2 \to Y$ is just the natural map. So to sum up: Every morphism in the diagram is defined canonically. Of course there are other choices possible, but no other choice does make sense. -I would like to make a digression which makes this cartesian square even more clear and deduces it from a more general result, namely that limits commute with limits. Besides, the general result will also yield other canonical isomorphisms which occur often in algebraic geometry. Typically, these isomorphisms are proven separatedly, but as you will see, they are all just corollaries of the following: - - -Lemma. In an arbitrary category, consider the following commutative diagram: -$\begin{matrix} X_1 & \longrightarrow & X_0 & \longleftarrow & X_2 \\\\ -\downarrow & & \downarrow & & \downarrow \\\\\ -S_1 & \longrightarrow & S_0 & \longleftarrow & S_2 \\\\ -\uparrow & & \uparrow & & \uparrow \\\\\ -Y_1 & \longrightarrow & Y_0 & \longleftarrow & Y_2 -\end{matrix}$ -Assuming that all the fiber products exist, then we have -$(X_1 \times_{S_1} Y_1) \times_{X_0 \times_{S_0} Y_0} (X_2 \times_{S_2} Y_2) = (X_1 \times_{X_0} X_2) \times_{S_1 \times_{S_0} S_2} (Y_1 \times_{Y_0} Y_2)$ - - -In order to make sense of the fiber products, we use, of course, the only possible maps. For example, the two squares on the left yield the map $X_1 \times_{S_1} Y_1 \to X_0 \times_{S_0} Y_0$. The Lemma may be memorized as follows: The horizontal fiber product of the vertical fiber products equals the vertical fiber product of the horizontal fiber products. -Now as for the proof of the Lemma, just use the Yoneda Lemma to reduce it to the case of the category of sets, where you can really see this equation immediately. It isn't necessary to draw any arrows and verify the universal property by hand; or rather you encode these arrows as elements. -The first corollary of the lemma is the "cancelling law": - -For morphisms $X \to T \to S$ and $Y \to S$, we have $X \times_T (T \times_S Y) \cong X \times_S Y$. - -Proof: Apply the lemma to: -\begin{matrix} X & \longrightarrow & T & \longleftarrow & T \\\\ -\downarrow & & \downarrow & & \downarrow \\\\\ -S & \longrightarrow & S & \longleftarrow & S \\\\ -\uparrow & & \uparrow & & \uparrow \\\\\ -S & \longrightarrow & S & \longleftarrow & Y -\end{matrix} -The second one is the "Magic square" of the initial question: - -For morphisms $X \to S$, $Y \to S$, $S \to T$, there is a cartesian square -$\begin{matrix} X \times_S Y & \longrightarrow & X \times_T Y \\\\ -\downarrow & & \downarrow \\\\ S & \longrightarrow & S \times_T S \end{matrix}$ - -Proof: Apply the lemma to: -\begin{matrix} S & \longrightarrow & S & \longleftarrow & X \\\\ -\downarrow & & \downarrow & & \downarrow \\\\\ -S & \longrightarrow & T & \longleftarrow & T \\\\ -\uparrow & & \uparrow & & \uparrow \\\\\ -S & \longrightarrow & S & \longleftarrow & Y -\end{matrix}<|endoftext|> -TITLE: What is the intuition behind the proof of the algebraic version of Cartan's theorem A? -QUESTION [6 upvotes]: I am trying to understand the idea behind the proof of GAGA. A crucial step is the following: -Theorem: Let $X=\mathbb{P}^r_{\mathbb{C}}$ (either as a variety or as an analytic space), and let $\mathcal{M}$ be a coherent sheaf on $X$. Then for $n>>0$, the twisted sheaf $\mathcal{M}(n)$ is generated by finitely many global sections. -In the algebraic case, this is Theorem 5.17 in Hartshorne chapter II. If one tries to read the proof of Theorem 5.17, one sees that it depends on Lemma 5.14, which in turn is a generalization of Lemma 5.3. Lemma 5.3 seems to me to be a completely algebraic lemma with no geometric intuition. It will disrupt the flow of the question to state it here, so I will put it at the end. My point is that I don't see any intuition in this statement -In the analytic case this is equivalent to Cartan's Theorem A. To quote the wikipedia page: "Naively, they imply that a holomorphic function on a closed complex submanifold $Z$ of a Stein manifold $X$ can be extended to a holomorphic function on all of $X$". I must confess that I have not read a proof of Cartan's Theorem A itself. But I would like to get some intuition about why it is true, and how it translates to the nilpotent proof found in Hartshorne... -Appendix -Lemma 5.3 in Hartshorne chapter II: -Let $X=Spec(A)$ be an affine scheme, let $f\in A$, let $D(f)\subset X$ be the corresponding open set, and let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. -a. If $s\in \Gamma(X,\mathcal{F})$ is such that its restriction to $D(f)$ is $0$, then for some $n>0$, $f^ns=0$. -b. Given a section $t\in \mathcal{F}(D(f))$ of $\mathcal{F}$ over the open set $D(f)$, then for some $n>0$, $f^nt$ extends to a global section of $\mathcal{F}$ over $X$. -The reason I call this a "nilpotent method" is because in complex algerbaic geometry $f^ns=0$ would imply that either $f=0$ or $s=0$. - -REPLY [7 votes]: Since we're looking for intuition, let's assume that $\mathscr F=\mathscr O_X$, the sheaf of regular functions. -The geometric meaning of this Lemma is that not-everywhere-defined regular functions are more like meromoprhic functions than holomorphic ones. In other words it says that regular functions can't have essential singularities. -Part (b) says regular functions (or sections of $\mathscr F$ in general) on $D(f)$ can be written as $g/f^n$ for some $n>0$ and some global regular function $g$. -Part (a) says that the expression $g/f^n$ is unique as much as possible: If there were $g$ and $h$ such that $g/f^n=h/f^m$ on $D(f)$ then $s=g/f^n-h/f^m$ is $0$ on $D(f)$ and hence by (a) there exists a $q>0$ such that -$$0=f^qs=f^q(g/f^n-h/f^m)$$ -i.e., -$$f^{m+q}g=f^{n+q}h$$ -as global regular functions, so the original function (or section) can be written as a fraction in essentially a unique way. (The ambiguity of multiplying both the numerator and the denominator by a power of $f$ is always there and unless we are working with a UFD we can't say that $g$ and $f$ should have no common divisors). -In other words, this Lemma says that the ring of regular functions on the complement of the zero set of a global regular function is simply the localization of the ring of global regular functions. Part (b) says that everything is a fraction and part (a) says that two different representation as a fraction gives the same element in the localization. If this sentence does not feel like a geometric description, then consider this: this is actually a geometric description of the process of localization by a single element. - -Finally let me point out that "nilpotent" method is not a good name, because $s$ does not have to be nilpotent for this. The condition only means that the support of $s$ is contained in the zero locus of $f$. This can happen even on smooth (complex) manifolds: Take the disjoint union of two manifolds and take a holomorphic function on each. Of course it can't happen on connected manifolds if we are talking about regular functions, but if we're talking about sheaves, then this only needs that the sheaf be not torsion free. For instance the structure sheaf of a proper closed submanifold would do. So, instead of "nilpotent" you should think "torsion" (zerodivisor).<|endoftext|> -TITLE: Fox differential calculus and the Alexander invariant of a link -QUESTION [13 upvotes]: I am teaching a course in knot theory, and I would like to describe the presentation of the Alexander module of a link obtained via Fox differential calculus. In doing this, I should prove the following fact. -Let $L$ be an $n$-component link in $S^3$ with complement $M$, and let $x_0\in M$ be a basepoint. Let $\widetilde M$ be the maximal abelian covering of $M$, and let $\widetilde{M}_0$ be the preimage of $x_0$ in $\widetilde{M}$. Fox differential calculus shows how to obtain a presentation matrix for $H_1(\widetilde{M},\widetilde{M}_0;\mathbb{Z})$ -(with coefficients in $\mathbb{Z}[t_1^{\pm 1},\ldots, t_n^{\pm 1}]$) from a presentation of $\pi_1(M)$. -The usual proof of this fact uses cellular homology, and proceeds more or less as follows: -1. Once a finite presentation of the group $G$ is given, if $X$ is the $2$-dimensional cellular complex associated to the presentation (with only one $0$-cell, one $1$-cell for every generator and one $2$-cell for every relation), then the matrix associated to the presentation of $G$ provides a presentation of $H_1(\widetilde{X},\widetilde{X}_0;\mathbb{Z})$; 2. Using Tietze's Theorem, one shows that the module presented by the Fox matrix of a presentation of a group does not depend on the chosen presentation; 3. One shows that the link complement $M$ admits a realization as a cellular complex with only one $0$-cell. -In fact, point (3) allows to provide a presentation of $\pi_1 (M)$ whose Fox derivatives compute $H_1(\widetilde{M},\widetilde{M}_0)$, by point (1). Then (2) allows us to use any other presentation to get the same result. -My question is: is it possible to prove that $H_1(\widetilde{M},\widetilde{M}_0)$ is presented by the Fox matrix of any presentation of $\pi_1 (M)$ without relying on the fact that a link complement admits a cellular structure with only one $0$-cell (and maybe without using cellular homology too much)? I feel that there should be a proof that does not use cellular homology, and only relies on Hurewicz Theorem. - -REPLY [7 votes]: Let $p:(\tilde{X},y)\rightarrow (X,x) $ be the universal cover of a CW-complex $X$. If -$x$ is the only $0$_cell, then every $1$-cell has a unique lift to the universal cover having -$y$ as its initial point, and these lifts form a basis for the $1$-chains over the group ring -of $\pi_1(X,x)$. The Fox derivative is an axiomatization of the computation of the boundary -of a two cell lifted to $\tilde{X}$ so that its initial point in a corresponding relator is $y$. -If you have more $0$-cells then the $1$-skeleton gives you information about the fundamental groupoid of $X$ restricted to the points which are the $0$-cells. As you want data about fundamental group, you could choose a maximal tree in the $1$-skeleton of $X$. The generators of $\pi_1(X)$ correspond to the edges of the $1$-skeleton that aren't in the tree. You can associate a relator to each two cell by keeping track of what order, and in what direction its boundary crosses those edges. The Fox Calculus should now work, because you could alternatively collapse the tree to a point, and not change the homotopy type of the $CW$-complex. -To compute the first homology of any cover as a module over $\mathbb{Z}$[\pi_1(X)]$ you just need to interpret the answer in the universal cover in the appropriate cover. -To effect your general proof of invariance, you just need to produce change of basis matrices corresponding to changing trees, and doing elementary moves on the two complex. -To avoid the cell complex use group cohomology. If $M$ is a left $R[G]$-module a crossed homomorphism is a map $\phi:G\rightarrow M$ satisfying $\phi(xy)=\phi(x)+x.\phi(y)$ for any $x,y\in G$. The crossed homomorphisms are an $R$-module. A principle crossed homomorphism is one of the form $\phi(g)=v-g.v$ for some $v\in M$. The quotient of the crossed homomorphisms by the principle homomorphsisms is $H^1(G,M)$. You are interested in this when $G=\pi_1(X)$ and $M=H_1(X)$. -Given a presentation of your group, with generators $x_i$ and relators $r_j$, there is a presentation of the crossed homomorphisms, as those crossed homomorphisms on the free group generated by the $x_i$ that annihilate the relators. It can be shown that the space of crossed homomorphisms on the free group are generated as an $R[F(x_i)]$ module by the Fox partial derivatives, and you are off and running, as you are really just interested in calculuating group cohomology.<|endoftext|> -TITLE: Twists of K-theory and tmf -QUESTION [12 upvotes]: I read in a paper by Christopher Douglas that third cohomology twists of $K$-theory may be interpreted as TMF-classes via a map $K(\mathbb{Z},3) \to TMF$, which is related to String orientations. How exactly is this map constructed? Could it be that there is an extension to higher twists, i.e. is there an extension to $BBU_{\otimes} \to TMF$? -EDIT: I know that $BBU_{\otimes}$ splits off $K(\mathbb{Z},3)$ as a factor. Therefore there is of course a map $BBU_{\otimes} \to TMF$, which factors over $K(\mathbb{Z},3)$. The corresponding classes of TMF, however, only see the ordinary third-cohomology twists. So, I reshould restate the second question as something like: Is there an extension $BBU_{\otimes} \to TMF$, which "sees" higher twists? - -REPLY [5 votes]: You can construct the map $K(\mathbb{Z},3) \to tmf$ as follows: first there is the String orientation of tmf, which you already mention. This is a map -$$ MString \to tmf$$ -Then String is by definition a $K(\mathbb{Z},2)$-fibration over Spin. This yields in particular a map -$$ K(\mathbb{Z},3) \to MString $$ -Then you can construct the map $K(\mathbb{Z},3) \to tmf$ as the composition of the above two maps. In order to extend this constuction you had to find a map $BBU_\otimes \to MString$. I think such a map does not exist apart from the one you describe, but I am not entirely sure.<|endoftext|> -TITLE: Extending a diffeomorphism of the sphere $S^2$ to the ball $D^3$ -QUESTION [34 upvotes]: A fundamental result in three-dimensional smooth topology, which in computer jargon we might refer to as "a primitive", is the statement that any ($C^\infty$) diffeomorphism of the two-sphere $S^2$ extends to a diffeomorphism of the closed three-ball $D^3$. Equivalently: $\mathrm{Diff}(S^2)$ is connected. This theorem was first proven by Munkres [Mich. Math. Jour. 7 (1960), 193-197]. Later, Smale proved the stronger result that $\mathrm{Diff}(S^2)$ has the homotopy type of $O(3)$ [Proc. AMS 10 (1959), 621-626]. Another proof of Smale's result is given by Cerf in the appendix to [Sur les difféomorphismes de la sphère de dimension trois ($Γ_4=0$), Lecture Notes in Mathematics, No. 53. Springer-Verlag, Berlin-New York 1968]. - -Question 1: Are there any other known proofs of the statement that any diffeomorphism of the two-sphere $S^2$ extends to a diffeomorphism of the closed three-ball $D^3$? - -There are two reasons I'm not fully happy with the proofs I cited above. Smale's proof and Cerf's proof show much more and use what looks to me like "too much machinery" for just the "$\mathrm{Diff}(S^2)$ is connected" statement, and, in particular, machinery which seems outside basic differential topology (maybe I'm wrong; I haven't gone into them in much detail). Munkres's proof has a number of back-references to another of his papers [Ann. Math. 72(3) (1960), 521-554], and corners need to be smoothed over and over and over and over again to get an honest smooth isotopy between a given diffeomorphism of $S^2$ and the identity. What is worse, it seems difficult to extract an algorithm from Munkres's proof (Lemma 1.1 looks non-constructive - I wouldn't know how to extract a concrete diffeomorphism out of its proof), which brings me to my second question: - -Question 2: How could I implement an extension of a smooth diffeomorphism of the two-sphere to the three-ball? To make things really concrete, let's say I had an image of the surface of the earth which I deformed by some strange diffeomorphism $f$ of $S^2$. How (by computer) could I smoothly deform it back to the usual picture of the earth? - -One dimension down, maybe one way to do it might be to "relax a diffeomorphism of a circle gradually using the heat equation" (see Greg Kuperberg's comment here). Does this work one dimension up? I couldn't figure this out, but I don't see an obvious obstruction- not in dimension three. Or maybe there's a slick way of implementing Munkres's proof by lifting an orientation-preserving diffeomorphism of $S^2$ to $\mathrm{Spin}(3)$ or something... I really have no idea. -Note, though, that other proofs that diffeomorphisms of $S^1$ extend to $D^2$ clearly seem to fail in dimension three... in particular, trying to use some sort of Alexander trick to comb all the "bad parts" of the diffeomorphism into a small disc and shrink that disc to a point will not give rise to a smooth isotopy. -Finally, Morris Hirsch says in a footnote on Page 38 of The Collected Papers of Stephen Smale: "Around this time [1959] an outline of a proof attributed to Kneser was circulating by word of mouth; it was based on an alleged version of the Riemann Mapping Theorem which gives smoothness at the boundary of smooth Jordan domains, and smooth dependence on parameters. I do not know if such a proof was ever published." - -Question 3: Was such a proof ever published? Is there anything else to be said about this proof outline? - -Edit: Actually, I'd like to add even a fourth question: - -Question 4: Are there any "second generation" detailed expositions of any of the above proofs? - -REPLY [9 votes]: Regarding (2), Smale's proof can be made algorithmic. Smale's proof gives a formula for the extension. The formula involves some bump functions, the derivative, a lift of a derivative to the universal cover of a circle, a straight-line homotopy, integration of a vector field and some linear algebra. All these things have effective numerical approximations so if you're happy with a numerical approximation to a diffeomorphism you can certainly algorithmically find one.<|endoftext|> -TITLE: Traces on Hecke algebras and the Jones polynomial -QUESTION [11 upvotes]: In his famous paper "Hecke algebra representations of braid groups and link polynomials," (Annals 1987), Jones uses a compatible family of traces $tr_z$ on the Iwahori-Hecke algebras $H(q,n)$ of type $A_n$ to construct the HOMFLY-PT polynomial, a polynomial invariant of links. In the paper there are a couple statements that are somewhat mysterious: - -(pg 336): This might also show how to use the other Hecke algebras (not of type $A_n$), and their rich representation theory, in some field related to knots...... (pg 343): Other Hecke algebras exist for other Coxeter-Dynkin diagrams and it would be nice to now if any of the ideas of this paper can be suitably modified for them. - -Question: Have people gone in this direction? Is there a reference? - -REPLY [12 votes]: The answer to both questions is positive (since mathematicians tend to leave no stone unturned). See for example: -Geck, Meinolf; Lambropoulou, Sofia. Markov traces and knot invariants related to Iwahori-Hecke algebras of type B. J. Reine Angew. Math. 482 (1997), 191–213. -What I don't know offhand is whether there is a useful up-to-date survey of the whole subject area, though I know of several surveys of knot theory.<|endoftext|> -TITLE: Conditioning on one term of a sum of random variables -QUESTION [6 upvotes]: Let $\theta$ be normally distributed with mean $\bar \theta$ and variance $s^2$. Let $Z$ be normally distributed with mean $0$ and variance $\sigma^2$, and chosen independently of $\theta$. Define $X = \theta + Z$. Clearly, $X$ has mean $\bar\theta$ and variance $s^2 + \sigma^2$. -Write $\zeta^2 = (\tfrac{1}{s^2} + \tfrac{1}{\sigma^2})^{-1}$. It is well-known that for normal random variables, $$\mathbb E(\theta|X) = \tfrac{\zeta^2}{\sigma^2} X + \tfrac{\zeta^2}{s^2} \bar\theta \qquad \mathrm{and} \qquad \operatorname{Var}(\theta|X) = \zeta^2 + \sigma^2.$$ -Note that the conditional mean is linear in the variable $X$, and the conditional variance does not depend on the conditioned value of $X$, simply that some conditioning occurred. These facts hold in wide generality. -Question: Let $\theta$ and $Z$ be independent random variables. What are the most general conditions on the distributions of $\theta$ and $Z$ so that the above conditional formulas hold? -While these formulas will hold for specific classes of distributions (e.g., Beta), Robert Israel points out below that these formulas are unlikely to hold generally. I would also be satisfied with some sort of "approximate" version of the formulas. Is there a nice way to quantify a distribution based on how poorly these formulas hold? - -Here's some motivation for this question. The random variable $\theta$ represents the "true" preference of society for party A over party B. Because of opinion polling, this is known to be within a few standard deviations $s$ of the value $\bar\theta$. On the other hand, each voter receives a noisy signal $X$, which provides some information about the true value $\theta$. The conditional formulas above quantify how much information she obtains from her signal. -For this application, I'm comfortable with $\theta$ being normally distributed, since the Central Limit Theorem is valid for opinion polls. However, I see no justification for the noise $Z$ being normally distributed instead of power law, for example. -For other applications, there's no reason to suppose that the true parameter $\theta$ is known with Gaussian certainty-- it too may be power law. - -REPLY [2 votes]: These equations are unlikely to be true for more general distributions, except in rather special circumstances. Certainly finiteness conditions on moments will not be enough. If $\Theta$ and $Z$ have densities $f_\Theta$ and $f_Z$, the general formula for the conditional expectation is -$$ E[\Theta | X=x ] = \frac{\int_{\mathbb R} \theta f_\Theta(\theta) f_Z(x-\theta)\ d\theta}{\int_{\mathbb R} f_\Theta(\theta) f_Z(x-\theta)\ d\theta}$$ -which in general won't have any particularly nice functional form.<|endoftext|> -TITLE: Stochastic Matrix: Second largest eigenvalue and second largest absolute value of eigen value -QUESTION [6 upvotes]: Setup -Let $A$ be a stochastic matrix. -Let the eigenvalues of $A$ be $1 = \lambda_1 \geq \lambda_2 \geq \lambda_3 ... \geq -1$. -Let $\lambda = \max_{x: x \perp 1} \frac{||Ax||}{|| x ||}$ -Question: -Besides $\lambda_2 \leq \lambda$, is there any relation between $\lambda$ and $\lambda_2$? In particular, I would love to see something of the form $\lambda \leq \lambda_2$. -Context: -Reading about expanders. Many of the proofs appears to prove upper bounds on $\lambda_2$, but I want upper bounds on $\lambda$, and it's not obvious to me: -(1) how an upper bound on $\lambda_2$ becomes an upper bound on $\lambda$ -or -(2) how to generalize some of these proofs. -Thanks! - -REPLY [5 votes]: The answer is due to Boyd, Diaconis, Sun & Xiao. If $A$ is a symmetric and bi-stochastic, then $\mu:=\max(\lambda_2,-\lambda_n)$ satisfies -$$\mu\ge\cos\frac\pi{n}.$$ -In addition, there exists such a matrix for which the equality holds. See Exercise 164 of my list http:\www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf . -P.S. Because you assume that the eigenvalues are real, I presume that you have in mind that the matrix is symmetric.<|endoftext|> -TITLE: A left inverse for the comultiplication on a Hopf von Neumann algebra -QUESTION [13 upvotes]: Edit: incorrect claim at end of earlier version; thanks to Matthew Daws for pointing this out in comments. -$\newcommand{\cM}{{\mathcal M}}\newcommand{\stp}{{\overline{\otimes}}}$The following technical question arose in some work I'm doing, which concerns traces on Banach algebras, but which has wandered into territory that I don't know well. -Let $(\cM,\Delta)$ be a Hopf von Neumann algebra: that is, - $\cM$ is a von Neumann algebra and $\Delta: \cM\to \cM\stp\cM$ is a coassociative, injective, normal $*$-homomorphism. - -Does there always exist a completely bounded, linear map $T:\cM\stp\cM\to \cM$ such that $T\Delta$ is the identity? If so, can we always choose $T$ to be normal? - -This works for many several examples, for instance when $\cM$ is injective as a von Neumann algebra [the image of $\Delta$ is then complemented in $B(H\otimes H)$ by a norm-one projection], or $\cM$ is a locally compact quantum group in the sense of Kustermans-Vaes [use the fundamental unitary and then slice]. or if the predual $\cM_*$ has a bounded approximate identity for the natural product induced by $\Delta_*$. - -REPLY [2 votes]: This is far from a full answer (but maybe it will inspire other answers). -I think all the cases Yemon gives, we only get a cb map, but we don't get normality. However, in some special cases, you do get a normal map. -Firstly, if $M=L^\infty(G)$ with $\Delta(f)(s,t) = f(st)$, then define $T_*:L^1(G) \rightarrow L^1(G\times G)$ by -$$T_*(f)(s,t) = f(st) f_0(s),$$ -where $f_0\in L^1(G)$ is some fixed, positive function with $\|f_0\|_1=1$. This is bounded, as -\begin{align} \|T_*(f)\|_1 &= \int_G \int_G |f(st) f_0(s)| \ dt \ ds -= \int_G \int_G |f(ss^{-1}t)| \ dt \ |f_0(s)| \ ds \\ -&= \int_G \|f\|_1 |f_0(s)| \ ds = \|f\|_1, \end{align} -using the left-invariance of the Haar measure. -Then the pre-adjoint $\Delta_*:L^1(G\times G)\rightarrow L^1(G)$ is -$$\Delta_*(f)(t) = \int_G f(s,s^{-1}t) \ ds,$$ -i.e. convolution (if you set $f=a\otimes b$). So then -$$\Delta_* T_*(f) (t) = \int_G T_*(f)(s,s^{-1}t) \ ds -= \int_G f(ss^{-1}t) f_0(s) \ ds = f(t).$$ -So $\Delta_* T_*$ is the identity, as required. -More generally, and as Yemon alludes to in a comment, if $(M,\Delta)$ is a compact Kac algebra, then it's operator biprojective (see Aristov's paper "Amenability and compact type for Hopf-von Neumann algebras from the homological point of view"). So we can choose $T$ to be cb and normal (and with various $M_*$ module properties). If we can choose $T$ with these module properties (but not necessarily normal) then $M$ is said to be operator biflat. For $VN(G)$ this was investigated by Aristov, Runde and Spronk, "Operator biflatness of the Fourier algebra and approximate indicators for subgroups". It seems to be unknown if $VN(G)$ ever fails to be operator biflat. Of course, this is a much stronger condition that Yemon asks for. -It seems surprising to me that we can just write down a suitable (normal) map $T$ for $L^\infty(G)$, but that it seems that Yemon's question is open for $VN(G)$. -Edit (29 Nov): (This is technical; I hope I got all the details correct). Suppose $G$ is a locally compact group such that $VN(G)$ admits a faithful normal trace (I think is true if $G$ is a separable SIN group, for example). Then let $\varphi$ be the Plancheral weight on $VN(G)$; let $\omega$ be a normal tracial state. Then $\psi=\varphi\otimes\omega$ is a semifinite trace on $VN(G\times G)=VN(G)\overline\otimes VN(G)$, and for $x\in VN(G)_+$, we have that $$ \psi(\Delta(x)) = \varphi((\iota\otimes\omega)\Delta(x)) = \omega(1) \varphi(x) = \varphi(x) $$ as $\varphi$ is invariant for $\Delta$ (I know this from the quantum group setting; see papers of Kustermans and Vaes). Let $(\sigma_t)$ be the modular automorphism group for $\varphi$; then $(\sigma_t\otimes\iota)\Delta(x) = \Delta(\sigma_t(x))$ for $x\in VN(G)$, and so as $\omega$ is a trace, we see that $\Delta(VN(G))$ is invariant for the modular automorphism group of $\psi$. By a theorem of Takesaki there is a normal conditional expectation $\epsilon:VN(G\times G)\rightarrow \Delta(VN(G))$ with $\psi(x) = \psi(\epsilon(x))$ for all $x$ in the definition ideal of $\psi$. -So in particular, this gives a positive answer (even with the "normal") condition in the separable SIN case, say.<|endoftext|> -TITLE: Fourier transforms of functions not in $L^2.$ -QUESTION [11 upvotes]: This is probably something five-year-old physicists know, but here goes: Is there a standard methodology for computing Fourier transforms of things like $\log |x|$? Wolfram Alpha will happily give an answer (involving a delta function), but actually trying to do this yourself (by parts) gives horribly divergent-looking terms (the question which actually came up had $x$ be a vector in $\mathbb{R}^3,$ where the divergent terms are even more horrible than in the one-dimensional case (I am referring to the technique of just cutting off the function at some large $R;$ there are obviously other techniques, like weighting the integrand by an exponential weight (so you are computing a combination of Fourier and Laplace transforms), then computing the analytic continuation at $0,$ but all these should give the same answer,and there should be a not-totally-ad-hoc way of doing this, one should think... - -REPLY [3 votes]: To find the Fourier transform of this and many other functions I enthusiastically recommend volume 1 of the magnificent treatise Generalized Functions, by Gelfand and coauthors. -This monograph contains so many mathematical gems and it pains me to notice that it is quasi - invisible to the Internet generation (By definition, you belong to the Internet generation, if you do no have a vivid memory of an era without E-mail.)<|endoftext|> -TITLE: Class number of PGL_2 -QUESTION [7 upvotes]: Hello. -Let $K = F_q(x)$ be the rational function field and let $G = \textbf{PGL}_{2,K}$. -For any finite and non empty set $S$ of valuations of $K$, -we refer to the subgroup of the adelic group $G(\Bbb{A})$: -$$ G(A_S) = \prod_{p \in S} G_p(K_p) \times \prod_{p \notin S} \underline{G}_p(\mathcal{O}_p) $$ -on which for any prime $p$, $K_p$ is the completion of $K$ at $p$, -$G_p = G \otimes K_p$, $\mathcal{O}_p$ is the ring of integers of $K_p$ and $\underline{G}_p$ is some matrix realization of $G_p$ in $\textbf{GL} \otimes \mathcal{O}_p$. -In particular, if $S$ contains only the single point at infinity we denote $G(\Bbb{A}_S)$ by $G(\Bbb{A}(\infty))$. -The adelic group is decomposed into double cosets: -$$ G(\Bbb{A}) = \bigcup_{i=1}^h G(\Bbb{A}(\infty)) x_i G(K) $$ -where $h=h(G)$ is finite and is called the class number of $G$. -As the universal covering of $G$ -- namely $\textbf{SL}_{2,K}$ -- -admits the strong approximation property, the subgroup $G(\Bbb{A}(\infty))G(K)$ is normal in $G(\Bbb{A})$ -and: $h(G) = (G(\Bbb{A}):G(\Bbb{A}(\infty)) G(K))$. -Could someone please tell me what is $h(G)$ in this case ? -Thank you, rony. - -REPLY [9 votes]: Let $H$ denote $SL_2$. -By strong approximation, see http://www.jstor.org/stable/1970924, $H(K_\infty)G(K)$ is dense in $H({\mathbb A})$. -Now $H({\mathbb A}(\infty))$ contains $H(K_\infty)$ and contains a unit-neighborhood. -Therefore we have $H({\mathbb A})=H({\mathbb A}(\infty))H(K)$. -We can conclude the same thing for $G$, if we can show that $H(R)$ surjects onto $G(R)$ and likewise for $R={\mathbb A}, {\mathbb A}(\infty), K$. -The adelic result can be deduced from the result for fields and for $R={\cal O}_p$. -By Hilbert's Theorem 90, for each field $k$ the Galois cohomology $H^1(k,GL_1)$ is trivial and so the exact sequence of Galois cohomology shows that the sequence -$$ -1\to GL_1(k)\to GL_2(k)\to PGL_2(k)\to 1 -$$ -is exact, which implies the claim for fields. -To verify the claim for $R={\cal O}_p$, we have to analyze the coordinate ring of $PGL_2$. -First, the coordinate ring of $GL_2$ over $R$ is -$$ -A_{GL_2}= R[x_1,x_2,x_3,x_4,y]/(x_1x_4-x_2x_3)y-1. -$$ -The coordinate ring of $PGL_2=GL_2/ GL_1$ is the ring of $GL_1$-invariants, where the action of $GL_1$ is given by -$$ -\lambda.f(x_1,x_2,x_3,x_4,y)=f(\lambda x_1,\lambda x_2,\lambda x_3,\lambda x_4,\lambda^{-2} y). -$$ -The ring of invariants is generated by all monomials of the form $x_ix_jy$ for $1\le i,j\le 4$. -Let now $\chi\in PGL_2({\cal O}_p)$. -Then $\chi$ is a homomorphism from $A_{PGL_2}$ to ${\cal O}_p$. -Every such can be extended to $A_{GL_2}\to K_p$ and we have to show that there exists an extension mapping to ${\cal O}_p$. -Pick any extension and denote it by the same letter $\chi$. -For the valuation $v$ on $K_p$ we have -$$ -0\ \le\ v(\chi(x_j^2y))=2v(\chi(x_j))+v(\chi(y)). -$$ -We are free to change $\chi(y)$ to $\chi(y)\pi^{2k}$ for any $k\in{\mathbb Z}$ if at the same time we change $\chi(x_j)$ to $\chi(x_j)\pi^{-k}$ and $\pi$ is a local uniformizer. -Thus we can assume $v(\chi(y))\in\{ 0,1\}$. -Then we conclude $v(\chi(x_j))\ge 0$ for every $j$ and so $\chi$ indeed maps into ${\cal O}_p$ as claimed. -This shows the result for ${\cal O}_p$. -Now let's put things together. We habe ${\mathbb A}^\times={\mathbb A}(\infty)^\times K^\times$, which is due to the fact that we are dealing with the curve ${\mathbb P}^1$ which has genus zero. Let $g\in GL_1({\mathbb A})$ then -$$ -g=d(x,1)y, -$$ -where $d(a,b)$ is the diagonal matrix with entries $a$ and $b$, $x$ is an idele and $y$ is in $SL_2({\mathbb A})$. By what we have shown, we have -$$ -g=d(x_\infty x_K,1)y_\infty y_k=d(x_\infty,1)\tilde y d(x_K,1)y_K, -$$ -where The $\infty$ indicates entries in ${\mathbb A}(\infty)$ and The $K$ indicates entries in $K$. The element $\tilde y$ is conjugate to $y_\infty$, therefore it no longer has entries in ${\mathbb A}(\infty)$, but it still lies in $SL_2$. Therefore, it can be decomposed again and finally we get $g=g_\infty g_K$. -We find that $h(G)$ is one. This however changes, if you take $K$ to be the rational function field of an arbitrary curve and ${\mathbb A}(\infty)$ the adeles which are unrestricted only at a given point.<|endoftext|> -TITLE: Two questions about vector spaces absent AC. -QUESTION [14 upvotes]: My questions are motivated by -this question -which asks, in the absence of AC, whether a subspace of a vector space with a basis must have a basis. - -Does every real vector space embed isomorphically into a vector space with a basis? - -If $V$ is a vector subspace of a vector space with a basis, then clearly the linear functionals on $V$ separate points. If $V$ is a vector space s.t. the the linear functionals on $V$ do not separate points, then by modding out the intersection of the kernels of all linear functionals you get a vector space that has no non zero linear functional. - -Is there a non zero real vector space on which there is no non zero linear functional? - -There are models of ZF in which every linear functional on every Banach space is continuous. In ZFC there are complete linear metric spaces on which every non zero linear functional is discontinuous. So I assume that (2) has a negative answer, which would imply that (1) also has a negative answer. - -REPLY [17 votes]: To complement François' answer, here's a fairly explicit example of a real vector space admitting no nonzero linear functional in a model of ZFDC + all sets of reals have the Baire property (which is equiconsistent with ZF). -The space is $\mathbb{R}^\mathbb{N} / E_1$, where $E_1$ is the equivalence relation of eventual agreement of real sequences. This can be given a real vector space structure in the obvious way. Now, if $f: \mathbb{R}^\mathbb{N} / E_1 \to \mathbb{R}$ is a linear function, it descends to a linear function $g: \mathbb{R}^\mathbb{N} \to \mathbb{R}$ by $g(x) = f([x]_{E_1})$. Since $g$ is Baire measurable and $E_1$-invariant, by generic ergodicity of $E_1$ it is constant on a comeager set. This constant has to be $0$. But then $g$ is a Polish group homomorphism and is automatically continuous, and thus equals $0$ everywhere. - -REPLY [12 votes]: Hans Läuchli [Auswahlaxiom in der Algebra, Comment. Math. Helv. 37, MR0143705, DOI:10.5169/seals-28602] has constructed a model of ZFA wherein there is vector space (over any given field, see comments) which is not finite dimensional but all of its proper subspaces are finite dimensional. In particular, there cannot be any nontrivial linear functionals since the kernel of such a linear functional would have codimension 1. -This can be transferred to ZF via the Jech-Sochor Embedding Theorem.<|endoftext|> -TITLE: How to prove projective determinacy (PD) from I0? -QUESTION [16 upvotes]: Martin and Steel (in 1987?) showed that if there are infinite many Woodin cardinals then every projective set of reals is determined (PD). -However, it is mentioned in many texts that in 1983/1984 Woodin proved PD from a very strong large cardinal axiom, I0. It is also said the proof is quite different from Martin-Steel's proof. -I would like to know a sketch of Woodin's proof. -Thank you. - -REPLY [6 votes]: Chapter 6 of the paper Large Cardinals and Projective Determinacy - -gives a proof of $PD$ from $I_0$.<|endoftext|> -TITLE: Bezout's Theorem for weighted homogeneous polynomials -QUESTION [14 upvotes]: Bezout's Theorem states that for two homogeneous polynomials $f(x,y,z), g(x,y,z)$ over an algebraically closed field of degrees $m,n$ respectively, such that the two polynomials do not share a common component, then the number of intersections of $f,g$ is equal to $mn$ counting multiplicity. Is there an analogue of this theorem for WEIGHTED homogeneous polynomials? That is, suppose that $w_1, w_2, w_3$ are three coprime positive integers, and for a given polynomial $h(x,y,z)$ let $e_1(h), e_2(h), e_3(h)$ denote the degrees of $x,y,z$ in $h$ respectively. We say that $h$ is weighted homogeneous of degree $d$ with weight $(w_1, w_2, w_3)$ if $h$ satisfies $w_1 e_1(h) + w_2 e_2(h) + w_3 e_3(h) = d$. If we allow $d$ to vary across all positive integers, then the resulting set of polynomials is the set of weighted homogeneous polynomials with weight $(w_1, w_2, w_3)$. -So my question is, is there an analogue to Bezout's Theorem in this setting? That is, is there a constant $W = W(w_1, w_2, w_3)$ which depends on $w_1, w_2, w_3$ such that if $f,g$ are two weighted homogeneous polynomials with weight $(w_1, w_2, w_3)$ with no common componets, then the number of intersections of $f,g$ is bounded by $W(w_1, w_2, w_3) \deg(f) \deg(g)$? - -REPLY [19 votes]: The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$. -In the case of weighted homogeneous polynomials one needs a Bézout theorem in the weighted projective plane $\mathbb{P}^2(w_1, w_2, w_3)$. Such a result can be found, for instance, in the paper by Bartolo, Martin-Morales and Ortigas-Galindo Q-resolutions and intersection numbers, Section 5. It turns out that the intersection number of two curves of equation $f=0$ and $g=0$ in $\mathbb{P}^2(w_1, w_2, w_3)$ is given by -$$\frac{1}{w_1 w_2 w_3} \deg_{\omega}(f) \deg_{\omega}(g),$$ -where $\deg_{\omega}$ denotes the weighted degree, see also auniket's comment below. -Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a singular variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are not Cartier. Consequently, one can obtain a rational intersection number, insted of an integer one. -For instance, let us consider $\mathbb{P}(1,1,2)$, which is isomorphic to a quadric cone in $\mathbb{P}^3$. If $x, y, z$ are the weighted homogeneous coordinates, the ruling of the cone is generated by $x=0$ and $y=0$; furthermore, the intersection numbers for any two curves $L_1$, $L_2 \subset \mathbb{P}(1,1,2)$ of equation $$\lambda_1x+\mu_1y=0, \quad \lambda_2x+\mu_2y=0 \quad (\lambda_i, \mu_i \in \mathbb{C})$$ -is equal to $\frac{1 \cdot 1}{ 1 \cdot 1 \cdot 2} = \frac{1}{2}$. -This happens because a line $L$ in the ruling is not a Cartier divisor (since it passes through the vertex $[0:0:1]$) but $2L$ is Cartier, being linearly equivalent to a conic, i.e. to a hyperplane section of the cone. Now two hyperplane sections intersect in two points, so we have $(2L_1)(2L_2)=2$, that is $L_1L_2=\frac{1}{2}$.<|endoftext|> -TITLE: invariant symmetric bilinear forms and Lie algebra cohomology -QUESTION [9 upvotes]: What are the most general conditions on a Lie algebra $\mathfrak{g}$ over a field $\mathbb{k}$ such that the space of invariant symmetric bilinear forms is isomorphic to $H^3(\mathfrak{g},\mathbb{k})$? -The isomorphism should look like this: $\langle \cdot,\cdot \rangle \to \langle \cdot, [\cdot,\cdot]\rangle$ and I've managed to prove the statement for $\mathfrak{g}$ semisimple compact, but since the question is purely algebraic I don't think the "correct" proof should involve integrals over $G$. ;-) - -REPLY [7 votes]: The map $\langle\cdot,\cdot\rangle\mapsto\langle\cdot,[\cdot,\cdot]\rangle$ is usually called Koszul homomorphism. Indeed Koszul showed that for a semisimple Lie algebra over a field of characteristic zero it is an isomorphism. -For an arbitrary Lie algebra (say over a field of characteristic zero), it always maps into the space of closed 3-forms $Z^3(\mathfrak{g})$, and thus defines by composition a natural homomorphism from the space $\mathrm{Sym}^2(\mathfrak{g})^\mathfrak{g}$ of invariant symmetric bilinear forms to the 3-cohomology space $H^3(\mathfrak{g})$ (let's call it "reduced Koszul homomorphism"), and there is an exact sequence in which the reduced Koszul homomorphism fits, see Neeb-Wagemann's appendix. In particular, if $H^1(\mathfrak{g},\mathfrak{g}^*)=H^2(\mathfrak{g},\mathfrak{g}^*)=0$ then the reduced Koszul homomorphism is an isomorphism (I don't have in mind any example for which this holds other than the semisimple ones). -It is not an isomorphism in general (neither injective nor surjective), for instance for an abelian Lie algebra it is the zero map; Magnin actually checked (see here) that the Koszul homomorphism is zero for every nilpotent Lie algebra up to dimension 7. Magnin also checks that the Koszul homomorphism itself is zero in many cases (this means that the invariant symmetric bilinear forms are only the obvious ones: those symmetric bilinear forms whose kernel contains $[\mathfrak{g},\mathfrak{g}]$). -This concerns specifically this natural map; in general I don't think there is much to say: to say that the vector spaces $\mathrm{Sym}^2(\mathfrak{g})^\mathfrak{g}$ and $H^3(\mathfrak{g})$ are isomorphic just means that they have the same dimension.<|endoftext|> -TITLE: Examples of forcing arguments which require an assumption in the ground model about the sizes of the power sets? -QUESTION [6 upvotes]: I would like to know examples of forcing arguments where in order to make the cardinal arithmetic go through one needs to assume something about the size of the continuum or other power sets. -The examples that I know of are forcing to add $\aleph_2$ many Cohen reals (which requires the assumption of CH in the ground model to show that the continuum in the extension is exactly $\aleph_2$), and the forcing in Easton's Theorem (which requires the assumption of GCH in the ground model). -Are there other examples of forcing arguments where one needs to know (if only to make the arguments go more smoothly) or needs to assume the sizes of the power sets in the ground model in order to produce the desired result in the extension? I am especially interested in examples where the assumption is more exotic than CH or GCH and also examples where the forced result in the extension is about something other than the size of the continuum or the size of other power sets. -Also, I was looking at whether or not one needs to assume anything about the sizes of the power sets to show that if a partial order P satisfies the $\kappa +$-chain condition then it preserves cardinals above $\kappa$. I did not see any place where an assumption on the sizes of the power sets is required, but I wanted to check if I had overlooked some hidden detail. Along with this, I'm also interested if an assumption on the sizes of power sets is needed (or helpful) for common forcing arguments like this one. - -REPLY [9 votes]: There are some odd examples coming from PCF theory, where one begins with assumptions about pcf structures that are not known to be consistent (even if one assumes consistency of large cardinals), and then do some forcing in order to make things happen. -For example, consider the following conjecture of Shelah: -"If $\mathfrak{a}$ is a set of regular cardinals greater than $|\mathfrak{a}|$, then ${\rm pcf}(\mathfrak{a})$ cannot have a weakly inaccessible accumulation point." -An affirmative answer to the above conjecture implies -\begin{equation} -{\rm cf}(\prod{\rm pcf}\mathfrak{a}, <)={\rm cf}(\prod\mathfrak{a},<) -\end{equation} -for every set of regular cardinals $\mathfrak{a}$ with $|\mathfrak{a}|<\min\mathfrak{a}$. -On the other hand, if the above conjecture fails for some $\mathfrak{a}$, one can do some mild forcing and get the consistency of -\begin{equation} -{\rm cf}(\prod{\rm pcf}\mathfrak{a}, <)\neq{\rm cf}(\prod\mathfrak{a}, <). -\end{equation} -Shelah has many arguments of this sort: he's got a whole family of pcf conjectures that are "linearly ordered" (none of which are known to be consistent) and he uses forcing to show that many other natural questions are (essentially) equivalent to one of his basic conjectures. -For a more substantial example, consider the question of whether every compact Hausdorff space can be partitioned into two pieces, neither of which contains a copy of the Cantor set. Assuming a supercompact cardinal, Shelah is able to force the existence of counterexample, but the resulting model satisfies CH. He is able to prove, however, that the existence of such a counterexample in a model where $2^{\aleph_0}=\aleph_2$ implies some weird pcf behavior that is not known to be consistent. Moreover (and this part speaks to your original question), if one assumes there is a model where this weird pcf behavior holds, then one can force the existence of a counterexample together with the continuum being $\aleph_2$. So in this case, the topological question is essentially equivalent to a question about pcf. (This is from his paper [Sh:682].) -Edit: I know your original question asked about assumptions on sizes of power sets, but I interpreted things a little broader to include assumptions about cardinal arithmetic.<|endoftext|> -TITLE: reductive group orbits in P(V)? -QUESTION [12 upvotes]: Say $G$ is a reductive group over $\mathbb{C}$. We can take a dominant highest weight $\lambda$ and look at the action of $G$ on $X = \mathbb{P} V(\lambda)$. The stabilizer of the class of the highest weight vector is a parabolic subgroup so the orbit is isomorphic to $G/P$. What about the other orbits in $X$? If $[v] \in X - G/P$ then is there a good description of $G.[v]$ or its closure? If this is hard for general $[v] \in X - G/P$ are there any conditions you can put on $[v]$ that make it easier? Is $G/P$ the only closed orbit? -Does anyone know of references that address these questions? - -REPLY [6 votes]: To supplement the answer given by Francois, I'd emphasize that the question is essentially algebraic (over an algebraically closed field of any characteristic) in the spirit of the Borel-Chevalley structure theory. In this general setting, the older Borel notes and his second edition book, along with my book and Springer's, provide similar treatments of Borel subgroups or larger parabolic subgroups in a reductive algebraic group. Along the way it turns out that only the quotients $G/H$ with $H$ parabolic can be projective varieties (so this condition characterizes parabolics). In particular, the kind of embedding in projective space described in the question only allows an orbit to be closed (hence projective) if the isotropy group is parabolic. (In my book, this is developed in 21.3.) -Aside from that, the study of orbits and invariants in such representation-theoretic situations has been actively pursued in a lot of papers, mostly concentrating on the classical characteristic 0 setting (where for instance reductive groups behave better). Some of the influential work has been done by Hanspeter Kraft, Claudio Procesi, Gerald Schwarz, along with the Russian school: Ernest Vinberg, Vladimir Popov, Dmitri Panyushev, etc.<|endoftext|> -TITLE: Vanishing constant term in powers of a Laurent polynomial -QUESTION [8 upvotes]: This is motivated by idle curiosity. I recently learned a result of Duistermaat and Van Der Kallen in "Constant terms of powers of a Laurent polynomial" which says that: - -If the constant term of $f^n$ vanishes for all $n\in \mathbb N$, where $f\in \mathbb C[x,x^{-1}]$ is a Laurent polynomial, then either $f\in \mathbb C [x]$ or $f\in \mathbb C[x^{-1}]$. - -Their proof uses complex analysis and I was wondering if one can prove the statement over an arbitrary field of characteristic zero using algebraic or combinatorial methods. Characteristic zero is of course necessary since $x^{p-1}+\frac{1}{x}$ satisfies the conditions in characteristic $p$. - -REPLY [5 votes]: I'll give an algebraic argument, which is I think essentially the same as Duistermatt's, but substitutes partial fractions and some valuation theory for complex analysis. K will be algebraically closed of characteristic 0; f will be in K[x,1/x]. We suppose f is not in K[x] or K[1/x]. Let r and -s be the largest and smallest exponent of x appearing in f; r and s are >0. -Lemma:---Let M be a finite extension of the field of fractions of K[[t]]. Extend the obvious valuation on the field of fractions to M. Then if a is in M with f(a)=t, ord f'(a) must be < 1. (To see this note that ord a is 0. Then a has a Newton-Puiseux expansion a0 +(a1)(t^p)+..., with a0 and a1 non-zero, and p a positive rational. The derivation D=d/dt extends to M, and 1=D(f(a)) is the product of f'(a) by (a1)(p)*(t^(p-1))+... So ord f'(a)=1-p which is < 1.) -Theorem 1:---Let S be a subset of the algebraic closure of K(z). Suppose that for each a in S, f(a)=1/z. Then the sum over S of the 1/(af'(a)) cannot be z. (To see this let t=1/z. Then K(z)= K(t) which imbeds in the field of fractions of K[[t]], and we may view S as a subset of a finite extension, M, of this field. By the lemma, each 1/(af'(a)) has ord > -1. But z=1/t has ord equal to -1.) -Now let c_n be the constant term in f^n, and W=1+... be the element sigma (c_n)(z^n) of K[[z]]. Combinatorialists know that a partial fraction argument shows that W is algebraic over L=K(z). Carrying out the partial fraction argument explicitly one finds: -Theorem 2:---There are a lying in a finite extension of L with each f(a) equal to 1/z, -such that zW is the sum of the 1/(af'a)). (So by Theorem 1, W is not 1, and c_1, c_2,... cannot all be 0.) -I'll sketch a proof of Theorem 2. Let U be the element (x^s)(1-zf) of L[x]. If a is a root of u, then f(a)=1/z. So by the lemma f'(a) is not 0. Then U'(a) is not 0, and U is separable. It follows that 1/(1-fz)=(x^s)/U is a sum of c/(x-a) where a runs over the roots of f(x)=1/z. It's easy to see that the c corresponding to a given a is -1/(zf'(a)). I now refer to Lemmas 3.6 and 3.7 of my article "Generating functions attached to some infinite matrices"---see arXiv 0906.1836 or Elec. J. of Comb, v.18 (1) 2011: -If we take U1=x^s and U2=U=(x^s)(1-zf) we are in the situation of Lemma 3.7. W is "the coefficient of x^0 in the element (U1)/(U2)". In the language of the lemma, W is the -sum of the l_0(c/(z-a)). The proof of Lemma 3.6 shows that l_0(1/(z-a)) is either 0 or -1/a. -So z l_0(c/(z-a)) is either 0 or 1/(af'(a)). This completes the proof.<|endoftext|> -TITLE: algebraic proof of Atiyah-Bott fixed point formula? -QUESTION [8 upvotes]: Hi, -Atiyah and Bott apparently proved the following theorem: - -Let $X$ be a smooth projective complex variety and $L$ a line bundle on $X$. -Let $f:X\to X$ be an automorphism of $(X,L)$ with finitely many fixed points $X^f$. -Then -$$ -\sum_{i=0}^{\dim X}(-1)^itr(f, H^i(X,L)) = \sum_{x\in X^f}\frac{tr(f,L_x)}{\det(1-T_xf)} -$$ -where $T_xf : T_xX\to T_xX$ is the derivative of $f$ at $x\in X$. - -Where can one find an algebraic proof of this result? -Thanks! - -REPLY [5 votes]: I am not sure this is the best place to learn the subject, but at least this book is an algebraic reference: -Riemann-Roch algebra By William Fulton, Serge Lang -more precisely VI \S 9 Lefschetz-Riemann-Roch . You can find your formula proven for an arbitrary vector bundle (not only a line bundle) under the name "fixed point formula". The machinery behind is quite heavy, tough, there is probably a more straightforward algebraic proof. - -REPLY [4 votes]: Notice that you must assume that the graph of $f$ intersects the diagonal tranversally (otherwise some determinants in the formula might vanish). This transversality condition is automatic if $f$ has finite order. With that assumption, the above formula is a special case of the "Woods hole" formula, which is proven using Grothendieck duality in SGA 5 (Springer Lecture Notes in mathematics 589), Appendix to Exp. III, Cor. 6.12, p. 131.<|endoftext|> -TITLE: morphism from a compact group to Z ? -QUESTION [32 upvotes]: I wonder if it there exists a topological compact group $G$ (by compact, I mean Hausdorff and quasi-compact) and a non-zero group morphism -$\phi : G \to \mathbb{Z}$ (without assuming any topological condition on this morphism). -For compact Lie groups, using the exponential map, the answers is no, but in general I don't know. - -REPLY [22 votes]: Sorry for resurrecting such an old question, but I think we can give a much simpler proof here. We'll reduce the problem from $G$ to the Bohr compactification $B\mathbf{Z}$ of $\mathbf{Z}$, then from $B\mathbf{Z}$ to the profinite completion $\hat{\mathbf{Z}}=\prod_p\mathbf{Z}_p$ of $\mathbf{Z}$, and then we'll argue directly. -Let $\phi:G\to\mathbf{Z}$ be a homomorphism and fix $x\in G$. The map $\mathbf{Z}\to G$ extending $1\mapsto x$ induces a map $B\mathbf{Z}\to G$ such that $1\mapsto x$, and thus we obtain a map $\phi':B\mathbf{Z}\to \mathbf{Z}$ such that $\phi'(1)=\phi(x)$. -Recall that to construct $B\mathbf{Z}$ one takes the dual of $\mathbf{Z}$, namely $\mathbf{R}/\mathbf{Z}$, strips the topology to get the discrete group -$\mathbf{R}_d/\mathbf{Z}\cong\mathbf{R}_d\times\mathbf{Q}/\mathbf{Z}$, then takes the dual again. The result is that $B\mathbf{Z} \cong B\mathbf{R}\times\hat{\mathbf{Z}}$. Since $B\mathbf{R}$ is divisible $\phi'$ must vanish on $B\mathbf{R}$. Since $\prod_{p\neq 2}\mathbf{Z}_p$ is infinitely $2$-divisibile and $\mathbf{Z}_2$ is infinitely $3$-divisible, $\phi'$ vanishes on $\hat{\mathbf{Z}}$. Thus $\phi'$ is identically $0$, so $\phi(x)=\phi'(1)=0$.<|endoftext|> -TITLE: Is the extension of the Abel-Jacobi map to the smooth locus of the minimal regular model of a curve an immersion? -QUESTION [8 upvotes]: Let $S$ be the spectrum of a discrete valuation ring with generic point $\eta$. Let $C/\eta$ be a smooth connected curve with an $\eta$-valued point, and let $\mathcal{C}/S$ be the smooth locus of the minimal proper regular model of $C$ over $S$. Let $N/S$ denote the Neron model of the Jacobian of $C_\eta$, and let $\alpha:C \rightarrow N_\eta$ denote the Abel-Jacobi map. -Now by the Neron mapping property, we obtain a (unique) extension of $\alpha$ to $\overline{\alpha}:\mathcal{C} \rightarrow N$. -Question: -Is the map $\overline{\alpha}$ necessarily an immersion? -If the genus is 1, we are OK! In general, I cannot think of a reason why this should be true, but I also cannot think of a counterexample. -I started thinking about this question from the point of view of Serre's book `Algebraic groups and class fields', ie as a question about line bundles on singular curves. I wanted to use such a result in my thesis, but in the end found an alternative approach. However, I am still curious... -Thank you for your time. - -REPLY [10 votes]: See my article: -http://arxiv.org/abs/math/9806173 -Best regards, Bas Edixhoven. -P.S. Thanks to Liu Qing who drew my attention to this.<|endoftext|> -TITLE: Question on geometric measure theory -QUESTION [16 upvotes]: I want to know the following is well-known or not: -Let X be a metric space with Hausdorff dimension $\alpha$. -Then for any $\beta < \alpha$, -X contains a closed subset whose Hausdorff dimension is $\beta$. - -REPLY [18 votes]: Let's do the case of complete metric space. Let $X$ be a complete metric space with Hausdorff dimension $\alpha < \infty$. Then of course $X$ is separable, as well. -We use a result of Howroyd [2] (following Marstrand [1] who did the real line). Let $0 < \beta < \alpha$. Then $H^\beta(X) = \infty$, the $\beta$-dimensional Hausdorff measure. By Howroyd's theorem ($H^\beta$ is semifinite), there is a Borel subset $A \subset X$ with $0 < H^\beta(A) < \infty$. Then since a finite Borel measure is regular, there is a Cantor set $B \subseteq A$ with $0 < H^\beta(B) < \infty$, so of course $B$ has Hausdorff dimension $\beta$. - -J. M. Marstrand, "The dimension of Cartesian product sets." Proc. Cambridge, Philos. Soc. 50 (1954) 198--202 -J. Howroyd, "On dimension and the existence of sets of finite positive Hausdorff measure." Proc. London Math. Soc. 70 (1995) 581--604<|endoftext|> -TITLE: A specific degeneration of a rank 2 bundle -QUESTION [5 upvotes]: I wish to know if there is a rank 2 vector bundle $E$ on $\mathbb{P}^1 \times \mathbb{P}^1$ such that $\mathbb{P}(E)$ when restricted to $\mathbb{P}^1 \times [0:1]$ is the $n$th Hirzebruch surface and when restricted to $\mathbb{P}^1 \times [x:y]$ is the $(n-2)$th Hirzebruch surface. - -REPLY [9 votes]: Let $F = O \oplus O(n-2,0)$ and denote the line $P^1\times[0:1]$ by $L$. Note that $F_{|L} = O \oplus O(n-2)$. Consider any surjective map $O_L \oplus O_L(n-2) \to O_L(n-1)$ (for example the one given by $u^{n-1}$ on the first summand and by $v$ on the second, where $(u:v)$ are the homogeneous coordinates on $L$). Consider the composition $F \to F_{|L} \to O_L(n-1)$ and let $E$ be its kernel. Note that for any line $L' = P^1\times[x:y]$ we have $E_{|L'} = F_{|L'} = O \oplus O(n-2)$. On the other hand, restricting to $L$ we obtain an exact quadruple -$$ -0 \to L_1i^*O_L(n-1) \to E_{|L} \to O_L \oplus O_L(n-2) \to O_L(n-1) \to 0 -$$ -where $i:L \to P^1\times P^1$ is the embedding. -Note that $L_1i^*O_L(n-1) = O_L(n-1)$, and the kernel of the rightmost map is $O_L(-1)$. Hence we have an exact triple -$$ -0 \to O_L(n-1) \to E_{|L} \to O_L(-1) \to 0. -$$ -Since there are no notrivial extensions, we see that $E_{|L} = O_L(-1) \oplus O_L(n-1)$. -So, the bundle $E$ gives what you need.<|endoftext|> -TITLE: Near points in several arithmetic progressions -QUESTION [11 upvotes]: We have 3 arithmetic progressions $na_1 + c_1, na_2 + c_2, na_3 + c_3$ with the reals $a_1,a_2,a_3$ linearly independent over the rationals. Is it always true that given $\epsilon > 0$ there are terms $x_1,x_2,x_3$, one from each progression, such that all three are within $\epsilon$ of each other ? -Same problem with any finite number of progressions. - -REPLY [3 votes]: For given $(a_1, a_2, a_3)$ and $(c_1, c_2 , c_3)$, the required condition is equivalent to: -$(c_2-c_1, c_3-c_2)$ is in the closure $\Gamma $ of the additive subgroup generated by $(a_1,0)$, $(-a_2,a_2)$, $(0,-a_3)$. But the latter need not be dense, even if $(a_1, a_2, a_3)$ are linearly independent over the rationals; for instance $(a_1, a_2, a_3):=(1+\lambda,1,1+1/\lambda)$ with $\lambda$ an irrational non quadratic, is a linear independent triple for which $\Gamma$ is a family of lines $\{(x,y)\, : \, y=\lambda x + n(\lambda+1)\, , n\in\mathbb{N}\}$.<|endoftext|> -TITLE: Stallings fibration theorem -QUESTION [6 upvotes]: Stallings' fibration theorem states that if we have a compact irreducible $3$-manifold $M^3,$ with -$G\rightarrow \pi_1(M^3) \rightarrow \mathbb{Z},$ and $G$ is finitely generated and is not of order $2,$ then $M^3$ fibers over a circle. The question is whether the last condition (that $G\neq \mathbb{Z}/2\mathbb{Z}$) is actually necessary (the answer is probably in Stallings' original paper, but I can't find it online). - -REPLY [7 votes]: This should follow from geometrization. The fundamental group of the manifold is $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}$, and geometrization should tell you that the manifold is then $\mathrm{RP}^2 \times S^1$. -I looked in Stallings's paper, and he says that it is a hard open problem, so this might be the only way.<|endoftext|> -TITLE: Groups with no bounds on the size of abelian subgroups without infinite ones -QUESTION [9 upvotes]: Is there a group $G$ in which every abelian subgroup is finite and there is no upper bound on the sizes of its abelian subgroups? -Let me say that a counterpart of the question above has posed by Paul Erd\"os in 1975 as follows: -Is there a group $G$ in which every subset consisting of pairwise non-commuting elements is finite and there is no upper bound on the sizes of such subsets? -The asnwer was given by B. H. Neumann in 1976 in negative. It was proved that groups $G$ satisfying the first assumption of the latter question must be center-by-finite and so the index of the center is a upper bound for sizes of subsets consisting of pairwise non-commuting elements. - -REPLY [10 votes]: Yes, such groups exist. -Consider the disjoint union of cyclic groups of odd order -$\mathcal{C}:=\{ \mathbb{Z}/(2n+1)\mathbb{Z} \mid n \in \mathbb{N} \}$. By a theorem of A. Ol'shanskii (see Thm 35.1 from his book "Geometry of defining relations in groups"), there exists an odd integer $n_0$ for which this collection of groups can be embedded into a countable $2$-generated simple group $G$ such that every proper subgroup of $G$ either is cyclic of order dividing $n_0$ or is conjugate to a subgroup of a group from $\mathcal C$. -Clearly $G$ is non-abelian, but every proper subgroup of $G$ is cyclic of finite odd order. Since each group from $\mathcal C$ is a subgroup of $G$, one sees that there is no bound on the orders of the cyclic subgroups. - -REPLY [7 votes]: One can repeat the construction of the free Burnside group from Olshanskii "Geometry of defining relations in groups" imposing relations $w^{n_w}=1$ where $n_w\gg 1$ odd, depending on $w$ instead of $w^n=1$. All the auxiliary properties of the free Burnside groups except finite exponent will hold. In particular all Abelian subgroups of it will be cyclic, but the orders of them will be unbounded. The proof carries verbatim. I am pretty sure somebody (a student of Olshanskii) has done it explicitly already: one can construct torsion groups satisfying many "strange" properties. A more general way is to use lacunary hyperbolic groups -(see our paper with Olshanskii and Osin in arXiv). There are many torsion lacunary hyperbolic groups, all Abelian subgroups there are cyclic but their orders unbounded.<|endoftext|> -TITLE: $\mathbb{Z}/2$ is to $\mathbb{Z}/3$ as K3 is to what? -QUESTION [14 upvotes]: I'd like to know "what" (say, in the classification of complex surfaces) the following complex manifold $X$ is: -Construction: Let $\Lambda$ be the hexagonal lattice in $\mathbb{C}$; that is, the lattice generated by $(1, \omega)$ where $\omega=e^{2i\pi/3}$ is a third root of unity. -Observe that the lattice $\Lambda^2\subseteq \mathbb{C}^2$ is invariant under the $\mathbb{Z}/3\subseteq SU(2)$ action -$\begin{pmatrix} -\omega & 0 \cr -0 & \omega^2 -\end{pmatrix}$. -I'm curious about the complex manifold $X$ obtained by quotienting $\mathbb{C}^2/\Lambda^2$ by $\mathbb{Z}/3$, and then blowing up at the $9=3^2$ singular points. -Remarks: - -Doing this with $\mathbb{Z}/4\subseteq SU(2)$ instead of $\mathbb{Z}/3$, and the square (i.e., generated by $(1, i)$) lattice rather than the hexagonal lattice, and the resulting $4=2^2$ singular points, is something I'm equally curious about and equally unable to answer. -Doing this with $\mathbb{Z}/2$ instead of $\mathbb{Z}/3$, and any lattice in $\mathbb{C}^2$ at all (since all are invariant under this action), and the resulting $16=4^2$ singular points, gives a K3 surface; this was my motivation for the question. -The exceptional divisor at each of the 9 blowups is a pair of $\mathbb{P}^1$'s intersecting at a point. - -REPLY [13 votes]: The complex manifold $X$ is a $K3$ surface with $9$ singular points of type $\frac{1}{3}(1,2)$, i.e. rational double points of type $A_2$. -In fact, let us denote by $A$ the abelian surface $\mathbb{C}^2 / \Lambda ^2$ and by $\pi \colon A \longrightarrow X$ the natural projection. Then the action of $\mathbb{Z}/3 \mathbb{Z}$ on $A$ is locally given by $$ \omega \cdot (x, y)=(\omega x, \omega^2 y).$$ Therefore the $1$-forms $dx$, $dy$, which give a basis of $H^0(\Omega^1_A)$, are not invariant, whereas the $2$-form $dx \wedge dy$, which gives a basis of $H^0(\Omega^2_A)$, is invariant. This means that $$p_g(X)=1, \quad q(X)=0.$$ -Moreover, since $X$ has only rational double points, one can write $$K_A=\pi^*K_X,$$ which implies $K_X^2=0$. Moreover $K_A$ nef implies $K_X$ nef, hence $X$ is a minimal model. -By Enriques-Kodaira classification it follows that $X$ is a $K3$ surface. -The case with group $\mathbb{Z}/4 \mathbb{Z}$ is similar: the quotient is a $K3$ surface whose singularities are $4$ points of type $\frac{1}{4}(1, 3)$, i.e. rational double points of type $A_3$. The minimal resolution of each of them is a tree made of $3$ smooth rational curves, each having self-intersection $(-2)$. -Remark. I called $X$ the quotient of $A$ by the finite group, without blowing up the singularities. Since $X$ has only rational double points, -this does not make any difference in the computation of invariants.<|endoftext|> -TITLE: A matrix algebra has no deformations? -QUESTION [22 upvotes]: I have often heard the slogan that "a matrix algebra has no deformations," and I am trying to understand precisely what that means. While I would be happy with more general statements about finite-dimensional semisimple algebras over non-necessarily algebraically closed fields, I am mainly interested in the case when the base field is $\mathbb{C}$, so the question is phrased in terms of matrix algebras over $\mathbb{C}$. I will state the question in three different contexts, in increasing order of interest to me, and say what I know about each one. -Formal Deformations -A formal deformation of an associative $k$-algebra $A$ is a $k[[h]]$-bilinear multiplication on $A[[h]]$ of the form -$$ a \cdot b = ab + m_1(a,b)h + m_2(a,b)h^2 + \dots $$ -where the first term is given by the original multiplication on $A$. This type of deformation is connected with the Hochschild cohomology $HH^\bullet(A)$. As I understand it, for a semisimple algebra $A$, the Hochschild cohomology vanishes in degree $\ge 1$, and this implies that a matrix algebra has no deformations in the formal sense. So I guess there isn't much of a question so far. -Deformation of Structure Constants -In this context, we fix a natural number $n$ and a vector space $V$ with a fixed basis $a_1, \dots, a_n$, and we consider associative algebra structures on $V$. The structure constants of an algebra $A$ with underlying vector space $V$ with respect to this basis are the complex numbers $c_{ij}^k$ such that -$$ a_i a_j = \sum_{k}c_{ij}^k a_k, $$ -and these numbers clearly determine the algebra. This gives us a point in $\mathbb{C}^{n^3}$. Associativity gives us some polynomial constraints on the structure constants, so we can think of the collection of $n$-dimensional associative algebras as some kind of algebraic variety in $\mathbb{C}^{n^3}$. (I know very little algebraic geometry, so please excuse/correct me if am using the wrong terminology.) Since one can take many different bases for an algebra, this variety overcounts associative algebras. -I don't know how this is proved, but as I understand things, semisimplicity is a Zariski-open condition in the variety of associative algebra structures, so semisimplicity is preserved under small deformations, and so a small deformation of $M_n(\mathbb{C})$ is still semisimple. -Question 1 - -Is the property of being a matrix algebra preserved under deformation of structure constants? - -Deformation of Relations -This is the most flexible setting, and the one in which I am most interested. Here we consider algebras with a fixed collection of generators and a fixed number of relations, and we allow the relations to vary smoothly. This allows the dimension of the algebra to vary as well. However, this is also the hardest notion to formalize. -An example will illustrate what I mean. Consider the family of algebras -$$ A_t = \mathbb{C}[x]/(tx^2 -x),$$ -where $t$ varies in $\mathbb{C}$. Clearly there is a drop in dimension at $t=0$, although generically the dimension is constant. A slightly more complicated example is given by the family $B_t$, where -$$ B_t = \mathbb{C} \langle x,y \mid x^2 = y^2 = 0, \quad xy + yx = 2t \rangle.$$ -In this example the dimension is constant, but at $t=0$ one has the exterior algebra $\Lambda(\mathbb{C}^2)$, while for $t \neq 0$ this is the Clifford algebra of $\mathbb{C}^2$ with respect to a nondegenerate bilinear form, and hence $B_t \simeq M_2(\mathbb{C})$ for $t \neq 0$. -Question 2 - -What is a good way to formalize this notion of deformation? - -A first guess would be something like this: fix a space of generators $V$, a number of relations $k$, and a base manifold $M$ (in my above examples $M=\mathbb{C}$). To limit things a bit, I guess I would like to consider only quadratic-linear relations. Allowing these relations to vary means giving a smooth function -$$F : M \to \mathrm{Gr}_k(\mathbb{C} \oplus V \oplus V^{\otimes 2})$$ -and considering the "bundle" (sheaf?) of algebras -$$ A_p = T(V)/ \langle F(p)\rangle, \quad p \in M $$ -over $M$. -However, I'm not sure if that completely captures what I want. I might also insist that the projection of each $F(p)$ onto $V^{\otimes 2}$ has no kernel (which is not the case for the first example I described, but is the case for the second). -Question 3 - -Given the setup just described (or an appropriately modified version of your choosing), what are some conditions on $M$, $F$ such that the property "$A_p$ is a matrix algebra" is local on $M$? - -For this question I am mainly interested in the cases when $M = \mathbb{R}$ or $M = \mathbb{C}$, although more general base spaces are of course interesting as well. -I have heard about something called the Azumaya locus which may be related to this, but I don't really know anything beyond the name. -I didn't realize this would be so long! If you're still with me, thanks for reading. - -REPLY [2 votes]: This isn't an exact answer to the question, but it might be of interest. The algebra of matrices might not have interesting deformations, but it does seem to have an interesting degeneration. More precisely, there is an algebra $A_t$ with $A_t \cong M_n(\mathbb C)$ when $t \not= 0$ such that $A_{t=0}$ is an interesting algebra. (See sections 1.1 and 2.3 of this paper by Allen Knutson and Paul Zinn-Justin.)<|endoftext|> -TITLE: Max-flow / min-cut and domino height function -QUESTION [8 upvotes]: In Conway's Tiling Groups, Bill Thurston shows how to get height functions for domino groups in two different settings. -Domino: $\langle x, y \, \mid x^2 y x^{-2}y^{-2}= y^2xy^{-2}x^{-1}=1 \rangle$ -Lozenge: $\langle a,b,c \, \mid aba^{-1}b^{-1} = bcb^{-1}c^{-1} = cac^{-1}a^{-1} = 1 \rangle $ -In both cases, there's an central extension sequence $1 \to \mathbb{Z} \to G \to \mathbb{Z}^2 \to 1 $. -Thurston says height function falls out of the Max-Flow Min-cut theorem (an explanation for max flow, if I understood that better I'd explain it here). -http://www.mccme.ru/~vadicgor/tilings_surface.png -I had trouble explaining how to get the height function f . The paper says you have to measure lengths of paths from one point on your boundary $\partial R$ to another by steps of positive elements. So $abc$ has height 3 while $aba^{-1}b^{-1}$ has height 0. What does positivity mean here with tiling groups and what does it have to do with max-flow/min-cut? -There seem to be another explanations for the existence of height functions related by cube complexes in geometric group theory. Where the height function is determined by the minimal surface bounding by a curve $\partial R$ . Here Thurston talks about a "contractible 3-manifold, but what do those surfaces look like in the case of dominos? - -So I guess I have a couple of questions. What is positivity doing here in the case of tiling groups? In the domino case what are the contractible 3-manifolds? How can we get these from minimal surfaces bounded by the region $\partial R$? -Mainly I'm looking for a clarification of Thurston's (brief) discussion of height functions. One coming from group theory and the other from discrete geometry. - -EDIT: I added the height function to a random tiling by dominos. Put various projections. -alt text http://www.freeimagehosting.net/newuploads/7bf09.gif - -REPLY [6 votes]: The answer is particularly nice for lozenge or domino tilings. There is a very elegant picture of "the space of tilings" of a certain region which makes the connection between height functions, max-flow-min-cut, the topology of the region and the respective Cayley graph very clear. I'm going to try to present the details here, but I'll also include some references in the end. It also helps a lot if you work out some examples by hand. -The main problem is, of course, being able to tell if a certain region can be tiled. If we restrict ourselves for a moment to simply connected regions, Conway-Lagarias gave a necessary condition which says that the word corresponding to the boundary should be the identity in the tiling group $G$. Let the Cayley graph of $G$ be $\Gamma$, which we can embedd as a lattice in $\mathbb R^3$ for lozenge or dominos. Now, the tilings themselves live in a plane region which is trianguated or quadriculated, and we can orient the edges of this triangulation/quadrangulation so it becomes $\Gamma_1$, the Cayley graph of $G_1$, which is a quotient of $G$ ($G_1$ is isomorphic to $\mathbb Z^2$ in both cases, but for lozenges we consider three generators, $(1,0),(0,1),(-1,-1)$). In both cases we can make the embeddings so that the projection $\mathbb R^3\to\mathbb R^2$ also projects $\Gamma\to \Gamma_1$. -Now a graph is a $1$-complex and once we introduce a tiling this turns the tiled region into a $2$-complex in the obvious way. This complex lifts up to translation to a unique surface in $\mathbb R^3$. Therefore tilings are in bijection with such surfaces up to translation. The height function for a tiling is just defined as the distance between a vertex of $\Gamma_1$ and the point on the surface which projects to it. Usually one fixes a point on the boundary to have height zero in order for the values at the other vertices to be well defined. -More concretely, for lozenges we have -$$G=\langle a,b,c \, | aba^{-1}b^{-1} = bcb^{-1}c^{-1} = cac^{-1}a^{-1} = 1 \rangle \simeq \mathbb Z^3,$$ -so $\Gamma$ is just the cubic lattice with positive directions given by $(0,0,1),(0,1,0)(1,0,0)$. The projection is on the $x+y+z=0$ plane so that $\Gamma_1$ is the skeleton of the triangular lattice with the corresponding edge orientations. Tilings of a region will correspond to a surface on the cubic lattice with fixed boundary. Notice that the height function as defined above increases by a unit when crossing an edge in the positive direction (Cayley graphs are oriented) and decreases by a unit otherwise. -For dominoes the picture is slightly different because the tiling group -$$G=\langle a, b \, | a^2 b a^{-2}b^{-1}= b^2ab^{-2}a^{-1}=1 \rangle$$ -is non-abelian. To embed it in $\mathbb R^3$ introduce a third generator $c=[a,b]$ so that -$$G=\langle a, b,c \, |ac=c^{-1}a,bc=c^{-1}b,ab=bac\rangle.$$ -Now every element in $G$ is in bijection with a word $c^wb^ya^x$. We represent this word as a vertex in $\Gamma$ by the lattice point $(x,y,z)$ where $z=4w$ if $x$ and $y$ are even, $z=4w+1$ if $x$ is odd and $y$ is even, $z=4w-2$ if $x$ and $y$ are odd and $z=4w-1$ if $x$ is even $y$ is odd. The projection of $\Gamma$ on the plane $z=0$ gives a square grid, and the tilings lift to surfaces with a fixed boundary, therefore defining a unique height function up to translation. Again the height function can be described locally by the difference at neighbouring vertices which will depend on the orientation of the edges in $\Gamma$. -Now since we are talking about simply connected domains, we can prove something nice from this picture. We can define some local moves (generally called flips) which in the case of lozenges switch the "full box" diagram with the "empty box" diagram, and for dominoes they switch two parallel horizontal dominoes with two vertical ones. If we assign to such moves a $3$-cell, this turns the collection of tilings into a contractible $3$-complex and therefore we see that one can transform any tiling into any other one just by a sequence of local moves. In fact with a little more work one can show that tilings form a distributive lattice where the minimum of two tilings is given by the lower convex hull of the two corresponding surfaces. -Now, what remains is the question of what happens with non-simply connected regions, and what does max-flow-min-cut have to do with all this? -Well, first we want to be able to tell as quickly as possible if a region can be tiled. This is what Thurston's algorithm does, and if the answer is yes it gives the minimal tiling in the sense described above. For the algorithm check the references below. A brief description is that one starts by building a candidate height function for a minimal surface inductively, by specifying the boundary first, and then assigning values to interior vertices greedily. There will be no tiling possible if we meet an inconsistency along the way. -The max-flow-min-cut theorem provides a characterization of height functions on the boundary which extend to the entire region, while at the same time giving some insight as to how one might naturally discover Thurston's algorithm. A height function defines a flow on the dual graph of $\Gamma_1$, while a cut in this dual graph corresponds to a directed path in $\Gamma_1$ joining two vertices on the boundary. We can define $d(u,v)$ to be a non-symmetric distance function, measuring the length of the minimum directed path from $u$ to $v$ in $\Gamma_1$. Max-flow-min-cut for the dual graph can be rephrased into a statement characterizing valid height functions: - -Theorem Given a simply-connected region $R$ in the square lattice. There exists a tiling if and only if there is a function $h$ on the lattice points of $R$ which satisfies $h(u)-h(v)\le d(u,v)$ for all $u,v$ on the boundary. - -This is also incidentally the condition under which Thurston's algorithm gives a positive answer. A similar statement will hold for lozenge tilings. A detailed discussion is given in this senior thesis of Matina Donaldson. All these statements have analogues for surfaces with arbitrary topology, but then one must consider height sections of a certain height bundle on the graph. Using the same techniques one can compute the number of connected components of the tiling space. For this general setting you can look at "Spaces of domino tilings" by Saldanha, Tomei, Casarin and Romualdo. The first two authors have another nice note "An overview of domino and lozenge tilings".<|endoftext|> -TITLE: Reference for the derived category of $X$, $[X/G]$ and $X/G$ -QUESTION [11 upvotes]: I'm trying to learn about derived categories of algebraic stacks. To be honest, as of now, I don't need anything fancy nor deep. In my setup I have a scheme $X$ (well, a smooth and projective variety over $\mathbb{C}$) and a finite group $G$ acting on it. I would like to understand simple things like: what does a coherent sheaf on $[X/G]$ look like? what do the (derived) pushforward functors between the various spaces do? and the pullbacks? -Any pointers would be greatly appreciated! - -REPLY [3 votes]: In case anyone is interested, I've been finding this thesis useful -http://tobias-lib.uni-tuebingen.de/volltexte/2007/2941/pdf/diss.pdf<|endoftext|> -TITLE: sections of morphisms of complex spaces -QUESTION [5 upvotes]: A smooth morphism of schemes $f: X \to Y$ admits an étale-local section through any point $x \in X$. -One might wonder if this fact is true in the more general context of complex spaces (i.e. things glued from analytic subsets of polydiscs much like algebraic varieties are glued from affine varieties). -A map $f: X \to Y$ between complex spaces is flat at $x \in X$ if $\mathcal O_{X,x}$ is flat as a $\mathcal O_{Y,f(x)}$-module, where $\mathcal{O}_{X,x}$ is the ring of germs of analytic functions at $x$. Then étale is flat (at all the points of the domain) and unramified. [One can find this definition in Grauert-Peternell-Remmert Several Complex Variables VII, Encyclopedia of Mathematical Sciences, vol 74] -It can't be true stated as it is though. Consider the Hopf surface: the quotient of $\mathbb{C}^2 - (0,0)$ by the action of $\mathbb Z$: $z \mapsto (1/2 z, 1/2 z)$. The map $(x,y) \mapsto (x : y)$ gives a projection $H \to \mathbb{P}^1$ with fibres elliptic curves. It can't have a meromorphic section because by composing with a projection to one of the axes one would get a meromorphic map from $\mathbb{P}^1$ to an elliptic curve,and since $\mathbb{P}^1$ is étale simply connected étale base change won't help. (I am not sure why taking étale base extensions won't help, but it seems to be the case) -I am going to be naive and arbitrary now and suppose that the reason that this counterexample works is that the fibres are "complicated". What if the fibres are, say, $\mathbb{C}^n$? -So my question is: given a smooth morphism $f: X \to Y$ of complex spaces such that all its fibres are isomorphic to $\mathbb{C}^n$, and a point $x \in X$, is it true that $f$ admits a section through $x$ defined on some Zariski open neighbourhood of $f(x)$, perhaps after an étale base change? -Do you have in mind a broader natural class of morphisms for which this statemnt works? -Update: This is actually a reminiscence of this old question. -I am intrested in families of vector spaces definable in the structure -of compact complex spaces (the latter are being extensively studied by -Rahim Moosa, see his survey "Model theory and complex geometry"). This means that I want to look at the following set of -objects: definable sets $X$ and $Y$ and a definable map $p: X \to Y$, -the maps $+: X \times X \to X$ and $\cdot: \mathbb{C} \times X \to X$ -and a section of $p$, $0: Y \to X$ such that restriction of these -maps on each fibre of $p$ define a vector space structure on it. Now a few words about what definable means. -A definable set is a constructible subset (in the analytic Zariski -topology) of a compact complex space. A meromorphic map between -complex spaces $X$ and $Y$ is an analytic subset $\Gamma \subset X -\times Y$ such that the projection on the first coordinate is onto and -is a biholomorphic map outside some proper analytic subset of -$X$. Note that this is not the same as just holomorphic map defined on -an open subset of $X$ (exponent is a holomorphic map from $\mathbb{C} -\supset \mathbb{P}^1$ to $\mathbb{P}^1$, but is not meromorphic, since -it's graph is not an analytic subset of $\mathbb{P}^1\times -\mathbb{P}^1$). A definable map is a piecewise meromorphic map, -meaning that there is a cover $\cup U_i = X$ and on each $U_i$ the map -coincides with a meromorphic map $\overline{U_i} \to Y$ for some -compact $\overline{U_i}$ into which $U_i$ embeds. -I want to prove that given a definable family of vector spaces $p: X -\to Y$ with fibres of constant dimension there is an analytic Zariski - open $U \subset X$ and a piecewise -meromorphic map $X|_U \to U \times \mathbb{C}^n$. Analytic Zariski -topology is like Zariski topology for algebraic varieties: analytic -subsets are closed sets. -Now to preserve sanity I will suppose for a momoent that $X$ is not -constructible but just an open subset of some compact complex analytic -space. -It seems that it is necessary first to be able to take analytic - Zariski local sections. The reason why I want to work with analytic - Zariski topology is that I need to produce a (piecewise) - meromorphic map, and seems very hard to construct one locally in the - finer complex topology - by looking at a local piece you don't know if it will extend to a meromorphic map on the whole domain. -The example with the Hopf surface shows that analytic Zariski local -sections are not always possible. I still hope that they are possible -in my restricted case (the fibres are something like $\mathbb{C}^n$), -maybe after an étale base extension. - -REPLY [3 votes]: I meant to write this as a comment, but it won't fit. -In my opinion, you are confusing the étale analytic with the étale algebraic topology. The étale analytic topology is essentially the same the same as the classical topology. The étale topology is meant to be a substitute for the classical topology over more general fields; but it only sees the "finite" part of the topology. -If you have a smooth morphism in the analytic category, it has local sections in the classical topology. The étale topology was introduced precisely to fix the fact that this is false for algebraic morphisms, if you interpret "local sections" as "local sections in the étale topology". In other words, it was introduced to make some version of the implicit function theorem work. -But you are doing the following. You have a smooth morphism of analytic manifolds $f \colon X \to Y$, where $Y$ has an additional structure of an algebraic variety, and you are asking local sections $V \to X$, where $V \to Y$ is an algebraic étale map. Again, this does not make any sense to me. Perhaps if you were to include some motivation for your question you might convince me that it is meaningful.<|endoftext|> -TITLE: Is the space of Hankel operators complemented in B(H)? -QUESTION [18 upvotes]: Let $H$ be $\ell^2({\mathbb N})$ and let $S:H\to H$ be the unilateral forward shift, so that $S^*S=I\neq SS^*$. Then a bounded operator $T:H\to H$ is Hankel if and only if it satisfies $TS=S^*T$. -Let $V$ be the space of all bounded Hankel operators on $H$. Does there exists a bounded linear projection from $B(H)$ onto $V$? -I thought I'd seen something on this question mentioned in Nikolskii's books on operator theory, but am currently having no luck finding anything on MathSciNet, nor via Google. The only thing I can find is a 1980 survey article by Steve Power, in which he remarks that it is "unlikely" that $V$ is complemented in $B(H)$; but there the trail appears to go cold. Certainly, the naive attempt to construct a projection by "averaging" doesn't seem to work. -Perhaps this question is known to be open? It feels like something that people would have worked on, since there are various negative results showing that certain other natural subspaces of $B(H)$ are uncomplemented. - -REPLY [22 votes]: The answer is no: there is no bounded projection from $B(H)$ onto $V$. For a proof, see for example Theorem 5.12 in Peller's book Hankel operators and their applications. -If you replace $B(H)$ by the Schatten class $S^p$ with $1\leq p <\infty$, the answer becomes yes. For $12$, and $1/\sqrt{p-1}$ if $p<2$). There is an interesting phenomenon at $p=1$: the natural projection is no longer bounded, but there is another one which is bounded. This is explained in chapter 5 of Peller's book.<|endoftext|> -TITLE: Coincidences amongst classifying spaces and Eilenberg Mac-Lane spaces -QUESTION [13 upvotes]: Given that $$\mathbb{R}P^{\infty} = B O(1) = K(\widehat{O(1)}, 1)$$ $$\mathbb{C} P^{\infty} = B U(1) = K( \widehat{U(1)}, 2)$$ is there any way to make sense of $$\mathbb{H}P^{\infty} = B Sp(1)$$ in a similar manner using the representation theory of the non-abelian group $Sp(1) \cong Spin(3) \cong SU(2)$? - -REPLY [4 votes]: This is not really an answer to the question posed but seems to be of relevance to people interested in the question (and is directly related to the case $BSU(2)\cong_{\mathbb{Q}}K(\mathbb{Z},4)$ mentioned by Mark Grant). There is a sequence of groups for which the classifying spaces are rationally products of Eilenberg-Maclane spaces: namely $BU(n)$. The $i$-th Chern class is an element of $H^{2i}(X,\mathbb{Z})$ and hence can be thought of as homotopy class of map to a $K(\mathbb{Z},2i)$ space. Therefore you get a map -$$c_1\times\cdots\times c_n\colon BU(n)\to \prod_{i=1}^nK(\mathbb{Z},2i)$$ -which turns out to be a rational homotopy equivalence. I learned this trick from Atiyah & Bott (http://www.jstor.org/stable/10.2307/37156), Section 2. I guess the same should work for $SU(n)$ when you leave out the $c_1$ factor.<|endoftext|> -TITLE: Higher dimensional Heegaard splittings? -QUESTION [10 upvotes]: Smooth (closed, connected, orientable) 3-dimensional manifolds are very special, in that for any 3-manifold $M$ there are two handlebodies, $V$ and $W$, of genus $g$ and an orientation reversing homeomorphism $f$ of their boundaries so that $M=V\ \cup_f W$. Such a decomposition is called a Heegaard splitting. -I want to know: Does this kind of symmetric handlebody decomposition extend into higher dimensions? More specifically, given an $n=2k+1$ manifold $M$, can we construct a $V$ and $W$ by attaching handles $D^i \times \small{D^{n-i}}\ (i\leq k)$ to an $n$-disk, and find an orientation reversing homeomorphism $f:\partial V\rightarrow\partial W$ so that $M=V\ \cup_f W$? Since $f$ is only continuous here, it might not provide a unique smooth structure for $M$; could we remedy this by requiring $f$ to be a diffeomorphism instead? After all, any exotic sphere $\Sigma\in\Theta_n$ can be constructed by gluing two copies of $D^n$ together with an orientation reversing diffeomorphism of the boundary (except possibly $n=4$?). - -REPLY [3 votes]: Any closed connected n-manifold admits a Morse function f with one critical point of index zero and one critical point of index n (see e.g. Matsumoto's "Introduction to Morse Theory", Theorem 3.35). If n is odd, you can slide all handles of index $< \frac{n}{2}$ to below $f^{-1}(\frac{1}{2})$ and all handles of index $> \frac{n}{2}$ to above $f^{-1}(\frac{1}{2})$, and "split along" the connected $n-1$-manifold $f^{-1}(\frac{1}{2})$. -Parenthetically, $4$--manifolds do have their own concepts of Heegaard diagrams. Namely, the union N of the 0 handle with all the 1-handles, together with frames closed curves on the boundary $\partial N$ along which we are to attach 2-handles. See e.g. Section 5.3 in Matsumoto's book for details.<|endoftext|> -TITLE: Classification of countable subgroups of the circle -QUESTION [8 upvotes]: Is there a classification of all countable subgroups of the circle $\mathbb{T} \simeq \mathbb{R}/\mathbb{Z}$? -It seems that there are quite a lot of them, e.g.: - -cyclic subgroups $\{a^n\colon n\in\mathbb{Z}\}$ -finite subgroups -subgroups of the form $\{k/2^n\colon k,n\in \mathbb{Z}\}$ or something similar -direct sums of the above... - -Equivalently, is there a classification of all compact monothetic groups? (Each countable subgroup of the circle can be realized as a group of eigenvalues of some probability-preserving transformation and vice versa). - -REPLY [8 votes]: "subgroups of the form $\{k/2n:k,n∈\mathbb{Z}\}$ or something similar [and] direct sums of the above..." -Are precisely the class of divisible subgroups. See http://en.wikipedia.org/wiki/Divisible_group for a full discussion and classification. Divisible groups are well-understood, and are always direct summands. As $\mathbb{R}/\mathbb{Z}$ is divisible, we can say a lot about its structure. -$\mathbb{Q}/\mathbb{Z}$ will be the torsion subgroup, which is the direct sum of all prufer groups. This contains all subgroups of the last three types which you list. -As $\mathbb{Q}/\mathbb{Z}$ is divisible, $\mathbb{R}/\mathbb{Z}$ is isomorphic to $\mathbb{R}/\mathbb{Q}\oplus \mathbb{Q}/\mathbb{Z}$. -But $\mathbb{R}/\mathbb{Q}$ is just a $\mathbb{Q}$ vector space. So it is a direct sum of copies of $\mathbb{Q}$. -This doesn't exactly answer your question, but it makes some points.<|endoftext|> -TITLE: Primes of the form $x^2+ny^2+mz^2$ and congruences. -QUESTION [20 upvotes]: This is a sequel of this question where I asked for which positive integer $n$ the -set of primes of the former $x^2+ny^2$ was defined by congruences (a set of primes $P$ is defined by congruences if there is a positive integer $d$ and a subset $A$ of $\mathbb{Z}/d\mathbb{Z}$ such that a prime $p$ is in $P$ if and only if $p$ mod $d$ is in $A$, up to a finite number of exceptions). I was taught there that the answer was "exactly when $n$ is idoneal", that there is finitely many idoneal numbers, and that all are known but perhaps one. - -When is the set of primes of the form $x^2+ny^2+mz^2$ ($x,y,z \in \mathbb{Z})$ defined by congruences? - -My motivation is not just an idle ternary generalization of the binary case. I really met this question while working on a problem concerning modular forms, and also the slightly more general question, given a fixed positive integer $a$: when is the set of primes $p$ such that $ap$ has the form $x^2+ny^2+mz^2$ ($x,y,z \in \mathbb{Z})$ defined by congruences? -I am well aware that since the set of integers represented by a ternary quadratic form is not stable by multiplication, it is much less -natural to ask the question for prime numbers instead of all positive integers than in the case of a binary quadratic form. -Yet this is really the question for primes that appears in my study (for about a dozen specific ternary forms, actually). -I have found a very interesting paper by Dickson (Ternary quadratic forms and congruences. Ann. of Math. (2) 28 (1926/27), no. 1-4, 333–341.) which solves the question for the integers represented by $x^2+ny^2+mz^2$: there is only a finite explicit numbers of $(n,m)$ such that this set of integers is defined by congruences (in the obvious sense). But the proof does not seem (to me) to be easily generalizable to primes. Other mathscinet research did not give me any more informations. -When I try to think to the question, I meet an even more basic (if perhaps slightli more sophisticated) question that I can't answer: - -When is the set of primes of the form $x^2+ny^2+mz^2$ ($x,y,z \in \mathbb{Z})$ Frobenian? (Is it "always"?) - -A set of primes $P$ is called Frobenian (a terminology probably introduced by Serre) if there is a finite Galois extension $K/\mathbb{Q}$, and a subset $A$ of Gal$(K/\mathbb{Q})$ stable by conjugacy such that a prime $p$ is in $P$ if and only if Frob${}_p \in A$, except for a finite number of exceptions. A set determined by congruences is a Frobenian set for which -we can take $K$ cyclotomic over $\mathbb{Q}$, which is the same by Kronecker-Weber as abelian over $\mathbb{Q}$. For a quadratic binary -quadratic form (for example $x^2+ny^2$), the set of represented primes is always Frobenian ($K$ can be taken as the -ring class field of $\mathbb{Z}[\sqrt{-n}]$, and $A=\{1\}$, as explained in Cox's book). But I fail to see the reason -(which may nevertheless be trivial) for which the same result would be true for a general ternary quadratic form. -I should add that for my specific ternary forms, I can show that the set is Frobenian, but I am not sure how to extend the argument to all ternary quadratic forms. -Finally, let me say that I would be interested in any book, survey or references on this kind of question (which surely must have been studied), and that I am also interested in analog questions for quaternary quadratic forms (which might be easier, -because of multiplicative properties related to quaternions). - -REPLY [4 votes]: Dear Joel, -I noticed your request for texts. The most informative chapter on positive ternaries, with the intent of predicting the represented integers, is in Dickson M.E.N.T. (1939). I have typed up a list of my books. For the moment, my websites are down, the host computer fried a power supply. So, I am including the link to my preprints on the arXiv, the papers with Alex Berkovich may be just the thing, overlap of modular forms and quadratic forms. The fundamental result is the weighted representation measure of Siegel. Again, as far as numbers integrally represented, the books of Jones, Watson, and Cassels are most helpful. I'm also including the Lattice website, although the emphasis there is classifying interesting lattices (positive forms) rather than finding the numbers represented (squared norms, often just called norms). I've included SPLAG and Ebeling, again I do not mainly use the lattice viewpoint, but there you go. -Carl Ludwig Siegel -Lectures on the Analytical Theory of Quadratic Forms (Second Term 1934/35) -Leonard Eugene Dickson -Studies in the Theory of Numbers (1930) -Modern Elementary Theory of Numbers (1939) -Burton Wadsworth Jones -The Arithmetic Theory of Quadratic Forms (1950) -George Leo Watson -Integral Quadratic Forms (1960) -John William Scott Cassels -Rational Quadratic Forms (1978) -Jean-Pierre Serre -A Course in Arithmetic (English translation 1973) -John Horton Conway -The Sensual Quadratic Form (1997) -Sphere Packings, Lattices and Groups (1988, with Neil J.A. Sloane) -Wolfgang Ebeling -Lattices and Codes (2nd, 2002) -Gordon L. Nipp -Quaternary Quadratic Forms (1991) -O. Timothy O'Meara -Introduction to Quadratic Forms (1963) -Yoshiyuki Kitaoka -Arithmetic of Quadratic Forms (1993) -Larry J. Gerstein -Basic Quadratic Forms (2008) -http://arxiv.org/find/math/1/au:+Jagy_W/0/1/0/all/0/1 -http://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/ -There are also some excellent, influential books by Lam. I sometimes ask him questions, the response is typically that he does not do forms over rings. So, among many threads that might be called quadratic forms, I put Lam in with the name Pfister. Again, I recently got involved with the lattice viewpoint, see SPLAG and Ebeling. The trick there is that it is possible to relate ideas such as covering radius to class number. This relationship is so easy that there really ought to be a short article on "here is how you do this, which you would never know by surveying the literature." But the entire matter is dismissed in a single paragraph on page 378 of SPLAG. When asking for help, I told Richard Borcherds that I sometimes wanted to write a book Here's how YOU can do quadratic forms, and he agreed that one can do a good deal with very little machinery.<|endoftext|> -TITLE: Unique existence and the axiom of choice -QUESTION [28 upvotes]: The axiom of choice states that arbitrary products of nonempty sets are nonempty. -Clearly, we only need the axiom of choice to show the non-emptiness of the product if -there are infinitely many choice functions. If we use a choice function to construct a mathematical object, the object will often depend on the specific choice function being used. So constructions that require the axiom of choice often do not provide the existence of a unique object with certain properties. In some cases they do, however. The existence of a cardinal number for every set (ordinal that can be mapped bijectively onto the set) is such an example. - -What are natural examples outside of - set theory where the existence of a - unique mathematical object with - certain properties can only be proven - with the axiom of choice and where the - uniqueness itself can be proven in ZFC (I - don't want the uniqueness to depend on - a specific model of ZFC)? - -The next question is a bit more vague, but I would be interested in some kind of birds-eye view on the issue. - -Are there some general guidelines to understand in which cases the axiom of choice can be used to construct a provably unique object with certain properties? - -This question is motivated by a discussion of uniqueness-properties of certain measure theoretic constructions in mathematical economics that make heavy use of non-standard analysis. -Edit: Examples so far can be classified in three categories: -Cardinal Invariants: One uses the axiom of choice to construct a representation by some ordinal. Since ordinals are canonically well ordered, this gives us a unique, definable object with the wanted properties. Example: One takes the dimension (as a cardianl) of a vector space and constructs the vector space as functions on finite subsets of the cardinal (François G. Dorais). -AC Properties: One constructs the object canonically "by hand" and then uses the axiom of choice to show that it has a certain property. Trivial example: $2^\mathbb{R}$ as the family of well-orderable sets of reals. -Employing all choice functions: Here one gets uniqueness by requiring the object to contain in some sense all objects of a certain kind that can be obtained by AC. Examples: The Stone-Čech compactification as the set of all ultrafilters on it (Juris Steprans), or the dual space of a vector space, the space of all linear functionals. (Martin Brandenburg) The AC is used to show that these spaces are rich enough. Formally, these examples might be categorized in the second category, but they seem to have a different flavor. - -REPLY [2 votes]: This may be a bit trivial an answer to this (old) question, but since a number of people have reacted with surprise when I mentioned this, and it seems to fall within the purview of what is being asked, let me state: -The Axiom of Choice is equivalent to the following statement: - -If $X$ is a set, $\sim$ an equivalence relation on $X$, and $I$ a set, then the obvious map $\Phi\colon X^I/(\sim^I) \to (X/{\sim})^I$ is a bijection. -(Here $X^I$ stands for the set of all $I$-indexed families of elements of $X$ and $\sim^I$ is the equivalence relation which holds between $(x_i)_{i\in I}$ and $(y_i)_{i\in I}$ in $X^I$ iff $x_i \sim y_i$ for all $i\in I$; and $\Phi$ takes the class of $(x_i)_{i\in I}$ to $([x_i])_{i\in I}$, where $[x]$ denotes the class of $x$ mod $\sim$.) - -Note that, independently of AC, $\Phi$ is injective by definition of $\sim^I$. -(The equivalence with AC of the above is almost obvious. If AC holds, then we can lift $(\bar x)_{i\in I}$ in $(X/{\sim})^I$ to a family $(x_i)_{\in I}$ in $X^I$ by choosing a representative for each class. Conversely, if the above holds and $(Z_i)_{i\in I}$ is a family of inhabited sets, let $X$ be their disjoint union, $\sim$ the equivalence relation for which $Z_i$ is the partition of $X$ so that $X/{\sim}$ is $I$, and use the above to lift the “identity” element of $(X/{\sim})^I = I^I$ to an element of $X^I$ to see that $\prod_{i\in I} Z_i$ is inhabited.) -Thus, in the above context, AC tells us that an injective map $\Phi$ is surjective, or, if we want, that every element in the target has a (necessarily unique) antecedent by $\Phi$.<|endoftext|> -TITLE: Invertible matrices over noncommutative rings -QUESTION [50 upvotes]: Let $A\in M_m(R)$ be an invertible square matrix over a noncommutative ring $R$. Is the transpose matrix $A^t$ also invertible? If it isn't, are there any easy counterexamples? -The question popped up while working on a paper. We need to impose that the transpose of certain matrix of endomorphisms is invertible, and we wondered if that was the same as asking if the matrix is invertible. - -REPLY [52 votes]: See: R.N. Gupta, Anjana Khurana, Dinesh Khurana, and T.Y. Lam, -Rings over which the transpose of every invertible matrix is invertible; J. Algebra 322 (2009), no. 5, 1627–1636 (MR). -Abstract: We prove that the transpose of every invertible square matrix over a ring $R$ is invertible if and only if $R/\text{rad}(R)$ is commutative. …<|endoftext|> -TITLE: Left orderable linear groups -QUESTION [9 upvotes]: Are all torsion-free finitely generated linear groups over $\mathbb{C}$ left orderable? In particular, are torsion-free congruence subgroups of $SL_n(\mathbb{Z})$ left orderable? - -REPLY [13 votes]: The answer is no for congruence subgroups of $SL(n,\mathbb{Z})$ for $n \geq 3$. This is a theorem of Dave Witte-Morris; see -MR1198459 (95a:22014) -Witte, Dave(1-MIT) -Arithmetic groups of higher Q-rank cannot act on 1-manifolds. (English summary) -Proc. Amer. Math. Soc. 122 (1994), no. 2, 333–340.<|endoftext|> -TITLE: When is it appropriate to entitle a paper "A note on..." or "On the ..."? -QUESTION [28 upvotes]: I rarely find modern research papers (on mathematics) that are less than 5 pages long. However, recently I came across a couple of mathematical research papers from the 1960/1970's that were very short (only 2-4 pages long). The authors of both papers solved very specific problems, and stopped writing (I guess) as soon as they were done. This made me realize that I very much like short papers! When possible, I will strive to do the same with my own future papers. -Sometimes one finds short papers that are entitled "A note on...". Since I am first of all not a native English speaker and also only a junior mathematician, I would like to know what you think about my questions/thoughts: -1) Is it a good (or bad?) idea to try to let the title of the paper reflect the fact that it is short? Are there any standard ways of doing this? -2) A paper which is entitled "A note on...", is it expected to be short? How short? Can a 50-pages long paper be "A note on..." or would that not be customary? -3) At least to me "A note on..." could give the impression that the paper is a survey article where no new material is presented, but I guess this need not be the case. When is it appropriate to entitle a paper "A note on..."? -4) When is it appropriate to entitle a paper "On the ..."? - -REPLY [6 votes]: Historically, it was quite common to start titles with "on," much more so than it is today. Many classic works of science have such titles, such as -"On the Origin of Species," by Charles Darwin; -"On Growth and Form," by D'Arcy Wentworth Thompson; -"De rerum natura," by Lucretius; -"De revolutionibus orbium coelestium," by Nicolaus Copernicus.<|endoftext|> -TITLE: A question about rejected journal submissions, similar results, and discrepancies between the order of submission and the order of publication -QUESTION [31 upvotes]: Today I just received the decision of my paper from a journal. The paper was submitted last December and my paper is kind of long (about 40 pages), so I think it's reasonable to take such a long time to receive the decision. Unfortunately, the paper was rejected. And I only received a very short report from a single referee (normally this journal requires two referees) saying that the same result of Prof. B using almost exactly same technique was proved in another paper which was published in another journal, journal A. Therefore the referee decided to reject my paper. So I try to look at this paper from journal A. What was surprising to me is that I have submitted my paper to journal A before I submitted my paper to the present journal and was rejected by journal A. Journal A thinks that the result of Prof. B is good which deserved publication in the journal. However, my paper was rejected. When I look at the submission date of the paper of Prof. B to journal A, I find that I submitted my paper to journal A before Prof. B submitted his paper. What is more surprising to me is that I check the decision letter of my paper from journal A, it seems to me that they even didn't send my paper to the referee to read it, and the paper was rejected within 3 weeks. -Now it seems to me that it's very difficult to get my paper published. So here are questions: Have you ever experienced this? If yes, what did you do to get your paper published? Or did you just give up? Actually I just got my PhD few years ago, and I think maybe that's one of the reason my paper haven't been treated seriously when I submitted to journal A. Any suggestions would be appreciated. Thanks! - -REPLY [8 votes]: Paul, One thing you may try to do is to send a letter to the editor of Journal B appealing the decision. From your description it appears that you submitted the paper to Journal B at about the same time as the other person submitted his paper to journal A so you can mention this fact and ask that your paper will be refereed on its own merits and not be rejected just based on the appearance of the other paper. -Of course, in order to be accepted your paper should be found to be correct and it is also possible that there are other distinctions between the two papers that contributed to the outcome. - -REPLY [8 votes]: The closest situation I saw was when a graduate student $X$ of a colleague $Y$ of mine started to work on some problem, solved it, but was very slow with writing the solution down. During that long and painstaking process, professor $Z$ sent to $Y$ his preprint with essentially the same result and proof. The situation was somewhat awkward but they finally agreed on $X$ and $Z$ publishing a joint paper. -That all could actually be sort of anticipated because both works were just applications of the methods from a paper that appeared shortly before this story to a slightly different problem. -Morals: -1) I do not think that this situation (when several people jump on the same recent idea and try to squeeze more from it at the same time) is unusual and it may be one of the main reasons why independent but nearly identical works are produced almost simultaneously. So, if your situation is like that, then it is ordinary rather than exceptional. -2) The submission dates do not really tell who was there first in any reliable way. -3) It never hurts to negotiate a bit with "the other guy" directly and see if the infamous "priority problem" can be solved peacefully to everyone's satisfaction. -4) If you really want to make a priority claim, it is a good idea to write a decent draft and put it in the public domain quickly (arXiv is the most obvious choice though you may also want to send it to a few experts who might be interested in your result in your opinion). -5) There is no point in having 2 nearly identical publications in (almost) the same journal or anywhere. -At last, I have to say that if you are any good, one unpublished paper won't kill you and if you aren't, one published paper will not save you, so whatever happens with this paper, take it easy and don't start a fight or, worse, a crusade. I doubt you'll be able to publish it anywhere once an almost identical work has appeared in print. The upside of it is that you now know someone else (B) who is interested in the things you are interested in. Try to make the best of it. -As to the open-close tag-of-war over this question, I agree that there is no mathematical content in it but, since it directly relates to the "social side" of our craft and is non-trivial, I'd let it stay.<|endoftext|> -TITLE: "Composition of Morita equivalences" or "Morita equivalence and the Nakayama functor" -QUESTION [8 upvotes]: This problem occured to me, when trying to find a Morita invariant for finite dimensional algebras. -Suppose $\Lambda$ and $\Gamma$ are two self-injective $k$-algebras ($k$ being a field) which are Morita equivalent. The Morita equivalence should be given by the $\Lambda-\Gamma$-bimodule $P$ and the $\Gamma$-$\Lambda$-bimodule $Q$. Each of the algebras has a Morita self-equivalence given by the bimodules $D\Lambda$ and $D\Gamma$, where $D$ is the standard $k$-duality. - -What can one say about the - commutativity of the two compositions, - i.e. is $$D\Gamma\otimes_{\Gamma} -Q\otimes_\Lambda - \cong -Q\otimes_\Lambda D\Lambda -\otimes_\Lambda - $$ in general? - -Further comment: The Morita self-equivalence can equivalently be given as $D \operatorname{Hom}_\Lambda(-,\Lambda)$. This is called the Nakayama functor. -I did not succeed in finding a counterexample nor giving a proof and I don't know what to expect. - -REPLY [2 votes]: Proposition 5.2 in J. Rickard's paper "Derived Equivalences as Derived Functors" seems to show what you want to show, however for standard derived equivalences and the left derived Nakayama functors. I.e., Rickard claims $$D\Gamma \otimes_\Gamma^{\mathbb L}Q \otimes_\Lambda^{\mathbb L}- \cong Q \otimes_\Lambda^{\mathbb L} D\Lambda\otimes_\Lambda^{\mathbb L}-$$ -as functors from $D^-(\textrm{Mod-}\Lambda)$ to $D^-(\textrm{Mod-}\Gamma)$. Your claim is, I believe, a corollary of this. The reason for that is that $D\Gamma \otimes_\Gamma^{\mathbb L}Q \otimes_\Lambda^{\mathbb L}-$ is isomorphic to $(D\Gamma \otimes_\Gamma Q) \otimes_\Lambda^{\mathbb L}-$ (since $Q$ is projective as a left $\Gamma$-module) and -$Q \otimes_\Lambda^{\mathbb L} D\Lambda\otimes_\Lambda^{\mathbb L}- \cong (Q \otimes_\Lambda D\Lambda)\otimes_\Lambda^{\mathbb L}-$, and I suppose this implies $D\Gamma \otimes_\Gamma Q\cong Q\otimes_\Lambda D\Lambda$ as $\Gamma$-$\Lambda$-bimodules.<|endoftext|> -TITLE: Are there standard examples of stable theories that are undecidable? -QUESTION [9 upvotes]: What is known about decidability of various first order theories studied in stable model theory, geometric model theory, o-minimality? For example, is there a natural example of an undecidable first order theory with nice stability properties? - -Many standard examples, such as ACF and hyperbolic groups, seem to decidable. -What about differentially closed fields ? Zariski geometries? What about theories obtained -via Hrushovski fusion? -I should add that there artificially constructed examples of non-decidable stable theories: say add infinitely many algebraic types coding an undecidable set. - -REPLY [12 votes]: Let X be your favorite undecidable set of primes. In a language with one unary function symbol, say f^p(x) = x iff p is in X and also that there is a unique cycle of length p. Add that f is a bijection. Now the prime model contains cycles of exactly the lengths in X and the theory is undecidable. But it is categorical in all uncountable power and therefore omega stable.<|endoftext|> -TITLE: How is called a semigroup... -QUESTION [7 upvotes]: Does anyone know, how is called a semigroup in which every equation $ax=b$ has only a finite set (maybe empty) of solutions? - -REPLY [7 votes]: Pedro Silva and I introduced what we called finite geometric type for a finitely generated semigroup. The definition was that the in-degree of each vertex of the right Cayley graph be finite. An easy induction on the length of a shows this is equivalent to xa=b has finitely many solutions for any fixed a,b. -This should really be called right finite geometric type and so you should have left finite geometric type. Actually we probably should have used proper since for finitely generated semigroups it means the Cayley graph is a proper metric space. -Update. It is proved by Ellis in http://www.ams.org/journals/tran/2001-353-04/S0002-9947-00-02704-5/S0002-9947-00-02704-5.pdf that an infinite semigroup T has finite geometric type if and ony if $\beta T\setminus T$ is a right ideal where $\beta T$ is the Stone-Czech compactification of T.<|endoftext|> -TITLE: can a common mortal understand why the affine line is not smooth in brave new algebraic geometry? -QUESTION [18 upvotes]: In the introduction to HAGII Toen and Vezzosi write that in brave new algebraic geometry (that is, algebraic geometry over the category of symmetric spectra) Z[T] is not smooth over Z. -I am told that this is due to the fact we allow negative homotopy groups (which do not occur with connective spectra, a setting for `ordinary' derived (but cowardly) algebraic geometry). -Can someone with only a vague understanding of derived geometry (like me) understand why this phenomenon occurs? Is it simply a case of extending the notion of smoothness incorrectly? -p.s. there seems to de a tag "derived algebraic geometr" can it be corrected? - -REPLY [24 votes]: I think the answer to your question is "yes". Toen-Vezzosi go over this in Proposition 2.4.15, but here is some version of why. -Away from characteristic zero, there's a sharp difference between being a "free" algebra (meaning, having some kind of universal mapping property) and looking like a polynomial algebra. This is because basically every construction has to be replaced with the derived version in order for them to give sensible answers. -For example, in characteristic zero you might take a vector space $V$ over $\mathbb{Q}$ and form the "free" algebra -$$ -Sym(V) = \bigoplus_{n \geq 0} Sym^k(V) -$$ -and this has the property that maps of commutative ring objects $Sym(V) \to A$ are the same as maps $V \to A$ of underlying objects. This still works even if $V$ is a chain complex. -However, the symmetric power functors aren't well-behaved integrally: there are maps of chain complexes $C \to D$ which are weak equivalences such that $Sym^k(C) \to Sym^k(D)$ isn't a weak equivalence. (These examples aren't particularly hard to find, either.) This means that we have to take the symmetric power -$$ -Sym^k(V) = (V^{\otimes k})_{\Sigma_k} -$$ -and replace it with the derived version: derived tensor product over whatever your base is, and derived functors of $\Sigma_k$-coinvariants. -Toen-Vezzosi mention the example which they call $\mathbb{F}_p[T]$, which is free on a 1-dimensional vector space over $\mathbb{F}_p$. Here, the derived functors of tensor product don't intrude, but the derived functors of coinvariants do, and they contribute a large amount (namely, the homology of the symmetric groups). In degree zero you're just getting a polynomial algebra on a single generator, but there is extra stuff in positive (homological) degree coming from the Dyer-Lashof operations. -EDIT: Sorry, this object is the "smooth" object. -The thing that looks like a "polynomial" object is the monoid algebra $\mathbb F_p[\mathbb N]$. On homology groups this looks like a polynomial algebra, but it doesn't have simple mapping properties that connect with the notion of smoothness. In particular, they give a description of the derived cotangent complex (which is some measure of how hard it is to build this object out of "free" objects) and it's rather large (certainly not concentrated in degree 0).<|endoftext|> -TITLE: Is it possible to solve the argument maximization problem $\arg\max_x \langle x,l \rangle −f_1(x)−f_2(x)$ via convex duality? -QUESTION [8 upvotes]: I am attempting to solve the argument maximization problem -$$\arg\sup_x \{ \langle x,l \rangle − f_1(x)−f_2(x) \} \ \ \ \ \ \ \ \ \ \ (1)$$ -where the functions $f_1$ and $f_2$ are concave but difficult to evaluate but their convex conjugates $f^∗_1$ and $f^∗_2$ are easy to evaluate. We can further assume that $f^∗_1$ and $f^∗_2$ are differentiable and that we can evaluate their gradients. Since the sum operation is dual to the infimal convolution (or epi-sum) operation -$$(g\#h)(x) = \inf_w \{ g(x−w) + h(w) \} $$ -the standard maximization problem is easy to compute by duality using the identity -$$\sup_x\{⟨x,l⟩−f_1(x)−f_2(x)\}=\inf_w \ \{ f^*_1(l−w) + f^∗_2(w) \}.$$ -Is it possible to compute the solution to problem $(1)$ is an analogous manner, making only calls to the conjugate functions $f^∗_1$ and $f^∗_2$ or their gradients? - -REPLY [2 votes]: I ultimately did find a solution to this problem. First, you compute the solution $\bar{w}$ to -$$\bar{w} \in \arg\inf_w f^*_1(\ell - w) + f_2^*(w).$$ -A solution $\bar{x}$ to (1) can then be computed by choosing -$$\bar{x} \in \partial f_1^*(\ell-\bar{w}) \cap \partial f_2^*(\bar{w})$$ -where $\partial f$ denotes the subgradient of $f$. A proof of this fact appears as part of Theorem 15 in this JMLR paper: -http://jmlr.org/papers/volume8/rifkin07a/rifkin07a.pdf<|endoftext|> -TITLE: Does $2^X=2^Y\Rightarrow |X|=|Y|$ imply the axiom of choice? -QUESTION [49 upvotes]: The Generalized Continuum Hypothesis can be stated as $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. We know that GCH implies AC (Jech, The Axiom of Choice, Theorem 9.1 p.133). -In fact, a relatively weak formulation: $|X|\le|Y|< 2^X\implies |X|=|Y|$ would already imply the axiom of choice, although in this case the proof is slightly longer. [Note: in Herrlich's book he refers to GCH stated above as "The Aleph Hypothesis" and the weak formulation is called GCH] -GCH itself is independent of the axiom of choice, we can have the axiom of choice and power sets can grow wildly, or just "a little bit". We can have the continuum function to be injective, but the continuum hypothesis can fail on a proper class of cardinals (for example $2^\kappa=\kappa^{++}$ for regular cardinals). -Let ICF (Injective Continuum Function) be the assertion: $$2^X=2^Y\implies |X|=|Y|.$$ -Question: Assuming ZF+ICF, can we deduce AC? -(I looked around Equivalents of the Axiom of Choice, but couldn't find much. It is possible that I missed this, though.) - -Edits: - -In an exercise in Jech he states that if there exists an infinite Dedekind-finite cardinal, then ICF does not hold. From the assumption that it holds we can deduce that there are no infinite D-finite sets. - -Note that if $f\colon X\to Y$ is a surjection then $A\subseteq Y\mapsto f^{-1}(A)$ is an injection from $P(Y)$ into $P(X)$. This means that ICF implies the Dual Cantor-Schroeder-Bernstein theorem: -Assume that $X$ and $Y$ have surjections from one onto the other, then there are injections between their power-sets therefore $2^X=2^Y$ and thus $|X|=|Y|$. - -We shall abbreviate Goldstern's variant of ICF as Homomorphic Continuum Function, or HCF: $$2^X\leq 2^Y\implies |X|\leq|Y|$$ -One major observation is that HCF implies The Partition Principle (PP), which states that $A$ can be mapped onto $B$ (or $B$ is empty) if and only if $B$ can be injected into $A$. This principle is quite an open choice principle, and it is unknown whether or not it implies AC in ZF. -To see that HCF implies PP, we observe the following: if $f\colon A\to B$ is surjective then the preimage map is an injection from $2^B$ into $2^A$, i.e. $2^B\leq 2^A$, from HCF it follows that $B\leq A$, i.e. there is $g\colon B\to A$ injective. - -REPLY [20 votes]: Here is a little progress towards AC. -Theorem. - ICF implies the dual Cantor-Schröder-Bernstein -theorem, that is $X$ surjects onto $Y$ and $Y$ surjects onto $X$, -then they are bijective. -Proof. You explain this in the edit to the question. If -$X\twoheadrightarrow Y$, then $2^Y\leq 2^X$ by taking pre-images, -and so if also $Y\twoheadrightarrow X$, then $2^X\leq 2^Y$ and so -$X\sim Y$ by ICF. QED -Theorem. ICF implies that there are no infinite D-finite -sets. -Proof. (This is a solution to the exercise that you mention.) If $A$ is infinite and -Dedekind-finite, then let $B$ be the set of all finite -non-repeating finite sequences from $A$. This is also D-finite, -since a countably infinite subset of $B$ easily gives rise to a -countably infinite subset of $A$. But meanwhile, $B$ surjects onto -$B+1$, since we can map the empty sequence to the new point, and -apply the shift map to chop off the first element of any sequence. -So $B$ and $B+1$ surject onto each other, and so by the dual -Cantor-Schöder-Bernstein result, they are bijective, -contradicting the fact that $B$ is D-finite. QED -Here is the new part: -Theorem. ICF implies that $\kappa^+$ injects into -$2^\kappa$ for every ordinal $\kappa$. -Proof. We may assume $\kappa$ is infinite. Notice that -$2^{\kappa^2}$ surjects onto $\kappa^+$, since every -$\alpha<\kappa$ is coded by a relation on $\kappa$. Since -$\kappa^2\sim\kappa$, this means -$2^\kappa\twoheadrightarrow\kappa^+$ and consequently -$2^{\kappa^+}\leq 2^{2^\kappa}$, by taking pre-images. It follows that -$2^{2^\kappa}=2^{2^\kappa}\cdot 2^{\kappa^+}=2^{2^\kappa+\kappa^+}$ and -so by ICF we get $2^\kappa+\kappa^+=2^\kappa$, which implies -$\kappa^+\leq 2^\kappa$, as desired. QED -This conclusion already contradicts AD, for example, since AD -implies that there is no $\omega_1$ sequence of distinct reals, -which violates the conclusion when $\kappa=\omega$. In particular, -this shows that ICF implies $\neg$AD, and so in every AD model, there are sets of different cardinalities, whose power sets are equinumerous.<|endoftext|> -TITLE: Galois theory for polynomials in several variables -QUESTION [27 upvotes]: I feel a bit ashamed to ask the following question here. - -What is (actually, is there) Galois - theory for polynomials in - $n$-variables for $n\geq2$? - -I am preparing a large audience talk on Lie theory, and decided to start talking about symmetries and take Galois theory as a "baby" example. I know that Lie groups are somehow to differential equations what discrete groups are to algebraic equations. But I nevertheless would expect Lie (or algebraic) groups to appear naturally as higher dimensional analogs of Galois groups. -Namely, the Galois group $G_P$ of a polynomial $P(x)$ in one variable can be defined as the symmetry group of the equation $P(x)=0$ (very shortly, the subgroup of permutations of the solutions/roots that preserves any algebraic equation satisfied by them). -Then one of the great results of Galois theory is that $P(x)=0$ is solvable by radicals if and only if the group $G_P$ is solvable (meaning that its derived series reaches $\{1\}$). -I was wondering what is the analog of the story in higher dimension (i.e. for equations of the form $P(x_1,\dots,x_n)=0$. I would naively expect algebraic group to show up... - -I googled the main key words and found this presentation: on the last slide it is written that - -the task at hand is to develop a - Galois theory of polynomials in two - variables - -This convinced me to anyway ask the question - -EDIT: the first "idea" I had -I first thought about the following strategy. Consider $P(x,y)=0$ as an polynomial equation in one variable $x$ with coefficients in the field $k(y)$ of rational functions in $y$, and consider its Galois group. But then we could do the opposite...what would happen? - -REPLY [9 votes]: (This should really be a comment I think, but I'm not highly rated enough to leave one, so please bear with me) -A Galois Theoretic condition for a polynomial in two variables to be solvable by radicals is found in the following paper: http://arxiv.org/abs/math/0305226. It seems to indicate that something similar can be done for higher variables. Perhaps I'll ask Jochen next time I see him about this.<|endoftext|> -TITLE: Can there exist a `natural' finitely generated group with an undecidable word problem? -QUESTION [18 upvotes]: There are naturally occurring groups that have undecidable algorithmic problems. For instance, $F_2\times F_2$ has undecidable generalized word problem (membership problem for subgroups) and there is a semidirect product $\Bbb Z^4\rtimes F_4$ with undecidable conjugacy problem. But to the best of my knowledge every finitely generated group with an undecidable word problem is directly constructed from a Turing machine variant or a non-recursive, recursively enumerable set. - - -Can there exist a naturally occurring finitely generated group with an undecidable word problem? - - -Of course this problem is not well-defined because `natural' is not well defined in this setting. It should somehow mean things built up from well studied classes of groups (hyperbolic groups, polycyclic groups, solvable groups, linear groups) via natural operations, e.g., extensions, amalgamations, maybe direct limit (I am not sure I like allowing this last option since some strange groups are direct limits of hyperbolic groups). - -REPLY [15 votes]: If you take a subgroup $H$ of $F_2\times F_2$ with undecidable membership problem and take the HNN extension of $F_2\times F_2$ where the stable letter commutes with $H$, you get a (finitely presented) group with undecidable word problem. I do not know how natural it is since $H$ encodes a Turing machine. The only known "natural" example of universal algebras with undecidable word problem are free modular lattices (Freese, 1980). -As a more natural examples you can take McKenzie-Thompson group that simulates all recursive functions or Valiev's (Boone-Collins) universal group that simulates all Turing machines, see comments below.<|endoftext|> -TITLE: Are there any algorithms for solving nonlinear matrix equations over $\mathbb{C}$? -QUESTION [5 upvotes]: Are there any algorithms for solving nonlinear matrix equations over $\mathbb{C}$? -I am especially interested in solving polynomial nonlinear matrix equations. -For instance, let $X$ be some matrix satisfying -$$X=A+BXC+DXEXF$$ -where $A,B,C,D,E,F$ are given matrices. -Of course, the equation could be in higher degree, such as -$$X=X^n+X^{n-1}+A$$ -Is there an algorithm that can solve this kind of matrix equations? - -REPLY [4 votes]: You could solve your equations by suitably extending the methods for solving Nonsymmetric Riccati equations, see e.g., - - -Nonsymmetric algebraic Riccati equations and Wiener-Hopf Factorization for M-Matrices by C.-H. Guo<|endoftext|> -TITLE: Motivation for Hall-Witt identity -QUESTION [18 upvotes]: I've wondered for a while about the (Hall-)Witt identity in group theory: - -$[[a,b^{-1}],c]^b \cdot [[b,c^{-1}],a]^c \cdot [[c,a^{-1}]],b]^a = 1$. - -(Here, $x^y$ means $y^{-1}xy$ and $[x,y]$ means $x^{-1}y^{-1}xy$.) Does anybody have any motivation for this? To me, it almost seems like it comes out of nowhere so that we can prove the three subgroup lemma or something. Is there some reason to expect a relation like this to hold, or a way of reducing it to simpler relations in a meaningful way? Perhaps we should expect something like this from the free-ness of the commutator subgroup of the free group on three letters? Or should we expect some analogue of the Jacobi identity to hold, and if so, why? - -REPLY [3 votes]: To add to Terry Tao's excellent description of some of the geometric ideas behind the Hall-Witt identity, I would draw attention to the interpretation given by Loday in his paper on Homotopical Syzygies (in Une dégustation topologique: Homotopy theory in the Swiss Alps, volume 265 of Contemporary Mathematics, 99 – 127, AMS, 2000; book on AMS site). In his section 1.4. in examining the syzygies / identities amongst the relations of the obvious presentation of the free Abelian group on three symbols, he shows how the commutators/ 2-syzygies are drawn as the faces of the Cayley graph considered as being embedded in a sphere (and thus appearing as an empty cube). There is a 3-syzygy needed to build the next stage of a resolution of the group and that is a 3-cell filling the cube. The boundary of that 3-cell can be read off as the Jacobi-Witt-Hall (JWH) identity (with slightly different conventions more as in Terry's article than in the original form in the question). -Thus the proof of freeness of the commutator subgroup of $F(x,y,z)$ does not give a useful basis for that subgroup and the commutators and their conjugates are not without relations between them. The JWH identity is one of those relations. The exploration of homotopical syzygies throws up some lovely calculations in general (and lots of nice pictures!) (Edit: I mean for quite general presentations.)<|endoftext|> -TITLE: Wanted: an example of a natural non-K\"ahler metric on a Kahler manifold -QUESTION [12 upvotes]: Let $X$ be a Kahler manifold. Associated to any hermitian metric $h$ on $X$ is a smooth real $(1,1)$-form $\omega = -\text{Im } h$, called the Kahler form of $h$. One of several equivalent conditions for the metric $h$ to be a Kahler metric is that $\omega$ is symplectic, or that $\text{d} \omega = 0$. -Morally speaking, this implies that Kahler metrics are rare. In a sense, they are contained in a proper subspace of the cone of hermitian metrics on $X$. -Question: Does anyone know an example of a manifold $X$ and a natural metric $h$ on $X$ which is not Kahler? -For extra credit: Can we find such an example where $X$ is compact? -I'll elaborate a bit on what I mean by "natural" and then provide some motivation for the questions. The following paragraphs are not meant to be mathematically rigorous, but rather heuristic, so please be gentle when you see any inaccuracies. -"Naturality": It is easy to give explicit examples of non-Kahler metrics on a Kahler manifold. Just take any Kahler metric $h$ and multiply it by a positive non-constant function $f$: the Kahler form of the new metric will then satisfy $\text{d} (f \omega) = \text{d} f \wedge \omega + f \text{d} \omega = \text{d} f \wedge \omega$. As $\omega$ is symplectic the wedge product $\text d f \wedge \omega$ can only be zero if $\text d f$ is zero, so the new metric is not Kahler. -This feels like cheating to me. It's like starting a book on linear algebra by defining vector spaces axiomatically and then only giving the trivial space as an example. The example does not advance our understanding in any significant way. -I would like to see an example where the metric $h$ arises in a geometric way or is somehow an obvious candidate for a metric on $X$. For example, consider a Hopf surface $X$, which arises as a quotient of $\mathbb C^2 \setminus \{0\}$ by a group $G$. The naive way to give an example of a metric on $X$ is to find a metric on $\mathbb C^2 \setminus \{0\}$ which is invariant under the action of $G$, and it is perfectly possible to give an explicit example of such a metric by some calculations (see [1] for an example). If only the Hopf surface were Kahler I would accept this as a "real" example. -Motivation: Given a hermitian metric $h$ there are several equivalent definitions of it being Kahler. One can say that its Kahler form is closed, that one can approximate the euclidean metric to the second degree in local coordinates, or that the Chern and Levi-Civita connections of $h$ are the same. This last condition is the one I like the most, because with good will one can interpret it as saying that the complex and Riemannian geometries defined by the metric are the same. -This is all well and good, and I feel I understand the different definitions and the links between them. However, given an explicit metric, I have absolutely no intuition for if it is Kahler or not. I can't look at a metric and just go "Aha!", I have to "fly blind" and calculate. -For example, take the Fubini-Study metric on $\mathbb P^n$. It can be obtained by considering a scalar product on $\mathbb C^{n+1}$ and saying that the scalar product of lines in that space is the "angle" between the lines (-ish). This is a very pretty and geometric way of obtaining a metric. Now, the only way I know to show that the metric obtained in this beautiful way is Kahler is by long and violent calculations. I can't give you an a priori plausibility argument for it being Kahler. The same is true for any explicit example of a Kahler metric on any manifold. -I see this as failure on my part, and a sign that I have not really understood Kahler metrics. I think that an explicit example of a natural non-Kahler metric would help me understand complex geometry better. - -[1] Examples of non-Kahler surfaces with explicit non-Kahler metric - -REPLY [4 votes]: Thanks Claudio, Francesco and José for your interesting answers. -This is an "answer in absence" of Demailly; he doesn't use this site, but I thought his remark was nice enough to share. What follows is only his sketch, there are details to fill in that I haven't yet had the time to take care of. -Let $X$ be a smooth projective variety, embedded in $\mathbb P^N$. Take $k$ big enough so that the vector bundle $T_X \otimes \mathcal O(k)$ is generated by its global sections. The Fubini-Study metric on $\mathbb P^N$ now gives a hermitian metric on $X$ by restriction, and on $\mathcal O(k)$ by taking powers of the determinant metric. These induce an $L^2$ metric on the global sections of $T_X \otimes \mathcal O(k)$. -If we tensor by $\mathcal O(-k)$, then we have a surjective bundle map -$$ H^0(X, T_X \otimes \mathcal O(k)) \otimes \mathcal O(-k) \to T_X \to 0. $$ -The $L^2$ metric and the metric induced by Fubini-Study on $\mathcal O(-k)$ now gives a metric on the tensor product on the left. This induces a quotient metric $h$ on $T_X$. Despite its algebraic origins, this metric $h$ should (almost) never be a Kahler metric. -Moreover, by applying approximation theorems of Tian, Demailly and others, one should be able to prove that these non-Kahler metrics are dense in the cone of hermitian metrics on $X$ -- i.e. starting from any metric on $X$ and using that to define the $L^2$ metric, it should be possible to fabricate a series of non-Kahler metrics as above which converges to the given metric. The process should in fact generalize and yield similar metrics on any holomorphic vector bundle over a projective manifold. -This should imply that any heuristic of the form "my metric is geometric/algebraic, thus Kahler" is doomed to failure.<|endoftext|> -TITLE: Hecke algebra and $H^*(G/B)$ -QUESTION [7 upvotes]: Given a complex reductive group, with Weyl group $W$, one can associate to it lots of "algebras of size $|W|$". For example $B$ equivariant functions on $G/B$ with convolution, grothendieck groups of categories associated with $G$, the group ring $\mathbb C[W]$... -So far they are all more or less specializations of the Iwahori-Hecke algebra! But there is another natural algebra of size $W$, namely the cohomology ring of the flag variety $H^*(G/B)$. So my question is, how is this ring related to the Hecke algebra? - -REPLY [6 votes]: The group algebra $\mathbb{C}[W]$ is the Borel-Moore homology of the union of conormals to the Schubert cells (the preimage of $\mathfrak{n}$ under the Springer resolution) endowed with a convolution structure; the coinvariant ring is its cohomology. These have the same dimension, since this union of conormals is a union of complex affine spaces, but there's no canonical isomorphism or duality between them, since this variety is horribly non-singular and non-compact.<|endoftext|> -TITLE: What is the structure of $SO(3)$ and its Lie Algebra? -QUESTION [6 upvotes]: First I want to give you some background how the question arised, before actually asking it. -Recently, in the context of quantum mechanics, I thought about the group $SO(3)$ and its Lie Algebra $so(3)$. Wherever I looked, I could only find a construction of $so(3)$ very concretely in terms of matrices, as being the tangent space of $SO(3)$ at the identity. When considering $SO(3)$ represented as $3\times3$ matrices, it follows directly (chain rule etc.), that elements in $so(3)$ are anti-symmetric, real $3\times3$ matrices and form a three dimensional vector space. Therefore (with the knowledge that they can be represented as matrices) we can take the commutator as being the Lie Bracket in $so(3)$, although in the Lie Algebra $so(3)$ a priori there is no product defined, which even rises the question why the commutator lies again in $so(3)$. Matrix multiplication in $SO(3)$ gives addition in $so(3)$ (chain rule). -It seems as if the Lie Bracket, as being the commutator, gets its definition from the fact, that we know elements in $so(3)$ can be represented as matrices. -Therefore my questions are the following: - -What is $SO(3)$ as abstract group? How can we get hold of it and present it without matrices? Especially with regard to the second question: -How to get the Lie Algebra out of $SO(3)$ in an algebraically satisfying way, i.e. without the explicit construction of matrices? - -Looking forward for interesting ideas! -Cheers, Niki -ps.: Although I am sure, if we take the ill of matrices, there should be at least a way of getting the commutator in $so(3)$ without multiplying matrices, but I only found the following argument for $R_i(t) \in SO(3) \ \forall t$ and $R_i(0) = id$: -\begin{eqnarray} -[\Omega_1,\Omega_2] = \frac{d}{dt} R_1(t)\Omega_2 R_1(t)^{-1} \mid_{t=0} \newline -\mbox{where } \frac{d}{dt}R_i(t) \mid_{t=0} = \Omega_i \newline -\mbox{with } R\Omega_i R^{-1} = \frac{d}{dt}R R_i(t) R^{-1} \mid_{t=0} -\end{eqnarray} - -REPLY [6 votes]: Abstractly you may think of $SO\left(3\right)$ as the group of rotations in Euclidean $3$-space, that is the group of linear transformations of $\mathbb{R}^3$ which preserve the norm $\left|\left(x,y,z\right)\right|^2=x^2+y^2+z^2$. Alternatively, you may think of elements of $SO\left(3\right)$ as orthonormal bases, $\hat{x},\hat{y},\hat{z}$, which corresponds to the rotation which sends the standard basis $\hat{i},\hat{j},\hat{k}$ to $\hat{x},\hat{y},\hat{z}$. -You're already familiar with the matrix definition of $so\left(3\right)$, which is skew-symmetric matrices with Lie bracket given by the commutator of matrices. Any such matrix looks like $$\left(\begin{array}{ccc}0 & -\omega_{z} & \omega_{y}\\\ \omega_{z} & 0 & -\omega_{x}\\\ -\omega_{y} & \omega_{x} & 0\end{array}\right)$$This corresponds to an "angular velocity" vector $\left(\omega_x,\omega_y,\omega_z\right)$ and the commutator of two skew-symmetric matrices corresponds to the cross product of their corresponding angular velocities. So you see that $so\left(3\right)$ is isomorphic to $\mathbb{R}^3$ with the Lie bracket given by the cross product. -In general for a Lie group $G$, one may define the Lie bracket on its Lie algebra $g=T_e G$ as follows. For $a\in G$, let $C_a:G\to G$, the commutator by $a$, be $C_a\left(b\right)=aba^{-1}$.The derivative at $e$, $\left(dC_a\right)_e:g\to g$, is the "adjoint representation" of $G$ on its Lie algebra, $Ad:G\to GL\left(g\right)$, $Ad\left(a\right)=\left(dC_a\right)_e$. Taking the derivative again gives a map $ad:g\to gl\left(g\right)$ and the Lie bracket on $g$ is $$\left[x,y\right]=ad\left(x\right)\left(y\right)$$So you see, you don't need to use matrix multiplication to abstractly define the Lie bracket of a Lie algebra, thought of as the tangent space to a Lie group.<|endoftext|> -TITLE: Number of spanning trees which contain a given edge -QUESTION [12 upvotes]: Suppose I have a connected graph $G$ and a fixed edge $e = \langle u, v \rangle \in G$, and I want to count the number of spanning trees that involve $e$. I really only want to estimate the fraction of spanning trees containing $e$ compared to the total number of spanning trees $G$, that is, I want to find a lower bound $c \leq \kappa(G \backslash e) / \kappa(G)$ -This lower bound should be in terms of the degrees of vertices $u,v$. Let $c(d,d')$ be the smallest possible value of $\kappa(G \backslash e) / \kappa(G)$ when the vertices have degree $d, d'$. What can one say about $c(d,d')$? -For example, if $d = 1$ or $d' = 1$, then $c(d,d') = 1$. (The edge must be part of a spanning tree of $G$.). -If $d = 2$, then $c(d,d') \geq 1/2$, as every spanning tree must involve one of the two edges incident on $u$, and the spanning trees using only $e$ are in bijection with the spanning trees using the other edge. -If $d = d' = 2$, then $c(d,d') \geq 2/3$; and so on. -Is there are general formula for $c(d,d')$? - -REPLY [19 votes]: The probability that an edge $e=(u,v)$ is part of a uniform spanning tree is equal to the resistance between $u$ and $v$ when the graph is considered as an electric network (see the book by Lyons with Peres, section 4.2). The bounds you get (in term of the degrees $d_u,d_v$) are -$$ \frac{1}{\min(d_u,d_v)} \le R_{eff}(u \leftrightarrow v) \le 1$$ -when you allow multiple edges, or -$$ \frac{(d_u-1)+(d_v-1)}{(d_u-1)+(d_v-1)+(d_u-1)(d_v-1)} < R_{eff}(u \leftrightarrow v) \le 1$$ -when the graph is simple, and these bounds are sharp.<|endoftext|> -TITLE: Yang Mills gradient/heat flow on 4-torus -QUESTION [7 upvotes]: The classic Donaldson-Kronheimer book (Geometry of 4-manifolds) uses the Yang Mills gradient flow (sometimes called heat flow) on $M$ all over the place, -$\frac{d A}{dt} = -\frac{\delta YM(A)}{\delta A}$ -where $YM(A)$ is the Yang Mills 'action' the integral of the curvature square, -$YM(A) = \int d^4x Tr F_{\mu\nu} F_{\mu\nu} > 0$ -The setting is quite general, either $M$ is a general 4-manifold or Kahler manifold and so all theorems, existence, uniqueness, etc, are quite general. -I'm wondering if there are further results somewhere for the specific case of $M = T^4$ the 4-torus. For example, is it known how the long time asymptotics look like in this case? Theorems about possible blow-ups? I'd think asymptotically the gradient flow drives the connection towards a critical point but is it known how it is approached, exponentially or polynomially in $t$? -Actually for $M=T^4$ I suspect $t^2 YM(A(t))$ goes to a constant for $t\to\infty$ as long as the initial condition for the flow is in a sufficiently small neighbourhood of the absolute minimum of $YM(A)$ but I can't prove it. What is certainly true is that $YM(A(t))$ is a decreasing function of $t$. - -REPLY [4 votes]: Donaldson and Kronheimer wrote their book by 1990. There were some further developments about long time behaviour of Yang-Mills flow on four manifolds by, among others, Struwe and collaborators. You may try starting with Schlatter's dissertation. -Crawling through MathSciNet reference links may get you "somewhere", but it is my impression that higher dimensional Yang-Mills heat flow is still somewhat of an open problem. Is there any reason why you expect $\mathbb{T}^4$ would be better behaved than, say, the unit ball?<|endoftext|> -TITLE: On the periods in the periodic table (or Why is a noble gas stable?) -QUESTION [25 upvotes]: Added 22, November: -I've succeeded in making the question entirely unintelligible with all my additions. -So I thought I would summarize it in the simplest form I could manage and add it to the title. The question is thereby somewhat narrower in scope than my original query, but I would be happy to have this focused version answered. -The main point is that the loose answer -(A) because the outer shell is filled -frequently heard is definitely wrong. If we take the usual definition of a shell, all the noble gases but Helium have just -the s and p subshells (corresponding to the representations $V_0\otimes S$ and $V_1\otimes S$) -filled in the outermost shell, and this is not a full shell once the atom is bigger than Argon. -The wikipedia article on noble gases offers an amusing formulation whereby, for -a noble gas, 'the outer shell of valence electrons is considered to be "full"' (my emphasis). -All this led to my initial confusion: I thought that the term 'shell' must mean something -else for multi-electron systems in a manner adapted to the answer (A). Such an alternative definition -does not seem to exist, and it is not at all obvious how to come up with -one in a non-tautological way. So I believe the essence of what - I am asking is whether or not there is -a natural mathematical explanation for the stability -of noble gases -that is more or less independent of experiments -confirmed by difficult computations using approximation schemes. -Anyone interested in the background and details is invited to read the incoherent paragraphs below, -or to look up a proper reference like Atkin's book on physical chemistry. - -Added, 21 November: -Having read some more here and there, I should make some terminological corrections, in case I mislead anyone with my ignorance. I will do so here, and leave the text below as written, in case the temporary confusion is helpful to other interested non-experts. -Firstly, as far as I can tell, the word 'orbital' does seem to refer to a wavefunction, not a representation. So the eigenfunctions in the second occurrence of the representation $V_1\otimes S$ will be called the $3p$-orbitals. That is, the individual wavefunction is an orbital, while the representation itself might be referred to as the so-and-so orbitals, in the plural. The definitive term for any given occurrence of an irreducible representation in $L^2\otimes S$ seems to be subshell. -Secondly, I finally read the wikipedia article on shells. It doesn't help much with my questions, but it does describe the convention regarding the term 'shell'. As far as I can tell, the shells are simply the direct sums of the following form: -K-shell: $1s$ -L-shell: $2s\oplus 2p$ -M-shell: $3s \oplus 3p \oplus 3d$ -N-shell: $4s\oplus 4p \oplus 4d\oplus 4f $ -O-shell: $5s\oplus 5p \oplus 5d\oplus 5f\oplus 5g $ -and so on. (In case you're wondering, the representations after $V_3\otimes S=f$ are labeled consecutively in the alphabet.) The key point is that, in this form, a shell does not consist of a grouping of subshells of similar energies when more than a few electrons are present. Rather, the conventional description of the phenomena says that more than one shell can be incomplete in an atom. For example, in the fourth row of the periodic table, starting with potassium (K) and up to copper (Cu), they say both the $M$-shell and the $N$-shell are incomplete. If this is confusing to you, I suggest you don't worry about it. I just wanted to point out that my question `what is a shell?' has a clear-cut answer in this usage. If we don't want to go against this convention, we need to stick to the version -`What determines a period?' -I'm very sorry for all the confusion. - -Added, 19 November: -After receiving the nice answers from Jeff and Antoine, I realized that I should sharpen the question -somewhat more. I like Neil's formulation, but I think I am asking something much more naive. -Allow me to start by providing more background for mathematicians whose knowledge is as hazy -as mine. As mentioned by Jeff and Antoine, the Hamiltonian for a system of $N$ electrons -moving around a nucleus of charge $Z$ looks like -$$H=-\frac{\hbar^2}{2m}\sum_{i=1}^N \Delta_i-\sum_{i=1}^N \frac{Ze^2}{r_i}+ \sum \frac{e^2}{r_{ij}}.$$ -In principle, one would like to understand the structure of - discrete spectrum solutions to the eigenvalue equation -$$H\psi=E\psi.$$ -One characterizes solutions using labels called 'quantum numbers'. In general, there is an objective label often -denoted $l$, the angular momentum quantum number. This comes from the fact that there is -an $SO(3)\times SU(2)$ symmetry, breaking up solutions into the irreducible representations $V_l\otimes S$, mentioned -earlier. All the vectors in a given irreducible representation must have the same energy level $E$, because -a basis for the space can be obtained from a highest weight vector by applying elements of $LieSO(3)$. -Any given $V_l\otimes S$ is called a subshell or an orbital (I'm a bit unclear about this because the term -'orbital' is also used for the individual eigenfunctions) and is described using the (historical) labels -$$V_0\otimes S \leftrightarrow s$$ -$$V_1\otimes S \leftrightarrow p$$ -$$V_2\otimes S \leftrightarrow d$$ -$$V_3\otimes S \leftrightarrow f$$ -The principal quantum number is determined as follows. We order all the orbitals by energy levels, -and the $n$-th time $V_0 \otimes S$ occurs, we call that subspace $ns$. However, the $n$-th time that -$V_1\otimes S$ occurs, we call it $(n+1)p$. In general, the $n$-th time $V_l\otimes S$ occurs, -the label is -$(n+l+1)$(whatever letter $l$ corresponds to). -For example, the orbital $3d$ is the first occurrence -of the representation $V_2\otimes S$. When there is just one electron in a $1/r$ potential, these $n$ really label -the order of the energy levels, and $d$ really occurs the first time in the third energy level. This is the reason for -the shift in labeling, even though this correspondence with energy levels breaks down for multi-electron systems. -In fact, in neutral atoms with $N=Z$, the energy levels are ordered like -$$1s<2s<2p<3s<3p<4s<3d <4p<5s<4d<5p<6s< \ldots$$ -(It's not supposed to be obvious how to fill in the $\ldots$. However, I've been told that knowing $4s<3d$ is essential to passing A-level chemistry in England.) -I hope it is clear from this that the orbitals (or the subshells) are completely well-defined and have a labeling -scheme that is a bit odd, but makes sense when the obvious translation is combined with history. -So the real question is -What determines a shell? -Some energy-non-decreasing consecutive - sequence of subshells is a shell. The shells then determine the rows of the periodic table (if we ignore the added complication -that $Z$ is also increasing as we move right and down). -Now, it is tempting to conclude that the standard -convention is a bit arbitrary. After all, the chemical similarity of the columns could possibly be explained simply by the -fact that -they have the same representation of $SO(3)\times SU(2)$ at the uppermost energy level, and the -same number of electrons in this representation -without any reference to an outermost shell at all. This pattern can be easily seen by the arrangement into blocks shown here. And then, unlike the $N=1$ case, -the energy levels in one shell are not even the same, just rather close to each other. -Unfortunately, the grouping into rows represents real phenomena. -This comes out very clearly, for example, in the graph of ionization energies, with the noticeable peaks -at the end of the rows immediately followed by precipitous drops. So I believe a bit of thought reduces a good deal of my original question to two parts: - -Is there some reason for a big gap in ionization energy when one moves to an $s$-orbital? -Can one show that two orbitals of the same type are necessarily separated by a huge energy -gap, so that it is highly unlikely for them to occur in the same shell? - -1 and 2 together make it natural that the dimensions we see in each shell will be of the form -2, 2+6, 2+6+10, 2+6+10+14, etc, -even in the general case. Of course, this doesn't say anything about how many times each combination is -likely to occur. -In any case, I hope there is a mathematical answer to these two questions that doesn't involve a full-blown programme in -hard analysis. -At the more speculative end, one might analyze a family of operators like -$$H_{\epsilon}=-\frac{\hbar^2}{2m}\sum_{i=1}^N \Delta_i-\sum_{i=1}^N\frac{Ze^2}{r_i}+\epsilon \sum \frac{e^2}{r_{ij}}.$$ -Is there a way to see the numbers we see occurring via the spectral flow of this family as we go from $\epsilon =0 $ to -$\epsilon=1$? But maybe this is just as difficult as giving a full account of the structures using analysis. -By the way, I certainly wouldn't like to complain, but it is a bit puzzling to me why some people regard this -question as inappropriate for the site. Since I don't keep too well in touch with cultural trends in the -mathematical world, maybe I am unaware of how much things have changed since I was a -Ph.D. student. In those days, a programme like -Prove the stability of matter, based only on the Schroedinger equation -was regarded as an example of an important -mathematical problem motivated by atomic structure, tackled by people like Fefferman and Lieb. The questions I ask here are hopefully much easier, but still research-level -mathematics to my mind. - -Original question: -This question prompts me to ask something more specific about the periodic table. -As far as I know, the main significance of the periodic table is that -The elements in the same column have similar chemical properties. -For example, the noble gases at the far right are all pretty stable on their -own. The explanation for this is that they have (in the neutral state) a full -outermost shell. Now my question is -Is there an explanation, at some reasonable level of mathematical rigor, of when a new shell -starts? -That is, where do the lengths of the periods -2, 8, 8, 18, 18, 32, 32 -come from? I confess I've been puzzled by this ever since my university physics course. -Allow me to pinpoint my confusion a bit more. I understand that there are some numbers that are important -in atomic structure, and these are 2, 6, 10, 14, and so on. This is because of the occurrence of -the representations -$$V_l\otimes S$$ -inside the Hilbert space for a single particle moving in a central potential. -These are the orbitals one hears about in physical chemistry courses. Here, the $V_l$ are the representations of $SO(3)$ of odd-dimension $2l+l$, while -$S$ is the standard two-dim representation of $SU(2)$. -Since all -the states in a single orbital have the same energy, it is natural that -the breaks will occur after some collection of orbitals are -all filled. -Thus, for a hydrogen atom, -the successive shells have dimensions -$2n^2$ -for $n=1, 2, 3, \ldots$ because the representations $V_l\otimes S$ for -$l=1, 2,\ldots, n-1$ -occur each with multiplicity one inside the $n$-th shell. -So if the periods in the table were of lengths -2, 8, 18, 32, -I would have vaguely assumed that the pattern of shells even in general looks like -that of the hydrogen atom. But of course, among the known elements, each period length that -occurs in the hydrogen atom is repeated twice, except for the first one. So a more -precise question is -Is there a reasonable mathematical explanation of this `multiplicity two' of the -periods? -The little I recall of discussions in standard -textbooks were quite unclear. There are various rules by the name of Hund's rule, -the Aufbau principle, and so on, but I couldn't gather from any of them -*Where the breaks should occur. * -What I do see is that periods end when the orbitals 1p, 2p, 3p, etc. get filled -following the Aufbau principle. (Here, p is the chemist's label for $V_1\otimes S$.) -So perhaps another version -of the question is -Is there some reason that the p-orbitals mark the end of the periods? -To be honest, I've never understood the Aufbau principle either, because I don't know - the rationale behind the principal quantum number for the larger atoms. That is, - the number 4 in the orbital 4p refers to the 4-th energy level in the case of - the hydrogen atom. But for larger nuclei, the $n $ in orbital '$np$' does not refer - to the energy level. (This discrepancy is in fact implied by the Aufbau principle.) So what is - the significance of the $n$ in general that enables them to play some role in - a physical principle? -I realize this question is becoming incoherent already. Nevertheless, I would very much - appreciate clarification on any sensible version of it at a level of mathematical - rigor of your choice. (I am not asking for any axiomatics.) Pointers to an accessible reference would be equally welcome. - -As with the earlier question, a word of explanation is in order on the decision to post this on - Math Overflow. I will draw upon an analogy I read long ago in an article of George Mackey's that - went something like this: Say my mother tongue is Korean. If I would really like - to use English fluently, it is probably best eventually to learn from native - speakers of English. On the other hand, if I would like a good translation into - Korean of English literature, it is better to consult an educated Korean who knows a lot - about English. Of course answers from real physicists will be very gratefully - received, especially if they bear in mind that the query comes from someone who - struggles against a serious language handicap. - -REPLY [6 votes]: I will rearrange the questions you asked into a language I prefer. As Jeff Harvey mentioned, there doesn't seem to be a mathematically precise solution, but I think there are "chemically rigorous" arguments in terms of screening. - -Why the do the high angular momentum orbitals (in a fixed principal energy level) have higher energy than low angular momentum orbitals? - -(revised 21 Nov.) One explanation is that in the no-Coulomb-repulsion limit, an electron in a high angular momentum orbital comes close to the nucleus much less often than an electron in a low angular momentum orbital. There is a nice set of pictures of orbital densities in the Wikipedia article on atomic orbitals where you can get an intuitive idea about the radial distribution. -When Coulomb repulsion is turned off, each electron sees a potential function $Z/r$, and the energy of an orbital depends only on the principal quantum number, not the angular momentum. Suppose you turn on Coulomb repulsion in an approximate way, by assuming you have an ion with $k$ electrons with some a priori charge density, and then add an electron. (In particular, we ignore any effect the extra electron has on the electrons already there, together with the fact that electrons don't really have well-defined individuality.) The screening effect then causes the potential energy function to interpolate between $(Z-k)/r$ far away and $Z/r$ near the nucleus. If we assume the wavefunctions of orbitals don't change too much under this perturbation of potential, we may attempt to compare the wavefunctions of two orbitals with the same principal quantum number but different angular momenta. Unless the interpolation is quite exotic, we should expect the orbital with more density near the center to have lower potential energy. - -Why is it that the 3d level is higher-energy than the 4s level, and more generally, why do the energy levels have a progression of (energy, angular momentum) values like $(2m,0), (m+1, m), (m+2, m-1), \ldots (2m,1)$ and $(2m+1,0), (m+2, m), \ldots, (2m+1,1)$? - -This is a quantitative version of the previous question, which I don't know how to answer in a satisfactory way. Experimentally, it seems that the d and f orbitals are energetically close to the corresponding s orbital, since one often sees s electrons stolen by seemingly higher-energy orbitals in cases like copper, silver, gold, palladium, etc. This suggests that numerical approximations need high precision. - -Why do rows end with full p orbitals? - -The choice of "shell being full" at angular momentum eigenvalue $\ell=1$ is a convention derived from the experimental fact that the corresponding elements are relatively chemically inert. To treat that mathematically, you have to approximate ionization energies and electronegativity (which requires some modeling of bonding behavior). It seems to happen that p-orbital electrons are hard to steal, and s-orbital electrons are not so well-attached. -A rough a posteriori explanation is that for fixed $n$, the energy gap between $(n,1)$ and $(n+1,0)$ is comparatively large and robust under perturbations like adding Coulomb repulsion terms of electrons of level less than $n$. This gap then becomes robust under change of row in the table, once you assume the energies of all other angular momentum eigenvalues fall into the gap between $(n,0)$ and $(n,1)$, as given in the previous question.<|endoftext|> -TITLE: How to see isometries of figure 8 knot complement -QUESTION [16 upvotes]: The figure 8 knot complement $M$ is the orientable double cover of the Gieseking manifold, which implies that $M$ has a fixed-point free involution. If we think of $M$ with its hyperbolic metric, this involution is an isometry. Is there some way of visualizing this isometry? -I know how to produce one relatively easy to see isometry of $M$. The fundamental group of $M$ is a two generator group. Lift geodesic representatives of a pair of generators to $\mathbb{H}^3$. These geodesics have a mutual perpendicular, and $180^{\circ}$ rotation about that geodesic descends to an involution of $M$. However, this map has fixed points, and I'd like to "see" one that doesn't. - -REPLY [3 votes]: In terms of 'seeing' the isometry $\rho$ needed to realize the Gieseking manifold as a quotient of the figure 8 knot complement $S^3-K$, we require a slight of hand, namely thinking of another name for the manifold. The reason for this is simple, the Gieseking manifold is non-orientable so $\rho$ must reverse orientation. However, there are no non-orientable manifolds covered by $S^3$, so we will not be able to 'see' $\rho$ acting on $S^3$. -Instead, we can think of $S^3-K$ by another name as follows. The figure 8 knot complement is fibered. In fact, it can be thought of as a once-punctured torus bundle with monodromy given by $\pmatrix{2 & 1\\ 1 & 1}$. It turns out $\rho$ acts on once-punctured torus bundle by a glide reflection (loosely translate 'half-way' up and reflect).<|endoftext|> -TITLE: Extension of the Hall-Witt identity -QUESTION [11 upvotes]: Although this question is mostly out of curiosity (as of now), I hope it is nevertheless suitable for MO. - -This very recent (and still open) question about the Hall-Witt identity led me to wonder: - - -Does some related generalization to 4 or more variables exist? - - -And also, - - -Does that then imply a 4 variable (or more) Jacobi like identity? - -REPLY [3 votes]: The following article can answer your question: -https://www.researchgate.net/publication/319957676_A_generalization_of_the_Hall-Witt_identity<|endoftext|> -TITLE: Complexity of Membership-Testing for finite abelian groups -QUESTION [9 upvotes]: Consider the following abelian-subgroup membership-testing problem. - -Inputs: - -A finite abelian group $G=\mathbb{Z}_{d_1}\times\mathbb{Z}_{d_1}\ldots\times\mathbb{Z}_{d_m}$ with arbitrary-large $d_i$. -A generating-set $\lbrace h_1,\ldots,h_n\rbrace$ of a subgroup $H\subset G$. -An element $b\in G$. - -Output: 'yes' if $b\in H$ and 'no' elsewhere'. - -Question: Can this problem be solved efficiently in a classical computer? I consider an algorithm efficient if it uses $O(\text{polylog}|G|)$ time and memory resources in the usual sense of classical Turing machines. Notice that we can assume $n= O(\log|G|)$ for any subgroup $H$. The input-size of this problem is $\lceil \log|G|\rceil$. -A bit of motivation. Intuitively it looks like the problem can be tackled with algorithms to solve linear systems of congruences or linear diophantine equations (read below). However, it seems that there are different notions of computational efficiency used in the context of computations with integers, such as: strongly versus weakly polynomial time, algebraic versus bit complexity. I am not an expert on these definitions and I can not find a reference that clearly settles down this question. - -Some possible approaches -The problem is closely related to solving linear system of congruences and/or linear diophantine equations. I briefly summarise these connection for the sake of completion. -Take $A$ to be the matrix whose columns are the elements of the generating set $\lbrace h_1, \ldots, h_n \rbrace$. The following system of equations -$Ax^{T}= -\begin{pmatrix} -h_1(2) & h_2(2) & \ldots & h_n(2)\\\\ -\vdots & \vdots & \cdots & \vdots\\\\ -h_1(m) & h_2(m) & \ldots & h_n(m) -\end{pmatrix}\begin{pmatrix} -x(1) \\\\ -x(2) \\\\ -\vdots \\\\ -x(n) -\end{pmatrix}= -\begin{pmatrix} -b(1) \\\\ -b(2) \\\\ -\vdots\\\\ -b(m) -\end{pmatrix} -\begin{matrix} -\mod d_1 \\\\ -\mod d_2 \\\\ -\vdots \\\\ -\mod d_m -\end{matrix}$ -has a solution if and only if $b\in H$. -If all cyclic factors have the same dimension $d=d_i$ there is an algorithm based on Smith normal forms that solves the problem in polynomial time. In this case, an efficient algorithm from [1] -finds the Smith normal form of $A$: it returns a diagonal matrix $D$ and two invertible matrices $U$ and $V$ such that $D=UAV$. This reduced the problem to solving the equivalent system system $DY=Ub \mod d$ with $D$ diagonal. We can decide efficiently if the system has a solution using the Euclidean algorithm. -The above example suggest that the problem can be solved efficiently using similar techniques in the general case. We can try to solve the system doing modular operations, or by turning the system into a larger system of linear diophantine equations. Some possible techniques to approach the problem that I can think of are: - -Computing the Smith normal forms of $A$. -Computing the row Echelon form of $A$. -Integer Gaussian elimination. - -REPLY [4 votes]: This question has been answered in CS.Theory Stack Exchange . Here I provide a brief summary of the discussion. -The answer to the problem is "yes". - -First, there is a simple efficient classical algorithm for testing-membership in an abelian subgroup in the prescribed set-up [2]. In short, the algorithm is the following. - -Algorithm -(a) Compute a generating-set of the orthogonal [1, 2] subgroup $H^{\perp}$ of $H$. -(b) Check whether or not the element $b$ is orthogonal to $H^{\perp}$. - -The algorithm is correct since, by definition [1, 2], $b$ belongs to $H$ if and only if $\chi_{b}(g_i)=1$ for all generators of $H^{\perp}$. -Moreover, there are efficient clasical algorithms to solve problems (a) and (b). The algorithm for (a) is based on Smith normal forms [1, 2]. (b) can be requires only evaluating the quantities $\chi_b(g_i)$ for all generators $g_i$ of the orthogonal [2]. Since there are a O(polylog($|G|$)) number of them and this can be done efficiently using modular arithmetic we are done . -In the particular case where all $d_i$ are of the form $d_i= N_i^{e_i}$ and $N_i, e_i$ are "tiny integers" then the problem belongs to $\text{NC}^3\subset \text{P}$ (cf. [2]). Tiny integers are exponentially small with the input size: $O(\log\log|A|)$. - -Confer the original discusion for more details.<|endoftext|> -TITLE: Question about Nyman-Beurling-Baez-Duarte Equivalent for Riemann Hypothesis -QUESTION [10 upvotes]: It is well known that the Riemann Hypothesis is true iff $$\lim_{N \rightarrow \infty}d_N=0$$,where -$$d_{N}^{2}=\inf_{A_N}\frac{1}{2 \pi}\int_{-\infty}^{\infty}\vert 1-\zeta A_N(1/2+it)\vert^2\frac{dt}{1/4+t^2}$$ -and the inf is over all the Dirichlet Polynomials of length N.I'm very interested in finding a minimizing polynomial $A_N=\sum_{n=1}^N\frac{a_n}{n^s}$, which is equivalent to minimizing a quadratic form. -I made several experiment with the method in this paper (Le critère de Beurling et Nyman pour l’hypothèse de Riemann: aspects numériques), and it is possible to extend this experiment to Dirichlet L-functions associated to primitive characters. -For a square-free number n, I expect that $$a_n\sim \mu(n)\left(1-\frac{\log n}{\log N}\right)$$ for Riemann zeta function. But the numerical computation shows that $$a_n\sim \mu(n)\left(1-c(n)\frac{\log n}{\log N}\right), $$where those coefficients c(n) behave quite irregularly. I drew histograms of c(n) and computed the average of c(n) over square-free numbers. The average of c(n) is far from 1. It is about 0.8. I made several experiments for primitive Dirichlet-L functions with small modulo q(q=3,4,5),$$ a_n\sim \chi(n)\mu(n)\left(1-c(n)\frac{\log n}{\log N}\right), $$ and the average of c(n) is still about 0.8 (the sum is over square-free number n, (n,q)=1). -Question: It is amazing that this minimizing polynomial is different from Selberg's mollifier (where c(n) is equal to 1). -Is it possible to give an explanation why the cofficients c(n) is close to 0.8 rather than close to 1?(RMT,etc.)? Is it possible that c(n) has limiting distribution? -p.s. -For Riemann zeta function, I chose N=5000, and for Dirichlet L-functions, I chose N=1000. It is beyond the capability of my laptop for larger N (etc,N=10000). - -REPLY [13 votes]: In this very nice recent paper by Bettin, Conrey and Farmer -http://arxiv.org/abs/1211.5191 -it is proven that assuming some very reasonable hypothesis concerning $\zeta(s)$ -that the optimal minimizing polynomial is in fact -$$ -\sum_{n \leq x} \frac{\mu(n)}{n^s} \cdot \bigg ( 1 - \frac{\log n }{\log x} \bigg ) -$$ -Now it is not certain that the minimizing polynomial is unique. I think you can -perturb this polynomial slightly and it will still be a minimizing polynomial. -Concerning the assumption of this paper: Any investigation concerning the optimal -Dirichlet polynomial in the Nyman-Beurling criterion will have to assume the -Riemann Hypothesis and the simplicity of the zeros (in light of the lower bound -obtained by Burnol). In the paper referenced above the simplicitly of the zeros -is assumed in a stronger quantitative form, -$$ -\sum_{T \leq \gamma \leq 2T} \frac{1}{|\zeta'(\rho)|^2} \ll T^{3/2 - \delta} -$$ -This is still a very reasonable conjecture since we in fact expect that -$$ -\sum_{T \leq \gamma \leq 2T} \frac{1}{|\zeta'(\rho)|^2} \sim \frac{6}{\pi^3} T -$$ -This is a conjecture of Gonek. Here, lower bounds which are just a constant factor of $1/2$ away from the -conjectured asymptotic have been obtained by Millinovich and Ng. -Possible explanation of the discrepancy with your numerical experiments: Even in the classical case, when you are mollifying the Riemann zeta-function, the optimal Dirichlet polynomial (the one that solves the optimization problem on the nose) has rather complicated arithmetic coefficients. However when you take the length of the Dirichlet polynomial to go to infinity, each coefficient tends to $(1 - \log n / \log x)$. One can then note that the polynomial with those coefficients is also a minimizing polynomial (it doesn't have to be, but luckily it is). I have once tried computing numerics with the two different polynomial. For small numbers they seem to behave quite differently when mollifying. I think what you are seeing here is basically a small numbers issue. This is something that happens a lot when dealing with the Riemann zeta-function or $L$-functions. I recommend the first few pages in the introduction to -http://arxiv.org/abs/0710.5176 -for a very clear explanation of another situation where the numerics are misleading.<|endoftext|> -TITLE: Growth of groups versus Schreier graphs -QUESTION [7 upvotes]: This question is motivated by this one What is the relation between the number syntactic congruence classes, and the number of Nerode relation classes? where it essentially asks to compare the growth of the syntactic monoid with the growth of the minimal automaton. A special case is the following. Let G be an infinite finitely generated group and H a subgroup such that G acts faithfully on G/H. How different can the growth of G and the Schreier graph of G/H be? -I know the Grigrchuk group of intermediate growth has faithful Schreier graphs of polynomial growth. - - -Are there groups of exponential growth with faithful Schreier graphs of polynomial growth? - - -Schreier graphs of non-elementary hyperbolic groups with respect to infinite index quasi-convex subgroups have non-amenable Schreier graphs so ths should be avoided. - -REPLY [5 votes]: This holds true, for example, for free groups. Actually, take $G$ to be a free product of three copies of $Z/2Z$, which has an index two subgroup which is rank 2 free. The Cayley graph for this group (which has undirected edges) is just a trivalent tree, with edges colored 3 colors by the generators, so that every vertex has exactly 3 colors (this is known as a Tait coloring). Any cubic graph with a Tait coloring corresponds to a Schreier graph of a (torsion-free) subgroup $H$ of $G$, which is the quotient of the Cayley graph of $G$ by the subgroup $H$ (one may choose a root vertex to correspond to the trivial coset). Closed paths starting from the root vertex correspond to elements of the subgroup $H$. -Choose a cubic graph with a Tait coloring which has linear growth and corresponds to a subgroup $H$ satisfying your condition ($G$ acts faithfully on $G/H$). This is equivalent to $\cap_{g\in G} gHg^{-1}=\{1\}$. For example, take a bi-infinite ladder, labeling the two stringers with matching sequences of colors, which then determine the colors of the rungs. By making these stringer sequences aperiodic, you can guarantee that $\cap_{g\in G} gHg^{-1}=\{1\}$. Changing the root vertex corresponds to changing the conjugacy class. In fact, we may choose stringer sequences which contain any word in $G$. Then putting a root at the endpoint of such a word, we guarantee that it is not in the corresponding conjugate subgroup of $H$.<|endoftext|> -TITLE: degenerating surface -QUESTION [6 upvotes]: Hi, -i have a sequence of immersed disc $u_n: \mathbb{D} \rightarrow \mathbb{R}^3$ which converge to a singular cover of the disc: $z^k$ for $k\geq 2$, moreprecisely $u_n \rightarrow z^k$ in $C^2(\mathbb{D})$. Of course the Gauss curvature of the image $\Sigma_n=u_n(\mathbb{B})$ blows up thanks to Gauss-Bonnet formula : $\int_{\Sigma_n}K= (1-k)2\pi + o(1)$. My questions are the following: -1) can the Gauss curvature be bounded form above? i.e the blow-up come only from necks and there is no pinching region... my feeeling is no, since you have to close the surface. -2) Extra bonus: same question with only a convergence in $C^2_{loc}(\mathbb{D}\setminus \{ 0\})$, here we allow the closing of the surface be made by a big sphere for example. - -REPLY [9 votes]: When $k$ is odd, there does exist such a family of immersions satisfying Paul's requirements. (When $k$ is even, Vitali Kapovich has shown, using a clever topological argument, that it's not possible to have such a family of degenerating immersions. Please see his answer for the details.) -Set $k=2m+1$, and consider the (complex) $1$-parameter family of maps $u_t:\mathbb{C}\to\mathbb{R}^3$ given by -$$ -u_t(z) = \bigl(Re(z^{2m+1}-(2m{+}1)t^2z),\ Im(z^{2m+1}+(2m{+}1)t^2z),\ - \tfrac{4m+2}{m+1} Re(t z^{m+1})\ \bigr). -$$ -These smooth maps converge smoothly to $u_0$ as $t\to0$, -and $u_t$ induces the metric -$$ -ds_t^2 = (2m{+}1)^2\bigl(|z|^{2m}+|t|^2\bigr)^2 |dz|^2. -$$ -Thus, $u_t$ is an immersion for $t\not=0$, -while $u_0(z) = \bigl(Re(z^{2m+1}),\ Im(z^{2m+1}),\ 0\ \bigr)$. -The family $u_t$ was constructed using the Weierstrass formula for minimal immersions, so the image $u_t({\mathbb{C}})\subset \mathbb{R}^3$ is an immersed minimal surface, and, as a result, the Gauss curvature is everywhere non-positive. In fact, for $t\not=0$, the Gauss curvature only vanishes at $z=0$, and then only when $m>1$. In particular, all of these degenerating immersions have curvature bounded above.<|endoftext|> -TITLE: Coherent sheaf with connection is locally free? -QUESTION [11 upvotes]: I think it is something like a folkore result that a coherent sheaf $\mathcal F$ on a smooth algebraic variety $X$ over $k$, which is equipped with a connection -$\nabla: \mathcal F \rightarrow \mathcal F \otimes \Omega^1_{X/k}$ -is already locally free. -Maybe one may weaken the assumptions, but I think the proof wouldn't alter very much. -I thought about how to prove it, but couldn't make it rigorous. In any case one should show that the stalks $\mathcal F_x$ are free, and this somehow must follow from the existence of the connection. -Addendum: It seems that the existence of a connection in some way rigidifies the underlying sheaf. -A related question in this respect is: in how much is a horizontal morphism $\phi: \mathcal F \rightarrow \mathcal F$ already rigidified by $\nabla$. E.g. if one knows that $\phi$ is the identity on $\mathcal F(x) \rightarrow \mathcal F(x)$, then is it so already around $x$? - -REPLY [8 votes]: See Prop. 8.8, p. 206 in N. Katz, "Nilpotent connections and the monodromy...", Publications Mathématiques de l'IHES, 39 (1970), p. 175-232.<|endoftext|> -TITLE: Elementary results with p-adic numbers -QUESTION [56 upvotes]: I'm giving a talk for the seminar of the PhD students of my math departement. I actually work on Berkovich spaces and arithmetic geometry but, of course, I cannot really talk about that to an audience that includes probabilists, computer scientists and so on. -I'd rather like to do an introduction to $p$-adic numbers and $p$-adic analysis. I think these kind of things come up to be really cool when you work on it even just for a short time, but I have the ambitious aim to show them something nice and elementary whose statement will be understood by everyone (of course the proof may also be really hard, but there I could give them just its general idea). -In other words the question is: if I had prepared something about Galois theory I would have finished with the application to resolubility of polynomial or compass and straightedge constructions; if it had been something about modular forms, it would have been for sure Fermat's last theorem; with 3-surfaces it would have been Poincaré conjecture and so on. What if it's about p-adic numbers or p-adic analysis? -I thought about results on valuations of roots of polynomials, but it seems to me already too complicated (par ailleurs since I'm introducing valuations at the beginning of the talk, it won't turn out to be an application to something they already knew). - -REPLY [4 votes]: I'm a bit surprised that nobody mentions the interesting fact that $\log(-1)=0$ in $\mathbb{Q}_2$. -This result could be stated without involving $p$-adic numbers: the partial sums of the series $\sum_{i=1}^\infty 2^i/i$ have $2$-adic valuations tending to infinity. -Same for the dilogarithm $\log_2$: one has $\log_2(-1)=0$ in $\mathbb{Q}_2$.<|endoftext|> -TITLE: Exceptional isomorphisms of Lie groups -QUESTION [16 upvotes]: It is known that in low dimensions certain exceptional isomorphisms arise between Lie groups. I have read about some of them in some papers, but I have not been able to find a "systematic" treatment of such isomorphims. By "systematic", I mean a presentation of a list of the most relevant of them (in the real and complex case, for instance) with some proofs and explanations. So, my question is, Does there exists any relatively complete reference on this topic? -Thanks in advance - -REPLY [12 votes]: The question goes back to paper A: Cartan's 1914 classification of real simple Lie algebras (Collected papers, Part I, Vol.1) with essentially complete answers for the Lie algebras . A complete discussion together with explicit isomorphisms for all the cases (both the Lie algebras and the Lie groups ) is in my book - B: Diff. Geometry, Lie Groups and Symmetric Spaces, pp. 517- 528. -Comments: On the first page of A (bottom) Cartan states that real forms of the same simple Lie algebra over C are in general commpletely determined by the signature of the Killing form. As shown in B p. 517 this is contradicted by so*(18) and so(12,6) as well as by su*(14) and su(9,5). Cartan's phrase "general" clearly and deliberately meant "almost all the time". However, at the end of the paper (p. 352-355) he identifies real forms on the basis of the signature statement (which would require some additional caution). As mentioned in B, p. 520 the isomorphism so*(8) ~ so(6,2) does not occur in Cartan's original list although in a later paper he shows that the corresponding symmetric spaces are isometric. -Generally the group isomorphism in B are computational although some have geometric interpretation like the isomorphism so(3)~su(2) which comes from the standard stereographic projection. -S. Helgason<|endoftext|> -TITLE: Topological proof of the Compactness Theorem in propositional logic without the Axiom of Choice -QUESTION [10 upvotes]: There is a well-known proof of the Compactness Theorem in propositional logic which uses the compactness of the space $\{0,1\}^P$, where $P$ is the set of propositional variables in consideration. In general, this compactness relies on the Tychonoff theorem which in turn requires the Axiom of Choice. Let me sketch it (in danger of boring the experts, but for reference it will be useful): -For a set $A$ of formulas in $P$, let $T(A)$ be the set of interpretations $P \to \{0,1\}$ which make the formulas in $A$ true. Then $T(\cup_i A_i)=\cap_i T(A_i)$ and each $T(A)$ is closed in $\{0,1\}^P$. Thus if $A$ is finitely consistent, then $T(A)$ is the directed intersection of nonempty closed sets, thus also nonempty (compactness). -My question concerns the case that $P = \{p_1,p_2,\dotsc\}$ is countable. Then it seems to me that it is provable in ZF that $\{0,1\}^P$ is compact: Use the homeomorphism to the closed cantor set $C \subseteq [0,1]$ and the compactness of $[0,1]$. From this we can conclude that the Compactness Theorem in propositional logic for countably many propositional variables is provable in ZF - if everything is correct so far. -Isn't this somehow counterintuitive? Or does this proof yield a constructive algorithm how to find an interpretation which makes all formulas in $A$ true provided one can find them for finitely many? - -REPLY [11 votes]: The proof is correct, but it does not provide an algorithm unless you have some additional information about $A$. If you untangle the proofs, you find the following pseudo-algorithm: Go through the propositional variables $p_i$ in order, adding each $p_i$ or its negation to $A$ "greedily", i.e., add $p_i$ if doing so leaves $A$ (which now contains the original $A$, all your previous additions, and $p_i$) finitely satisfiable, and otherwise add $\neg p_i$. An easy induction shows that this preserves finite satisfiability. At the end, your decisions tell you what truth value to give each $p_i$, and it's easy to check that $A$ is satisfied by this truth assignment. The reason this is only a pseudo-algorithm rather than an algorithm is that there is, in general, no way to decide what happens at any step. You'd need to able to decide whether a certain set of formulas (the original $A$ plus decisions) is consistent, and that might not be algorithmically decidable, even if the original $A$ was a computable set of formulas. - -REPLY [3 votes]: In P. Johnstone "Notes on Logic and Set theory" there is a proof of completeness (and then compactness) of propositional logic for a countable set of base atomic proposition, avoiding the choise axiom (in the exercise Johnstone prove it in general by Zorn Lemma). -If you want I send you the outline of the proof. \ -THE OUTLINE: -Let $S\subset Bool(P)$ a set of propositions (where $Bool(P)$ is the set of propositional expressions building from a set of primitives proposition alphabet $P=${p, q, r, s..} this is the free Boolean algebra on the set $P$). A valuation is a funtion $v: P\to ${$1$}, any valuation as a unique natural extension to $S$ that indicate still by the some letter $v: S\to ${$1$}. We write $S\models s$ if any valuation $v$ such that is 1 (true) on all elements of $S$ is $1$ also on $s$. We write $S\vdash s$ is there is a deduction of $s$ from $S$. We call $S$ inconsistent if $S\vdash \bot$ and consistent if $S\vdash \bot$ is false. -Then $S\models \bot$ means that any valuation $s$ is $0$ (false) on each members of $S$ (where $\bot$ is the atomic expression "false"). -LEMMA1. -If $S\models \bot$ then $S\vdash \bot$ (i.e. $S$ is inconsistent). -PROOF (in the case of $P$ countable): We have to suppose $S$ consistent, and show a valuation $v$ on $P$ such that $v(s)=1$. -Observe that for $t\in Bool(P)$ we have that either $S\cup${$t$} or $S\cup${$\neg t$} is consistent (if $S\cup${$t$}$\vdash \bot$ from deduction lemma follow that $S\vdash (t\to \bot)$ i.e. $S\vdash \neg t$ (we can define $ \neg t:= t\to \bot$), and then $S\models \neg t$ then is $S$ is consistent (hypothesis) so is $S\cup${$\neg t$}). -Then we enumerate the elements of $Bool(P)$ and one to one add it to $S$ is preserve the consistence, ot add its negation in opposite. We have a set $S'\supset S$ which is consistent (is $S'\vdash \bot$ then exist a dedution of $\bot$ from $S'$ that involve a finite number of elements of $S'$), and such that for any $t\in Bool(P)$ or $t$ ot $\neq t$ belong to $S'$. Let $v(t)=1\ if\ t\in S'$ and $v(t)=0\ if\ t\not\in S'$ . We claim that $v$ is a valutation i.e. preserve the boolean operations: clearly $v(\bot)=0$, and its enought show that $v(s\to t)= (v(s)\to v(t))$, we shall consider three cases: -1) $v(t)=1$ i.i $t\in S'$ Then we cannot have $\neg (s\to t)\in S'$ since $t\vdash (s\to t)$ and $S'$ is consistent. So $s\to t)\in S'$ i.e. $v(s\to t)=1= v(s)\to v(t))$. -2) $v(s)=0$, then $\neg s\in S'$, and since $\neg s \vdash (s\to t)$ (exercise) as in $(1)$ we must have $v(s\to t)=1= v(s)\to v(t))$. -3) $v(s)=1,\ v(t)=0$ . In this case we have $s\in S'$ and $t\not\in S'$ so since {$s,\ s\to t$}$\vdash t$ we have $(s\to t)\not\in S'$ hence $v(s\to t)=0= v(s)\to v(t))$. -Theorem of Completeness: $S\models t\ iff\ S\vdash t$. -PROOF. The "$if$" part is obvious. Suppose $S\models t$, from {$t, \neg t$}$\vdash \bot$ we have $S\cup${$\neg t$}$\models \bot$, by lemma we have $S\cup${$\neg t$}$\vdash \bot$, by deduction lemma we have -$S\vdash (\neg t)\to \bot$ then $S\vdash t$ . -Theorem of Compactness: is $S\models t$ then $S\vdash t$ then exist a finite $S'\subset S$ such that $S'\vdash t$ and then $S'\models t$.<|endoftext|> -TITLE: Numerical coincidence involving the number 1663 -QUESTION [5 upvotes]: The numerical coincidence -$\displaystyle \frac{1663e^2}{3} \approx 2^{12}$. -showed up in a comment of this mostly-unrelated question. -Numerically, it's not a surprise that $e^2$ is close to a rational number whose numerator and denominator are in this range — similarly good approximations to most numbers can be obtained by truncating the continued fraction at the desired level of accuracy. What's much more of a surprise to me is the appearance of a 12th power in this expression. Is there a good explanation for having such a smooth number here? E.g. see the j-invariant explanation for Ramanujan's observation that -$e^{\pi\sqrt{163}}$ -is close to an integer, or the Pisot number explanation for the fact that even powers of the golden ratio are close to integers. - -REPLY [3 votes]: It can be related to the simpler -$$\frac{13e^2}{3}\approx 2^5$$ -because -$$ \frac{3}{e^2} \approx \frac{1663}{2^{12}} = \frac{13·2^7-1}{2^{12}}= \frac{13}{2^5}-\frac{1}{2^{12}}$$ -Further decomposition into alternating sign bits is -$$\frac{1663}{2^{12}} = \frac{1}{2}-\frac{1}{2^3}+\frac{1}{2^5}-\frac{1}{2^{12}}$$<|endoftext|> -TITLE: interlacing roots/eigenvalues results and modern analogues -QUESTION [17 upvotes]: Is there any relation between these theorems on interlacing roots? - -The roots of $f(x), f'(x)$ interlace (if all the roots of $f(x)$ are real and have real coefficients). -The eigenvalues of an $n \times n$ matrix $A$ and it's minor $n-1\times n-1$ matrix $A_{ii}$ interlace (is a real symmetric matrix, or Hermitian). - -Is there a more modern version of these results? - -**EDIT**: Picturing Terry Tao's answer, let $f(x)= (x-1)(x-2)\dots(x-9)(x-10)$. For aesthetic reasons, the roots are perturbed slightly. -A plot of $x \mapsto (f(x), f'(x))$ for 1 < x < 7. This figure is symmetric, and not very representative. - -REPLY [2 votes]: A rather tenuous connection, but I'll throw it out and hope somebody can build a better answer from this (in short, too long for a comment): -It is well known (or it should be!) that one can use the Sturm sequence construction for a polynomial $p(x)$ with all roots real and its derivative to construct a symmetric tridiagonal companion matrix (that is, the symmetric tridiagonal matrix whose characteristic polynomial is $p(x)$). See Fiedler's paper for details. -One could then apply Cauchy's interlacing theorem on the tridiagonal matrix just constructed. As a matter of fact, the characteristic polynomials of successive leading submatrices of a tridiagonal form a Sturm sequence. - -To expand on the comment I gave regarding orthogonal polynomials: consider the monic orthogonal polynomials $p_n(x)$ satisfying the difference equation -$$p_{n+1}(x)=(x-c_n)p_n(x)-d_np_{n-1}(x),\qquad p_{-1}(x)=0,p_0(x)=1$$ -One can associate a symmetric tridiagonal matrix with this recursion, called a Jacobi matrix: -$$\begin{pmatrix}c_0&\sqrt{d_1}&&\\\sqrt{d_1}&\ddots&\ddots&\\&\ddots&&\sqrt{d_{n-1}}\\&&\sqrt{d_{n-1}}&c_{n-1}\end{pmatrix}$$ -whose characteristic polynomial is $p_n(x)$. It can be seen that the characteristic polynomials of the leading $1\times1, 2\times 2, \dots$ submatrices are $p_1(x),p_2(x),\dots$ We see here the correspondence between the Sturm sequences for orthogonal polynomials and tridiagonal matrices.<|endoftext|> -TITLE: An optimization problem involving sum of binomial coefficients upto some value -QUESTION [5 upvotes]: I would like to minimize $f(s, n, \epsilon)$ with respect to $s$ where -$$f(s,n,\epsilon) = \left( 1 + \frac{n}{2^s} \right)\frac{1}{s} \sum_{k=0}^{\lfloor s\epsilon \rfloor} {s \choose k}~.$$ -Note that $0 < \epsilon < \frac{1}{2}$ and $n > 0$. -Clearly optimal $s$ is going to be a function of $\epsilon$ and $n$, which might be ugly. However, I think $s^*$ should be close to $\log_2{n}$, based on the intuition from the problem giving rise to this, but I cannot find any rigorous argument for this choice of $s^*$. Any hint or idea is highly appreciated. -The following upper bound might be helpful: -$$ \sum_{k=0}^{\lfloor s\epsilon \rfloor} {s \choose k} \le 2 ^ {H(\epsilon)s } $$ -where $H(\epsilon) \equiv -\epsilon \log \epsilon -(1-\epsilon) -\log(1-\epsilon)$; the entropy of a Bernoulli dist. with probability $\epsilon$. -I don't know of any clean lower bound. Any idea? -Thanks for your time in advance. - -REPLY [4 votes]: I assume you have $n\to\infty$. If $\epsilon<\frac12$ and $s\to\infty$ then $$\sum_{k=0}^{\lfloor \epsilon s\rfloor} \binom{s}{k}$$ is very close to a geometric progression at the big end. Together with Stirling's approximation (ignoring the floor), this gives -$$\sum_{k=0}^{\lfloor \epsilon s\rfloor} \binom{s}{k} - \approx \frac{1-\epsilon}{1-2\epsilon}\binom{s}{\lfloor \epsilon s\rfloor} - \approx \frac{1}{\sqrt{2\pi s}}\frac{1-\epsilon}{1-2\epsilon} (\epsilon(1-\epsilon))^{-s-1/2}. -$$ -Now you can look for the minimum wrt $s$ by differentiating. I don't get a closed form but it seems that the minimum is around $s=\log_2(n)+O(1)$. -ADDED: Note that the function is not continuous. Due to the floor function, it jumps up by a ratio close to $(1-\epsilon)/\epsilon$ as $\epsilon s$ passes an integer. So there are very many local minima. However, the global minimum ought to be close to the minimum of the function without the floor.<|endoftext|> -TITLE: Does S^6 have the structure of an algebraic variety? -QUESTION [6 upvotes]: More accurately, the question should be: Is it known that $S^6,$ the 6-dimensional sphere, is $not$ a (proper) complex algebraic variety, or algebraic space? And is there a reference? It's easy to see that it cannot be projective (as $H^2=0$), but I don't see how it can violate the usual general properties of algebraic varieties [e.g. by Chow's lemma one can get a projective (and smooth, by resolution of singularities) variety lying over and birational to it, but the cohomology groups can get larger when we go up...]. Maybe one needs some theory of classification of 3-folds (if there is one such theory). -Knowing that it cannot be algebraic will somehow make the analogous question of complex structure much sharper (by the way, it does have an almost complex structure). -Thank you. - -REPLY [22 votes]: Suppose that $X$ is a smooth complete positive-dimensional algebraic space over $\mathbb C$. Then $\mathrm H^2(X, \mathbb Q)$ can not be 0. In fact, every algebraic space contains an open dense subscheme. Let $U \subseteq X$ be an open affine subscheme; by a well known-result, the complement $C$ of $U$ has pure codimension 1; consider $C$ as a divisor, with its reduced scheme structure. I claim that the cohomology class of $C$ can not be 0. Let $Y \to X$ a birational morphism, where $Y$ is a smooth projective variety; this exists, by Chow's lemma for algebraic spaces, and resolution of singularities. The pullback of $C$ is a non-zero effective divisor on $Y$, so its cohomology class is not 0. Since formation of cohomology classes of divisors is compatible with pullbacks, this implies the result. -This says very little about the problem of existence of a complex structure. Complete algebraic spaces, or, if you prefer, Moishezon manifolds, are not so distant from projective varieties; it seems clear to me that if $S^6$ has a complex structure, this will be a very exotic animal, very far away from the world of algebraic geometry. -[Edit]: why does $C$ have codimension 1? This is Corollaire 21.12.7 of EGAIV, when $X$ is a scheme. In general you can reduce to the case of a scheme by considering an étale morphism $Y \to X$, where $Y$ is an affine scheme. This map is affine, because $X$ is separated, so the inverse image of $U$ in $Y$ is affine. - -REPLY [5 votes]: For whatever it's worth: $S^6$ is a ($6$-dimensional) homogenous nearly Kähler manifold. These were classified by Butruille. In particular, they have a canonical almost complex structure. I assume one should be able to prove that this almost complex structure is not a complex structure. -[EDIT, thanks Spiro] Apparently this does not imply that one may conclude that there is no complex structure on $S^6$, so this is not much help. -There are also some interesting uniqueness results for connections on $S^6$ in Fukami and Ishihara's paper that might be helpful for this. -[EDIT, thanks David] For a short time there was a preprint on arXiv claiming that $S^6$ is a complex manifold. See discussion here.<|endoftext|> -TITLE: A problem concerning $L^2([0,1]\times[0,1])$ -QUESTION [10 upvotes]: Trying to solve a conjecture in differential geometry, I am leaded to the following problem (which may seem weird to a analyst). I wonder if anyone know some techniques that happen to solve it. -Let $f$ be $\mathbb{Z}^2$-periodic positive-real-valued function on $\mathbb{R}^2$, which you may assume to be as soomth as you like. Let $h$ be another such function without the positivity assumption. We suppose that both $f$ and $h$ are unit vectors in $L^2([0,1]\times[0,1])$ (defined with the usual Euclidean measure), and they are orthogonal. -Problem: Is it always possible to write $h$ as a sum of two functions $a,b\in L^2([0,1]\times[0,1])$ with the following conditions: -(1) $\int_0^1 f(x,y)a(x,y)dx=0$ for any $y$ -(2) $\int_0^1 f(x,y)b(x,y)dy=0$ for any $x$ -(3) $\int_0^1\int_0^1a(x,y)b(x,y)dxdy\geq 0$ -Remark: Using Fourier expansion, the proof is easy if $f(x,y)$ is a product such as $\cos(2\pi mx)\sin(2\pi ny)$ (although this $f$ is not positive). In fact in this case we may take $a$ and $b$ to be orthogonal, instead of having inner product $\geq 0$. -But for general $f$, the analysis of Fourier coefficients seems hard. - -Addendum: As fedja mentioned below, the above assertion is not always true. So I feel like to mention here another assertion which a consequence of the above problem. You can also try to prove it directed. This would provide the last ingredient that I need -to solve some conjecture. -For unit vectors $f, h_1, h_2, h_3, h_4\in L^2([0,1]\times[0,1])$ such that each $h_i$ is orthogonal to $f$, we define -$$ -I(f,h_1,h_2,h_3,h_4)=\int_0^1[\sum_{i=1}^4(\int_0^1fh_idy)^2]^{1/2}dx\times \int_0^1[\sum_{i=1}^4(\int_0^1fh_idx)^2]^{1/2}dy -$$ -we consider $I$ as a functional depending on a point $f$ in the unit sphere of $L^2([0,1]\times[0,1])$ and four unit tangent vectors of the unit sphere at $f$. -Then the assertion is that the maximal value of $I$ is $2$. -(The value $2$ is achieved when, for instance, $f=1$ and $h_1=\sqrt{2}\cos(2\pi x)$, $h_2=\sqrt{2}\sin(2\pi x)$, $h_3=\sqrt{2}\cos(2\pi y)$, $h_4=\sqrt{2}\sin(2\pi y)$.) -A first attempt to prove this is to apply Cauchy-Schwarz two times, one can then get $I\leq 4$, which is not enough. Then I came up with the idea: If the $h_i$'s can be written as $a_i+b_i$ satisfying above conditions (1), (2) and (3), then the proof goes through. - -REPLY [6 votes]: Alas, Cauchy-Schwarz is the best you can do. Indeed, imagine $n$ squares with side $1/n$ arranged along the diagonal of the unit square. Now let $f$ be $\sqrt n$ on each such square and $0$ outside (well, you said "positive", but since it is not quantitative, it is just as good as "non-negative"). Let $h_i$ be $\pm\sqrt n$ on each such square and $0$ outside with $\pm$ chosen to satisfy orthogonality relations (which is possible with $n=16$, say). Now, all inner integrals are $1$ in absolute value, so you get exactly $4$ instead of the $2$ you were looking for.<|endoftext|> -TITLE: What is growth of ass. algebra with 3 generators and relation a1a2a3 + a2a3a1 +a3a1a2 - a1a3a2 - a2a1a3 -a3a2a1 ? -QUESTION [5 upvotes]: Consider ass. algebra with 3 generators a1 a2 a3 and relation: - a1a2a3 + a2a3a1 +a3a1a2 - a1a3a2 - a2a1a3 -a3a2a1 = 0. -i.e. $$ \sum_{ s \in S_3} (-1)^{sgn (s)} a_{s(1)} a_{s(2)} a_{s(3 )} = 0.$$ -Informal questions: how far this algebra is from commutative polynomial algebra k[a1 a2 a3] ? -What is known about this algebra ? -Formal question: what is the Hilbert series of this algebra ? -==== -Some reformulations of the defining condition: -Consider 3 Grassman variables $\psi_i$ i = 1,2,3. -Define $$ \psi = a_1 \psi_1 + a_2 \psi_2 + a_3 \psi_3$$ -The condition above is equivalent to -$$ \psi ^3 =0 $$. -For commutative algebras we clearly have $\psi ^2=0$. So this algebra extends the commutative ones in certain sense. -So I wonder how far it is away from commutative one ? -Yet another way to reformulate the defining relation is the following:denote by "M" 3* matrix: -M = -a1 a1 a1 -a2 a2 a2 -a3 a3 a3 -The condition above is the same as $det^{column} M =0$ where column-determinant is used i.e. first take elements from the first column, second from the second and so forth. -====== -Some example. -Consider $E_{ij}$ - "elementary matrices" i.e. n*n matrix with zeros everywhere except position (i,j) where we put 1. -Take for example $E_{11} , E_{21} , E_{31} $, -Observation: -they satisfy the above relation. -More generally one can take $E_{i p} , E_{j p} , E_{k p} $ (important the second index is the same). -This means that the algebra above admits a homomorphism to universal enveloping of $E_{11} , E_{21} , E_{31} $. Universal enveloping algebras are very close to commuttive (at least their size is the same). So it suggests that in general such algebra is close to commutative, -but probably this is wrong... -====== -It seems Roland Berger discusses similar alegbras at section 3 of -http://arxiv.org/abs/0801.3383 -as far as I can understand he proves that such algebra is N-Koszul (i.e. generalization of Koszul duality to non-quadratic algebras). -But I cannot get far in his theory. - -REPLY [4 votes]: (edited a bit to cover a few questions about the notation) -A slight simplification of David Speyer's argument: his argument using Groebner bases explains that is we degenerate the relation into $a_1a_2a_3=0$, the resulting algebra has the same Hilbert series. Now, the latter algebra $B$ has a very economic resolution of the trivial module by free right modules: - $$ -0\to span_k(a_1a_2a_3)\otimes_k B\to span_k(a_1,a_2,a_3)\otimes_kB\to B\to k\to 0 - $$ -(the leftmost differential maps $a_1a_2a_3\otimes 1$ to $a_1\otimes a_2a_3$, the next one maps $a_i\otimes 1$ to $a_i$). Computing the Euler characteristics, we get H_B(t)(1-3t+t^3)=1$.<|endoftext|> -TITLE: Noether normalization vs. normalization of varieties -QUESTION [6 upvotes]: As far as I can tell, Noether normalization uses the term "normalization" in the English sense, that something has been given a standard form. And as such it's not intrinsically related to normalization in the sense of "take the integral closure in the ring of fractions". I would leave the matter at that, except that in book after book, the two are introduced and studied in very close proximity. Am I missing something? - -REPLY [5 votes]: What is the statement of Noether normalization theorem you are referring to? -The one I know is the following, that you can find for instance in Shafarevich's book "Basic Algebraic Geometry I", page 65. - -Theorem (Noether normalization). For an irreducible, affine variety $X$ there exists a finite map $$\varphi \colon X \longrightarrow \mathbb{A}^n$$ to an affine space. - -That means that any integral domain $A$ which is finitely generated over the field $k$ is integral over a subring isomorphic to a polynomial ring. -In particular, let $X \subset \mathbb{A}^{n+1}$ be any degree $d$ hypersurface. Then its "normal form" is given by a monic equation -$$z_{n+1}^d + f_{d-1}z_{n+1}^{d-1}+ \ldots + f_1 z_{n+1}+f_0=0,$$ -with $f_i \in k[z_1, \ldots, z_n]$, and the map $\varphi \colon X \longrightarrow \mathbb{A}^n$ in Noether theorem is simply the projection onto the first $n$ coordinates $z_1, \ldots, z_n$. -This shows how the two concepts of "normalization" are closely related.<|endoftext|> -TITLE: What is a symmetric monoidal $(\infty,n)$-category? -QUESTION [17 upvotes]: This question arose from reading Jacob Lurie's "Classification of topological field theories" paper. In that paper, he uses complete $n$-fold Segal spaces as a model for $(\infty,n)$-categories, but he "glosses over" the question what a symmetric monoidal $(\infty,n)$-category is (a notion that is central for the whole theory). So here is the question: - -What structure on an $n$-fold complete Segal space does turn it into a symmetric monoidal $(\infty,n)$-category, and where can I read about it? - -REPLY [30 votes]: There are many (equivalent) definitions for the notion of symmetric monoidal $(\infty,n)$-category. One approach is based on the observation that a monoidal category can be identified with a bicategory having only a single object. You can define a monoidal -$(\infty,n)$-category to be an $(\infty,n+1)$-category with a specified object, -such that all other objects are isomorphic to it (in the complete Segal space model, this means that the space of objects should be connected). Similarly you can define a braided monoidal $(\infty,n)$-category to be an $(\infty,n+2)$-category equipped with a distinguished object satisfying a simple connectivity condition, and so on and so forth. You get to the symmetric monoidal case by taking the homotopy inverse limit (that is, a symmetric monoidal -$(\infty,n)$-category is a collection of pointed $(\infty,n+k)$-categories, each of which is obtained by "looping" the next one). You might find this definition convenient in the context of bordism categories, since they are naturally related in this way (if you ``loop'' the a bordism category of $d$-manifolds, you get a bordism category of $d+1$-manifolds: and this is sensible even when $d$ is negative). -Alternatively, you can define a symmetric monoidal $(\infty,n)$-category to be a commutative monoid in the setting of $(\infty,n)$-categories. There are many ways to formalize this. Since you're asking about specific models, let's suppose you start with some model category $\mathbf{A}$ for the homotopy theory of $(\infty,n)$-categories (higher-dimensional Segal spaces, for example). If you have a simplicial model category, you can do as Charles suggested and take algebras for some $E_{\infty}$-operad in simplicial sets. If you'd prefer not to mention operads, you can just copy Segal's definition of a $\Gamma$-space: take the category of functors $F$ from pointed finite sets into $\mathbf{A}$, and equip it with a model structure that enforces the relevant Segal condition. -As Martin mentions in his answer, there is an extensive discussion of the case $n=1$ in my book, and also of commutative monoids in an arbitrary $(\infty,1)$-category (of which this is a special case, since the collection of all $(\infty,n)$-categories can be regarded as an $(\infty,1)$-category). I don't know of references that address your question more specifically (though I would not be surprised if there were some).<|endoftext|> -TITLE: Is the set of undecidable problems decidable? -QUESTION [9 upvotes]: I would like to know if the set of undecidable problems (within ZFC or other standard system of axioms) is decidable (in the same sense of decidable). Thanks in advance, and I apologize if the question is too basic. - -REPLY [8 votes]: For every consistent recursively axiomatizable theory $T$ (one of which $ZFC$ is widely believed to be) there exists an integer number $K$ such that the following Diophantine equation (where all letters except $K$ are variables) has no solutions over non-negative integers, but this fact cannot be proved in $T$: - -\begin{align}&(elg^2 + \alpha - bq^2)^2 + (q - b^{5^{60}})^2 + (\lambda + q^4 - 1 - \lambda b^5)^2 + \\ -&(\theta + 2z - b^5)^2 + (u + t \theta - l)^2 + (y + m \theta - e)^2 + (n - q^{16})^2 + \\ -&((g + eq^3 + lq^5 + (2(e - z \lambda)(1 + g)^4 + \lambda b^5 + \lambda b^5 q^4)q^4)(n^2 - n) + \\ -&(q^3 - bl + l + \theta \lambda q^3 + (b^5-2)q^5)(n^2 - 1) - r)^2 + \\ -&(p - 2w s^2 r^2 n^2)^2 + (p^2 k^2 - k^2 + 1 - \tau^2)^2 + \\ -&(4(c - ksn^2)^2 + \eta - k^2)^2 + (r + 1 + hp - h - k)^2 + \\ -&(a - (wn^2 + 1)rsn^2)^2 + (2r + 1 + \phi - c)^2 + \\ -&(bw + ca - 2c + 4\alpha \gamma - 5\gamma - d)^2 + \\ -&((a^2 - 1)c^2 + 1 - d^2)^2 + ((a^2 - 1)i^2c^4 + 1 - f^2)^2 + \\ -&(((a + f^2(d^2 - a))^2 - 1) (2r + 1 + jc)^2 + 1 - (d + of)^2)^2 + \\ -&(((z+u+y)^2+u)^2 + y-K)^2 = 0. -\end{align} - -The set of numbers with this property is not recursively enumerable (let alone decidable). But curiously enough, one can effectively construct an infinite sequence of such numbers from the axioms of $T$. -The equation is derived from Universal Diophantine Equation by James P. Jones.<|endoftext|> -TITLE: adelic quadratic forms -QUESTION [6 upvotes]: The classification of quadratic forms over local and global fields is well understood. But what about quadratic forms over adele rings? Let G = GL(n,A), where A is the adele ring of a global field F. Let S be the set of symmetric matrices in G. Let G act on S by g * s = g s (transpose g). What are the orbits? Even for n = 1 and F = Q, not every orbit contains a rational point. - -REPLY [2 votes]: EDIT: I have never seen this one, but it is said to cover some of the same ground as Kneser (1961): A. Weil, Sur la theorie des formes quadratiques (1962). -ORIGINAL: I did not notice this earlier...for what it may be worth, the introduction of adeles into the consideration of quadratic forms may well originate with Martin Kneser, Darstellungsmasse indefiniter quadratische Formen, Math. Z. volume 77 (1961) pp 188-194. Also MR0140487. -Similar in spirit, Rainer Schulze-Pillot, Darstellungsmasse von Spinorgeschlectern tern$\ddot{a}$rer quadratische Fromen, J. Reine Angew. Math. vol. 352 (1984), pp 114-132. Also MR758697.<|endoftext|> -TITLE: Theorem of Bryant in higher dimensions -QUESTION [18 upvotes]: I have the following question. I read about Bryant's theorem which says that: any real-analytic 3-dimensional Riemannian manifold $(Y,g)$ with real-analytic metric $g$ can be isometrically embedded as a special Lagrangian submanifold of some Calabi-Yau manifold $(X, \Omega, \omega)$. My question is: does this result also hold in dimensions greater than 3? Or is there any possibility to establish this? Thanks in advance. -Mira - -REPLY [60 votes]: First, the hypotheses of the theorem I proved require that $Y$ be compact and oriented, in addition to requiring that $g$ be real-analytic. -Second, the method I used (the Cartan-Kähler Theorem) extends, essentially without modification, to higher dimensions as long as $Y$ is compact and parallelizable and $g$ is real-analytic. -Real-analyticity is certainly necessary, since a minimal submanifold of a real-analytic Riemannian manifold (such as a Calabi-Yau manifold in any dimension) is necessarily real-analytic itself. -By contrast, not all special Lagrangian submanifolds of a Calabi-Yau are parallelizable. Thus, parallelizability is not necessary in general, but I don't know how to remove that hypothesis in the existence proof. For example, I do not know whether every real-analytic metric on $S^4$ is obtainable by embedding it as a special Lagrangian in some $4$-dimensional Calabi-Yau.<|endoftext|> -TITLE: Fastest Algorithm to Compute the Sum of Primes? -QUESTION [24 upvotes]: Can anyone help me with references to the current fastest algorithms for counting the exact sum of primes less than some number n? I'm specifically curious about the best case running times, of course. I'm pretty familiar with the various fast algorithms for prime counting, but I'm having a harder time tracking down sum of prime algorithms... - -REPLY [10 votes]: I'll put in a plug for my original paper with Lagarias and Odlyzko, as well as a recent paper by Bach, Klyve and Sorenson: http://www.ams.org/journals/mcom/2009-78-268/S0025-5718-09-02249-2/home.html Computing prime harmonic sums Math. Comp. 78 (2009), 2283-2305. Although the algorithms of Lagarias and Odlyzko (with the extra algorithm of Odlyzko-Schonhage for computing good approximations to a bunch of values of $\zeta(s)$) are asymptotically the best, having complexity $O(x^{1/2 + \epsilon})$ the combinatorial algorithms will probably work best for any reasonable ranges. Though one should look at J. Buethe, J. Franke, A. Jost, and T. Kleinjung, "Conditional Calculation of pi(10^24)", Posting to the Number Theory Mailing List, Jul 29 2010. A little more detail (that's all I have right now) is contained in the talk of David Platt: www.maths.bris.ac.uk/~madjp/junior%20talk.pdf . The idea in the combinatorial method is the following: -Suppose that $f(n)$ is a completely multiplicative function of the positive integers. In the case of calculating $\pi(x)$, we take $f(n) = 1$. In the case of calculating the sum of the primes $\le x$ we take $f(n) = n$. In the case of Calculating $\pi(x,a,q)$ we take $f(n)$ to be a Dirichlet character mod $q$. -Define $\phi(x,a)$ as the sum of $f(n)$ for all integers $n \le x$ which are not divisible by the first $a$ primes. Clearly $\phi(x,0) = \sum_{n \le x} f(n)$. We have the recursion: -$\phi(x,a+1) = \phi(x,a) - f(p_{a+1})\phi(x/p_{a+1},a).$ -Imagine that you have a labeled tree, where the labels are $\pm f(k) \phi(x/k,b)$ for some $k$ and $b$. You can expand any node in the tree you want by applying the above formula. This has the property that it leaves the sum of the values at the leaves the same. So the idea is to start with a tree consisting of one node labeled $\phi(x,\pi(\sqrt x))$ and keep expanding until either $b=0$ or $x/k < $ some cutoff. If you accumulate all such nodes you can evaluate all of them by sieving (and a special data structure -- see the Lagarias-Miller-Odlyzko paper for details), and the optimal value for the cutoff is $x^{2/3}$ perhaps multiplied by some logarithmic factors. -Here's the reference to my paper with Lagarias and Odlyzko: Computing $\pi(x)$: The Meissel-Lehmer Method -Author(s): J. C. Lagarias, V. S. Miller, A. M. Odlyzko -Source: Mathematics of Computation, Vol. 44, No. 170 (Apr., 1985), pp. 537-560<|endoftext|> -TITLE: For which algebras does \{Differential Operators\} satisfy a PBW-like theorem? -QUESTION [9 upvotes]: Let $k$ be a commutative ring, $A$ a commutative $k$-algebra, and for some other part of why I'm asking this question I only care about the case when $k \supseteq \mathbb Q$. Recall the following notion, I think originally due to Grothendieck: -Definition (differential operator): Let $D : A\to A$ be $k$-linear. Define $s_nD : A^{\otimes n} \to A$ by: -$$ s_nD(a_1,\dots,a_n) = \sum_{I \subseteq \lbrace 1,\dots,n\rbrace} (-1)^{|I|}\; \left( \prod_{i\not\in I} a_i\right) \;D\left(\prod_{i\in I}a_i\right) -$$ -One says that $D$ is an $n$th order differential operator if $s_{n+1}D = 0$. -Examples: $s_0D = D(1) \in A$. $s_1D(a) = D(a) - aD(1)$. $s_2D(a,b) = D(ab) - aD(b) - bD(a) + abD(1)$. -Remark: $s_nD$ is symmetric in the $a_i$s. If $D$ is an $n$th order differential operator, then $s_nD(-,a_2,\dots,a_n)$ is a derivation, for $a_2,\dots,a_n$ fixed. -Thus if $D$ is an $n$th order differential operator, $s_nD$ is a symmetric polyderivation. It deserves the be called the principal symbol of $D$. It measures the failure of $D$ to be an $(n-1)$th order differential operator. - -Question: For which algebras $A$ (i.e. what are natural, checkable conditions) does the following "PBW theorem" hold: - $$ s_n: \lbrace n\text{th order differential operators}\rbrace \to \lbrace \text{symmetric }n\text{-polyderivations} \rbrace $$ - surjective for every $n$? - -Examples: This PBW theorem holds for $A = k[x_1,\dots,x_n]$ and $A = k [\\![ x_1,\dots,x_n ]\\!]$ and $A = \mathscr C^\infty(M)$ where $M$ is a smooth manifold. This PBW theorem failes for $A = k[x]/x^2$, as the space of symmetric biderivations is one-dimensional spanned by $x \frac{\partial}{\partial x}\otimes \frac{\partial}{\partial x}$, whereas every second-order differential operator is also a first-order differential operator. Edit: I don't know a characteristic-$0$ example for which PBW theorem fails, but I don't expect it to always hold. -Remark: I would expect that algebras $A$ for which the PBW theorem holds are those for which $\operatorname{spec}(A)$ is "smooth" in the appropriate sense, but I don't know if this is "smooth" in other usual senses of the word. - -REPLY [7 votes]: I think a sufficient condition is: - -the $A$-module $Der_k(A)$ of $k$-derivations - of $A$ is projective. - -A hint for the proof you may find in an old paper of G.S. Rinehart: Differential Forms on General Commutative Algebras (just have to notice that $(A,Der_k(A))$ is an example of a $(k,A)$-Lie algebra - today known as Lie-Rinehart algebra, or Lie algebroid). Especially: you might want to have a look at Theorem 3.1 and its proof.<|endoftext|> -TITLE: Union of a object (a set) in the Elementary Theory of the Category of Sets -QUESTION [5 upvotes]: I see the Todd Trimble article "Elementary Theory of the Category of Sets" on catlab. -I ask: how make (in the categorical setting) the usual union of a set $\cup X=${$y |\exists x\in X: y\in x$}? -This object is (strongly) no natural (no amenable to a functor and natural transformation). -I ask this because the "Union axiom" is one of the ZF theory of sets. - -REPLY [5 votes]: There are several nice answers already, but François's answer makes me think that the following point might also be relevant. -It's a known fact that the ETCS axioms do not imply the existence of a coproduct -$$ -\mathbb{N} + P(\mathbb{N}) + P(P(\mathbb{N})) + \cdots -$$ -where $P$ means power set. (At least, they don't unless they're inconsistent.) So if that's the kind of "union" you had in mind, it's not always provided by ETCS. -However, you can add the following axiom scheme to ETCS (and indeed, ETCS plus this axiom scheme is equivalent to ZFC). I'll say it a bit informally; a precise statement is in Section 8 of - -Colin McLarty, Exploring categorical structuralism, Philosophia Mathematica 12 (2004), 37-53. - -So: suppose you have a set $I$ and a family $(X_i)_{i \in I}$ of sets specified by a first-order formula. The axiom states that the coproduct $\sum_{i \in I} X_i$ exists. How does it state this? By saying that there are a set $X$ (to be thought of as $\sum_{i \in I} X_i$) and a map $p: X \to I$ (to be thought of as the evident projection) such that for each $i \in I$, the fibre $p^{-1}(i)$ is isomorphic to $X_i$. -One moral of the variation between our answers: "union" and "disjoint union" are more different concepts than they might at first appear.<|endoftext|> -TITLE: A product approximation to the Taylor series of the exponential -QUESTION [24 upvotes]: I recently came across the following in something I'm working on, and I'd never seen it before. Consider -\begin{align*} -f_1(x) &= (1+x)^{1/1} \\\ -f_2(x) &= (1+x)^{2/1} (1+2x)^{-1/2} \\\ -f_3(x) &= (1+x)^{3/1} (1+2x)^{-3/2} (1+3x)^{1/3} \\\ -f_4(x) &= (1+x)^{4/1} (1+2x)^{-6/2} (1+3x)^{4/3} (1+4x)^{-1/4} \\\ -& \cdots \\\ -f_n(x) &= \prod_{k=1}^n (1+kx)^{(-1)^{k-1}\binom{n}{k}/k}. -\end{align*} -Think of these as formal power series with coefficients in $\mathbb{Q}$. It turns out that these series approximate the Taylor expansion for the exponential, in the sense that -$$ f_n(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} +O(x^{n+1}).$$ -(This isn't too hard to prove, using a logarithmic derivative: -\begin{align*} -1-f_n'(x)/f_n(x) &= \sum_{k=0}^n (-1)^k\binom{n}{k}(1+kx)^{-1} -\\\ -&= \sum_j \biggl(\sum_k (-1)^k\binom{n}{k}k^j\biggr) (-x)^j. -\end{align*} -The coefficient of $x^j$ calculates the number of surjections $\{1,\dots,j\}\to \{1,\dots,n\}$ (up to a sign), and so is $0$ if $j -TITLE: Higher-dimensional Catalan numbers? -QUESTION [24 upvotes]: One could imagine defining various notions of higher-dimensional Catalan numbers, -by generalizing objects they count. -For example, because the Catalan numbers count the triangulations of convex polygons, -one could count the tetrahedralizations of convex polyhedra, or more generally, triangulations of -polytopes. Perhaps the number of triangulations of the $n$-cube are similar to the Catalan -numbers? -Or, because the Catalan numbers count the number of below-diagonal monotonic paths -in an $n \times n$ grid, one could count the number of monotonic paths below -the "diagonal hyperplane" in an $n \times \cdots \times n$ grid. -I would be interested in references to such generalizations, which surely have been considered—I am just not looking under the right terminology. -I would be particularly grateful to learn of generalizations that share some of the -Catalan number's ubiquity. Thanks for pointers! - -REPLY [2 votes]: One relatively direct generalization is to interpret the $n$-th Catalan number as the multiplicity of the trivial representation in the $2n$-th tensor power of the standard $2$-dimensional representation of $SU_2(\mathbf{C})$, and to consider instead the multiplicity of the trivial representation in the $kn$-th tensor power of the standard $k$-th dimensional representation of $SU_k(\mathbf{C})$. -(On the other hand, I don't known which other properties of the Catalan numbers will have any analogue here...)<|endoftext|> -TITLE: Bounding the absolute sum of entries of the inverse of a 0-1 matrix -QUESTION [14 upvotes]: I have a non-singular square 0-1 matrix and I want to bound the sum of absolute values of its inverse as a function of n (or the vector 1-norm). -Asymptotic results are also useful. -Does anyone know any result that can help me? -Thank you, -ifog - -REPLY [14 votes]: An answer is given in: -N. Alon and V. H. Vu, Anti-Hadamard matrices, coin weighing, -threshold gates and indecomposable hypergraphs, J. Combinatorial -Theory, Ser. A 79 (1997), 133-160. -Indeed C above is essentially sqrt n<|endoftext|> -TITLE: Geometrical meaning of semi-stable reduction? -QUESTION [20 upvotes]: So let $R$ be a discrete valuation ring and let $X$ be a scheme which is proper and flat over $R$. Let $X_s$ denote the special fiber of $X$. -So intuitively, when somebody says that a curve $X$ is semistable I kind of equate this in my mind with the property that $X_s$ has only ordinary double points as singularities. -Q1: So in general (i.e. in higher dimension) what is the geometrical meaning for a scheme -to be semistable? -On the Galois representation side we have a very precise definition of what semistable means using Fontaine's ring $\mathbf{B}_{st}$. -Q2 If there is a precise answer to Q1, is there a good reference (more on the intuitive side than on the technical side) where one shows the equivalence (under suitable assumptions) that the geometrical definition coincides with the galois representation one? - -REPLY [15 votes]: Just to elaborate on Francois Brunault's answer: it is not true in general that having semistable etale cohomology implies having semistable reduction (just as it is not true that having crystalline etale cohomology implies having good reduction). So the implication only goes in one direction. -Added in response to comment below: E.g. if $E$ is an elliptic curve over $\mathbb Q_p$ -with good reduction, and $P$ is an $E$-torsor with no $\mathbb Q_p$-rational point, then $P$ will not have good reduction. It will not have semi-stable reduction either, since it has -potentially good reduction (it obtains a rational point over some extension of $\mathbb Q_p$, and hence becomes isomorphic to $E$ over that same extension). The etale cohomology ($\ell$-adic or $p$-adic) of $P$ coincides with that of $E$, and so $P$ is an example of a variety over $\mathbb Q_p$ with semistable (indeed crystalline) Galois action on its $p$-adic etale cohomology, which does not have semistable reduction.<|endoftext|> -TITLE: Compare three 2-categories of (Lie) groupoids -QUESTION [6 upvotes]: Lie groupoids are groupoids with smooth structures. There is a nature 2-category of Lie groupoids: Lie groupoids, smooth functors of Lie groupoids, smooth natural transformations of smooth functors. However this is usually not the correct one, partially because weak equivalence (fully-faithful and essential surjective in the smooth setting) of Lie groupoids cannot be inverted. We need more morphisms between Lie groupoids. There are three (or 2.5) ways to define the more morphisms: -1) Span: Span of Lie groupoid morphism G<<--K-->H such that the left leg is a weak equivalence. - Equivalence of such spans: G<<--K1-->H and G<<--K2-->H are equivalent if there exists G<<--K3-->H with K1<<--K3-->>K2 such that all triangles 2-commute. -1') Span:Span of Lie groupoid morphism G<<--K-->H such that the left leg is a weak equivalence and the map between objects are surjective submersion. This is a slightly different version of the 1). -2) Bibundle: right principal bibundle of Lie groupoids. - Equivalence of bibundles: equivariant map of bibundles (this is in fact a diffeomorphism). -Modulo equivalent relation in three cases we obtain three categories. A classic result tell us that they are isomorphic. -One could go a bit further, it is possible to define three 2-categories (i.e. bi-categories, in fact (2,1)-categories, 2-morphisms are invertible), and naturally there are functors between them. The construction is given in Hellen Colman http://www.springerlink.com/content/3472617rj6178271/. -One expect also that they are 2-equivalent 2-categories. Recall that a equivalent functor of 2-categories must be locally equivalent and surjecttive-up-to-equivalence on objects (See Leinster "Basic Bicategories"). This means that 2-morphisms must be the same. However it seems that 2-morphisms of (1) and (2) are different. The 2-morphisms induced from (2) to (1) must be and strict isomorphism (namely, K1 and K2 are isomorphic). What's wrong with my reasoning? - -REPLY [8 votes]: These three bicategories ARE equivalent. To go from (2) to (1) it's helpful to introduce a (2)': -Let (2)' have the same objects, and morphisms as (2) except each principal bundle $P$ for $G$ over $H$, is equipped with a choice of local sections of the map $P \to H_0$ (so this is equivalent to (2)). If $U$ is the cover of $H_0$ over which these local sections are defined, this can easily be seen to the same data as a map $H_U \to G$, where $H_U$ is $H$ pulled back along the map $\coprod U_i \to H_0$. But now if $\alpha:P \to P'$ is an isomorphism and we have two different choices of covers of $H_0$, $U$ and $U'$ over which they admit local sections, then this induces a $2$-cell in (1) between the maps $H_U \to G$ and $H_U' \to G$ in the bicategory of fractions. These $2$-cells are a bit complicated, see the work of Pronk. This two cell is represented by the two (canonically equivalent) induced smooth functors $H_{U \cap U'} \to G$. -You should note that there is ANOTHER bicategory to which these are all equivalent, namely the bicategory of differentiable stacks, which are stacks over the category of smooth manifolds, which admit a surjective representable submersion from a manifold.<|endoftext|> -TITLE: Modulus of Continuity -QUESTION [13 upvotes]: I originally posted this question on math.stackexchange (https://math.stackexchange.com/questions/83182/modulus-of-continuity-take-2), but it's been a few days and I haven't received any correct answers. -Let $\rho: \mathbb{R}^+ \to \mathbb{R}^+$ be a continuous nondecreasing function such that $\rho(t) = 0$ if and only of $t = 0$. If you can answer my question in the special case $\rho(t) = Ct$ where $C$ is a constant then it will probably be possible to adapt your construction to the general case. -Say that a function $f: X \to \mathbb{R}$ on a metric space has modulus of continuity $\rho$ at a point $x_0 \in X$ if $|f(x) - f(x_0)| \leq \rho(d(x,x_0))$ for every $x \in X$. For example, a function has modulus of continuity $Ct$ at $x_0$ if and only if it is Lipschitz with Lipschitz constant $C$ at $x_0$. -Question If $X$ is a compact metric space without isolated points, is it true that the set $S$ of all continuous functions on $X$ which have modulus of continuity $\rho$ at some point of $X$ is nowhere dense in $C(X)$ equipped with the supremum norm? -Note that the "some point" that I am referring to is not fixed and may depend on the function. The complement of $S$ is the set of all functions which do not have modulus of continuity $\rho$ at any point. -To prove that the answer is affirmative for a given $X$ one must be able to construct functions of arbitrarily small norm which oscillate arbitrarily rapidly. For example, if $\rho(t) = Ct$ and $X = [0,1]$ then one can use a piecewise linear function such that the slope of each linear piece is larger than $C$ in absolute value. However, I don't see how to generalize this idea to an arbitrary compact metric space without isolated points. - -REPLY [10 votes]: All right, I don't think that anything I'll say below will be new to Paul. However it'll address Vaughn's concerns (to the extent he expressed them in the comments by the moment of this writing). -Of course, the ultimate purpose is to show that for a given modulus of continuity $\rho$, the set $U$ of continuous functions $f$ that have that modulus of continuity at at least one point $x$ in the sense that there exists $\delta>0$ (depending on $x$) such that $|f(x)-f(y)|\le\rho(d(x,y))$ for all $y$ satisfying $d(x,y)\le \delta$ is small. -However, there is no chance to show that it is nowhere dense because you can take any function and flatten it a bit near any value $v$ it takes, i.e., to consider $g(x)=\min(f(x)+\varepsilon,v)+\max(f(x)-\varepsilon,v)-v$. This function will be constant near the point where $f=v$ and differ from $f$ by at most $\varepsilon$. So, every open set contains a "bad" function. -The right words to use instead of "nowhere dense" are "of the first category". Note that if $f$ has modulus of continuity $\rho$ at some point locally, then it has modulus of continuity $A\rho$ at the same point globally for some large integer constant $A$. Now fix $A$ and consider the set $F_A$ of functions $f$ such that there exists a point $x\in X$ at which $f$ has global modulus of continuity $A\rho$. Note that the union of $F_A$ contains $U$. -Note that each $F_A$ is closed: if $f_n\in F_A$ converge to $f$ uniformly and $x_n\in X$ are the corresponding bad points, then any accumulation point $x$ of the sequence $x_n$ is bad for $f$. -It remains to show that $F_A$ contains no open set. Let $f$ be any function in $C(X)$. Let $\lambda>0$. Choose $\delta>0$ so small that $|f(x)-f(y)|\le\lambda$ whenever $d(x,y)\le 4\delta$ and $A\rho(2\delta)<\lambda$ as well. Now run the construction in my remark only denote the resulting function $g$ and put $g=0$ on $A$ and $g=4\lambda$ on $B$. -Consider $h=f+g$. On one hand, it differs from $f$ by $4\lambda$ or less. On the other hand, for each point $x\in X$, the ball of radius $2\delta$ centered at $x$ contains a point $a\in A$ and a point $b\in B$. If $h$ had global modulus of continuity at $x$, we would have $|h(a)-h(b)|\le 2A\rho(2\delta))<2\lambda$. On the other hand, this difference is at least $4\lambda-|f(a)-f(b)|\ge 3\lambda$. Thus, every open ball in $C(X)$ intersects the complement of $F_A$.<|endoftext|> -TITLE: Singular chains in Spivak's Calculus on Manifolds -QUESTION [9 upvotes]: On page 90 of Calculus on Manifolds, Spivak defines the -pull-back $f^*\omega$ of a differential form by a differentiable -map by the usual formula. On page 97, he defines a singular -$k$-cube as a continuous map $c:[0,1]^k\to\mathbb{R}^n$. Finally, -on page 101, he defines the integral of a differential $k$-form -against a singular $k$-cube by the formula -$\int_c\omega=\int_{[0,1]^k} c^*\omega$. -I don't see, however, -where he defines the pull back $c^*\omega$ of a differentiable form by -a continuous map (or how he could). -So my questions are: (a) Is this an error (or am I missing -something)? (b) Have other people noticed it? - -REPLY [2 votes]: Continuity is not enough. However Lipschitz continuity would do. Lipschitz continuous maps are differentiable almost everywhere and you can define the pullback almost everywhere. For more on this point of view, have a look at a book on geometric measure theory, e.g., Frank Morgan's: Geometric Measure Theory. A Beginner's Guide. -Another beautiful reference is the classic Geometric Integration Theory by H. Whitney (now available in Dover)<|endoftext|> -TITLE: Is the following construction of the 0-Hecke monoid (well) known? -QUESTION [11 upvotes]: Let W be a Coxeter group with Coxeter generators S. The corresponding 0-Hecke monoid H(W) has generating set S, the braid relations of W and the relations that each element of S is an idempotent. If one specializes the Hecke algebra associated to W to q=0 one gets the monoid algebra of H(W) (replace generators by their negatives to see this). It is also the monoid generated by foldings across the walls of the fundamental chamber of the Coxeter complex of W. It has been studied by a number of people and is sometimes called the Springer-Richardson monoid. -Margolis and I came across the following construction of it and I wanted to know if it is known. Let P(W) be the power set of W. It is a monoid with usual set product: $AB=\lbrace ab\mid a\in A, b\in B\rbrace$. Let $I(w)$ be the principal Bruhat ideal generated by $w\in W$, e.g., $I(s)=\lbrace 1,s\rbrace$ for $s\in S$. Then the principal Bruhat ideals form a submonoid of P(W) isomorphic to H(W). - - -Question: Does this construction appear explicitly in the literature and what is a reference? - -REPLY [6 votes]: Here is one such reference: -Representation and classification of Coxeter monoids -by S. V. Tsaranov -European Journal of Combinatorics, Volume 11 Issue 2, Mar. 1990 -http://dl.acm.org/citation.cfm?id=84891<|endoftext|> -TITLE: When two singularities $\mathbb C^n/G$ and $\mathbb C^n/G'$ are the same? -QUESTION [7 upvotes]: Let us consider two singularities $\mathbb C^n/G$ and $\mathbb C^n/G'$, where $G$ and $G'$ are finite subgroups of $\mathrm{GL}(n,\mathbb{C})$ acting linearly. -It is easy too see, that a different groups may give the same singularities. For example, $G'=G/\langle g\rangle$, where $\langle g\rangle$ is a subgroup of $G$ generated by a reflection $g$. - -Which groups $G$ and $G'$ give the same singularity? - -REPLY [15 votes]: The answer is provided by the following - -Prill's isomorphism criterion. Let $G_1, G_2 \subset \textrm{GL}(n, \mathbb{C})$ be two finite subgroups, $n \geq 2$. Assume moreover that they are small, i.e. without pseudo-reflections. Then the two germs of quotient singularities $$(\mathbb{C}^n/G_1, 0), \quad (\mathbb{C}^n/G_2, 0)$$ are analitically isomorphic if and only if $G_1$ and $G_2$ are conjugated in $\textrm{GL}(n, \mathbb{C})$. - -The reference is -D. Prill: Local classification of quotients of complex manifolds by discontinuous groups, Duke Mathematical Journal 34 Number 2 (1967), 375-386. -See also this paper, Theorem 1.2 page 126.<|endoftext|> -TITLE: Fourier transform of x2 invariant measure -QUESTION [9 upvotes]: Let $T:\mathbb{R}/\mathbb{Z}\rightarrow \mathbb{R}/\mathbb{Z}$ be the map defined by $T(x)=2x$, and suppose that $\mu$ is a $T$ invariant and ergodic Borel probability measure on the space, which is not a finite sum of point masses. What if anything can be said about the decay of the Fourier transform of $\mu$, along the sequence of odd integers (to take into account Andreas's comment below)? -For example is it true that $\hat{\mu}(2n+1)\ll |2n+1|^{-c}$ for some $c>0$? -If it helps to answer the question (or in case the answer is no in general) I would also be happy to know, what can be said under the additional hypothesis that $T$ has positive measure theoretic entropy with respect to $\mu$? - -REPLY [4 votes]: There was an interesting paper posted on the arxiv this morning, Invariant Measures on the Circle and Funcitonal Equations, by Chris Deninger, which seems to essentially answer my question. Corollary 9 of the paper states that a sequence of complex numbers $(c_m)_{m\in\mathbb{Z}}$ is the sequence of Fourier coefficients of a (uniquely determined) times-$n$ invariant special measure (defined below) if and only if: - -$c_{-m}=\overline{c}_m$ for all $m\in\mathbb{Z}$, -$c_{mn}=c_m$ for all $m\in\mathbb{Z}$, and -the series $$\sum_{m=0}^\infty |b_m|^2$$ converges, where $(b_m)$ is defined by -$$\sum_{m=0}^\infty b_mz^m=\exp\left(-c_0-2\sum_{m=1}^\infty c_mz^m\right).$$ - -A finite signed Borel measure $\mu$ is a special measure if $\mu=\alpha \lambda +\nu$ where $\alpha\in\mathbb{R},~\lambda$ is Lebesgue measure, and $\nu$ is a positive singular Borel measure. So in particular any ergodic times-$n$ invariant measure is a special measure and the result above applies.<|endoftext|> -TITLE: Tiling by regular simplices -QUESTION [12 upvotes]: The plane can be tiled without gaps by congruent two-dimensional regular simplices (i.e., equilateral triangles). The three-dimensional Euclidean space cannot be tiled by congruent three-dimensional simplices (i.e., equilateral tetrahedra). Can this be done for higher dimensions? I have a vague recollection that I saw a proof that for some dimensions the tiling exists and that the existence is somehow related to the divisibility of the dimension by four, but this may be false. Can anybody please formulate the correct statement and provide either a proof or a reference? -Thanks! - -REPLY [19 votes]: Senechal's survey paper has a beautiful historical overview and some standard references. Recently, we proved in this paper that one cannot tile $\Bbb R^d$, $d\ge 4$, with congruent copies of an acute simplex (this answers your question as well). Interestingly, existence of an acute simplex in $\Bbb R^3$ which tiles the space is open since Sommerville's classification does not allow reflections (see e.g. this paper). Curiously, any polytope which tiles $\Bbb R^3$ or $\Bbb R^4$ must be scissors congruent to a cube, even if the tiling is aperiodic (see here).<|endoftext|> -TITLE: elementary embeddings -QUESTION [14 upvotes]: Fact 1: If $M$ and $N$ are transitive models of ZF with the same ordinals, and $M \prec N$, then $M = N$. -Fact 2: If $M$ and $N$ are transitive models of ZFC with the same ordinals, and $j: M \to N$ is an elementary embedding that does not move ordinals, then $j$ is the identity map and $M=N$. -Question: Is it consistent that $M$ and $N$ are transitive models of ZF with the same ordinals, $j: M \to N$ is an elementary embedding that does not move ordinals, and $M \not= N$? Perhaps with $j(\mathbb{R}^M) \not= \mathbb{R}^M$? - -REPLY [21 votes]: Yes, this is possible. In fact, this is equiconsistent with ZF. For example, let ${\mathbb P}={\rm Add}(\omega,\omega_1)$ be the forcing that adds $\omega_1$ Cohen reals with finite conditions. Note that the definition of ${\mathbb P}$ is absolute between (transitive) models that agree on the value of $\omega_1$. -By standard arguments, any ${\mathbb P}$-generic $G$ is isomorphic to a product $G_0 \times G_1$, where $G_0$ is ${\mathbb P}$-generic over $V$ and $G_1$ is ${\mathbb P}$-generic over $V[G_0]$. Similarly, $G$ can be decomposed as $G'\times G''$ where $G'$ adds countably many Cohen reals and $G''$ is ${\mathbb P}$ generic over $V[G']$. -Let ${\mathbb R}_0 = {\mathbb R}^{V [G_0]}$ and ${\mathbb R}_1 = {\mathbb R}^{V [G]}$. In $V [G]$ we can define a nontrivial $j : L({\mathbb R}_0)\to L({\mathbb R}_1)$ that is the identity on the ordinals. The point is that any element of $L({\mathbb R})$ is definable from an ordinal and a real, so we can see it as a "term" $\tau(x,y)$ in an appropriate language, interpreted in $L({\mathbb R})$, when $x=\alpha$ and $y=r$ are an ordinal and a real, respectively. -[We can avoid talking about terms, of course, but it is perhaps a bit easier to see what follows if we insist in mentioning them.] -The fact above tells us that we have not much choice on how to define $j$, namely, we must have $$ j(\tau^{L({\mathbb R}_0)}(\alpha,r))=\tau^{L({\mathbb R}_1)}(\alpha,r) $$ for any term $\tau$, any ordinal $\alpha$, and any real $r\in{\mathbb R}_0$. -To see that $j$ works, note that there is a recursive way to assign to each formula $\varphi(\vec x,\vec y)$ a formula $\psi_\varphi(\vec x,\vec y)$ in such a way that (provably in ZF), for any ${\mathbb P}$-generic $H$, in $V[H]$ we have that $L({\mathbb R})\models\varphi(\vec \alpha,\vec r)$ iff $V[H]\models \psi_\varphi(\vec \alpha,\vec r)$ for any finite sequence of ordinals $\vec\alpha$ and any finite sequence of reals $\vec r$. -Now, and this is the key, there is a countable subset $X\subset\omega_1$ such that $r\in V [G_0\upharpoonright X]$, and we can find ${\mathbb P}$-generics $G'$ over $V[G_0\upharpoonright X]$ and $G^*$ over $V[G_0\upharpoonright X][G']$ such that $V[G_0\upharpoonright X][G'] = V [G_0]$ and $V[G_0\upharpoonright X][G^*]=V [G]$. -But then $L({\mathbb R}_0)\models \varphi(\vec \alpha, r)$ iff $V [G_0] \models\psi_\varphi(\vec\alpha,r)$ iff there is a $p\in G'$ such that in $V[G_0\upharpoonright X]$, $p$ forces that $\varphi(\vec\alpha,\vec r)$ holds, iff the empty condition forces it (this is homogeneity of ${\mathbb P}$, and is really the point of the whole thing) iff there is a $q\in G^*$ that forces it iff $V[G]\models\psi_\varphi(\vec\alpha,r)$ iff $L({\mathbb R}_1)\models\varphi(\vec \alpha,r)$. -Of course, it follows immediately from this that $j$ is well-defined, elementary, and the identity on ordinals. -The argument comes more or less verbatim from my thesis, but it is probably folklore. Certainly, A. Miller has very similar arguments, which makes me suspect he knew this ages before I asked myself the question. One can also use random forcing or several other forcing notions with minor variations. -One reason why I like this argument is because it gives us a cute way of proving that choice fails in $L({\mathbb R})$ after adding $\omega_1$ Cohen reals. On the other hand, adding countably many Cohen reals to $L$ is the same as only adding one, and the resulting $L({\mathbb R})$ is just $L[r]$ for some real $r$, so choice holds there.<|endoftext|> -TITLE: Inverting an asymptotic series -QUESTION [5 upvotes]: I have the first few terms of a series of the form, -$y(x)=\ln(x)+x+a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\cdots$. -Knowing that the inverse $x(y)$ exists, I am looking for method to write x in terms of y (at least the first few terms of the expansion). Does anybody know how I could achieve this? -Thanks to a mathematician much greater than I, I know that this is certainly possible in the case, the $x$ term is not present in the expansion of $y$ (i.e. $y(x)=\ln(x)+a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+O(\frac{1}{x^3})$). It turns out in this case $x(y)$ can be written as a series expansion in powers of $e^{-y}$. But I can't seem to be able adapt that method to handle this new case. -Thanks for reading. - -REPLY [4 votes]: There is a general technique for doing this, found in expositions dealing with transseries. One example is my own... -Transseries for beginners, Real Analysis Exchange 35 (2010) 253--310 -see Problem 4.2.<|endoftext|> -TITLE: one-parameter subgroup and geodesics on Lie group -QUESTION [17 upvotes]: Hi, -Given a Matrix Lie Group, I would like to know if the one-parameter subgroups (which can be written as $\exp^{tX}$) are the same as the geodesics (locally distance minimizing curves). Geodesics depends on the metric used so perhaps a more precise formulation of this question is: -Given a Matrix Lie Group, under what metric the one-parameter subgroups are the same as the geodesics ? -Thanks, -Frank - -REPLY [3 votes]: I may have a similar answer with an alternative derivation. Given a group $G$ with Lie algebra $\mathfrak{g}$ and right-invariant inner product $\langle A,B\rangle_g=\langle Ag^{-1},Bg^{-1}\rangle_e$ where $A,B\in T_gG$, $Ag^{-1}=(R_{g^{-1}})_*A$, $Bg^{-1}=(R_{g^{-1}})_*B$ and $e$ being the neutral element. Right-translations of a group are generated by left-invariant vector fields which are therefore Killing vector fields. Let us assume that $M_I\in\mathfrak{g}$ for $I\in\{1,\cdots,\dim{G}\}$ is a basis of the Lie algebra. This means we have the left-invariant vector fields -\begin{align} -X_I=gM_I=L_gM_I\in T_gG\,. -\end{align} -It is a well-known fact that the inner product of (affinely parametrized) geodesics $u: [0,1]\to G: t\mapsto u(t)$ have conserved quantities -\begin{align} -C_I&=\langle X_I,\dot{u}(t)\rangle_{u(t)}\\ -&=\langle uM_I,\dot{u}\rangle_u\\ -&=\langle uM_Iu^{-1},\dot{u}u^{-1}\rangle_e\,. -\end{align} -For a Lie group with right-invariant metric, knowing the $\dim{G}$ conserved quantities $C_I$ and initial point of a geodesics $u(0)$ determine the geodesics uniquely. -Let us check what the condition for a 1-parameter subgroup $u(t)=e^{tA}$ with $A\in\mathfrak{g}$ is to be a geodesic, or equivalently to have the $C_I$ being conserved. Plugging $u(t)$ into above formula gives -\begin{align} -C_I=\langle e^{tA}M_Ie^{-tA},A\rangle_e\,. -\end{align} -We can take the derivative $d/dt$ on both sides and find -\begin{align} -0=\langle e^{tA}[A,M_I]e^{-tA},A\rangle_e\,. -\end{align} -This formula is equivalent to $\frac{d}{dt}\lVert e^{tM_I}Ae^{-tM_I}\rVert=\langle[M_I,A],A\rangle=0$ that Claudio wrote above and that Denis guessed. It's just an alternative derivation without making reference to the Killing form or its properties...<|endoftext|> -TITLE: Who proved that a group of polynomial growth has growth exactly polynomial? -QUESTION [7 upvotes]: I need to put a reference about the classical result that a f.g. group of polynomial growth has growth which is exactly polynomial. -Talking personally with people and also here in A question about groups of intermediate growth, it seems that the result is attributed to Gromov and Pansu. In particular Gromov proved that a group of polynomial growth is virtually nilpotent and Pansu proved, in a paper in Ergodic Th & Dyn Systems that goes back to 1983, that a nilpotent group has exactly polynomial growth (is this right?) -Now, reading the introduction of http://de.arxiv.org/PS_cache/arxiv/pdf/1011/1011.5266v2.pdf, Bartholdi and Erschler say that the result is due to Guivarch (without reference) and Bass (in 1972). -So I am now little confused... who proved this theorem? -I guess that the problem is that nobody proved that theorem in this form. In this case, can anybody tell me a short history of the result, in such a way not to write wrong things? -Thank you in advance, -Valerio - -REPLY [6 votes]: Guivarc'h certainly did prove such a result, see -MR0302819 (46 #1962) -Guivarc'h, Yves -Groupes de Lie à croissance polynomiale. (French) -C. R. Acad. Sci. Paris Sér. A-B 272 (1971), A1695–A1696. -22E15 -As did Bass: -MR0379672 (52 #577) -Bass, H. -The degree of polynomial growth of finitely generated nilpotent groups. -Proc. London Math. Soc. (3) 25 (1972), 603–614.<|endoftext|> -TITLE: Is an identity that is true for matrix Lie groups true for all Lie groups? -QUESTION [12 upvotes]: Many identities for Lie groups are more easily proved for matrix groups. A non-trivial example is the equation -$$ -\frac{d}{dt}\vert_{t=0} \exp(-X)\exp(X+tY) = \frac{1-e^{-\operatorname{ad} X}}{\operatorname{ad} X} Y. $$ -My question is if it is always sufficient to prove identities in the case of matrix algebras. For example, is there some sort of density argument using a topology on the space of Lie groups that makes matrix groups dense? Or perhaps there is an analogue of the notion of "permanence of identities" used to prove statements about matrices. - -EDIT: Admittedly my question is vague, and as I responded to a comment, I am really looking for a heuristic that works in a large set of ``natural" cases. -Here is a good example of what I'm after is this mathoverflow question about showing if $\alpha$ and $\beta$ are one parameter subgroups of a Lie group then $(\alpha \beta)'(0) = \alpha'(0) + \beta'(0)$. This is very easy to prove for matrix groups since the product rule carries over. It is not hard to prove for general Lie groups (see e.g. my answer to that question) but it is more difficult in my opinion. - -REPLY [4 votes]: Here is a general result on promoting identities for Lie groups from linear groups to all connected groups. -Let $\mathcal{LG}$ be the category of connected Lie groups and local isomorphisms ($=$ coverings) as morphisms. -Assigning to a group $G$ the topological space $G^m \times \mathfrak{g}^n$ defines a functor -from $\mathcal{LG}$ to topological spaces, with the property that it maps all coverings to coverings. It makes sense to ask for a natural subset $U(G) \subset G^m \times \mathfrak{g}^n$. In the example -from the original question, $U= \mathfrak{g}^2$, -in Tom's example, it is an algebraic set, but it could be the set of all $X \in \mathfrak{g}$ with $ad$-spectral radius bounded by $1$, -for example. Now let $(F_i)_G:U(G) \to G$ (one could look at maps $\to \mathfrak{g}$ as well, which is slightly easier), -$i=0,1$, be continuous natural transformations. -If you unwind this definition, $F_i$ is a function that is composed out of the data that are available for all Lie groups: the group operations, -the identity element, vector space operation and Lie bracket on $\mathfrak{g}$, the adjoint representation, all operations in the matrix algebra -$End (\mathfrak{g})$ (as taking power series, characteristic polynomials etc), the exponential map and taking derivatives of curves through the -identity element in $G$ (no attempt to make a complete list). -Theorem: ''With the above notations, assume that - -$F_0=F_1$ holds for all linear groups. -$U(G)$ is path-connected for all $G$ (recall that I assumed all groups to be connected); -For each covering $H \to G$, the map $U(H) \to U(G)$ is surjective; -There exists a natural transformation $u:\ast \to U$ of functors such that $F_0 (u)=F_1(u)$ holds for all connected Lie groups. - -Then $F_0=F_1$ is true for all connected Lie groups.'' -Proof: ''Denote by $P(G)$ the statement that the Theorem holds for $G$. For each connected Lie group $G$, there exist, by Ado's theorem, -a Lie group $H$, a linear Lie group $L$ and coverings $H \to G$ and $p:H \to L$. Naturality and assumption $3$ show the implication $P(H) \Rightarrow P(G)$ and -the tricky part is $P(L) \Rightarrow P(H)$. Observe that $p \circ (F_0)_H = (F_0)_L \circ p_U = (F_1)_L \circ p_U = p \circ (F_1)_H$; -the second equality is assumption $1$. Therefore, $(F_0)_H$ and $(F_1)_H$ are both lifts of the map $(F_0)_L \circ p_U$ through the covering -$H \to L$. By assumption $2$, they have to coincide once they coincide at one point, which is the content of assumption $4$. qed.'' -The proof makes clear that one can restrict to the subcategory of connected Lie groups whose Lie algebra is isomorphic to one of a fixed set $S$ -of Lie algebras, -say $S=\{\mathfrak{sl}_2 (\mathbb{R})\}$. -The set $U$ in Toms example has at least two components: the boring one is $\{(0,0\}$ and other one contains those $(X,Y)$ that generate -$\mathfrak{sl}_2 (\mathbb{R})$ (this is probably connected). It is assumption $3$ that fails for this component. -The proof also shows that the target of the transformations $F_i$ can be more generally any functor from $\mathcal{LG}$ that maps local -isomorphisms to coverings.<|endoftext|> -TITLE: Picard groups of (fiber) products -QUESTION [6 upvotes]: Let us work in the "nice" situation where $X,Y,Z$ are smooth complex algebraic varieties, not necessarily compact. Assume that the fiber product $W:= X \times_Z Y$ is also smooth. What can we say about the Picard group of $W$? -More precisely, assume that $Pic(X)=Pic(Z)=0$. -1) Can we deduce that $Pic(W)=Pic(Y)$? -2) If 1) is not true in general, can we draw the conclusion if $Z$ is a point and thus $W=X \times Y$? -3) If 1) is not true in general, can we draw the conclusion if $Y \to Z$ is a finite étale cover? - -REPLY [5 votes]: A weakening of 2) is an exercise (III.12.6) in Hartshorne: Let $X$ be an integral projective scheme over an algebraically closed field $k$ and assume that $H^1(X, \mathcal O_X) = 0$. Let $T$ be a connected scheme of finite type over $k$. Then $\textrm{Pic}(X) \times \textrm{Pic}(T) \cong \textrm{Pic}(X \times T) $ under the obvious morphism. However, it is not true in general that $\textrm{Pic}(X) \times \textrm{Pic}(T) \cong \textrm{Pic}(X \times T) $ for two arbitrary $k$-schemes $X$ and $T$, cf exercise IV.4.10 in Hartshorne. -In general, you might be interested in exercises III.12.4 and III.12.5 of Hartshorne as well; they give more results on Picard groups in this context.<|endoftext|> -TITLE: Non-analytic function with convergent Taylor series everywhere -QUESTION [8 upvotes]: Is there a smooth function on an interval in $\mathbb R$, not analytic on any subinterval, whose Taylor series at every point has positive radius of convergence? The Fabius function might be an example, but this is questionable since the n'th derivative has maximum $2^{\sigma(n)}$, where $\sigma(n)=\frac{n(n+1)}{2}$, which is not quite good enough using a crude estimate for radius of convergence if there are points where many derivatives are close to the maximum. - -REPLY [3 votes]: We define $$\Psi(x)= -\sum_{k\ge 0} 2^{-k}\psi_{\sigma_k}(x-x_k),\quad -\psi_{\sigma}(y)=\exp{-{\vert x\vert}^{-\frac{1}{s-1}}}, -$$ -where $(x_k)_{k\ge 0}$ is dense in $\mathbb R^d$ and $(\sigma_k)_{k\ge 0}$ is decreasing and valued in $[s_1,s_0]\subset(1,+\infty)$. -That function is good explicit substitute to Fabius function since it is smooth and nowhere analytic: even better, it is multidimensional and its analytic wave-front-set is all the cotangent space (minus the zero section). To prove this use Gevrey classes. -It seems likely that the radius of convergence of the Taylor series is positive on a dense subset of $\mathbb R^d$. Anyhow it is a good candidate. -Bazin.<|endoftext|> -TITLE: Probing a manifold with geodesics -QUESTION [27 upvotes]: Supposed you stand at a point $p \in M$ on a smooth 2-manifold $M$ -embedded in $\mathbb{R}^3$. -You do not know anything about $M$. -You shoot off a geodesic $\gamma$ in some direction $u$, -and learn back the shape of the full curve $\gamma$ as it sits in $\mathbb{R}^3$. -(One could imagine a vehicle traveling along $\gamma$, sending back $xyz$-coordinates at regular time intervals; assume -$t \rightarrow \infty$.) -For example, if the geodesic happens to be closed, your probe might -return the blue curve left below: - -          - - -                                                       -(Based on an image created by -Mark Irons.) - -I would like to know what information one could learn about $M$ -from such geodesic probes. -I am interested in the best case rather than the worst case. -For example, you might learn that $M$ is unbounded, if you are -lucky enough to shoot a geodesic to infinity. -In particular, - -Are there circumstances (a manifold $M$, a point $p$, directions $u$) - that permit one to definitively conclude that the genus of $M$ - is nonzero, by shooting (perhaps many) geodesics from one - fixed (well-chosen) point $p$? - -I believe that, if one knew all the geodesics through every point, -then there are natural circumstances under which the metric -is determined -[e.g., "Metric with Ergodic Geodesic Flow is Completely Determined by Unparameterized Geodesics." -Vladimir Matveev and Petar Topalov. -Electronic Research Announcements of the AMS. -Volume 6, Pages 98-101, 2000]. -But I am more interested what can be determined from a single point $p$ (and many directions $u$). -Thanks for thoughts/pointers! -(Tangentially related MO question: -Shortest-path Distances Determining the Metric?.) - -REPLY [21 votes]: There's a different kind of answer to this that you might be interested in: Suppose that, when you fire off a probe along a unit speed geodesic starting at $p\in M$, you record the direction $\theta$ in which you sent it, and the probe reports the inertial forces it is experiencing, i.e., it sends back a running report on the curvature and torsion of the curve it is traveling along. Thus, you get to record these two data as functions $\kappa(t,\theta)$ and $\tau(t,\theta)$ of $t$, the time since the probe was launched, and $\theta$, the direction in which it was sent. -The question, then, is "Can you recover the metric on the surface $M$ from the data $\kappa$ and $\tau$?" -Not surprisingly, the answer is yes in the generic situation. If, for example, you are in the situation in which $\tau\not=0$, you find (by computing with the structure equations) that the induced metric must be of the form -$$ -g = dt^2 + f(t,\theta)^2\ d\theta^2, -$$ -where -$$ -f(t,\theta) -= \frac{\sqrt{|\tau(t,\theta)|}}{2\ \tau(t,\theta)} -\int_0^t\frac{\kappa_\theta(\rho,\theta)}{\sqrt{|\tau(\rho,\theta)|}}\ d\rho\ . -$$ -(Obviously, there will be some singularity issues at places where $\tau$ vanishes, but, generically, this isn't a problem. The case in which $\tau$ vanishes identically, such as when $p$ is a pole of rotational symmetry of the surface $M$, has to be treated separately.) -Since you can recover the curvature from $f$ by the formula $K = -f_{tt}/f$, you could detect when, for example, the surface $M$ is locally convex, and so forth. -As for computing the Euler characteristic, since $K\ dA = -f_{tt}\ dt\wedge d\theta -= -d\left(\ f_t\ d\theta\ \right)$, it follows that, if you could figure out the star-shaped domain (in good cases, of the form $0\le t < T(\theta)$ for some piecewise differentiable $2\pi$-periodic function $T$ ) that maps one-to-one and onto the complement of the cut locus of $p$, then you could compute the Euler characteristic as -$$ -\chi(M) = \frac{-1}{2\pi}\int_0^{2\pi} f_t\bigl(T(\theta),\theta\bigr)\ d\theta. -$$ -How numerically stable all these calculations are, I don't know. (I also don't know how hard it would be to figure out or approximate $T$.) Since you know that the result is an integer, though, you might be able to tolerate a reasonable amount of numerical error.<|endoftext|> -TITLE: Is strong approximation difficult? -QUESTION [24 upvotes]: Recently a colleague and I needed to use the fact that the natural map $SL_2(\mathbb{Z}) \rightarrow SL_2(\mathbb{Z}/N\mathbb{Z})$ is surjective for each $N$. I happily chugged my way through an elementary proof, but my colleague pointed out to me that this is a consequence of strong approximation. -After browsing through some references (including Ch. 7 of Platonov and Rapinchuk, and one of Kneser's papers) I now cheerfully agree with my colleague. Indeed, as P+R describe, strong approximation is "the algebro-geometric version of the Chinese Remainder Theorem". -But what about the proofs? To this particular novice in this area, the proofs seemed rather difficult -- involving a variety of cohomology calculations, nontrivial theorems about Lie groups, and much else besides. I confess I did not attempt to read them closely. -But perhaps this is because the theorems are proved in a great deal of generality, or perhaps such arguments are regarded as routine by experts. My question is this: - -Is the proof of strong approximation genuinely difficult? - -For example, can a reasonably simple proof be given for $SL_n$ which does not involve too much machinery, but which illustrates the concepts behind the general case? - -REPLY [8 votes]: I am quite surprised that it seems to be so poorly known, but an elementary proof of Strong Approximation for $SL_n$ over a Dedekind domain can already be found in Bourbaki (Algebre Commutative, VII, $\S$2, n.4). Essentially, the idea is to deduce it from the Chinese Remainder Theorem, by means of elementary matrices, as hinted in Ralph's answer. Actually, I would be curious to know a bit more about the history of this case of Strong Approximation: who and when did prove it first?<|endoftext|> -TITLE: Constructive proof of algebraic elements forming a subfield -QUESTION [5 upvotes]: Let $F \leqslant E$ be a field extension. -If $a, b \in E$ are algebraic over $F$ then $a+b$ and $ab$ are algebraic as well. There is an short proof of this using the extension $E(a,b)$: -$[E(a,b):E]$ is finite so all elements are algebraic otherwise powers non-algebraic would form an infinite linearly independent set. In particular $a+b \in E(a,b)$ and $ab \in E(a,b)$ are algebraic. -But this proof doesn't construct the actual polynomial. Is there a constructive proof or any reason for such a proof not to exist? - -REPLY [4 votes]: This question keeps getting asked on forums, e.g.: -http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=202382 and www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=180471 . -This fact is a particular case of a more general fact, which claims that if $A$ is a subring of a commutative ring $B$, and if $a$ and $b$ are two elements of $B$ integral over $A$, then $a+b$ and $ab$ are integral over $A$, too. This is a standard result in commutative algebra, and usually proven constructively: e.g. Corollary 2.1.11 in Swanson-Huneke yields it. -PS. This might be clear to you, but there is no chance to get the minimal polynomial of either $a+b$ or $ab$ by any algorithm. What you can get are polynomials, both of degree $\left[E\left(a\right):E\right]\cdot\left[E\left(b\right):E\right]$, that have $a+b$ resp. $ab$ as roots. These can be obtained as resultants in a way similar to Felipe's, or by executing any of the above-mentioned constructive proofs as programs.<|endoftext|> -TITLE: Does there exist infinitely many prime knots? -QUESTION [7 upvotes]: I'm not a topologist and I just saw the definition of prime knot a while ago. -Today I'm somewhat supprised to realize that I don't even know if there are infinitely many prime knots? If this question is not completely trivial then I'm hoping to see a "proof from the book". -(A related question is, is the decomposition of a composite knot into prime knots unique? I would hope so but I don't have a strong reason to support that.) - -REPLY [2 votes]: Yes. -Kauffman and Lopes found some very elementary families: https://arxiv.org/pdf/1604.02510.pdf .<|endoftext|> -TITLE: Vocabulary of 19th Century analytic projective geometry: What are "order" and "dimension"? -QUESTION [14 upvotes]: I am trying to understand the following introductory passage in an early lecture by the philosopher/mathematician Gottlob Frege because I am interested in how Frege conceived of the role of geometric intuition in mathematical reasoning. - -One of the most far-reaching advances made by analytic geometry in more recent times is that it regards not only points but also other forms (e.g., straight lines, planes, spheres) as elements of space and determines them b means of coordinates. In this way we arrive at geometries of more than three dimensions without leaving the firm ground of intuition. The geometry of straight lines for example is a four dimensional one, and so is the geometry of spheres. But there is a difference between the two, in that a sphere can always be determined unequivocally by four numbers, whereas it would seem that this is not possible in the case of straight lines. we determine a straight line by an equation between six quantities with a quadratic equation holding between them. We express this peculiarity of the geometry of straight lines by calling it a second order one, whereas the geometry of spheres is of the first order. -The geometry of pairs of points in the plane, with which we will here be concerned, is four-dimensional and of the third order. - -How are we to understand "dimension" and "order" in this passage? I guessed that the dimension is the number of coordinates required (six for lines) minus the order of the equation required to hold between the coordinates (two for lines). To get a better grip on the basic idea I worked the algebra on the equations for lines in three dimensional space. -$x = x_{0} + at$ -$y = y_{0} + bt$ -$z = z_{0} + ct$ -So that: -$t = \frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}$ -And: -$abct^{3} = xyz - xyz_{0} - xy_{0}z + xy_{0}z_{0} - x_{0}yz + x_{0}yz_{0} + x_{0}y_{0}z - x_{0}y_{0}z_{0}$ -Substituting the original equations into that, and doing a bunch of algebra, gives a quadratic in t with coefficients expressed in $x_{0}, y_{0}, z_{0}, a, b, c$. So I'm trying to make sense of Frege's description of the geometry of lines is dimension four and order two by thinking of the dimension as the number of coordinates minus the order of the equation required for those coordinates to characterize a line. But, I can't quite make that fit with what he says about the geometry of spheres. -Frege's early work in geometry is influenced by the analytic projective geometry in the style of Plucker, using homogeneous coordinates. I've consulted some internet sources and used the undergraduate textbook "Modern Geometries" to better understand homogeneous coordinates, but I have been unsuccessful finding definitions of dimension and order for this context. What does Frege mean by dimension and order in the above quotation? - -REPLY [6 votes]: As Francesco Polizzi remarks, the idea is to identify $\text{Sym}^2(\mathbb{P}^2)$ with a cubic hypersurface in $\mathbb{P}^5$. Frege's lecture (see link in comments) actually goes on to explain how this is done. The cubic hypersurface is (in the six variables $s_1,s_2,s_3,t_1,t_2,t_3$), -$\text{det} \begin{pmatrix} -s_1 & t_3 & t_2 \cr -t_3 & s_2 & t_1 \cr -t_2 & t_1 & s_3 -\end{pmatrix}=0$ -and the identification of $\text{Sym}^2(\mathbb{P}^2)$ with this hypersurface is -$\{[x_1:x_2:x_3],[y_1:y_2:y_3]\}\mapsto [x_1y_1:x_2y_2:x_3y_3:\frac{1}{2}(x_2y_3+x_3y_2):\frac{1}{2}(x_3y_1+x_1y_3):\frac{1}{2}(x_1y_2+x_2y_1)]$. -Quite possibly, someone more knowledgeable than me can motivate this!<|endoftext|> -TITLE: Complement of a subspace which is a cartesian product -QUESTION [10 upvotes]: Let $H$ be a Hilbert space and $U$ a closed subspace of $H\times H$ . -Does then exist closed subspaces $V$ and $W$ of $H$ such that $H\times H = -U \oplus (V\times W)$ ? -See also Perturbations of an operator that disconnect the spectrum . - -REPLY [8 votes]: Let me expand my comment: the answer is yes, and in general $V$ and $W$ can be constructed as follows: let $p_1,p_2:U \to H$ denote the restrictions to $U$ of the two coordinate projections. Then $V$ is the image of the spectral projection $1_{[0,1/2]}(p_1p_1^*)$ and $V$ is the image of the spectral projection $1_{[0,1/2[}(p_2 p_2^*)$ (note that one interval is open at $1/2$, whereas the other is close). One can replace the two $1/2$'s by $\delta$ and $1-\delta$, but the choice of $\delta = 1/2$ gives the best constants. -This construction is perhaps clearer in the particular case when $U$ is a graph (i.e. $U \cap (\{0\} \times H) = \{0\}$, or equivalently $U = \{ (x,Tx),x \in D\}$ for a closed operator $T$ of domain $D \subset H$). One can then easily reduce to the case when $T$ is densely defined, and (using the polar decomposition), $T$ is self-adjoint. By the spectral theorem, we can assume that there is a measure space such that $H = L^2(X,\mu)$ and $T$ is the multiplication operator by some function $f:X \to \mathbb R^+$. Then $V$ (resp. $W$) is the space of functions in $L^2(X,\mu)$ that are zero outside of $I = f^{-1}([1,\infty[)$ (resp. $J = f^{-1}([0,1[) = X \setminus I$). For $(a,b) \in H \times H$ the corresponding decomposition in $U \oplus V \times W$ is then $(a 1_J + b/f 1_I , a f 1_J + b 1_I) + ((a-b/f)1_I,(b- a f)1_J)$.<|endoftext|> -TITLE: Signs in the unstable homotopy groups of spheres -QUESTION [17 upvotes]: Let $\mathbb{HP}^2$ denote the quaternionic projective plane. According to -A note on $\mathcal{E}(\mathbb{HP}^n)$ for $n\leq 4$, N. Iwase, K-I. Maruyama, S. Oka, Math. J. Okayama Univ. 33 (1991) , 163-176. -any homotopy self equivalence of $\mathbb{HP}^2$ is homotopic to the identity on the 4-cell. That seems to mean that the composition -$$S^7 \overset{\nu}\longrightarrow S^4 \overset{-1}\longrightarrow S^4 \longrightarrow \mathbb{HP}^2$$ -is not nullhomotopic, where $\nu$ is the Hopf map, and $-1$ is a map of degree $-1$. This in turn seems to show that $(-1)\circ \nu$ is not in $\langle \nu \rangle$, and in particular is not $-\nu$ (negatives are taken by composing with a map of degree $-1$ in the source sphere). -In the stable homotopy groups of spheres, the composition product is graded-commutative, so $(-1) \circ \nu = \nu \circ (-1) = -\nu$. Have I made a mistake, or is it really true that this fails unstably? -If it is true, how can I detect the nontrivial class $\nu + (-1)\circ \nu \in \pi_7(S^4)$? - -REPLY [5 votes]: I like this discussion: I just wanted to add that of course a similar thing happens for $\mathbb{O} P^2$ as well. The homotopy theory is very nicely discussed by Baues in section 9 of On the group of homotopy equivalences of a manifold .<|endoftext|> -TITLE: Creating Models of $ZFC$ -QUESTION [9 upvotes]: I know of only 2 main techniques to create a model of $ZFC$. The first one is creating a model which is an extension of $V$: this is forcing. The second technique is that of inner model theory and looking at subclasses of $V$. Do all methods to generate models of $ZFC$ fall in the these 2 categories or are there other radically different ways to generate of a model? - -REPLY [14 votes]: As Carl Mummert mentioned, the usual ways of constructing models — completeness and compactness — are also available for the construction of models of set theory. The models constructed in this way are usually not (externally) wellfounded but they have found some interesting uses. -For example, Joel David Hamkins proved the consistency of the Maximality Principle in this way. This principle asserts that any sentence $\phi$ that holds in some forcing extension and then in all further forcing extension — such a sentence is called forceably necessary — already holds in the ground model. He did this by showing that there is no sentence such that $\phi$ and $\lnot\phi$ are both forceably necessary. Therefore, the forceably necessary sentences form a consistent theory, which therefore has a model by completeness. [A simple maximality principle, JSL 68 (2003), 527–550. arXiv] -Another example (that I learned from Joel David Hamkins, but might be due to Sol Feferman) is the relative consistency with ZFC of the existence of some $\delta$ such that $V_\delta$ is an elementary submodel of $V$. The consistency of this follows from the reflection principle, which states that if $\phi(\bar{a})$ is any formula of set theory, then there is a closed unbounded class of $\delta$ such that $V_\delta \vDash \phi(\bar{a})$ iff $V \vDash \phi(\bar{a})$. It follows that every model of set theory satisfies every finite fragment of the said theory, therefore the whole theory is satisfiable by compactness. -End-extensions of models is another way by which new models can be constructed. This is an unusual method because it adds new ordinals and it can still form wellfounded models in some circumstances. This extends a well-known method of McDowell and Specker for models of arithmetic. Jack Silver, Jerome Keisler, and Ali Enayat have done significant work on this. In the opposite direction, there are various ways of cutting down models of set-theory to obtain new models. One such example is Cohen's minimal model of set theory, which is $L_\delta$ for the smallest ordinal $\delta$ (possibly equal to $Ord$) such that $L_\delta \vDash \mathrm{ZFC}$. [A minimal model for set theory, Bull. AMS 69 (1963), 537–540. MR0150036] (Note that these are not always inner-models since they often have fewer ordinals and they are generally not definable.)<|endoftext|> -TITLE: Homotopy type of the self-homotopy equivalences of a bouquet of spheres -QUESTION [13 upvotes]: Before I state the questions I have in mind, let me give some background. If one considers $S^2$ then it is known due to Kneser that $\textrm{Homeo}^{+}(S^2)$ has the homotopy type of $SO(3)$. By Smale's work we also know that $\textrm{Diff}^{+}(S^2)$ is homotopic to $SO(3)$. However, when we work in the homotopy category this changes. Later Hansen considered $\textrm{Aut}_0(S^2)$, the connected component of identity in the space of all self homotopy equivalences. He showed that its homotopy type is that of $SO(3)\times \mathbf{\Omega}$, where $\mathbf{\Omega}$ is the universal cover of the connected component of the constant loop in the double loop space $\Omega^2 S^2$. -Although this is a very nice and interesting fact, it's turns out to be hard to generalize his method of proof for higher spheres. For one, the homotopy type of $\textrm{Diff}(S^n)$ is not really known for $n\geq 7$ if I'm not mistaken. For another, Hansen crucially uses the fact that $S^2$ is the base of the usual fibration $SO(2)\to SO(3)\to S^2$ and the fact $SO(2)=S^1$ has no higher homotopy. This indubitably fails for higher spheres! -Question 1 What is known about the homotopy type of $\textrm{Aut}_0(S^n)$ for $n\geq 3$? -I should say that the rational homotopy type is fairly easily calculable via Sullivan's minimal models. So, I'm looking for a bit more here. -Question 2 What is the homotopy type of the identity component of self homotopy equivalences of $\vee_k S^2$, a bouquet of $2$-spheres? -Of course, one can ask this question for higher spheres but bearing in mind question 1, I decided I would be happy with an answer for $S^2$. If it helps, the homotopy groups of $\vee_k S^2$ can be calculated by Hilton-Milnor theorem. This fits in a long exact sequence of groups associated to -$\textrm{Aut}_0^\ast(\vee_k S^2)\to \textrm{Aut}_0(\vee_k S^2)\to \vee_k S^2$ -where the last map is evaluation of an automorphism at the common point of the bouquet and $\textrm{Aut}^\ast_0(\vee_k S^2)$ consists of based maps. But this doesn't seem to lead anywhere! - -REPLY [9 votes]: Regarding question 1 the following is well known. -For a reasonably nice space $X$ (say, a finite connected CW-complex) let $Aut_0(X)$ be the monoid of self-equivalences of $X$ homotopic to the identity and let $Aut_0^\bullet(X)$ be the submonoid fixing a basepoint. Then there is a natural fibration given by the evaluation map $Aut_0^\bullet(X)\to Aut_0(X)\to X$. This gives rise to the fibration $X\to BAut_0^\bullet(X)\to BAut_0(X)$ which is the universal fibration with fiber $X$. -For $X=S^n$ it's easy to see that $Aut_0^\bullet(S^n)\cong Map_0^\bullet(S^n, S^n)\cong Map_0^\bullet(S^{n-1}, \Omega S^n)\cong \ldots Map_0^\bullet (S^0, \Omega^n(S^n))=\Omega^n(S^n)$. In particular, $\pi_k(Aut_0^\bullet(S^n))\cong \pi_{n+k}(S^n)$. Combining this with the fibration $Aut_0^\bullet(S^n)\to Aut_0(S^n)\to S^n$ this in principle gives you homotopy groups of $Aut_0(S^n)$. In practice though I don't think this is very useful away from rational coefficients or for small values of $k$ where you have stability. -I'm not familiar with the general theory for spaces other than spheres. As I mentioned in my comment, rationally the situation is in principle well understood. But even for rational coefficients the homotopy structure of $Aut_0(X)$ is complicated when $X$ is a wedge of spheres. As I mentioned, rational homotopy groups of $Aut_0(X)$ grow exponentially. See "The monoid of self-homotopy equivalences of some homogeneous spaces" by Félix and Thomas for a careful proof of that. -One general fact that I know is a result of Dror, Dwyer and Kan that when $X$ is a nilpotent finite $CW$-complex then $BAut_0(X)$ has finite type, i.e. it has the homotopy type of a CW complex with a finite number of cells in each dimension.<|endoftext|> -TITLE: area of triangle from coefficients of its cubic? -QUESTION [11 upvotes]: Three points $z_1$, $z_2$, $z_3$ on the complex plane are given by the coefficients $a_k$'s of the cubic polynomial $f(z)=(z-z_1)(z-z_2)(z-z_3)=\sum_{k=0}^3 a_k z^k$. How does one express the (signed) area $V$ of the triangle with vertices $z_1$, $z_2$, $z_3$ in terms of $a_k$'s and $\overline{a}_k$'s? One is tempted to try to expand $V^2$ in the symmetric functions in the roots of $f(z)\overline{f}(z)$, as well as these of $f(z)$ and of $\overline{f}(z)$, e.g. starting from -$$ -V=\frac{\sqrt{-1}}{4}\det -\begin{pmatrix} -1& 1&1 \\ -z_1& z_2& z_3 \\ -\overline{z}_1& \overline{z}_2& \overline{z}_3\end{pmatrix}, -$$ -(this not so well-known formula can be found in R.Deaux, Introduction to the Geometry of Complex Numbers, Ungar, New York, 1956, pp.59-60), but this rather calls for some kind of joint invariants of $f$ and $\overline{f}$ to be used. -Any pointers etc. are much appreciated. -Added: the motivation comes from a moment problem: suppose one is given a part of the sequence $\mu_n=\int_\Delta t^n dx dy$, where $t:=x+\sqrt{-1}y$, and wants to find the triangle $\Delta$, e.g., its vertices $z_i$'s. P.Davis in his paper "Triangle formulas in the complex plane" (Math.Comp. 18(1964)) shows that the first 4 moments $V=\mu_0$,...,$\mu_3$ determine $\Delta$; this is one more parameter than needed to determine the $z_i$'s. We can do better, but were unsure how $V$ depends upon $\mu_1$,...,$\mu_3$, which boils down to this very question. - -REPLY [13 votes]: NB: Note that my $a_k$ have different signs from those defined in the question. For me, -$$ -(z - z_1)(z-z_2)(z-z_3) = z^3 - a_1\ z^2 + a_2\ z - a_3, -$$ -so that $a_k$ is the $k$-th elementary symmetric function of the $z_i$. This doesn't really affect the answer in any significant way. -While I don't think that the final result relating $V$ to the $a_k$ and $\bar a_k$ is that interesting or useful, people might want to know a way to derive it. Here is what I did: -Start with the formula -$$ -V=\frac{\sqrt{-1}}{4}\det -\begin{pmatrix} -1&1&1 \\ -z_1& z_2& z_3 \\ -\overline{z}_1&\overline{z}_2&\overline{z}_3\end{pmatrix}, -$$ -and note that multiplying the matrix on the right by its transpose yields -$$\begin{pmatrix} -1&1&1 &\\ -z_1&z_2&z_3\\ -\bar z_1& \bar z_2 &\bar z_3 \end{pmatrix} -\begin{pmatrix} -1&z_1&\bar z_1\\ -1&z_2&\bar z_2\\ -1&z_3&\bar z_3 \end{pmatrix} -=\begin{pmatrix} -3&a_1&\bar a_1 \\ -a_1&{a_1}^2-2a_2& S \\ -\bar a_1&S&{\bar a_1}^2-2\bar a_2 -\end{pmatrix} -$$ -where $S = z_1\bar z_1 + z_2\bar z_2 + z_3\bar z_3$. Thus, one has the polynomial relation -$$ -R := 16V^2 - 3S^2 + 2 a_1\bar a_1 S + {a_1}^2{\bar a_1}^2 -- 4{a_1}^2\bar a_2 - 4{\bar a_1}^2a_2 + 12 a_2 \bar a_2 = 0. -$$ -It remains to find a relation between $S$ and the $a_k$ and $\bar a_k$. To do this, note that $S$ will be a root of the polynomial -$$ -Q := \prod_{\pi\in S_3} \bigl(S - z_1{\bar z_{\pi(1)}} - z_2{\bar z_{\pi(2)}} - z_3{\bar z_{\pi(3)}}\bigr). -$$ -Note that $Q$ is a polynomial of degree $6$ in $S$ that is symmetric in the $z_i$ and the $\bar z_i$ separately. Hence, $Q$ can be regarded as a polynomial of degree $6$ in $S$ with coefficients that are polynomials in the $a_k$ and $\bar a_k$. The resulting expression for $Q$ as a polynomial in $S$, the $a_k$, and the $\bar a_k$ has $66$ terms. (Actually expressing $Q$ this way is not easy by hand. However, it's very easy to implement the algorithm for writing a symmetric polynomial in three variables as a polynomial in the elementary symmetric functions in those variables on a computer, which is what I did.) -Finally, $R$ and $Q$ are polynomials in $S$ with coefficients that are polynomials in the variables $V, a_1,a_2,a_3,\bar a_1,\bar a_2,\bar a_3$. Set -$$ -P(V,a_1,a_2,a_3,\bar a_1,\bar a_2,\bar a_3) := \text{Resultant}_S(R,Q). -$$ -Then $P=0$ is the desired relation. Computation (using Maple) shows that it has the following properties: $P$ is irreducible, is even and of degree $12$ in $V$, is of total degree $20$, and contains $598$ monomial terms, with typical integer coefficient in the millions. -Added remark: The formulae simplify dramatically if one assumes that the centroid of the triangle is at $z=0$, i.e., that $a_1 = \bar a_1 = 0$. (This is analogous to the way that the cubic formula itself simplifies when one removes the quadratic term. Moreover, this can always be easily arranged in the usual way by translation.) When $a_1 = \bar a_1 = 0$, we have -$$ -R = 16V^2 - 3S^2 + 12 a_2 \bar a_2 -$$ -and -$$ -\begin{array} -\\ -Q &= S^6-6{a_2}{\bar a_2}S^4-27{a_3}{\bar a_3}S^3+9{a_2}^2{\bar a_2}^2S^2\\ - &\qquad {} +81{a_2}{a_3}{\bar a_2}{\bar a_3}S - -27{a_3}^2{\bar a_2}^3 - 27{a_2}^3{\bar a_3}^2 - 4{a_2}^3{\bar a_2}^3. -\end{array} -$$ -The formula for $P$ is still not that nice, though; it has $17$ terms, with coefficients in the millions. Rather than input it here, I'll just recommend that those interested compute the resultant of this reduced $R$ and $Q$ with respect to $S$ to get it. -Another added remark (the real case): If, in addition, one assumes that the $a_k$ are real (and that $a_1=0$), then the polynomial $P$ factors as -$$ -16V^2 (16V^2{+}9{a_2}^2)^2 -(4096V^6{+}4608{a_2}^2 V^4{+}1296{a_2}^4V^2{-}19683{a_3}^4{-}2916{a_2}^3{a_3}^2). -$$ -In particular, note that $V=0$ is always a root and that $P=0$ has a positive real root $V$ if and only if $27{a_3}^2+4{a_2}^3$ is positive (and $a_3\not=0$), i.e., if and only if the original cubic has only one real (nonzero) root, just as one would expect.<|endoftext|> -TITLE: Localizability of differential operators a la Grothendieck -QUESTION [5 upvotes]: Hello, -Maybe this question is trivial, so sorry -Let $A$ be a (comm. with 1) $k$-algebra, where $k$ is a ring (comm. with 1). -Then we can define the module of differential operators $D^{\leq n} (A)$, a submodule of $End_k (A,A)$ (endomorphisms of vector spaces). $D^{\leq -1} = 0$, and then inductively $D^{\leq n} = \{ d | [d,a]\in D^{\leq n-1}\}$. -We have a lemma: -Lemma. Let $f \in A$. Then for every $d \in D^{\leq n}(A)$ we can find unique $e \in D^{\leq n}(A_f )$, such that $l\circ d = e \circ l$, where $l: A \to A_f$ is the localization map. -I think that I know how to prove the lemma, by induction on the order of diff. op. (just need to see how to apply operators to fractions). It gives us a map $D^{\leq n}(A) \to D^{\leq n}(A_f)$. -Question 1. Under which assumptions on $A/k$ this map $D^{\leq n}(A) \to D^{\leq n}(A_f)$ is a localization map (i.e. becomes an isomorphism after tensoring (say on the left, it does not matter) with $A_f$)? -Question 2 (my real question). If $A/k$ is finitely generated, or finitely presented, is this a localization map? -Somehow, I am having trouble with the "surjectivness" part. Maybe there is some reference? -Thank you, -Sasha - -REPLY [2 votes]: Two observations (with $k$ a field of characteristic zero): - -If $A$ is a domain over a field $k$, then elements of $D(A)$ extend to elements of $D(A_S)$ for all multiplicatively closed sets $S$. -Your questions become easier if you ask instead about the subalgebra $\Delta(A)$ of $D(A)$ generated by $A$ and derivations: then the answer is yes to your two questions. Now, if $A$ is finitely generated and regular, then $D(A)=\Delta(A)$, so in this case the answer is yes for $D(A)$ too. - -You'll find this in the last chapter of McConnell and Robson's book on noetherian rings. -(Finally, $D^{\leq n}(A)$ is an $A$-bimodule which is not symmetric, so tensoring with $A_f$ on one side or the other is not the same thing)<|endoftext|> -TITLE: Uniqueness of compactification of an end of a manifold -QUESTION [23 upvotes]: Let $M$ be an $n$-dimensional manifold (smooth or topological). I call $\bar{M}$ a compactification of $M$ if it is an $n$-dimensional compact manifold with boundary $\partial \bar{M}$, an $(n-1)$-dimensional manifold, such that $M$ is the interior of $\bar{M}$. I understand that not every manifold $M$ has such a compactification. The obstructions have been discussed in some previous MO questions (22441, 34602). Essentially, no compactification can exist if the ends of the manifold are too "wild". -However, I am more interested in how two compactifications are related to each other, provided they exist. For simplicity, let me assume that $M$ has only one end and a compactification exists. If $\bar{M}_1$ and $\bar{M}_2$ are both compactifications, what can be said about the relationship between them and between $\partial\bar{M}_1$ and $\partial\bar{M}_2$? Does there exist some kind of "universal" compactification $\bar{M}_U$ from which both $\bar{M}_1$ and $\bar{M}_2$ could be obtained? If these questions are easier to answer in the topological category, then I would be quite happy with just that information. -Here's an example. Consider $M=\mathbb{R}^n$ ($n\ge 2$). It has one end. An obvious compactification is to consider it as the interior of a closed ball $\bar{M}_1=\bar{B}^n$, so the boundary is a sphere $\partial\bar{M}_1=S^{(n-1)}$. However, I can apply a homotopy to $\bar{B}^n$ which maps the boundary into itself and restricts to a homoemorphism on the interior. Composing this operation with the embedding of $M$ in $\bar{M}_1$ I get a different compactification $\bar{M}_2$. By construction, there is a map $\bar{M}_1 \to \bar{M}_2$, which is a homeomorphism between the interiors but only a homotopy between the boundaries. In particular, the homotopy could blow up a point on $\partial\bar{M}_1$ to a closed set with open interior. Then two curves that had the same end point on the boundary of $\bar{M}_1$ could be mapped to curves with distinct end points on the boundary of $\bar{M}_2$. Based on this example, I would naively guess that $\mathbb{R}^n$ can be compactified by adding an $S^{(n-1)}$ boundary and all other compactifications can be obtained by applying a homotopies to any one element of this class of compactifications. So this compactification could be considered universal. How far is this intuition from reality? -I'm aware of resources like L. Siebenmann's thesis and the book Ends of Complexes by Ranicki and Hughes. Unfortunately, I don't have enough algebraic and topological background to immediately see whether they contain the answer or how to dig it out if they do. So precise suggestions for where to look would also be helpful! - -REPLY [19 votes]: Suppose $\overline{M}_i$, $i=0,1$, are compact smooth manifolds with boundary whose interiors are diffeomorphic: let $\psi$ be such a diffeomorphism, and $M$ for either interior (identified via $\psi$). For both manifolds one can find a smooth collar $c_i : \partial \overline{M}_i \times [0, 1) \hookrightarrow \overline{M}_i$, as they are compact. By shrinking $c_0$ (and re-parametrising) we can suppose it has image inside that of $c_1$ (restricted to the interior, compared via $\psi$), and then we obtain an embedding -$$\partial \overline{M}_0 \times [\tfrac{1}{2}, 1) \hookrightarrow \partial \overline{M}_1 \times (0, 1) \subset \overline{M}_1 \times [0, 1).$$ -The complement of this embedding has two components, precisely one of which is compact and is a cobordism $W : \partial \overline{M}_0 \leadsto \partial \overline{M}_1$. -This is an $h$-cobordism, as each of the $\partial \overline{M}_i \times [0, 1)$ and $W$ are homotopy equivalent to the space -$$\mathcal{E}(M) := \mathrm{holim}_{K \subset M \text{compact}} M \setminus K$$ -as there are evident cofinal subdiagrams indexed by $K = M \setminus c_i(\partial \overline{M}_i \times [0, \epsilon))$ for either $i$, which are homotopically constant. -Thus, the two boundaries $\partial \overline{M}_i$ are $h$-cobordant (so diffeomorphic if they are simply-connected). If the boundaries are not simply-connected, then I imagine that this $h$-cobordism can be non-trivial, as in Benoît Kloeckner's discussion.<|endoftext|> -TITLE: Can we always find such an irreducible polynomial of degree n where degree(p(x)-x^n)<= n/2? -QUESTION [11 upvotes]: Consider unitary polynomials of degree $n$ over $GF(2)$. That is, polynomials of the form $p(x) = \sum_{i=0}^n a_i x^i$ where $a_i\in GF(2)$ and $a_n=1$. -Can we always find such an irreducible polynomial $p(x)$ of degree $n$ where $\textrm{degree}(p(x)-x^n)\leq n/2$? -Example: $p(x)=1+x^2+x^3+x^5+x^{400}$ is an irreducible polynomial where $\textrm{degree}(p(x)-x^n)=5\leq 400/2$. -Further question: If I assume that $n$ is sufficiently large, say $n\geq 32$, is a tighter bound possible (e.g., $\textrm{degree}(p(x)-x^n)\leq n/4$)? - -REPLY [2 votes]: I leave my earlier post below where I mistakenly understood the original problem because of a bad illustrative example given there. Here I only indicate the explicit formula for the number $N_ m(p)$ of monic polynomials of exact degree $m$ irreducible over $GF(p)$: -$$ -N_m(p)=\frac1m\sum_{d\mid m}\mu(d)p^{m/d}. -$$ -This formula is again from Prasolov's Polynomials, and it seems to be absent in other posts and comments. - -I assume that the irreducibility is discussed over $\mathbb Z$, otherwise $x^{400}+x^5+x^2+1$ has zero $x=1$ in $GF(2)$. -Victor Prasolov in Section 8 of his book Polynomials discusses the irreducibility -of trinomials and quadrinomials, mostly based on the work [W. Ljunggren, On the -irreducibility of certain trinomials and quadrinomials, Math. Scand. 8 -(1960), 65--70]. One of the results from there is as follows. -Theorem. Let $n\ge2m$, $d=\text{gcd}(n,m)$, $n_1=n/d$ and $m_1=m/d$. Then the -polynomial -$$ -g(x)=x^n+\epsilon x^m+\epsilon', \quad\text{where}\ \epsilon,\epsilon'\in\lbrace\pm1\rbrace, -$$ -is irreducible except for the following cases in which $n_1+m_1\equiv0\pmod3$: -(a) $n_1$ and $m_1$ are odd and $\epsilon=1$; -(b) $n_1$ is even and $\epsilon'=1$; -(c) $m_1$ is even and $\epsilon=\epsilon'$. -In all three cases (a)--(c), $g(x)$ is a product of a certain irreducible polynomial -and $x^{2d}+\epsilon^m{\epsilon'}^nx^d+1$. -Corollary. -If $n\not\equiv2\pmod3$, then $x^n+x+1$ is irreducible. -If $n\equiv2\pmod3$, then $x^n+x^2+1$ is irreducible. -In other words, there is an irreducible degree $n$ polynomial $g(x)$ of the wanted form such that $\deg(g(x)-x^n)\le2$, and this bound cannot be further improved.<|endoftext|> -TITLE: A homotopy commutative diagram that cannot be strictified -QUESTION [21 upvotes]: By a "homotopy commutative diagram," I mean a functor $F: \mathcal{I} \to \mathrm{Ho}(\mathrm{Top})$ to the homotopy category of spaces. By a "strictification," I mean a lifting of such a functor to the category $\mathrm{Top}$ of topological spaces. I am curious about simple instances where such a "strictification" does not exist: that is, there are obstacles to making a diagram that commutes only up to homotopy strictly commute. -For example, it is known that a homotopy coherent diagram in topological spaces can always be strictified. In other words, if one imposes homotopy commutativity, but also keeps track of all the homotopies and requires that they satisfy compatibility conditions with one another, then one can strictify. More precisely, a homotopy coherent diagram $\mathcal{I} \to \mathrm{Top}$ can be described as a simplicial functor $\mathfrak{C} \mathcal{I} \to \mathcal{Top}$, where $\mathfrak{C}\mathcal{I}$ (notation of HTT) is a "thickened" version of $\mathcal{I}$ where the usual relations in $\mathcal{I}$ only hold up to coherent homotopy. Then it is a theorem of Dwyer and Kan that the projective model structures on homotopy coherent diagrams (i.e. $\mathrm{Fun}(\mathfrak{C} \mathcal{I}, \mathrm{SSet})$) and on commutative diagrams (i.e. $\mathrm{Fun}(\mathcal{I}, \mathrm{SSet})$ are Quillen equivalent under some form of Kan extension. -Part of this question is to help myself understand why homotopy coherence is more natural than homotopy commutativity. One example I have in mind is the following: let $X$ be a connected H space, and suppose $X $ is not weakly equivalent to a loop space. Then one can construct a simplicial diagram in the homotopy category given by the nerve construction applied to $X$ (as a group object in the homotopy category). This cannot be lifted to the category of spaces, because then Segal's de-looping machinery would enable us to show that $X$ was the loop space of the geometric realization of this simplicial space. -However, this example is rather large. Is there a simple, toy example (preferably involving a finite category) of a homotopy commutative diagram that cannot be strictified? - -REPLY [11 votes]: A good place to look for finite toy examples is in the theory of group extensions with non-abelian kernel. -A homomorphism $G \to Out(H)$ is the same as a functor $G \to Ho(Top)$ sending $*$ `to $BH$, where $G$ is viewed as a category with one object $*$ and $G$ as morphisms. By the Dwyer-Kan obstruction theory (noting that the diagram is centric) the obstructions to lifting this to a diagram in Top lies in $H^3(G;Z(H))$. -A closer inspection shows that this obstruction class is the same as the obstruction in $H^3(G;Z(H))$ to constructing an extension -$$1 \to H \to ? \to G \to 1$$ -It is possible to calculate this obstruction group in a number of cases: -Eilenberg-MacLane show in the 1947 Annals paper "Cohomology Theory in Abstract Groups II" that given a group $G$ and an abelian group $C$ with an action of $C$, and a class $v$ in $H^3(G;C)$ then there exist a group $H$ with center $C$ and a morphism $G \to Out(H)$ realizing the class $v$. (A special case is proven in MacLane's book Homology, and there is an exposition on some of this in Brown's book on group cohomology.) Note that in this construction $H$ may be infinite even if $G$ and $C$ are not. -It is also possible to give completely finite examples: Given a group H there is a universal obstruction in $H^3(Out(H);Z(H))$ associated to this group, determining whether the extension -$$1 \to H \to ? \to Out(H) \to 1$$ -exists. Other obstruction classes will be a pull-back of this one (in particular it has the best chance of being non-zero). -The topological interpretation of this obstruction class is that it is the unique k-invariant of the space $BAut(BH)$, where Aut means the space of self-homotopy equivalences. (Recall that group extensions correspond to fibrations, and fibrations with fiber $F$ are classified by maps into $BAut(F)$, and this is a good way to understand group extensions with non-abelian kernel.) -Now, it is possible to calculate this class explicitly for small groups $H$. It of course vanishes when H is abelian. It also turns out to vanish for $H = D_8$ or $Q_8$, but is actually non-zero for $D_{16}$ and $Q_{16}$. -[As an aside I can say that the H^3 class always vanishes for compact connected Lie groups, which is a theorem of de Siebenthal from the 1950's. Kasper Andersen and I proved that this generalizes to p-compact groups (in a G&T paper from 2008), which was why Kasper and I examined the case where $H$ finite.] -[As a further aside I can say that there are a lot of similar obstruction questions over relatively small categories such as the orbit category and corresponding centric diagrams, which occur in e.g., in the theory of p-compact groups and p-local finite groups, when addressing certain uniqueness questions -- here lucily the obstructions are usually zero, although this is usually not so easy to prove, and in particular don't seem to follow from "formal" arguments.]<|endoftext|> -TITLE: regular homotopy -QUESTION [6 upvotes]: Hello. I am trying to give a seminar in my University about the Whitney-Graustein Theorem. There are many elementary proofs for that including Whitney's paper. The conclusion is that the connected components ($π_0$) of regular immersions $S^1 \rightarrow R^2$ are equal to $Z$ (mod regular homotopies). Is there an elementary way to find the fundamental group of the space of immersions ? -There are many books and papers that treat fundamental group of mapping spaces including Smale's,Michor's etc, but they are far from elementary and the audience are undergraduates. -Any idea would be much appreciated - -REPLY [3 votes]: See theorem 2.10 (with elementary proof) for the case of rotation idex $\ne 0$ of the paper: -Peter W. Michor; David Mumford: Riemannian geometries on spaces of plane curves. J. Eur. Math. Soc. (JEMS) 8 (2006), 1-48. pdf -For rotation index $=0$ (with a somewhat surprising answer) see the paper: -Hiroki Kodama, Peter W. Michor: The homotopy type of the space of degree 0 immersed curves. Revista Matemática Complutense 19 (2006), no. 1, 227-234. pdf<|endoftext|> -TITLE: Given a polynomial f, can there be more than one constant c such that every root of f(x)-c is repeated? -QUESTION [26 upvotes]: The question -Let $f$ be a nonconstant polynomial over $\mathbb{C}$. Let's say that a point $c \in \mathbb{C}$ is unusual for $f$ if every root $x$ of $f(x) - c$ is repeated. Can $f$ have more than one unusual point? -Short remarks - -There can be exactly one unusual point, e.g. if $f(x) = x^2$. There can be none, e.g. if $f(x) = x^3 + 3x$. -There are at most $\deg(f) - 1$ unusual points, since every unusual point is the image under $f$ of a critical point. -The hypotheses "nonconstant" and "over $\mathbb{C}$" could be varied. I added them to rule out cases like the following: over a field $k$ of characteristic $p$, every point of $k$ is unusual for $x^p$ (since the derivative of $x^p$ is $0$). I'd be happy to change the hypotheses to "nonconstant polynomial over an algebraically closed field of characteristic 0". Maybe something like "polynomial whose derivative is nonzero, over an algebraically closed field" would also be sensible. - -Weak reason to expect the answer to be "no" -Perhaps there's a very short answer to my question: I could be overlooking something elementary. But in case it's not so easy, I'll give a flimsy argument for why we might expect the answer to be "no" - that is, for why we might expect every polynomial to have at most one unusual point. -My question is equivalent to: is there a nonconstant polynomial $f$ over $\mathbb{C}$ for which $1$ and $-1$ are both unusual? For $1$ to be unusual means that every root of $f(x) - 1$ is also a root of $f'(x)$. Writing $d = \deg(f)$, this is equivalent to -$$ -(f(x) - 1) \mid f'(x)^d. -$$ -Similarly, for $-1$ to be unusual means that $(f(x) + 1) \mid f'(x)^d$. Both together are equivalent to -$$ -(f(x)^2 - 1) \mid f'(x)^d, -$$ -that is, -$$ -(f(x)^2 - 1)\cdot g(x) = f'(x)^d -$$ -for some $g(x) \in \mathbb{C}[x]$. This forces $\deg(g) = d(d - 3)$ (and so $d \geq 3$). -So, can we find $f$ and $g$ satisfying the last displayed equation? Comparing coefficients, what we have here is a system of $d(d - 1) = d^2 - d$ equations in $(\deg(f) + 1) + (\deg(g) + 1) = d^2 - 2d + 2$ unknowns. There are $d - 2$ more equations than unknowns, and $d \geq 3$, so a first guess is that it can't be done. - -REPLY [37 votes]: Encouraged by Pierre-Yves Gaillard here is a quick solution to the original problem. It is based on Tom Leinster's answer which in turn is an elaboration of Noam Elkies's comment to Qiaochu Yuan's answer. -Assume $a\neq b$ are unusual for $f(x)$. Then $f'(x)^2$ is divisible by $f(x)-a$ and $f(x)-b$, hence also by their product. This would show $\deg(f')\geq\deg(f)$, a contradiction.<|endoftext|> -TITLE: To what extent does Poincare duality hold on moduli stacks? -QUESTION [8 upvotes]: Poincare duality gives us, for a smooth orientable $n$-manifold, an isomorphism $H^k(M) \to H_{n-k}(M)$ given by $\gamma \mapsto \gamma \frown [M]$ where $[M]$ is the fundamental class of the manifold, and $\frown$ is the cap product. -This allows us (for example) to, given an embedding $f : X \to Y$ of manifolds to push forward integration: -$$ -\int_X f^*\omega = \int_Y \omega \smile (f_*[X])^\vee -$$ -which is a simple application of the push-pull formula: $f_*(a \frown f^*b) = (f_*a) \frown b$. -To what extent do results like this hold for stacks such as $\overline{M}_{g,n}(X,\beta)$? I know that these are not in general nice, that they can have components of a variety of dimensions, etc. They do, however, have virtual fundamental classes of pure dimension which behave in many ways like the fundamental classes that we would want. -My question is then the following: -Is there a Poincare duality for such stacks, using the virtual fundamental class? Does the push-pull formula hold in such a case? -Bonus question: What about moduli stacks with reduced obstruction theories, such as for studying Gromov-Witten invariants of K3 and Abelian surfaces? Is there any chance that it is still true there? - -REPLY [6 votes]: I know of one very limited case, which will almost certainly not satisfy you, where the "virtual cohomology" mentioned by Angelo exists, with a version of Poincare duality. -Presumably the generality one would like to address this question is for a moduli space (or stack) $M$ carrying a perfect obstruction theory (let me ignore the stackyness everywhere below). Let me restrict right away to virtual dimension 0, and in fact further to the case when the obstruction theory is symmetric (in the sense of Behrend, Behrend-Fantechi). Note that this already excludes Kontsevich moduli spaces $\bar M_{g,n}(X,\beta)$, even for $X$ a Calabi--Yau threefold, but we are still OK with DT- or other sheaf-theoretic moduli spaces. -In this case, we of course have Behrend's result that locally $M$ is cut out in a smooth ambient space $N$ by the zeros of an almost closed one-form. Now specialize further, and assume that in fact $M$ is cut out by the zeros of a closed one-form. Finally, specialize to the case when in fact globally $M=Z(df)\subset N$; here $N$ is a smooth variety, $f\colon N\to {\mathbb C}$ is a smooth function, and $M$ is cut out by the zeros of the exact one-form $df$. Note that there are still reasonably interesting examples, such as the Hilbert scheme of points on ${\mathbb C}^3$ or related toric or quivery examples. -In this case, it seems that the correct "virtual cohomology" to consider is $H^*(M,\phi_f)$, the cohomology of $M$ with coefficients in the perverse $\mathbb Q$-sheaf $\phi_f\in{\rm Perv}_{\mathbb Q}(M)$ of vanishing cycles of $f$ (appropriately shifted). This is called critical cohomology by Kontsevich-Soibelman. For example, this cohomology has the correct Euler characteristic, namely the integral of the Behrend function on $M$. It also carries a mixed Hodge structure, and the Euler characteristic of the weight filtration (which will differ from the degree filtration because the Hodge structure is not pure) gives what seems to be the right "quantization" (or "refinement") of the numerical invariant. -Now the point is that by standard theory, $D_M\phi_f\cong\phi_f$, where $D_M$ is the Verdier duality functor on $M$. This will give you a kind of Poincare duality -$$H^n(M,\phi_f)\cong H^{-n}_c(M,\phi_f)$$ -albeit between ordinary and compact-support cohomology, since $M$ is of course almost always noncompact. But at least (using the appropriate shifts) it is elegantly symmetric around zero, which is what we would expect in the virtual dimension 0 case. -I don't really know how well this generalizes; almost nothing is known (to me) about gluing, functoriality, etc.<|endoftext|> -TITLE: How are these number-theoretical constants actually distributed? -QUESTION [5 upvotes]: I'm very curious about this and would be really grateful for any help or comments in this direction. If we consider any of the following number-theoretical constants: -1)The various singular series arising from any given system $\Psi: \mathbb{Z}^{d}\rightarrow \mathbb{Z}^t$,$d,t \geq 1$ in the Green-Tao paper "Linear Equations in Primes" (http://arxiv.org/PS_cache/math/pdf/0606/0606088v2.pdf): -$$ -\prod_{p}\beta_p. -$$ -2)The Hardy-Littlewood/Bateman-Horn constants arising in the Hardy-Littlewood Conjecture F/ general Bateman-Horn conjecture, say for $f\in \mathbb{Z}[x]$ irreducible and $n_{p,f}$ the number of solutions to $f(n) \equiv 0 \bmod p$ in $\mathbb{Z}/p\mathbb{Z}$, the constant -$$ -C(f) = \prod_{p}\left(\frac{p-n_{p,f}}{p-1}\right), -$$ -and also the case (also covered by Bateman-Horn) where there is more than one polynomial (thus including the Hardy-Littlewood $k$-tuple conjecture) -3)The analogous constants that would arise naturally from maybe even a combination of (1) and (2) (Is this possible? I think the last slide of http://www.dpmms.cam.ac.uk/~bjg23/papers/icm-handout.pdf already talks about this.), so that one asks for systems $\Psi:\mathbb{Z}^d \rightarrow \mathbb{Z}^t$, $d,t\geq 1$, $\Psi = (\psi_1,\dots,\psi_t)$, where this time the $\psi_i$ need not be linear(!) (But here I'm only asking about the constants, not the actual conjecture.) -The question is, is there an easy way to see whether these constants are dense on any part of the real line or not? If I give you a fixed positive real number, and say you can pick any system from (3) that you want, or even something more general than (3), how would you go about systematically picking the system so that its constant is $\epsilon$-close to the given real number? Thank you very much. - -REPLY [2 votes]: This paper by Kowalski also looks highly relevant: -http://arxiv.org/abs/0805.4682?context=math<|endoftext|> -TITLE: Up-to-date version of Principia Mathematica? -QUESTION [11 upvotes]: Background: I found this interesting translation of Godel's On formally undecidable propositions of Principia Mathematica and related systems I that, along with translating it into English, uses more modern and understandable symbols, includes some hyper text links throughout the paper, and, if I'm not mistaken, makes the language a little more readable. This seems like a good and noble undertaking: taking an important paper and revamping it to make it more accessible, or at least less intimidating for modern readers. -So while looking through it, I couldn't help but think of another work that could benefit from such a treatment, Russell and Whitehead's Principia Mathematica. Every review I have read about it is something like "It lead to a lot of important developments in logic, but is mostly a historical curiosity. And besides, it is incredibly difficult to read, so putting forth the necessary effort is probably not worth it." -Here is the question: does such a thing exist, and would such a paper be worthwhile? Is there anything in the Principia that would be of interest to modern mathematicians. (Finding some points of interest and reproducing them seems like a better idea than trying to reproduce all three volumes.) Godel's theorem and his proof are still interesting, in spite of the fact that there are many other (better?) ways of getting to the Incompleteness theorems. -Another example of this sort of revision is The Annotated Turing. -I'd appreciate any knowledge of such a source, or any insight into why one wouldn't be too valuable, or really any thoughts on looking at Principia Mathematica at all. - -REPLY [12 votes]: In his Stanford Encyclopedia of Philosophy article "The Notation of Principia Mathematica", Bernard Linsky makes the following claim: - -This translation is offered as an aid to learning the original notation, which itself is a subject of scholarly dispute, and embodies substantive logical doctrines so that it cannot simply be replaced by contemporary symbolism. Learning the notation, then, is a first step to learning the distinctive logical doctrines of Principia Mathematica. - -The point is that Principia was not intended simply to be a development of mathematics in type theory: it was intended to make a philosophical argument that mathematics could be carried out using only "logic". Thus translating PM so that the underlying mathematical principles are more clearly described would miss the point that there are not supposed to be any underlying mathematical principles, only "logical" ones. This differs sharply from Gödel's paper, which was intended to be mathematical (the result can be viewed as just a particular type of combinatorics or number theory) rather than philosophical.<|endoftext|> -TITLE: Nonseparable disintegration theory: references -QUESTION [20 upvotes]: I mean a theorem of the following kind. Let $A$ be a C*-algebra, and let $\pi: A\to B(H)$ be its representation. Then there exist a set $P$ with a positive measure $\mu$, a field of Hilbert spaces such that $H\simeq \int_P H_p d\mu(p)$, and irreducible representations $\pi_p: A\to B(H_p)$ such that $\pi=\int_P \pi_p d\mu(p)$. -In classical references (Dixmier/Takesaki/Kadison...) both $A$ and $H$ are assumed to be separable. Is there a canonical reference for the nonseparable case? -I have found two articles, not counting particular cases: S. Teleman On reduction theory. {\it Rev. Roumaine Math. Pures Appl.} {\bf 21}, no.~4 (1976), 465--486. and R. Henrichs Decomposition of invariant states and nonseparable C*-algebras. Publ. Res. Inst. Math. Sci. 18, 159-181 (1982). Both use definition of fields of Hilbert spaces given by W. Wils in Direct integrals of Hilbert spaces I. {\it Math. Scand.} {\bf 26} (1970), 73--88. -Both prove the theorem above (Henrichs for the unital case), with one main difference: in Teleman's version, $P$ is a subset of pure states of $A$, but $\mu$ may not be regular (not every set is approximated by compacts from inside). In Henrichs', $\mu$ is regular but one and the same irrep can repeat, even for every $p$. -In the history of this question there were lots or erroneous articles, so I treat these two also with caution. I've gone through Teleman's proof(because it is self-contained). It seems correct, but it turns out that $\pi_p$ may be zero, and this is not indicated in the paper. Through Henrichs I didn't go in detail. He relies on a rarely used theorem of Tomita, for which he however gives an independent proof. -So this is my question: do you use this theory, and if yes, what authors do you refer to? - -REPLY [4 votes]: A related problem (if you look at it from the viewpoint of disintegration of states in C*-algebras) is the problem of disintegration of measures in a nonseparable setting. There is a recent paper which revisits this old problem by M. Kosiek and K. Rudol, "Fibers of the $L^\infty$ Algebra and Disintegration of Measures". Archiv der Mathematik 97 (2011) 559-567. Supposing it is also correct, it may help...<|endoftext|> -TITLE: Is this a subcase of the fundamental lemma? -QUESTION [9 upvotes]: Let $F$ be a local field and $G= GL(n,F)$. -Assume that $\gamma$ is an element of $G$ and $G_\gamma$ is its centralizer. -The orbital integral is defined as -$$ O_\gamma^G( \phi) = \int\limits_{G_\gamma \backslash G} \phi( g^{-1} \gamma g) d g.$$ -We can assume wlog that $\gamma$ is elliptic. Can we lift the elliptic orbital integral on $G$ to an elliptic orbital integral on $G' =GL(n,F')$, where $F' = F[X] / det(X-\gamma)$? -More precisely, can we find an $T:C_c^{\infty}(G) \rightarrow C_c^\infty(G')$ such that -$$ O_\gamma^G(\phi) = O_\gamma^{G'} ( T\phi).$$ -I think for explicit computations, it is much more convenient to work on $G'$, where $\gamma$ is diagonalizable nad $G'_\gamma$ can be chosen to be the diagonal matrices. -What is the Reference? What is the Buzzword for this? - -REPLY [15 votes]: I will offer some words on this, but only because no-one else has; I was holding out hoping that one of the more automorphic people would chip in. It might be worth taking much of the below with a pinch of salt. -So I've been trying to penetrate the "fundamental lemma" literature myself, and let me begin by showing my hand and saying that my current impression, that may be wrong, is that "the fundamental lemma" is not a well-defined mathematical statement, it is a principle that applies in many situations, and I think that mathematics is open to the situation where there will be precise mathematical statements formulated in the future, when people are studying situation $X$, and that people will call these statements "the fundamental lemma for $X$". For example, I think that perhaps in 50 years' time when people are proving general base change (rather than cyclic base change) there may be a "general base change fundamental lemma". -So this of course immediately raises the question -- what did Ngo Bao Chau do? Well, my perception is that he has proved the version of the fundamental lemma that shows up in the theory of endoscopy. -As I've already tried to make clear, I am not 100% sure on all of this. But let me muddle on regardless. My impression is that what the game looks like is this. There is the general functoriality principle, which says that if I have two connected reductive groups $G$ and $H$ over a global field, and a map between the $L$-group ${}^LH\to{}^LG$, then there should be some sort of way of transferring automorphic representations on $H$ to automorphic representations on $G$. Already this is a "principle" rather than a precisely-formulated statement, because I think there are lots of issues with packets and multiplicities, that certainly I don't understand, when trying to make a precise statement, and I am not sure I've seen a precise statement in the literature that ties up all the loose ends that I want to see tied up, other than some very weak ones that just demand compatibility at the unramified places and don't care about multiplicities at all. -But it turns out, when attempting to e.g. study the zeta functions of some Shimura varieties, that Langlands needed some very special cases of the functoriality principle, where $H$ was an endoscopic subgroup for $G$, and here he postulated a strategy for proving functoriality using the trace formula. The idea is that you try and stabilise the trace formula for $G$, whatever that means, and this involves, amongst other things, figuring out some way of transferring functions locally and matching the orbital integrals that come up. The upshot is that you relate the trace formula for $G$ to the trace formula for all the endoscopic subgroups for $G$. -So my perception is that, in its initial state, the situation was this: $G$ was a connected reductive group, now over a local field, $H$ an endoscopic subgroup (and part of the data of this endoscopic situation is that you're given a map ${}^LH\to{}^LG$) and there is supposed to be a "transfer" map, that maps functions on $G$ to functions on $H$. Even the existence of the transfer map is not at all clear in general, I don't think. For example Labesse-Langlands have to do some calculations (only a few pages, but some work) to prove that one can transfer functions from $SL(2)$ to a subtorus (this is the simplest example of endoscopy, I think). -So my impression is that the general notion of moving from functions on one group, locally, to functions on another, is called "transfer of functions". My understanding is that the transfer map is not at all well-defined, that sometimes one can characterise the image (as being functions whose orbital integrals vanish on some certain subgroup), and that at the end of the day the precise relationship you want between the function and its transfer can be quite complicated. I think that in general you want the orbital integral of one function to be the orbital integral of the other multiplied by a "fudge factor" whose definition is the key point of a 100-page paper by Langlands and Shelstad. One can already see these fudge factors in the $SL(2)$ case with Langlands-Labesse. -My understanding of what the fundamental lemma is, is the following: in the situation where $G$ and $H$ are unramified, one extra condition you could put on the "transfer of functions" map is that the identity element for the unramified Hecke algebra for $G$ gets sent to the identity element for the unramified Hecke algebra for $H$. Hence in this situation, "the fundamental Lemma", I think, boils down to the assertion that a certain volume equals a fudge factor that it takes 100 pages to define, multiplied by another volume. -I'm slowly getting to the point :-) I think I can answer one of your questions at least -- the "buzzword" you're looking for is not "fundamental lemma" but "transfer of functions" or perhaps "local transfer of functions". I think. -However, my understanding is that the situation you are looking at is not an "endoscopic situation". In particular I don't think that reading the complete works of Ngo Bao Chau will give you an answer to your question. You have groups $G$ and $G'$ and they're both $GL(n)$ but over different fields, so if I were trying to prove a hard global theorem and I needed the type of transfer that you're looking for, I would probably not be trying to stabilise the trace formula (indeed I think the trace formula for $GL(n)$ is already stable and that "$GL(n)$ has no endoscopy" in some sense) -- I would probably be trying to prove base change. -Now here are, for me, some BIG problems I would fear when attempting to try and get an answer to your question from the literature that I know about. -The first is that, when trying to prove global base change for $GL(n)$ for a global cyclic extension $L/K$, I attempt to find a relation between the trace formulas for $GL(n)$ over $L$ and over $K$ and then I attempt to start matching up terms etc etc, and the problem is that one trace formula is a sum over conj classes in $GL(n,L)$ and the other is a sum over $GL(n,K)$. As far as I know, people don't know how to relate these two sets in a natural way. So what they do is they relate conjugacy classes in $GL(n,K)$ to twisted conjugacy classes in $GL(n,L)$. I think the local story looks like this: say $E/F$ is now local and has cyclic Galois group. Given $\gamma$ in $GL(n,E)$ they take its "norm" in the most naive way (multiply it by its Galois conjugates) and get an element of $GL(n,F)$ and in this nice cyclic situation they can inject twisted conj classes ($x\sim \sigma(g)xg^{-1}$ with $\sigma$ a generator of the Galois group) in $GL(n,E)$ with conj classes in $GL(n,F)$. Using this trick they want to relate the usual trace formula for $GL(n,K)$ with a twisted trace formula for $GL(n,L)$, and to do this the transfer of functions they require is a twisted transfer! In particular, they need a machine which, given a function $a$ on $GL(n,E)$ spits out a function $b$ on $GL(n,F)$ such that the orbital integrals of $b$ equal certain twisted orbital integrals for $a$, possibly again multiplied by some fudge factors, but I believe that in this base change situation the fudge factors (which are I think not covered by the Langlands-Shelstad monster because this is not an endoscopic situation) are all 1 anyway. -The upshot is that in the Arthur-Clozel book, proving cyclic base change for $GL(n)$, you will see a definition of transfer of functions, which looks formally a bit similar to the thing you write above, but there are certain crucial differences: -1) if $a$ transfers to $b$ then an orbital integral for $b$ will equal a twisted orbital integral for $a$, rather than an orbital integral. -2) The twisted orbital integral for $a$ will be attached to an element $\gamma$ and the orbital integral for $b$ that it equals will be attached not to $\gamma$ but to its naive norm (which is a well-defined conj class, I believe, in $GL(n,F)$) -3) $E/F$ will ALWAYS be cyclic (or perhaps more generallu they will allow $E=F^n$ so $E$ is not actually a field but it's still etale over $F$). -Now I look at the question you're asking, and there are of course two ways of approaching it: the first is to try and get your hands dirty and write down the map between functions yourself. But it sounds to me like you're hoping that you can take another approach -- to get what you want from the literature. And what I am really scared by is that although what you write looks to me superficially like transfer of functions, you are in a situation where $F'$ is not in general a cyclic Galois extension of $F$, so you can throw away Arthur-Clozel, and you are not I think in an endoscopic situation either, so you can also throw away Ngo Bao Chau, and unfortunately I personally do not know what is left. Of course this will largely be my own ignorance and probably there are people who have thought about "transferring" in its own right, independent of links to functoriality. But I am now not sure where to point you. -Aah, the joyful tones of my daughter, who has apparently just woken up. What timing she has! I've gotta go, but I've said all I can say anyway.<|endoftext|> -TITLE: Does there exist a finite-index subgroup of SL2Z with all cusps irregular? -QUESTION [10 upvotes]: Recall that if $\Gamma$ is a finite-index subgroup of $\operatorname{SL}_2(\mathbf{Z})$, then a cusp of $\Gamma$ is an orbit of $\Gamma$ on the set $\mathbf{P}^1_{\mathbf{Q}}$. If $-1\notin \Gamma$, then for any cusp $c$, the stabilizer of $c$ in $\Gamma$ (well-defined up to conjugacy) is an infinite cyclic group generated by an element conjugate in $\operatorname{SL}_2(\mathbf{Z})$ to either $\begin{pmatrix} 1 & h \\ 0 & 1\end{pmatrix}$ or $\begin{pmatrix} -1 & h \\ 0 & -1 \end{pmatrix}$ for some $h \in \mathbf{N}$ (the width of the cusp $c$), and we say that $c$ is regular in the first case and irregular in the second. -My question, as in the title, is: - -Does there exist a finite-index subgroup $\Gamma \le \operatorname{SL}_2(\mathbf{Z})$ for which every cusp is irregular? - -(It would suffice to find a finite-index normal subgroup with at least one irregular cusp, since the cusps of a normal subgroup are either all regular or all irregular; but I don't know if that helps!) - -REPLY [4 votes]: Just to record what I mentioned in the comments, you may argue as follows: -Let $X$ be the orbifold $\mathrm{SL}_2(\mathbb{Z})\backslash \mathbb{H}^2$. The curve $X$ has two singular points, one a cone point $x$ of order $2$ (the image of the fixed point of an elliptic of order two), and a cone point $y$ of order $3$ (the image of the fixed point of an elliptic of order three). -There's a $6$--fold branched cover $T$ of $X$ which is smooth and diffeomorphic to a punctured torus. (First take a $3$--fold cover $X_3$ of $X$ branched over $x$. Then take the $2$--fold cover of $X_3$ branched over the three preimages of $y$ to obtain $T$.) -If $\Gamma$ is the subgroup of $\mathrm{SL}_2(\mathbb{Z})$ corresponding to $T$ (of index twelve to avoid the center), then we may write $\Gamma = \langle a , b \rangle$ so that the parabolic conjugacy class is represented by the commutator $[a,b]$, and a simple computation reveals that $\mathrm{trace}([a,b]) = - 2$. (As mentioned in my comment, you don't need to compute $a$ and $b$ to compute their trace.)<|endoftext|> -TITLE: The Lagrangian formulation of mechanics without going through variational principles. -QUESTION [5 upvotes]: In some texts on classical mechanics and not only, the Euler--Lagrange equations of motion are directly obtained as solution of variational problems. -On the other side, sometimes reading about hamiltonian mechanics, one find the expression that this latter formulation is preferred to the lagrangian one because of it does completely avoid the appeal to variational principles. -This observation suggested to myself the following question: -Is the variational approach to the Euler--Lagrange equations the only one viable? -If not, is there some reason that explain why the geometry of the Euler-Lagrange eqns is much more hidden than the geometry of the Hamilton eqns? - -I was searching for suggestion of reading for best tackle this question. - -As usual any feedback is welcome. - -REPLY [3 votes]: I'm just going to make a basic point, so apologies if this is completely obvious to you (it's also contained in Igor Khavkine's much more thorough answer). -Let $Q$ be the configuration manifold of the system. The corresponding cotangent bundle $T^*Q$ has an intrinsic symplectic form $\omega=-d\Theta$, where $\Theta$ is the tautalogical one-form on $T^*Q$. For a Hamiltonian $H:T^*Q\rightarrow \mathbb{R}$, Hamilton's equations can be expressed in terms of the Hamiltonian vector field $X_H$ (defined by $i_{X_H}\omega=dH$). Note $\omega$ is intrinsic to the phase space $T^*Q$, and doesn't depend on the Hamiltonian $H$. -Now given a Lagrangian $L:TQ\rightarrow\mathbb{R}$, and corresponding Legendre transform $\mathbb{F}L:TQ\rightarrow T^*Q$, and assuming here for simplicity that $\mathbb{F}L$ is a diffeomorphism ("L is hyperregular"), one can use $\mathbb{F}L$ to pull everything back to $TQ$. $TQ$ becomes a symplectic manifold, with symplectic form $\omega_L=(\mathbb{F}L)^*\omega$, and the Euler-Lagrange equations are just the equations for the flow of the Hamiltonian vector field $X_E$ defined by $i_{X_E}\omega_L = dE$, where $E=(\mathbb{F}L)^*H = H\circ\mathbb{F}L$ is the energy function on $TQ$. I guess the main point though is that $TQ$ is not intrinsically a symplectic manifold. The symplectic form $\omega_L$ depends also on the choice of Lagrangian. This is one possible answer to why the geometric formulation is more common on the Hamiltonian side, and appears to be `hidden' on the Lagrangian side: the geometry of $TQ$ (as it pertains to the E-L eqns) is tied up with the particular Lagrangian $L$, whereas the geometry of $T^*Q$ is independent of the particular Hamiltonian $H$. -I'd also recommend any of the books by Jerry Marsden mentioned in Spiro Karigiannis' answer as the best place to learn this (my notation is consisent with his). -Edit: I should clarify that by `Legendre transform' above I mean (in coordinates) the map $p_i(x, \dot{x}) = \frac{\partial L}{\partial \dot{x}^i}(x, \dot{x})$. This is standard in the literature, but it differs from the classical meaning $H(x, p) = p\dot{x}-L(x, \dot{x})$ (where $p_i = \frac{\partial L}{\partial \dot{x}^i}$).<|endoftext|> -TITLE: Restriction of characters of hyperoctahedral groups. -QUESTION [9 upvotes]: The hyperoctahedral group $H_n$ has several descriptions; as a wreath product; as signed permutation matrices; as the Weyl group of type $B_n$ or $C_n$. In all these descriptions it is apparent that the symmetric group $S_n$ is a subgroup. -I would like to know the Frobenius characters of the restrictions of the irreducible -characters of $H_n$ to $S_n$. I imagine this is known but not to me. -The irreducible characters of $H_n$ are indexed by pairs of partitions $\alpha$,$\beta$ -such that the total number of boxes is $n$. There is a well developed combinatorial theory involving bitableaux as well as an analogue of the Robinson-Schensted correspondence. The character theory is described in I. G. MacDonald "Symmetric functions and Hall polynomials" -Chapter I, Appendix B in terms of symmetric functions. -I am also aware of the question 48532 which asks about the restriction map for the inclusion of $H_n$ in $S_{2n}$. - -REPLY [4 votes]: Even though you found the answer, I'd like to point to the following reference, where I learned this result, at a time when the book by Macdonald (note the captitalization) did not yet have an appendix B to chapter I (i.e., before its second edition): -A. V. Zelevinsky, Representations of Finite Classical Groups A Hopf algebra approach; Lecture Notes in Mathematics 869 (1981). -The answer to your question is contained in the Proposition in section 7.10 (page 105), which says (somewhat indirectly) that more generally for the wreath product of $S_n$ and a finite abelian group $G$, the irreducible representations correspond to families of partitions indexed by the irreducible representations of $G$ and of total size $n$, and that the restriction to $S_n$ of such a representation is obtained by multiplying together, in the combined Grothendieck ring $R(S)$ of the symmetric groups, the irreducible representations associated to those partitions. The multiplication in $R(S)$ is defined by induction from a product of small symmetric groups into the containing symmetric group $S_n$, and so a the level of Frobenius characters becomes multiplication of Schur functions.<|endoftext|> -TITLE: Is the Axiom of Union independent of the rest of ZF? -QUESTION [12 upvotes]: Short version: Is the axiom of union independent of the rest of axioms of ZF? -NO) Tourlakis (2003) says in p. 177 that the axiom of union can be derived from the rest of ZF if an appropriate version of collection axiom[*] is chosen. The quote is: -«Bourbaki (1966b) adopts the axiom of pairing, but adopts collection version (2), and proves both separation and union» -YES) In the other hand, I have read here and here something like "$H_{\kappa}$ is a model for ZF-Union+¬Union", where $\kappa$ was $\beth_\omega$ or a singular cardinal. -Any reference on the subject would be highly appreciated. I apologize in advance if the question is too basic (not a mathematician!). Also, I have googled it and followed some false trails before asking here. Thanks. -[*] The appropriate version of collection is apparently weaker (or equivalent at most) than the collection axiom that he is adopting in his text. I think the statement is: -$(∀x)(∃z)(∀y)({\mathcal P} [x, y] → y ∈ z) → (\forall A)(\exists B)(\forall y)(y\in B\leftrightarrow (\exists x\in A){\mathcal P}[x,y])$ -I have translated the notation from III.8.12 and III.2 (obviating any reference to ur-elements). -EDIT: Thank you very much for the answers, they were really helpful. - -REPLY [6 votes]: Let me add a nice remark I just became aware of: In - -Greg Oman. On the axiom of union, Arch. Math. Logic, 49 (3), (2010), 283–289. MR2609983 (2011g:03122), - -Oman clarifies precisely which unions can be proved to exist in $\mathsf{ZFC}-\mathrm{Union}$: $\bigcup x$ exists iff $\{|y|\colon y\in x\}$ is bounded. In particular, $A\cup B$ exists for any sets $A,B$. -See this MSE question for details. The proof uses in essential ways both choice and replacement. -(Curiously, I do not know of an original reference for the fact that $\mathsf{ZFC}-\mathrm{Union}$ does not suffice to prove the existence of infinite unions, the usual argument being the one in the body of the question, and in Andreas's answer. It would be nice to have the reference, so it can be added here.)<|endoftext|> -TITLE: Alexander John Thompson - Logarithmetica Britannica -QUESTION [5 upvotes]: Alexander John Thompson was the author/computer the nine-volume Logarithmetica Britannica published between 1924 and 1952. He was born in Plaistow, Essex, England, in 1885. He was still a member of the Mathematical Tables Committee of the BAAS in 1965. -Does anybody have a date of death? -Thanks for any insight. -Cheers, Scott - -REPLY [13 votes]: I accessed a copy of the 1968 official full death certificate and the occupation of this Alexander John Thompson is stated as retired 'Statistical Officer' at the 'General Register Office' which was his occupation according to the LOCOMAT article referenced. So this confirms the death details as follows. -June 17th 1968 at 36 Link Lane, Wallington, (in county of Surrey at time of death, now in London Borough of Sutton). -regards -David -A note on the source as requested. -In the UK to obtain a death certificate copy (or birth, marriage) an application with payment of £9.25 has to be made to the UK General Register Office (GRO) online or local register office. -https://www.gov.uk/order-copy-birth-death-marriage-certificate -Ancestry sites like www.findmypast.co.uk with a paid for small credit give the registry index details and these details are used to order. -Because of the cost(!) it’s a service I only use very occasionally but does potentially provide a firm answer to a query that the ancestry sites may not unambiguously resolve. -I can send a certificate scan via email for personal use to anyone if interested which has more info’. -I think the sort of citation for a formal reference would be: -Alexander John Thompson, certified copy of death certificate (age at death 83), General Register Office, UK. Entry – Sutton, Surrey, 1968, Apr-Jun, vol. 05E, p.192.<|endoftext|> -TITLE: What are the best known results for the stable homotopy groups of spheres? -QUESTION [17 upvotes]: There are a number of proposed ways to compute the stable homotopy groups of spheres. One can rather peculiarly consider stable (co)homotopy of an Eilenberg Maclane spectrum as a generalised (co)homology theory and use the Atiyah–Hirzebruch spectral sequence (in the same way one sometimes uses the Serre spectral sequence knowing information about the $E_{\infty}$ page to deduce it about the $E_{2}$ page). Another approach is to use the Adams spectral sequence. Here one plays off the rigidity of the cohomology of generalised Eilenberg Maclane spectra against this failing in general. This leads to a spectral sequence which converges to the p-part (where $p$ refers to taking cohomology in $\mathbb{Z}/p\mathbb{Z}$) of the stable homotopy group of spheres. A variant is to do this with some (nice enough I guess) generalised cohomology theory which leads to the Adams–Novikov spectral sequence. I think there are probably quite a few other methods which are used to calculate these. My questions are: - -What are the best results on this? I see here it says that the best known result as of 2007 was up to the 64th stem. -Which method gives the best known results? -What stops us here? Do we simply not yet know the differentials around 64 in the Adams spectral sequence? - -REPLY [8 votes]: You could look at this talk: http://neil-strickland.staff.shef.ac.uk/talks/oslo_talk.pdf -About the first third of the talk is a map of various different approaches to the stable and unstable homotopy groups of spheres. However, several approaches are missing: - -Kochman's approach via the AHSS for $\pi_*(BP)$, as in Christian Nassau's answer. -The similar approach that you mentioned, via the AHSS for $\pi_*(H/p)$ -The superstable EHPSS, which is essentially the AHSS for the pro-spectrum $\mathbb{R}P_{-\infty}^\infty$ together with the theorem that that pro-spectrum is $2$-adically equivalent to $S^{-1}$. (There is also an analog at odd primes using $B\Sigma_p$ in place of $\mathbb{R}P^\infty$.) This should enable one to build a bridge between Kochman-style calculations and the EHP sequence and the Goodwillie tower, but I'm not aware that anyone has worked that out. - -(I should mention that Mark Behrens has independently worked out most of the ideas mentioned in the remainder of the talk that I linked to, with much more success than I had.)<|endoftext|> -TITLE: Cardinality of cofinal set of normal functions $f \colon \omega_1 \to \omega_1$ -QUESTION [5 upvotes]: What is the cardinality of the set $F$ of all normal functions $f \colon \omega_1 \to \omega_1$, where $\omega_1$ is the first uncountable ordinal? -What is the least cardinality of a subset of $F$ such that every function in $F$ is bounded by some element of the subset? - -REPLY [9 votes]: For both questions, the answer does not change when you remove the word "normal" from the question. -For the first question: There is a 1-1 map $f\mapsto N(f)$ that assigns to each function $f$ a normal function $N(f)$. $N(f)(\alpha)$ just adds up all values of $f$ below $\alpha$. (Or better: of $f+1$, to make it strictly increasing.) So there are $2^{\aleph_1}$ many of them. -For the second question: Let $\mathfrak d(\kappa)$ be the smallest number functions needed to dominate all functions from $\kappa$ to $\kappa$. - -James Cummings and Saharon Shelah, Cardinal invariants above the continuum, Ann. Pure Appl. Logic 75 (1995), no. 3, 251–268, https://doi.org/10.1016/0168-0072(95)00003-Y, arXiv:math/9509228, MR1355135 - -shows that, just like the continuum functions $\kappa\mapsto 2^\kappa$, also the "dominating" function $\kappa\mapsto \mathfrak d(\lambda)$ can have quite arbitrary behaviour. In particular, both $\mathfrak d(\aleph_1)=2^{\aleph_1}$ and $\mathfrak d(\aleph_1)< 2^{\aleph_1}$ are consistent. -(This specific result for $\aleph_1$ may be older, though.)<|endoftext|> -TITLE: What is a twisted modular operad? -QUESTION [10 upvotes]: I find Getzler and Kapranov's article Modular Operads difficult to understand. Can anyone explain what a (twisted) modular operad is conceptually, or what the underlying idea behind the concept of a modular operad is? For example, how can we write down the explicit structure map in a twisted modular operad? - -REPLY [6 votes]: Here is a long and belated answer, mostly written so that I can work out some details I should have worked out long ago. The short version of it is the following: when $g > 0$, one needs the notion of a twisted modular operad in order to describe the precise sense in which the "gravity operad" is an operad. I do hope all signs below are correct. -The "standard" motivation for introducing twisting is that it is needed for the higher genus generalization of the bar transform (the "Feynman transform"): twisted modular operads appear naturally as "Feynman duals" of ordinary operads. The following is a different perspective. -Remark: Since the gravity operad was introduced by Getzler, who also proved it dual to the hypercommutative operad, and since the gravity operad is the only twisted modular operad discussed in any detail in Getzler-Kapranov's paper, it seems plausible that this was one of their main motivating example for introducing the notion of twisting. -The starting point is the situation in genus zero, and Getzler's paper "Operads and moduli of genus 0 Riemann surfaces". We consider the collection of spaces $\newcommand{\M}{\overline{M}} \M_{0,n}$, which form a cyclic operad in the category of algebraic varieties. Taking cohomology we get a cyclic co-operad $H^\bullet(\M_{0,n})$. So far nothing surprising. The surprise is that even though the spaces $M_{0,n}$ do not form an operad in any natural sense, their cohomologies still do - this is the "gravity" operad. Moreover, the two cohomology operads are dual in a precise sense (which is purely a genus zero phenomenon), but I will not talk about that. -The magic words that make the cohomology of $M_{0,n}$ an operad are "Poincaré residue". For this read Deligne, "Hodge II", pp 31-32, or Peters-Steenbrink, pp 92-93. The short version is that the Poincaré residue is defined whenever you have an open subvariety $U \subset X$ where $X$ is smooth and the complement is a simple normal crossing divisor. Let the divisor be $D_1 \cap \cdots \cap D_N$, let $D_I = \bigcap_{i\in I} D_i$, and let $D_I^\circ$ denote the interior of the intersection (the complement in $D_I$ of all other components $D_j$). Then it is a map $H^\bullet(U) \to H^{\bullet-|I|}(D_I^\circ)$. Apply this to $X = \M_{0,n}$ and $U = M_{0,n}$: the set of possible intersections $D_I$ is exactly the set of all stable trees with $n$ legs, and $D_I^\circ$ is the corresponding product of open moduli spaces $M_0(\Gamma) = \prod_{v \in \mathrm{Vert}(\Gamma)} M_{0,n(v)}$ (where $n(v)$ = valence of the vertex). -Hence we have co-composition maps $H^{\bullet+|\mathrm{Edge}(\Gamma)|}(M_{0,n}) \to H^\bullet(M_{0}(\Gamma))$, in particular -$$ H^{\bullet+1}(M_{0,n+n'}) \to H^\bullet(M_{0,n+1}) \otimes H^\bullet(M_{0,n'+1}),$$ -and we would like to say that this makes the cohomology of $M_{0,n}$ a co-operad. There is one obvious problem here, which is that there is a degree shift in the definition. This is not such a big deal: we can get rid of it by declaring instead that the collection -$$ \{ H^{\bullet-1}(M_{0,n})\}$$ -should form a co-operad of graded vector spaces. -But even this does not give us a co-operad. The issue is in the very definition of Poincaré residue: it is only defined up to an ordering of the boundary divisors, which potentially introduces a sign ambiguity. This is dealt with explicitly in Deligne who twists everything by a kind of orientation sheaf $\varepsilon^n$ to get things defined canonically. However, there is a simple operadic solution also to this problem: the correct statement is instead that the collection -$$ \{ H^{\bullet-1}(M_{0,n}) \otimes \mathrm{sgn}_n\}$$ -does form a cyclic co-operad of graded vector spaces. Here $\mathrm{sgn}_n$ is the sign representation of the symmetric group. -So far so good. Now we wish to generalize this to higher genera. Again the spaces $\M_{g,n}$ form a modular operad, and their cohomologies $H^\bullet(\M_{g,n})$ a co-operad. We would like to play the same game again to get a co-operad structure on the cohomology of $M_{g,n}$. The self-glung gives us maps -$$ H^{\bullet+1}(M_{g+1,n-2}) \to H^{\bullet}(M_{g,n})$$ -that we would like to use to define the modular co-operad structure. Hence we are led to introducing a second degree shift, now depending on the genus: one might hope that the spaces -$$ \{ H^{\bullet+g-1}(M_{g,n}) \otimes \mathrm{sgn}_n\}$$ -will form a suitable modular co-operad. This is unfortunately not true! The self-gluing does not work with this definition, as the $\mathbb{S}_2$-action on the two points that get identified is not the right one. In fact no strategy like the one used in genus zero will work in this situation. -What one should then do is to define one's way out of the situation. Instead of an "ordinary" operad $\newcommand{\P}{\mathcal{P}}\P$ with structure maps -$$ \P(\Gamma) \to \P(n)$$ -where $\Gamma$ is a rooted tree with $n$ inputs, or an "ordinary" modular operad with structure maps $\P(\Gamma) \to \P(g,n)$ where $\Gamma$ now has genus $g$ and $n$ legs, one introduces a "twisting" $\newcommand{\D}{\mathfrak{D}}\D$ such that there are maps -$$ \D(\Gamma) \otimes \P(\Gamma) \to \P(g,n).$$ -Then $\D(\Gamma)$ is required to depend functorially on $\Gamma$ in certain ways making it meaningful to talk about associativity and equivariance conditions. One has defined the notion of a twisted modular operad. -The game we played in genus zero can now be re-interpreted: instead of saying that $ \{ H^{\bullet-1}(M_{0,n}) \otimes \mathrm{sgn}_n\}$ is a cyclic co-operad, we can say that the usual cohomology groups $ \{ H^{\bullet}(M_{0,n})\}$ form a cyclic $\D$-co-operad, where $\D(\Gamma)$ is given by a suspension for each edge on the graph, and tensoring with the sign representation of the symmetric group acting on the set of edges: in other words, $\D(\Gamma) = \mathrm{Det}(\mathrm{Edge}(\Gamma))^{-1}$, using the terminology introduced in Getzler-Kapranov's paper. This more abstract (but in a sense much more natural) definition now works without any changes also in higher genus!! -In this framework one can also give a nice explanation of why these suspensions and sign changes worked in genus zero. The correct cocycle to twist with was $\D(\Gamma)$ defined in the preceding paragraph. But $\D$ is cohomologous, in an appropriate sense, to the cocycle $D(\Gamma) = \mathrm{Det}(H_1(\Gamma))^{-1}$, which is trivial when restricted to trees. In fact $\D$ and $D$ differ by a coboundary, and this coboundary is precisely given by putting a suspended copy of the sign representation in each spot $(g,n)$. So we have recovered exactly the recipe for making $H^\bullet(M_{0,n})$ into a co-operad that we wrote down in an ad hoc way above.<|endoftext|> -TITLE: Does -I belong to Weyl group? -QUESTION [6 upvotes]: Let $\Phi$ be an irreducible root system, with positive roots $\Phi^+$ relative to the base $\Delta$. -If $W$ is the Weyl group, how can I determine if $-I$ belongs to $W$? Equivalently how can I see if the (unique) longest element in $W$ is $-I$? - -REPLY [7 votes]: The mnemonic I use: if the diagram has a natural involution, then $-w_0$ induces it, otherwise $w_0 = -1$. The only place this fails is in $D_n$, where one can switch the antlers, but shouldn't always, so how to remember when? In $D_4$, there's no natural involution, so that helps me remember that in $D_{2n}$, the involution is trivial.<|endoftext|> -TITLE: Ring of algebraic integers in a quadratic extension of a cyclotomic field -QUESTION [8 upvotes]: Hello, -I have a question which arose when trying to classify orders of certain algebras. -We know that if $K=\mathbb{Q}(\zeta)$ is any cyclotomic field, and $\zeta$ is an $n$-th root of unity (for some number $n$), then the ring of algebraic integers in $K$ is exactly $\mathcal{O}_K=\mathbb{Z}[\zeta]$. -Consider now the following quadratic extension $L=K(t)$ where $t$ satisfies the equation $$t^2 = \omega(1-\xi)$$ where $\xi$ is a $p$-th root of unity, where $p|n$ is some odd prime number, and $\omega$ is some unit in $\mathcal{O}_K=\mathbb{Z}[\zeta]$ -I would like to ask the following question about the ring of integers $\mathcal{O}_L$ of $L$: -does $\mathcal{O}_L$ contains elements of the form $X=\frac{1}{2}(a+bt)$ with $a,b\in \mathcal{O}_K$ and such that $2\nmid a$ and $2\nmid b$? The trace of such an element is $a\in\mathcal{O}_K$, but its determinant is $\frac{1}{4}(a^2-\omega(1-\xi)b^2)$, and I do not know if there are $a$ and $b$ in $\mathcal{O}_K - 2\mathcal{O}_K$ for which we will get an integral expression. -I will appreciate your help, -Thanks, -Udi. - -REPLY [4 votes]: UPDATE: Previous argument was flawed. Here is what can salvage. -I can show there is no solution with $n$ an odd prime, or with $n$ odd and $\omega$ cyclotomic. -Let $\sigma$ denote complex conjugation. -Let $\zeta$ be a primitive $n$-th root of unity for $n$ odd and let $K = \mathbb{Q}(\zeta)$. -Lemma: Let $n$ be an odd prime and let $u$ be any unit of $K$, or let $n$ be odd and let $u$ be a cyclotomic unit of $K$. We have $\sigma(u)/u = \zeta^k$ for some integer $k$. -Proof: First, note that $\sigma(u)/u$ is an algebraic integer and all its Galois conjugates have norm $1$. So, by a result of Kronecker, it is a root of unity, and must be of the form $\pm \zeta^k$. Our goal is to show that the minus sign is impossible. -If $u$ is a cyclotomic unit, then it is a product of terms of the form $(1-\zeta^a)/(1-\zeta)$ and an explicit computation shows that the sign is positive. -Now, suppose that $n$ is an odd prime. So, suppose that $\sigma(u)/u = - \zeta^k$. Since $k$ is only determined modulo the odd number $n$, we may assume that $k$ is even. Replacing $u$ by $\zeta^{-k/2} u$, we have $\sigma(u)/u = -1$. -But, for any algebraic integer $v$ in $K$, we have $\sigma(v) \equiv v \mod 1-\zeta$. So $\sigma(u) \equiv u \mod 1- \zeta$ and (since $u$ is a unit) we have $\sigma(u)/u \equiv 1 \mod 1-\zeta$. Putting these together, we deduce that $1 \equiv -1 \mod 1-\zeta$. Since $n$ is an odd prime, $1-\zeta$ is a prime which does not divide $2$, a contradiction. $\square$ -The error in the earlier version was forgetting that $1-\zeta$ can itself be a unit when $n$ is not prime. (In fact, this occur whenever $n$ is a square free non-prime.) And this unit, of course, violates the lemma. Not sure whether the original statement might still be true in these cases. -Now suppose that we have a solution to -$$\omega \equiv 1-\zeta^m \mod 4$$ -for $m$ a proper divisor of $n$ and $\omega$ a unit. (I am using Franz's rephrasing.) -We hit both sides with $u \mapsto \sigma(u)/u$. By the lemma, we have $\sigma(\omega)/\omega = \zeta^j$ for some $j$. Also, $\sigma(1-\zeta^m)/(1-\zeta^m) = - \zeta^{-m}$. -So -$$\zeta^j \equiv - \zeta^k \mod 4$$ -This equation is not true (using again that $n$ is odd), so we have a contradiction. $\square$<|endoftext|> -TITLE: How is this observation related to Koszul duality? -QUESTION [5 upvotes]: Let $X$ be a smooth variety, $\mathcal D$ the sheaf of algebraic differentail operators, $\Omega$ the algebraic deRham complex and $\mathcal M$ a quasi coherent $\mathcal O_X$-module. -Now there is a bijection between $\mathcal D$-module structures on $\mathcal M$ and $\Omega$ dg-module structures on $\Omega \otimes_{\mathcal O_X} \mathcal M$. -For example given a $\mathcal D$-module structure on $\mathcal M$ we can define the corresponding differential by imposing the rule $d(m)(X)=X.m$ on $\Omega\otimes_{\mathcal O_X} \mathcal M$. (X denotes a vectorfield and $m$ a local section of $\mathcal M$. -A similar statement holds if one takes instead $\mathcal D=U(\mathfrak g)$ the universal envelope of a Lie-algebra and $\Omega=\bigwedge \mathfrak g^*$ the standard complex. -Now in both of there cases $\mathcal D$ and $\Omega$ are Koszul-dual, meaning that there are equivalences between carefully defined versions of their "derived categories". -Yet formulations of Koszul duality I am aware of, seem not to extend the above correspondence. Roughly speaking they are $\mathcal M \mapsto Hom(\Omega, \mathcal M)$ instead of $\mathcal M \mapsto \Omega \otimes \mathcal M$. -So my questions are: -What does the above observation have to do with Kosul duality? -Is there a formulation of Koszul duality extending the above correspondence? - -REPLY [7 votes]: There are, basically, two answers to your question that I am aware of. -The first one amounts to saying that there is no difference between tensoring with $\Omega$ and taking $Hom$ from $\Omega$. More precisely, the difference between these things corresponds to the difference between the left and right $\mathcal D$-modules: the former are supposed to be tensored with $\Omega$, while for the latter one takes $Hom$ from $\Omega$. Indeed, the $\mathcal O$-algebra $\Omega$ being Frobenius, tensoring with it only differs from taking the $Hom$ from it by the twist with the line bundle of top forms and the homological shift. This twist just transforms left $\mathcal D$-modules into right $\mathcal D$-modules. The three functors form a commutative triangle (up to the shift). -Similarly, if your Lie algebra $\mathfrak g$ is finite-dimensional, tensoring $\mathfrak g$-modules with the cohomological or the homological standard complex of $\mathfrak g$ is almost the same functor, the twist with the one-dimensional $\mathfrak g$-module of top exterior forms on $\mathfrak g$ and the homological shift being the only difference between the two. -The second answer purports to construct the kind of duality that you are looking for in the more general situations when $\Omega$ is no longer Frobenius. The first question that you can ask yourself in this case is, what would be the functor in the opposite direction, i.e., the adjoint functor to tensoring with $\Omega$ over $\mathcal O$? The answer is, it is the functor $Hom_{\mathcal O}(\mathcal D,{-})$. -The latter functor looks somewhat problematic when the variety $X$ is not affine, as the internal $Hom$ from a noncoherent quasi-coherent sheaf may be not a well-behaved operation. Perhaps this problem can be dealt with, but at the moment I do not know how to do it. -When $X$ is affine, however, the theory of derived $D$-$\Omega$ duality for the functors $\Omega\otimes_{\mathcal O}{-}$ and $Hom_{\mathcal O}(\mathcal D,{-})$ is developed in Appendix B to my AMS Memoir "Two kinds of derived categories, Koszul duality, and comodule-contramodule correspondence", http://arxiv.org/abs/0905.2621. In the case of a (possibly infinite-dimensional) Lie algebra $\mathfrak g$, this is done in Section 6.6 of the same paper. -Basically, you want to look on the module-contramodule side of the commutative triangle of "Koszul triality", as presented in my paper. In particular, the appropriate category of $\Omega$-modules that you want to consider (when $\Omega$ is infinite-dimensional, e.g., $\Omega=\bigwedge\mathfrak g^\ast$ and $\mathfrak g$ is infinite-dimensional) is that of contramodules (modules with the infinite summation operations). And the appropriate version of the derived category of $\Omega$-modules is the contraderived category.<|endoftext|> -TITLE: Free $\mathbb{Z}_2$-actions match at some point -QUESTION [7 upvotes]: I have in front of me a proof of this lemma: - -If $f$ and $g$ are free $\mathbb{Z}_2$-actions on $S^1$, then $f(x)=g(x)$ for some $x \in S^1$. - -A $\mathbb{Z}_2$-action on the unit circle $S^1$ is a homeomorphism $f \;:\; S^1 \rightarrow S^1$ -such that $f(f(x))=x$ for all $x \in S^1$; -and $f$ is free if $f(x) \neq x$ for all $x \in S^1$. -The proof (in a paper I'm refereeing) is clear but somewhat laborious. It would be nice to -either have a succinct proof, or a reference, rather than a detailed proof from first principles. -Has anyone seen this before? Thanks! - -REPLY [13 votes]: we can clearly assume that $f(z)=-z$ in the standard metric on $S^1\subset \mathbb C$ (as we can assume that $f$ is isometric with respect to some Riemannnian metric on $S^1$). Then $g(z)=z\cdot e^{i\alpha(z)}$ with $0<\alpha(z)<2\pi$. If $\alpha(z_0)<\pi$ then $\alpha(g(z_0))>\pi$ and there is a point $z_1$ with $\alpha(z_1)=\pi$ by the intermediate value theorem.<|endoftext|> -TITLE: When does the blow-up of $CP^2$ at N points embed in $CP^4$? -QUESTION [19 upvotes]: Write $X_N$ for this blow up. Place the N points in 'general position' as needed. Then $X_6$ embeds in $CP^2$ as a smooth cubic surface. (See, eg, Griffiths and Harris.) But there is no other $N$ (except $N=0$) -for which $X_N$ embeds in $CP^3$. -(Proof: The topology of the blow-up -disagrees with that of a smooth surface of degree $d$ in $CP^3$. (Gompf-Stipsisz p. 21.) On the other hand, $X_N$ embeds in $CP^5$ simply because any -smooth algebraic surface $X$ so embeds. (Harris, `Algebraic Geometry, a first course', p. 193.) -Embarrassingly, I don't even know the answer for $N=1$ where $X_1$ is the 1st Hirzebruch surface! (I'm betting it does embed.) -Motivation: This question began in an attempt to better understand the 27 lines on the cubic -and my initial surprise at how the construction described in GH of $X_6$ yielded -a smooth surface in $CP^3$, and how all such surfaces arise through that construction by varying the 6 points. I am hoping answers might help me understand the moduli of blow-ups as I move the N points about the plane, and orient me as a novice to algebraic surfaces. - -REPLY [26 votes]: For $N=1$ the answer is yes: the embedding into ${\mathbb P}^4$ is given by the linear system of conics through the blown up point (the image has degree $d=3$). -For $N=5$, the system of cubics through the 5 points gives an embedding ($d=4$). -ADDED: here are 2 slightly less obvious examples: -For $N=8$ one can take quartics with an assigned double point and 7 simple base points ($d=5$). -For $N=10$ take the quintics with 3 assigned double points and 7 simple base points ($d=6$; I did not check all the details here, because it's very boring, but I'm sure that it works). -In general, giving a satisfactory answer to your question seems very hard. There is a numerical equality, - the so-called "double point formula" (Hartshorne, "Algebraic geometry", p.434), which is satisfied by all smooth surfaces of ${\mathbb P}^4$: -$$d^2-10d-5HK+12\chi-2K^2=0,$$ -where $H$ is the hyperplane section, $d=H^2$ is the degree, $K$ is the canonical divisor and $\chi$ the Euler characteristic of ${\mathcal O}_{X_N}$. -In our case the formula becomes: -$$d^2-10d-5HK+2N-6=0.$$ -In addition there is result by G. Ellingsrud and C. Peskine [Invent. Math. 95 (1989), no. 1, 1--11] saying that only finitely many components of the Hilbert scheme of smooth surfaces in ${\mathbb P}^4$ contain smooth rational surfaces. - So in principle it should be possible to classify all the smooth rational surfaces in ${\mathbb P}^4$. In practice, it is known that the degree is $\le 76$, -[Cook, An improved bound for the degree of smooth surfaces in P4 not of general type. -Compositio Math. 102 (1996)] and it is conjectured that $d\le 15$ (examples with $d=15$ do exist). There are also papers by several authors (Ranestad, Schreyer, Popescu, and others) that classify the smooth rational surfaces of ${\mathbb P}^4$ of degree $\le 11$. In these papers you can find examples of the kind you are looking for. For instance there are examples with $d=10$ and $N=18$.<|endoftext|> -TITLE: How "frequent" are smooth projective varieties with (anti-)ample canonical bundle? -QUESTION [7 upvotes]: How "frequent" are smooth projective varieties $X$ with (anti-)ample canonical bundle $\omega_X = \bigwedge^d \Omega^1_{X/k}$? -E.g. for curves $C/k$, the canonical bundle is ample iff the genus $g(C) > 1$ and anti-ample iff $g(C) = 0$. What is the situation like in the higher dimensional case? -(This question is inspired from How "frequent" are smooth projective varieties with trivial canonical bundle?) - -REPLY [25 votes]: Just as in the case of curves, there is in general a trichotomy of cases: -Let $X$ be a smooth projective variety. Then $X$ is built from pieces $Y$ with - -$\kappa(Y)<0$ -$\kappa(Y)=0$ -$\kappa(Y)=\dim Y$. - -Here $\kappa$ denotes the Kodaira dimension. If the anti-canonical is ample, i.e., $X$ is a Fano, then it belongs to the first class, if the (anti-)canonical is trivial it belongs to the second and if the canonical is ample, then it belongs to the third. Otherwise one can perturb these cases birationally or by taking a finite quotient to get other examples in these classes. -"Built" means the following: -Any such $X$ is birational to $\widetilde X$ that has an iterated fiber space structure with general fiber a Fano variety (for each fiber space) and the target a variety that has $\kappa\geq 0$. Then this target has an iterated fiber space structure with general fiber $\kappa=0$ (for each fiber space) and target that has $\kappa=\dim$ (i.e., it is a variety of general type). (The first series of fibrations come from the MMP and they are sometimes called Mori fiber spaces (each step individually) and the second is the Iitaka fibration). -As far as how "frequent" the cases are, it seems that the curve case kind of tells us the relative frequency of each compared to the others. Of course we have less explicit data in higher dimensions, but as Jason (and Balázs answering the linked question) mentioned there are results to suggest that we should expect a similar distribution. -In the curve case it is interesting that the same trichotomy appears from various points of view and in some of these we can put some quantitative measure on their relative frequency: - -Topology: the fundamental groups in each case: trivial, abelian, non-abelian -Arithmetic: the group of rational points in each case (this has to be taken with a grain of salt, but it is instructive): non-finitely generated, finitely generated, finite. -Differential Geometry: curvature in each case: positive, flat, negative or if you like parabolic, elliptic, hyperbolic. -Algebraic Geometry: the dimension of the moduli space in each case: $0$, $1$, $3g-3$ - -Now one can go out and try to work out what happens to these classifications in any of these disciplines. As far as I know what (little) evidence we have suggests that the same kind of ratio occurs always: The $\kappa<0$ case is relatively small, the $\kappa=0$ is a little larger, but even combined they are nowhere near the "frequency" of the last case, which is accordingly named general type. -One interesting thing is that even though about 100% of varieties is of general type, most of those that we know explicitly are not. This is not a contradiction though. The fact that we can easily write down a description of a variety either by their equation or by a construction makes them special, so it is not surprising that these are not "general". -At the same time, by a different measure rational curves are quite ubiquitous. Abstractly there is only one smooth projective rational curve, but it appears everywhere. If you consider any birational morphism between smooth varieties, then the exceptional divisors will always be covered by rational curves. So one might argue that they are quite frequent.<|endoftext|> -TITLE: History of the notation $\mathbb Z_n$ -QUESTION [15 upvotes]: This question was motivated by Martin's comment in Free $\mathbb{Z}_2$-actions match at some point -When was the notation $\mathbb Z_n$ introduced for $n$-adic integers and by whom? When was it introduced for integers modulo $n$ and by whom? -I tried searching on Google without much luck. -Added. Martin brings up the related point when was the subscript notation $A_f$ introduced for localizing a ring at the monoid generated by $f$ (which is a third possible interpretation of $\mathbb Z_n$)? - -REPLY [11 votes]: I fished around in Google scholar and found so many examples that I don't feel like listing any of the links. Nonetheless, a clear picture emerges of an answer that I found a bit surprising: The notation $\mathbb{Z}_p$ for the $p$-adic integers evolved in three separate parts. I should also explain that the real science of etymology is about the evolution of words or notation, not just "when did it first happen". -The subscript notation not only for the $p$-adic integers, but more generally for $p$-adic completions, already appears in several papers in the 1930s and 1940s. For instance Carl Ludwig Siegel says in 1941, "$R$ is the field of rational numbers, $R_p$ the field of $p$-adic numbers, where $p$ denotes any prime number, $R_\infty$ the field of real numbers; moreover $J$ is the ring of integral numbers and $J_p$ the ring of $p$-adic integers". Of course, no one would use this notation today! -The use of $Z$ for the integers has a semi-separate history. I even found an old paper, but more recent than this one by Siegel, that used $Z$ for the integers but $R$ for the $p$-adic integers, with no subscript. -Generally the notation for $p$-adic integers and $p$-adic numbers standardized at $Z_p$ and $Q_p$ in the 1950s. Quite possibly Bourbaki, Algebra, deserves credit for standardizing $Z$ and $Q$ for integers and rationals. -Blackboard bold notation ($\mathbb{Z}$ and $\mathbb{Q}$) came last, at least in print. Despite its name, it's no longer obvious to me that blackboard bold actually first came from blackboards or from typewriters. It's sometimes also credited to Bourbaki, but this seems to be wrong. There is a historical account by Lee Rudolph (in comp.text.tex) that credits certain typewriter models in the 1960s for producing blackboard bold typography for the integers, etc. If that is where it started, then the notation seemed to catch on fairly quickly, although there were holdouts that used ordinary bold for decades after that. (But, before blackboard bold was fashionable, it wasn't even standard to make the set of integers bold $\mathbf{Z}$ instead of just $Z$.) -As an aside, the collision of notation between the $p$-adic integers and the integers mod $p$ is unfortunate. I really prefer to write $\mathbb{Z}/n$ for the integers mod $n$, because it is then written exactly as it reads. Also, partly since it is such a commonly used object, I see no need for extra parentheses, or an extra $\mathbb{Z}$, and certainly just using an $n$ subscript is bad. I'm optimistic that this notation is the way of the future and it would be an interesting separate question in history of notation. -(Sorry, I didn't see the entire string of comments before I wrote all of this. The comments make most of these remarks, but it seems useful to combine them into one historical summary.)<|endoftext|> -TITLE: "Orthogonal complement" in $\mathbb{Z}_q^n$ -QUESTION [5 upvotes]: Let $W$ be the finite $\mathbb{Z}$-module obtained from $\mathbb{Z}_q^n$ with addition componentwise where $\mathbb{Z}_q$ is the integers mod $q$. Let $V$ be a submodule of $W$. Let $V^{\perp} = \{w \in W : \forall v \in V \quad w \cdot v = 0 \}$ where $w\cdot v = w_1v_1 + \ldots + w_nv_n$. Is it true that ${(V^{\perp})}^{\perp} = V$ for all $q \geq 2$? -According to Wikipedia, this holds for finite dimensional inner product space, but I wish to know whether it holds in $\mathbb{Z}_q^n$ where $\cdot$ isn't an inner product. - -REPLY [12 votes]: Yeah, it's true. Since $\mathbb{Z}/q$ is a principal ideal ring, there is an extension of the Euclidean algorithm to matrices that puts any matrix in Smith normal form. It means that after an automorphism of $(\mathbb{Z}/q)^n$, any submodule $V$ can be put into a standard form in which it is generated by vectors of the form $d_k e_k$, where $e_k$ is a standard basis vector, $d_k$ is a divisor of $q$, and each $k$ only appears at most once. In that case you can check directly that $(V^\perp)^\perp$ is no larger than $V$. -(I'm taking the question in the more interesting case in which $q$ might not be prime.)<|endoftext|> -TITLE: Why does inner model theory need so much descriptive set theory (and vice versa)? -QUESTION [19 upvotes]: I am curious about how much descriptive set theory is involved in inner model theory. -For instance Shoenfield's absoluteness result is based on the construction of the Shoenfield tree which projection is $\aleph_1$-Suslin. Also the Schoenfield tree is homogeneous, meaning the direct limit $M_x$ of the ultrapowers by the measures $\mu_{x\upharpoonright n}$ is wellfounded. The measures $\mu_{x\upharpoonright n}$ are defined on the sections of the tree. We also have that $L(\mathbb{R}) \vDash AD$ is equiconsistent with $ZFC+$ there are infinitely many Woodin cardinals. Descriptive set theory talks a lot about homogeneously Suslin sets and homogeneous trees (they pave the way for determinacy results) but these concepts seem themselves to be very important for inner model theory (just a simple fact: a set $X$ is homogeneously Suslin iff $X$ is continuously reducible to the wellfoundedness of towers of measures). The Martin Solovay tree is what gives $\Sigma^1_3$ absoluteness between $V$ and a generic extension $V[G]$ assuming measurability. Also, the Kechris-Martin Theorem has a purely descriptive theoretic proof and a purely inner model theoretic proof. A theorem of Woodin states that $(\Sigma^2_1)^{Hom_{\infty}}$ sentences are absolute for set forcing if there are arbitrarily large Woodin cardinals. -My question is why are there so many links between descriptive set theory and inner model theory? I would love to hear from an expert about the intuition as to what is really going on. The relationship between both field does not seems "ad hoc", it appears as though there is very deep beautiful and natural structure. I apologize in advance for any vagueness in my question. Thx. - -REPLY [11 votes]: Philip gave a nice answer but let me add some points to it. -The thing is that while IMT and DST were far apart back in the days, over the years it became clear that part of DST and IMT deal with exactly the same problem. -The above sentence may not be completely objective, but at any rate, that seems to be the point of view of some people who do inner model theory (including mine). The main goal of IMT is to construct models that are correct about the universe. How you measure this correctness? Well, there is Levy hierarchy and there is a more refined and more useful version of it, namely the Wadge hierarchy. So to make the question more precise, what we do in inner model theory is that we try to construct and analyze canonical models called mice that capture the levels of the Wadge hierarchy. Back in the days, it wasn't so clear that there is such a deep relationship between Wadge hierarchy (or their more refined forms, the Universally Baire sets, or homogeneously Suslin sets) and the large cardinal hierarchy. Now, the connection is much more clear and transparent. -DST is really inevitable. The technical problem that people doing IMT are trying to solve is the construction of $\omega_1+1$ iteration strategies. These iteration strategies allow one to build trees of height $\omega_1$ and then it is guaranteed that any such tree must have a branch. But wait a minute, in ZFC alone, there are trees of height $\omega_1$ with no branches. So how are we going to make sure that these strategies will build only trees of height $\omega_1$ for which there are branches. The answer must be that the strategy has various DST like properties, like it is Universally Baire or hom Suslin and etc (it is a nice exercise to show that such strategies are indeed nice). -At any rate, when doing IMT, running to DST seems to be inevitable. It is probably not so true if you are doing DST (by DST we mean Moschovakis' book not the directions it went since 90s). But still there are many theorems you can prove under AD using IMT and no proof avoiding IMT is known. For instance, every regular cardinal below Theta is a measurable cardinal, a fact proven using IMT and no proof avoiding IMT is known. -The reason that the subjects are so close, though, is just that they study the same object (namely the Wadge hierarchy) from different point of views. DST does it using recursion theoretic methods (coding lemma, pointclass arguments and etc) and IMT does it by translating the Wadge hierarchy into hierarchy of iteration strategies, which is what IMT studies.<|endoftext|> -TITLE: Inverse of a matrix over a non-commutative ring -QUESTION [15 upvotes]: What's the best algorithm to invert a matrix of non-commutative elements? In my case I have a matrix of matrices. -From first principals by equating the elements of M * M' to I (where M' is the inverse) I've worked out the inverse for a 2x2 Matrix: -$$ -M=\begin{bmatrix} - A &B \\ - C &D \\ -\end{bmatrix} -$$ -$$ -M^{-1}=\begin{bmatrix} - (A-BD^{-1}C)^{-1} &(C-DB^{-1}A)^{-1} \\ - -D^{-1}C(A-BD^{-1}C)^{-1} &-B^{-1}A(C-DB^{-1}A)^{-1} \\ -\end{bmatrix} -$$ -In the above, you only need to perform 4 inversions: B, D and the components of the top row. The bottom row elements are obtained by multiplying the top row by 2 existing matrices. -One further optimisation of this is to choose to 2 "easiest" matrices to invert since you can transpose and/or swap the columns around and obtain the final result by transposing and/or swapping the rows. -As pointed out in the answers, one can use Schur complements to obtain similar / better formulae and recursively apply them to reach the result. -I've implemented the Doolittle LU Decomposition ensuring it respects multiplication order. Is there a better approach? It seems the concept of "pivot" is not really applicable (or at least more complex) when the elements are matrices, which is why I chose the simple Doolittle approach. - -REPLY [9 votes]: If yours matrix is "generic" (i.e. You do not suspect there are some specific algebraic relation between elements) then (as far as I know) there is nothing better than just use LU decomposition carefully putting elements on the right or on the left. (I mean LU can be easyly generalized to noncommutative case but formulas will be a little more complicated). -Actually the formulas which you write in 2*2 are very instructive ! -The expressions A - B D^{-1} C called "Schur complements" see -http://en.wikipedia.org/wiki/Schur_complement -You can obtain inversion of non-commutative matrix in this way - just consider that D - is $(n-1) \times (n-1)$ matrix since you did not use commutativity in yours formulas - they will work in this case also. -So you can inductively go on till $n=1$ - so you obtain the inversion of yours matrix. -Actually this is the LU algorithm. - -There some math involved for specific matrices with non-commutative entries. -For example if and only if [a,c] = [b,d] =0 and [a,d] = [b,d], -than you can see that the inverse matrix can be given by the same formula as in commutative -case -$$ \frac{ 1 }{ad-cb} d ~~~\frac{ -1 }{ad-cb}b $$ -$$ \frac{ -1 }{ad-cb}c ~~~\frac{ 1 }{ad-cb}a $$ -Similar fact holds true for $n\times n$ matrix if each 2 by 2 submatrix satisfy the relations above. -We propose to call such matrices "Manin matrices", since Yurii Manin first considered them at 1988-89. For such matrices basically all commutative facts holds true, despite they are quite far from commuative ones. -http://arxiv.org/abs/0901.0235 - -Another probably most famous examples are - "quantum group" matrices. -Here one requires ac=q cb, etc... for some number "q" , -then there are q-determinant and also inverse can be written by the usual formula -substituting determinants by q-determinants... -There are many other examples super-matrices, Capelli matrices, matrices satisfying "reflection equation"... but the systematic theory is not developped. - -REPLY [5 votes]: There is a theory of determinants of matrices over non-commutative (in particular, free) rings. It was mainly developed by Gelfand and Retah. I think the first paper is this: Gelfand, I. M., Retakh, V. S. Theory of noncommutative determinants, and characteristic functions of graphs. Funktsional. Anal. i Prilozhen. 26 (1992), no. 4, 1--20, 96; translation in -Funct. Anal. Appl. 26 (1992), no. 4, 231–246 (1993) . In particular, they discuss a connection between inverses of matrices ovr non-commutative rings and these "quasi-determinants".<|endoftext|> -TITLE: finiteness of torsion points of an abelian variety over a totally real field？ -QUESTION [9 upvotes]: Ribet has shown the following result: if $A$ is an abelian variety over a number field $E$, then the torsion subgroup of $A(E^{cyc})$ is finite. Here $E^{cyc}$ is the union $\bigcup_nE(\mu_n)$, $\mu_n$ being the set of n-th roots of unity. -Here we have used $E^{cyc}=E\mathbb{Q}^{ab}$. I would like to know if such finiteness result holds for other fields. For example, if $A$ is a CM abelian variety over a number field $E$, and let $F$ a totally real number field, with $F^{ab}$ the maximal abelian extension of $F$, can we expect the torsion subgroup of $A(EF^{ab})$ to be finite, too? Or should one expect the torsion subgroup of $A(EF^{ab})$ to be much smaller that the total torsion part of $A((EF)^{ab})$? The latter is infinite by a theorem of Zarhin, and I wonder if the torsion part can be somehow measured, and cut out some smaller subgroup. -Thanks! - -REPLY [4 votes]: This is Corollary 2 in Show-Wu Zhang's Equidistribution of points of small heights on abelian varieties, Annals of Math. 147 (1998), no. 1, 159--165. -Observe that $\mathbf Q^{\mathrm{cyc}}=\mathbf Q^t(i)$, where $\mathbf Q^t$ -is a totally real extension of $\mathbf Q$ (generated by all $\cos(\pi/n)$, $n>0$). -Consequently, your assertion follows from the equidistribution theorem of Szpiro, Ullmo and Zhang, as applied to the Weil restriction of scalars $A'=R_{E(i)/\mathbf Q}A$. Indeed, if the set of torsion points of $A(E(i)\mathbf Q^t)$ were Zariski dense, one would be able to define a dense sequence of torsion points in $A'(\mathbf Q^t)$. The corresponding sequence of probability measures on $A'(\mathbf C)$ supported by their Galois orbits would be supported on $A'(\mathbf R)$ but converge (by the S-U-Z theorem) to the Haar measure on $A'(\mathbf C)$. -The general case can be deduced from this one by passing to an appropriate -abelian subvariety. -Note that Zhang attributes this result to Zarhin (Duke Math. J. 54 (1987)).<|endoftext|> -TITLE: Good reference for globally formulated calculus of variations on Riemannian manifolds? -QUESTION [15 upvotes]: I have been working on a formulation of the calculus of variations on Riemannian manifolds, formulated globally using [pullback] vector bundles and tensor bundles and their induced covariant derivatives (specifically regarding maps of type $ \phi\colon M\to S $ between Riemannian manifolds $ \left(M,g\right) $ and $ \left(S,h\right) $, where $ M $ may be required to be compact, and the relevant energy functionals are action functionals having Lagrangian depending on $ T\phi\in\Gamma\left(\phi^* TS\otimes T^{*}M\right) $ and covariant derivatives of $ T\phi $). -Of all the non-elementary references I've found, the exposition is restricted to the energy functional whose critical points are harmonic maps. I am in the process of finding all the relevant papers by James Eells, as was suggested to me by Arthur Fischer, though the papers I've found so far deal exclusively with the harmonic map case. The monograph “Variational Problems In Geometry” by Seiki Nishikawa covers a lot of relevant material, but is mainly based on local coordinate calculations, and also only addresses the harmonic map case. I have found other relevant papers and textbooks along the same lines. -My question is: What are the best references for this type of calculus of variations? I wish to know what has been already published regarding global formulation using vector bundles with covariant derivatives/linear Ehresmann connections (geared mainly towards calculating the first and second variations), so that I may tie my work in with known results and identify particular areas that could be expanded/improved upon. -For reference, the texts/papers I've so far found relevant to or inspiring my work include: Marden and Hughes “Mathematical Foundations of Elasticity”, Jeff M Lee “Manifolds and Differential Geometry”, Yuanlong Xin “Geometry of Harmonic Maps”, David Bleecker “Gauge Theory and Variational Principles”, Richard Palais “Foundations of Global Nonlinear Analysis”. - -REPLY [2 votes]: I had forgotten about this question I had asked until I stumbled upon it again today. The work I mentioned is here -- https://arxiv.org/abs/1212.2376 -- for anyone interested. -The punchline of the paper is the covariant Euler-Lagrange equation for maps between Riemannian manifolds, using a "strongly typed" global tensor calculus formalism in order to cleanly handle what would otherwise be horrendous coordinate calculations.<|endoftext|> -TITLE: What is the intuition of connections for cubical sets? -QUESTION [6 upvotes]: I am beggining to do some work with cubical sets and thought that I should have an understanding of various extra structures that one may put on cubical sets (for purposes of this question, connections). I know that cubical sets behave more nicely when one has an extra set of degeneracies called connections. The question is: Why these particular relations? Why do they show up? Precise references would be greatly appreciated. - -REPLY [3 votes]: As far as higher cubical categories are concerned, a connection will allow you to literally rotate a face, i.e. turn a face of one type into a face of another type, in an invertible way. -In short it materializes an equivalence between the different types of faces into special degenerate cubes. -The 2d case for example is fairly simple as one can either turn horizontal arrows into vertical arrows or vice versa. -One advantage of a connection is therefore that it allows one to speak of commutative n-cubes in an n-tuple category with connection. To do so, you can take an n-cube, apply connections until you only have non trivial faces of one type. Then check whether the obtained cube is an identity or not. It turns out that it does not depend on the way you chose to apply the connection, if your cube gives an identity cube with one face rearrangement, it will with another. It is, to my understanding the essence of Brown and AlAlg's equivalence between cubical categories with connections and globular categories. -So for cubical categories it is very restrictive, which is also why they are so friendly. But I am not sure about the impact on cubical sets. You surely will find good material in Tim and Ronnie's references.<|endoftext|> -TITLE: Solving the cubic by "radicals" in characteristics 2 and 3 -QUESTION [25 upvotes]: This question has no justification other than a bit of fun. -We all know that the cubic is solvable by "radicals" ($\root2\of{}$ and $\root3\of{}$) in characteristics $\neq2,3$. The formula was discovered by the Italians in the 16th century (see here). -In characteristic $2$, there should be a similar formula involving $\wp_2^{-1}(\ )$ and $\root3\of{}$, and in characteristic $3$ there should be a formula involving $\root2\of{}$ and $\wp_3^{-1}(\ )$. -By $\wp_2^{-1}(a)$ and $\wp_3^{-1}(a)$ I mean a root of the polynomials $\wp_2(T)=T^2-T-a$ and $\wp_3(T)=T^3-T-a$ respectively, which give all cyclic extensions of degree $2$ and $3$ respectively. -Has somebody worked out these formulæ ? -Edit. I have accepted one of the answers --- the choice was difficult --- but I'm still curious as to whether these formulæ can be found somewhere in the literature. - -REPLY [16 votes]: I asked an undergraduate (Dubravka Bodiroga at Hood College) to work these results out last summer. Here is her cubic formula in characteristic 3 (paraphrasing from something she sent me): -Consder the polynomial -$$ -x^3 - a_1 x^2 + a_2 x - a_3, -$$ -where the coefficients belong to a commutative ring in which $3=0$. Assume moreover that $a_1$ is invertible. Let $b$ be a solution to -$$ -b^2 = -\frac{a_2^3}{a_1^6}+\frac{a_2^2}{a_1^4}-\frac{a_3}{a_1^3}, -$$ -and let $\beta$ be a solution to $y^3 - y - b = 0$. Then -$$ -x^3 - a_1 x^2 + a_2 x - a_3 = (x - (u\beta^2+v))(x-(u(\beta+1)^2+v))(x-(u(\beta+2)^2+v)), -$$ -where $u=-a_1$ and $v=a_1 -\frac{a_2}{a_1}$. -If one inverts the procedure, letting -$$ -x_i = u(\beta+i)^2 + v, -$$ -then ([Parson: assuming I haven't scrambled her indices]) -$$ -b= \frac{(x_0 - x_1)(x_1 - x_2)(x_0 - x_2)}{(x_0+x_1+x_2)^3}, -$$ -and -$$ -y= \frac{x_2 + 2x_1}{x_0+x_1+x_2} = -\frac{2\times x_2+1\times x_1+0\times x_0}{x_0+x_1+x_2}. -$$ -I believe she also worked out the (simpler) details for the cubic formula in characteristic $2$, but I could not find them just now. She used a heuristic method of Euler and B\'ezout to find the formulas, an exposition of which one can find in Tignol's book on Galois theory. She then solved for the auxiliary quantities $b$ and $y$ in terms of the $x_i$ to see what Lagrange would have made of her solution procedure.<|endoftext|> -TITLE: Irreducibility for multivariate polynomial polynomial generated from sum of irreducible polynomial in one variable -QUESTION [6 upvotes]: Could any tell me if a multivariate polynomial generated from the sum of irreducible single variable polynomial is irreducible? -For example, f(x)=x^2+2x+2, g(x)=x^2+3x+3, h(x)=x^3+2x^2+2x+2 all of them are irreducible, then what about f(x,y,z) = f(x)+g(y)+h(z)? - -REPLY [12 votes]: For two variables, see W. Feit, - Some consequences of the classification of finite simple groups, in - The Santa Cruz Conference on Finite Groups, Proc. Sympos. - Pure Math. 37, American Mathematical Society, Providence, - RI, 1980, pp. 175-181. The result is the following. A polynomial $f(x)\in\mathbb{C}[x]$ is indecomposable if - whenever $f(x)=r(s(x))$ for polynomials $r(x),s(x)$, then either - $\deg r(x)=1$ or $\deg s(x)=1$. Suppose that $f(x)$ and $g(x)$ are - nonconstant indecomposable polynomials in $\mathbb{C}[x]$ such that - $f(x)-g(y)$ factors in $\mathbb{C}[x,y]$. Then either $g(x)=f(ax+b)$ for some - $a,b\in \mathbb{C}$, or else - $$ \deg f(x) = \deg g(x) = 7,\ 11,\ 13,\ 15,\ 21,\ \mathrm{or}\ - 31. $$ - Moreover, this result is best possible in the sense that for - $n= 7,11,13, 15, 21$, or $31$, there exist indecomposable polynomials - $f(x)$, $g(x)$ of degree $n$ such that $f(x)\neq g(ax+b)$ and - $f(x)-g(y)$ factors. The proof uses the classification of finite simple groups!<|endoftext|> -TITLE: Citation of a paper with a proof you would like to improve -QUESTION [23 upvotes]: Imagine that you are in the following situation: You write up a proof which eventually gets published. There you need a result which is not so well-known but it is contained in another paper P; therefore you just cite it. You read P and come to the conclusion: It's awful. You need plenty of time to insert the details or even correct it. It may also happen that the proof is somewhat too complicated because in your situation it is much easier. Maybe you have found a shorter proof, but based on the ideas in P. Now what do you do? Several options come into my mind: - -Just cite the paper without any further explanation. -Cite the paper but give a short hint how to simplify the arguments. -Cite the paper but give a more elaborate explanation of the arguments. -Write up the details of the proof of the desired result in your situation and remark somewhere that it was inspired by the paper P. - -For each option there are pros and cons. For example, you don't want to blow up your proof with material which does not seem be so important. Also, you don't want to bore your readers. This favors the first options. On the other hand, you might want to be sure that the readers understand the argument and don't have to read P. This favors the last options. What do you think, which option is your favorite and why? Also, are there other appropriate options? - -REPLY [25 votes]: Improving existing proofs is an important and undervalued part of mathematics. We don't just want to know whether something is true; we want to know why it's true. So I think that if you have a better proof of something, you should find a way to share it with the world. -Here are a couple of thoughts about the practicalities, to add to Andrew's suggestion about the nLab. -First, you could put the simplified proof into an appendix to your paper. I quite like appendices, as both a reader and a writer. Used well, they help to keep the main part of the paper flowing, while providing crucial details to those who want them. -Second, it's entirely possible that the author of [Awful 2009] will referee your paper. So whatever you write, you need to keep them sweet. I think this also favours the appendix option.<|endoftext|> -TITLE: Primitive elements in a free group of rank three -QUESTION [10 upvotes]: It is well-known that the fundamental group of a twice-punctured torus is a free group of rank three. -I see that there is no one-to-one correspondence between the homotopy classes of essential simple loops on twice-punctured torus and the conjugacy classes of primitive elements in a free group of rank three. -Do we know which primitive elements in a free group of rank three represent simple loops on a twice-punctured torus? - -REPLY [6 votes]: There are 3 types of simple closed curves on a twice punctured torus. -The first type are two isotopy classes of curves which are isotopic to the two -boundary components. These are primitive and separating. -The second type is separating curves which cut the torus into -a once-punctured torus and a pair of pants. These curves are not -primitive, since they are commutators (although I assume this -is not what you mean when you observe there is no 1-1 correspondence -between primitive curves and simple closed curves). -The third type is non-separating curves which are primitive. These -curves have the property that they are sent to their inverse -(when keeping track of orientation) under the elliptic involution (which -exchanges the two boundary components). -So this property is certainly a restriction: a primitive curve conjugate -to a simple closed curve -which is not isotopic to a boundary component must be sent to -its inverse under the elliptic involution. I think this symmetry -condition is not sufficient, since the subgroup of $Aut(F_3)$ -commuting with this involution is larger than the mapping torus of -the twice-punctured torus.<|endoftext|> -TITLE: Quivers of selfinjective algebras. -QUESTION [8 upvotes]: Let's say a quiver $Q$ is covered by cycles if each of it’s arrows can be included in an oriented cycle. -It's easy to prove that if a path-algebra with relations $KQ/I$ (where $I$ is an admissible ideal) is selfinjective then $Q$ is covered by cycles. -The following question occured to me: is the opposite true? -Is it true that if a quiver $Q$ is covered by cycles then there exist an admissible ideal $I\triangleleft KQ$ such that the algebra $KQ/I$ is selfinjective? -I found an answer on my own. It's true indeed. But the proof is quite difficult. Maybe this result is already known? - -REPLY [7 votes]: One way to prove a finite-dimensional algebra is self-injective is to find an isomorphism of modules $A\cong A^*$, that is a non-degenerate bilinear form such that $(ab,c)=(a,bc)$. Of course, such a form is uniquely determined by the linear map $t(a)=(a,1)$ (you can recover the form by $(a,b)=t(ab)$). -Thus, for any such linear map on an algebra, we have a canonical quotient on which it is non-degenerate (you can check that the kernel of $(-,-)$ is a 2-sided ideal), which is self-injective if it is finite dimensional. -Of course, if we consider a path algebra $kQ$, and consider a linear map that kills all paths of length $\geq N$, this will obviously have finite dimensional self-injective quotient. -Thus, we need to see that if $Q$ is covered by cycles, we can find such a form whose kernel is an admissible ideal. We can do this in a very silly way: pick any set of cycles $C_i$ which cover the graph and pick $a_i$ non-zero elements of $k$. Let $t$ be the linear map which sends any path that traces out $C_i$ (with any starting point) to $a_i$ (you also need to add a term which gives non-zero values on all isolated vertices). The kernel of this contains all paths longer than all $C_i$, so it is finite dimensional. On the other hand, for any non-zero linear combination of idempotents (length 0 paths) and length 1 paths, one can always take pairing against a path either tracing one of the cycles starting at one of the vertices (if idempotents appear) or completing one of the cycles (if length 1 paths appear). Thus, none of these lie in the kernel, and we are done.<|endoftext|> -TITLE: About integer polynomials which are sums of squares of rational polynomials... -QUESTION [14 upvotes]: I have the following question for which I haven't been able to find any reference or proof. -Suppose we know that a univariate polynomial $P(X)$ with integer coefficients is the sum of squares of two polynomials with rational coefficients. -Is it true that $P(X)$ must also be the sum of squares of two polynomials with -integer coefficients? -For example, take $P(X)=50X^2+14X+1$, then we see that $P(X)=(5X+3/5)^2+(5X+4/5)^2$, but it is also $X^2+(7X+1)^2$. -I would greatly appreciate any help pointing me into the right direction. -Thanks in advance, and regards, -Guillermo - -REPLY [9 votes]: In fact, if $P(x)$ is a polynomial with integer coefficients and if every arithmetic progression contains an integer $n$ for which $P(n)$ is a sum of two rational squares, then $P(x) = u_1(x)^2 + u_2(x)^2$ identically, where $u_1(x)$ and $u_2(x)$ are polynomials with integral coefficients. This follows from a theorem of Davenport, Lewis, and Schinzel; see the Corollary to Theorem 2 in Polynomials of certain special types (Acta Arith. IX, 1964, 107--116). -(In my restatement of their result, I use that being a sum of two rational squares is equivalent to being a sum of two integer squares. This is easy to prove directly from the characterization; alternatively, it follows from a lemma in Serre's book, attributed to Davenport--Cassels, used to prove the three squares theorem. Also, Davenport, Lewis, and Schinzel seem to have an argument similar to Gjergji's implicitly in mind in their proof of the Corollary above. So Gjergji's answer is the "real" one; but maybe this paper will interest others.)<|endoftext|> -TITLE: Somewhat general question that includes: "Do quasi-isomorphic cdgas have quasi-isomorphic spaces of derivations?" -QUESTION [6 upvotes]: Question: Given two quasi-isomorphic dg commutative algebras (over a field of characteristic zero, if you like), to what extent do their various homological geometric data agree? -Example: Given a dg commutative algebra $A$, there is a dg Lie algebra $\operatorname{Der}(A)$ defined by understanding the notion of "derivation" internal to the category of dg vector spaces. If $A$ and $B$ are quasi-isomorphic dg commutative algebras, are $\operatorname{Der}(A)$ and $\operatorname{Der}(B)$ quasi-isomorphic dg Lie algebras? Note that any homomorphism $f: A \to B$ defines a (dg) vector space of "derivations relative to $f$", which is a bimodule for the Lie algebras $\operatorname{Der}(A)$ and $\operatorname{Der}(B)$ and receives maps as one-sided modules from each of these; one would expect these maps to be quasi-isomorphisms if $f$ is. -Remark: I intend my question to be somewhat open ended. As such, I would accept an answer that points me to the appropriate literature. - -REPLY [4 votes]: If you have two cdgas which are cofibrant (so built out of free cdgas and their cones via an iterated sequence of pushouts) and quasi-isomorphic then their homological invariants agree (one of the properties of cofibrant models is that any quasi-isomorphism between them admits a homotopy inverse). Vitali's example illustrates the problem in general: the minimal model for $H^* (S^2)$ is cofibrant, but $H^* (S^2)$ with trivial differential is not. -To clarify things a bit further, as Tom pointed out in the comments above, when we talk about derivations we usually have a map $A\rightarrow B$ of simplicial commutative rings (in characteristic zero we can use commutative dgas) and a $B$-module $M$. Since you didn't mention this data I assumed we were taking $A$ to be the unnamed field of characteristic zero and $B$ to be augmented over $A$ so that we could take the module $M$ to be $A$ (I think this is a common situation). Now $Der_A (B;A)$ is contravariantly functorial in $B$ as an augmented $A$ algebra and takes homotopic maps to the same map. So if I have maps of augmented $A$ algebras $B\rightarrow C\rightarrow B\rightarrow C$ such that the composite of each two maps is homotopic to the identity (such as when we have a quasi-isomorphism between two cofibrant $A$-algebras) I can apply $Der_A (-,A)$ to the sequence and obtain an isomorphism between $Der_A(B,A)$ and $Der_A(C,A)$. -You might find it helpful to read Quillen's 1970 paper: On the (co-)homology of commutative rings. These ideas are explained and generalized there.<|endoftext|> -TITLE: an easy example of valuation ring which is not noetherian？ -QUESTION [6 upvotes]: Is there an easy example of valuation ring which is not noetherian？ - -REPLY [2 votes]: Construction of valuation domains of Krull dimension $>1$: -Let $O\neq K:=\mathrm{Frac}(O)$ be a valuation domain. Consider the natural map $h:O\rightarrow k$, -where $k$ is the residue field of $O$. Let $\overline{O}$ be a valuation domain -of $k$. Then $O^\prime:=h^{-1}(\overline{O})\subseteq O$ is a valuation domain -of $K$ with the following properties: - -$\mathrm{Spec}(O)\subset\mathrm{Spec} (O^\prime)$, -$O=O^\prime_M$, where $M$ is the maximal ideal of $O$, -$O^\prime/M\cong\overline{O}$. - -In particular: $O^\prime$ is never noetherian.<|endoftext|> -TITLE: Is $SL(n,\mathbb{Z})$ a CAT(0) group? -QUESTION [11 upvotes]: Is it possible to find a CAT(0) space on which the matrix group $SL(n,\mathbb{Z})$ acts properly discontinuously and cocompactly? Note: when the cocompactness is dropped , it is possible. - -REPLY [24 votes]: If n=2, yes: it acts on its Bass-Serre tree. -If n>2, no: your group contains distorted elements, i.e. elements conjugated to a proper power of themselves (look at unipotent matrices). -Such an element will have zero displacement length, which is impossible for an infinite order element in a group acting discretely cocompactly. -For this and much more, see the monograph of Bridson and Haefliger. -For even more restrictions on SL_n actions, see Theorem 1.14 in: -Caprace-Monod, -Isometry groups of non-positively curved spaces: structure theory -Journal of Topology 2 No. 4 (2009), 661–700<|endoftext|> -TITLE: Determining the exceptional set in the theorem of Ax & Kochen -QUESTION [5 upvotes]: Ax & Kochen [1] proved that for every $d\in\mathbb{N}$ there exists a finite set $A(d)$ such that for every prime $p\not\in A(d),$ every homogeneous polynomial of degree $d$ over $\mathbb{Q}_p$ in at least $d^2+1$ variables has a nontrivial zero. - -Is there an effective procedure for determining $A(d)$? -For what values of $d$ is $A(d)$ known? Ax & Kochen mention that the special cases $d\in\{2,3,5,7,11\}$ were known but not if $A$ was known for those cases. - -[1] James Ax and Simon Kochen, "Diophantine problems over local fields I.", American Journal of Mathematics 87 (1965), pp. 605–630. -[2] Simon Kochen, "The model theory of local fields", Lecture Notes in Mathematics 499 (1975), pp. 384–425. - -REPLY [15 votes]: Scott Brown (Mem. AMS, 1978) gave a bound for the largest prime $p_0(d)$ lying in $A(d)$. So we know that for every $d\in {\Bbb N}$, one has -$$p_0(d)\le 2^{2^{2^{2^{2^{d^{11^{4d}}}}}}}.$$ -Good! In addition, one knows that $A(d)$ is empty for $d=1$ (no prizes), $d=2$ (classical) and $d=3$ (Demyanov and Lewis, independently, about 1950). For $d=5,7,11$ and no other values, there is work of Laxton and Lewis (pre-dating Ax and Kochen) which has been made effective more recently by Leep and Yeomans, Knapp, Heath-Brown and Wooley. Thus we know that $p_0(5)\le 13$ (Heath-Brown, 2010), and $p_0(7)\le 883$ and $p_0(11)\le 8053$ (Wooley, 2008).<|endoftext|> -TITLE: Continuous choice of Hahn-Banach extensions -QUESTION [6 upvotes]: QUESTION RETRACTED - My original argument was fundamentally mistaken (mixing up lower and upper semi-continuity). Sorry (and thanks for the useful comments) -I need, and (unless I am seriously mistaken) can prove, the following: -Let $E \subseteq F$ be an (isometric) inclusion of Banach spaces, and let $E^*_1$, $F^*_1$ denote the closed unit balls of their respective duals. Then the restriction map $F^*_1 \to E^*_1$ admits a weak$^\*$-continuous section from $E^*_1$ to $F^*_1$. -On the unit sphere this section gives you norm-preserving extensions. -This must be known - can anyone provide a reference? (am not a Banach space theorist) - -REPLY [3 votes]: After Phil's answer and the ensuing discussion, the remaining question can be formulated as: -Estimate $\lambda$ s.t. if $F\subset E$ and $E/F$ is finite dimensional, then every norm one linear operator $T$ from $F$ into a $C(K)$ space can be extended to an operator from $E$ into the $C(K)$ space which has norm at most $\lambda$. -I want to point out that $\lambda$ cannot be less than two. I think that it is known that $\lambda$ can be anything larger than two, but I did not find a reference after a quick search. I called Morry Zippin's attention to the problem; maybe he will comment. Or maybe this was proved after he worked on almost complementation; perhaps by Jesus Castillo and colleagues. I more or less thought through that $\lambda$ can be anything larger than three. -Before saying why $\lambda$ cannot be less than two, let's further reformulate the problem. The second dual of every $C(K)$ is $1$-injective, so the problem reduces to the case where $F$ is a $C(K)$ space and $T$ is the identity operator on $F$. That is, we need to estimate how well complemented a $C(K)$ must be in a superspace in which the $C(K)$ space has finite codimension. In fact, the superspace of $C(K)$ can be assumed to be in $C(K)^{**}$, but that is not particularly useful in getting the lower bound on $\lambda$. -Consider the $C(K)$ space $c$, the space of convergent sequences under the sup norm. (I'll treat the real case, but the complex case is only slightly more complicated.) The superspace $E$ is the span in $\ell_\infty$ of $c$ and the sequence $x$ defined by $x(n)= (-1)^n$. Let $P_n$ be the projection onto the span of the first $n$ unit vectors in $\ell_\infty$ and let $P$ be any projection from $E$ onto $c$. Write $Px = a\cdot 1 + y$ with $y$ in $c_0$ and where $1$ is the constant one function. Now for every $n$, the norm of $x-2P_n x$ is one, and $P(x- 2P_n x)=a 1 + y - 2P_n x$. Since $y$ is in $c_0$, letting $n\to \infty$ gives $\|P\| \ge |a|+2$. The choice $a=0$, $y=0$ of course gives a projection of norm two. -EDIT Nov. 29, 2011: Zippin's article in Handbook of the Geometry of Banach Spaces, vol. 2, contains information about almost complementation (differently named) including many open problems.<|endoftext|> -TITLE: étale cohomology with values in an abelian scheme is torsion? -QUESTION [6 upvotes]: Let $A/X$ be an abelian scheme. Is $H^n(X,A)$ torsion for $n > 0$? -Perhaps this can be proved analogously as Proposition IV.2.7 of Milne's Étale cohomology (where it is proved that the cohomological Brauer group of a quasi-compact scheme is torsion) exploiting the Kummer sequence. For integral affine schemes and the multiplicative group one can also prove this using that the Brauer group injects into the Brauer group of the fraction field, which is torsion as a Galois cohomology group. - -REPLY [3 votes]: In [Milne, Arithmetic Duality Theorems], II.5, it is proved that $H^r(U,\mathcal{A})$ is torsion for $U$ the ring of integers of a number field or a complete smooth curve over a finite field, respectively.<|endoftext|> -TITLE: When is the tensor product of two fields a field? -QUESTION [125 upvotes]: Consider two extension fields $K/k, L/k$ of a field $k$. - A frequent question is whether the tensor product ring $K\otimes_k L$ is a field. The answer is "no" and this answer is often justified by some particular case of the following result: -Proposition Given a strict field extension $k \subsetneq K$ , the tensor product $K\otimes_kK$ is not a field. -Proof The multiplication $m:K\otimes_kK\to K:x\otimes y \mapsto xy$ cannot be injective for dimension reason, hence it has a kernel which is a non-zero ideal of the ring $K\otimes_kK$ and thus that ring cannot be a field. -Corollary If the extensions $K/k, L/k$ contain finite subextensions $k\subsetneq K'\subset K, k \subsetneq L'\subset L$ which are $k$-isomorphic ( $K' \stackrel {k}{\simeq} L'$), then $K\otimes_k L$ is not a field. -The most powerful and beautiful tool in this context is Grothendieck's underrated result ( often attributed to Sharp who redicovered it ten years after Grothendieck! cf. this answer in math.stackexchange generalized ten years later by Sharp who suppressed Grothendieck's hypothesis that $K\otimes_k L$ should be noetherian): -Theorem (Grothendieck-Sharp ) The Krull dimension of the tensor product of the field extensions $K/k, L/k$ is given by the formula -$$ \dim_{\mathrm{Krull}}(K\otimes_k L) = \min(\operatorname{trdeg}_k(K),\operatorname{trdeg}_k(L)) $$ -This shows that we can only hope that $K\otimes_k L$ will be a field if at least one of the extensions $K,L$ is algebraic over $k$. An example where we do obtain a field is when the extension fields $K,L$ are finite dimensional over $k$ with relatively prime dimensions. -[To see this, embed $K$ and $L$ into an algebraic closure $\overline k$ of $k$ and notice that the canonical morphism $K\otimes_k L\to K\cdot L\subset \overline k$ is an isomorphism because it is surjective and because $K\cdot L$ has the same dimension as $K\otimes_k L$ by the relative primeness assertion] -A fairly general criterion for obtaining a field is the following. -A sufficient condition The tensor product $K\otimes_k L$ is a field if the three conditions below simultaneously hold: - -At least one of $K,L$ is algebraic over $k$. -At least one of $K,L$ is primary over $k$ -At least one of $K,L$ is separable over $k$ -Proof -The ring $K\otimes_k L$ is zero-dimensional by 1) and Grothendieck's formula. -Once divided by its nilpotent radical it is a domain by 2). -However, by 3), its nilpotent radical is zero. -So $K\otimes_k L$ is a zero-dimensional domain, hence a field. - -[Reminder: a field extension $E/k$ is primary if the algebraic closure of $k$ in $E$ is purely inseparable over $k$. In that case for any field extension $F/k$ the quotient $E\otimes_k F/Nil (E\otimes_k F)$ is a domain. In other words $Spec(E\otimes_k F) $ is irreducible.] -I feel that all these results are a little fragmentary and my not very precise question is , as you have guessed : -Question Is there a general procedure for deciding whether the tensor product $K \otimes_k L$ of two field extensions is a field? -Bibliography Grothendieck's result is to be found in EGA IV, Quatrième partie, page 349 , Remarque (4.2.1.4). This is in the Errata et Addenda to the volume! -Edit Since linearly disjointness keeps getting mentioned in the comments, let me insist that it makes no sense to say that $K$ and $L$ are linearly disjoint unless they are provided with embeddings into an extension $E$ of $k$. -For example take $K=L=k(x)$ ($x$ an indeterminate over $k$) and consider the extension $k \subset E=k(y,z)$, the function field in two indeterminates over $k$. - If you embed $K$ (resp. $L$) into $E$ by sending $x\mapsto y$ (resp.$x\mapsto z$), the images will be linearly disjoint. -However if you embed $K$ (resp. $L$) into $E$ by sending $x\mapsto y$ (resp. $x \mapsto y$), the images will be equal and certainly not linearly disjoint. -However the $k$-algebra $k(x)\otimes_k k(x) $ does not care about all these embeddings: Grothendieck has decreed that it is not a field, and that's it. -(Our friend Pete Clark has a section on these questions in his extremely well-written online notes, page 65. According to Pete, that section was inspired by an exchange he had concerning a question asked by our other friend Andrew Critch ) -New edit: Is all this a real problem? Since we know so many conditions ensuring that $K\otimes_k L$ is a field and so many conditions ensuring that it isn't, I wonder if someone could come up with a tensor product of extensions $K\otimes_k L$ for which MO users couldn't (immediately) say whether it is a field or not. -I would be very happy to consider such a challenge as an answer, to upvote it and possibly to accept it. -Edit (April 24th, 2016):Apologies to Sharp -Due to EGA's abstruse cross-reference system I had missed that Grothendieck's formula is proved by him only under the supplementary hypothesis that $K\otimes_k L$ is noetherian. -It is indeed Sharp who first proved that formula in complete generality, without any noetherian hypothesis: - -Rodney Y. Sharp, The Dimension of the Tensor Product of Two Field Extensions, Bulletin of the London Mathematical Society 9 Issue 1 (1977) pp 42–48, doi:10.1112/blms/9.1.42 - -REPLY [33 votes]: This is the most complete treatment I could come up with. Let $k \subseteq K^{\operatorname{sep}} \subseteq K^{\operatorname{alg}} \subseteq K$ and $k \subseteq L^{\operatorname{sep}} \subseteq L^{\operatorname{alg}} \subseteq L$ be the separable algebraic and algebraic closures of $k$ in $K$ and $L$. The result is the following. - -Theorem. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then - -$K \otimes_k L$ is irreducible if and only if $k \subseteq K^{\operatorname{sep}}$ and $k \subseteq L^{\operatorname{sep}}$ are linearly disjoint with respect to one (equivalently, every) choice of embedding into $\bar k$. -$K \otimes_k L$ is reduced if and only if the intersection of $\ker(\Omega_k \otimes_k K \to \Omega_K) \otimes_k L$ and $K \otimes_k \ker(\Omega_k \otimes_k L \to \Omega_L)$ inside $\Omega_k \otimes_k (K \otimes_k L)$ is generically trivial. -$K \otimes_k L$ has dimension $0$ if and only if $k \subseteq K$ or $k \subseteq L$ is algebraic. - -In particular, $K \otimes_k L$ is a field if and only if all three criteria hold. - -Here we say a property holds generically if it holds for the localisation at all minimal primes. The second criterion is not very nice in general, but if $K \otimes_k L$ has dimension $0$ we can remove the word 'generically', so we get a rather clean criterion. This applies in particular in the case the OP is interested in, for $K \otimes_k L$ to be a field. On the other hand, in Example 3 we show that 'generically' in part 2 of the theorem cannot be removed, even in the case $K = L$. -In Proposition 3 below we give a more conceptual (but more technical) criterion equivalent to the one stated here. -Statements 1, 2, and 3 of the theorem generalise without difficulty to finite tensor products of fields. The statement is the 'obvious generalisation' of the version for binary tensor products. For example, in 1 we need that $K_1^{\operatorname{sep}} \otimes_k \ldots \otimes_k K_n^{\operatorname{sep}} \to \bar k$ is injective for one (equivalently, any) choice of embeddings into $\bar k$. In 2, we need that generically the sum of the subspaces is direct. In 3, we need that at most one of the extensions has a transcendental part. -Remark. We will use the algebro-geometric properties of field extensions: a field extension $k \subseteq \ell$ is - -algebraic if and only if it is integral (obvious); -separable if and only if it has geometrically reduced fibres (Tag 030W); -separable algebraic if and only if it is integral with geometrically reduced fibres (combine 1 and 2); -purely inseparable (algebraic) if and only if it is radicial (by definition, see Tag 01S3); -primary if and only if it has geometrically irreducible fibres (partially explained in Tag 037Q); -regular if and only if it has geometrically integral fibres (combine 2 and 5). - -This is useful because not all rings we encounter are fields, and the algebro-geometric properties are preserved by base change. In fact all these properties descend under faithfully flat morphisms (see [EGA IV$_2$, Prop 2.6.1(v)] for radicial, [EGA IV$_2$, Prop 2.7.1(xvii)] for integral, and obvious for all other properties), so $k \to K$ has one of the properties above if and only if $L \to K \otimes L$ does. -Below we will treat parts 1, 2, and 3 of the theorem separately; see Lemma 1, Proposition 3, and Lemma 4. Also have a look at examples 1, 2, 3, and 4, because they provide some insight into the types of behaviours you can expect. - - -1. Irreducibility - - -The precise version of part 1 of the theorem is as follows. -Lemma 1. Let $k \subseteq K$ and $k \subseteq L$ be field extensions, and let $\bar k$ be a separable algebraic closure of $k$. Then the following are equivalent: - -$K \otimes_k L$ is irreducible; -$K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is irreducible; -$K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is a field; -for every pair of embeddings $i \colon K^{\operatorname{sep}} \to \bar k$ and $j \colon L^{\operatorname{sep}} \to \bar k$, the subfields $i(K)$ and $j(L)$ of $\bar k$ are linearly disjoint; -for one pair of embeddings $i \colon K^{\operatorname{sep}} \to \bar k$ and $j \colon L^{\operatorname{sep}} \to \bar k$, the subfields $i(K^{\operatorname{sep}})$ and $j(L^{\operatorname{sep}})$ of $\bar k$ are linearly disjoint. - -Proof. Note that $K^{\operatorname{sep}} \to K$ and $L^{\operatorname{sep}} \to L$ are primary, hence flat with geometrically irreducible fibres. Then the same goes for $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}} \to K \otimes_k L$, so (1) $\Leftrightarrow$ (2) follows from [EGA IV$_2$, Prop. 4.5.7] (or modify the proof of Tag 038F). Since $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is reduced of dimension $0$, it is a field if and only if it is irreducible, proving (2) $\Leftrightarrow$ (3). -If $i(K^{\operatorname{sep}})$ and $j(L^{\operatorname{sep}})$ are linearly disjoint for one choice of $i$ and $j$, then -$$i \otimes j \colon K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}} \to \bar k$$ -is injective, so $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is a field. The converse is also clear, showing (3) $\Leftrightarrow$ (4) $\Leftrightarrow$ (5). $\square$ -We will generalise properties 4 and 5 in Further remarks and examples to arbitrary field extensions. In the separable algebraic case, it has a reinterpretation in terms of Galois theory: if $i \colon K \to \bar k$ and $j \colon L \to \bar k$ is a choice of embeddings, then we get closed subgroups $H_K$ and $H_L$ of $G_k = \operatorname{Gal}(\bar k/k)$. The condition on linear disjointness then means that for any open subgroups $U_K \supseteq H_K$ and $U_L \supseteq H_L$, we have $[G_k : U_K \cap U_L] = [G_k : U_K] \cdot [G_k : U_L]$. -Remark. One is tempted to believe that $K$ and $L$ are linearly disjoint if and only if $i(K) \cap j(L) = k$ for every pair of embeddings $i \colon K \to \bar k$ and $j \colon L \to \bar k$, but this is not true: -Example 1. Let $k \subseteq M$ be an $S_5$-extension; for example $k = \mathbf C(\sigma_1,\sigma_2,\sigma_3,\sigma_4.\sigma_5)$ and $M = \mathbf C(x_1, x_2, x_3, x_4,x_5)$, where -$$\prod_{i=1}^5(X-x_i) = X^5 - \sigma_1 X^4 + \sigma_2 X^3 - \sigma_3 X^2 + \sigma_4 X - \sigma_5.$$ -Let $K$ be the fixed field of $(12)$ and $L$ the fixed field of $(12345)$. -Then $K \cap L$ is fixed by $(12)$ and $(12345)$, hence equals $k$. What's more, any conjugate of $K$ is the fixed field of a transposition, and any conjugate of $L$ is the fixed field of a $5$-cycle, hence their intersection is still $k$ since any pair of a transposition and a $5$-cycle generate $S_5$. However, $K$ and $L$ cannot be linearly disjoint, because $[K:k] = 60$ and $[L:k] = 24$, but they're both contained in a degree $120$ extension of $k$. - - -2. Reducedness - - -We will use repeatedly that if $k \subseteq K$ and $k \subseteq L$ are separable, then they are geometrically reduced, hence $K \otimes_k L$ is reduced by Tag 034N. In particular, if $\operatorname{char} k = 0$ there is nothing to discuss, so assume $\operatorname{char} k = p > 0$. -For any field $k$ of characteristic $p$, write $\Omega_k = \Omega_{k/\mathbf Z} = \Omega_{k/\mathbf F_p}$ for the module of absolute differentials. Although my criterion was originally based on the idea of $p$-bases (Tag 07P0), I was ultimately unable to prove it in that language. The proof uses the (naive) cotangent complex $\mathbf L_{B/A}$ (see Tags 00S0 and 08R6) instead, where again we write $\mathbf L_A$ for $\mathbf L_{A/\mathbf F_p}$. -Lemma 2. Let $k \subseteq \ell$ be a field extension in characteristic $p > 0$, and let $A$ be a successive extension -$$\ell = A_0 \subseteq A_1 \subseteq \ldots \subseteq A_r = A$$ -where $A_i = A_{i-1}[X_i]/(X_i^p - x_i)$ for some $x_i \in A_{i-1}$. Then the following are equivalent: - -$A$ is a field; -$H_1\mathbf L_A = 0$; -the boundary map $d \colon H_1\mathbf L_{A/k} \to \Omega_k \otimes_k A$ is injective. -the boundary map $d \colon H_1\mathbf L_{A/\ell} \to \Omega_\ell \otimes_\ell A$ is injective. - -Proof. For any field $F$ of characteristic $p$, the extension $\mathbf F_p \subseteq F$ is separable as $\mathbf F_p$ is perfect. Therefore, we have $H_1\mathbf L_F = 0$ (see for example Tag 07E5, but it also follows from Tag 08Q1, localisation, and the étale case of Tag 0D0M). Thus $H_1\mathbf L_A = 0$ if $A$ is a field, showing (1) $\Rightarrow$ (2). -If $A$ is not a field, then let $i$ be the largest index such that $A_i$ is a field, so by the above $H_1 \mathbf L_{A_i} = 0$. Then $A_{i+1}$ is not a field, meaning that $x_{i+1}$ is a $p$-th power in $A_i$. Then the long exact cotangent sequence (Tag 00S2) for $\mathbf F_p \subseteq A_i \subseteq A$ reads -$$0 \to H_1 \mathbf L_A \to H_1 \mathbf L_{A/A_i} \stackrel d\to \Omega_{A_i} \underset {A_i}\otimes A \to \ldots.$$ -Hence $H_1 \mathbf L_A \neq 0$ since $X_{i+1}^p - x_{i+1}$ maps to $-dx_{i+1} = 0$ under $d$, showing (2) $\Rightarrow$ (1). -Finally, the long exact cotangent sequence for $\mathbf F_p \subseteq k \subseteq A$ reads -$$0 \to H_1\mathbf L_A \to H_1\mathbf L_{A/k} \stackrel d\to \Omega_k \underset k\otimes A \to \Omega_A \to \Omega_{A/k} \to 0.$$ -This shows (2) $\Leftrightarrow$ (3), and replacing $k$ by $\ell$ gives (2) $\Leftrightarrow$ (4). $\square$ -Proposition 3. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then the following are equivalent: - -$K \otimes_k L$ is reduced; -$H_1 \mathbf L_{K \otimes_k L}$ is generically trivial; -the intersection of $\ker(\Omega_k \otimes_k K \to \Omega_K) \otimes_k L$ and $K \otimes_k \ker(\Omega_k \otimes_k L \to \Omega_L)$ (in $\Omega_k \otimes_k (K \otimes_k L)$) is generically trivial. - -Proof. We can reduce to the case that $k \subseteq K$ and $k \subseteq L$ are finitely generated, since tensor product, formation of $\Omega$ and $\mathbf L$, and computation of kernel commute with filtered colimits, and nilpotents are defined over a finitely generated subring. -By 'baby Serre's criterion' (Tag 031R), we need to check that $K \otimes_k L$ is $S_1$ and $R_0$. But $S_1$ is automatic, for example by Tag 0339, so it suffices to check $R_0$. Thus, we see that $K \otimes_k L$ is reduced if and only if every localisation $A$ of $K \otimes_k L$ at a minimal prime $\mathfrak p$ is a field. -We claim that every such $A$ is of the form described in Lemma 2. Indeed, let $k \subseteq K' \subseteq K$ be a subextension such that $k \subseteq K'$ is separable and $K' \subseteq K$ is purely inseparable; for example by taking $K'$ to be the elements separable over a given transcendence basis. Then $K' \otimes_k L$ is reduced and $K' \otimes_k L \to K \otimes_k L$ is radicial and faithfully flat (in particular a bijection on irreducible components by Tag 01S4). -If $\mathfrak q$ is the preimage of $\mathfrak p$ in $K' \otimes_k L$ and $\ell$ is the localisation of $K' \otimes_k L$ at $\mathfrak q$, then $\ell$ is a field since $\mathfrak q$ is minimal and $K' \otimes_k L$ reduced. The map $\ell \to A$ is a map as described in Lemma 2, since $K' \subseteq K$ is of that form. Thus from Lemma 2 we see that $A$ is a field if and only if $H_1 \mathbf L_A = 0$. Since formation of $\mathbf L_A$ commutes with localisation by Tag 00S7, we see that $K \otimes_k L$ is reduced if and only if $H_1\mathbf L_{K \otimes_k L}$ is generically trivial, proving (1) $\Leftrightarrow$ (2). -For (2) $\Leftrightarrow$ (3), the exact sequences -\begin{alignat*}{1} -0 & \to H_1\mathbf L_{K/k} & \stackrel d\to \Omega_k \underset k\otimes K & \to \Omega_K & \to \Omega_{K/k} & \to 0,\\ -0 & \to H_1\mathbf L_{L/k} & \stackrel d\to \Omega_k \underset k\otimes L & \to \Omega_L & \to \Omega_{L/k} & \to 0. -\end{alignat*} -for $\mathbf F_p \subseteq k \subseteq K$ and $\mathbf F_p \subseteq k \subseteq L$ identify $H_1\mathbf L_{K/k}$ and $H_1\mathbf L_{L/k}$ as the kernels of the maps on differentials. Choosing a presentation of $K \otimes_k L$ by tensoring presentations for $K$ and $L$ over $k$, we see that -$$\mathbf L_{K \otimes_k L/k} \simeq \left( \mathbf L_{K/k} \underset k\otimes L \right) \oplus \left( \mathbf L_{L/k} \underset k\otimes K \right),$$ -and the connecting map $H_1\mathbf L_{K \otimes_k L} \to \Omega_k \otimes_k (K \otimes_k L)$ is given by $(d,d)$ with respect to this decomposition. In particular, the connecting map for $K \otimes_k L$ is generically injective if and only if the images of the connecting maps for $K$ and $L$ have generically trivial intersection. $\square$ -Remark. However, it is not true that $K \otimes_k L$ is a field if and only if there are no elements that become a $p$-th power in both; see Example 2 below (even in the finite purely inseparable case). In fact, $K \otimes_k K$ may be reduced even if $k \subseteq K$ is inseparable; see Example 3 below (a transcendental primary inseparable extension). -On the other hand, this last phenomenon does not happen for algebraic inseparable extensions $k \subseteq K$, since $k[x]/f \otimes_k k[x]/f \subseteq K \otimes_k K$ clearly picks up nilpotents if $f$ is inseparable. See also Example 4 at Further remarks and examples for an example of the interplay between the separable algebraic part and the purely inseparable part. -Example 2. Let $k = \mathbf F_p(x,y,z)$, let $K = k[u,v]/(u^{p^2}-z, v^p-xu^p-y)$ be the field adjoining a $p^2$-th root $u$ of $z$ and a $p^2$-th root $v$ of $x^pz+y^p$ (this example is due to Sweedler, as far as I can tell), and let $L = k(x^{1/p}, y^{1/p})$. Then -$$K \underset k\otimes L = \frac{\mathbf F_p(x^{1/p}, y^{1/p}, z^{1/p^2})[v]}{(v^p-xz^{1/p}-y)} \cong \frac{\mathbf F_p(x^{1/p}, y^{1/p}, z^{1/p^2})[X]}{(X^p)},$$ -since $xz^{1/p}+y$ is a $p$-th power in $\mathbf F_p(x^{1/p}, y^{1/p}, z^{1/p^2})$. Thus, $K \otimes_k L$ is not reduced. -The elements in $k$ that become $p$-th powers in $K$ are given by -$$k \cap K^p = \mathbf F_p(x^p,y^p,z).\label{Eq become p-th power}\tag{1}$$ -Indeed, the inclusion $\supseteq$ is obvious. For the converse, consider the chain of field extensions -$$K^{p^2} = k \cap K^{p^2} \subseteq k \cap K^p \subseteq k \cap K = k.$$ -We have $[k:K^{p^2}] = [k:k^{p^2}]/[K^{p^2}:k^{p^2}] = (\operatorname{tr. deg.} k)^2/[K:k] = p^3$, and $x^p$ is in $k \cap K^p$ but not $k \cap K^{p^2}$ since $dx \neq 0 \in \Omega_K$ (see computation below). Thus, $[k:k \cap K^p] \leq p^2$, which gives equality in (\ref{Eq become p-th power}) by a dimension count. -By a similar (but easier) computation, we get -$$k \cap L^p = \mathbf F_p(x,y,z^p).$$ -Thus, the only elements in $k$ that become $p$-th powers in both $K$ and $L$ are already $p$-th powers in $k$ to begin with. -What's going on is that $\ker(\Omega_k \otimes_k K \to \Omega_K)$ is not defined over $k$, and a fortiori cannot be spanned by elements of the form $dx_i$ for $x_i \in k$. Indeed, $\Omega_k = k \cdot dx \oplus k \cdot dy \oplus k \cdot dz$, and one computes -$$\Omega_K = \frac{K \cdot dx \oplus K \cdot dy \oplus K \cdot du \oplus K \cdot dv}{u^p\ dx + dy}.$$ -The kernel of $\Omega_k \otimes_k K \to \Omega_K$ is generated by $dz$ and $u^p\ dx + dy$, which is not defined over $k$. -Example 3. Let $k = \mathbf F_p(s,t)$, and let $K$ be the fraction field of $k[x,y]/(sx^p + ty^p - 1)$ (this example is due to Mac Lane, according to MO user nfdc23). Then $K$ is geometrically irreducible since -$$\bar k[x,y]/(sx^p + ty^p - 1) = \bar k[x,y]/(ux+vy-1)^p,$$ -where $u^p = s$ and $v^p = t$, which is a thickening of $\bar k[x,y]/(ux+vy-1) = \bar k[x]$. In particular, $k \subseteq K$ is primary, i.e. $k$ is algebraically closed in $K$. -The kernel of $\Omega_k \otimes_k K \to \Omega_K$ is generated by $df = x^p\ ds + y^p\ dt$, where $f = sx^p + ty^p - 1$. Again this is not defined over $k$, and in this case not even over an algebraic extension of $k$. -If we take two copies of $K$, then the elements -\begin{align*} -df_1 &= x_1^p\ ds + y_1^p\ dt\\ -df_2 &= x_2^p\ ds + y_2^p\ dt -\end{align*} -agree at the locus $V((x_1y_2-y_1x_2)^p)$, which is a strict closed subscheme. Thus the intersection of the kernels is generically trivial (since $K$ is the function field of a regular but not geometrically regular curve over $k$, so the diagonal is not an irreducible component), so $K \otimes_k K$ is reduced. -(In a previous version of this answer, I erroneously stated a version of the criterion for reducedness that did not take this type of example into account.) -On the other hand, $K \otimes_k K \otimes_k K$ cannot be reduced, because we get $3$ relations $df_1$, $df_2$, and $df_3$ involving only $2$ variables $ds$ and $dt$. - - -3. Zero-dimensionality - - -Lemma 4. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then $K \otimes_k L$ is $0$-dimensional if and only if one of $k \subseteq K$ and $k \subseteq L$ is algebraic. -Proof. Immediate by Grothendieck–Sharp. -This can actually be seen directly as well: clearly the tensor product is $0$-dimensional if one of $k \subseteq K$ and $k \subseteq L$ is algebraic. Conversely, if $K$ and $L$ both contain an element transcendental over $k$, then this gives embeddings $k \subseteq k(x) \subseteq K$ and $k \subseteq k(x) \subseteq L$. View $K$ and $L$ as extensions of $k(x)$ and choose $k(x)$-linear embeddings $i \colon K \to \Omega$ and $j \colon L \to \Omega$ into an algebraically closed extension $k(x) \subseteq \Omega$. Then $x \otimes 1 - 1 \otimes x$ is in the kernel $\mathfrak p$ of $i \otimes j \colon K \otimes_k L \to \Omega$. -The maps $k(x) \to K$ and $k(x) \to L$ are flat, hence the same goes for $k(x) \otimes_k k(x) \to K \otimes_k L$. The preimage of $\mathfrak p$ under this map contains $x \otimes 1 - 1 \otimes x$, hence is a nonzero ideal. But $k(x) \otimes_k k(x)$ is a domain, so going down for flat morphisms (see Tag 00HS) shows that $\mathfrak p$ cannot be minimal. Hence, $K \otimes_k L$ has positive dimension. $\square$ - - -Further remarks and examples - - -Here is an interesting example of the interplay between the separable and purely inseparable parts, in light of statement 2 of the precise version of the theorem (note that already the separable part disqualifies this example from producing a field, but the point is only the reducedness statement now): -Example 4. Let $k = \mathbf F_p(\sigma_1,\sigma_2)$, and consider the extension $M = \mathbf F_p(x_1,x_2)$ given by $(X-x_1)(X-x_2) = X^2 - \sigma_1X + \sigma_2$. This is a $\mathbf Z/2$-Galois cover (even if $p = 2$, but feel free to assume $p > 2$). Let $K = \mathbf F_p(x_1^{1/p},x_2)$ and $L = \mathbf F_p(x_1,x_2^{1/p})$. -Then $K^{\operatorname{sep}} = L^{\operatorname{sep}} = M$, and $M \otimes_k M \cong M^{\mathbf Z/2}$ given by $m_1 \otimes m_2 \mapsto (m_1\sigma(m_2))_{\sigma \in \mathbf Z_2}$. That is, on the factor $M_0$ the inclusions are both given by $x_i \mapsto x_i$, and on the factor $M_1$ the inclusion from $L^{\operatorname{sep}}$ swaps $x_1$ and $x_2$. -Then $K \otimes_k L^{\operatorname{sep}} = M_0(x_1^{1/p}) \times M_1(x_1^{1/p})$, and $K^{\operatorname{sep}} \otimes_k L \cong M_0(x_2^{1/p}) \times M_1(x_1^{1/p})$ since the embedding $L^{\operatorname{sep}} \to M_1$ sends $x_2$ to $x_1$. We see that $K \otimes_k L$ is not reduced, because it contains $M_1(x_1^{1/p}) \otimes_{M_1} M_1(x_1^{1/p})$. -In terms of differentials, the relations $\sigma_1 = x_1+x_2$ and $\sigma_2 = x_1x_2$ give $d\sigma_1 = dx_1 + dx_2$ and $d\sigma_2 = x_1\ dx_2 + x_2\ dx_1$. The kernel of $\Omega_k \otimes_k K \to \Omega_K$ is spanned by -$$dx_1 = \frac{d\sigma_2 - x_1\ d\sigma_1}{x_2-x_1},$$ -and the opposite happens for the kernel of $\Omega_k \otimes_k L \to \Omega_L$. These subspaces of $\Omega_k \otimes_k K$ and $\Omega_k \otimes_k L$ are not defined over $k$, so their intersection does not show up over $k$. In the factor $M_0$ of $K \otimes_k L$ they are linearly independent (note that indeed $M_0(x_1^{1/p}) \otimes_{M_0} M_0(x_2^{1/p})$ is a field), but in the other factor $M_1$ they give the same subspace because $x_2$ and $x_1$ are swapped in $L^{\operatorname{sep}} \to M_1$. -We end with the generalisation of Lemma 1 to arbitrary field extensions: -Lemma. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then $K \otimes_k L$ is a field if and only if for every pair of embeddings $i \colon K \to \Omega$ and $j \colon L \to \Omega$ into an algebraically closed field $\Omega$, the images $i(K)$ and $j(L)$ are linearly disjoint (i.e. $i \otimes j \colon K \otimes_k L \to \Omega$ is injective). If $K$ and $L$ are algebraic over $k$, it suffices to take $\Omega$ an algebraic closure of $k$, and it suffices to take one pair $(i,j)$ of embeddings. -Proof. Since -$$\operatorname{Hom}_k(K,\Omega) \times \operatorname{Hom}_k(L,\Omega) = \operatorname{Hom}_k\left(K \underset k\otimes L, \Omega\right),$$ -the result follows since a commutative ring $R$ is a field if and only if every map $R \to \Omega$ to a(n algebraically closed) field is injective. -This proves the first statement, and the final statement is Lemma 1. $\square$<|endoftext|> -TITLE: Grothendieck's manuscript on topology -QUESTION [53 upvotes]: Edit: Infos on the current state by Lieven Le Bruyn: http://www.neverendingbooks.org/grothendiecks-gribouillis -Edit: Just in case anyone still thinks that Grothendieck's unpublished manuscripts are (by his letter) entirely out of sight: Declared as "national treasure", they seem to be in principle accessible (+ Thanks to Jonathan Chiche who points - see his comment below - that it is not so clear if that idea was made a reality by now): http://www.liberation.fr/sciences/2012/07/01/le-tresor-oublie-du-genie-des-maths_830399 -On p. 185 - 186 of the 3rd volume of Winfried Scharlau's Grothendieck biography, a handwritten text from 1986 by Grothendieck on foundations of topology, different from the concepts of topoi or tame topology, is shortly described. Scharlau doubts if it could be turned into a readable text, but perhaps someone knows the texts and has ideas about it? -Edit: Acc. to Winfried Scharlau's book, Grothendieck described his work in a letter to Jun-Ichi Yamashita as: "some altogether different foundations of 'topology', starting with the 'geometrical objects' or 'figures', rather than starting with a set of 'points' and some kind of notion of 'limit' or (equivalently) 'neighbourhoods'. Like the language of topoi (and unlike 'tame topology'), it is a kind of topology 'without points' - a direct approach to 'shape'. ... appropriate for dealing with finite spaces... the mathematics of infinity are just a way of approximating an understanding of finite agregates, whose structures seem too elusive or too hopelessly intricate for a more direct understanding (at least it has been until now)." Scharlau gives a copy of one page of the manuscript (at p. 188) and obviously has a copy of the complete text and remarks (on p. 199) that Grothendieck wrote a in 1983 letter about that theme to Z. Mebkhout. -Edit: In the meantime I could read a letter by Grothendieck about that, a summary: He started thinking from time to time about that ca. in the mid-1970's, the motivation was roughly that dissatisfaction with the usual topology which he expressed in the Esquisse, and looking at stratifications of moduli-"spaces" is his new starting point. Maybe, but not expressed in the letter or the Esquisse, the ubiquity of moduli problems in algebraic geometry (e.g. expressed in the beginning of Lafforgue's text ) is an other motivation. He describes his guiding ideas on new foundations of topology as more complicated than the guiding ideas behind the new foundations of algebraic geometry of EGA, SGA. A main test of his concepts now would be a "Dévissage"-theorem on "startified obstructions"(?) in terms of equivalences of categories. He has a precise heuristic formulation of that which helped him to find a "dévissage" corresponding to Teichmueller groups (probably what now is called "Grothendieck-Teichmueller group"?) which are related to stratifications "at infinity" of Deligne-Mumford moduli stacks. - -REPLY [10 votes]: In the light of past events ("Les Archives Grothendieck"), we now have: -Vers une Géométrie des Formes (1986) - -I. Vers une géométrie des formes (topologiques) : notes manuscrites (05/06/1986). -Cote n° 156-1 (26 p.) -II. Réalisations topologiques des réseaux : notes manuscrites (06/06/1986). -Cote n° 156-2 (18 p.) -III. Réseaux via découpages : notes manuscrites (08/06/1986). -Cote n° 156-3 (40 p.) -IV. Analysis situs (première mouture) : notes manuscrites (10/06/1986). -Cote n° 156-4 (88 p.) -V. Algèbre des figures : notes manuscrites (14/06/1986). -Cote n° 156-5 (48 p.) -VI. Analysis situs (deuxième mouture) : notes manuscrites (18-20/06/1986). -Cote n° 156-6 (93 p.) -VII. Analysis situs (troisième mouture) : notes manuscrites (23-26/06/1986). -Cote n° 156-7 (113 p.) -VIII. Analysis situs (quatrième mouture) : notes manuscrites (26/06-04/07/1986). -Cote n° 156-8 (126 p.) -IX et IX bis. [Ateliers] : notes manuscrites (05-15/07/1986). -Cote n° 156-9 (139 p.) - - -Project of transcription.<|endoftext|> -TITLE: When do tensor products of C*-algebras commute with colimits? -QUESTION [10 upvotes]: Let $I$ be a filtered poset, which you should think of as being huge. Let $A_i$ be an $I$-diagram of $C^{\star}$-algebras and let $A$ be the colimit of this diagram; if necessary, we can also assume that all structure maps $A_i \rightarrow A_j$ are inclusions. Given a nuclear $C^{\star}$-algebra $N$, is it then true that tensoring with $N$ commutes with the colimit, i.e. that $A \otimes N$ is the colimit of $A_i \otimes N$? In particular, I am interested in the case $N = C(X)$ for some compact space $X$. -I expect this to be true (and seem to have a proof for C(X) at least), but have not been able to find anything in the literature, so a reference would be welcome. - -REPLY [6 votes]: If $N$ is exact and the tensor products are minimal then $A\otimes N$ is the colimit of the $A_i\otimes N$'s. Say the connecting maps are $\phi_{i,j}$. Then to check that $A\otimes N$ -is the colimit of $\{A_i\otimes N,\phi_{i,j}\otimes\mathrm{id}_N\}$ two properties must be verified: -(1) the union of the ranges of the maps $\phi_{i,\infty}\otimes \mathrm{id}_N$ is dense in $A\otimes N$, -(2) for each $i$, $\ker (\phi_{i,\infty}\otimes \mathrm{id}_N)=\overline{\bigcup_{j>i} \ker({\phi_{i,j}\otimes \mathrm{id}_N})}$. -The first property is straightforward, since the span of the elementary tensors $\phi_{i,\infty}(a)\otimes n$ is dense. The second property follows from the fact that $\ker(\phi_{i,\infty}\otimes \mathrm{id}_N)=\ker(\phi_{i,\infty})\otimes N$ and the correspoding property (2) for the colimit of the $A_i$'s. -That $\ker(\phi\otimes \mathrm{id}_N)=\ker(\phi)\otimes N$ follows, if $\phi$ is surjective, by exactness of $N$, if $\phi$ is injective, by the minimality of $\otimes$ -(the minimal tensor product behaves well with respect to inclusions), and the general case is a composite of those two.<|endoftext|> -TITLE: Generalizing Representation Theory of Finite Groups to Module Theory -QUESTION [8 upvotes]: My question is essentially this: which parts of the representation theory of finite groups are really just applications of module theory, and which are not? Here is an example of each case. Induction of representations (at least for finite groups) is simply extension of scalars for modules over the group ring. Thus the definition of induction and its basic properties can be equally well studied for suitable class of modules with no mention of groups or representations. On the other hand, I have heard Mackey's irreducibility criterion listed as a classic example of a theorem that has no "good" module theoretic generalization because it involves double cosets, and it is very difficult to generalize double coset structure to modules. -Based on these examples I can break my question up into two, more specific questions: -1) What are some other examples of representation theoretic theorems that do or don't generalize well to a module theoretic setting? -2) When, in your opinion, is it useful for a representation theorist to work with the module theory, and when does this merely introduce unnecessary abstraction. - -REPLY [5 votes]: I would say Clifford's theorem is a group theoretic result since it involved restricting representations to normal subgroups, which doesn't make sense for algebras. A number of theorems about the character table are at least not so natural in the module theory language. The fact that the dimensions of simple modules divide the order of the group and related algebraic integer results are group specific. -Things about faithful representations are group specific (or at least Hopf algebra specific) like that the tensor powers of any faithful rep contains all simples as constituents.<|endoftext|> -TITLE: Is there a standard notation for a "shift space" in functional analysis? -QUESTION [5 upvotes]: I'm writing up some notes on the nLab about things like embedding spaces and infinite spheres and similar things (can't link to them yet as I haven't put them up yet). One aspect that crops up time and time again is the contractibility of some big space, such as an infinite sphere, and this almost always boils down to some special property of whatever topological vector space is sitting in the background. -This special property is the existence of a "shift map" which acts pretty much like the obvious shift map on a sequence space. So I'm going to refer often to pairs $(V,S)$ where $V$ is a locally convex topological vector space and $S \colon V \to V$ is a "shift map". A little more precisely, we want to have an isomorphism $V \cong V \oplus \mathbb{R}$, so that $S \colon V \to V$ is the inclusion of the first factor, with certain properties, the main one being that $\bigcap S^k V = \{0\}$. The obvious notation is that $(V,S)$ is a shift space, and that $V$ is a shiftable space, but if there's an already existent notation then I should use that. -So my question is that: is there a standard notation for any of these concepts? The map itself, the space that admits the map, and the pair. -A closely related concept that I'll also use a bit could be termed a split space. This would be a locally convex topological vector space $V$ with an isomorphism $V \cong V \oplus V$. So: same question for that. -Edit: As Bill Johnson hasn't heard of these, I've written the relevant pages. It may be that I've included some detail there that I didn't put here. If any further information comes to light, I'll edit the pages accordingly. - -Shift space and split space (this defines them) -Embedding space (this uses them) - -REPLY [2 votes]: I don't know of any standard notation for such things, Andrew. $V\oplus V$ is often called the square of $V$, so for the second one I would just say that $V$ is isomorphic to its square. -Being shiftable looks much stronger for separable Banach spaces than being isomorphic to hyperplanes. Is it really stronger? Is $\ell_p \oplus \ell_r$ shiftable when $p\not= r$?<|endoftext|> -TITLE: Ordinals and complexity classes -QUESTION [10 upvotes]: What is the least recursive ordinal $\alpha$ such that there is no algorithm in complexity class $\mathsf{P}$ which implements a well-ordering of $\mathbb{N}$ with order type $\alpha$? (where the size of input is the total number of digits in numbers being compared) -Is it true that no well-ordering of $\mathbb{N}$ with order type $>\alpha$ can be implemented using an algorithm in $\mathsf{P}$? -Has connection between ordinals and complexity classes been studied? Can you recommend any books or papers related to this topic? - -REPLY [3 votes]: You may want to check "Dynamic Ordinal Analysis" by Arnold Beckmann which is an attempt to define a finer notion to classical ordinals that can be used ti distinguish between complexity classes.<|endoftext|> -TITLE: Burnside problem for hyperbolic groups? -QUESTION [5 upvotes]: Let $\Gamma$ be a nonelementary hyperbolic group. Then is it true that $\Gamma^{p}:=<\gamma^{p}| \gamma \in \Gamma>$ is a finite index subgroup of $\Gamma$? Here $p$ is a prime number. What is known about this problem? What can we say when $p$ is just an odd number(not necessarily prime) or just a postive integer? - -REPLY [14 votes]: A. Yu. Olshanskii in the paper "Periodic quotient groups of hyperbolic groups." ((Russian) Mat. Sb. 182 (1991), no. 4, 543--567; translation in Math. USSR-Sb. 72 (1992), no. 2, 519–541) proved that for every torsion-free non-elementary hyperbolic group $G$ there is a number $N \in \mathbb{N}$ such that for any odd $n \ge N$ the quotient $G/G^n$ is infinite. -In a more recent article, Ivanov and Olshanskii ("Hyperbolic groups and their quotients of bounded exponents". Trans. Amer. Math. Soc. 348 (1996), no. 6, 2091–2138) proved a similar statement for an arbitrary non-elementary hyperbolic group $G$ (torsion is allowed): there is $n=n(G) \in \mathbb{N}$ such that $G/G^n$ is infinite. In this case one cannot say that $G/G^k$ is infinite for any sufficiently large odd $k$, because if the group $G$ is generated by elements of, say, order $3$, then for any $k$ not divisible by $3$, $G^k=G$.<|endoftext|> -TITLE: Example of two structures -QUESTION [6 upvotes]: This is probably a very trivial question, still I don't seem to find an answer. -I'd like to see an example (in some language) of two countable structures $\mathcal{M}_1 $ and $ \mathcal{M}_2 $ with $$ \mathcal{M}_1 \equiv\mathcal{M}_2 $$ and the property that there is no elementary embedding from either one to the other. -Maybe also another one: the same setting as above, but with existing embeddings, just no elementary embeddings. -Are there such examples? - -REPLY [3 votes]: For each $n \geq 1$, let $F_{n}$ be the free group on $n$ generators and let $\mathbb{Q}^{n}$ be the direct product of $n$ copies of the additive group of the rationals. Since $\mathbb{Q}^{2} \equiv \mathbb{Q}$ and $F_{2} \equiv F_{3}$, it follows that $F_{2} \times \mathbb{Q}^{2} \equiv F_{3} \times \mathbb{Q}$. Clearly $F_{3} \times \mathbb{Q}$ embeds into $F_{2} \times \mathbb{Q}^{2}$ and it is easily seen that $F_{2} \times \mathbb{Q}^{2}$ does not embed into $F_{3} \times \mathbb{Q}$. On the other hand, using the proof that elementary subgroups of free groups are free factors, it follows that $F_{3} \times \mathbb{Q}$ cannot be elementarily embedded into $F_{2} \times \mathbb{Q}^{2}$. -Just for fun, I will also show that there exists an uncountable family of pairwise non-embeddable elementarily equivalent finitely generated simple amenable groups ... but fail to provide a single explicit example of a pair of such groups. Let $\mathcal{G}$ be the Polish space of f.g. groups. Using recent work of Grigorchuk-Medynets, there exists a Borel reduction $\varphi: 2^{\mathbb{N}} \to \mathcal{G}$ from $E_{0}$ to $\cong$; say, $x \mapsto G_{x}$. Let $L$ be the language of group theory and let $\psi: \mathcal{G} \to \mathcal{P}(L)$ be the Borel map $G \mapsto Th(G)$. Then there exists a fixed complete theory and a comeagre $X \subseteq 2^{\mathbb{N}}$ such that $Th(G_{x}) = T$ for all $x \in X$. Consider the Borel subset $Z = \varphi(X) \subseteq \mathcal{G}$. Define a Borel coloring $\theta: [Z]^{2} \to 2$ by $\theta(G,H) = 0$ iff $G$, $H$ are incomparable with respect to embeddability. Then there exists a Cantor set $C \subseteq Z$ such that $\theta$ is constant on $[C]^{2}$. Since each f.g. group has only countably many f.g. subgroups, it follows easily that $C$ is an uncountable family of pairwise non-embeddable elementarily equivalent finitely generated simple amenable groups.<|endoftext|> -TITLE: Identifying finite-index subgroups of SL2Z from generators -QUESTION [9 upvotes]: (This question is reposted from math.SE, since it's sat there for a while with no answers. Apologies if it's not considered research-level, but I'm not a group theorist myself.) -Suppose I have a finite set of elements $x_1, \dots, x_n$ of the modular group $\operatorname{SL}_2(\mathbf{Z})$. Is there is an algorithm that will determine in finitely many steps whether or not the subgroup generated by $x_1, \dots, x_n$ has finite index? - -REPLY [9 votes]: Yes, there is an algorithm. First find the intersection $U$ of $H=\langle x_1,..., x_n\rangle$ with the free subgroup of index 12 in $SL_2(\mathbb{Z})$ (the free group has two generators $a,b$). Let it be generated by words $u_1,...,u_m$. Consider the Stallings graph associated with $U$. It is a finite labeled graph where every edge is labeled by $a,b,a^{-1}$ or $b^{-1}$ and no two edges sharing the initial/termnal vertex have the same label. The index is finite if and only if every vertex of that graph has degree 4. See, for example, Margolis, S.; Sapir, M.; Weil, P. -Closed subgroups in pro-V topologies and the extension problem for inverse automata. -Internat. J. Algebra Comput. 11 (2001), no. 4, 405–445 and the references there.<|endoftext|> -TITLE: A profinite group which is not its own profinite completion? -QUESTION [31 upvotes]: Is there a profinite group $G$ which is not its own profinite completion? - -Surely not, I thought. But upon looking into it, I found that there is a special name given to a $G$ which is its own profinite completion, namely "strongly complete". And a recent (2003) hard theorem (which according to Wikipedia uses the classification of finite simple groups) due to Nikolov and Segal asserts that, if $G$ is finitely generated (as a topological group), then it is "strongly complete". -So the $G$ I'm looking for cannot be topologically finitely generated. An equivalent question to the above is: - -Is there a profinite group $G$ which admits a non-open subgroup of finite index? - -Now here's my problem; the only exposure to profinite groups I've had has been in the context of number theory, absolute Galois groups, local fields, etc. In particular, the only non-topologically-finitely-generated profinite group I'm aware of is the absolute Galois group of a number field, say $\mathbb{Q}$. But I reckon the Krull topology demands that the finite index subgroups of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ be open. -Maybe there is a more 'exotic' example of such a $G$... - -REPLY [5 votes]: It's hard to pick many non-finitely-generated pro-$p$ groups which are isomorphic their own profinite completions. Partially as non-finitely-generated pro-$p$ groups aren't generally that easy to construct. -My paper here http://arxiv.org/abs/1101.3005 outlines the construction of infinitely many topologically non-isomorphic pro-$p$ groups isomorphic to $\prod_i C_{p^{i}}$. These will thus all have the same profinite completion, but are non-isomorphic.<|endoftext|> -TITLE: Adjunctions between derived functors -QUESTION [6 upvotes]: Given an adjunction $F\dashv G$ between functors between Abelian categories, we know that $F$ is right exact and $G$ is left exact so there are derived functors $LF$ and $RG$ between (bounded above, respectively below) derived categories. What can one say about the existence of an adjunction $LF\dashv RG$? - -REPLY [3 votes]: It seems this question was considered in: -William W Adams, Marc A Rieffel. -Adjoint functors and derived functors with an application to the cohomology of semigroups -Journal of Algebra, V. 7, N 1, 1967, 25-34<|endoftext|> -TITLE: non-integrable subadditive ergodic theorem -QUESTION [8 upvotes]: Dear MO_World, -I have (another) question about relaxing the assumptions in the sub-additive ergodic theorem. Apologies if this is something I should know already... -There are a number of statements of the Kingman sub-additive ergodic theorem and its extensions. Here is a fairly typical version: -Let $T\colon X\to X$ be a measure-preserving transformation of a probability space. Let -$(f_n)$ be a sequence of measurable functions from $X$ to $[-\infty,\infty)$ satisfying $f_{n+m}(x)\le f_n(x)+f_m(T^nx)$ for every $m,n\in\mathbb N$ and $x\in X$. Suppose further that $f_1^+$ is integrable. Then $f_n(x)/n$ is convergent for almost every $x$. -There are also statements about convergence in the mean in certain situations, but I focus on pointwise convergence. -My general question is: When can you remove the integrability assumption? -In the following example the conclusion holds, even though the hypothesis fails. -Consider a distribution on the positive reals so that if $(X_n)$ is a sequence of iid random variables with the given distribution, then $\mathbb E(\min(X_1,X_2,\ldots,X_n))=\infty$ for all $n$. (As an example, consider a random variable that takes the value $2^{k^3}$ with probability $2^{-k}$ for each $k\ge 1$. Let $M_n$ be $\min(X_1,X_2,\ldots,X_n)$. Then $\mathbb P(M_n=2^{k^3})\approx 2^{-nk}$ so that $\mathbb E M_n=\infty$. -Let $\Omega= \lbrace 2^{k^3}\colon k\ge 1 \rbrace ^\mathbb Z$ equipped with the Bernoulli probability measure arising from the distribution above. -Now define the sequence $f_n(\omega)$ as follows: $f_{2n}(\omega)=0$; $f_{2n+1}(\omega)=\min_{0\le j\le 2n+1}\omega_j$. This sequence of functions is clearly sub-additive. Also $0\le f_n(\omega)/n\le \omega_0/n$, so that $f_n/n\to 0$ everywhere. -My specific question is as follows: - -Suppose that $(f_n)$ is a sub-additive sequence of functions taking values in $[-\infty,\infty)$. If we assume that $\limsup f_n(x)/n \lt \infty$ a.e., then does it follow that $f_n(x)/n$ converges almost surely? [ The above example shows that we cannot expect convergence in norm, even if the $f_n$ are non-negative ] - -REPLY [5 votes]: Having read Terry's answer, I found a counterexample of my own. It's quite a bit simpler (although less nice in the sense that the liminf is $-\infty$ rather than 0). -It's actually an additive process rather than sub-additive one. Define a transformation $T$ on $X=\mathbb N^\mathbb Z\times\lbrace 0,1\rbrace$ by $T(x,0)=(x,1)$ and $T(x,1)=(Sx,0)$ where $S$ is the shift. Define the function $f(x,0)=-x_0$ and $f(x,1)=x_0$. Let the measure on $X$ be $\mu\times c$ where $c$ is counting measure and $\mu$ is a horrendously non-integrable iid process on $\mathbb N^\mathbb Z$. -If you look at the values of $f$ along an orbit, you see the values $-x_0,x_0,-x_1,x_1,-x_2,x_2,\ldots$ if you start from an `even' point or $x_0,-x_1,x_1,-x_2,x_2,\ldots$ if you start from an odd point. In the first case the partial sums $S_nf(x)$ are always non-positive and take the value 0 infinitely often, so that $\limsup S_nf(x)/n=0$. In the second case, the partial sums $S_nf(x)$ are bounded above by $x_0$ and take this value infinitely often, so that again $\limsup S_nf(x)/n=0$. -On the other hand, in the first case, summing $2n+1$ terms, $S_{2n+1}f/(2n+1)=-x_n/(2n+1)$. If the distribution is sufficiently nasty (e.g. the $x_i$'s take the value $3^n$ with probability $2^{-n}$), you get $\liminf S_nf/n=-\infty$. Similarly in the second case, summing $2n$ terms, you see that $\liminf S_nf/n=-\infty$.<|endoftext|> -TITLE: Solutions to the eikonal equation -QUESTION [7 upvotes]: Theorem. Let V be a $C^\infty$ function on a riemannian manifold $M$ and $p$ be a nondegenerate local minimum with $V(p)=0$. Then there is a unique positive function $\varphi \in C^\infty(U)$ such that $\varphi$ solves the eikonal equation -$$\|\mathrm{grad} \varphi \|^2 = V.$$ -Here, $U$ is some open neighborhood of $p$. -I found this statement (at least a quite similar one) in a physics paper without a real proof, just some motivation. It seems highly nontrivial and somehow I am struggling to find a proof anywhere. Can someone give me a good reference? -\Edit: I forgot to write down the (for uniqueness obviously essential) condition that $\varphi$ is positive. Sorry everyone! - -REPLY [10 votes]: Note: I have realized that, using the Stable Manifold Theorem, one can prove the smoothness of the solution $\phi$ that I describe below. Thus, I am modifying my answer to incorporate that. -Local existence and uniqueness of a smooth solution near $p$ satisfying $\phi(p)=0$ and $\phi\ge0$ near $p$ follows from an application of the Stable Manifold Theorem. Here is the argument. -This is a local question, so we might as well assume that $M=\mathbb{R}^n$, that $p=0$, that $g = g_{ij}dx^idx^j$ satisfies $g_{ij}(0)=\delta_{ij}$, and that the function $V$ has a Taylor expansion $V = h_{ij}x^ix^j + O(|x|^3)$, where $(h_{ij})$ is a symmetric, positive definite matrix. We are looking for a closed $1$-form $d\phi = f_i\ dx^i$ near $x=0$ so that $\phi$ satisfies the equation $g^{ij}f_if_j = V$ and, at the same time, satisfies $\phi(0)=0$ and $\phi\ge0$ near $x=0$. -Let $p_i$ be the coordinates on $T^*\mathbb{R}^n$ such that the canonical $1$-form has the expression $p_i\ dx^i$. Then the graph of $d\phi$, described by equations $p_i = f_i(x)$, will be a Lagrangian submanifold for the $2$-form $dp_i\wedge dx^i$ and will lie in the zero locus of the Hamiltonian $H(x,p) = g^{ij}(x)p_ip_j - V(x)$. Therefore, it will be a union of integral curves of the Hamiltonian vector field -$$ -X_H = 2g^{ij}p_i\frac{\partial\ \ }{\partial x^j} -+ \left(\frac{\partial V}{\partial x^k} -- \frac{\partial g^{ij}}{\partial x^k}p_ip_j\right)\frac{\partial\ \ }{\partial p_k}. -$$ -This graph will have to pass through the unique singular point of $X_H$, i.e., $x = p = 0$ (since $\phi$ clearly must have a critical point at $x=0$ because it vanishes there and is nonnegative nearby), and the linear part of $X_H$ at $x=p=0$ is -$$ -Y_H = 2\delta^{ij}p_i\frac{\partial\ \ }{\partial x^j} - + 2h_{ij}x^i\frac{\partial\ \ }{\partial p_j}. -$$ -The unstable manifold of $Y_H$ is the $n$-dimensional submanifold defined by $p_i = L_{ij} x^j$, where $L$ is the (unique) symmetric positive definite square root of $(h_{ij})$. The stable manifold of $Y_H$ is defined by $p_i = -L_{ij}x^j$. It follows from the Stable Manifold Theorem that $X_H$ has a smooth $n$-dimensional unstable submanifold $N_+$ given by $p_i = f_i(x) = L_{ij}x^j + O(|x|^2)$ and a smooth $n$-dimensional stable submanifold $N_-$ given by $p_i = -L_{ij}x^j + O(|x|^2)$. -From the dynamics of $X_H$, it is clear that the only possibility for $d\phi$, when $\phi$ satisfies the above conditions and is at least $C^2$, is to have its graph be $N_+$. Conversely, taking $d\phi = f_i(x)\ dx^i$ where $p_i = f_i(x)$ is the (necessarily Lagrangian) unstable manifold of $X_H$ and fixing the additive constant by requiring that $\phi(0)=0$ does give a smooth solution to the original equation.<|endoftext|> -TITLE: Categorical Brouwer-Heyting-Kolmogorov interpretation -QUESTION [8 upvotes]: Let $\mathcal{L}$ be the language of intuitionistic propositional logic generated by some atomic propositions $t_1, t_2, \ldots$. The Lindenbaum–Tarski algebra of $\mathcal{L}$ can be regarded as a bicartesian closed category in which there is an arrow $p \to q$ if and only if there is a proof of $q$ assuming $p$. Unfortunately it is a somewhat dull category, as there is at most one arrow between any two objects. -Question. Is there a categorification of the Lindenbaum–Tarski algebra which enables a category-theoretic form of the Brouwer–Heyting–Kolmogorov interpretation of intuitionistic propositional logic? In particular, - -Objects should be propositions. -Arrows should be (equivalence classes) of proofs. -The coproduct should be disjoint, at least for the coproduct of two distinct atomic propositions. -The terminal object should be indecomposable, so that the disjunction property is validated (i.e. an arrow $\top \to p \lor q$ is either an arrow $\top \to p$ or an arrow $\top \to q$). - -It feels like the free bicartesian closed category generated by the atomic propositions is the most likely candidate, and it can be concretised by the Yoneda embedding into the presheaf topos: then we would have a genuine BHK interpretation, i.e. interpreting a proposition as the ‘set’ of its proofs. This has probably been well-studied, in which case I would appreciate any references to the literature. - -REPLY [2 votes]: Perhaps this paper would be good to look at: -ERIK PALMGREN (2004). A categorical version of the Brouwer–Heyting–Kolmogorov interpretation. Mathematical Structures in Computer Science, 14 , pp 57-72 doi:10.1017/S0960129503003955<|endoftext|> -TITLE: Limit shape for fixed-perimeter lattice polygons -QUESTION [11 upvotes]: Let $P$ be a simple polygon defined by $n$ unit-length segments -connecting lattice points of $\mathbb{Z}^2$. -I have two operations that preserve the perimeter of $P$. -The first is the "pop" of a corner either inward or outward; -the second is pushing in a 3-edge "tab" at one spot and immediately pushing out -a tab at another. Both operations are only applied if -they preserve simplicity (non-self-intersection): - -          - - -My question is: - -Q. What is the limit shape of $P$ as $n \rightarrow \infty$ - and the operations are applied randomly for sufficiently many iterations? - -I realize this question is imprecise. Perhaps the limit shape depends -on the frequency balance between the two operations, which I have -not specified. -In addition, I am not quite sure how to best to describe a shape. -But even an intuitive answer would be useful: Will the shape -tend toward a convex-like blob, or toward a more spidery shape? -(My guess is the latter.) -I was hoping to get a sense with small experiments as below, but -it may require much larger $n$ and many more iterations for -a clear pattern to emerge. -I would appreciate any help in sharpening the question, -answering it in some form, or pointers to related literature -that could shed light on the issue. -Thanks! - -Animation of 200 operations, starting from a crenellated square. The green dot is the centroid of the vertices. - -REPLY [2 votes]: Sometimes models like this can simplified e.g. loop erased random walk and self-avoiding walk. Your model may even be integrable if you allow for a small amount of self-intersections. - -There are certainly lots of 2D growth models (e.g. Laplacian growth) and they all tend to fall into the KPZ universality class. However, we ought to prove that your model also falls in this class. And maybe there already exists theorem which accomplishes as much. The notes certainly say "ballistic deposition" (as of 2012) is unknown. [1] -It may be helpful to use a small amount of topology here. Your curve is always contractible so that -$$ \gamma \equiv 0 \in H_0(\mathbb{R}^2) $$ -and I'm phrasing it in this rather pretentious way in order emphasize that your model is equivalent to adding squares at random from the interior inside the boundary or vice versa. -Homology is clearly not refined enough to detect curvature (in the "hydrodynamic" limit). I already suspect there can be no limit shape at all. If I start with a perfect circle, we're gently adding bits of curvature everywhere, there's no reason why these should perfectly cancel out globally.<|endoftext|> -TITLE: $\kappa$-coloring of $\mathbb{R}^2$ and triangle with area 1 -QUESTION [7 upvotes]: What is the largest cardinal number integer $\kappa$ such that every $\kappa$-coloring of $\mathbb{R}^2$ contains a triangle with area 1 and all vertices of the same color? - -REPLY [10 votes]: R. Graham in "On Partitions of $\mathbb E^n$" , J. Combinatorial Theory Ser. A, 28 (1980), 89–97, proves that for any finite coloring of $\mathbb R^n$ and any positive $a\in \mathbb R$ there is a monochromatic simplex of volume $a$. -The question is originally due to Gurevich from the 70's. Graham's proof has been simplified, see for example the paper "Monochromatic simplices of any volume" by A. Dumitrescu, M. Jiang, Discrete Mathematics, Vol. 310, 4, 2010, 956-960. - -REPLY [4 votes]: I think that you can always find such a triangle for any finite $\kappa$. Not sure how to prove that though. -There are countably infinite colorings that do not contain such a triangle. Partition the plane into a checkerboard of unit squares and color each square a different color. Then, the largest monochromatic triangle has area $0.5 < 1$.<|endoftext|> -TITLE: Cohomology vanishing for tensor powers of tangent bundle on the flag variety -QUESTION [9 upvotes]: Let $X$ denote the flag variety of a semi-simple group $G$ (in characteristic 0) -and let $T_X$ denote its tangent bundle. I would like to ask the following question(s): -1) Is it true that for any $n\geq 0$ we have $H^i(X,T_X^{\otimes n})=0$ for $i>0$? -2) More generally, let $\lambda$ be a dominant weight of $G$ and let $\mathcal O(\lambda)$ -be the corresponding line bundle on $X$. Is it true that -$H^i(X, T_X^{\otimes n}\otimes \mathcal O(\lambda))=0$ for $i>0$? -When tensor powers of $T_X$ are replaced by symmetric powers, this is known to be true -(for example it is proved in a paper of Kumar, Lauritzen and Thomsen). - -REPLY [4 votes]: It's been almost 9 years, so maybe you are no longer interested in this question, but in a recent(-ish) preprint with Maxim Smirnov we have shown that the answer to the first question is no in general, even when restricted to the exterior powers of the tangent bundle (which describes the Hochschild cohomology of $X$). -In Hochschild cohomology of generalised Grassmannians - we have mostly focused on the case of $G/P$ where $P$ is a maximal parabolic subgroup, and discuss when there is such vanishing for exterior powers. But we also give an explicit family of examples (namely for the symplectic Grassmannian $\operatorname{SGr}(3,2n)$) which has an $\mathrm{H}^1$, which is what you are looking for. -If you care about the full flag variety, rather than generalised Grassmannians, then there are also counterexamples, including in type A, starting from $\mathrm{SL}_6$. This one was obtained using Macaulay2 by Allen Knutson.<|endoftext|> -TITLE: Class field theory using only ideles of norm 1 -QUESTION [6 upvotes]: I am a total non-expert, so the answer to this question may be obvious, but here goes. -In Chevalley's formulation of CFT we get Artin maps $J_k \rightarrow Gal(L/k)$, where $J_k$ is the group of all ideles of $k$. However, we know there is a nice subgroup $J_k^1$ of the ideles obtained by taking only those satisfying the product formula $\prod_{v} |x_v| = 1$. Note that this still contains all the principal ideles, still surjects onto $I_k$ and has additional attractive properties like the compactness of $J_k^1/k^*$. Is there a way to set up CFT using quotients of this nicer group, and if so, what are the advantages of working with the superficially more unwieldy $J_k$? -Thanks. - -REPLY [9 votes]: I am no expert either, but here is what I think. If $k$ is a global field of characteristic zero (i.e. an algebraic number field), then one can work with $J_k^1$ instead of $J_k$ without losing (or changing) anything. This is because the kernel of the Artin map contains a subgroup $N$ of $J_k$ isomorphic to the multiplicative group of positive reals, and $J_k$ factors (non-canonically) as $J_k\cong N\times J_k^1$. In other words, in characteristic zero the Artin map does not see the norm of an idele. If $k$ is a global field of finite characterstic (i.e. a function field over a finite field $\mathbb{F}_q$), then the situation is markedly different. In this case the image of $J_k^1$ under the Artin map equals $Gal(L/k_0)$, where $k_0$ is the union of all constant field extensions of $k$ (i.e. the compositum of $L$ and the algebraic closure of $\mathbb{F}_q$ in the algebraic closure of $k$). More precisely, the norm of an idele precisely tells how its image under the Artin map acts on $k_0$: if the norm is $q^{-r}$, then the action on the algebraic closure of $\mathbb{F}_q$ is $x\mapsto x^{q^r}$.<|endoftext|> -TITLE: Asymptotic question about time ordered exponentials -QUESTION [8 upvotes]: Let $A(t)$ be a smooth function from $[-1,1]$ to the $n \times n$ complex matrices. Define the time ordered exponential -$$\prod_{-1}^1 \exp(A(t) dt)$$ -as in this question, as the limit of Riemann products $\prod_{i=1}^n \exp(f(t^{\ast}_i) \ \delta t_i)$. -The actual quantity I am interested in is -$$B(r) = \prod_{-1}^1 \exp(r A(t) dt)$$ -as $r \to \infty$. -As $r \to 0$, there is a known power series expansion for $B(r)$ called the Magnus series. As $r \to \infty$, I would expect there to be something like the stationary phase approximation, but I haven't been able to find it or figure it out. -I should mention that in my situation, $A(t)$ obeys -$$A(-t) = A(t)^{\ast} \quad (\dagger)$$ -where $\ast$ is conjugate transpose. Condition $(\dagger)$ implies that $B(r)$ is Hermitian. I don't know whether this is helpful in any way. - -REPLY [3 votes]: I think I might see what was confusing me. This is really a comment, but it's too long for the comment thread. As my example, let's take -$$A(t) = \frac{1}{1+t^2} \begin{pmatrix} 2 & t \\ -t & -2 \end{pmatrix}$$ -So we want to solve the differential equation $U'(t) = r A(t) U(t)$, where $U$ is a $2 \times 2$ matrix with initial condition $U(-1) = \mathrm{Id}$. -We can actually compute the eigenvalues of $A(t)$ explicitly: They are $\sqrt{4-t^2}/(1+t^2)$. We compute $\int_{-1}^1 \pm \sqrt{4-t^2}/(1+t^2) dt \approx \pm 3.03022$. So your formula, as I understand it, is -$$U(1) = e^{3.03022 r} u_1 v_1^T + e^{-3.03022 r} u_2 v_2^T + \cdots$$ -where $u_i$ and $v_i$ are the eigenvectors of $A(-1)$ and $A(1)$. -What I think was confusing me is that it is somewhat misleading to call this the leading terms. The later terms in the series look like $e^{3.03022 r} r^{-k} (\mbox{stuff})$, right? So they actually dominate the $e^{-3.03022 r}$ term. - -I wish I weren't having so much trouble getting good numerical data, it would probably clear up my confusion a lot. In the meantime, here is why I am worried. -Let $A(t)$, $B(t)$ and $C(t)$ be three $2 \times 2$ matrix-valued functions as above, with $A(1)=B(1)=C(1)$ (and hence the same at $-1$.) Let $X(r)$, $Y(r)$ and $Z(r)$ be the parallel transport from $-1$ to $1$ be the differential equations $\phi'(t) = r A(t) \phi(t)$, $\phi'(t) = r B(t) \phi(t)$ and $\phi'(t) = r C(t) \phi(t)$. As I understand it, your method gives asymptotic expansions -$$X(r) \approx U \begin{pmatrix} e^{x_1 r} & 0 \\ 0 & e^{x_2 r} \end{pmatrix} V \quad Y(r) \approx U \begin{pmatrix} e^{y_1 r} & 0 \\ 0 & e^{y_2 r} \end{pmatrix} V \quad Z(r) \approx U \begin{pmatrix} e^{z_1 r} & 0 \\ 0 & e^{z_2 r} \end{pmatrix} V \quad (1)$$ -where I have the SAME matrices $U$ and $V$ in each cases, because they depend only on the eigenvectors of $A(1)=B(1)=C(1)$ and of $A(-1)=B(-1)=C(-1)$. -Am I right about $(1)$? -If so, here is the issue. Look at the quadratic form -$$\det(x X(r) + y Y(r) + z Z(r)) \approx \det(U) \left( e^{r x_1} x + e^{r y_1} y + e^{r z_1} z \right) \left( e^{r x_2} x + e^{r y_2} y + e^{r z_2} z \right) \det(V).$$ -The matrix of this form has leading terms -$$\begin{pmatrix} -\exp(r(x_1+x_2)) & & \\ -\exp(r\max(x_1+y_2, x_2+y_1)) & \exp(r(y_1+y_2)) & \\ -\exp(r\max(x_1+z_2, x_2+z_1)) & \exp(r\max(y_1+z_2, y_2+z_1)) & \exp(r(z_1+z_2)) \\ -\end{pmatrix}$$ -as long as the approximations in $(1)$ are good enough that we don't get extra cross terms. -Unless I am very confused, I can construct $A(t)$, $B(t)$, $C(t)$ such that this quadratic form looks like $x^2+y^2+z^2 + (e^r+e^{-r}) (xy+xz+yz)$. And there are no real numbers $(x_1, x_2, y_1, y_2, z_1, z_2)$ with $x_1+x_2=y_1+y_2=z_1+z_2=0$ and $\max(x_1+y_2, x_2+y_1)=\max(x_1+z_2, x_2+z_1)=\max(y_1+z_2, y_2+z_1)=1$. So something is wrong...<|endoftext|> -TITLE: Reference request: gluing manifolds along pieces of boundary -QUESTION [10 upvotes]: I've been asked for a reference for the following construction and since I didn't know one, I thought I'd ask here if anyone did. -Consider two smooth manifolds with boundary of the same dimension, $M$ and $N$. Suppose that we have a submanifold $S$ of the boundaries $\partial M$ and $\partial N$ of codimension $0$ ($S$ may have boundary as well). Then glue $M$ and $N$ together along $S$, smoothing out the corners (corresponding to the boundary of $S$) if necessary. To do this properly, one would have to add a collar to each of $M$ and $N$ which is "broken" at the boundary of $S$ (thus making manifolds-with-corners) and glue those. -Is there a standard reference for this that works through the details? - -REPLY [5 votes]: This was a bit too long for a comment, so I am posting it as an answer. You are sort of asking two things: - -How to turn your manifolds M and S into an appropriate manifold with corners together with an appropriate notion of collar? -How to then glue these to obtain a new manifold? - -To do (1) you'll need some assumptions on S, probably including compactness. In many cases though it might be clear that you can choose such collars. - In that case you might be interested in Theorem 3.5 from my 2009 dissertation (arXiv:1112.1000, page 140). There I show that even if the collars are not specified, the glued manifold is still unique up to (non-canonical) diffeomorphism fixing S and restricting to the identity outside a neighborhood of S. In fact the construction shows that there is a canonical contractible family of these diffeomorphisms (and so there is a canoncial isotopy class of diffeomorphisms). -I used this to build one version of the 2-category of cobordisms, where you need to glue along parts of the boundary in the manner you describe, but where you also don't want to mod out by diffeomorphisms too early. -When S is a component of the boundary, you can find this result here: -James R. Munkres, Elementary differential topology, Lectures given at Massachusetts Institute of Technology, Fall, vol. 1961, Princeton University Press, Princeton, N.J., 1966. -I basically adapted this proof to cover the case of gluing manifolds along a portion of the boundary.<|endoftext|> -TITLE: Asymptotic dimension of graph manifold groups -QUESTION [5 upvotes]: Does every non-geometric graph manifold have fundamental group of asymptotic dimension 3? - -This is affirmed in http://arxiv.org/abs/0909.1098 for closed graph manifolds, but I am interested in non-closed graph manifolds as well. -Notice that the asymptotic dimension of such groups is always at least 2 (obvious) and at most 3 (by a result of Bell and Dranishnikov). - -REPLY [5 votes]: I should have seen much earlier that it's 2. In fact, every non-closed graph manifold has a finite sheeted cover that fibers over the circle, see here. The fundamental group of such cover is an HNN extension of a free group, so that it has asymptotic dimension at most 2 by the main result of this paper.<|endoftext|> -TITLE: Shannon-McMillan-Breiman Theorem -QUESTION [7 upvotes]: Does anyone know of an easy proof of Shannon-McMillan-Brieman Theorem? -Thanks - -REPLY [3 votes]: In addition to Igor's answer, there's also: -D. Ornstein and B. Weiss, "The Shannon-McMillan-Breiman theorem for a class of amenable groups", Israel J. Math. 44 (1983), 53-60.<|endoftext|> -TITLE: Fast (subquadratic) evaluation of a class of N degree polynomials over N points -QUESTION [5 upvotes]: Let $(x_1 \ldots ,x_n) \in \mathbb{R}^n$ and $f_i = \Pi_{j=1, j \neq i }^n ( x_i - x_j )$ -I'm trying to evaluate $(f_1, \ldots, f_n)$. A trivial algorithm runs in $\mathcal{O}(n^2)$ but given the very specific form of the problem, there's got to be something faster. Maybe I've overlooked something simple, maybe a fourier transform is in order... What are your thoughts? - -REPLY [2 votes]: The following may be of some help, if you haven't thought of it already. Let $V$ be the Vandermonde matrix with $(i,j)$th entry $x^{i-1}_{j}$, $i,j=1,\ldots,n$. Its inverse $W$ has $(i,1)$th entry -$$ -(-1)^{n} \frac{x_1 \ldots x_{i-1} x_{i+1} … x_n}{f_i}. -$$ -Hence, to find $f_1,\ldots,f_n$ we need to solve $V \alpha = e_1$, where $e_1=(1,0,\ldots,0)$. Thus, the question is, how fast can one solve a system of equations with a Vandermonde matrix? There's discussion on this topic in Nick Higham's book "Accuracy and stability of numerical algorithms" (see chpt 22). It appears that there are algorithms that are $O(n (log n)^2)$. However, it's noted that these may not be numerically stable or practical.<|endoftext|> -TITLE: explicit lower bounds on $|L(1,\chi)|$ -QUESTION [10 upvotes]: Does anyone know of an explicit effective lower bound for $|L(1,\chi)|$, where $\chi$ is an odd complex (primitive) Dirichlet character? -I know of Landau's paper Uber Dirichletsche Reihen mit komplexen Charakteren, where he bounds $$ |L(1,\chi)|>\frac{1}{c \log(q)},$$ -where $q$ is the conductor of $\chi$, but the constant $c$ he gets is on the order of $e^{50}$, and is totally useless for computations. -I know of many papers dealing with quadratic characters but very few that address complex characters (explicitly). - -REPLY [10 votes]: This is discussed on page 47 of Narkiewicz's new book (Rational Number Theory in the 20th Century); see -http://books.google.ca/books?id=3SWNZaDM6iMC&lpg=PP1&dq=rational%20number%20theory%20in%20the&pg=PA47#v=onepage&q&f=false -Reference [4268] is to -Metsankyla, T.: Estimations for L-functions and the class numbers of certain imaginary cyclic fields, Ann. Univ. Turku, Ser. AI 140, 1--11 (1970) -[3995] is -Louboutin, Stéphane(F-CAEN) -Minoration au point 1 des fonctions L et détermination des corps sextiques abéliens totalement imaginaires principaux. (French) [Lower bound at the point 1 of L-functions and determination of the principal totally imaginary abelian sextic fields] -Acta Arith. 62 (1992), no. 2, 109–124. -and [338] is -Barrucand, Pierre; Louboutin, Stéphane(F-CAEN) -Minoration au point 1 des fonctions L attachées à des caractères de Dirichlet. (French) [Lower bound at the point 1 of L-functions associated with Dirichlet characters] -Colloq. Math. 65 (1993), no. 2, 301–306.<|endoftext|> -TITLE: Why do we use $\varepsilon$ and $\delta$? -QUESTION [31 upvotes]: My understanding (from a talk by Rob Bradley) is that Cauchy is responsible for -the now-standard $\varepsilon{-}\delta$ formulation of calculus, introduced in his -1821 Cours d’analyse. Although perhaps instead it was introduced by Bolzano a few years -earlier. My question is not about who was first with this notation, but -rather: - -Why were the symbols $\varepsilon$ and $\delta$ used? - -Why not, say, $\alpha$ and $\beta$? -(Imagine how different our mathematical discourse would be...) -Are there appropriate (French) words beginning with 'e' and/or 'd' that determined the choice? -Or perhaps Cauchy used up $\alpha,\beta,\gamma$ for other purposes prior to introducing $\delta,\varepsilon$? Does anyone know? - -REPLY [30 votes]: Thanks to H. M. Šiljak for finding the 1983 Amer. Math. Monthly -paper by Judith Grabiner, which I feel settles the question (at least for $\epsilon$). -Here is a longer quote encompassing that which H.M. excerpted: - -Mathematicians are used to taking the rigorous foundations of the calculus as a - completed whole. What I have tried to do as a historian is to reveal what went into - making up that great achievement. This needs to be done, because completed wholes by - their nature do not reveal the separate strands that go into weaving them—especially - when the strands have been considerably transformed. In Cauchy's work, though, one - trace indeed was left of the origin of rigorous calculus in approximations—the letter - epsilon. The $\epsilon$ corresponds to the initial letter in the word "erreur" (or "error"), and - Cauchy in fact used $\epsilon$ for "error" in some of his work on probability [31]. It is both - amusing and historically appropriate that the "$\epsilon$," once used to designate the "error" in - approximations, has become transformed into the characteristic symbol of precision - and rigor in the calculus. As Cauchy transformed the algebra of inequalities from a tool - of approximation to a tool of rigor, so he transformed the calculus from a powerful - method of generating results to the rigorous subject we know today. -[31] Cauchy, Sur la plus grande erreur à craindre dans un résultat moyen, et sur le - système de facteurs qui rend cette plus grande erreur un minimum, Comptes rendus - 37, 1853; in Oeuvres, series 1, vol. 12, pp. 114–124. - - -A further finding by H. M. Šiljak (linked in a comment above), -verifying that Cauchy did indeed use both $\epsilon$ and $\delta$:<|endoftext|> -TITLE: What is the smallest diameter ring a non-convex polyhedron can pass through in 3-space? -QUESTION [5 upvotes]: The question is mostly in the title: - -What is the smallest diameter ring a non-convex polyhedron can pass through in 3-space? - -Imagine I have some non-convex polyhedron $P$, and I would like to find the smallest diameter ring that it can pass through in 3-space, undergoing any necessary rotations as it does so. Is there an efficient way to calculate $D_{ring}$? Pressing my luck, can I find the set of rotations for $P$ as it passes through the ring? - -REPLY [5 votes]: Just a side remark on convex polyhedra: Each of the regular polyhedra except the cube can pass -through a circle of radius smaller than the smallest-radius cylinder circumscribing the polyhedron. -This is proved in Tudor Zamfirescu's delightful paper, "Convex polytopes passing through circles" -(PDF link). There is quite a nice (non-algorithmic) literature on this problem.<|endoftext|> -TITLE: The (co)monadicity theorem relative to a presheaf topos -QUESTION [8 upvotes]: Recall the following theorem of Linton: -A functor $U:E\to \operatorname{Set}$ is monadic if - -$U$ has a left adjoint, -$E$ admits kernel pairs and coequalizers, -A parallel pair $R \rightrightarrows S$ in $E$ is a kernel pair if and only if its image under $U$ is so, and -A morphism $A\to B$ in $E$ is a regular epimorphism if and only if its image under $U$ is so. - -According to the nLab page, this can be modified such that we may replace the category $\operatorname{Set}$ in the above with the category $\operatorname{Set}^C$ for any small category $C$. However, I'm having a bit of a time finding a place that proves this. -I'd be extremely happy with a reference and will accept any answer with such a reference. - -REPLY [7 votes]: The result you are after is Theorem 1.2 of "A Monadicity Theorem" by Borceux and Day, published in the Bulletin of the Australian Mathematical Society in 1977, and the paper is available online. If you have access to Mathscinet reviews but not the paper you can even see the result stated in detail in the review by Kelly. -Their Theorem 1.2 is exactly as you state it above for E over Set with the exception that Set is replaced by an arbitrary category with kernel pairs and coequalisers, and so can be any presheaf category.<|endoftext|> -TITLE: Getting nervous refereeing a paper -QUESTION [43 upvotes]: I am refereeing my first paper and I'm quite excited! But inexperienced and I would like to ask an advice to the Maths Community of MO. Let me tell you that I have already read Refereeing a Paper, but it seems that my question is quite different. Roughly: - -What is the point after which you get nervous while refereeing a paper? - -Specifically: - -I've found many English mistakes. (Well, I'm not a native English and so I can understand. So I am not nervous yet) -I've found some maths inaccuracies like "let A be any set"... and then I have discovered that the proof of the first result works only for finite sets. (OK, those are only inaccuracies - I am not nervous yet) -There are many references like "we use the notation of [X]", "this result is proved in [X]", where [X] is a BOOK, without specifying a precise section, or the number of the result... should I get this book and read all to find out the correct references? - just thinking of it, makes me a bit nervous.. -(most importantly). There is a mathematical more serious mistake. Something that might be fixed, but not obviously (my definition of obvious is three evenings, in this case). I am not saying that the paper is completely wrong but that now... now I'm getting nervous! - -Now, taking into account that the person who asked me to referee this paper told me: be selective, we accept only 20% of submitted papers, - -what should you do in these cases? Reject? Ask for a revision? Not getting nervous and try to see if the rest of the piece is good (I'm quite a good guy and I'm doing that at the moment)? - -Of course I will talk with the editor, but I also would like to know more opinions that might be helpful in future. -Update: Thank you very much to everybody for the numerous and helpful comments. -Valerio - -REPLY [13 votes]: I like the comment of Francis Crick in `What mad pursuit' that one should always start a referee report with praise, but of course not hold back on the detailed criticism. -I have found that a referee can also help the author by asking for clarification in the Introduction of the aims of the paper, what does it accomplish in relation to previous work in the area, and what questions are still open. The author may also not have got the title right in relation to the expectations of readers: some of my early papers had lousy titles which not not convey easily what was in the paper. -It is useful to know that the referee is not responsible for the correctness of the paper: that is the responsibility of the author. Of course, if the referee has doubts on the correctness, these must be raised. -I am unhappy about remarks such as "there are doubts about the interest of this area", since some papers are of interest precisely because they go against the so-called `mainstream', which, to mix a metaphor, flaps about like a sail in a gusty wind. -The job of referee is very important, and I wish you luck and interest in this.<|endoftext|> -TITLE: Building a polyhedron from areas of its faces -QUESTION [10 upvotes]: Is there a known algorithm which, given a finite multiset (unordered list) of integers $A$, returns a yes/no answer for "Is there a polyhedron such that the multiset of areas of all its faces is exactly $A$?"? -Is there a known general algorithm for $n$-dimensional polytopes? - -REPLY [17 votes]: I can answer your question with the specialization to convex polyhedra and polytopes. -Specializing further to $\mathbb{R}^3$, the result is that - -$n \ge 4$ positive real numbers are the face areas of a convex polyhedron - if and only if the largest number is not more than the sum of the others. - -I wrote up a short note -establishing this: "Convex Polyhedra Realizing Given Face Areas," arXiv:1101.0823. -The result relies on Minkowski's 1911 theorem, which perhaps you know: - -Theorem (Minkowski). Let $A_i$ be positive faces areas and $n_i$ distinct, - noncoplanar unit face normals, - $i=1,\ldots,n$. - Then if $\sum_i A_i n_i = 0$, there is a closed convex polyhedron - whose faces areas uniquely realize those areas and normals. - -This theorem reduces the problem to finding orientations $n_i$ so that vectors of -length $A_i$ at those orientations sum to zero. And this is not difficult. -Here is Figure 3 from my note from which you can almost infer the construction: -            - -Minkowski's theorem generalizes to $\mathbb{R}^d$ and so does an analog of the above claim -(but I did not work that out in detail in the arXiv note). -In terms of an algorithm, the decision question is linear in the number $n$ of facet areas, -and even constructing the polyhedron is linear in $\mathbb{R}^3$, -and likely $O(dn)$ in $\mathbb{R}^d$ (but again, I didn't work that out). -But you don't mention the word "convex" in your post, so perhaps you are interested -in nonconvex polyhedra and polytopal complexes?<|endoftext|> -TITLE: uniformization theorem via ricci flows -QUESTION [6 upvotes]: Hi, I have a question about the positive Euler characteristic case. My question is: why is it so difficult as compared to the zero and negative cases? I am more interested in a pictorial/intuitive answer as compared to a very rigorous analytic answer. In other words, I want to get an intuitive "feel" of what goes wrong..... -Thanks in advance. - -REPLY [2 votes]: Given a closed surface $M$ with Euler characteristic $\chi$ and a fixed -pointwise conformal class $\mathcal{C}$, consider the problem of finding a -metric $g\in\mathcal{C}$ with $R(g)\equiv r\doteqdot\operatorname{sign}(\chi -)$. If $r=-1,$ then there is a unique such metric. If $r=0,$ then such a $g$ -is unique up to scaling; one can see this from the fact that if $g=e^{u}h$, -then $R_{g}=e^{-u}\left( R_{h}-\Delta_{h}u\right) $. On the other hand, the -group of conformal diffeomorphisms of $S^{2}$ is a noncompact group of linear -fractional transformations. So pulling back such a $g$ on $S^{2}$ yields a -noncompact set of such metrics. So, on the torus, the noncompactness is due -only to scaling, whereas on the sphere, the noncompactness is insidious. -For $r=-1$, Hamilton used the potential $f$ defined by $\Delta f=r-R$ and -proved that $H=R-r+\left\vert \nabla f\right\vert ^{2}$ satisfies -$(\frac{\partial}{\partial t}-\Delta)H=-2|\operatorname{Ric}+\nabla^{2}% -f-\frac{r}{2}g|^{2}+rH$. Since $r<0$, with consequent higher derivative estimates, this -implies that the modified flow $\frac{\partial}{\partial t}% -g=-2(\operatorname{Ric}+\nabla^{2}f-\frac{r}{2}g)$ converges exponentially -fast in each $C^{k}$-norm to an expanding gradient Ricci soliton. This expander, since it is on a closed manifold, must -have constant curvature, which in turn, implies that the original normalized -Ricci flow converges to a metric $g_{\infty}$ with $R(g_{\infty})\equiv r,$ -where $g_{\infty}$ is in the same conformal class as the initial metric -$g\left( 0\right) $. -See also the paper of X. Chen, P. Lu and G. Tian.<|endoftext|> -TITLE: Maximum thickness of three linked Euclidean solid tori -QUESTION [16 upvotes]: Consider three circles of radius $1$ in $\mathbb{R}^3$, linked with each other in the same arrangement as three fibers of the Hopf fibration. Now thicken the circles up into non-overlapping standard round Euclidean solid tori of equal thickness. Allowing the tori to move, there will be some maximum thickness (distance from the core circle to the boundary of the solid torus) before the tori must overlap each other. -I'm interested in the case that the three tori have $3$-fold rotational symmetry, as in the first image here (I think it might be possible to do better with a less symmetrical configuration). - (source) -If realised as a physical object this should have the interesting properly that the only way in which the tori can move is rotation along their axes. -I have a numerical approximation to the best arrangement of the tori, given as follows: -Set up a coordinate system as in the second image, with one circle (of radius $1$) centered on the $x$-axis, at distance $r$ from the origin, and rotated by angles $\theta$ and $\omega$ from the axes. The vectors $U$ and $V$ give the orientation of the circle, and are given by -$$U = (\cos(\omega), \sin(\omega), 0), \qquad V=(-\sin(\omega)\sin(\theta), \cos(\omega)\sin(\theta), \cos(\theta)).$$ -The other two circles are copies of this one, rotated by $2\pi/3$ and $4\pi/3$ about the $z$-axis. The approximation I have is $r=0.4950, \omega = 0.0000, \theta = -0.8561$, with resulting distance between circles of $0.64576$ (and so torus thickness is $0.32884$). These are accurate to around $4$ decimal places. The arrangement in the picture is this approximation. -Questions: - -Why does it appear that $\omega=0$? It doesn't seem obvious to me that this should be true. Is there a symmetry argument? -Is there a closed form solution? - -A closed form solution is probably too much to hope for. In particular, this paper: Finding the distance between two circles in three-dimensional space shows that there is no closed form solution for the distance between arbitrary pairs of circles in $\mathbb{R}^3$. But maybe there is some symmetry argument that helps? -Edit: Better estimates: $r=0.4950197, \theta=-0.8560281$. By Ian Agol's answer, $w=0$. Here is another mysterious symmetry in the numerical solution. Set up a coordinate system on the surface of each torus. Each torus is parameterised by $\alpha, \beta\in[-\pi,\pi)$. The parameter $\alpha$ is in the longitude direction, with 0 at the vector $U$ and $V$ nearest in the positive direction. The parameter $\beta$ is in the meridian direction, with 0 at the biggest longitude (i.e. on the outside of the torus), and the direction $U\times V$ closest in the positive direction. With these coordinates, we can plot the points of contact with the other two tori ($\alpha$ on the x-axis, $\beta$ on the y-axis): - (source) -The (numerically approximated) positions of the points are: -$(-2.941921822296, -1.2298655866392636),$ -$(-1.9117269877782, 2.941921878383725),$ -$(1.9117269877782, -2.941921878383725),$ -$(2.941921822296, 1.2298655866392636)$ -Why does the number $2.941921$ appear in both $\alpha$ and $\beta$ coordinates? - -REPLY [9 votes]: I think the minimizer should have dihedral symmetry. I'll give a -heuristic explanation for this suggestion. -Consider two linked solid tori of the same shape, such that -the outer diameter is less than twice the inner diameter of -the hole. If two such tori are tangent at a single point, -then one may separate them by a translating each torus -in the direction of a vector pointing to one side -of the tangent plane at the tangency point. -Now assume that the two linked tori are related by a rotation -of order three, rotating one torus to the other. If there -is only one tangency point, then one may choose one vector -pointing to one side of the tangency plane, which is rotated -to the other side of the tangency plane by the rotation. Move each torus by a small translation in the direction of its -vector, then by the above observation, they will be separated. -A third torus obtained by taking the inverse rotation will -also be separated, by symmetry. -Thus, a tight configuration must have two tangencies between -each solid torus. What I believe is that two tangent linked isometric tori -with two tangency points should be related by an involution rotation -exchanging the two (and exchanging the tangency points). -If this is true, then in a tight configuration, there should -be an extra dihedral symmetry, which would imply $\omega=0$.<|endoftext|> -TITLE: A percolation problem -QUESTION [26 upvotes]: Let's consider the 2-dimensional integer lattice $\mathbb{Z}^2$ for simplicity. In "ordinary" bond percolation, there is a parameter $p \in [0,1]$, and each edge is on with probability $p$. Consider now the following model : all edges are present, but each is to be given an orientation either away from or toward the origin (this is well-defined in $\mathbb{Z}^2$). Hence, each edge is oriented away from $(0,0)$ with probability $p$, and toward $(0,0)$ with probability $1-p$. The basic question is whether we percolate when $p > 1/2$, i.e.: if $p > 1/2$, is there a non-zero probability of there being an oriented path from $(0,0)$ to infinity ? The intuition behind this is that a biased 2-D random walk is known to be transient. I have seen other models of "oriented percolation" but not this one, so my question is really whether there are any results on this model ? By the way, it is known we don't percolate at $p = 1/2$. Here the model coincides with one that appears for example in several of Grimmett's books, where the bias is in a certain direction (north and east), rather than away from $(0,0)$. There it is also apparently open whether one percolates for $p > 1/2$. - -REPLY [2 votes]: Not sure if that is what Linusson does, but it is not difficult to see that the cluster distribution at the origin at $p=1/2$ is exactly the same as for ordinary percolation at $p=1/2$ (in fact, that holds in any graph, in particular in $\mathbb Z^3$ there is percolation wpp at $p=1/2$; there is no use of duality). -Here is a coupling argument. Start with standard percolation, and explore the cluster of the origin by looking at every edge you need one by one. On the directed side of the coupling, if the edge is open, put it in the orientation that allows you to proceed, and if it is closed, put it in the reverse orientation. That builds an orientation on a random set of bonds, which you can complete into a configuration in the whole plane. -In particular, the average size of the cluster at $p=1/2$ is infinite. Maybe there is even Russo-Seymour-Welsh and so on, anyway it looks very critical. And there is "too much symmetry" in that for every $x$ in the plane, the cluster of $x$ has the same distribution in both models. -But indeed without monotonicity it is difficult to see how to proceed ...<|endoftext|> -TITLE: Gromov-Hausdorff distance between p-adic integers. -QUESTION [20 upvotes]: What is the distance in the sense of Gromov-Hausdorff between $\mathbb{Z}_{p_1}$ and $\mathbb{Z}_{p_2}$ with the usual p-adic metrics? -I got stuck and simply have no idea how to deal with such questions: I've got two metric trees and have to observe somehow all embeddings to all spaces which seems a bit intractable. - -REPLY [12 votes]: The Gromov--Hausdorff distance is good only to define topology; -i.e., one should not care about distance between particular spaces. -But since you insist, I will answer an easier question which is closely related. -There is a modified distance $d'_{GH}(X,Y)$ defined as infimum of all numbers $\varepsilon>0$ such that there are maps $f_1\colon X\to Y$ and $f_2\colon Y\to X$ such that -$$|f_i(x)-f_i(y)|\ge |x-y|-\varepsilon.$$ -This distance $d^\prime_{GH}$ is equivalent to $d_{GH}$ -and it is usually easier to find value $d^\prime_{GH}$ -If $ p < q < p^2$ then it is easy to see that -$$ d^\prime_{GH} ( \mathbb Z_{p},\mathbb Z_{q}) = \tfrac{p-1}{p}. $$ -Further, if $ p^2 < q < p^3$ then -$$ d^\prime_{GH} ( \mathbb Z_{p},\mathbb Z_{q}) = \tfrac{p^2-1}{p^2}$$ -and so on.<|endoftext|> -TITLE: Mumford-Tate group and Galois representations -QUESTION [6 upvotes]: Could someone please point me towards a proof of why the image of a Galois representation on the Tate-module of an abelian variety is limited by its Mumford-Tate group? - -REPLY [12 votes]: This is an immediate consequence of Theorem 2.11 and Proposition 2.9 in Deligne's 'Hodge cycles on abelian varieties' (notes by Milne available here: http://www.jmilne.org/math/Books/DMOS.pdf -2.11 shows that every Hodge cycle on an abelian variety $A$ over a field $k$ embeddable in $\mathbb{C}$ is absolutely Hodge, and 2.9 shows that all absolutely Hodge cycles are defined over a finite extension of $k$. So it follows that all Hodge cycles on an abelian variety have canonical $l$-adic realizations that are defined over the same finite extension of $k$. In particular, an open sub-group of the Galois group fixes all Hodge cycles; that is, it maps into the Mumford-Tate group of $A$.<|endoftext|> -TITLE: Twisted de-Rham cohomology and Eilenberg-Mac Lane spaces -QUESTION [11 upvotes]: One can define twisted cohomology theories via bundles of classifying spaces. In particular, given a cohomology theory $h^{*}$ and a corresponding $\Omega$-spectrum $E_{n}, \varepsilon_{n}$, we can consider on a space $X$ a bundle with fiber $E_{n}$, and define a twisted version of $h^{n}$ as the set of homotopy classes of sections of such a bundle. If the bundle is trivial, we recover the ordinary cohomology theory. -Let us consider a manifold $X$ and its de-Rham cohomology. I suppose there are no problems in defining the Eilenberg-MacLane spaces $K(\mathbb{R}, n)$, even if $\mathbb{R}$ is not countable. For a fixed odd-dimensional form $H$, we can define the (even and odd) twisted de-Rham cohomology groups, via the twisted coboundary $d + H\wedge$, for $d$ the de-Rham differential. -Is there a way to relate the two previous approaches? In particular, given an odd-form $H$, is there a suitable bundle of real Eilenberg-MacLane spaces, whose homotopy classes of sections correspond to the twisted cohomology classes defined via $d + H\wedge$? - -REPLY [5 votes]: You should turn your d+H into a flat superconnection of degree one. Here is the example which works for K-theory. You consider the $\mathbb{Z}$-graded complex $\Omega(M)[b,b^{-1}]$ with $b$ of degree $-2$ and define the superconnection by $d+bH$ for the closed $3$-form $H$. -There is an equivalence between the $\infty$-categories of -such superconnections and bundles of chain complexes (representations of the singular complex of your underlying manifold), see Block-Smith. If you then apply the Eilenberg-MacLane equivalence between the categories of chain complexes and $H\mathbb{Z}$-modules, then you get a bundle of $H\mathbb{Z}$-modules. This (or the bundle of its $\infty$-loop spaces) is what you are lokking for. Actually, all these steps are equivalences so that you can go backwards.<|endoftext|> -TITLE: K-theory and regular rings -QUESTION [5 upvotes]: Let $R$ be a noetherian (commutative) ring. It is a well-known fact that for $R$ regular, $K$-theory of (finitely generated) projective modules and $K$-theory of arbitrary (f.g.) modules agree. Does the converse hold, i.e., suppose the natural map -$$K_i(Proj R-Mod) \rightarrow K_i(R-Mod)$$ -is an isomorphism for all $i \geq 0$ (or just for $i=0$, maybe). Is $R$ regular then? - -REPLY [9 votes]: I don't know how to answer that question, but suppose we strengthen the hypothesis by making it apply also to all polynomial rings $R[t_1,\dots,t_n]$ over $R$, for $n \ge 0$. Then because $K_i(R-Mod) \to K_i(R[t]-mod)$ is an isomorphism (Quillen, Theorem 8, Higher Algebraic K-theory:I) for all noetherian rings $R$, it follows that $K_i(R) \to K_i(R[t_1,\dots,t_n])$ is an isomorphism for all $n$ --- that is called $K_i$-regularity of $R$. The paper "$K$-regularity, cdh-fibrant Hochschild homology, and a conjecture of Vorst", J. Amer. Math. Soc. 21 (2008), no. 2, 547–561, by Cortiñas, Haesemeyer, and Weibel available at http://www.math.uiuc.edu/K-theory/0783/chwpost.pdf shows that $K_i$-regularity of $R$ for all $i$ implies regularity of $R$ if $R$ is an algebra essentially of finite type over a field of characteristic 0.<|endoftext|> -TITLE: How this set of functions is ordered? -QUESTION [8 upvotes]: Notation: - -$k, m, n$ are non-negative integers -$f, g, h$ are functions $\mathbb{N} \to \mathbb{N}$ -$f^k$ is $k$-th iterate of the function $f$: $f^0(n)=n, f^{k+1}(n)=f^k(f(n))$ -$f \prec g$ means eventual domination: $\exists_m \forall_{n>m} f(n) < g(n)$. - -Let $S$ be the minimal set of functions $\mathbb{N} \to \mathbb{N}$ such that - -$n \mapsto 0, n \mapsto 1, n \mapsto n, n \mapsto n+1 \in S$, and -If $f, g, h \in S$ then $n \mapsto f^{g(n)}(h(n)) \in S$. - -Is $S$ well-ordered by $\prec$? If yes, then what is the corresponding ordinal? Otherwise, what is the supremum of ordinals corresponding to well-ordered subsets of $S$? Can you suggest an algorithm implementing $\prec$? -Update: nontrivial lower and upper bounds on the ordinal would be also appreciated. - -REPLY [7 votes]: Here is a trivial way in which it is not well-ordered: the ordering used is not total (we have a preorder, rather than a total order). This is because we can construct a function $Z(n)$ given by $Z(0)=1$ and $Z(n)=0$ for $n>0$. Simply take, in the combination law, $f(n)=0$, $g(n)=n$, and $h(n)=1$. Then $Z$ and $0$ are distinct but equivalent. -Edit: Actually, here's a further (but more problematic) way in which it's not totally-ordered. We can construct $E(n)$ where $E(n)=0$ if $n$ is even and $E(n)=1$ if $n$ is odd. Simply take, in the combination law, $f=Z$, $g(n)=n$, and $h(n)=0$. If we instead take $h(n)=1$, we get $D(n)$ where $D(n)=1$ if $n$ is even and $E(n)=0$ if $n$ is odd. But then neither of $E$ and $D$ eventually dominates the other, in the stronger sense that they are not eventually equal, either. -So even if we take equivalence classes (two functions being equivalent if they're eventually equal), we will still only have a partial order. I suppose we could ask if it's a well-founded partial order... (it's certainly not a well-partial order in the stronger sense of that term, I will post a proof here later). -Edit again: OK, here is an example of an infinite antichain. Note that this still might be a well-founded partial order. (Later: It isn't, see below.) -I will show that all periodic functions are in $S$; then for primes $p$, we can take $E_p(n)=1$ if $p\mid n$ and $E(n)=0$ otherwise, and these will form an infinite antichain. I will do this by inductively showing that for any $m\ge 1$, all functions of period $m$ are in $S$. -First observe that $S$ is closed under pointwise addition, since given $g,h\in S$ we can apply the combination law with $f(n)=n+1$ and the given $g$ and $h$. Thus all constant functions are in $S$ and so the case $m=1$ is proved. So suppose it is true for $m$, and we want to prove it for $m+1$. -Again, since $S$ is closed under pointwise addition, it suffices to construct each of the functions $E_{m+1,k}$ given by $E_{m+1,k}(n)=1$ if $n\equiv k \pmod{m+1}$ and $E_{m+1,k}(n)=0$ otherwise. Actually, it suffices to construct a function $M$ such that: - -$M$ is periodic with period $m+1$ -There is a unique congruence class $a$ mod $m+1$ which is mapped to $0$ by $M$. - -Once we have this, since $S$ is closed under compostion (take $g=1$ in the combination law), and contains all functions of the form $n\mapsto n+k$ (either by addition or composition), we can make $E_{m+1,a-k}$ by taking $E_{m+1,a-k}(n)=Z(M(n+k))$. -It remains to construct $M$. By the inductive hypothesis, we may construct $L$ given by $L(n)=(n-1) \bmod{m}$. Then we can construct $L'$ given by $L'(n)=L(n)$ for $n>0$ and $L(0)=m$ by taking $f(n)=m$, $g=Z$, and $h=L$ in the combination law. Now we just take $f=L'$, $g(n)=n$, $h=0$ in the combination law to get $M$: $L'(0)=m$, and for $0\lt n\le m$, $L'(n)=n-1$, meaning that iterating $L'$ starting from $0$ yields an $m+1$-cycle, with $1$ serving the role of $a$ above. -Section below updated to now include an explicit construction -Edit yet again: Using the above and a similar construction, you can also get that eventually periodic functions are in $S$, and using that, you can get that taking a function in $S$ and modifying it at only finitely many places yields another function in $S$ (just take, in the combination law, $h$ is your old function, $f$ is a function containing your replacement values at the appropriate spots, and $g$ is a function that is $1$ on the appropriate spots and $0$ elsewhere). -Proof: It suffices to show that for each $k$, the function $Z_k$ given by $Z_k(n)=1$ if $n=k$ and $Z_k(n)=0$ otherwise lies in $S$. So take a periodic function $J$ such that $J(0)=k+1$, $J(n)=n-1$ for $1\le n\le k$, and $J(k+1)=k+1$. Then construct $Y(n)$ by taking in the combination law, $f=J$, $g(n)=n$, and $h(n)=k$. Then for $n\le k$, $Y(n)=n-k$, and for $n>k$, $Y(n)=k+1$. In particular, $Y(n)=0$ iff $n=k$, so we can take $Z_k=Z\circ Y\in S$. -Actually, the above provides an example of -- well, it's not actually an infinite descending sequence in the order he actually defined, but it is an infinite descending sequence in the obvious nonstrict modification (which is what I have really been implicitly using all along -- otherwise the distinction between "equivalent" and "incomparable" doesn't really make sense!). Simply take $E_m$ as above and observe that $E_{2^{k+1}}\lt E_{2^k}$. Of course, this is a slightly different notion of "less than", so the question of whether one can find an infinite descending sequence with the original notion of "less than" remains unanswered. -Final edit: Here I will construct an infinite descending sequence that works even with Vladimir's original, stricter, ordering, showing that it does not, in any way, well-order this set. (However, it seems to me to be plausible that Vladimir's idea is correct if $0$ is excluded from $\mathbb{N}$; as Gerald points out, everything I'm doing is based around trickery with $0$, and indeed you can prove that in the absence of zero, every function constructible this way must be either constant, the identity, or strictly monotonic and satisfing $f(n)\gt n$.) -I will show that for all $k\ge 1$, the function $F_k(n)=\max(0,2(n-k))$ lies in S. (Of course, we already know it's in $S$ for $k\le 0$ by other means!) Then $F_1 \succ F_2 \succ \ldots$ forms an infinite descending chain. First, define $A_k(n)$ by $A_k(n)=0$ if $n=2k-1$ and $A_k(n)=n+2$ otherwise; then $A_k\in S$ by applying the combination rule with $f(n)=0$, $g=Z_{2k+1}$, and $h(n)=n+2$ (since $n=2k-1$ iff $n+2=2k+1$). Then construct a function $G_k(n)$ by applying the combination rule with $f=A_k$, $g(n)=n$, and $h(n)=1$. Then for $n\le k-1$, $G_k(n)=2n+1$, $G_k(k-1)=2k+1$, $G_k(k)=0$, and $G_k(n)=2(n-k)$ for $n\ge k$. This is really enough, but for completeness, define $E'(n)=n$ for n even and $E'(n)=0$ for n odd; this lies in $S$ by applying the combination rule with $f(n)=0$, $g=D$ from above, and $h(n)=n$. Then $F_k=E'\circ G_k\in S$, and we are done. -And now I really better not edit this anymore or I think it'll become CW! But I think I've answered the $0$-included case pretty thoroughly now; if I figure out anything about the case when $0$ is excluded, I'll make that a separate answer. -(Edit many years later: Now that automatic CW is no longer a thing, I'm going back and editing this post to be slightly more readable. :) )<|endoftext|> -TITLE: Non-uniqueness of solutions of the heat equation -QUESTION [9 upvotes]: For the heat equation $(\partial_t-\partial_x^2)f(t,x)=0$ defined on $[0,T)\times(-\infty,\infty)$, to obtain uniqueness of the initial value problem, usually it is required to limit the growth of the potential solution at infinity, i.e. $|f(t,x)|<\exp(c\cdot x^2)$. -My question is, if we do not impose any such conditions, is uniqueness no longer valid? In particular, is there a well known example of a function $f(t,x)$ that satisfies the heat equation on $[0,T)\times(-\infty,\infty)$, $f(0,x)=0$, but $f$ is not identically zero? -What if we relaxed the conditions a little bit, and only required that $f$ satisfies the heat equation in $(0,T)\times(-\infty,\infty)$ and is continuous on $[0,T)\times(-\infty,\infty)$, is there an example in this case? - -REPLY [14 votes]: Tychonoff in his 1935 paper Théorèmes d'unicité pour l'équation de la chaleur proved uniqueness if the solutions are not too large, and gave an example to show that the solution is not unique in general. His counterexample grows extremely rapidly for large x.<|endoftext|> -TITLE: Center of the category of special $\lambda$-rings -QUESTION [5 upvotes]: Recall that the center $\mathrm{Z}(C)$ of a category $C$ is the monoid of endomorphisms of $\mathrm{id}_C$. Thus $\eta \in \mathrm{Z}(C)$ is given by a familiy of endomorphisms $\eta_x : x \to x$, where $x \in C$, such that for all morphisms $x \to y$ the obvious diagram commutes. The center of the category of rings is trivial (see here). -In the category of special $\lambda$-rings $\lambda\mathrm{Ring}$, we have for every $q \in \mathbb{N}_{\geq 1}$ the Adams-Operation $\Psi^q$ which is known to be a $\lambda$-ring endomorphism for every special $\lambda$-ring and is compatible with $\lambda$-ring homomorphisms (see these notes by Darij Grinberg). Besides, we have $\Psi^p \circ \Psi^q = \Psi^{pq}$ and $\Psi^1 = \mathrm{id}$. This shows that there is a homomorphism of monoids -$(\mathbb{N}_{\geq 1}, *) \to \mathrm{Z}(\lambda\mathrm{Ring})$, $q \mapsto \Psi^q$ -It is easy to see that it is injective. It is also surjective? If not, what do we have to add to get $\mathrm{Z}(\lambda\mathrm{Ring})$? - -REPLY [4 votes]: Your map is surjective too. -The free (special) $\lambda$-ring on one generator is a polynomial algebra of the form $F=\mathbb{Z}[\lambda^1(x),\lambda^2(x),\lambda^3(x),\dots]$. (This is well-known; I think Donald Yau proves it in his book on $\lambda$-rings.) The set of endomorphisms of the forgetful functor $\lambda\mathrm{Ring}\to \mathrm{Set}$ corresponds to the underlying set of $F$, so $Z(\lambda\mathrm{Ring})\subset F$. -One way to proceed from here is to use the fact that $A\mapsto A\otimes \mathbb{Q}$ is a functor $\lambda\mathrm{Ring}\to \mathbb{Q}\backslash\lambda\mathrm{Ring}$, and that $\lambda$-rings containing $\mathbb{Q}$ are nothing more that commutative $\mathbb{Q}$-algebras equipped with Adams operations. It should be easy to see that -$Z(\lambda\mathrm{Ring})\subseteq Z(\mathbb{Q}\backslash \lambda\mathrm{Ring})$, since $F\subseteq F\otimes\mathbb{Q}$, and that $Z(\mathbb{Q}\backslash\lambda\mathrm{Ring})=\mathbb{N}$. -Added. -Let me write $F\{x_1,x_2,\dots\}$ for the free (special) $\lambda$-ring on generators $x_1,x_2,\dots$. Then $F\{x_1,x_2\}\approx F\otimes F$, since coproducts in $\lambda$-rings are tensor products. There is a comultiplication $\Delta\colon F\{x\}\to F\{x_1,x_2\}$ (i.e., $F\to F\otimes F$) defined by sending $x\mapsto x_1+x_2$; it makes $F$ into a Hopf algebra. The map $\Delta$ encodes how polynomials in $\lambda$-operations act on sums, so we see that $\psi^k(x)\in F$ is primitive: $\Delta(\psi^k(x))=\psi^k(x_1)+\psi^k(x_2)$. -The subgroup $P\subset F$ corresponds precisely to the set of polynomials $f(\lambda^1,\lambda^2,\dots)$ such that $f(x+y)=f(x)+f(y)$ in any $\lambda$-ring. -I want to identify $P$ with the $\mathbb{Z}$-linear span of the $\psi^k(x)$. It is a little easier to identify the subgroup of primitives in $F\otimes \mathbb{Q} \approx \mathbb{Q}[\psi^1(x),\psi^2(x),\psi^3(x),\dots]$ as the $\mathbb{Q}$-linear span of the $\psi^k(x)$ (for instance, using structure theorems for Hopf algebras; $F\otimes\mathbb{Q}$ is primitively generated as a Hopf algebra.). -There is a second comultiplication $\Delta'\colon F\{x\}\to F\{x_1,x_2\}$ sending $x\mapsto x_1x_2$, which encodes how operations act on products. We want the elements inside $P$ (or just $P\otimes \mathbb{Q}$) which are grouplike with respect to $\Delta'$, (i.e., $\Delta'(u)=u\otimes u$). We already know that the $\psi^k(x)\in P$ have this property, so we just need to show that if a linear combination of $\psi^k(x)$'s is grouplike wrt to $\Delta'$, then it is just a single $\psi^k(x)$, which is relatively elementary.<|endoftext|> -TITLE: Resubmitting a paper -QUESTION [6 upvotes]: (Hopefully this question is not too "soft": I asked it at math.stackexchange.com recently and was advised to ask it here instead). -I submitted a paper to quite a low-rated journal (impact factor less than 0.3) a while ago, partly because I didn't think that my results were particularly good, but also because at the time I was largely ignorant of the significance of journal rankings. After many months "with editor" it has now finally been sent to a referee, but I'm slightly regretting choosing this journal, as recent feedback has given me the impression that my paper is publishable in a much better journal, and I am at the point in my career (starting to look for my first post-doc position) where I really need some good additions to my publication record. -Now that someone is possibly taking the time to read it, would it be considered bad form to withdraw the paper (or wait until the review comes back, assuming it is favourable), and submit it elsewhere? And is it worth the risk? When considering job applications, how much emphasis do people actually put on where a person's papers have been published, as opposed to how many have been published? -Any advice much appreciated. - -REPLY [16 votes]: Research mathematics is a small world. It is rather likely that the people who are currently editing and refereeing your paper will play a significant role in the development of your career in the next 5 years or even more. They might not take too kindly to the withdrawal, especially if they think that the move was not really warranted (i.e., if they don't think that your result is really much too good for the journal you submitted). So think to yourself: is this a risk worth taking? Is your result that badly served by the current journal? -As for the impact factor, have you really tried to find out what people think about those? Deans and other paper pushers might put a lot of store by them, and so it's definitely something to keep in mind in the long run, but your colleagues' opinion of a journal is not based on a somewhat artificial numerical construct, so for your immediate future, the actual name and reputation of the journal is much more important than a somewhat arbitrary numerical score. (It is a fact that the math journals with the highest impact factor are not necessarily the most prestigious ones.) - -REPLY [9 votes]: A kind of compromise solution: leave your paper where it is, and, given that you now realize that your result is worth more publicity, write a short note (like 1 page or so) where you quote your paper and describe its main results without proofs (possibly, adding small improvements or remarks). A journal expressly aimed to this end is e.g. Comptes Rendus Mathematique, but many other do it as well; your advisor may give you a good hint.<|endoftext|> -TITLE: Where might I find a scanned handwritten copy of Ramanujan's second letter to Hardy? -QUESTION [18 upvotes]: I am giving a lecture to undergraduates on the lovely identity $$1 + 2 + 3 + 4 + \cdots = -\frac{1}{12}.$$ -Ramanujan wrote in his second letter to Hardy (courtesy Wikipedia), -"Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich's Infinite Series and not fall into the pitfalls of divergent series. … I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter. …" -Does anyone know where I might find a scanned copy of this or closely related material, in Ramanujan's handwriting? Bruce Berndt kindly pointed me to a typeset version, and much else related of extreme interest -- but I would love to use the original in my lecture, if possible. -Thank you! - -REPLY [2 votes]: This article, by Stephen Wolfram from his book Idea Makers, has 9 photo images of Ramanujan's letter to Hardy. Not quite sure if it is the whole of his second letter, but it has the famous -1/12 formula on it.<|endoftext|> -TITLE: Effective topos and computability in topological spaces -QUESTION [12 upvotes]: The classical computability theory taking place in $\mathbb{N}$, can be extended to more general spaces, like $T_0$ second countable topological spaces $(X, \mathcal{O}, v)$ where $\mathcal{O}$ is a countable basis of $X$ and $v:\mathbb{N} \rightarrow \mathcal{O}$ a total surjection. Then we say that $f:X \rightarrow Y$ is computable if given any enumeration of all basic open sets containing $x$, one can compute (in the sense of Turing) an enumeration of all basic open sets contaning $f(x)$. Equivalently we can state that for any basic open set $O$ of $Y$, we have that $f^{-1}(O)$ is a recursively enumerable open in the sence that $f^{-1}(O)=\cup_n O_n$ where $O_n$ is a recursively enumerable sequence of basic open sets of $X$ -This brings some sort of a generalization of Turing reduction, when $X$ and $Y$ are both the Cantor space. -Now, I am wondering if it is possible to have a categorical point of view on all this. My research brought me to Topoi, the effective topos and the recursive topos. I have some difficulties to understand what is the effective topos (as I have to understand what is the recursive topos). -Before I spend a lot of time to study them, I would like to know if there are somehow related to what I've stated above (i.e., does one of them contains in some way the computable functions between topological spaces ? is there only continuous functions between topological spaces in these Topos ? can we even talk about topological spaces in these topos ?) or are hey something completely different ? -I apologize if my question seems too general. Any answer will be appreciated :) - -REPLY [2 votes]: Without certain additional assumptions, the two characterizations of computability in topological spaces are not equivalent also in the case when uniformity is required in the second characterization. Indeed, the following property follows then from computability in the second sense: a partial recursive function $\iota$ exists such that, whenever $O$ is a basic open set of $Y$, $j$ is an index of $O$ in the given surjection of $\mathbb{N}$ onto the basis of $Y$, and $f^{-1}(O)$ is non-empty, then the number $j$ belongs to the domain of $\iota$, and the set with index $\iota(j)$ in the given surjection of $\mathbb{N}$ onto the basis of $X$ is a subset of $f^{-1}(O)$. On the other hand, computability in the first sense can be present without having this property. To construct an example for this, we can proceed as follows. We first construct such a partition of $\mathbb{N}$ into disjoint two-elements sets of the form {$n_0,n_0+1$},{$n_1,n_1+2$},{$n_2,n_2+3$},{$n_3,n_3+4$},$\ldots$ that no recursive function exists whose value at $j$ belongs to {$n_j,n_j+j+1$} for all $j$ in $\mathbb{N}$. Then we take $X=Y=\mathbb{N}$ with the discrete topology, its basis consisting of all singletons in $\mathbb{N}$, but with different surjections of $\mathbb{N}$ onto this basis. For the space $X$, let both numbers $n_i$ and $n_i+i+1$ be indices of the basic set {$i$}, and in the case of $Y$, let $j$ be the only index of {$j$}. Let $f(x)=x$ for all $x$ in $\mathbb{N}$. Then $f$ is computable in the first sense, since, given any enumeration $k_0,k_1,k_2,\ldots$ of the set {$n_x,n_x+x+1$}, we have the equality $x=|k_l-k_{l+1}|-1$, where $l$ is the first $m$ such that $k_m\ne k_{m+1}$. However, $f$ has not the above-formulated property. -For assumptions which guarantee the equivalence of the two characterizations, cf. Theorem 3.3 in M. Korovina's and O. Kudinov's paper "Towards Computability over Effectively -Enumerable Topological Spaces", Electron. Notes Theor. Comput. Sci., 221 (2008) 115--125, -https://doi.org/10.1016/j.entcs.2008.12.011 (Unfortunately, the formulation of the theorem needs a correction. For the validity of its conclusion some additional assumption is needed. For instance, it is sufficient to add the assumption that $\alpha i$ is empty for some $i$). -Remark. A partition of $\mathbb{N}$ with the properties used in the above counter-example can be made as follows. We take a sequence $k_0,k_1,k_2,\ldots\,$ of natural numbers which is dominated by no recursive function and satisfies $k_{l+1}>k_l+2l+1$ for all $l$ in $\mathbb{N}$. Then we form subsets $C_0,C_1,C_2,\ldots$ of $\mathbb{N}^2$ so that $C_0$ consists of the pairs $(k_l,k_l+2l+1)$ for $l=0,1,2,3,\ldots$, and, for all $r$ in $\mathbb{N}$, $C_{r+1}$ is obtained from $C_r$ by adding a pair $(\bar{n},\bar{m})$ of natural numbers such that $\bar{n}<\bar{m}$ and next three conditions are satisfied: -1. The numbers $\bar{n}$ and $\bar{m}$ occur in no pair from $C_r$. -2. Whenever $(n,m)\in C_r$, the inequality $\bar{m}-\bar{n}\ne m-n$ holds. -3. If $r$ is even then $\bar{n}$ is the least natural number which occurs in no pair from $C_r$, otherwise $\bar{m}-\bar{n}$ is the least positive integer different from all differences $m-n$, where $(n,m)\in C_r$. -Let $C$ be the union of the sets $C_0,C_1,C_2,\ldots$ Then, for any $j$ in $\mathbb{N}$, exactly one pair $(n,m)$ in $C$ exists such that $m-n=j+1$. We set $n_j=n$ for this pair.<|endoftext|> -TITLE: Non-trivial class number at some finite level in the cyclotomic $\mathbf{Z}_p$-extension of $\mathbf{Q}$? -QUESTION [11 upvotes]: An MSc student asked me if I knew an example of a prime $p$ and some finite layer $K_n$ in the cyclotomic $\mathbf{Z}_p$-extension of $\mathbf{Q}$ (so $[K_n:\mathbf{Q}]=p^n$) which had non-trivial class group. My gut feeling was that fixing a $p$ and then going up the tower was a bad idea in the sense that going along might find a counterexample quicker, so I tried going along instead, but my computer is having trouble looking at the bottom level when $p$ starts getting bigger than about 30. So now I'm just confused. Presumably this question has been raised before? Can anyone enlighten me with a counterexample or reassurance that this is a standard open question? - -REPLY [12 votes]: I found the notes of Coates' seminar I alluded to in my comment above. He said the following: -For $n \ge 1$, let $h(n)$ denote the class number of the unique cyclic degree $n$ extension contained in the compositum of all the cyclotomic $\mathbb{Z}_p$-extensions for $p \mid n$. It is apparently not too difficult to show that if $n \mid m$, then $h(n) \mid h(m)$, and that if $n$ is a power of $p$, then $h(n)$ is prime to $p$. -Then Weber (not Kronecker, but close!) has conjectured that $h(2^n) = 1$ for all $n$, and this is known for $n \le 5$ and for $n=6$ conditional on GRH. More generally, it is a folklore conjecture that $h(n) = 1$ for all prime power values of $n$. -Horie, Fukuda and Komatsu have recently shown that $h(62)$ is divisible by 31, the first known example of an integer $n$ where $h(n) > 1$. -So the verdict seems to be that your student's question is a well-known open problem.<|endoftext|> -TITLE: Piecewise constant functions with zero average -QUESTION [9 upvotes]: Consider $0=t_0\leq t_1\leq...\leq t_n=1$, $f_0,...,f_{n-1}\in\mathbb{Z}$ and $F:[0,1]\to\mathbb{R}$ be such that -1) $F\equiv f_i$ on the interval $(t_i,t_{t+1})$, for all $i=0,...,n-1$, -2) $\displaystyle \int_0^1 F(t) dt=\sum_{i=0}^{n-1}(t_{i+1}-t_i)f_i=0$. -Does there exist an arbitrarily large prime number $p$ and a positive integer $k=k(p)$ such that $q:=p^k$ satisfies -$\displaystyle \sum_{i=1}^{q-1} F\left(\frac{i}{q}\right)=0$ ? -I know that the answer is YES when all the $t_j$'s are rational number: if $t_j=\frac{p_j}{q_j}$, then it suffices to choose $q\equiv 1$ mod $\mathrm{lcm}(q_1,...,q_{n-1})$. -Any idea for the general case? - -REPLY [6 votes]: Thanks to the comments of George Lowther and Greg Martin (for which I am most grateful), I can now show that the answer is YES for infinitely many primes $q$. -Theorem 1. Let $t_1\dots,t_{n-1}$ be any finite set of real numbers.Then for any $\epsilon>0$ and any integer $r>0$ there are infinitely many primes $q\equiv 1\pmod{r}$ such that -$$\|(q-1)t_i\|<\epsilon,\qquad i=1,\dots,n-1.\tag{1}$$ -Here $\|x\|$ stands for the distance of $x$ to the nearest integer. -Proof. Without loss of generality, the numbers $1,t_1,\dots,t_{n-1}$ are linearly independent over $\mathbb{Q}$. Indeed, we can express each of them as a $\mathbb{Z}$-linear combination from a suitable basis $\frac{1}{s},t_1^*,\dots,t_{m-1}^*$ of their $\mathbb{Q}$-linear span, where $s>0$ is an integer. Then the statement for $t_1,\dots,t_{n-1}$ follows from the statement for $t_1^*,\dots,t_{m-1}^*$ (with $\mathrm{lcm}(r,s)$ in place of $r$). When the elements of $1,t_1,\dots,t_{n-1}$ are linearly independent over $\mathbb{Q}$, -the statement follows from the stronger result that the vectors $(qt_1,\dots,qt_{n-1})$ are dense modulo $1$, as $q$ runs through the primes such that $q\equiv 1\pmod{r}$. This result is essentially due to Vinogradov, with technical improvements by Vaughan and Harman. See Theorem 4 in Harman: Diophantine approximation with primes, J. London Math. Soc. (2) 39 (1989), 405–413. Well, Harman does not have the condition $q\equiv 1\pmod{r}$, but it seems straightforward to incorporate it. -Theorem 2. Let $t_i$, $f_i$, $F$ be as in the original question but without the assumption $f_i\in\mathbb{Z}$. Then for any $\epsilon>0$ there is a prime $q$ such that -$$\left|\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)\right|< -\epsilon.$$ -In particular, if $F$ is integer valued and $\epsilon=1$, then the left hand side is zero. -Proof. Assume that $\epsilon>0$ is sufficiently small, namely -$$\epsilon<\|t_i\|,\qquad i=1,\dots,n-1.\tag{2}$$ -By Theorem 1, there is a prime $q$ such that (1) holds. -Observe that -$$\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)=\sum_{i=0}^{n-2}f_i\bigl([qt_{i+1}]-[qt_i]\bigr)+f_{n-1}\bigl(q-1-[qt_{n-1}]\bigr),$$ -where $[x]$ stands for the integral part of $x$. Here we used that no $\frac{j}{q}$ coincides with any $t_i$, as follows from (1) and (2). We subtract -$$0=(q-1)\int_0^1 F(t)dt=\sum_{i=0}^{n-1}f_i\bigl((q-1)t_{i+1}-(q-1)t_i\bigr),$$ -then with the notation -$$\tilde t_i:=[qt_i]-(q-1)t_i,\qquad i=0,\dots,n-1,$$ -we get -$$\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)=\sum_{i=0}^{n-2}f_i\bigl(\tilde t_{i+1}-\tilde t_i\bigr)-f_{n-1}\tilde t_{n-1}=\sum_{i=1}^{n-1}(f_{i-1}-f_i)\tilde t_i.$$ -By (1) we can write $(q-1)t_i=n_i+e_i$ with $n_i\in\mathbb{Z}$ and $|e_i|<\epsilon$. Then -$$qt_i = n_i+t_i+e_i,\qquad i=1,\dots,n-1,$$ -whence by (2), that is by $\epsilon<\min(t_1,1-t_{n-1})$, we have -$[qt_i]=n_i$, so that $\tilde t_i=-e_i$. It follows that -$$\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)=\sum_{i=1}^{n-1}(f_i-f_{i-1})e_i,$$ -whence -$$\left|\sum_{j=1}^{q-1}F\left(\frac{j}{q}\right)\right|<\epsilon\sum_{i=1}^{n-1}|f_i-f_{i-1}|.$$ -The right hand side can be made arbitrary small, so we are done.<|endoftext|> -TITLE: Dedekind spectra -QUESTION [10 upvotes]: Is there a class of ring spectra that corresponds to and/or extends the class of Dedekind rings from traditional algebra? Is there a notion of "ring of integers" of a ring spectrum? Additionally, is there a notion of an ideal class group of a Dedekind ring spectrum (Picard group)? -Thanks -PS This question was already asked on math.stackexchange, but I have heard there are some people on this site working on this sort of thing. - -REPLY [17 votes]: In trying to generalize concepts from algebra to spectra, there are several issues that come into play. -In order for a concept in stable homotopy theory to be intrinsically meaningful it generally needs to be invariant under weak equivalence - whatever the appropriate notion of "weak equivalence" is (of spectra, of commutative ring spectra, etc). There are multiple reasons for this. On one hand, it's the homotopy category rather than the category of spectra that is "algebraic" enough to support generalizations like this. On the other hand, there is the practical consideration that there are many different models for spectra (e.g. symmetric spectra, orthogonal spectra, EKMM spectra, various diagram categories); if a concept isn't meaningful from the point of view of homotopy theory, it may have entirely different meanings in different models. -In addition, a concept may have several different directions of generalization. You could generalize an algebraic concept to one that's defined in terms of homotopy groups; this is easy to define and check, but tends to be less interesting and not satisfied in some principal cases of interest. You could try to phrase things in terms of categorical properties, and express a generalization that way; in order for this to be sensible you generally have to replace all concepts by their appropriate "derived" notions (derived pullback, derived invariants under a group action, etc), which makes it difficult to work with concepts that have almost no exactness properties. You could do something ad-hoc. -For these reasons, it's not a straightforward procedure. It's often a good idea to have some examples in mind or be looking for an application, rather than just generalizing for its own sake. A handy test for how difficult it will be is to try and determine a generalization for differential graded modules and algebras first. -Here are some of the pieces that show up in the definition of a Dedekind domain. - -Integral domains. I don't really know a useful generalization that doesn't involve being an integral domain on homotopy groups. This leaves out a lot of interesting examples - there is a large zoo of regularity conditions in algebraic geometry are not satisfied by a large class of ring objects in homotopy theory. -Fields of fractions. Inverting elements - and localization in general - is something that works well in homotopy theory, and tends to give the expected results. -Integral closure. This one is much more difficult, because it involves solutions of an equation, and trying to "adjoin" elements in the fraction field. As David White mentioned, the concept of being a "subobject" is one that doesn't translate well, and so there's not a straightforward way to take an element in the homotopy of the fraction field and adjoin it to the base ring. In general it is very difficult to construct commutative ring objects with prescribed properties. -Rings of integers. See above. If you figure a out a useful notion for this, I'd love to hear from you. -Ideals. Again, there isn't an intrinsic meaning to "subobject" or "quotient". There are generalizations of the concept of an "ideal", but all the ones that I'm aware of boil down to an ideal being, by definition, something that gives you a map out to another ring. More problematically, because taking the "quotient by an ideal" usually involves a mapping cone/cofiber, being an ideal isn't a property of a map $I \to R$ of modules - it is all the extra data that allows you to construct a ring structure on $R/I$. In addition, for $R$ commutative, ideals as an associative ring and ideals as a commutative ring become separate concepts. -Principal ideals. A principal ideal, ideally, would be generated by an element in homotopy that you want to take the quotient by. Simply put, given an element in homotopy you may not be able to construct such an ideal even in cases that look amenable. If you can construct an ideal so that there is a quotient associative algebra, there are likely to be many different choices of quotient algebra structures. Being able to construct a quotient commutative algebra is an entirely different, much harder problem that often doesn't have a solution, and when we hope or expect it to have a solution we often can't prove it. There are decades-old conjectures about some of these. -Dimension. To define dimension you usually need prime ideals. There are useful definitions of dimension that use thick subcategories of the homotopy category of perfect complexes - see the work of Paul Balmer in particular. However, it is much harder to translate "deep" results about dimension into homotopy theory. More seriously, a heavy ratio of the interesting examples we know don't satisfy anything like a Noetherian property. - -Having said all of this, the subject is in flux and we understand more as time goes on. -The Picard group exists very generally, for some notion of "exists". A general definition was given in a paper by Hopkins, Mahowald, and Sadofsky. For a strictly commutative ring spectrum $R$, an invertible module $M$ is one such that there exists an object $N$ such that $M \wedge_R N \simeq R$. The category of such $M$ is the Picard groupoid; if the homotopy category is essentially small then there is an associated Picard group. This has applications in a number of areas including computational applications. You can also interpret $RO(G)$-graded homotopy groups in equivariant homotopy theory in terms of some part of the Picard group - but $RO(G)$-graded homotopy predates this work by a very significant margin.<|endoftext|> -TITLE: Free group actions on complex projective spaces -QUESTION [8 upvotes]: Hi, -is there a free group action of the cyclic group $\mathbb{Z}/n\mathbb{Z}$ on the infinite dimensional projective space $\mathbb{CP}^\infty$ for every $n\in \mathbb{N}$? And if there is one, how does it work? -Thanks - -REPLY [10 votes]: I believe I can give a complete answer. -First, let me collect several earlier comments by myself, Dylan Wilson and Alain Valette. When $N<\infty$ then there can not be a free action of $\mathbb Z_n$ on $\mathbb{CP}^N$ when $n>2$. Indeed, if there were such an action then the square of the generator would act trivially on cohomology. It would therefore have a positive Lefschetz number and hence a fixed point. -(This also means that for $n>2$ no free $\mathbb{Z}_n$ action on $\mathbb{CP}^\infty$ can leave any $\mathbb{CP}^N$ with $N<\infty$ invariant). -When $n=2$ then such an action is possible if and only if $N$ is odd. This action easily generalizes to a $\mathbb Z_2$ action on $\mathbb{CP}^\infty$ (with any definition of $\mathbb{CP}^\infty$ ) with the generator acting on $\mathbb{S}^\infty$ by -$(z_1,z_2,z_3,z_4,\ldots)\mapsto (-\bar z_2,\bar z_1, -\bar z_4,\bar z_3,\ldots)$. This action normalizes the diagonal $S^1$ action and thus descends to an action on $\mathbb{CP}^\infty$ which is easily seen to be free. -Before proceeding further let's discuss the fact that we have two competing definitions of $\mathbb{CP}^\infty$. They are homotopy equivalent to each other but that's not enough for this problem. Indeed, if we are only interested in the question up to homotopy equivalence then the answer is trivially "yes" by Borel construction. -The first definition (what Alain calls topologist's definition) is that $\mathbb{CP}^\infty$ is the direct limit of $\mathbb{CP}^k$s under canonical inclusions $\mathbb{CP}^1\hookrightarrow \mathbb{CP}^2\hookrightarrow \mathbb{CP}^3\hookrightarrow\ldots$. -The other (analyst's) definition is $\mathbb{CP}^\infty_H:=\mathbb{S}_H^\infty/\mathbb S^1$ where $\mathbb{S}^\infty_H$ (here $H$ stands for Hilbert) is the unit sphere in $l_2$. -With the second definition we have that $\mathbb{CP}_H^\infty$ and $\mathbb{CP}_H^\infty\times \mathbb{S}_H^\infty$ are homotopy equivalent (since $\mathbb{S}_H^\infty$ is contractible) and hence are homeomorphic since any 2 homotopy equivalent $l_2$-manifolds are homeomorphic (see section k-11 here). -Moreover, I believe the same works for the first (topological) definition of $\mathbb{CP}^\infty$ as well. In that case $\mathbb{CP}^\infty$ is not modeled on $l_2$ but rather on $\mathbb C^\infty$ (the direct limit of $\mathbb C^k$). However, if I understand the definitions correctly, for such spaces we again have that homotopy equivalence of $\mathbb{CP}^\infty$ and $\mathbb{CP}^\infty\times \mathbb S^\infty$ implies homeomorphism (same reference as before, see here for the specific chapter on infinite dimensional manifolds and relevant definitions) since the spaces involved are $\mathcal C$-absorbing. -To summarize, with whatever definition of $\mathbb{CP}^\infty$ free actions of $\mathbb Z_n$ on it exist because of Borel construction together with the fact that in infinite dimension the Borel construction does not change homeomorphism type for relevant classes of infinite dimensional manifolds. -Lastly, let me add that as observed by Alain, with either definition such action can not be $\mathbb C$-linear for any $n$ (my example for $n=2$ is of course $\mathbb R$-linear only). See his answer for more details.<|endoftext|> -TITLE: Elliptic genus for manifolds with boundary -QUESTION [9 upvotes]: Let M be a closed spin manifold of dimension $d$. One form of the elliptic genus of $M$ is -$$ F(q)=q^{-d/8} \hat A(M) {\rm ch} \otimes_{k=1/2,3/2,\cdots} \Lambda_{q^k}T \otimes_{\ell=1}^\infty S_{q^\ell}T [M] $$ -where the notation follows that of E. Witten, ``The Index of the Dirac Operator in Loop Space." -The coefficient of $q^{n/2-d/8}$ is the index of a Dirac operator $D_n$ which acts on sections of -$S \otimes T_{R_n}$ where $S$ is the spinor bundle and $T_{R_n}$ is the bundle associated to -a representation $R_n$ of $Spin(d)$ with the first few representations being -$$ R_0=1, \qquad R_1=T, \qquad R_2=\Lambda^2 T \oplus T $$ -where $T$ is the fundamental (vector) representation. -I'm interested in the generalization of the elliptic genus to manifolds with boundary. In the actual -application I'm interested in one eventually takes the boundary to infinity to obtain a noncompact -manifold, but I'd be happy to understand the situation for a compact manifold with boundary first. -The index of the Dirac operator in such a situation acquires boundary corrections of the form -$$ CS[ \partial M] - \frac{1}{2}(\eta(0)+h) $$ -where $h$ is the number of zero modes of the Dirac operator on $\partial M$ and $\eta(0)$ is the -$\eta$ invariant. In the examples I'm interested in I believe the Chern-Simons contributions $CS[\partial M]$ vanish. -Summing up these boundary contributions to the index of $D_n$ weighted by -$q^{n/2-d/8}$ leads to a "boundary" contribution to the elliptic genus on manifolds with boundary -with the "bulk" contribution given by $F(q)$. My questions are whether this variant of the elliptic genus has been studied and if so where, whether this leads to interesting invariants of manifolds with boundary, and whether the modular properties of the bulk and boundary contributions are known. - -REPLY [4 votes]: I studied such modular invariants in my thesis, Modular invariants for manifolds with boundary, http://www.tdx.cat/handle/10803/3071 , http://www.tdx.cat/bitstream/handle/10803/3071/migc1de2.pdf , http://www.tdx.cat/bitstream/handle/10803/3071/migc2de2.pdf , and never find the time to get back to them, they are very interesting.<|endoftext|> -TITLE: The Correlation of the Mobius Function and Dirichlet Characters. -QUESTION [6 upvotes]: Let $\chi$ be a Dirichlet character, and define $\phi_\chi (n)$ so that it satisfies $$\sum_{n=1}^\infty \phi_\chi (n)n^{-s}=\frac{\zeta(s-1)}{L(s,\chi)}.$$ -In other words -$$\phi_{\chi}(n)=\sum_{d|n}\mu\left(\frac{n}{d}\right)\chi\left(\frac{n}{d}\right)d=\left(\text{Id}*\mu\chi\right)(n).$$ - -My question is, how large can $\phi_\chi(n)$ be? More precisely, what is the smallest function $f$ such that for all $n$ and $\chi$ $$\frac{\phi_{\chi}(n)}{n}=\sum_{d|n}\frac{\mu(d)\chi(d)}{d}\ll f(n).$$ - -It is not hard to see that $\frac{\phi_{\chi}(n)}{n}\ll \log n$ for all $n$ so $f\ll \log n$. For Euler's $\phi$ function, the first term is the main contributor, as $\phi(n)\leq n$ we know that $\frac{\phi(n)}{n}\leq 1$ for all $n$. Since $\chi$ has norm $1$, and the sums $\sum_{n\leq x}\mu(n)\chi(n)$ are small, we might -conjecture that $\frac{\phi_\chi (n)}{n}\leq 1$. -However this is not so. We can find a character such that $\mu$ and $\chi$ function have a lot of correlation, enough to make the sum of size $\sqrt{\log\log n}$. This is outlined by the following construction: - -Let $n$ be the product of all primes $p=3+4k$ where $p\leq M$, and let $\chi$ be the unique quadratic character modulo $4$. Then choose whether or not to remove the prime $3$ from this product as to force the equivalence $n\equiv 1 \pmod{4}$. For each divisor $d$ of $n$, if $\omega(d)$ is even, then $d\equiv 1\pmod{4}$ so that $\chi(d)\mu(d)=1$, and if $\omega(d)$ is odd, then $d\equiv 3\pmod{4}$ so that $\chi(d)\mu(d)=1$ yet again. This means that $$\frac{\phi_\chi (n)}{n}=\sum_{d|n} \frac{1}{d}\gg \sqrt{\log M}\gg\sqrt{\log \log n}.$$ The second last $\gg$ follows from the fact that if $A$ is the set of integers composed only of primes congruent to $3$ modulo $4$, then $\sum_{n\leq M,\ n\in A} \frac{1}{n}=\sqrt{\log M}$, and the last $\gg$ follows from the fact that $\log n =\theta(M;4,3)$. - -Any references to papers which might deal with this sort of sum is greatly appreciated, -Thanks, - -REPLY [7 votes]: Up to a multiplicative constant the answer is what you discovered, $f(n)=\sqrt{\log\log n}$. Obviously you can assume that $\chi$ is not a principal character since in that case you get something less than $1$. Then after writing -$$\sum_{d|n}\frac{\mu (d)\chi(d)}{d}=\prod_{p|n}\left(1-\frac{\chi (p)}{p}\right)$$and taking the logarithm of the RHS the problem boils down to finding an upper bound for the real part of the sum -$$\sum_{p|n}\frac{-\chi (p)}{p}.$$ -Note that only the real part of this expression is important to us, since we are going to exponentiate in the end to get back to the original problem (i.e. we are only interested in bounding the modulus of the resulting complex number). -Since $\chi$ is not principal, the worst thing that could happen in the previous sum is to have half of the values equal to $-1$ (e.g. use the Legendre symbol modulo some prime $q$) and then suppose that $n$ is a product of all primes up to some point which fall in the residue classes $d$ for which $\chi (d)=-1$. Keep in mind that you can't have more than $1/2$ of the values equal to $-1$, because of the orthogonality relations. -By Dirichlet's Theorem this accounts for half of the primes, evenly distributed in the various residue classes, and translating everything back to the original formulation gives the upper bound $\sqrt{\log\log n}$. If you want to see the details of how to carry out this final calculation then I can provide them, but from your example it looks like you already understand how that part of the argument works. -Final note: As Greg pointed out below this bound is not uniform in $\chi$. The best bound in general is $\log\log n$, which is achieved by letting $n$ run through the sequence of primorials and simultaneously letting $\chi_n$ run through a sequence of characters which take the value $-1$ at all primes dividing $n$.<|endoftext|> -TITLE: finding the $n$ closest pairs between $2n$ points -QUESTION [9 upvotes]: Given $2n$ points $x_1, x_2 \ldots x_{2n}$ and a distance $d_{i,j}$ defined between them, how can I best find the set $P$ of mutually exclusive pairs $(i,j)$ such that the sum of their distances -$$ -\sum_{(i,j) \in P} d_{i,j} -$$ -is minimised? The definition of $d_{i,j}$ is open and the function could be convex. The motivation for this problem is practical. How can I pair of 30 pictures say into most similar pairs? -I apologise in advance for the choice of tags on this post. I have been out of maths proper for a long time. - -REPLY [10 votes]: Actually, the OP is on a slightly different problem: the Euclidean minimal matching problem, where the complexity should be a lot better. See -The Open Problems Project, Problem 6 -(we will probably hear more from Joseph...). -EDIT I also found this paper -which seems to be much simpler than Edmonds, while not being particularly 2D-oriented.<|endoftext|> -TITLE: Why is there no formula for partial sums of some simple series? -QUESTION [14 upvotes]: I'm pretty sure that the sequences like $F_n=\sum_{k=1}^n \frac 1k$ are not traces of elementary functions on positive integers (take any reasonable definition of "elementary" you want, just make sure that all high school formulae are there). However, all proofs of non-elementarity I know make heavy use of differential fields and I do not see what and how to differentiate in this discrete setting. Any ideas, suggestions, or references? -P.S. I posted it on AoPS as well but then decided that there may be a slightly better chance to get an answer here :) - -REPLY [2 votes]: I am not sure what you mean by "traces", but there is certainly a theory of non-elementarity in the difference equation setting. One of the canonical early papers is: -MR1187234 (94a:39006) -Petkovšek, Marko(SV-LJUB) -Hypergeometric solutions of linear recurrences with polynomial coefficients. (English summary) -J. Symbolic Comput. 14 (1992), no. 2-3, 243–264. -And there is also Wilf-Zeilberger and the book of Petkovsek and Zeilberger (A=B) which probably covers this in greater detail. The methods are still those of differential algebra.<|endoftext|> -TITLE: Is the class of inner-anodyne morphisms right-cancellative with respect to the of the class of monomorphisms? -QUESTION [6 upvotes]: Recall: -Given a category $A$, and two classes of morphisms $S,S'$, we say that $S$ is right-cancellative with respect to $S'$ if for any pair of maps $f\in S, g\in S'$ such that $gf$ is defined, we have the implication $gf\in S \Rightarrow g\in S$. -Recall that the class of inner-anodyne morphisms in the category of simplicial sets is defined to be the class $\operatorname{llp}(\operatorname{rlp}(E))$, where $E$ denotes the set of inner-horn inclusions $\iota^n_k:\Lambda^n_k \hookrightarrow \Delta^n$ for $0 -TITLE: A question on the product of element orders of a finite group -QUESTION [6 upvotes]: Let $G$ be a finite group of order $n$ and $\psi(G)$ be the sum of element orders of $G$. Then $\psi(G)\leq\psi(C_n)$, where $C_n$ is the cyclic group of order $n$ (see "Sums of element orders in finite groups", Comm. Algebra 37 (2009), 2978-2980). Is it true a similar inequality for the product of element orders of $G$? - -REPLY [25 votes]: Denoting the order of $g$ by $o(g)$, you can show that for any decreasing function $f$ the following inequality holds -$$\sum_{g\in G}f(o(g))\geq \sum_{g\in \mathbb Z/n\mathbb Z}f(o(g)).$$ -This is because one can actually construct a bijection $\sigma:G\to\mathbb Z/n\mathbb Z$ which satisfies $$o(\sigma(g))\geq o(g)$$ for all $g\in G$. The main ingredient is a classical theorem of Frobenius saying that when $k$ divides the order of a group, the number of elements of order dividing $k$ is divisible by $k$, then proceed by induction. An application of this exact idea is for example problem 10775 on the American Math Monthly. For your question we just need $f(x)=-\log x$.<|endoftext|> -TITLE: Viète's generalized infinite product yielding other converging values? -QUESTION [5 upvotes]: I took Viète's infinite product for $\frac{2}{\pi}$: -$\displaystyle \dfrac{2}{\pi} = \dfrac{\sqrt2}{2} . \dfrac{\sqrt{2+\sqrt2}}{2} . \dfrac{\sqrt{2+\sqrt{2+ \sqrt2}}}{2} \dots$ -and made it generic: -$\displaystyle v = |\dfrac{\sqrt{z}}{z} . \dfrac{\sqrt{z+\sqrt{z}}}{z} . \dfrac{\sqrt{z+\sqrt{z+ \sqrt{z}}}}{z} . \dots |$ -Checked whether other converging values for $v$ exist and found three more (by trial & error): -$z=-1$ yielding $1.29425...$ -$z=i-1$ yielding $0.92741...$ -$\displaystyle z=i+ \frac13 (1+ \sqrt[3]{(28-3 \sqrt{87})}+\sqrt[3]{(28+3 \sqrt{87})})$ yielding $0.64801...$ -Tested the necessary (but not sufficient) convergence of the 'final' factor to 1 by using: -$\displaystyle a = \sqrt{z+\sqrt{z+ \sqrt{z} \dots}}$ -then solving $a$ via: -$a = \sqrt{z + a}$ -and indeed $\displaystyle |\frac{a}{z}|=1$ for all these three values. -How could this be proven? Are there more converging values $v$? -Are these values for $v$ connected in any way e.g. on a circle in the complex plane or to other mathematical constants (did check Plouffe's, Wolfram Math and Sloane's but without any result)? -Thanks! - -REPLY [6 votes]: If $a$ satisfies $a=\sqrt{z+a}$ then $a/z = 1/(a-1)$ which usually does not have modulus $1$. So for $1/(a-1) = e^{i\theta}$ we get $z = e^{-2i\theta}+e^{-i\theta}$.<|endoftext|> -TITLE: Cyclotomic Polynomials in Combinatorics -QUESTION [15 upvotes]: I am searching for a combinatorial significance of cyclotomic polynomials. The only examples I got are a paper by Neville Robbins http://www.emis.de/journals/INTEGERS/papers/a6/a6.pdf and two recent papers by Gregg Musiker and Victor Reiner http://arxiv.org/PS_cache/arxiv/pdf/1012/1012.1844v2.pdf and http://combinatorics.cis.strath.ac.uk/fpsac2011/proceedings/dmAO0161.pdf but these do not essentially give a clear picture. I am interested in examples that relate cyclotomic polynomials to foundations of combinatorics (if these exist) or if someone can give a direct combinatorial interpretation of coefficients of cylotomic polynomials that'll be quite helpful. - -REPLY [4 votes]: A recent paper of mine on permutation enumeration that uses cyclotomic polynomials is I. M. Gessel, Reciprocals of exponential polynomials and permutation enumeration, Australasian J. Combin. 74 (2) (2019), 364–370.<|endoftext|> -TITLE: Placing points on a sphere so that no 3 lie close to the same plane -QUESTION [17 upvotes]: Motivation -I am working with arbitrary parallelopiped tilings given by projection from a higher dimensional space. The collection of tiles, and some properties of the higher dimensional space are specified by a finite collection of vectors forming the edges of the tiles. The tiles themselves are parallelopipeds given by all choices of three vectors from this set. A patch of such a tiling is shown below. - -The tricky thing is trying to find sets of vectors where none of the parallelopipeds are nearly flat. This lead me to a more general question, a sort of dual problem to packing circles on a sphere. -Question -Given $V$ a set of 3d vectors in general position, so that no two lie on the same line and no three lie on the same plane. Without loss of generality, we may assume they are unit vectors. We can consider $V_P$, the set of planes generated by all pairs of vectors in $V$. What arrangements of vectors $V$ will maximise the smallest angle between two planes in $V_P$? - -REPLY [5 votes]: This is not an answer, just a remark. -I find your question interesting even for small values of $n=|V|$. -For example, if $n=4$, then choosing $V$ as the vertices of a regular tetrahedron, -the $\binom{4}{2}=6$ planes $V_P$ determine six normal lines that define a cuboctahedron: - -           - - -The angles between these lines/planes is either $90^\circ$ or -about $63.6^\circ$ (or $116.4^\circ$) $60^\circ$ (or $120^\circ$). -Now, the optimal packing of six lines is known (Conway, Hardin, Sloane) to be the six diameters of the icosahedron. -The minimum angle determined by those diameters is a bit larger, if I've calculated -correctly: $63.4^\circ$. -So: Is there an arrangement of four vectors $V$ that yields this optimal line packing? -Answer: No, $60^\circ$ is the optimal for $n=4$. See Henry Cohn's argument in the comments. - -Here is Edmund's suggestion for $n=6$: $V$ is given by half the vertices of an icosahedron (blue), -which generate -15 normal lines (red) passing through the midpoints of the icosahedron's 30 edges, -with the normal lines separated by $49.7^\circ$. - -           - - -The tips of the normal lines form the vertices of an icosidodecahedron.<|endoftext|> -TITLE: Is there a periodic table for knots? -QUESTION [15 upvotes]: When I see knot tables, I have two feeling: ah, it's beautiful, and... painful. -I don't see how knots are ordered in the knot table, the way to go from one knot of a certain crossing number to another seems to be completely random. But I would guess there are some order? For example, why are the Perko pair put next to each other even before people knew they are the same? -In short, if the word "periodic table" seems confusing, my real question is, how are the knots in knot table ordered? - -REPLY [3 votes]: I'm surprised no one mentioned the story by William Thompson, also known as lord Kelvin, which originally thought of atoms or elements as knots! That would literally put them in a periodic table :). -I learned this from this fascinating talk by Haynes Miller, which I quote the relevant part - -Thomson’s vision was that perhaps atoms consisted -of tiny vortex rings, persistent topological singularities in the aether. Different elements -might correspond to different knot types!—so perhaps the hydrogen atom was a simple -loop (the “unknot”), helium was the trefoil knot, . . . . These rings vibrated, putting them in -different energy states. And perhaps the mystery of the formation of molecules could be -explained by the linking of several knots to form links.<|endoftext|> -TITLE: On a compact manifold, what kind of function can be the Jacobian of a diffeomorphism? -QUESTION [10 upvotes]: I could not answer or find references of this question, even for the following special case: -On $S^2$ (the two-sphere equiped with the standard Riemannian metric), is every positive smooth function with integral $1$ the Jacobian of some diffeomorphism? -An equivalent formulation of the question is: On $S^2$, is every positive smooth probability measure the translate of the standard one by some diffeomorphism? - -REPLY [17 votes]: Here is an "answer-version" of my comment: -Yes, this is true in general. The reference I know is Moser's 1965 paper "On the volume elements on a manifold" (http://www.jstor.org/stable/1994022). -Specifically, let $M$ be a compact connected orientable manifold, and let $\sigma$ and $\tau$ be smooth volume forms on $M$ both with integral 1. Then there exists a diffeomorphism $\varphi:M\to M$ such that $\varphi^*\tau=\sigma$. -The orientability hypothesis isn't really necessary (just use densities rather than volume forms; see Moser's footnote (2)).<|endoftext|> -TITLE: Geometric picture of invariant differential of an elliptic curve -QUESTION [9 upvotes]: What is the geometric meaning of $\omega=dx/(2y+a_1x+a_3)$ for an elliptic curve? -This question is an adjunct to MO Q1 on formal laws and L-series. Silverman (Q1) and Darmon (pg. 6) state: -The invariant holomorphic differential form (Neron differential) attached to an elliptic curve is -$\omega=dx/(2y+a_1x+a_3)$. -(Ancilliary question: Relation to Weierstrass's elliptic functions?) -I'd like to broaden the question as a community wiki to ask, "What are some interesting manifestations of this one-form in various families of elliptic curves?" -E.g., J. Hoffman in Topics in Elliptic Curves and Modular Forms (revised 2013) gives for the Jacobi quartic family of elliptic curves -$\omega=dx/(1+2\kappa x^{2}+x^{4})^{1/2}=\sum_{n=0}^{\infty}L_{n}(\kappa)x^{2n}dx$ -with $L_{n}(\kappa)$ the Legendre polynomials. - -REPLY [9 votes]: A paper by John Tate (pg. 1 and 2) gives a clear derivation of the diff. form: -Reparametrize the elliptic curve -$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4 x+a_6$ -with $p(z)=x+(a_1^2+4a_2)/12$ and $p^{'}(z)=2y+a_1x+a_3$ to obtain -$(p^{'})^2=4p^3-g_2p-g_3$, defining the Weierstrass elliptic fct., and -$\omega=dp(z)/p^{'}(z)=dz=dx/(2y+a_1x+a_3)$. -Per Dan's comment, a coordinate transformation of $x=u^2x^{'}+r$ and $y=u^3y^{'}+su^2x^{'}+t$ -leaves $\omega^{'}=u\,\omega$. -Given $\sigma=p(z)$ and the inverse $z=p^{-1}(\sigma)$, -$dz=(p^{-1}(\sigma))^{'}d\sigma=(p^{-1}(\sigma))^{'}p^{'}(z)dz$, so -$(p^{-1}(\sigma))^{'}=1/p^{'}(z)$ and $dz=d\sigma/p^{'}(z)=\omega$. -The amplitwist interpretation of differentiation and inversion presented by Tristan Needham in his book Visual Complex Analysis provides a geometric interpretation of these differential relations. -Consider as an analogy $P(\theta)=sin(\theta), P^{'}(\theta)=cos(\theta),\,and\, P^2+(P^{'})^2=1$. - -Update 3/13/21: -Delineating the relationships between a compositionally inverse pair of complex functions/formal power series $f(z)$ and $f^{-1}(w)$ with $f(0)=0$ collates the analogous relationships among the invariant differential above and the associated formal group law (FGL) and Dirichlet/L series as noted in Silverman's post and basic geometric constructs (the original intent of my question). I'll reference H below for "Three lectures on formal groups" by Hazewinkel. -For the inverse pair of $w=f(z)$ and $z=f^{-1}(w)$, -$$dz=(f^{-1}(w))' \; dw=(f^{-1})'(w) \cdot f'(z) \; dz,$$ -so -$$(f^{-1})'(w)=\frac{1}{f'(z)}$$ -for $(z,w) = (f^{-1}(w),f(z))$. -This analytic relation has the equivalent geometric interpretation in the case of real variables and curves that a tangent of the curve $y=f(x)$ at the point $(x,y)=(z,f(z))=(f^{-1}(w),w)$ is the reflection through the line $y=x$ of the tangent to the curve $y=f^{-1}(x)$ at the point $(x,y)=(f(z),z)= (w,f^{-1}(w))$; that is, more simply, the two curves are reflections of each other through the quadrant bisector $y=x$. -Let -$$ g(z) = \frac{1}{f'(z)} = (f^{-1})'(f(z)),$$ -a morphed flow/evolution equation since -$$(f^{-1})'(z) = g(f^{-1}(z))$$ -(see OEIS A145271 for flow equations and, in particular, a Riccati equation for $g(x)$ a quadratic in A008292). -Then -$$dw = df(z) = f'(z) \; dz = \frac{df^{-1}(w)}{(f^{-1})'(w)} = \frac{dz}{(f^{-1})'(f(z))} = \frac{dz}{g(z)} = \omega,$$ -and -$$f(z) = \int_0^{z} \frac{1}{g(u)} du = \int_0^{z} \frac{1}{(f^{-1})'(f(u))} du = \int \omega .$$ -Check with $ w = f(z) = \ln(1+z)$ and $z = f^{-1}(w) = e^w-1$ and, therefore, $g(z) = \frac{1}{f'(z)} = (f^{-1})'(f(u)) = 1+z$. -(For the invariant differential for elliptic curves above and in Silverman's post, see p. 61 of H.) -Now define the associated infinitesimal generator/vector, as suggested by the flow equation above, -$$ D_{w} = \frac{d}{dw} = \frac{d}{df(z)} = \frac{1}{f'(z)}\; \frac{d}{dz} = (f^{-1})'(f(z)) \; \frac{d}{dz} = g(z) \frac{d}{dz} $$ -(all derivatives are partial derivatives w.r.t. the designated variable). -Then -$$e^{(t \; g(z) \; D_z )} \; z = e^{(t \; D_{w}) } \; f^{-1}(w) = f^{-1}(w + t) = f^{-1}(f(z) + t),$$ -and -$$e^{(t \; g(z) \; D_z )} \; z \; |_{z=0}= f^{-1}(t).$$ -Substitutions give the FGL -$$F(x,y) = f^{-1}(f(x) + f(y)),$$ -so -$$ \frac{dF(x,y)}{dy} \; |_{y=0} = (f^{-1})'(f(x)) \cdot f'(0) = g(x)$$ -if $f'(0) = 1$. -For $g(z)$ a quadratic, the elliptic FGL of Examples 3 and 63 of Buchstaber and Bunkova's "Elliptic formal group laws, integral Hirzebruch genera, and Krichever genera" apply (see also the A008292 formulas dated Sep 18 2014). The associated Riccati equation is related to soliton solutions of the Kdv equation presented in my response to the MO-Q "Is there an underlying explanation for the magical powers of the Schwarzian derivative?" The quadratic $g(z)$ is related to the ratio of the Schwarzian derivatives of an inverse pair of functions, one of which is the integral of the soliton solution, and to the velocity of the soliton. -Now characterize $f(z)$ as a logarithmic series (with $a_n =1$ for a canonical FGL), -$$f(z) = -\ln(1-a.z) = \sum_{n \ge 1} \frac{(a.z)^n}{n} = \sum_{n \ge 1} a_n \; \frac{z^n}{n}$$ -(see p. 55 of H on the use of the term logarithm in the FGL formalism). -Then -$$qf'(q) = \frac{q}{g(q)} = \frac{a.q}{1-a.q} = \sum_{n \ge 1} a_n q^n \; ,$$ -and, with $q = e^{-t}$, the normalized Mellin transform -$$\int_{0}^{\infty} \frac{e^{-t}}{g(e^{-t})} \; \frac{t^{s-1}}{(s-1)!} \; dt = \sum_{n \ge 1} \; a_n \; \int_{0}^{\infty} \;e^{-nt} \; \frac{t^{s-1}}{(s-1)!} \; dt = \sum_{n \ge 1} \; \frac{a_n}{n^s} $$ -generates an associated Dirichlet series (see p. 53 of H). For the Riemann zeta function, $g(z) = 1-z$. -The tangent plays multiple roles in the multiplicative and compositional inversions above, so one should expect to encounter free probability, symmetric function theory, Hopf algebras, and the associated geometry and combinatorics as well. For more on this, see H, the recently revised "Formal Groups, Witt vectors and Free Probability" by Friedrich and McKay, and refs/links in the OEIS.<|endoftext|> -TITLE: Riemann mapping theorem and smoothness on the boundary -QUESTION [10 upvotes]: Let $U\subset \mathbb C$ be open, bounded, simply connected, with $C^\infty$ boundary. Apply the Riemann mapping theorem to get a bilolomorphic isomorphism -$$ -f:U\to \mathbb D -$$ -between $U$ and the unit disc $\mathbb D:=\{z\in \mathbb C:|z|<1\}$. - -How can I see that $f$ extends to a $C^\infty$ map from the closure of $U$ to the closure of $\mathbb D$? - -REPLY [2 votes]: This is well-known result by Kellogg (O. Kellogg: On the derivatives of harmonic functions on the boundary, Trans. Amer.Math. Soc. 33 (1931), 689-692.), and Warschawski (On the higher derivatives at the boundary in conformal mapping,} Trans. Amer. Math. Soc, {\bf 38}, No. 2 (1935), 310-340.), where they prove even more, that the if the boundary is C^{n,\alpha}, then the conformal parametrization is C^{n,\alpha} up to the boundary.<|endoftext|> -TITLE: Representing a product of matrix exponentials as the exponential of a sum -QUESTION [14 upvotes]: In Proof of a conjectured exponential formula, R. C. Thompson (1986) [edit: apparently, assuming Horn's conjecture] proved that if $A$ and $B$ are Hermitian matrices, then there exist unitary matrices $U$ and $V$, such that -$$ e^{iA}e^{iB} = e^{i (UAU^*+VBV^*)}.$$ -I was wondering whether a similar result holds without the $i$? More precisely, my question is: - -Let $A$ and $B$ be Hermitian matrices. Do there exist unitary matrices $U$ and $V$ such that - $$e^{A/2}e^{B}e^{A/2} = e^{UAU^* + VBV^*}$$ - - -EDIT -Denis Serre has pointed out an inconsistency in my statement above, in the sense that Thompson's result was based on an the hypothesis that Horn's conjecture is true. Now that this conjecture is indeed known to be true, Thompson's claim may officially be regarded as a theorem. - -REPLY [6 votes]: This follows from a result of Klyachko. Klyachko proved: - -Let $\alpha$, $\beta$ and $\gamma$ be three vectors in $\mathbb{R}^n$. Then the following are equivalent: -(1) There exist Hermitian matrices $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ with $\mathfrak{a}+\mathfrak{b}= \mathfrak{c}$, and with eigenvalues $\alpha$, $\beta$ and $\gamma$ respectively. -(2) There exist invertible matrices $A$, $B$ and $C$ with $AB=C$ and singular values $e^{\alpha}$, $e^{\beta}$ and $e^{\gamma}$ respectively. Here $e^{\alpha}$ etcetera mean termwise exponentiation. - -I will show that $(2) \implies (1)$ implies your statement (and is basically equivalent to it). I'll call your Hermitian matrices $X$ and $Y$, to leave the letters $(A,B,C)$ clear. -Let $A=e^{X/2}$, let $B=e^{Y/2}$ and let $AB=C$. Let $e^{\alpha}$, $e^{\beta}$ and $e^{\gamma}$ be the singular values of $A$, $B$ and $C$. So the eigenvalues of $C C^{\ast}$ are $e^{2 \gamma}$, and we note that $C C^{\ast} = e^{X/2} e^Y e^{X/2}$. -Using $(2) \implies (1)$, let's find Hermitian $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ with eigenvalues $2 \alpha$, $2 \beta$ and $2 \gamma$ and $\mathfrak{a}+\mathfrak{b} = \mathfrak{c}$. Then $C C^{\ast}$ and $e^{\mathfrak{c}}$ are Hermitian with the same eigenvalues and, conjugating by a unitary matrix, we can arrange that $C C^{\ast} = e^{\mathfrak{c}}$. -Now, $X$ and $\mathfrak{a}$ are both Hermitian with eigenvalues $2 \gamma$, so we can find unitary $U$ with $\mathfrak{a} = U X U^{\ast}$. Similarly, we can find unitary $V$ with $\mathfrak{b} = V Y V^{\ast}$. So $\mathfrak{c} = U X U^{\ast} + V Y V^{\ast}$ and $e^{X/2} e^Y e^{X/2} = e^{U X U^{\ast} + V Y V^{\ast}}$ as desired.<|endoftext|> -TITLE: Numbers of intersection points and lines -QUESTION [11 upvotes]: Hello, -I don't know if this question has already been posted, I have made a little search with keywords and did not found it, sorry if I missed anything. -Is it possible to characterize the set of pairs of integers ($l$,$i$) such that one can draw $l$ lines on the euclidean plane with exactly $i$ intersection points? -It is quite trivial to see is that given $l$, an upper bound for $i$ is $l(l+1)/2$. -More generally, given $l$, any additive decomposition of $l$ of the form $\underset{j=1}{\overset{k}{\sum}} l_i$ provides a value for $i$ which is $\underset{i=1}{\overset{k}{\sum}} $ $\underset{j>i}{\overset{k}{\sum}} l_i l_j$ if we fix for any $i \in [1,k]$ exactly $l_i$ parallel lines such that there is no intersection of three or more lines at the same point. -It is not difficult to see that there are pairs ($l$,$i$) that are not of this form. -For instance, if you try all decompositions of 6, you may draw 6 lines with 5, 8, 9, 11, 12, 13, 14 or 15 intersection points with this method, but 7 and 10 are missing (they can be obtained with intersection points of three lines). -Here is a link containing some observations (http://www.ics.uci.edu/~eppstein/junkyard/how-many-intersects.html). As far as I know, this is the only place where this problem has been seriously considered, but it is quite old and maybe lacks of results. So any additional comment will be welcomed :) -Just a final remark, thanks to some projective properties, this question is the same as finding $c$ circles sharing a common point with exactly $i+1$ intersection points. Don't know if this can help. - -REPLY [6 votes]: I asked Jon Lenchner, an expert on point-line incidences, and he told me the -question (in dual form) was posed in -Grünbaum's 1971 book -Arrangements and Spreads, -and fully answered in a paper by -Peter Salamon and -Paul Erdős: -"The solution to a problem of Grünbaum," -Canad. Math. Bull. 31: 129-138 (1988). -Here are its first two sentences: - -In the paper below we characterize for large $n$ the possible values of the - number of connecting lines determined by a set of $P_n$ points in the plane, - where a connecting line is any straight line containing at least two points of $P_n$. - This solves a problem posed by Grünbaum [5,6] which asks for the sequence - of all integers $m$ with the property that some configuration of $n$ - points determine exactly $m$ lines. - -([6] is Arrangements and Spreads; [5] is Erdős's earlier partial solution.) -They obtain exact expressions "for the lower end of the continuum of values leading -down from $\binom{n}{2}-4$." "The possible values...can be seen to bear a strong resemblance -to physical spectra." -The lower end of the continuum grows as $n^{3/2}$ (with constant 1). -Here are two figures from the paper:<|endoftext|> -TITLE: Explicit formula for Riemann zeros counting function -QUESTION [31 upvotes]: I've often seen it stated (in vague terms) that there's a Fourier duality between the set of prime numbers and the set of nontrivial Riemann zeta zeros. -Because there are various explicit formulae whereby prime counting functions can be expressed as infinite sums of sinusoidal functions over the set of zeta zeros, I had always assumed that the same thing would work in the opposite direction. That is, I imagined there was a well-known formula whereby the nontrivial-zeta-zeros-counting function could be expressed as an infinite sum of sinusoidal functions over the set of primes. -However, I've been unable to find such a formula. The only explicit formula I've found for this counting function looks like this: -$$f(E) = \frac{1}{\pi} \Im(\ln(\Gamma(1/4 + iE/2)) - \frac{E}{2\pi}\ln(\pi) + \frac{1}{\pi} \Im(\ln(\zeta(1/2 + iE)) + 1$$ -where there's no explicit involvement of the primes. -Is anyone aware of a formula of the type I'm seeking? I'm quite sure such a thing must exist, since a programmer friend recently looked at this and got curious enough to "illegally" substitute the (truncated) Euler product expression for zeta in the above (it doesn't converge in the critical strip, hence such a substitution is not mathematically valid), just to see what the resulting function would look like. His plots of -$$N_m(E) = \frac{1}{\pi} \Im(\ln(\Gamma(1/4 + iE/2)) - \frac{E}{2\pi}\ln(\pi) - \frac{1}{\pi} \sum_{p < m} \Im(\ln(1-p^{-1/2-iE})) + 1$$ -in three colours (corresponding to $m = 100$, $1000$ and $10000$, and where $E$ varies from $4$ to $60$) can be seen here: - -The positions of the zeta zeros are clearly visible, so even though this isn't a valid formula, I imagine that there must be something like this which is valid. - -REPLY [2 votes]: The Riemann zeta zeros counting function as a sum of the von Mangoldt function (or expanded with the Möbius function): - -Mathematica program that exports 120 gif images to your Documents folder. -(*Integral*)(*start*) -Clear[k, n, s, i, d, r, q, f, a, b, integral, c, nn, n, k]; -nn = 120; -delta = 1/10; -k = 16; -q = 10; -c = 1; -rr = 60; -f[s_, d_] = - N[-I*MoebiusMu[ - d]*(If[d == 1, 0, -((d^(1 - s) 1^-s)/(Log[d] + Log[1]))] + - Sum[-((d^(1 - s) n^-s)/(Log[d] + Log[n])), {n, 2, k}] + - ExpIntegralEi[-(-1 + s) Log[d] - (-1 + s) Log[ - k]] + (d^(1 - s) k^-s)/(2 (Log[d] + Log[k])) + - Sum[Sum[-(BernoulliB[2*(n + 1)]/((2*(n + 1))!))*(Abs[ - StirlingS1[2*n + 1, i]]) d k^(-2*n - 1) Gamma[1 + i, - s Log[k*d]] Log[k*d]^(-1 - i), {i, 0, 2*n + 1}], {n, 0, - q - 1}])]; -integral[a_, b_, d_] = f[b, d] - f[a, d]; -Print["Counting to ", nn]; -rrr = N[Table[RiemannSiegelTheta[t]/Pi, {t, 0, rr, delta}]]; -Monitor[iii = - Table[Table[ - Re[Sum[integral[1/2 + I*0, 1/2 + I*t, d]/n^c, {d, Divisors[n]}]]/ - Pi, {t, 0, rr, delta}], {n, 1, nn}];, n] -Monitor[Table[ - Export[StringJoin["image", ToString[i], ".gif"], - Show[Plot[(RiemannSiegelTheta[t] + Im[Log[Zeta[1/2 + I*t]]])/Pi + - 1, {t, 0, 60}, PlotStyle -> {Red, Thickness[0.004]}, - ImageSize -> Large], - ListLinePlot[rrr - Sum[iii[[j]], {j, 1, i}], - PlotStyle -> Thickness[0.004], DataRange -> {0, rr}, - ImageSize -> Large], - Graphics[ - Text[Style["Counting Riemann zeta zeros using", Large], {21, - 12}]], Graphics[ - Text[Style[ - StringJoin[ToString[i], - " terms of the von Mangoldt function."], Large], {23, - 10}]]]], {i, 1, nn}], i] -(*end*)<|endoftext|> -TITLE: Truncated Exponential Series Modulo $p$: Deeper meaning for a Putnam Question. -QUESTION [14 upvotes]: Apparently B6 of the Putnam this year asked: - -Suppose $p$ is an odd prime. Prove that for $n\in \{0,1,2...p-1\}$, at least $\frac{p+1}{2}$ of the numbers $\sum^{p-1}_{k=0} k! n^{k}$ are not divisble by $p$. - -With some rearrangements, this is equivalent to showing that $$E_p(z):=\sum_{k=0}^{p-1} \frac{z^k}{k!}$$ has at most $\frac{p-1}{2}$ zeros. A proof of this is at the end. -My question is: Can we improve the bound for the number of zeros? Also is there a deeper connection here with other parts of mathematics motivating this problem? - -Proof of problem: Consider $$Q(z)=z^{p}-z+\sum_{l=0}^{p-1}\frac{z^{l}}{l!}.$$ Then for each integer $Q(n)=E(n).$ However, $$Q^{'}(z)\equiv E^{'}(z)-1=E(z)-\frac{z^{p-1}}{(p-1)!}-1\equiv E(z)+z^{p-1}-1.$$ Then, if $Q(n)=0$ for $n\neq0$ , we must also have $Q^{'}(n)=0$ so that $n$ is a double root of $Q(n).$ Since $\deg Q(n)=p$, we see that at most half of the integers $n\in\{ 1,2,\dots,p-1\}$ satisfy $E(n)=0.$ Since $E(0)=1$, we conclude the desired result. - -Remark: This was asked in a slightly different form on math stack exchange. I felt the answer I posted there was inadequate there, and I personally became more curious while attempting to answer the question. - -REPLY [9 votes]: Thinking about this problem again, I found the following simpler explanation which does not invoke orthogonal polynomials and uses the nonvanishing of only one Hankel determinant. The determinant arises via Padé approximants but the exposition below is self-contained. -Suppose in general that $a_0 \neq 0$, $a_1, a_2, \ldots, a_{q-1}$ are elements of a finite field $F$ of $q$ elements for which the polynomial -$$ -A(X) = \sum_{i=0}^{q-1} \phantom. a_i X^i -$$ -vanishes at all but $t$ nonzero field elements, say $x_1,x_2,\ldots,x_t$, with $2t < q-1$. Then -$$ -A(x) = \frac{P(x)}{Q(x)} (1 - x^{q-1}) -$$ -for some polynomials $P,Q$ of degree $t$, where -$$ -Q(X) = \prod_{m=1}^t (X-x_m) = \sum_{i=0}^t \phantom. q_i X^i -$$ -for some field elements $q_i$. Thus $A(X)$ is within $O(X^{q-1})$ of the power series about $X=0$ of the degree-$t$ rational function $P(X)/Q(X)$. For any $t' \in [t, \phantom. (q-1)/2)$ and $n \in (0, \phantom. q-1-2t')$ it follows that the square Hankel matrix -$$ -(a_{n+i+j})_{i,j=0}^{t'} = \left( -\begin{array}{ccccc} -a_n & a_{n+1} & a_{n+2} & \cdots & a_{n+t'} \\ -a_{n+1} & a_{n+2} & a_{n+3} & \cdots & a_{n+1+t'} \\ -\vdots & \vdots & \vdots & & \vdots \\ -a_{n+t'} & a_{n+t'+1} & a_{n+t'+2} & \cdots & a_{n+2t'} -\end{array} -\right) -$$ -of order $t'+1$ is singular, because the nonzero column vector -$(0,0,\ldots,0,q_d,q_{d-1},q_{d-2},\ldots,q_1,q_0)^{\rm T}$ [with $t'-t$ initial zeros] is in the kernel. Thus a single invertible matrix of this form implies that there are more than $t' \geq t$ nonzero $x\in F$ at which $A(x) \neq 0$. -To apply this to the Putnam problem, take $q=p$ and $a_i = i!$, and set $t' = t = \frac12(p-1) - 1$ and $n=1$. The resulting Hankel matrix $((i+j+1)!)_{i,j=0}^t$ is invertible mod $p$ thanks to the formula $\prod_{k=0}^t k!(k+1)!$ for its determinant. [This formula can be obtained from properties of the Laguerre orthogonal polynomials, but also has an elementary direct proof: see the second solution of this problem at the Putnam directory.] Hence $A(x) \neq 0$ for at least $t+1 = (p-1)/2$ nonzero values of $x \bmod p$, QED. -It is also known that the Hankel matrix $(a_{i+j})_{i,j=0}^t$ is invertible, else we'd have $P/Q \equiv P_1/Q_1 \bmod X^{2t}$ for some $P_1,Q_1$ of degree less than $t$, whence $P/Q = P_1/Q_1$ identically and $P/Q$ is not in lowest terms. This lets us connect the above solution with the orthogonal polynomials evaluated at $X^{-1}$ that arose in our previous solution. Indeed, define a bilinear pairing $\langle\cdot,\cdot\rangle$ on $F[X]$ by $\langle P,Q \rangle = I(PQ)$, where $I: F[X] \rightarrow F$ is the linear form taking each $X^i$ to $a_i$. The restriction of this pairing to the polynomials of degree at most $t$ is nondegenerate because its Gram matrix is the Hankel matrix we just proved invertible. But $G(X) := X^t Q(1/X) = \sum_{i=0}^t q_i X^{d-i}$ is orthogonal to $X^j$ for each $j = 1,2,\ldots,t+1$. Hence $XG$ is orthogonal to every polynomial of degree at most $t$, and is therefore a multiple of the orthogonal polynomial of degree $t+1$ for our inner product (which is unique up to scaling thanks to the nondegeneracy of the pairing). It follows that in this case the orthogonal polynomial of degree $t+1$ must vanish at zero and at $t$ other elements of $F$. This does not happen often for classical orthogonal polynomials, but (as noted in my previous answer) there is at least one infinite family of examples, the Čebyšev polynomials of the second kind $U_{(q-1)/2}$ when $q \equiv 3 \bmod 4$.<|endoftext|> -TITLE: Partitions and finite sets -QUESTION [10 upvotes]: If we consider a partition ${\mathcal X}=(X_1,\dots,X_n)$ of a finite set $X$ and a partition ${\mathcal Y}=(Y_1,\dots,Y_m)$ of another finite set $Y$, then $(X_i\times Y_j)_{i=1,...,n,j=1,...,m}$ is a partition of $X\times Y$. I will call this partition the "product partition" ${\mathcal X}\times{\mathcal Y}$. -On the converse, given a partition ${\mathcal Z}$ of a finite set $Z$ with finite cardinal $n=pq$ ($p$ and $q$ different from $0$ or $1$), how can one know if this partition is isomorphic to a non trivial product partition (i.e. if there exists two sets $X$ and $Y$ whose cardinals at least $2$ and endowed with partitions ${\mathcal X}$ and ${\mathcal Y}$ (respectively) and a bijection of $X\times Y$ on $Z$ such that the images of the elements of ${\mathcal X}\times{\mathcal Y}$ by this bijection are exactly the elements of ${\mathcal Z}$)? -In the very particular case where one knows that $Z$ is a finite commutative group and that ${\mathcal Z}$ is the set of classes $gH$ for some subgroup $H$ of $Z$, then using the classification of finite commutative groups, one obtains that ${\mathcal Z}$ is a product partition. -This question seems complicated (at least for me!) in the general case. Does someone know a partial answer to my question? In some particular case? - -REPLY [3 votes]: Here are comments but not a solution. Boris points out that the problem reduces to this: given a list of positive integers $c_1 \dots c_n$ are there lists $a_1 \cdots a_u$ and $b_1\cdots b_v$ so that the products $a_ib_j$ are the $m$ values. One problem is finding the values $u,v$ with $uv=n$ let us assume that $u$ and $v$ are given (for example if $n=899$ then we would have to have $u=29$ and $v=31$ since we desire that $u,v \ge 2$) -There are many necessary conditions which might rule out most lists of numbers but that does not mean that there could not be difficult cases. If we assume that the lists are ordered (and why not) then - -$a_1b_1=c_1$ and -$a_ub_v=c_n.$ -this might or might not limit possibilities. Any $c_k$ with few factorizations limits the possibilities. Also the sum of the $c_k$ may limit things because -$\sum a_i\sum b_j=\sum_1^nc_k$ - -If we look at the $\binom{n}{2}$ ratios $\frac{c_{s}}{c_t}$ then we must have $\binom{u}{2}$ values (which will be the ratios $\frac{a_{x}}{a_{y}}$) which occur at least $v$ times (and involve $2v$ distinct indices) and similarly $\binom{v}{2}$ values which appear at least $u$ times each. -It is worth looking at the special case that the $c$ values are all powers of a single prime $p.$ That makes it easier to have integer ratios and values with many factors. This is also relevant to the general case. Let $p^{\gamma_k}=c_k$ or merely be the (known) power of $p$ dividing $c_k.$ Then we will need $u+v$ values $\alpha_i$ and $\beta_j$ so that the $n$ sums $\alpha_i+\beta_j$ are the $\gamma_k.$ -Particularly relevant to this case is that me may as well assume that the $c$ values have no common divisor (so in the one prime case we may as well assume $\gamma_1=\alpha_1=\beta_1$.) This is because we can divide out any factor $f$ common to all the $c$ values, solve the reduced problem (if possible) by $a^{'}_ib^{'}_j=c^{'}_k$ then chose any factorization $f=gh$ and let $a_i=ga^{'}_i$ and $b_j=hb^{'}_j$ -further comment Replace each integer $c=2^e3^f5^g\dots$ by a monomial $x_2^ex_3^fx_5^g\dots$ where the $x_{*}$ are formal variables. Then add. This reduces the problem to factoring a multinomial with positive integer coefficients into two of the same type. Then the single prime case is factoring a polynomial (into factors with positive coefficients). There are algorithms for that.<|endoftext|> -TITLE: inverse limits of group algebras and profinite groups -QUESTION [13 upvotes]: For an inverse system {$G_i$} of finite groups, and a fixed field $\mathbb{k}$, one can consider the corresponding group algebras $\mathbb{k}[G_i]$. The latter form an inverse system of $\mathbb{k}$-algebras {$\mathbb{k}[G_i]$} (unless I miss something obvious). Is it true that the inverse limit of {$\mathbb{k}[G_i]$} is the group algebra $\mathbb{k}[G]$, for $G=\lim\limits_\leftarrow${$G_i$} ? -In my case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$, and $G_i$ are abelian $p$-groups, if this helps. -Added: I see that the answer is much less trivial than I expected. What about the simplest case, perhaps, when $G_i=\mathbb{Z}/p^i\mathbb{Z}$, for $i\geq 1$, and thus $G$ is the additive group -of $\mathbb{Z}_p$? Are there fields for which the complete group algebra $[[\mathbb{k}G]]$ is easy to describe (particularly interesting for me would be the case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$.) ? [This is answered by Simon Wadsley in the comment below.] -P.S. This is a spill-over of an innocently looking final year project on invertible circulant matrices of an undergraduate student of mine---I am unfamiliar with profinite things... - -REPLY [8 votes]: Although the reference to Ribes and Zalesski (given in another answer) is excellent, another very good starting point for this area is: -A. Brumer, Pseudocompact algebras, profinite groups and class formations , J. Alg, 4, (1966), 442–470. -The key point is the pseudocompactness of the result. Brumer goes into the homological algebra of these algebras. Of course, in the 45 years since that was published there have been a lot of advances, but the basic theory is very well explained there.<|endoftext|> -TITLE: Presheaves on a complete Segal space -QUESTION [6 upvotes]: Let C be an $(\infty,1)$-category, incarnated as a complete Segal space, hence in particular a bisimplicial set. Is there a model structure on the slice category of bisimplicial sets over C which presents the $(\infty,1)$-presheaf category of C? Ideally, such a model structure would be Quillen equivalent to the contravariant model structure over a quasicategory incarnation of C, and to the projective model structure for simplicial presheaves on a simplicial-category incarnation of C. - -REPLY [5 votes]: Yes. Let $W$ be a complete Segal space, thought of as a simplicial "space" $(W_q)$. The fibrant objects of your model category will be the fibrations $f:X\to W$ such that for each simplicial operator $\delta:[q]\to [p]$ with $\delta(q)=p$, the evident map from $X_p$ to the pullback of -$$X_q \xrightarrow{f} W_q \xleftarrow{\delta} W_p$$ -is a weak equivalence of spaces. (Edit: in fact, it suffices to require the evident map to the pullback to be a weak equivalence only for $\delta:[0]\to[p]$ with $\delta(0)=p$.) -I worked out some of this years ago, but never finished it; somebody should do this (or perhaps someone has already?). Lurie has done pretty much exactly the same thing in the context of quasi-categories, in HTT.<|endoftext|> -TITLE: Gromov-Hausdorff convergence for locally finite metric spaces -QUESTION [8 upvotes]: This question might be very easy, but I am little confused by the Gromov-Hausdorff convergence. -My situation is the following: I have a fixed set $X$ which is finite or countable; on it I have locally finite metrics $d_n$ and $d$. For the application that I have in mind, I've found that a good notion of convergence of the sequence of spaces $(X,d_n)$ to the space $(X,d)$ would be described by the uniform convergence of $d_n$ to $d$. Therefore, I am wondering if this convergence is equivalent to Gromov-Hausdorff's convergence. - -Question: Are the following statements equivalent: - -$d_n\rightarrow d$ uniformly -For any point $x\in X$, the sequence of pointed locally compact metric spaces $(X,d_n,x)$ converges to $(X,d,x)$ in Gromov-Hausdorff sense? - - -It would sound strange for my intuition if they turn out to be different, but I am in trouble to write down a proof, basically because I am quite new in the definition of Gromov-Hausdorff convergence and I am pretty confused/scared by all these isometric embeddings one should consider. -Thank you in advance for any help, -Valerio - -REPLY [10 votes]: 1) does imply 2). An alternative equivalent definition of Hausdorff-Gromov distance is as follows. A correspondence between two sets $X, Y$ is a subset of $X\times Y$ which intersects each horizontal and each vertical fiber. Then, given two metric spaces $(X,d)$, $(Y,e)$, their Hausdorff-Gromov distance is -$$ -\frac12 \inf_R \max_{(x,y), (x',y')\in R} |d(x,x')-e(y,y')|, -$$ -where the infimum is over all correspondences (This is Theorem 7.3.25 in the book of Burago-Burago-Ivanov). -When the underlying space is the same, the diagonal correspondence gives that the Hausdorff-Gromov distance between $(X,d)$ and $(X,e)$ is at most half the uniform distance between the metrics $d$ and $e$ (likewise for pointed metric spaces). In particular, 1) implies 2).<|endoftext|> -TITLE: Do subalgebras of C(X) admit a description in terms of the compact Hausdorff space X? -QUESTION [10 upvotes]: In light of the well-known theorem of Gelfand that, bluntly put, ends up saying that unital abelian C*-algebras are the 'same' as compact Hausdorff topological spaces, I tried to compile a dictionary of concepts between these two objects. More specifically, given a compact Hausdorff space $X$, I ask in what manner are topological properties of $X$ encoded into $C(X) := C(X, \mathbb{C})$? And, conversely, in what way do algebraic properties of the latter manifest topologically in the former? Here is the elementary list I was able to gather: -$\cdot$ $C(X)$ has $2^n$ idempotent elements $\Leftrightarrow$ $X$ has $n$ connected components -$\cdot$ $C(X)$ separable $\Leftrightarrow$ $X$ metrizable -$\cdot$ $C(X)$ isomorphic to $C(Y)$ $\Leftrightarrow$ $X$ homeomorphic to $Y$ -$\cdot$ continuous functions from $g:X \to Y$ induce *-homomorphisms $\hat g: C(X) \to C(Y)$ and vice-versa -$\cdot$ there is a bijective correspondence between ideals of $C(X)$ and open -sets of $X$ -What do subalgebras of $C(X)$ correspond to? If this is not a well-posed question please tell me why. Subalgebras are a very natural substructure to consider and yet I am at a loss as to how it translates over. -If you have any additions (or corrections) to the above dictionary, please share them. - -REPLY [4 votes]: This may be an interesting add-on for Martin's answer. In this paper Pavlov and Troitskii show that an inclusion of commutative $C^*$-algebras $C(X) \to C(Y)$ (with $X$ and $Y$ compact Hausdorff), which allows a positive unital conditional expectation $E \colon C(Y) \to C(X)$ that satisfies an index condition, corresponds via Gelfand duality to a branched covering $p \colon Y \to X$. The latter means that $p$ is a continuous closed and open surjection with boundedly many preimages $p^{-1}(x)$ at every $x \in X$,<|endoftext|> -TITLE: On Sketches and Institutions -QUESTION [7 upvotes]: There seems to be two competing(?) formalisms for specifying theories: sketches (as developped by Ehresmann and students, and expanded upon by Barr and Wells in, for example, Toposes, Triples and Theories), and the setting of institutions. -But I sometimes get a glimpse that sketches are really a very nice way to specifiy a good category of signatures, while institutions are much more model-theoretic. But in works on institutions, the category of signatures is usually highly under-specified (which is quite ironic, really). -So my question really is: what is the relation between Sketches and Institutions? -A subsidiary question is, why do I find a lot of work relying on institutions, but comparatively less on sketches? [I am talking volume here, not quality.] Did sketches somehow not prove to be effective? - -REPLY [3 votes]: So, saying that sketches compete with institutions is not correct because the former is an instance of the latter. The actual content of the question is probably this. -There are two types/styles/paradigms of predicate logic: elementwise (eg, ordinary FOL) and sortwise, or categorical, logic. In the latter, predicates are just sorts, whose intended internal structure (set of tuples) is given by a family of projection arrows pi: P-->Ai (i=1,2..n for n-ary predicate P \subset A1x..xAn) declared to be jointly monic to exclude duplication of tuples. Importantly, a theory interpretation t: T1-->T2 can map sorts Ai in T1 to predicates Qj in T2. Such "mixing" theory interpretations are not considered in the elementwise logic. Thus, the competition is between two styles of predicate logic: sortwise vs. elementwise. Below I'll comment on this, but first let's clarify the institution part of the question. -There are many elementwise logics (Horn, FOL, etc), each one forms an institution. There are many sortwise logics (formed by sketches of different types), each one also forms an institution (models are sketch morphisms m:T-->U from a theory sketch T to some "semantic" sketch U, usually extracted from a "semantic" category like Set). The actual Jacques' question is (Q1): why a typical work on institution theory is normally motivated by elementwise logic examples, and never by sketches? The dual of this question is (Q2): why institutions are so rarely mentioned/used in the categorical logic literature? -(Q1). Institutions were invented by Goguen and Burstall to unify the diversity of elementwise logics; I think that they never considered sortwise logics in their papers. The institution theory has been heavily used by the algebraic specification community, whose main motivating example is algebras considered elementwise (like in universal algebra rather than in categorical algebra). Sortwise logics simply did not appear in their contexts. -(Q2). For a categorical logician, the institution framework may seem unnecessarily too abstract. I mean that abstract institution's functors, sen: Sig-->Set and mod:Sig^op-->Cat, have quite concrete origin for a sketch logic. Since both sentences (diagrams) and models are arrows, functors sen and mod are given by, respectively, post- and pre-composition of these arrows with signature moprhisms. On the other hand, the institution theory does not take into account the concreteness of models (they have underlying sets), which is crucial for "real" model theory (and algebraic logic a la Henkin-Monk-Tarski). -Now about "competition" sortwise vs. elementwise logic. Different contexts may need one or the other, or both. For example, in my application area -- model-driven software engineering --- sortwise setting is very convenient and actually widely sped in practice (of course, implicitly :). However, only considering universal predicates (limits, colimits,...) would be too heavy and very inconvenient. What is needed is the possibility to consider arbitrarily sortwise predicates in syntax, but specify their semantics elementwisely (with FOL or the like) rather than by universal properties. This idea leads to a version of Makkai's generalized sketches described in [1]. The paper discusses advantages of generalized sketches over Ehresnmann's sketches in software engineering applications, and shows that generalized sketches form an institution. -[1] Diskin and Wolter, A Diagrammatic Logic for Object-Oriented Visual Modeling. ENTCS, Volume 203 Issue 6, November, 2008<|endoftext|> -TITLE: Banach-Mazur applied to a Hilbert space -QUESTION [16 upvotes]: The Banach-Mazur theorem says that every separable Banach space is isometric to a subspace of $C^0([0;1],R)$, the space of continuous real valued functions on the interval $[0;1]$, with the sup norm. -If we apply this to $\ell^2(R)$, then we see that $C^0([0;1],R)$ has a subspace which is a Hilbert space for the sup norm. -My question is can one write down explicitly such a subspace of $C^0([0;1],R)$? -I'm just curious, that's all. - -REPLY [8 votes]: By "explicit", I was really asking for a sequence of continuous functions... - -Ah, OK. Take a Peano curve $(f,g)$ from $[-1,1]$ to $[-1,1]^2$. Now define inductively $F_1=f$, $F_{k+1}=F_k\circ g$. Note that for any finite sequence of points $y_j\in[-1,1]$ ($j=1,\dots, n$), we can find $x\in[-1,1]$ such that $F_j(x)=y_j$ (induction: the last $n-1$ functions determine $g(x)$ after which $f(x)$ is free to choose). Now just define $H_1=F_1$, $H_{k+1}=F_{k+1}\sqrt{1-H_1^2-\dots-H_k^2}$. This is your sequence. -It just remains to construct the Peano curve "explicitly". It is not going to be nice, so an elementary formula is out of question. However, you can take the Fourier series $\sum_{k} v_k\cos(2\pi A_k x)$, where $v_k$ is a sequence of vectors in $\mathbb R^2$ and $A_k$ is a sequence of integers, with properly chosen (explicit) $v_k$ and $A_k$ to get the image to contain the unit square. Now just truncate the values to $[-1,1]$ in the usual way.<|endoftext|> -TITLE: Extremal properties of the determinant for matrices with entries in a fixed subset of $[-1,1]^{n^2}$? -QUESTION [9 upvotes]: Given a multiset $S\subset [-1,1]^{n^2}$, we set -$$m(S)=\min\vert \det(M)\vert$$ -where the minimum is over all matrices with entries forming the multiset $S$ -and -$$a(n)=\max m(S)$$ -where the maximum is over all multisets with $n^2$ elements in $[-1,1]$. -Obviously $a(2)=2$ by considering $S=\lbrace 1,1,1,-1\rbrace$. -I know nothing else (except for the trivial bounds $0 < a(n)\leq n^{n/2}$). -Even the computation of $a(3)$ (or of a good lower bound on $a(3)$) -seems quite a feat to me. - -REPLY [2 votes]: I don't know the answer to your question, but the following is a bit long for a comment. -We can always assume one of the matrix elements is 1 since rescaling all elements so that the largest element equals 1 increases the minimal determinant by the scaling factor raised to the power $n$. It seems likely that any solution will have many 1s—of course less than $2n$, but possibly order $n$ of them—although I can't see how to prove this. For $n=3$ the matrix with elements in $[-1,1]$ of largest determinant is -$$ -\begin{bmatrix} --1 & 1 & 1\\ -1 & -1 & 1\\ -1 & 1 & -1 -\end{bmatrix}. -$$ -This provides an upper bound of 4, which is slightly better than Hadamard's bound of about 5.2, but still apparently far too high. -It's possible to get a very slight improvement on Kevin Costello's lower bound for $n=3$. Consider the multiset $\{1,1,1,1,1,a,b,c,d\}$ with -$$ -a=0.19552006830186067389, -$$ -$$ -b=-0.47998412524185001, -$$ -$$ -c=-0.898326460649234689, -$$ -$$ -d=-0.248885944004550461. -$$ -It has minimal determinant 0.185913849057346968. A hill-climbing procedure repeatedly arrives at solutions with five 1s. If the procedure is modified to assume five 1s, it readily finds solutions very close to the one above. The hill-climbing solutions (approximately) satisfy the property that the four matrices -$$ -\begin{bmatrix}1 & 1 & a\\ 1 & d & 1\\b & 1 & c\end{bmatrix},\quad -\begin{bmatrix}1 & 1 & 1\\ 1 & 1 & a\\d & b & c\end{bmatrix},\quad -\begin{bmatrix}1 & 1 & a\\ 1 & c & 1\\d & 1 & b\end{bmatrix},\quad -\begin{bmatrix}1 & 1 & a\\ 1 & 1 & d\\b & c & 1\end{bmatrix}, -$$ -all have (minus the) minimal determinant. Imposing equality of these determinants, and maximizing their magnitude, yields the precise values above. (They are some messy algebraic numbers.)<|endoftext|> -TITLE: Algebraic K-theory of the group ring of the fundamental group -QUESTION [20 upvotes]: I know of two places where $K_{*}(\mathbb{Z}\pi_{1}(X))$ (the algebraic $K$-theory of the group ring of the fundamental group) makes an appearance in algebraic topology. -The first is the Wall finiteness obstruction. We say that a space $X$ is finitely dominated if $id_{X}$ is homotopic to a map $X \rightarrow X$ which factors through a finite CW complex $K$. The Wall finiteness obstruction of a finitely dominated space $X$ is an element of $\tilde{K_{0}}(\mathbb{Z}\pi_{1}(X)) $ which vanishes iff $X$ is actually homotopy equivalent to a finite CW complex. -The second is the Whitehead torsion $\tau(W,M)$, which lives in a quotient -of $K_{1}(\mathbb{Z}\pi_{1}(W))$. According to the s-cobordism theorem, if $(W; M, M')$ is a cobordism with $H_{*}(W, M) = 0$, then $W$ is diffeomorphic to $M \times [0, 1]$ if and only if the Whitehead torsion $\tau(W, M)$ vanishes. -For more details, see the following: -http://arxiv.org/abs/math/0008070 (A survey of Wall's finiteness obstruction) -http://www.maths.ed.ac.uk/~aar/books/surgery.pdf (Algebraic and Geometric Surgery. See Ch. 8 on Whitehead Torsion) -My question is twofold. -First, is there a high-concept defense of $K_{*}(\mathbb{Z}\pi_{1}(X))$ as a reasonable place for obstructions to topological problems to appear? I realize that $\mathbb{Z}\pi_{1}(X)$ appears because the (cellular, if $X$ is a cell complex) chain groups of the universal cover $\tilde{X}$ are modules over $\mathbb{Z}\pi_{1}(X)$. Is it the case that when working with chain complexes of $R$-modules, we expect obstructions to appear in $K_{*}(R)$? -Second, is there an enlightening explanation of the formal similarity between these two obstructions? (Both appear from considering the cellular chain complex of a universal cover and taking an alternating sum.) - -REPLY [17 votes]: To add to Tim Porter's excellent answer: -The story of what we now call $K_1$ of rings begins with Whitehead's work on simple homotopy equivalence, which uses what we now call the Whitehead group, a quotient of $K_1$ of the group ring of the fundamental group of a space. -On the other hand, the story of $K_0$ of rings probably begins with Grothendieck's work on generalized Riemann-Roch. What he did with algebraic vector bundles proved to be a very useful to do with other kinds of vector bundles, and with finitely generated projective modules over a ring, and with some other kinds of modules. -I don't know who it was who recognized that these two constructions deserved to be named $K_0$ and $K_1$, and viewed as two parts of something larger to be called algebraic $K$-theory. But Milnor gave the right definition of $K_2$, and Quillen and others gave various equivalent right definitions of $K_n$. -Let me try to lay out the parallels between the topological significances of Whitehead's quotient of $K_1(\mathbb ZG)$ and Wall's quotient of $K_0(\mathbb ZG)$. My main point is that both of them have their uses in both the theory of cell complexes and the theory of manifolds. -The Whitehead group of $G$ is a quotient of $K_1(\mathbb ZG)$. Its significance for cell complexes is that it detects what you might call non-obvious homotopy equivalences between finite cell complexes. An obvious way to exhibit a homotopy equivalence between finite complexes $K$ and $L$ is to by attaching a disk $D^n$ to $K$ along one half of its boundary sphere and obtain $L$. Roughly, a homotopy equivalence between finite complexes is called simple if it is homotopic to one that can be created by a finite sequence of such operations. The big theorem is that a homotopy equivalence $h:K\to L$ between finite complexes determines an element (the torsion) of the Whitehead group of $\pi_1(K)$, which is $0$ if and only if $h$ is simple, and that for any $K$ and any element of its Whitehead group there is an $(L,h:K\to L)$, unique up to simple homotopy equivalence, leading to this element in this way, and that this invariant of $h$ has various formal properties that make it convenient to compute. -One reason why you might care about the notion of simple homotopy equivalence is that for simplicial complexes it is invariant under subdivision, so that one can in fact ask whether $h$ is simple even if $K$ and $L$ are merely piecewise linear spaces, with no preferred triangulations. This means that, for example, a homotopy equivalence between compact PL manifolds (or smooth manifolds) cannot be homotopic to a PL (or smooth) homeomorphism if its torsion is nontrivial. (Later, the topological invariance of Whitehead torsion allowed one to eliminate the "PL" and "smooth" in all of that, extending these tools to, for example, topological manifolds without using triangulations.) -But the $h$-cobordism theorem says more: it applies Whitehead's invariant to manifolds in a different and deeper way. -Meanwhile on the $K_0$ side Wall introduced his invariant to detect whether there could be a finite complex in a given homotopy type. Note that where $K_0$ is concerned with existence of a finite representative for a homotopy type, $K_1$ is concerned with the (non-)uniqueness of the same. -Siebenmann in his thesis applied Wall's invariant to a manifold question in a way that corresponds very closely to the $h$-cobordism story: The question was, basically, when can a given noncompact manifold be the interior of a compact manifold-with-boundary? Note that there is a uniqueness question to go with this existence question: If two compact manifolds $M$ and $M'$ have isomorphic interiors then this leads to an $h$-cobordism between their boundaries, which will be a product cobordism if $M$ and $M'$ are really the same. -One can go on: The question of whether a given $h$-cobordism admits a product structure raises the related question of uniqueness of such a structure, which is really the question of whether a diffeomorphism from $M\times I$ to itself is isotopic to one of the form $f\times 1_I$. This is the beginning of pseodoisotopy theory, and yes $K_2$ comes into it. -But from here on, the higher Quillen $K$-theory of the group ring $\mathbb Z\pi_1(M)$ is not the best tool. Instead you need the Waldhausen $K$-theory of the space $M$, in which basically $\mathbb Z$ gets replaced by the sphere spectrum and $\pi_1(M)$ gets replaced by the loopspace $\Omega M$. It's a long story!<|endoftext|> -TITLE: State of the art for Gersten's conjecture for K-theory? -QUESTION [15 upvotes]: Does anyone know (of a reference to) under what restrictions on the regular scheme $X$ it is known that we have an exact sequence -$$0 \to \mathcal{K}_n(X) \to \bigoplus_{x \in X^{(0)}} K_n(k(x)) \to \bigoplus_{x \in X^{(1)}} K_{n - 1}(k(x))$$ -where $\mathcal{K}_n$ is the Zariski sheaf associated to $K_n$? -More specifically, I would like the following to be true. -1) The above sequence is exact when $X$ is a (separated noetherian) regular scheme of dimension one. -2) $\mathcal{K}_n(X) \to \bigoplus_{x \in X^{(0)}} K_n(k(x))$ is injective for all regular (separated noetherian) schemes. -And I would like these to be true with the Nisnevich sheafification of $K_n$ as opposed to the Zariski one (I'm aware that in many cases the two sheafifications are the same so I guess another part of the question is in which cases $(K_n)_{Zar} = (K_n)_{Nis}$). - -REPLY [3 votes]: On the arXiv there is a paper of Satoshi Mochizuki on "Gersten’s conjecture for commutative discrete valuation rings" from the year 2007. In this paper Mochizuki claims to prove the Gersten conjecture for all commutative DVRs. But the paper didn't appear in any journal, apperently. Does anyone know if it has been verified?<|endoftext|> -TITLE: Proper morphism sending coherent to coherent -QUESTION [8 upvotes]: Hello, -Is there a proof that the push forward by a proper morphism of Noetherian schemes sends coherent sheaves to coherent ones, without passing in the argument through projective morphisms? -Thank you, -Sasha - -REPLY [4 votes]: Gerd Faltings, Finiteness of coherent cohomology for proper fppf stacks, J. Algebraic Geometry 12 (2003) 357–366<|endoftext|> -TITLE: Does the right adjoint of a Quillen equivalence preserve homotopy colimits? -QUESTION [10 upvotes]: Call a diagram $E$ in a model category a homotopy colimit diagram if the morphism $$\mathrm{hocolim}~E\to \mathrm{colim}~ E$$ is a weak equivalence. A homotopy colimit is defined as the categorical colimit of a cofibrant replacement of the diagram in the projective model structure and this is where the morphism comes from. -Let $F:C\rightleftarrows D:G$ be a Quillen equivalence between model categories $C$ and $D$. The (Edit: derived!) left adjoint $F$ preserves homotopy colimits, i.e. if $E$ is a homotopy colimit diagram in $C$, then $F\circ Q\circ E$ is a homotopy colimit diagram in $D$ where $Q$ denotes a cofibrant replacement. - -Does the (Edit: derived!) right adjoint $G$ preserve homotopy colimits if the adjunction is a Quillen equivalence? -To be more precise, if $E$ is a homotopy colimit diagram in $D$, is $G\circ R\circ E$ is a homotopy colimit diagram in $C$ where $R$ denotes a fibrant replacement? - -I suppose that this is true since the notion of homotopy colimit should depend only on the homotopy category and not on the model, I guess, but I cannot think of an argument. - -REPLY [5 votes]: The homotopy colimit functor $Ho(D^I)\rightarrow Ho(D)$ is the left adjoint of the constant diagram functor $Ho(D)\rightarrow Ho(D^I)$. Quillen equivalences induce Quillen equivalences between diagram categories, you you can replace $D$ with $C$, hence you're done by uniqueness of adjoints. -PS Don't worry about the fact that $D^I$ may not be a model category if $D$ is not cofibrantly generated. You can work with weaker axioms and convenient replacement of the notion of Quillen equivalence.<|endoftext|> -TITLE: Quotients of f.p. amenable groups -QUESTION [8 upvotes]: Can you give me an example of a finitely generated infinitely presented amenable group which is a quotient of a finitely presented amenable group? - -REPLY [12 votes]: Take the finitely presented solvable group $G$ with undecidable word problem, constructed by Kharlampovich. That group has infinite center that is a direct product of infinite number of cyclic group. The center has uncountably many subgroups $N_\alpha$, each normal in $G$, most groups $G/N_\alpha$ are not finitely presented but amenable. The description of an easier construction of Kharlampovich's group is in our survey Kharlampovich, O. G. Sapir, M. V. -Algorithmic problems in varieties. Internat. J. Algebra Comput. 5 (1995), no. 4-5, 379–602. Another, easier, example, is Abels' group and its quotients by central subgroups. Abels' group can be found here: H. Abels, An example of a finitely presented solvable group, Homological group theory (Proc. Sympos., Durham, 1977), London Math. Soc. Lecture Note Ser., 1979, p. 205–211. It has been mentioned on MO before (see Cornulier's answers).<|endoftext|> -TITLE: Unirational implies rationally connected -QUESTION [5 upvotes]: It is evidently a well-known fact that a unirational variety $X$ over an algebraic closed field (i.e. there is a dominant rational map from $\mathbb P^n$ to $X$) is rationally connected (by which I mean that any two points can be joined by a chain of rational curves). Numerous authors on birational geometry seem to state this as a remark, but don't indicate how one might prove it. The only proofs I have found of this fact (i.e. Fulton's Intersection Theory book example 10.1.6 and the paper of Samuel he quotes there) use the completion of local rings and power series. I was wondering if there was a purely algebraic (i.e. without completions) proof of this result. -In particular, by blowing $\mathbb P^n$ at the indeterminancy locus of the rational map to $X$ we get a commutative diagram involving a birational, projective, surjective morphism from $\tilde{\mathbb P^N}$ to $\mathbb P^n$, our original rational map from $\mathbb P^n$ to $X$, and a projective, surjective morphism $\tilde{\mathbb P^n} \rightarrow X$, so if we can show that the blowup is rationally connected then mapping to $X$ will give us our chain of rational curves connecting any two points of $X$. This reduces to the following affine case: We are then left with the case of showing that if $\pi: T\rightarrow \mathbb A^n$ is the blow-up of $\mathbb A^n$ along a subcscheme Z, with exceptional divisor $E$, and $t\in E$, then there is a morphism $h: \mathbb A^1\rightarrow T$ with $h(0)=t$ but $h(\mathbb A^1)$ not contained in $E$. It is here that I was wondering if people knew of a way to procede without using power series as Fulton and Samuel do. -I would also be interested in other proofs of this result. - -REPLY [3 votes]: Assume that $X$ is unirational. Then we have a dominant rational map -$$\phi:\mathbb{P}^{n}\dashrightarrow X$$ -Let $x,y \in X$ be two general points and let us consider two points $p,q$ in $\pi^{-1}(x),\pi^{-1}(y)$ respectively. Take the line $L$ generated by $p$ and $q$. Then $\phi_{|L}:L\rightarrow X$ is a finite morphism and its image $C = \phi(L)$ is a rational curve through $x$ and $y$. So $X$ is rationally connected.<|endoftext|> -TITLE: $c_0$-direct sum of $\mathcal{K}(\mathcal{H})$ -QUESTION [7 upvotes]: Let $\mathcal{K}(\mathcal{H})$ be the C*-algebra of compact operators on a Hilbert space $\mathcal{H}$. I am interested in the ($c_0$-)sum -$A=\sum \mathcal{K}(\mathcal{H})$ -of countably many copies of this algebra. -Is it *-isomorphic to $\mathcal{K}(\mathcal{H})$ itself? Or at least as a Banach space? - -REPLY [11 votes]: Yes, there is a Banach space isomorphism. The $c_0$ sum of $\mathcal{K}(H)$ is clearly isometrically isomorphic to its own $c_0$ sum and contains $\mathcal{K}(H)$ as a norm one complemented subspace, so by the Pelczynski decomposition method it is enough to observe that the $c_0$ sum of $\mathcal{K}(H)$ embeds into $\mathcal{K}(H)$ as a complemented subspace. Write $H$ as the orthogonal direct sum of orthogonal infinite dimensional subspaces $H_n$ and $P_n$ the corresponding orthogonal projections; then $\mathcal{K}(H_n)\subset \mathcal{K}(H)$ isometrically in an obvious way. Moreover, if $T_n$ is in $\mathcal{K}(H_n)$ then $\sum_{n=1}^N P_nT_n P_n$ in $\mathcal{K}(H)$ has norm the maximum of $\|T_1\|,\dots \|T_N\|$. This gives an isometric embedding of the $c_0$ sum of $\mathcal{K}(H)$ into $\mathcal{K}(H)$. You get a norm one projection onto this subspace of $\mathcal{K}(H)$ by defining $P(T)=\sum P_nTP_n$.<|endoftext|> -TITLE: Haar measure on infinite dimensional Lie groups? -QUESTION [7 upvotes]: Hi. Is there a Haar measure or equivalent on infinite dimensional Lie groups? I've been playing around with $Diff(S^1)$, and at least a direct approach seems quite hopeless. It goes something like this: -Def. element on the group by "Euler coordinates", -$g \doteq \prod\limits_{i=-\infty}^{\infty} e^{\omega^i X_i}$, with $\left[ X_i ,X_j \right] = (j-i)X_{i+j}$. -Now I could define a (left invariant) Maurer-Cartan form as $\Omega_L \doteq g^{-1} dg = X_i \otimes \theta^i$, where $\theta^i = \mathcal L^i_j d\omega^j$. Then the Haar measure is -$d\mu (g) \doteq ||\mathcal L || \bigwedge\limits_i d\omega^i$. -Elements of $\mathcal L$ can be written as -$\mathcal L^i_j = \left( \prod\limits_{n=\infty}^{j+1} \exp(-\omega^n adX_n) \right)^i_j$ -Clearly the determinant $||\mathcal L ||$ will be horrible... is there any hope for a manageable explicit expression? I couldn't find any literature on the subject (yet), so I'd appreciate any hints to the right direction. -EDIT: umm and of course the whole question of existence of such a measure should probably addressed... -EDIT 2: I realized that the question setup is a bit misleading: I'm actually looking for a measure on the Virasoro group (with zero central charge), i.e. the Lie group corresponding to the algebra above... maybe the Shavgulidze measure has something to do with it, I don't know... - -REPLY [7 votes]: There is something called the Shavgulidze, or the Malliavin-Shavgulidze measure on Diff of smooth manifolds. You can find a discussion in Differentiable measures and the Malliavin calculus (p. 397, available on google books). It is not quite invariant, but quasi-invariant.<|endoftext|> -TITLE: Algebraic varieties and UFD -QUESTION [6 upvotes]: Given an affine algebraic variety $V$ such that $\Gamma(V,\mathcal{O}_V)$ is a UFD, its sheaf of ring can be determined easily since one can show that: -$$\Gamma(D(f_1) \cup \cdots \cup D(f_n),\mathcal{O}_V) \simeq \Gamma(D(h),\mathcal{O}_V),$$ -where $h=\mathrm{gcd}(f_1, \ldots, f_n)$. -It is then natural to ask whether we can characterize the set of affine algebraic varieties $V$ such that $\Gamma(V,\mathcal{O}_V)$ is a UFD? Are there any geometric interpretations? -Thanks you! - -REPLY [5 votes]: A noetherian domain is a UFD if and only if it is normal and the divisor class group is zero. See, for example, exercise 1D here.<|endoftext|> -TITLE: A metabelian quotient of a free group -QUESTION [8 upvotes]: I don't know much about free groups (excepted the very basics), and the following question may be trivial, although it isn't to me. -Let $F$ be a free group with $n$ generators $x_1,\dots,x_n$. Consider the -'augmentation' map $a:F \rightarrow \mathbb{Z}$ that sends $x_i$ to $1$ for $i=1,\dots,n$, -and let $A = \ker f$. So $A$ is a free group as a subgroup of the free groups $F$, of infinite rank I presume. Is it easy to describe a free family of generators of $A$? This would surely allow to answer my question, which is: - -Can we describe the action of $\mathbb{Z}$ on $A^{ab}$ ? - -Here, $A^{ab} = A /D(A)$ is the abelianization of $A$ (with $D(A)$ its derived subgroup), -and the action in question is the one given by the short exacts sequence: -$$1 \rightarrow A/D(A) \rightarrow F/D(A) \rightarrow \mathbb{Z} \rightarrow 1$$ -The group $F/D(A)$ is the metabelian group of the title. Of course, describing the action is -essentially equivalent to describing the group $F/D(A)$, since the extension splits. -(The question feels elementary to me, but at the same time I feel helpless to solve it, because I don't know how to recognize when a family of elements in a free group is free.) -Now the truth is that the real question I need is when $F$ is a free pro-$p$ group -with $n$ generators instead of a free group, $a$ is the continuous map from $F$ to $\mathbb{Z}_p$ that sends the generators to $1$, and $D(A)$ is the closed derived subgroup. But I believe (perhaps naively) that the solution of the discrete problems will easily give a solution of its pro-$p$ analog, and that the discrete problem is more natural. -This question in turn comes from my trying to understand the structure of the maximal metabelian quotient of some pro-$p$ Galois group of number fields with prescribed ramification. In some cases such a group is the $p$-adic $F/D(A)$ considered in this question. - -REPLY [7 votes]: To find generators of $A$, use the Nielsen-Schreier method. It is very easy in that case: http://en.wikipedia.org/wiki/Nielsen%E2%80%93Schreier_theorem - -REPLY [4 votes]: Do you mean the augmentation map on the group ring $\mathbb{Z}[F]$ (and in the pro-$p$ case, the completed group ring $\mathbb{Z}_p[[F]]$?) I ask only because this augmentation map comes up frequently and significantly in the study of large number-theoretic Galois groups. -Assuming this is the case (and apologies for misinterpreting if not -- hopefully the answer will still be of some use to you), there is a tremendous amount of machinery set up for dealing exactly with questions of this sort -- probably the best starting place is the phrase "pro-p Fox Differential Calculus." (And so, indeed, your intuition that solving the discrete problem turns out to provide the correct pro-$p$ analog is correct. It was Iwasawa who carefully established the fundamental analogy here. In fact, thanks to the topology of $\mathbb{Z}_p$, in some ways the pro-p Fox calculus is nicer than the discrete version.) In particular, if you filter the group ring $\mathbb{Z}_p[[F]]$ by powers of the augmentation ideal (the group-ring version of your $A$), you land upon the sequence of "dimension subgroups" of F. -These subgroups have shown up repeatedly in the analysis of pro-$p$-groups arising in the study of large Galois groups arising from restricted ramification questions (as appears to be the case for you). A couple of the highlights of the theory are the work of Vogel and Morishita interpreting number-theoretic analogs of the a priori knot-theoretic notion of Milnor invariants, refined versions of Golod-Shafarevich-type inequalities, and perhaps most relevant for your question, work of Arrigoni (e.g., "On Schur $\sigma$-groups") which I think explicitly answers questions of your type. For a more fundamental reference, see Koch's "Galois theory of $p$-extensions.") -Sorry to be mostly hand-wavey -- I'm away from good references at the moment.<|endoftext|> -TITLE: Acceptability and Soundness of J-structures. -QUESTION [7 upvotes]: I would like an example of a J-structure $(J^A,B)$ which is not acceptable and one that is not 1-sound. -Edit:Let us recall that a structure $J^A_\alpha$ is acceptable if for every limit ordinal $ \xi<\alpha $. $J^A_{\xi+\omega}\models \vert \xi\vert\leq \vert \tau\vert $, whenever $\tau<\xi$ and satisfies ${\mathcal P}(\tau)\cap J^A_{\xi+\omega}\not \subset J^A_\xi $ -A structure $J^A_\alpha$ is 1-sound if the 1-standard parameter is a very good parameter. -(I'm using the notation in Zeman's article in the handbook of set theory) -It is known that $J_\alpha$ are acceptable and sound for every ordinal $\alpha$. Moreover, being acceptable and sound is needed for almost all basic results concerning the $J^A_\alpha$ hierarchy. - -REPLY [4 votes]: For an amenable $(J,B)$ which is not 1-sound, take a non-constructible real $x$ such that $\aleph_1^L = \aleph_1^{L[x]}$ (and let's say $V = L[x]$ so this is the true $\aleph_1$). Set $B = \lbrace\omega_1+n:n \in x\rbrace$. Then $(J_{\omega_1+1},B)$ is amenable and $x$ is $\Sigma_1(J_{\omega_1+1},B)$ (with parameter $\omega_1$). The $\Sigma_1$-projectum is therefore $1$, so $(J_{\omega_1+1},B)$ cannot be $1$-sound because $J_{\omega_1+1}$ is uncountable and the available parameter set is countable. -(My notes say that this example is from Lee Stanley, but they don't say where I found it. If anybody knows where this is from, please leave a comment.)<|endoftext|> -TITLE: Are subgroups of hyperbolic groups quasiisometrically embedded ? -QUESTION [9 upvotes]: Given a finitely generated subgroup of a finitely generated hyperbolic group. Is it true that the inclusion of each subgroup is a quasiisometric embedding ? -The first example for a group that does not have this property is a Baumslag-Solitar group $BS(1,m)= \langle a,b| bab^{-1}=a^m\rangle$. We have $a^{m^k}=b^kab^{-k}$ for each $k$. This shows for example that the inclusion of the subgroup generated by $a$ is not a quasiisometric embedding. -Then one can consider the class of groups with the property that the inclusion of any subgroup is a quasiisometric embedding. Has this class been studied? - -REPLY [9 votes]: To add to the examples given, let G be the fundamental group of a closed hyperbolic 3-manifold that fibers over the circle, and let F be the fiber subgroup. Then F is a surface subgroup that is not quasiisometrically embedded.<|endoftext|> -TITLE: Categorifying the free monoid and non-commutative generating functions -QUESTION [8 upvotes]: I am a complete novice in the art of categorification, so this may not be a great question. -Background. The groupoid $\mathbf {FSet}$ of finite sets and bijections categorifies the natural numbers and the functor category $\mathbf {FSet}^{\mathbf {FSet}}$ (called the category of combinatorial species) categorifies generating functions. - If $X$ is a finite set, the slice category $\mathbf {FSet}/X$ of finite sets over $X$ (or $X$-indexed families of finite sets) categorifies the free commutative monoid on $X$, the forgetful functor $\mathbf {FSet}/X\longrightarrow \mathbf {FSet}$ categorifies word length (or degree of a monomial) and I believe the functor category $(\mathbf {FSet}/X)^{\mathbf {FSet}/X}$ categorifies generating functions in commuting variables $X$. - - -Question. Is there a categorification $\mathbf C$ of the free monoid on a finite set $X$ together with a ''forgetful" functor $\mathbf C\longrightarrow \mathbf {FSet}/X$ categorifying the abelianization map such that $\mathbf C^{\mathbf C}$ categorifies generating functions in non-commuting variables $X$. - - -I assume the answer will be some sort of "free" monoidal category. Please take into account that I am a category-friendly mathematician, but not a category theorist, when formulating your answer. -Vague motivation. Is there a categorification of the -Chomsky-Schützenberger theorem on context-free grammars and algebraic power series? -I realize that species categorify exponential generating fuctions and Chomsky-Schützenberger is about ordinary generating functions, but this shouldn't matter. -Update.. The free monoidal category is no good because it only has identity morphisms. It doesn't even work to recover the 1-generated case. - -REPLY [2 votes]: I think you've been misled by the standard description of combinatorial species. It's cleaner to think of (exponential) generating functions in one variable as being categorified by analytic functors, namely endofunctors of the form -$$X \mapsto \bigsqcup_{n \ge 0} F_n \otimes_{S_n} X^{\otimes n}$$ -where $X$ is an object in a symmetric monoidal category $M$ with infinite coproducts and finite colimits, although these endofunctors won't behave as expected unless the monoidal product distributes over colimits. For example, pointwise coproduct corresponds to adding generating functions, pointwise tensor product corresponds to multiplying generating functions, and composition corresponds to composition. -A straightforward categorification of (again, exponential) generating functions in more than one variable is then analytic functors of more than one variable, e.g. in two variables these take the form -$$X, Y \mapsto \bigsqcup_{n, m \ge 0} F_{n, m} \otimes_{S_n \times S_m} X^{\otimes n} \otimes Y^{\otimes m}.$$ -The relationship to species comes from describing convenient choices of "coefficients" $F_n, F_{n, m}$, which I didn't specify above. A common situation is that the coefficients live in a symmetric monoidal category $V$ such that $M$ is both enriched and tensored over $V$. Then $M$ naturally acquires an action by analytic functors with coefficients given by $V$-valued species, or equivalently presheaves $S^{op} \to V$, where $S$ denotes the groupoid of finite sets and bijections. For analytic functors in two commuting variables the coefficients come from presheaves $S^{op} \times S^{op} \to V$, etc. Here you should think of $S$ as the free symmetric monoidal category on an object, $S \times S$ as the free symmetric monoidal category on two objects, etc. -For noncommuting variables and ordinary generating functions, replace "symmetric monoidal" with "monoidal" everywhere above, except that I think extra conditions are required for the resulting functors to be closed under composition (e.g. $V$ should probably still be symmetric monoidal and it should act "centrally" on $M$). Coefficients are now given by $V$-valued presheaves on the free monoidal category on however many objects. -It's important to note that $S$ and $V$ have nothing to do with each other. The misleading thing about the classical case of combinatorial species is that $S$ and $V$ look very similar; the more general theory shows how they diverge. A very important special case is when $V$ is some variant of vector spaces; then $V$-valued species are Schur functors, which subsequently naturally act on any symmetric monoidal linear category with enough colimits.<|endoftext|> -TITLE: Reference request for translating from Top to C*-alg -QUESTION [32 upvotes]: Some recent questions on MO (for example, Do subalgebras of C(X) admit a description in terms of the compact Hausdorff space X?) have been about Gelfand duality — namely, that the categories of compact Hausdorff spaces with continuous maps, and commutative unital $\newcommand{\Cstar}{{\rm C}^*}\Cstar$-algebras with unital $*$-homomorphisms, are anti-equivalent. Thus one can "translate" properties about compact spaces over to $\Cstar$-algebras. This can lead to a sort of "dictionary", see for example page 3 of Várilly's book on noncommutative geometry (Google Books link). - -Does anyone know a reasonably definitive reference for proofs of such dictionaries, in a self-contained form?? - -I'm guessing that perhaps such a thing doesn't exist, as these results are folklore (and are easy to prove really—given the statement, the proofs often form nice exercises). As one is really just studying compact spaces via the category of compact spaces with continuous map, might there be a category theory book which is suitable? -Actually, I am more interested in the non-unital case. Rather than working with proper maps, I instead want to follow Woronowicz. Define a "morphism" between $\Cstar$-algebras $A$ and $B$ to be a non-degenerate $*$-homomorphism $\phi:A\rightarrow M(B)$ from $A$ to the multiplier algebra of $B$, where "non-degenerate" means that $\{ \phi(a)b \mathbin{\colon} a\in A,b\in B \}$ is linearly dense in $B$. Then the category of commutative $\Cstar$-algebras and morphisms is anti-equivalent to the category of locally compact spaces and continuous maps. One can then form a similar dictionary — but here I think the proofs can be a bit trickier (or maybe just they use slightly less standard topology). - -Does anyone know a reasonably definitive reference in this more general setting? - -REPLY [10 votes]: Gert Pedersen wrote "In a careless moment a C*-algebraist -might be quoted for saying that there is a covariant functor between the categories of commutative C*-algebras with morphisms and the category of locally compact Hausdorff spaces with continuous maps." (Morphisms of Extensions of C*-Algebras: Pushing Forward the Busby Invariant, by Eilers me and Pedersen.) -Either of these correspondences is valid: - -proper continuous maps $\leftrightarrow$ proper *-homomorphisms from A to B -continuous maps $\leftrightarrow$ nondegenerate *-homomorphisms from A to M(B). - -I misunderstood the exercise on Page 44 of Wegge-Olsen is wrong. This set of the discussion below.<|endoftext|> -TITLE: Using MAGMA for Group Theory -QUESTION [7 upvotes]: I've just started a PhD in Group Theory and need to use the computer programme MAGMA. I wonder if anyone could help me with a couple of (probably very basic things). - -I need to produce a Hasse diagram for subgroups of a given group containing a given Sylow subgroup of the group. In MAGMA I can use the command Subgroups(G:OrderMultipleOf:=??) to obtain all subgroups of a group G which contain a Sylow subgroup, however is there a command I can use so that for a given group G and a Sylow subgroup S, I can produce all subgroups of G containing S. -As I'm very new to MAGMA, does anyone know of any good books, publications or websites aiding someone to use MAGMA for group theoretical purposes. - -Thanks in advance for your help. -David - -REPLY [4 votes]: Although you say you'd prefer not to use GAP, producing a Hasse diagram is very easy in GAP, at least with the right packages. -You'll need the xgap GAP package; and either the xgap binaries, which requires an X Windows system (easiest done with Linux or a similar Unix-like system), or else Gap.app, which requires a Mac. -Once you have these installed, start xgap/Gap.app, and follow these steps: -Type "GraphicSubgroupLattice(SymmetricGroup(4));" -In the window that pops up, go to the Subgroups | All Subgroups menu. -The Hasse diagram of the subgroup lattice will appear. -It's also quite easy to show parts of the subgroup lattice -- essentially, you can take any list of subgroups and show the inclusion relations. To do this: -Type "GraphicSubgroupLattice(G);" as before. -Compute the list of subgroups you want to display. It should be the output of the last GAP command. -Go to the Subgroups | Insert Vertices menu. -The Hasse diagram of the subposet consisting of subgroups from your list will appear. -There's probably a comparably easy way to show Hasse diagrams in MAGMA. (But I'm telling you what I know...) -Xgap is usually included with GAP, and is also available from: -http://www.gap-system.org/Packages/xgap.html -Gap.app is available from: -https://cocoagap.sourceforge.io/<|endoftext|> -TITLE: Does there exist a subset of $\mathbb{R}^2$ which is "very small" and "very big" in the specified way? -QUESTION [8 upvotes]: Does there exist a set $M \subset \mathbb{R}^2$ which has the following two properties: - -Forall $x \in \mathbb{R}$ the set $\{y \in \mathbb{R} \mid (x,y) \in M\}$ is countable. -Forall $y \in \mathbb{R}$ the set $\{x \in \mathbb{R} \mid (x,y) \notin M\}$ is countable. - -REPLY [6 votes]: Here's an easy proof of the equivalence of the statement to CH (I think). One direction is just what Emil said. For the other direction: -Suppose continuum is $\geq \aleph_2$. Restrict attention to an $\aleph_2$-sized subset of $\mathbb{R}$, and suppose we had a relation with the properties you want on that subset. (Note: if there is a relation with those properties on $\mathbb{R}^2$ then it will retain those properties when we restrict to a smaller set.) -Take $\aleph_1$ many $x$-coordinates from this set; each of them only has countably many $y$'s that it gets paired with in the relation, so in total there are at most $\aleph_1$ many $y$'s that get paired with any of these $x$'s. So, take some $y$ which doesn't get paired with any of these $x$'s. (There will be such a $y$ because we have $\aleph_2$ many $y$'s in total.) This $y$ has at least $\aleph_1$ many $x$'s that it does not get paired with; and this contradicts the properties of the relation. -Ramiro, I'm guessing this is the same argument Sierpinski gave?<|endoftext|> -TITLE: Centralizers of non-iwip elements of $Out(F_n)$ -QUESTION [10 upvotes]: Does there exist an infinite order element $\phi\in Out(F_n)$, for some or all $n\geq 3$, which is not iwip but has finite index in its centralizer? How about an element such that all its non-zero powers have this property? - -Motivation: iwip elements are Morse (i.e., roughly speaking, all quasi-geodesics connecting points on a given orbit stay close to it), and a standard way of proving that an element is not Morse is showing that it has infinite index in its centralizer, or one of its powers do. An element in the mapping class group of a closed surface is Morse if and only if it is pseudo-Anosov. - -REPLY [7 votes]: There is a Nielsen-Thurston type method due to Feighn and Handel which is useful for approaching this question. The method is laid out in the papers arXiv:math/0612702 and arXiv:math.GR/0612705 by Feighn and Handel, "The recognition theorem for $Out(F_n)$ and "Abelian subgroups of $Out(F_n)$". It is an outgrowth of the train track and relative train track machinery developed in earlier papers of Bestvina, Feighn, and Handel, particularly math.GT/9712217, part I of their series on the Tits alternative for $Out(F_n)$. -As Ashot suggests, take $F_4 = \langle a,b,c,d \rangle$ and take $\Phi \in Aut(F_4)$ to preserve $\langle a,b,c\rangle $ so that the action on $\langle a,b,c\rangle $ has no periodic conjugacy classes. This implies that $\Phi(d) = u d^{\pm 1} v$ for words $u,v$ in $a,b,c$. One could take $\Phi(d)=d$ as Ashot suggests but I prefer $u$ and/or $v$ to be nontrivial for reasons explained below. Any such $\Phi$ is a "principal automorphism", which is guaranteed by having a fixed subgroup (when $u$, $v$ are trivial) or by having no fixed subgroup but having three or more attracting periodic points in the boundary of the free group (when $u$ and/or $v$ are nontrivial). Furthermore, if you throw in the condition that the action of $\Phi$ on the $a,b,c$ rose is a train track map and has no periodic Nielsen paths, then $\Phi$ is the ONLY principal automorphism in its outer automorphism class $\phi \in Out(F_n)$ up to conjugation by inner automorphism. -The centralizer $C(\phi)$ in $Out(F_n)$ has to act on the "principal data" of $\phi$, meaning that it has to permute the conjugacy classes of fixed subgroups, and it has to permute orbits of attracting periodic points in the boundary. Furthermore, $\phi$ has an attracting lamination $\Lambda$ that is supported in the $a,b,c$ subgroup, and there is an association between certain attracting periodic points in the boundary and certain rays of the lamination $\Lambda$, and this association implies that $C(\phi)$ preserves $\Lambda$ and so has a well-defined stretch factor homomorphism for $\Lambda$. The gist of the Recognition Theorem in this situation is that an element of $C(\phi)$ is more-or-less determined by all of this "principal data". The situation is a little more complicated to analyze when there is a nontrivial fixed subgroup, so for that reason I actually prefer the opposite case where $u$ and/or $v$ is nontrivial, which for the class of examples under consideration implies that there is no fixed subgroup for any automorphism representing $\phi$ (this is one place where $Out(F_n)$ departs from $MCG(S)$, for if a mapping class is not pseudo-Anosov then after passing to a power there MUST be a principal automorphism having nontrivial fixed subgroup). So in that case, $C(\phi)$ acts on the set of orbits of attracting periodic points, this is a finite set, and so the action has a finite index kernel; the kernel of that action has a stretch factor homomorphism to $Z$, and the kernel of THAT homomorphism is trivial.<|endoftext|> -TITLE: Test for Homogeneity of Symplectic Manifolds? -QUESTION [7 upvotes]: How can one tell whether a given (finite-dimensional) symplectic manifold is homogeneous (that is, admits a transitive group of symplectomorphisms)? -Note that it is of little help to observe that a homogeneous symplectic manifold must be a covering space of a coadjoint orbit of some Lie group, since this provides no effective test applicable to a given example. Note also that in the Riemannian case, there is a well-known local obstruction to the existence of a transitive group action by sutomorphisms: the covariant derivative of the curvature must vanish. But since all symplectic manifolds of dimension n are locally symplectomorphic, there can be no such obstruction in this case, and any analogous constraint must be simultaneously sensitive to the symplectic form and the topology of the manifold. - -REPLY [2 votes]: See -How transitive are the actions of symplectomorphism groups ?<|endoftext|> -TITLE: Infinite Coxeter groups with a non-trivial finite conjugacy class? -QUESTION [7 upvotes]: Let $(W,S)$ be a Coxeter system, where $S$ is finite. Assume that $W$ has an infinite number of elements. -Is it true that conjugacy classes of elements of non-central elements of $S$ have always an infinite number of elements? -Or maybe it is better to ask if there there exists examples of infinite Coxeter groups $(W,S)$ in which the conjugacy class of some non-central element $s\in S$ is finite. -EDIT: Sam Nolen gave me an example in which $(W,S)$ is reducible. So the natural addendum to the original question would be to assume that $(W,S)$ is irreducible. - -REPLY [10 votes]: The conjugacy class of a reflection in an infinite irreducible Coxeter group is always infinite. This follows from a result of mine, which was earlier proved by Kleiner and Pelley in the case of a symmetrizable integer Cartan matrix: - -Let $W$ be an infinite irreducible Coxeter group, with generating set $S = \{s_1, s_2, \cdots, s_n\}$. Then the word $s_1 s_2 \cdots s_n s_1 s_2 \cdots s_n \cdots s_1 s_2 \cdots s_n$, consisting of any number of repetitions of $s_1 s_2 \cdots s_n$, is reduced. - -Now, recall the following criterion for a word to be reduced: Let $x_1$, $x_2$, ..., $x_N$, be any sequence of elements of $S$. - -The word $x_1 x_2 \cdots x_N$ is reduced if and only if the $N$ group elements $x_1$, $x_1 x_2 x_1$, $x_1 x_2 x_3 x_2 x_1$, ... $x_1 x_2 \cdots x_N \cdots x_2 x_1$ are all distinct. - -Write $c$ for $s_1 s_2 \cdots s_n$. Then, combining these two results, we see that $s_1$, $c s_1 c^{-1} = s_1 s_2 \cdots s_n s_1 s_n \cdots s_2 s_1$, $c^2 s_1 c^{-2}$, $c^3 s_1 c^{-3}$, etcetera are all distinct. So the conjugacy class of $s_1$ is infinite. Since the ordering of $S$ was arbitrary, every element of $S$ has infinite conjugacy class. - -I had earlier read the question as asking whether every conjugacy class in an infinite irreducible Coxeter group was infinite. My answer to that remains below. Thanks to Swiat Gal for pointing out the misreading. -This is not true; the class of a translation in an affine group provides a counterexample. -The simplest such counterexample is the group generated by $a$ and $b$ subject to $a^2=b^2=1$. The conjugacy class of $ab$ is $\{ ab, ba \}$, as you can verify by seeing that $a (ab) a^{-1} = b (ab) b^{-1} = ba$ and $a (ba) a^{-1} = b (ba) b^{-1} = ab$. Geometrically, one can think of this as the group of maps $\mathbb{R} \to \mathbb{R}$ of the form $x \mapsto \pm x + k$, for $k \in \mathbb{Z}$. The generators $a$ and $b$ are $x \mapsto -x$ and $x \mapsto -x+1$. -More generally, every affine Coxeter group is of the form $W_0 \ltimes \mathbb{Z}^r$ for some finite group $W_0$. If you take a group element of the form $(e, v)$ with $v$ a nonzero element of $\mathbb{Z}^r$, its conjugacy class will be $\{ (e, gv): g \in W_0 \}$, which is finite. Geometrically, you should think of this as the group of maps $\mathbb{R}^r \to \mathbb{R}^r$ of the form $x \mapsto gx+v$ where $v$ ranges through $\mathbb{Z}^r$ and $g$ ranges through a finite subgroup $W_0$ of $GL_r$. Then the conjugacy class of a translation $x \mapsto x+v$ is finite.<|endoftext|> -TITLE: Logarithmic Integral of n^Zeta Zeroes and certain nested sums of the fractional part function -QUESTION [7 upvotes]: Below is an approach I've been exploring for connecting the prime counting function with the logarithmic integral and expressing the error term between the two. I find it beguiling, but I've largely run out of ideas for further developing this approach. -Hence this question. What I'd really, really appreciate is either 1) references to papers or research covering the general space of ideas I'm exploring here, 2) an explanation for why this approach is going to be a dead end, or especially 3) ideas for manipulating or further exploring my equation (4) below that I might have overlooked. [Edit]I'm specifically interested in techniques I might be unfamiliar with from the study of the divisor summatory function $D_k$, which is at the heart of (4). [End Edit] -Preliminary -Start by summing Linnik's identity from 2 to n. -$\displaystyle\sum_{j=2}^n 1 - \frac{1}{2}\sum_{j=2}^n \sum_{k=2}^{\lfloor\frac{n}{j}\rfloor} 1 + \frac{1}{3}\sum_{j=2}^n \sum_{k=2}^{\lfloor\frac{n}{j}\rfloor}\sum_{l=2}^{\lfloor\frac{n}{j k}\rfloor} 1 - \frac{1}{4}...=$ -$\pi(n) + {\frac{1}{2}}\pi(n^{\frac{1}{2}}) + {\frac{1}{3}}\pi(n^\frac{1}{3})+...$ -(1) -where $\pi(n)$ is the prime counting function. The nested sums on the left only need to be computed up to a depth of $\log_2 n$ before they start equalling 0. -Next, approximate the left hand side by replacing sums with integrals, starting integration at 1 rather than 2, and removing floor functions from upper bounds. This immediately gives the logarithmic integral. -$\displaystyle\int_{1}^n dx - \frac{1}{2}\int_{1}^n \int_{1}^{\frac{n}{x}} dy dx + \frac{1}{3}\int_{1}^n \int_{1}^{\frac{n}{j}}\int_{1}^{\frac{n}{x y}} dz dy dx - \frac{1}{4}... = li(n) - \log\log n - \gamma$ -(2) -where $li(n)$ is the logarithmic integral and $\gamma$ is the Euler-Mascheroni constant. (You can find a fairly straightforward sketch of the derivations of both (1) and (2) here) -Now, Riemann's explicit formula for the prime counting function, with a touch of term rearrangement and with $\rho$ the Zeta Zeroes, is of course -$li(n) - \pi(n) - {\frac{1}{2}}\pi(n^{\frac{1}{2}}) - {\frac{1}{3}}\pi(n^\frac{1}{3})-...=\displaystyle\sum_{\rho} li(n^\rho) + \log 2 - \int_x^{\infty}\frac{dt}{t(t^2-1)\log t}$ -(3) -Question -Define the following $O( \log \log n)$ function -$E(n) = \log 2 - \int_n^{\infty}\frac{dt}{t(t^2-1)\log t} + \log \log n + \gamma$ -Then, applying both (1) and (2) to (3) gives -$\displaystyle \sum_{\rho} li(n^\rho) + E(n) =$ -$\displaystyle - \frac{1}{2}(\int_{1}^n \int_{1}^{\frac{n}{x}} dy dx - \sum_{j=2}^n \sum_{k=2}^{\lfloor\frac{n}{j}\rfloor} 1 )$ -$\displaystyle+ \frac{1}{3}(\int_{1}^n \int_{1}^{\frac{n}{j}}\int_{1}^{\frac{n}{x y}} dz dy dx - \sum_{j=2}^n \sum_{k=2}^{\lfloor\frac{n}{j}\rfloor}\sum_{l=2}^{\lfloor\frac{n}{j k}\rfloor} 1)$ -$\displaystyle- \frac{1}{4}(\int_{1}^n \int_{1}^{\frac{n}{j}}\int_{1}^{\frac{n}{x y}}\int_{1}^{\frac{n}{x y z }} dw dz dy dx - \sum_{j=2}^n \sum_{k=2}^{\lfloor\frac{n}{j}\rfloor}\sum_{l=2}^{\lfloor\frac{n}{j k}\rfloor}\sum_{m=2}^{\lfloor\frac{n}{j k l}\rfloor} 1)$ -$\displaystyle +\frac{1}{5}...$ -(4) -So here is, really, my main question: can anything interesting be done with the right hand side of (4)? I've largely hit a dead-end here. -[Edit] -What I'm looking for especially is any smart ideas about different ways of working with, or breaking up, or transforming those various nested sums, particular in ways that might take advantage of symmetries or connections between them. -As Eric Naslund mentions in his equation (2) below, these nested sums are closely related to, and can be expressed in terms of, the divisor summatory function. As a consequence, this opens up the door to the possibility of applying Voronoi summation or the Dirichlet Hyperbola method in connection with these nested sums, for example. I'm curious if there are any other smart techniques from the study of the divisor summatory function that might be brought to bear on (4). -[End Edit] -Extra -I have explored, a bit, one approach with (4), to inconclusive results. -The following are all equal -$\displaystyle\sum_{j=2}^n \sum_{k=2}^{\lfloor \frac{n}{j} \rfloor} 1 = \sum_{j=2}^n \lfloor \frac{n}{j} \rfloor - 1 = \sum_{j=2}^n \frac{n}{j} - 1 - \sum_{j=2}^n ${$\frac{n}{j}$} -where {n} is the fractional part function. This approach can be generalized to arbitrary depths of nested sums. $\sum_{j=2}^n \frac{n}{j} - 1$ and its generalization appear to be relatively smooth curves, capable of relatively tight approximation, or that's my hunch based on empirical results. Anyway, relying on this, (4) can be rewritten as -$\displaystyle \sum_{\rho} li(n^\rho) + E(n) =$ -$\displaystyle - \frac{1}{2}(\int_{1}^n \int_{1}^{\frac{n}{x}} dy dx - (\sum_{j=2}^n \frac{n}{j} - 1) + \sum_{j=2}^n ${$\frac{n}{j}$}) -$\displaystyle+ \frac{1}{3}(\int_{1}^n \int_{1}^{\frac{n}{j}}\int_{1}^{\frac{n}{x y}} dz dy dx - (\sum_{j=2}^n \sum_{k=2}^{\lfloor\frac{n}{j}\rfloor}\frac{n}{j k} - 1) + \sum_{j=2}^n \sum_{k=2}^{\lfloor\frac{n}{j}\rfloor}${$\frac{n}{j k}$}) -$\displaystyle- \frac{1}{4}(\int_{1}^n \int_{1}^{\frac{n}{j}}\int_{1}^{\frac{n}{x y}}\int_{1}^{\frac{n}{x y z }} dw dz dy dx - (\sum_{j=2}^n \sum_{k=2}^{\lfloor\frac{n}{j}\rfloor}\sum_{l=2}^{\lfloor\frac{n}{j k}\rfloor}\frac{n}{jkl}-1) + \sum_{j=2}^n \sum_{k=2}^{\lfloor\frac{n}{j}\rfloor}\sum_{l=2}^{\lfloor\frac{n}{j k}\rfloor}${$ \frac{n}{jkl} $}) -$ \displaystyle +\frac{1}{5}...$ -(5). -If you sum up just the terms that look like $\sum_{j=2}^n \frac{n}{j} - 1$, you get a smooth curve that bears a strong resemblance to the Logarithmic Integral - you can see part of it here. I'm very curious how to approximate these terms, and what relationship they bear to the logarithmic integral. So that's another question of mine. -Finally, if you sum up just the fractional part sums in (5), you get this function. At least empirically, it appears that all the discontinuities of $\sum_\rho li(n^\rho)$ are confined to just these terms. All of which leads to my final question - are there any interesting tools or approaches that can be applied to these fractional part sums to make any interesting observations? -I hope this question isn't too vague or broad for MO. - -REPLY [4 votes]: Your above identity stems from $$\text{li(x)}-\Pi(x)+\text{small}=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} \log \left((s-1)\zeta(s)\right)\frac{x^s}{s}ds,$$ and a Taylor expansion of the logarithm. What follows is not an exact answer to your question, it is not clear what can or cannot be done with a particular identity. Instead, we give an alternate derivation which lives more in the realm of Dirichlet series and contour integrals, but I think it will shed some light upon the problem. The identity you gave above is essentially an expansion of the logarithm in the classical identity above. -Notation: Let $\Pi(x)=\sum_{n\leq x}\frac{\Lambda(n)}{\log n}$ denote Riemann's Pi function, that is $$\Pi(x)=\pi(n)+\frac{1}{2}\pi\left(n^{\frac{1}{2}}\right)+\frac{1}{3}\pi\left(n^{\frac{1}{3}}\right)+\cdots,$$ and define $$H_{k}(x):=\sum_{j_{1}=2}^{x}\sum_{j_{2}=2}^{\lfloor\frac{x}{j_{1}}\rfloor}\cdots\sum_{j_{k}=2}^{\lfloor\frac{x}{j_{1}\cdots j_{k-1}}\rfloor}1$$ to be the number of integer points under hyperbola with entries greater than $2$, and let $$I_{k}(x)=\int_{1}^{x}\int_{1}^{\frac{x}{y_{1}}}\cdots\int_{1}^{\frac{x}{y_{1}\cdots y_{k-1}}}1\text{d}y_{1}\cdots\text{d}y_{k}$$ be the same area when integrated. -Lets begin by looking at the generalized divisor problem. Using Perrons formula, for any $a>1$ we have that $$D_{k}(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\zeta(s)^{k}\frac{x^{s}}{s}ds,\ \ \ \ \ \ \ \ \ \ (1)$$ where $D_{k}(x)$ is the sum of the generalized divisor function. $D_{k}(n)$ is exactly the number of integer points under the $k$-dimensional hyperbola $x_{1}\cdots x_{k}=n,$ but our sums start at $2$ instead of $1$ so we need to modify things slightly. To get the sum of the integer points whose smallest entry is larger than $2,$ we have to subtract away the $k$ different $(k-1)$-dimensional hyperbolas, and look at $D_{k}(n)-kD_{k-1}(n).$ However this overcompensates, and we have to add back in the $\binom{k}{2}$ different $(k-2)$-dimensional hyperbolas. Continuing in this way, we see that -$$\sum_{j=0}^{k}\binom{k}{j}(-1)^{j}D_{k-j}(n)=H_{k}(n) \ \ \ \ \ \ \ \ \ \ (2)$$ -and hence equation (1) and (2) together imply that $$H_{k}(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\left(\zeta(s)-1\right)^{k}\frac{x^{s}}{s}ds. \ \ \ \ \ \ \ \ \ \ (3)$$ -Similarly, in your note you proved that $$I_k(x)=x\sum_{j=1}^{k}\frac{\left(\log x\right)^{k-j}}{(k-j)!}(-1)^{j},$$ and noticing the similarities in expansions, we may rewrite this as $$I_k(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} \frac{x^s}{s} \frac{1}{(s-1)^k}ds.\ \ \ \ \ \ \ \ \ \(4)$$ -Using (3) and (4), it follows the individual terms are $$\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} \frac{x^s}{s} \left((\zeta(s)-1)^k-\frac{1}{(s-1)^k}\right)ds.$$ Summing over $k$, and using absolute convergence to switch orders, we have that $$\sum_{k=1}^\infty \frac{(-1)^k}{k}\left(H_k(x)-I_k(x)\right)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\sum_{k=1}^\infty \frac{(-1)^k}{k}(\zeta(s)-1)^k \frac{x^s}{s}ds+$$ $$\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\sum_{k=1}^\infty \frac{(-1)^k}{k}(s-1)^{-k} \frac{x^s}{s}ds.$$ -Removing the series expansion for the logarithm, this is -$$=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\left(-\log\zeta(s)\right)\frac{x^{s}}{s}ds+\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}-\log\left(\frac{s-1}{s}\right)\frac{x^{s}}{s}ds.$$ And using Perrons formula once again, since $$-\log\zeta(s)=\sum_{k}\frac{\Lambda(n)}{\log n}n^{-s},$$ we have shown that $$\sum_{k=1}^\infty \frac{(-1)^k}{k}H_k(x)=\Pi(x),$$ and on the other hand the integral $$\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} \log(s-1)\frac{x^s}{s} ds$$ gives rise to the $\text{li}(x)$ term.<|endoftext|> -TITLE: Lagrangian submanifold containing a curve -QUESTION [6 upvotes]: Suppose $(M,\omega)$ is a compact symplectic manifold and $C$ a closed curve in it. Is there a Lagrangian submanifold containing $C$? -I have a sequence of $J_i$-holomorphic maps from a disk to $M$, and all the maps are identical on the boundary of the disk. So, if I know that the image of the boundary is contained in a Lagrangian submanifold, I can apply Gromov convergence. - -REPLY [6 votes]: Yes, such a Lagrangian submanifold does exist. The isotropic neighborhood theorem (see for instance p. 24 of Weinstein's Lectures on Symplectic Manifolds), together with the fact that all symplectic vector bundles over the circle are trivial, implies that any closed curve in a symplectic $2n$-manifold has a neighborhood symplectomorphic to the standard product $(S^1\times(-\epsilon,\epsilon))\times B^{2n-2}(r)$, via a symplectomorphism mapping the curve to $S^1\times \{0\}\times\{0\}$. Here $B^{2n-2}(r)$ denotes the standard symplectic ball of dimension $2n-2$ and some small radius $r$. Now let $T$ be a Lagrangian torus in $B^{2n-2}(r)$ which contains the origin (for instance you could use a product of small-radius-cirles containing the origin in each of the $n-1$ factors of $\mathbb{R}^2$ in $\mathbb{R}^{2n-2}$). Then the image of $S^1\times 0\times T$ under the Weinstein symplectomorphism will be a Lagrangian submanifold containing your curve.<|endoftext|> -TITLE: R-matrices, crystal bases, and the limit as q -> 1 -QUESTION [11 upvotes]: I am seeking references for precise statements and rigorous proofs of some facts about the actions of quantum root vectors and $R$-matrices on crystal bases for finite-dimensional representations of quantum groups. I am very new to crystal bases, so I would also appreciate corrections if my questions are not well-formulated. I am putting the questions first, followed by the motivation for those who are curious. -Questions -Do you know references (with proofs) for the following statements: - -Let $V$ be a finite-dimensional $U_q(\mathfrak{g})$-module. Then the divided powers $E_{\beta}^{(t)}, F_\beta^{(t)}$ of the quantum root vectors have matrix coefficients given by Laurent polynomials in $q$, with respect to the global crystal basis for $V$. -Let $V,W$ be finite-dimensional $U_q(\mathfrak{g})$-modules. Then the matrix coefficients of the $R$-matrix $R_{V,W}$ are Laurent polynomials in $q$, with respect to the tensor product of the global crystal bases for $V,W$. - -I believe I have proofs for these statements, but it would be nice to just reference something definitive instead of writing the proofs out myself. -Background -Let $U_q(\mathfrak{g})$ be the quantized enveloping algebra of $\mathfrak{g}$ for $q$ not a root of unity, with generators $E_i,F_i,K_i$ corresponding to the simple roots of $\mathfrak{g}$. Using an action of the braid group of $\mathfrak{g}$ on $U_q(\mathfrak{g})$ one can define quantum root vectors $E_\beta,F_\beta$ for all positive roots $\beta$. (This depends on a choice of decomposition of the longest word of the Weyl group, so assume that we have fixed such a decomposition.) -Let $R_{U,V}$ be the action of the $R$-matrix on $U \otimes V$, (as in Chari-Pressley or Klimyk-Schmudgen, say) so $\tau \circ R_{U,V}$ is the braiding. -I would like to make sense of the statement that $R_{V,W} \to \mathrm{id}_{V \otimes W}$ as $q \to 1$ (and hence the braiding tends to the flip as $q \to 1$). This is not trivial because for different $q$'s, the operators $R_{V,W}$ are really operators on different vector spaces. This is where the crystal bases come in. -As I understand it, a crystal basis for a module has the property that the matrix coefficients of the generators $E_i,F_i$ of $U_q(\mathfrak{g})$ (and divided powers of the generators) are given by universal Laurent polynomials in $q$ whose coefficients are independent of $q$. Using this basis we can think of all of the algebras $U_q(\mathfrak{g})$ for various $q$'s acting on the same vector space. The point is that Laurent polynomials are continuous and well-defined at $q=1$, i.e. they are specializable to $q=1$. -Taking the tensor product of the crystal bases for $V$ and $W$, we can think of all of the $R$-matrices for various $q$'s acting on the same space as well, and it makes sense to ask if this family of $R$-matrices is continuous in $q$, and if so, whether it can be extended to $q=1$. -The formula for the action of the $R$-matrix is a big sum of products of operators of the form -$$ \frac{1}{[t]_{q_\beta}!} E_\beta^t \otimes F_\beta^t$$ -with coefficients given by Laurent polynomials in $q$. Putting the $q$-factorial under, say, the $E_\beta^t$ term gives the divided power $E_\beta^{(t)}$. If the quantum root vectors and their divided powers act by Laurent polynomials, then the $R$-matrix does as well, and hence everything in sight is continuous in $q$, can be specialized to $q=1$, and it is clear that at $q=1$ the $R$-matrix is just the identity. - -REPLY [4 votes]: I never found a precise reference for the statement about the R-matrix, so I ended up writing it up myself. The precise statements and proofs can be found in $\S 4.1$ of my paper with Alex Chirvasitu, Remarks on quantum symmetric algebras, available here.<|endoftext|> -TITLE: Spherical building of an exceptional group of Lie type -QUESTION [10 upvotes]: I've read that one of Tits' original motivations for studying buildings was that he wanted to give a unified description of algebraic groups that would allow the definition of exceptional groups such as $E_6$, $E_7$, $E_8$, $F_4$ and $G_2$ over any field. -Now, Chevalley managed to construct the corresponding groups of Lie type over finite fields; but is there a way one can instead first construct the buildings and then define the group as a particular subgroup of the automorphism group of the building? -What do these buildings look like? The building for $\mathrm{SL}_3(\mathbb{F}_2)$, for instance, is the Heawood graph (I can't embed images yet); one can count the 28 apartments making up this complex, which are just hexagons with alternating colors. This graph is the incidence graph for the Fano plane $ \mathbb{P}^2/\mathbb{F}_2 $. One recovers the full group $ \mathrm{SL}_2(\mathbb{F}_2) $ as the group of type-preserving automorphisms of this graph (preserving the distinction points vs lines of the geometry, i.e. the colours in the above graph). -One also recovers the corresponding BN-pair: pick any chamber (any edge) and any apartment containing it (a hexagon with alternating colours); then B consists of the subgroup of automorphisms that fix the chosen chamber, and N consists of the subgroup of automorphism preserving the chosen apartment. Obviously, one could use Chevalley's construction to get a BN-pair and then get back the corresponding building, but I'm really looking for something that starts with the building and gets back the corresponding group. -Are there similar descriptions for the exceptional groups and their corresponding buildings? -(I'm aware that such neat pictorial representations aren't really going to be possible, given that already $G_2(3)$ has order of several million.) -(Crossposted from Math Stackexchange.) - -REPLY [8 votes]: In the case of groups of rank 2, such as your examples $\mathrm{SL}_3(\mathbb{F}_2)$ or $\mathsf{G}_2(3)$, the building is rather easy to describe (either as an incidence geometry or as a bipartite graph); it is a so-called generalized polygon. -Spherical buildings of higher rank are completely determined by their rank $2$ residues, but perhaps that point of view is not explicit enough for your purposes. There exist several more "geometric" characterizations of the exceptional buildings in the literature, notably by people like Arjeh Cohen and Bruce Cooperstein; see for instance Cohen and Cooperstein - A characterisation of some geometries of Lie type. -The split buildings of type $\mathsf{G}_2$, $\mathsf{F}_4$ and $\mathsf{E}_6$ can also be described in terms of algebraic structures (but then the corresponding groups can also more easily be described in terms of this structure). For instance, the split groups of type $\mathsf{F}_4$ are precisely the automorphism groups of split Albert algebras (these are certain 27-dimensional non-associative algebras); the corresponding building can be described in terms of isotropic subspaces of this Albert algebra. A similar description is possible for $\mathsf{E}_6$, but is much more difficult for $\mathsf{E}_7$ and to the best of my knowledge unknown for $\mathsf{E}_8$.<|endoftext|> -TITLE: Is there a known method for finding the minimum bridge index of a knot? -QUESTION [5 upvotes]: It is easy to establish an upper bound $n$ for the bridge index of a knot by producing a diagram with the knot in $n$-bridge position. - -Is there a known method to produce a reasonable lower bound on the bridge index? - -For example, knot 11a1 has at most bridge index $4$, because a $4$-bridge position of this knot can be drawn: DT code (42, 66, 44, 56, 54, 46, 64, 40, -38, 86, 94, 76, 100, 80, 82, 98, 74, 96, 84, -110, -108, -68, -104, -22, -34, -32, -24, -102, -26, -30, -36, -20, -106, 90, 122, 60, 50, -116, -10, -6, -120, -70, -112, -14, -2, -4, -12, -114, -72, -118, -8, 78, 92, 88, 16, 18, 62, 48, 52, 58, -28) gives one such diagram. What is known about methods to eliminate the possibility that this knot has a $2$-bridge or $3$-bridge projection? -Update -After posting this question I found a $3$-bridge projection (and because the knot is not rational, this is the lowest possible projection), given by DT code (12, 16, 58, 60, 14, -92, -90, -94, -32, -40, 120, 102, 108, 112, 98,116, 124, 106, 104, 122, 118, 100, 110, -26, -34, -38, -22, -42,-30, -96, -28, -44, -24, -36, 50, 18, 56, 62, 46, 64, 54, 20, 52, 66, 48, 128, 126, 114, -6, -74, -80, -86, -68, -88, -78, -76, -8, -4, -72, -82, -84, -70, -2, -10), but I am still interested in what is already known about establishing lower bounds more generally. Ryan Budney's suggestion of the Heegaard genus is a good one, but I haven't found a reference that shows this bound to be sharp. - -REPLY [9 votes]: This is really a comment on Ryan's answer, but I don't have enough rep. Here are two relevant papers: -http://arxiv.org/abs/0710.1262 (Algorithmically Detecting the bridge number for hyperbolic knots, by Alex Coward) -and -http://arxiv.org/abs/0709.3534 (Robin Wilson's Meridional Almost Normal Surfaces in Knot Complements, Algebr. Geom. Topol. 8 (2008) 1717-1740, https://doi.org/10.2140/agt.2008.8.1717) -I know of some people who are working on other ways to get lower bounds for bridge number in terms of other invariants.<|endoftext|> -TITLE: Cohomology of a space with local coefficients and singular cohomological dimension -QUESTION [5 upvotes]: If $X$ is a space with (singular) cohomological dimension n, i.e. $H^i(X;\mathbb{Z})=0$ for all $i>n$, (may be cohomology with rational coefficients). -Is it true that $H^i(X;\mathcal{L})=0$ for all $i>n$, where $\mathcal{L}$ is a system of local coefficients on $X$? -Can someone give some reference? - -REPLY [10 votes]: The answer is no. It's easiest to give examples arising from group cohomology (so the spaces are $K(\pi,1)$'s). The reason for this is that if $G$ is a nontrivial group, then there must exist some local coefficient system $M$ such that $H^i(G;M) \neq 0$ for some $i$. Indeed, let $H < G$ be a nontrivial cyclic subgroup, and let $M$ be the coinduction of the trivial $\mathbb{Z}$-module from $H$ to $G$. Then Shapiro's lemma (see here) says that -$$H^i(G;M) = H^i(H;\mathbb{Z}),$$ -which has to be nonzero for some $i$. -Dramatic examples of groups with the property you seek are so-called "acyclic groups". These are groups $G$ such that $H^i(G;\mathbb{Z})=0$ for all $i \geq 1$. For a nice survey of such groups, see the last section of -MR1967745 (2004c:20001) -Berrick, A. J.(SGP-SING) -A topologist's view of perfect and acyclic groups. Invitations to geometry and topology, 1–28, -Oxf. Grad. Texts Math., 7, Oxford Univ. Press, Oxford, 2002. - -REPLY [3 votes]: How about $X=\Bbb RP^2\times L^2_3$, where $L^2_3$ is the cone of the $3$-fold cover $S^1\to S^1$. Here $H^4(X;\Bbb Z)\simeq\Bbb Z/2\otimes\Bbb Z/3=0$, but $H^4(X;\mathcal L)\simeq\Bbb Z\otimes\Bbb Z/3$, where $\mathcal L$ is the pullback of the orientation sheaf of $\Bbb RP^2$. -As a side remark, "singular cohomological dimension" is not something that people normally do, perhaps because singular cohomology is not Brown representable, or because of the Barratt-Milnor example of a compact subset of $\Bbb R^3$ (in fact it is just the one-point compactification of $\Bbb R^2\times\Bbb Z$) which has infinite "singular cohomological dimension". If you do care about spaces not homotopy equivalent to CW-complexes, you can look up some books on traditional dimension theory, which deal with usual (that is, Cech) cohomological dimension. For spaces homotopy equivalent to CW-complexes, there's no distinction because all ordinary cohomology theories coincide.<|endoftext|> -TITLE: Algorithmic war -QUESTION [5 upvotes]: No, not the war on drugs, but the game of War considered in -Does War have infinite expected length? -As noted in that discussion, the game of war can go on forever, but my question is: can it be decided in polynomial time whether a given configuration leads to a periodic or a finite game (the question is decidable, since you can just look at a sequence of $n n!$ moves (where $n$ is the initial number of cards), but $n n!$ is not so small.) -EDIT To answer Joel's very good question: the setup is: the two adversaries have decks $A$ and $B,$ both face down. they flip their top cards, call them $a_1$ and $b_1.$ If $v(a_1) > v(b_1)$ ($v()$ is the value), we put $a_1$ on top of $b_1,$ turn the stack of two cards upside down and add them to the bottom of $A$'s deck (and similarly if $v(b_1) > v(b_k).$ If $v(a_1) = v(b_1)$ we flip two more cards $a_2, b_2$ put them on top of $a_1, b_1$ respectively. If $v(a_2) > v(b_2)$ we put the stack of $a$s on top of the stack of $b$s, flip the resulting $4$-stack upside down, and add it to the bottom of the $A$ stack. If $v(a_2) = v(b_2)$ we continue as before. If we keep getting equal values, and one of the players runs out of cards, the other player wins. If both players run out of cards simultaneously, the game is declared a draw. - -REPLY [2 votes]: Here is an observation which might be put to use by someone more clever than I. -The obvious fact is that if player A has all the cards of top value greater than the n+1st card, then during the play A keeps those n cards in Igor Rivin's version of the game. So when does A acquire all cards having the same value as the (n+1)th card? For many games, it is quite likely that A will acquire those cards, so the question now becomes: for which patterns is B able to keep his top value card indefinitely. This version should be should be tractable, although I don't know enough yet to say that it is decidable quickly. Probabilistically, I would say almost never. -Gerhard "Ask Me About Random Guessing" Paseman, 2011.12.08<|endoftext|> -TITLE: A categorical characterization of the lexicographic order -QUESTION [9 upvotes]: In $Pos$ (the category of partial ordered sets and order preserving maps) there is the categorical product of two objects, but on the set product there is (naturally) also the lexicographic order. I ask: -has this latter some kind of categorical universal property? Or a categorical (external) characterization? - -REPLY [3 votes]: As David Roberts said, I post the solution for preorders: -for two preorders $(X,\leq),\ (Y,\leq)$, let $(X,\leq)\times_l(Y,\leq)$ the lexicographic preorder (priority to the first factor $(X,\leq)$). -We observe that a preorder morphism $(A,\leq)\to (X,\leq)\times_l (Y,\leq)$ is a set map $(f, g): A\to X\times Y$ such that $f: (A,\leq)\to (X,\leq)$ is a preorder morphism, and $g: A\to Y$ a set map (we can view it as a morphism $g: (A,\leq)\to ch(Y)$ where $ch(Y)$ is the chaotic preorder on the set $Y$ (dont exists the functorial analogy for orders)) and If $a\leq a'$ and $f(a)=f(a')$ then we have $g(a)\leq g(a')$. -Reciprocally if we have the data maked as: a couple of morphisms like $f: (A,\leq)\to (X,\leq)$, $g: (A,\leq)\to ch(Y)$ such that "If $a\leq a'$ and $f(a)=f(a')$ then we have $g(a)\leq g(a')$", then we get the (unique) morphisms $(f, g): (A,\leq)\to (X,\leq)\times_l (Y,\leq)$. -Let $U: Preord\to Set$ the forgetfull functor from the preorders to sets category, this has a right adjoint funtor $Ch: Set\to Preord$ (the chaotic preorder), now the phrase "$a\leq a'$ and $f(a)=f(a')$" is described by the categorical relation $r_l: R_A(\leq)\cap P\subset A\times A$ where $R_A(\leq)$ is the preorder relation on $A$ and $P$ the pullback of $f\times f$ by the diagonal morphism $\Delta_X\subset X\times X$. Then $X\times_lY$ is universal (initial) for the objects $A$ with two morphisms $f: A\to X$, $g: A\to Ch(U(Y,\leq))$ such that $(g\times g)\circ r_l: R_A(\leq)\cap P\to Y\times Y $ as a (unique) factorization to the preorder relation $R_Y(\leq)\subset Y\times Y$ of $(Y,\leq)$. -now I realized it was just an easy exercise in translation of categorical logic, that is from the the formula to categorical diagrams.<|endoftext|> -TITLE: Are higher categories useful? -QUESTION [54 upvotes]: Of course, personally, I think the answer is a big Yes! -However once, a while ago, while giving a talk about higher category theory, I was asked a question about whether higher category theory was useful outside of the realm of higher category theory itself. I was asked if there was anything that can be proven using higher category theory that couldn't be proven without it? -I think it is a somewhat common to experience this sort of resistance to higher categories, and I think this is a fair question, at least if you also allow for insights gleaned from a higher categorical perspective that would not have been possible or obvious otherwise. -I have a small handful of answers for this question, but I certainly don't feel like I know most of the applications, nor the best. I thought it could be useful to compile a big list of applications of higher category theory to other disciplines of mathematics. - -Question: What are useful applications of higher categories outside the realm of higher category theory itself? Are there any results where higher categories or the higher categorical perspectives play an essential role? - -Here I want "higher category" to be interpreted liberally, including various notions of n-category or $(\infty,n)$-category. I am not picky. -I also want to interpret "essential" to just mean that it would be hard to imagine getting the results or insights without the use of higher categories, not in some precise mathematical sense. But, for example, saying "homotopy theory is just an example of the theory of $(\infty,1)$-categories" doesn't really count. -The usual big-list rules apply: This is community wiki, and please just one application per answer. - -REPLY [12 votes]: All examples so far involve only invertible k-morphisms for k≥3, -i.e., they can be described as (∞-)bicategories. -I would like to give an example of an interesting tricategory (with non-invertible 3-morphisms): -The tricategory of conformal nets, as constructed by Bartels, Douglas, and Henriques. -This tricategory is used by Douglas and Henriques to give an algebraic description of string structures -in the same vein as the existing algebraic descriptions of spin structures and orientations.<|endoftext|> -TITLE: A characterization of the blow-up -QUESTION [7 upvotes]: It is well-known that there is a universal characterization of blow-ups. However, this one is not so easy to use. According to Hartshorne-Theorem II.8.24. The blow-up of a nonsingular $X$ along a nonsingular subvariety $Y$ have three properties: -1. The blow-up $\tilde{X}$ is nonsingular. -2. The blow-up restricts to the exceptional divisor $\tilde{Y}$ is the projective bundle $P$ of the normal bundle of $Y$ in $X$. -3. The normal sheaf $N_{\tilde{Y}/\tilde{X}}$ corresponds to the $P(-1)$. -My question is that assume we have a projective birational morphism $f: \tilde{X}\to X$ with exceptional locus $Y$ on $X$. We assume both $X$ and $Y$ are nonsingular as before (even projective). Let $\tilde{Y}$ be the inverse image sheaf of $Y$. If $f$ satisfies the above three properties, must it be the blow-up of $X$ along $Y$? Especially, I want to see an example for a non-blowup having property 1,2. Thank you so much. - -REPLY [8 votes]: Example (property 1 fails, but property 2 is satisfied) -Look for $f$ as the blow up of an ideal sheaf $\mathscr I$, so $\widetilde X=\mathrm{Proj}_X(\oplus_d \mathscr I^d)$. Then the pre-image of the subscheme $Z\subset X$ defined by the ideal $\mathscr I$ is given by $\widetilde Z=\mathrm{Proj}_Y(\oplus_d \mathscr{I^d/I^{d+1}})$. Now if $X$ is Cohen-Macaulay and $Z$ is a complete intersection in $X$, (i.e., $\mathscr I$ is generated by a regular sequence), then $\mathscr{I/I^2}$ is locally free and -$\mathscr{I^d/I^{d+1}}\simeq \mathrm{Sym}^d(\mathscr{I/I^2})$ and hence $\widetilde Z\simeq \mathbb P(\mathscr{I/I^2})$. -Property #3 is kind of a red herring. The $(-1)$-twist is almost automatic, it comes from the construction of the blow up of $\mathscr I$. -Finally, here is a simple concrete example: -Let $X$ be a plane (or any smooth surface) and $\mathscr I=(x^2,y^2)$ where $x,y$ are local coordinates at a point. The blow up will be the surface with a pinch point (locally around the interesting singularity defined by $x^2z=y^2$) with the singular line contracted to a point. I think it is relatively easy to check that this satisfies properties #2 and #3. - -To round things up Mike Roth in the comments below gives a nice example of a blow up along a non-smooth subvariety such that the resulting variety is actually smooth.<|endoftext|> -TITLE: Algorithmic decidability of equality in the ring of periods -QUESTION [8 upvotes]: Suppose two elements of the ring of periods are given by their systems of polynomial inequalities with rational coefficients. Is there a known algorithm deciding their equality? Is it known if their equality is algorithmically decidable at all? -It is quite obvious that if they are not equal, then it can be shown after finite number of steps just by evaluating them with big enough precision. The difficult part is when they are equal. - -REPLY [7 votes]: There is no known algorithm for that, and it is unknown if such an algorithm exists.<|endoftext|> -TITLE: Example of Möbius inversion on integer partition poset -QUESTION [5 upvotes]: We know that integer partitions form a poset (actually we can define more than one partial orders on it), and so we can have some kind of Möbius function on it and consequently Möbius Inversion Formula (correct me if I am wrong). -I am looking for some examples of applications of such kind of formula if it exists, and whether we can view Hook length formula as a manifestation of it or not. Either you can give an example or point out some literature to look into... - -REPLY [13 votes]: The Möbius function of the poset $P_n$ of partitions of the integer $n$ ordered by refinement is not well-behaved, as is discussed e.g. in: -Günter Ziegler, On the poset of partitions of an integer, J. Combin. Theory Ser. A 42 (1986), no. 2, 215--222. -where the poset is shown not to be Cohen-Macaulay for $n\ge 19$ and not to have Möbius function on intervals that alternates in sign for $n\ge 111$. Thus, its order complex (nerve) on intervals is not always homotopy equivalent to a wedge of top dimensional spheres. Wanting nonetheless to understand the topology of the order complex for this poset and more general posets of multiset partitions was the original motivation for my work with Eric Babson on discrete Morse theory for posets, but we only obtained partial results in this direction, focusing mainly on $\mu_{P_n} (\hat{0},\hat{1})$ rather than arbitrary intervals. -If you are (more likely) interested in partially ordering the number partitions by shape containment rather than by refinement, then this is Young's lattice, and it is indeed well-behaved, as described e.g. at the wikipedia article on "Young's lattice". Young's lattice is a distributive lattice, hence each interval is shellable, and each interval has Möbius function equaling $0, 1, $ or $-1$. Specifically, it is 0 except for $\mu (\lambda_1,\lambda_2 )$ where the skew shape $\lambda_2 \setminus \lambda_1$ consists of single boxes touching at most at corners. -As far as applications, I would be looking in the theory of symmetric functions and in representation theory. There seems to be some discussion of this and references in the wikipedia article on "Young's lattice".<|endoftext|> -TITLE: what does BG classify? i.e. what is a principal fibration? -QUESTION [14 upvotes]: I'm looking for cold hard facts about just what $BG$ classifies, if $G$ is any grouplike topological monoid. I have some vague idea that $[X,BG]$ is in bijection with equivalence classes of "principal fibrations" over $X$. What exactly is a principal fibration? -May's Classifying Spaces and Fibrations looks like a good source, but I'm having trouble teasing out whether his notion of $G\mathcal U$-fibration (which he proves is classified by $BG$) is equivalent to the notion of a fibration $E \rightarrow B$ with a fiberwise right $G$-action giving weak equivalences $G \rightarrow E_b$, $g \mapsto xg$ for each point $x$ in the fiber $E_b$. (I can show the $\Rightarrow$ but not the $\Leftarrow$. Maybe there needs to be another condition in my naive description.) -Also, any grouplike monoid $G$ is weakly equivalent to $\Omega BG$, so "principal fibrations," whatever they are, should correspond (in some sense I want to make precise) to pullbacks of the path-loop fibration over $BG$. - -REPLY [13 votes]: In view of the references to my Memoir, Classifying spaces and fibrations, -in other answers, I guess I should answer too. The requested answer is -implicit but not quite explicit there. Fix a grouplike topological monoid -$G$. Maybe assume for simplicity that its identity element is a nondegenerate -basepoint (no loss of generality by 9.3). Define a $G$-torsor to be -a right $G$-space $X$ such that the $G$-map $G\longrightarrow X$ that sends -$g$ to $xg$ is a weak equivalence for $x\in X$. Let $\mathcal{G}$ be the category of -$G$-torsors and maps of $G$-spaces between them. The Memoir defines a -$\mathcal{G}$-fibration in terms of the $\mathcal{G}$-CHP, which is equivalent to having a $\mathcal{G}$-lifting function. Cary asks whether that notion is equivalent to an -a priori weaker notion. The answer depends on what ``equivalence'' means. -The Memoir insists on $\mathcal{G}$-fibrations, but it defines an equivalence (6.1) -to be a $\mathcal{G}$-map over the base space, where a $\mathcal{G}$-map only has to be a map of $G$-torsors on fibers. (More precisely, it takes the equivalence relation generated by such maps.) Using $\mathcal{G}$-fibrations allows one often to replace -that notion of equivalence by the nicer one of $\mathcal{G}$-fiber homotopy equivalence. But with the -equivalence relation as given, any $\mathcal{G}$-map that is a quasifibration is equivalent to a $\mathcal{G}$-fibration, by the $\Gamma$-construction in Section 5. -Therefore the classification theorem remains true allowing all $\mathcal{G}$-spaces that are quasifibrations, which of course includes Cary's preferred notion of a -$G$-fibration.<|endoftext|> -TITLE: How to do integrals involving two Bessel functions and another function? -QUESTION [9 upvotes]: I often encounter the integrals in the following form: -$\int_0^\infty{\rm Bessel}(ax)\cdot{\rm Bessel}(bx)\cdot f(cx)dx$, -where Bessel can be $J$, $N$, $H^{(1)}$, $H^{(2)}$, $I$, or $K$; and $f(x)$ can be $\sin(x)$, $e^x$, etc. For example, -$\int_0^\infty K_\nu(ax)I_\nu(bx)\cos(cx)dx=\frac{1}{2\sqrt{ab}}Q_{\nu-1/2}(\frac{a^2+b^2+c^2}{2ab})$,$\qquad{\rm Re}(a)>|{\rm Re}(b)|$, $c>0$, ${\rm Re}(\nu)>-1/2$ -I already know the result of this integral because it is in Gradshtein & Ryzhik's book. Sometimes the integration is with respect to the order of the Bessel function. -Both Mathematica 8 and Maple 15 cannot do this kind of integrals. When the integral involves two Bessel functions or two other special functions, Mathematica and Maple usually cannot do even if the integral has a closed-form result. My questions are as follows: - -Is there any general theory about how to do this kind of integrals? (I wonder how the authors of that book did the above integral.) -I know there is a Mathematica package "HolonomicFunctions." It seems that this package can help, but it does not seem very straightforward to obtain the final result. This package can verify the integrals with already known results, but can it do new integrals as above? Are there any better ways to deal with these integrals by computer? - -REPLY [5 votes]: Take a look at "A Treatise on the Theory of Bessel Functions" by Watson. There is a long chapter on integrating Bessel functions over the infinite range $0-\infty$. -In addition, I think a Mellin transform approach could very well get you what you want. The idea is to find Mellin transforms of the functions in the integrand (most often separated into two parts) and use Parseval's theorem to write the integral as a contour integral. Then one can often move the contour over the poles of the integrand and generate a series representation of the integral, which can sometimes be identified as some known special function. -You could look at the paper http://www.risc.jku.at/publications/download/risc_3924/AlgorithmicMellinTransform.pdf which briefly describes the method and shows a computer algebra technique for getting the final result. There is also a great, simple book by Fikoris called "Mellin Transform Method for Integral Evaluation." -Good luck, -Tom<|endoftext|> -TITLE: Largest possible volume of the convex hull of a curve of unit length -QUESTION [26 upvotes]: What is the largest possible volume of the convex hull of an open/closed curve of unit length in $\mathbb{R}^3$? - -REPLY [7 votes]: Here is an image of the optimal open convex curve. -Taken from Open Problems from CCCG 2012, -based on this paper, which cites Nudel'man (1975): - -Paolo Tilli. - "Isoperimetric inequalities for convex hulls and related questions." - Trans. Amer. Math. Soc. 362 (2010), 4497-4509.<|endoftext|> -TITLE: What is Ricardo Pérez-Marco's eñe product? Does it explain his statistical results on differences of zeta zeros? -QUESTION [29 upvotes]: The number theory community here at University of Michigan is abuzz with talk of this paper recently posted to the arxiv. If you haven't seen it already, the punch line is that the global differences of imaginary parts of zeta zeros show an unusual tendency not to be close to imaginary parts of zeta zeros. "Riemann zeros repel their deltas." -The people I have talked to have said that this simple observation appears to be completely new and potentially groundbreaking. Here are my questions: -1) Has anyone independently verified these statistics? (See my histograms below, and also the excellent graphs in Jonathan Bober's answer.) -2) The author discusses the "eñe product" as a source of a mathematical explanation for his observation, and he refers to his own unpublished manuscripts in which he introduces this product. Are these manuscripts available? Does anyone have any reference which defines the eñe product? -Finally, any insights into Marco's results would be much appreciated. -The histograms (produced in Mathematica) are of the deltas of the first 10000 zeros of $\zeta(s)$ falling in $[100, 110]$ and $[190, 200]$, respectively. The purple bars show the locations of the zeros. - -REPLY [12 votes]: My student, Brad Rodgers, has just posted a paper on the arXiv at http://arxiv.org/abs/1203.3275 which proves a partial result towards the repulsion effect (that differences of two imaginary parts of Riemann zeroes tend to avoid another Riemann zero), in the spirit of Montgomery's partial result towards his pair correlation conjecture (i.e. this repulsion effect can be detected when tested against sufficiently band-limited test functions, assuming RH). -Ultimately, the reason for this repulsion lies in the obvious approximate formula -$$ |\Lambda(n)|^2 \approx \Lambda(n) \log n$$ -where $\Lambda$ is the von Mangoldt function. If one compares this with the explicit formula, which is formally of the form -$$ \Lambda(n) = 1 - \sum_\rho n^{\rho-1} + \ldots$$ -one begins to see the negative correlation between differences of imaginary parts of zeroes $\rho$ (which show up in the expansion of $|\Lambda|^2$) and in the imaginary parts of zeroes themselves. (Making this intuition rigorous, though, is somewhat non-trivial, requiring manipulations similar to those in Montgomery's original paper to deal with the fact that the explicit formula as given above is only convergent in a very weak sense.) -EDIT: it is likely that a similar analysis would also explain why Riemann zeroes correlate with differences of zeroes of other functions. For instance, starting from $|\Lambda(n) \chi(n)|^2 \approx \Lambda(n) \log n$, one can predict that differences of imaginary parts of zeroes of a Dirichlet L-function should repel away from the imaginary part of zeroes of zeta.<|endoftext|> -TITLE: On the Paley-Wiener theorem -QUESTION [7 upvotes]: Is there any even Schwartz function whose restriction to $[0,\infty)$ is monotone and whose Fourier transform is compactly supported? In other words, is there any entire analytic function satisfying the condition of the Paley-Wiener theorem that is even on the real line and whose restriction to $[0,\infty)$ is monotone? - -REPLY [5 votes]: The answer is yes : -Let $h$ be an even real valued Schwartz function whose Fourier -transform has compact support. Then choose $f(y) = \int_{-\infty}^y -x h(x)^{2} dx$ .<|endoftext|> -TITLE: What is the probability for a random algebraic cycle to be homologically trivial? -QUESTION [6 upvotes]: Someone recently asked me about the Hodge conjecture. As I understand it the conjecture asserts the existence of many non trivial algebraic cycles. The difficulty comes from the fact that we don't have a process on how to define "interesting" algebraic cycles. -So I asked myself a very naive question. What is the probability for a random algebraic cycle to be homologically trivial? -If $X\subset \mathbb{P}^n$ be a smooth variety, we can chose random homogenous polynomials $f_1,\ldots, f_r$ and consider an algebraic sub-variety $Z = \{x \in X ~|~ f_1 = \ldots = f_r = 0\}$. Is there anything interesting we can say about the random cohomology class $cl(Z)$ (after imposing some conditions obviously)? Is there some good reference about probabilistic treatment of algebraic cycles? If not, is such an approach considered as doomed for fail and for which reasons? Difficulty to define a good parameter space and non trivial probability measure on it comes to mind. - -REPLY [3 votes]: If $X \subset \mathbb{P}^N$ is a projective variety and $Z$ is a nonempty effective algebraic cycle, then the cohomology class of $Z$ is always nonzero. Proof: Let $\dim X = n$ and let $\dim Z = k$. Then, by Bertini, a generic $\mathbb{P}^{N-k}$ in $\mathbb{P}^N$ meets $Z$ in a finite number of points. Let $H$ be such a $\mathbb{P}^{N-k}$ and let $W = X \cap H$. Then the classes of $W$ and $Z$ have nontrivial cup product in $H^{\ast}(X)$. The key point here is that, when two algebraic varieties of complementary dimensions meet in a set of dimension zero, that intersection always has positive multiplicity. -So nothing like the construction you suggest in your final paragraph can give trivial classes -- you need to either work on a non-projective variety or allow negative terms in your cycles.<|endoftext|> -TITLE: Exact sequence of monoids -QUESTION [15 upvotes]: What is the right definition of an exact sequence of monoid homomorphisms? -I can't seem to find a consistent in my searches; indeed Balmer (Remark 2.6, - http://www.math.ucla.edu/~balmer/Pubfile/Prod.pdf ) -calls it a "slippery notion". -Among other things, I'll know that it's a good definition if I can take an exact sequence of groups and forget inverses and obtain an exact sequence of monoids. -One possible answer is just to take the definition for groups: a sequence -$$A \xrightarrow{f} B \xrightarrow{g} C $$ -of monoid homomorphisms is exact if $\mathrm{ker}\ g = \mathrm{img}\ f$, i.e., -$$ \{ y \in B : g(y) = 1 \} = \{ f(x) : x \in A \}. $$ -But, according to Bergman (3.10.6--7, http://math.berkeley.edu/~gbergman/245/Ch.3.pdf), -this definition of kernel does not determine the image. Instead, we look at the \emph{kernel congruence} -$$ K_g = \{(y,z) \in B \times B : g(y)=g(z)\}. $$ -The set $K_f$ defines a \emph{congruence} on $A$, an equivalence relation compatible with the operation on $A$, i.e., if $(y,z),(y',z') \in K_g$ then $(yy',zz') \in K_g$. -So then how should we define the image $I_f$ so that our sequence is exact if and only if $I_f=K_g$. One proposal is given by Arturo Magidin (https://math.stackexchange.com/questions/18387/is-there-an-analogue-of-short-exact-sequences-for-semigroups): he defines -$$ I_f = (f(A) \times f(A)) \cup \Delta(B) $$ -where $\Delta(B)=\{(b,b) : b \in B\}$. -This notion, though probably useful in some ways, has a grave defect. Let -$$ A \xrightarrow{f} B \xrightarrow{g} C, $$ -be an exact sequence of groups, so -$$ \mathrm{ker} g = \{y \in B : g(y) = 1\} = \mathrm{img} f = \{f(x) : x \in A\}. $$ -I would like to verify that the sequence is exact as a sequence of monoids, so $I_f=K_g$. We have $(y,z) \in K_g$ if and only if $g(yz^{-1}) = 1$ since g is a group homomorphism, if and only if $yz^{-1} \in K = \mathrm{ker} g$ if and only if $y = kz$ for some $k \in K$. Thus -$$ K_g = \{(kz,z) \in B \times B : k \in K, z \in B \}. $$ -On the other hand, suppose g is not injective and that f is not surjective (not an -atypical situation). Then there exists $z \in B$ such that $z \not\in f(A)$ -and there exists $k \in K$ such that $k \neq 1$. Then $(kz,z) \in K_g$ by -definition, but $(kz,z) \not\in I_f$: it does not belong to $f(A) \times f(A)$ -since $z \not \in f(A)$ and does not belong to $\Delta(B)$ since $kz \neq z$. So -$K_g \not\subseteq I_f$. -I can see why one must add the diagonal to $f(A) \times f(A)$, since one -wants a reflexive relation on B. It follows that $I_f$ is an -equivalence relation. I would think that one should go farther and -take the image to be the congruence closure of $I_f$, i.e., the submonoid -$\overline{I}_f$ of $B \times B$ generated by $f(A) \times f(A)$ and $\Delta(B)$. Then it is clear that $(kz,z) = (k,1)(z,z)$ in $\overline{I}_f$, and so -$K_g \subseteq \overline{I}_f$. Conversely, it is clear that $\overline{I}_f$ is -a subset of $K_g$, so then I'm happy again. -But this definition has a different problem: an exact sequence -$$ A \xrightarrow{f} B \to 1$$ -of monoids is exact if $f$ is surjective but not conversely: the inclusion map -$$ \mathbb{N} \xrightarrow{f} \mathbb{Z} $$ -of the natural numbers in the integers has $\overline{I}_f = \mathbb{Z} \times \mathbb{Z}$ but is not surjective. Arturo Magidin pointed out to me in an e-mail that this map $f$ is an epimorphism in the category of monoids; maybe I'll have to accept that. -Can anyone shed some light on this matter? I'd already be pretty happy with a reference. - -REPLY [2 votes]: For the special case of commutative monoids, or more generally semimodules over a semiring, my related preprints on arXiv might be helpful (see below). For arbitrary monoids, the general definition is similar, however it is difficult to apply since the notion of the cokernel of a morphism of monoids is really "slippery": -http://arxiv.org/abs/1111.0330 -http://arxiv.org/abs/1210.4566<|endoftext|> -TITLE: Removable base loci for non-projective varieties -QUESTION [5 upvotes]: The theorem of Zariski-Fujita says the following: Given a line bundle $L$ with base locus $B$ on a projective variety $X$. If $L_{|B}$ is ample on $B$, then $L$ is semiample, i.e. $L^{\otimes m}$ is generated by global sections for some $m>0$. -Does this remain true if we only assume $X$ is complete? - -REPLY [4 votes]: Short answer: No. -In his original paper, Fujita posed the question whether we can weaken the assumptions in the theorem. (Fujita 1983, 1.16) There has been one paper written since then with an attempt at improvement. -Let $R$ be a commutative Noetherian ring, $X$ a scheme proper over $R$, and $\mathcal{L}$ a line bundle on $X$. Define -$H^p_\*(\mathcal{F},\mathcal{L}) = H^p(X, \mathcal{F}\otimes\text{Sym}\ \mathcal{L})=\bigoplus_{n\geq0} H^p(X,\mathcal{F}\otimes \mathcal{L}^{\otimes n})$, a module over the graded ring $\Gamma_\*(\mathcal{L}) = H^0(X,\text{Sym} \ \mathcal{L})$. -Then there is -Theorem. (Schröer, 2001) Let $B$ be the stable base locus of $\mathcal{L}$. The following are equivalent: - -$\mathcal{L}$ is semi-ample, -For each coherent sheaf $\mathcal{F}$ and $p \geq 0$, $H^p_\*(\mathcal{F},\mathcal{L})$ is finitely generated over $\Gamma_\*(\mathcal{L})$. -For all coherent ideal sheaves $\mathcal{I} \subset \mathcal{O}(B)$, the module $H^1_*(\mathcal{I}, \mathcal{L})$ is finitely generated over $\Gamma_\*(\mathcal{L})$. - -Schröer uses this generalization of Zariski-Fujita to characterize contractible curves in 1-dimensional families, where considers $X$ complete (i.e., proper over a base field) in a couple instances. -I think any other attempt will run into the same problem: the best you can do is a cohomological characterization. -Keeler (2003) worked in the setting of schemes with filters of line bundles, rather than a single individual line bundle. (This is again a cohomological condition.) He proved a generalization of Serre's vanishing theorem and some other results in Fujita's paper, but not a very promising lead on your question: -Proposition. (Keeler, 2003) Let $X$ be a complete variety, $\mathcal{L}$ a line bundle on $X$, and suppose the base locus is zero-dimensional or empty. Then $\mathcal{L}$ is nef. -There is also the paper of Ein (2000) where he finds a more elegant proof of Zariski-Fujita using Koszul complexes. - - -Ein, L., Linear systems with removable base loci, Comm. Algebra 28, n. 12 (2000), 5931–-5934 -Fujita, Takao, Semipositive line bundles, J. Fac. Sci. Univ. Tokyo 30 (1983), 353--378 -Keeler, Dennis S, Ample filters of invertible sheaves, J. Algebra 259 (2003), 243--283 -Schroer, Stefan, A characterization of semiampleness and contractions of relative curves, Kodai Math. J. Volume 24, Number 2 (2001), 207--213.<|endoftext|> -TITLE: What is the main goal of a paper, really? -QUESTION [22 upvotes]: My question, motivated by idle curiosity while sitting in LaGuardia airport, is the following. You've just proved nice result A and it is time to write the paper. What is the real goal of the paper? -(a) to get the reader understand why result A is true. -(b) to convince the reader result A is true. -In some cases it may be feasible to do both. But often there is some tension between (a) and (b). To understand a result one needs global understanding. One may need to do some ugly computations in coordinates or deal with pictures that LaTex doesn't handle. -On the other hand to check the truth of a paper it is often easier to have local understanding because humans can only keep so much info in their brain at once. So if (b) is your goal you will work hard to break things up into smaller certifiable statements. That way once the reader has been convinced that Lemma X is true they can work with the statement and forget why it is true. To make it easier to check a proof one is often led to invent new formalisms or language and find coordinate free arguments. This can affect negatively (a) because you may not see the forest because of the trees. -I'd like to get the community's opinion. Usual Community Wiki rules are applicable. -Edit in hope of reopening. Perhaps if I make things more specific I can get the kind of answer I wanted and things would be less subjective. Suppose I have 2 proofs that finite sets $X$ and $Y$ have the same cardinality. One is proof is a relatively easy computation of the sizes of each set using known identities with binomial coefficients, Stirling numbers, etc. Any decent referee would follow it. The other proof is an involved bijection between $X$ and $Y$ whose details would be involved to check. Space considerations in the journal do not allow for both proofs. Which one should I submit? - -REPLY [7 votes]: Put (a) in the introduction, and explanations between theorems and lemmas. Then use proofs to achieve (b).<|endoftext|> -TITLE: Is there a high-concept explanation of the dual Steenrod algebra as the automorphism group scheme of the formal additive group? -QUESTION [19 upvotes]: Recall that for any space $X$, the cohomology $H^*X$ (always, in this post, with $\mathbb{Z}/2$-coefficients) has an action of the Steenrod algebra $\mathcal{A}$; that is, a natural morphism $\mathcal{A} \otimes H^*X \to H^*X$. This is not a morphism of algebras, but $\mathcal{A}$ has a Hopf algebra structure such that the appropriate diagrams involving comultilication on $\mathcal{A}$ and $H^*X$ commute. -Consider now a space with finitely generated homology in each degree. Then the Steenrod algebra -acts on homology by duality, and dualizing this gives that the dual Steenrod algebra $\mathcal{A}^{\vee}$ coacts on the completed cohomology ring; that is, there is a map -$$ H^*X \to H^*X \hat{\otimes} \mathcal{A}^{\vee}$$ -which is in fact a morphism of rings, and makes appropriate diagrams commute for products in $X$. It follows that when $X$ is a H space, and $H^*X$ is a Hopf algebra, then we have a coaction of $\mathcal{A}^{\vee}$ on the completed cohomology ring $H^{\star \star}X$ (or something like that). Anyway, the upshot of this is that $\mathrm{Spec} \mathcal{A}^{\vee}$ is a (noncommutative) group scheme because of the Hopf algebra structure, and what we really have is an action of this group scheme (in some sense, at least) on the formal scheme $\mathrm{Sppf} H^{\star \star}(X)$. -In the case where $X = \mathbb{RP}^\infty$, then the formal scheme just described is the formal additive group. Milnor's paper "The Steenrod algebra and its dual" shows that $\mathrm{Spec} \mathcal{A}^{\vee}$ is precisely the automorphism group scheme of this (i.e. a polynomial ring on variables in each power of $2$ minus one). -This is established by computation in Milnor's paper. -Q1: Is there a high-concept explanation of why this should the case? -Q2: What's the analog in characteristic $p \neq 2$? - -REPLY [15 votes]: What I'd like to say is, to some degree, commentary and expansion on what some people have said in the comments above. -I feel that the answer to (Q1), about whether there is a "high-concept" explanation for the dual Steenrod algebra, is "no" at the current point in time. Even further, I feel like the attempt to do so right now would be misleading. We can assign some high-concept descriptions to phenomena in terms of formal group laws, but trying to give a high-concept reason why the Steenrod algebra takes the form it does would be misinterpreting the role of the Steenrod algebra. -We construct the stable homotopy category from the category of topological spaces. Based on its construction, we can say a lot of things in older and more modern language. It's a tensor triangulated category; it's the homotopy category of a symmetric monoidal stable model category; it's the homotopy category of a stable $\infty$-category; it's some kind of "universal" way to construct a stable category out of the category of topological spaces. Based on this, we can describe a lot of the foundational properties of the stable homotopy category and the category of spectra. However, there are a ton of categories that satisfy basically the same properties. Without doing more work, we don't understand anything about specifics that distinguish the stable homotopy category from any other example. -These kinds of computations originate with the Hurewicz theorem, the Freudenthal suspension theorem, Serre's computation of the cohomology of Eilenberg-Mac Lane spaces and his method for computing homotopy groups iteratively, and Adams' method of stepping away from Postnikov towers and upgrading Serre's method into the Adams spectral sequence. Before these, you would have no reason to necessarily believe that the stable homotopy category isn't equivalent to, say, the derived category of chain complexes over $\mathbb{Z}$ or some other weird differential graded algebra. Before you have these, you don't have Milnor's computation of the homotopy groups of $MU$ or $MU \wedge MU$, and you don't have Quillen's interpretation in terms of the Lazard ring. -In short, understanding the Steenrod algebra and its dual are prerequisites for all the qualitative things we understand about the stable homotopy category to distinguish it from another example. Right now there is not a door to the stable homotopy category that comes from formal-group data, as much as we would like one. As a result, I guess I feel like trying to assign a high-concept description to the Steenrod algebra is backwards right now. -The fact that often computations come first and the conceptual interpretations second is something that gives the subject some of its flavor. Are there high-concept explanations why vector bundles should have Stiefel-Whitney classes? Why complex vector bundles should have Chern classes? Why the "quadratic" power operations in mod-$2$ cohomology should generate all cohomology operations, and why all the relations are determined by those coming from composing two? -This is definitely not to say that such a description wouldn't be desirable. It would be very desirable to have a more direct route from concepts to the stable homotopy category, because constructing objects that realize conceptual descriptions can be very difficult.<|endoftext|> -TITLE: Are there any "related rates" calculus problems that don't feel contrived? -QUESTION [58 upvotes]: I just finished teaching a freshman calculus course (at an American state university), and one standard topic in the curriculum is related rates. I taught my students to answer questions such as the following (taken, more or less, from the textbook): - -A man starts walking due north at 5 ft/sec from a Point A. Ten seconds later, a woman starts walking south at 4 ft/sec from a point 20 ft due east of Point A. How fast are they moving apart when the woman has been walking for ten seconds? -A 6' man walks away from a 20' lamppost at a speed of 5 ft/sec. How fast is the distance between the tip of his shadow and the top of the post changing when he is 40' away? -A baseball player runs from first base to second at 20 ft/sec, and simultaneously another baseball player runs from third base to home at the same speed. How fast are they approaching each other after one second? - -To put my question bluntly: Who cares? -My students do, but only because they know these questions will appear on their exams. The baseball question (or something very similar) is actually an exercise in Stewart, and I struggled in vain to imagine a situation in which the manager of a baseball team would need to know the answer. -This is in stark contrast to many other topics addressed in first-year calculus -- optimization, basic differential equations, etc. -- which are realistic models of questions of natural interest in business, biology, etc. Basically, all the related rates questions seemed to be cooked up in response to the fact that calculus students now knew a method to solve them. -My question is in the title. Can anyone share any related rates questions which don't seem quite as contrived, and which might naturally seem interesting and motivated to a typical class of college freshmen? -Thank you! - -REPLY [3 votes]: The following example is contrived, but I created it with the same frustration at the boring and repetitive nature of most related rates problems. Since I found this MO question by doing a google search for a "really interesting related rates problem," I'm recording this here for other instructors looking for juicier problems. -Question: What is the rate of change of the width of a shadow as you walk away from a street lamp considered as a point light source which is higher than your head? You can model yourself as a rectangle is you'd like! -The answer to this question was surprising to me. You can also ask for the rate of change of the trapezoidal shadow's area.<|endoftext|> -TITLE: Examples of CAT(0)-groups -QUESTION [6 upvotes]: My question is the following: -Let M be a simply connected Riemannian manifold whose sectional curvatures -are all nonpositive and let G be a group. Suppose that G acts in M properly discontinuous and -cocompactly by isometries. Is G a CAT(0) group? - -REPLY [7 votes]: The answer is yes, since $M$ is a $CAT(0)$ space, and the group is quasi-isometric to it. See (for example) Jim Cannon's article in Bedford Keane Series: -The theory of negatively curved spaces and groups -J Cannon - t. Bedford, M. Keane, and C. Series. Oxford University …, 1991<|endoftext|> -TITLE: Where was Riemann Existence first proven? -QUESTION [5 upvotes]: Nowadays the standard reference for Riemann's Existence theorem is SGA1, where the proof heavily relies on Serre's GAGA. I imagine that the theorem is much older, as its name suggests, and that its original proof is quite different. I thought it would be instructive for me to look at how this theorem was viewed in the pre-SGA era. -Question -Where does a proof of Riemann's Existence Theorem original appear? Or better yet, where is a readable summary of it in English (or somewhat less preferably in French)? - -REPLY [9 votes]: The original source for the modern version is -H.Grauert and R. Remmert, Komplexe Räume, Math.Ann. 136 (1958), 245–318. -I don't know of an exposition in English. -The version due to Riemann was just for algebraic curves. This is covered in many sources; my favorite is Narasimhan's book "Compact Riemann Surfaces".<|endoftext|> -TITLE: Is the first differential Pontryagin class a morphism of stacks? -QUESTION [13 upvotes]: In Cech Cocycles for Characteristic Classes, Jean-Luc Brylinski and Dennis McLaughlin provide explicit formulas for Cech cocycles for characteristic classes of real and complex vector bundles, and show how to refine these to Cech-Deligne cocycles for differential characteristic classes in differential cohomology. -When the compact Lie group $G$ involved is enough connected, e.g., if one is interested in the second Chern class of a principal $SU(n)$-bundle, the Brylinski-McLaughlin formula drastically simplifies, and it can be shown that in this case it actually gives a morphism of stacks from the classifying stack of principal $G$-bundles with connections to the (higher) stack of principal $U(1)$-$n$-gerbes with connections (here $n+2$ is the degree of the characteristic class involved). -This stacky interpretation is emphatised, e.g., in the follow-up Cech cocycles for differential characteristic classes - An $\infty$-Lie theoretic construction, by Urs Schreiber, Jim Stasheff and myself, where it is obtained (as the title suggests) via integration of $L_\infty$-algebras to higher Lie groups. For this approach, the connectivity of $G$ is essential. For instance one can see that the first fractional differential Pontryagin class $\frac{1}{2}\hat{p}_1$ is a morphism of stacks from the stack of princiapl Spin bundles with connection to the 3-stack of $U(1)$-2-gerbes with connection, and this precisely reproduces Brylinski-McLaughlin construction, the but one cannot see the Brylinski-McLaughlin cocycle for the first differential Pontryagin class $\hat{p}_1$ for princiapal $SO$-bundles with connections via Lie integration: $SO$ is not enough connected to allow this. This pheneomenon is in a sense not surprising: for instance the "identity" morphism from the Lie algebra of $O(2)$ to the Lie algebra of $SO(2)$ cannot be integrated to a morphism of Lie groups from $O(2)$ to $SO(2)$ due to "lack of connectivity" reasons. -Yet the fact that a particular technique fails does not mean that a statement is false. So here is my question: is there a natural interpretation of Brylinski-McLaughlin cocycle for the first differential Pontryagin class $\hat{p}_1$ as a morphism of stacks (from $SO$-bundles with connections to $U(1)$-2-gerbes with connections)? or is there a natural interpretation of Brylinski-McLaughlin cocycle for the second differential Chern class $\hat{c}_2$ as a morphism of stacks from $U$-bundles (not $SU$!) with connections to $U(1)$-2-gerbes with connections)? -My feeling is that once there are no topological obstruction (i.e., once the characteristic classes are defined at the non-very-conected level, as in these cases), the morphism of stacks (which surely exists at the highly connected level) descends from the higher connected cover of $G$ involved to the original $G$. So, for instance since $\frac{1}{2}p_1$ is not an integral class for $SO$-bundles one can not make $\frac{1}{2}\hat{p}_1$ descend from principal Spin-bundles with connections to princiapal $SO$-bundles with connections; but since $p_1$ is an integral class for $SO$-bundles, then it should be possible that $\hat{p}_1$ descends. But all my attemps towards a rigorous proof of this have failed so far, so I've begun thinking that I may be wrong, and that $\hat{p}_1$ can be given no natural interpretation as a morphism of stacks after all. -Any suggestion, reference or criticism concerning this problem is welcome. - -REPLY [8 votes]: Yes, every differential characteristic class is a stack morphism. -The point is that there exist universal differential characteristic classes. These are not easy to describe since they involve a notion of differential cohomology of classifying spaces. One way is to use Urs Schreiber's approach. At least in the case of degree four classes one can alternatively use the theory of multiplicative bundle gerbes. -In this context, I'd use the following working definition: -Definition: A universal degree four differential characteristic class on a Lie group $G$ is a multiplicative $U(1)$-bundle gerbe over $G$ with (multiplicative) connection of curvature $H$. -Here, $H$ is the "canonical" 3-form on $G$, it is defined using a bilinear symmetric linear form on the Lie algebra of $G$ - I think any such form is fine. -Multiplicative bundle gerbes have been introduced in the paper (1); they represent classes in $H^4(BG,\mathbb{Z})$. The notion of a multiplicative connection is subtle, I'd refer to Definition 1.3 of my paper (2). The point is that the "naive" definition is too strict and leaves essentially no space for examples. In particular, while a multiplicative gerbe over $G$ can be seen as a 2-gerbe over $BG$, a multiplicative connection is NOT a connection (in the ordinary sense) on this 2-gerbe. -Example 1: If $G=Spin(n)$, the basic gerbe $\mathcal{G}$ over $G$ carries a canonical connection of curvature $H$ and a canonical multiplicative structure. It is the universal differential half first Pontryagin class, $\frac{1}{2}\widehat{p_1}$. It underlies the definition of string connections I have proposed in my paper (3). -Example 2: If $G = SO(n)$, the bundle gerbe $\mathcal{G}$ descends together with its connection along the projection $Spin(n) \to SO(n)$, but its multiplicative structure does not descend. Instead, only the multiplicative bundle gerbe $\mathcal{G}^2 := \mathcal{G} \times \mathcal{G}$ descends together with its connection and its multiplciative structure. The descended gerbe over $SO(n)$ is the universal differential first Pontryagin class, $\widehat{p_1}$. Descent theory for multiplicative gerbes, together with obstructions is discussed in (4), see Table 1 at the end of the paper. -Now suppose that $\mathcal{G}$ is a universal differential characteristic class, $X$ is a smooth manifold and $P$ is a principal $G$-bundle with connection $A$ over $X$. The Chern-Simons 2-gerbe $\mathbb{CS}_P(\mathcal{G})$ is a $U(1)$-bundle 2-gerbe over $X$, see (1). A connection on $\mathbb{CS}_P(\mathcal{G})$ is constructed from the Chern-Simons 3-form $CS(A)$ and the multiplicative connection on $\mathcal{G}$; here the condition that the curvature of $\mathcal{G}$ is $H$ is important. This construction is described in detail in Section 3.2 of (2). It has the following properties: -Theorem: - -if $\xi_P: M \to BG$ is a classifying map for $P$, then $[\mathbb{CS}_P(\mathcal{G})] = \xi_P^*[\mathcal{G}] \in H^4(M,\mathbb{Z})$. -a connection-preserving bundle morphism $P_1 \to P_2$ (covering some smooth map $X_1 \to X_2$ between base manifolds) induces a morphism -$$ -\mathbb{CS}_{P_1}(\mathcal{G}) \to \mathbb{CS}_{P_2}(\mathcal{G}) -$$ -between bundle 2-gerbes. - -The first statement is Theorem 3.2.3 in (2). It means that on the level of the underlying (non-differential) characteristic class the construction is just pullback. The second statement follows by inspection of the definition of $\mathbb{CS}_P(\mathcal{G})$. It means precisely that we have a stack morphism -$$ -\mathbb{CS}(\mathcal{G}): Bun_G^{\nabla} \to 2\text{-}Grb_{U(1)}^{\nabla}. -$$ -References: - -"Bundle gerbes for chern-simons and wess-zumino-witten theories" -"Multiplicative bundle gerbes with connection" -"String connections and Chern-Simons theory" -"Polyakov-Wiegmann formula and multiplicative gerbes"<|endoftext|> -TITLE: When is a reflective subcategory of a topos a topos? -QUESTION [16 upvotes]: Suppose $\mathcal{E}$ is a topos and $\mathcal{F}\subseteq \mathcal{E}$ is a reflective subcategory with reflector $L$, say. Under what conditions is $\mathcal{F}$ a topos? -A well-known sufficient condition for this is that $L$ be left exact. But this is certainly not necessary. For instance, let $f\colon C\to D$ be a functor such that $f^*\colon Set^D \to Set^C$ is fully faithful. A sufficient condition for this is given in C3.3.8 of Sketches of an Elephant — for every $d\in D$ the category of $c\in C$ with $d$ exhibited as a retract of $f(c)$ must be connected, and every morphism of $C$ must be a retract of the $f$-image of some morphism of $C$. These conditions do not imply that $\mathrm{Lan}_f$ is left exact, but nevertheless they allow us to identify $Set^D$ with a reflective subcategory of $Set^C$, and of course both are toposes. -Is there any general sufficient condition for a reflective subcategory of a topos to be a topos which includes this case? - -REPLY [9 votes]: Let me pay no attention to size issues: -Denote the adjunction by $R$ right adjoint to $L$. Equip $\mathcal{E}$ with the canonical topology $J$ (so generated by jointly surjective epimorphisms), so that we have $Sh_J(\mathcal{E}) \simeq \mathcal{E}.$ Denote the induced sheafication functor $a:Set^{\mathcal{E}^{op}} \to \mathcal{E}.$ Now consider the Yoneda embedding $y:\mathcal{F} \hookrightarrow Set^{\mathcal{F}^{op}}.$ Since $\mathcal{F}$ is reflective in $\mathcal{E}$, it is cocomplete, so we can left-Kan extend the identify functor of $\mathcal{F}$ along Yoneda, to get a functor $a_\mathcal{F}:Set^{\mathcal{F}^{op}} \to \mathcal{F}$ which, by construction is left-adjoint to the Yoneda embedding.. EDIT: Since $\mathcal{E}$ is a topos, and therefore total, so is the reflective subcategory $\mathcal{F}.$ So we get to get a functor $a_\mathcal{F}:Set^{\mathcal{F}^{op}} \to \mathcal{F}$ which is left-adjoint to this Yoneda embedding. However, it is also canonically equivalent to $L \circ a \circ R_{!}$ since this composite is colimit preserving and along representables is the identity. So, it follows that $\mathcal{F}$ is a (Grothendieck) topos if and only if the composite $L \circ a \circ R_{!}$ is left-exact. -Note: By one of the comments I made below, $a \circ R_!$ is left-exact, so $\mathcal{F}$ is a topos if and only if $L$ preserves those finite limits of the form $\rho:\Delta_{aR_!(C)} \Rightarrow a \circ R_! \circ G,$ with $G:D \to Set^{\mathcal{F}^{op}}$ a finite diagram.<|endoftext|> -TITLE: Why are Jucys-Murphy elements' eigenvalues whole numbers? -QUESTION [21 upvotes]: The Jucys-Murphy elements of the group algebra of a finite symmetric group (here's the definition in Wikipedia) are known to correspond to operators diagonal in the Young basis of an irreducible representaion of this group. As one can see from the Wikipedia entry, all of the elements of such a diagonal matrix (in other words the operator's eigenvalues) are integers. -I'm looking for a simple way of explaining this fact (that the eigenvalues are wholes). By simple I mean without going into more or less advanced representation theory of the symmetric group (tabloids, Specht modules etc.), so trying to prove the specific formula given in Wikipedia is not an option. (I'm considering the Young basis as the Gelfand-Tsetlin basis of the representation for the inductive chain of groups $S_1\subset S_2\subset \ldots\subset S_n$, which is uniquely defined thanks to this chain's spectrum's simplicity, not as a set of vectors in correspondence with the standard tableaux.) -In fact, I'm trying to prove the first statement ($a_i\in \mathbb{Z}$) of proposition 4.1 in this article. - -REPLY [3 votes]: Awfully sophisticated proof for the fact :) -Just to relate it with the question about Knizhnik-Zamolodchikov equation: -Find polynom p(z) with values in C[S_n] such that p'(z) = \sum_i (Id+(1i))/(z-i) p(z). [Knizhnik-Zamolodchikov equation for S_n] -Consider the following KZ ODE: -$ p'(z) = \sum_{i=2...n} \frac{ Id + \pi( (1i) )}{z-z_i} p (z) $ -As it is discussed in MO-question above it is known to have polynomial solution. -The reside at infinity is equal to $Res=-\sum_{i=2...n} { Id + \pi( (1i) )}$. -Which is our beloved JM-element up to sign and n*Id. -Hence its eigenvalues must be non-positive integers (this is obvious since -at infinity the solution looks like $(1/z)^{Res}, so in order to be polynomial in z -they must be non-positive ints). -Hence we are done. -Moreover we got that eigs are greater or equal -n (as David Speyer proved directly above).<|endoftext|> -TITLE: Can we find strong fixed-points in the fixed-point lemma of Gödel's incompleteness theorem, that is, where the fixed point is syntactically identical to its substitution instance rather than merely provably equivalent to it? -QUESTION [22 upvotes]: At Graham Priest's talk for the CUNY set theory seminar yesterday, an issue arose concerning the possibility (or impossibility) of a stronger-than usual form of the arithmetic fixed-point lemma often used to prove the Gödel incompleteness theorem. -The fixed-point lemma, treated in previous MO -questions, is commonly described as asserting that for any formula $\varphi(x)$ in the language of arithmetic, with a free variable $x$, there is a sentence $\psi$ such that $PA\vdash\psi\leftrightarrow\varphi({\ulcorner}\psi{\urcorner})$, where ${\ulcorner}\psi{\urcorner}$ denotes the Gödel code of $\psi$. Stronger versions of the fixed-point lemma weaken PA to the theory of Robinson's Q or make the conclusion for more complex webs of fixed points, and these strengthenings are well-known. The fixed-point lemma leads easily to the first incompleteness theorem, since with it one can produce a sentence $\psi$ asserting its own unprovability, and such a statement cannot be provable (or else we would prove something false) and hence is both true and unprovable. -The stronger-than-usual form of the fixed-point lemma desired in the context of one of the speaker's arguments was to find a sentence $\psi$ that was not only provably equivalent to $\varphi({\ulcorner}\psi{\urcorner})$, but was syntactically identical to this assertion. That is, what was needed was what we might call a strong fixed-point, a sentence $\psi$ syntactically of the form $\varphi({\ulcorner}\psi{\urcorner})$. -This appears to be impossible for any of the standard methods of Gödel coding, since for each of these the code ${\ulcorner}\psi{\urcorner}$, as a number, is much larger than the length of $\psi$ as a formula. For example, the code of a sequence of length $k$ is typically much larger than $k$. For such coding, if we understand $\varphi(n)$ to be $\varphi(1+\cdots+1)$, where the substituted term is $n$ many $1$s added, it could never be that $\psi$ is literally the same as $\varphi({\ulcorner}\psi{\urcorner})$, since the latter assertion has length strictly longer than ${\ulcorner}\psi{\urcorner}$, and this is larger than the length of $\psi$. Carl Mummert made a similar point in the latter part of his answer to the MO question mentioned above. -My question is: Is there some method of Gödel coding which would support the strong fixed-point lemma? -And if not, how would one prove such a thing? -Graham suggested that in Gödel's original work, the version of the Gödel sentence produced had the strong fixed-point property, of being literally the same as the assertion that this sentence was not provable. -It occurred to me that if one makes the move beyond what is now considered the usual language of arithmetic $\{+,\cdot,0,1,\lt\}$, for example by adding the function symbols for primitive recursive functions, then one can achieve something very like a strong fixed point. (And perhaps this is close to what Gödel did?) -Theorem. For any formula $\varphi(x)$ with one free variable, there is a closed term $t$ in the language of arithmetic augmented by primitive recursive functions, whose value is precisely ${\ulcorner}\varphi(t){\urcorner}$. Thus, $\varphi(t)$ is a strong fixed point, in the sense that it asserts that $\varphi$ holds of a term whose value is its own Gödel code. -Proof. Let $\text{Sub}(x,y)$ be the primitive recursive function such that $\text{Sub}({\ulcorner}\alpha(x){\urcorner},k)={\ulcorner}\alpha(k){\urcorner}$, substituting $k$ as a term $1+\cdots+1$ in each free occurence of $x$ in $\alpha(x)$. Let $n={\ulcorner}\varphi(\text{Sub}(x,x)){\urcorner}$ and let $t=\text{Sub}(n,n)$, as a syntactic term, using $1+\cdots+1$ to represent $n$. The actual value of the term $t$ can be obtained by evaluating $\text{Sub}(n,n)$, which in light of the definition of $n$, is precisely ${\ulcorner}\varphi(\text{Sub}(n,n)){\urcorner}$, and this is the same as ${\ulcorner}\varphi(t){\urcorner}$, as desired. QED -This lemma essentially provides strong fixed points in the expanded language; but one difference is that we don't have $\psi=\varphi({\ulcorner}\psi{\urcorner})$, where the number is substituted as a term of the form $1+\cdots+1$ of the right length, but rather we have $\psi=\varphi(t)$, where $t$ is a term whose actual value is ${\ulcorner}\psi{\urcorner}$. -Can one prove an analogue of this theorem using terms in the ordinary language of arithmetic $\{+,\cdot,0,1,\lt\}$? - -REPLY [17 votes]: The existence of a Gödel-numbering that supports a strong fixed point was claimed by Kripke in his famous essay Outline of a Theory of Truth, Journal of Philosophy vol. 72 pp.690–716 (the only online copy of the paper I could locate is on JSTOR, so those of you with an academic connection can easily access it). -On p.693, second paragraph, Kripke makes it clear that he has a proof of the existence of strong fixed points, but he writes "The argument must be omitted from this outline". -Thankfully, Albert Visser has provided a detailed exposition of the existence of strong fixed points (in the usual language of arithmetic) in his majestic 2002 paper Semantics and the Liar Paradox, Handbook of Philosophical Logic, vol. 10 pp.159-245. -An online copy of Visser's paper is available on Googlebooks; see pp.168-170 for the details of nonstandard Gödel-numbering that supports a strong fixed point. -Here is the link to Visser's paper on Googlebooks: -http://books.google.com/books?id=wwXfHT5ka_8C&pg=PA149&dq=%22handbook+of+philosophical+logic%22+%22semantics+and+the+liar+paradox%22&hl=en&sa=X&ei=6t2ET5SXHebk0QGiv8nGBw&ved=0CDYQ6AEwAA#v=onepage&q=%22handbook%20of%20philosophical%20logic%22%20%22semantics%20and%20the%20liar%20paradox%22&f=false<|endoftext|> -TITLE: On J. T. Condict's Senior Thesis on Odd Perfect Numbers -QUESTION [16 upvotes]: I am trying to locate a copy of J. T. Condict's senior thesis on odd perfect numbers: -J. Condict, On an odd perfect number's largest prime divisor, Senior Thesis, Middlebury College (1978). -I am sure a soft copy would have been archived somewhere. Would anybody know where (in the Internet) that archive is? - -REPLY [68 votes]: This is Jim (Condict) Grace, the author. I just saw this post. I'm sorry to hear that Middlebury may have lost their copy. I'm not sure where mine is, but I'll keep an eye out for it. (It may be at a family house 1,000 miles away but I'll look for it at Christmas when I visit.) If I find it, I'll scan and post it somewhere, and I'll comment again here with a link for it. As far as I know there is no existing soft copy. I typed it originally on Middlebury's timesharing academic computer but I long since lost my ability to read the mag tape reel backup -- and subsequently parted with the tape. -When I finished the thesis I sent another hard copy as a courtesy to Peter Hagis Jr. at Temple U., since he and McDaniel had done the research on which I based my work. Hagis subsequently cited my work in a published paper, and it may be that others have just copied this citation without seeing the original. -The main innovation in my thesis over previous work is that I wrote a FORTRAN program to automate some of the algebraic manipulations that had previously been done by hand, making it practical to extend the previous result upwards. It still took several months on Middlebury's academic 16-bit PDP11-45 computer to finish all the computations. I managed to install the program to start as a low-priority background job whenever the computer restarted, and I programmed it to pick up from where it had left off. -If you have any other questions about the thesis, let me know and I might be able to answer them. :) Cheers, Jim -My answer to Arnie Dris's question below is too long to be entered as a comment, so I'm putting it here: -I'm happy to help as I am able, although I've spent the 35 years since my BA thesis as a software developer and my number theory is a bit rusty! Also I don't have access to my thesis, or immediate access to the prior work it was based on. But I may still be able to answer your questions. The number theory part of my work was as explained by Peter Hagis Jr. and Wayne L. McDaniel in "On the largest prime divisor of an odd perfect number." II, Math. Comp. 29 (1975), 922��924. I did not attempt to add or subtract from the number theory analysis they had done. My only innovation was to automate by computer some of the algebra they had done by hand, which made it more practical to get a higher result. -My memory of this is a bit fuzzy, but I seem to recall that there were two rather tedious steps in the proof. The first involved some algebraic manipulations which Hagis and McDaniel did by hand and I did by computer. The second involved some more straightforward computations which they did by computer and so did I. If you have access to their paper, I expect that you will see what I mean as you read it. If this doesn't make sense on reading their paper, please let me know. I do live near a college that I expect would have the Mathematics of Computation, so I could go and reread their article. It might help to improve my memory! ;) Cheers, Jim<|endoftext|> -TITLE: Morita equivalence of DG algebras? (reference needed) -QUESTION [17 upvotes]: A stupid question: whether ther exists a universally recognized definition of when two differential graded algebras should be Morita equivalent? I mean a sort of equivalence which would incorporate the usual Morita equivalence of algebras (when there is no differential) and the usual homotopy equivalence of differential graded algebras induced by an $A_\infty$/-quasi-isomorphism. In particular, I would like to believe, that the cyclic cohomology of such algebras should be canonically isomorphic via a generalised trace map. -I believe, this should exist, but a short search on Google rendered only papers on derived Morita equivalence, i.e. as much as I can judge, on equivalence between derived categories of algebras (non-differential). If I am mistaken and this is actually what I need, please, do explain it to me! Or if you know a good reference on the question I am asking, please, do share it with me! - -REPLY [2 votes]: Just a comment on "For example, every linear invariant that I know of (algebraic K-theory, cyclic (co)homology, Hochschild (co)homology,...) is not only invariant under Morita equivalence but also under derived Morita equivalence." : -Given an algebra $A$ (view it as trivially d.g.), the (say left) global dimension of $A$ is a Morita invariant, but it is not a derived-Morita invariant. On the positive, the global dimension to be finite or infinite is a derived Morita invariant.<|endoftext|> -TITLE: Is there a Plancherel Theorem for Gowers norms? -QUESTION [11 upvotes]: In the process of counting arithmetic sequences in sets, the Gowers norms -$$ ||f||\_{U^s[N]}^{2^s} = \frac{1}{N^s} \sum_{\vec{h} ,\\, n } \Delta_{h_1}\dots\Delta_{h_s}f(n) $$ -where the sum is $ \vec{h} \in\mathbb{Z}^s_N$ and $n \in \mathbb{Z}_N $. Here the ``discrete derivative" $\Delta_h f(n) = f(n+h)\overline{f(n)}$ so the $h=0$ terms correspond to $L^2$ norm. -To find arithmetic sequences in the set $E$ want to take norms of characteristic functions $1_E$ with $E \subset \mathbb{Z}_N$ and $|E| = \delta N$. -Letting $N \to \infty$, if $||1_E - \delta||\_{U^s} $ is small -$$ \sum_{x,d \in \mathbb{Z}_N} 1_E(x)1_E(x+h)\dots 1_E(x+sh) \sim \delta^{s+1}N^s $$ -the number of arithmetic sequences is comparable with that of a random set. -The inverse conjecture for Gowers norms states the only obstructions to small Gowers' $U^s$ norms is correlation with s-step nilsequences. - -Is there a way to ``decompose" a set into its nilsequence contributions? This could be analogous to how an $L^2$ function $f: S^1 \to \mathbb{C}$ decomposes into its Fourier series $f(x) = \sum a_n e^{2\pi i n x}$. -This might not be very well-defined. In that case, what steps could I take to make it meaningful. Also, these formulae are taken from Tamar Ziegler's slides. -Looks like there might be problems if your set $E \subset \mathbb{Z}$ and then you have to decide how to approximate $E$ as subsets of $\mathbb{Z}/N\mathbb{Z}$ where $N$ is large. Maybe this is why they use ultralimits. - -REPLY [4 votes]: This is a very natural and good question. If you want to widen the applications of Gower's norms and higher Fourier analysis to topics like discrete isoperimetry, and applications to probability and TOC (where ordinary Fourier analysis is useful) and perhaps also to improved bounds for error-correcting codes, then having analogs for Parseval's formula may be crucial. -Balazs Szegedy's paper "On higher order Fourier analysis" also have a sort of Perseval's formula for higher Fourier. There is a chapter there on higher order Fourier decompositions and higher order dual groups.<|endoftext|> -TITLE: How to efficiently vacuum the house -QUESTION [13 upvotes]: Let $P$ be a polygon (perhaps with no acute angles inside) and let $L$ be a line segment. The segment may move through the area inside $P$ in straight lines, orthogonal to $L$, or it may pivot on any point on $L$ (while remaining entirely within $P$). -Let $S$ be a legal sequence of pivots and straight motions for $L$ in $P$, and say $S$ covers $P$ if applying the motions $S$ to $L$, passes over every part of the area in $P$. The covered area of $S$ is the cumulative area passed over by $L$. - -How can we compute the minimum number of pivots over all covering sequences $S$? -How can we compute the minimum covered area over all covering sequences $S$? - -Better formulations of the problem are welcome. -For a rectangular polygon, $h\times w$ and a line segment of length $l$, with $h < w$, and $l < h$. So in the best sequence I can think of the number of pivotes is $\lceil h/l \rceil$, and the second question is just a sum of areas of semicircles of radius $l$, plus the rectangular overlap from the last strip. -___________________________ -| l--> | 1 pivot at each end -| <--l | -| l--> | -| <--l | 2 pivots if h is not integer multiple of l ---------------------------- - -REPLY [18 votes]: This problem and variants were extensively explored in the paper -"Optimal Covering Tours with Turn Costs," by -Esther M. Arkin, Michael A. Bender, Erik D. Demaine, Sandor P. Fekete, Joseph S. B. Mitchell, Saurabh Sethia, -SIAM Journal on Computing, volume 35, number 3, 2005, pages 531–566. (Link to preliminary 2003 arXiv version). - -We prove that the covering tour problem with turn costs is NP-complete, even if the objective is purely to minimize the number of turns, the pocket is orthogonal (rectilinear), and the cutter must move axis-parallel. The hardness of the problem is not apparent, as our problem seemingly bears a close resemblance to the polynomially solvable Chinese postman problem; ... - -Almost every variant is NP-complete/hard, so the concentration has been on approximations. -To give a sense of the bewildering variety of results, here is a table from the above paper: -  - - -This paper has 45 references, and surveys related milling and lawnmowing results.<|endoftext|> -TITLE: Definability of ground model -QUESTION [11 upvotes]: I have seen mentioned that by a result of Laver, the ground model is definable in any set forcing extension (using parameters). -Does the same hold for class forcing? If it does, in order to establish this fact, does one need to assume more properties about the class-forcing notion, apart from the obvious preservation of the ZFC axioms? -References would also be welcome, since my only current source is the (admittedly short) passage on class forcing in Jech's book. - -REPLY [16 votes]: Let me address the two questions that have arisen here. -First, the question is whether there a significant collection of class forcing notions for which the Laver theorem continues to hold? The answer is yes. -Theorem. (Hamkins) $\ $ If an extension $V\subset W$ of models of ZFC exhibits the $\delta$-approximation and cover properties, then $V$ is definable in $W$ using parameters in $V$. -Unfortunately, this particular theorem is not published explicitly as such. Nevertheless, the proof that Laver gives of the ground model definability theorem in his paper Certain very large cardinals are not created in small forcing extensions (prepublication version), a proof he credits to me, actually achieves this stronger version of the theorem. (The terminology is defined in my paper Extensions with the approximation and cover properties have no new large cardinals, as well as in Laver's paper.) -The way it happened was this. Laver had proved his remarkable ground model definability theorem in the case of set forcing and had contacted me about it---without sending me his proof, which I have never seen---when he noticed, he said, a resemblance in his argument to an argument that Woodin and I had used in our paper Small forcing creates no new strong or Woodin cardinals, a method that was also the basis for my paper on the approximation and cover properties linked above. I was immediately excited about Laver's idea, and in reply I sent him a proof of the stronger theorem above. He responded that this was clearly the right argument even in the case of set forcing, and he proceeded to use my proof in his paper, graciously crediting it to me. Meanwhile, Woodin had independently concluded the set-forcing case implicitly in some of his work. Fascinated with the ground model definability result, Jonas Reitz and I shortly formulated the Ground Axiom, on which Jonas wrote his dissertation, and these ideas formed the basis of our subsequent joint work with Gunter Fuchs on set-theoretic geology. -The key point here in relation to the question is that the collection of class forcing notions that do exhibit the $\delta$-approximation and cover properties is rather extensive and includes many of the standard class forcing notions, such as the forcing of the GCH, progressively closed Easton iterations, the usual forcing of V=HOD, the Laver preparation and many others. Any forcing notion that can be factored as forcing of size $\delta$ followed by $\leq\delta$-closed forcing has the $\delta^+$-approximation and cover properties. So for example, adding a Cohen real and then doing countably closed forcing will have the $\omega_1$-approximation and cover property. -The second question is, does Laver's ground model definability theorem hold for all class forcing notions? -For this, as mentioned in the comments, the answer is no. Consider the following argument, a version of which I recently heard from Carolin Antos-Kuby, a student of Sy Friedman, during a recent trip to Vienna, and this may be due to Sy. Let $\mathbb{P}$ be the Easton support class product forcing over V that adds a Cohen subset to every regular cardinal, and let $V[G][H]$ be $V$-generic for $\mathbb{P}\times\mathbb{P}$, which is actually forcing equivalent to $\mathbb{P}$, since adding two Cohen sets is the same as adding one. Consider the model $V[G]$, which is a ground model of $V[G][H]$, but which we will show cannot be definable there by parameters. Suppose $A$ is any set in $V[G][H]$ and suppose toward contradiction that $\varphi(x,A)$ defines the relation $x\in V[G]$ in $V[G][H]$, forced by some condition $p$. Far above $A$ and the support of $p$, however, we may apply an automorphism $\pi$ of the product forcing $\mathbb{P}\times\mathbb{P}$ that swaps a factor of $G$ on some large coordinate with the corresponding factor of $H$, but fixes the lower part of the forcing, and hence fixes the name of $A$ and the condition $p$. It follows that the definition $\varphi(x,A)$ should also define $V[\pi(G)]$, which by design is not the same as $V[G]$, a contradiction. So there can be no such definition.<|endoftext|> -TITLE: Switching tenure-track position -QUESTION [9 upvotes]: I have a tenure-track position at a research university (among top 20 in USA) in a decent city. This has been my fourth year as a tenure-track. I still have not heard from my department regarding the promotion. I do have very good publication record (over 20 published papers in top journals, including Annals and Inven. Math papers), and a decent teaching record. Is it advisable for me to apply for other jobs? I am happy with my current university, but I am very worried about the prospect of getting tenure. I would be very grateful for any comments and suggestions. - -REPLY [6 votes]: I have my doubts about adding the third answer to this question, but here goes: -You should be in touch with your chair about whether you will get tenure. If the chair and various bigwigs in your department are remotely competent and there is even the slightest danger of you not getting tenure, they should be on your case about it; departments really do not want to have people go up for tenure and fail. It makes them look bad to both the administration and to job candidates. -So what you should do is talk to your chair and say "Hey, we haven't really talked about me going up for tenure. Do you have any concerns about my case? Is there anything I should be doing to prepare?" In all likelihood, your mind will be set at ease, and if not, well, then that's information you're in a better position for having.<|endoftext|> -TITLE: An approximate converse of discrete uncertainty principle -QUESTION [5 upvotes]: Let $f:\mathbb{Z}_n \rightarrow \{0, 1\}$ and let's normalize the Fourier transform $\hat{f}$ so that $\|\hat{f}\|_2 = \|f\|_2$, i.e. -$$\hat{f}(\xi) = \frac{1}{\sqrt{n}}\sum_{x \in \mathbb{Z}_n}{f(x)e^{-2\pi i x \xi/n}}$$ -Also let $\hbox{supp}(f) = \{x \in \mathbb{Z}_n: f(x) \neq 0\}$. -What I am calling the discrete uncertainty principle is the following statement: - -If $|\hbox{supp}(f)| > 0$ then $|\hbox{supp}(f)| \cdot |\hbox{supp}(\hat{f})| \geq n$. - -This inequality is tight for the Dirac comb. Also, for $n$ a prime number a much stronger inequality is true: $|\hbox{supp}(f)| + |\hbox{supp}(\hat{f})| \geq n + 1$ (again as long as $f$ is not the constant 0 function). -The uncertainty principle states that if $f$ is is "concentrated" then $\hat{f}$ is "spread-out". I am interested in the existence of a weak converse, i.e. is it true in some approximate sense that if $f$ is very spread out then $\hat{f}$ is fairly concentrated. -Here is a possible theorem statement that I would like to be true: - -Let $f:\mathbb{Z}_n \rightarrow \{0, 1\}$ and let $\hat{f}$ be define as above. Is it true that for any $f$ s.t. $\|f\|_2^2 \geq \sqrt{n}$ there exists a set $S \subseteq \mathbb{Z}_n$ s.t. $|S| \leq \sqrt{n}$ and - $$\sum_{\xi \in S}{|\hat{f}(\xi)|^2 \geq \|\hat{f}\|_2^2 - \sqrt{n}} = \|f\|_2^2 - \sqrt{n}$$ - -Note that since the range of $f$ is $\{0, 1\}$, $\hbox{supp}(f) = \|f\|_2^2$. Note also that the condition that $\|f\|_2^2 \geq \sqrt{n}$ is redundant given the error factor of $\sqrt{n}$. On the other hand, some error factor is necessary, given the strong inequality for $n$ a prime number that I mentioned above. -The reasons I have for guessing this statement are that - -I want it to be true (for my application) :) -I have checked it by brute-force enumeration for $n \leq 23$. - -Is there any statement of this form known? Or is it obviously false? - -REPLY [6 votes]: (My previous comment, converted to an answer as requested.) -If one sets $f$ to be the random 0-1 valued function, then from the Chernoff inequality one sees that with non-zero probability, one has $\hat f(0) = \sqrt{n}/2 + O(1)$, $\|f\|_2^2 = n/2 + O(\sqrt{n})$ and $\hat f(\xi) = O(\log n)$ for all $\xi \neq 0$, so the Fourier transform is basically maximally dispersed, so there is no concentration at anywhere near the scale suggested. -If $n$ is prime, one can obtain a deterministic version of this example (without the losses of $\log n$) by taking $f$ to be the indicator function of the quadratic residues, and then using Gauss sums. -Informally, "most" functions (drawn from, say, a gaussian measure) will be more or less uniformly spread out in phase space, which implies that the function and its Fourier transform will both be spread out uniformly as well. Concentration (either in physical space or frequency space) is the exception rather than the rule.<|endoftext|> -TITLE: Mumford-Tate groups of abelian varieties with potentially good reduction everywhere -QUESTION [5 upvotes]: Let $A$ be an abelian variety defined over a number field, and let $MT(A)$ be its Mumford-Tate group. It is a conjecture of Morita that if $MT(A)$ is anisotropic-mod-center (that is, it has no $\mathbb{Q}$-rational unipotent elements), then $A$ has potentially good reduction. The basic example is when $MT(A)$ is a torus: in this case, this conjecture is one of the first results one proves in the theory of complex multiplication of abelian varieties. -One way that Morita's conjecture is formulated is as follows: The condition on $MT(A)$ ensures that the Shimura variety attached to it is compact, and the conjecture says that the Shimura variety in fact has proper reduction at any finite prime; a transcendental condition has been transported to the world of arithmetic. Strictly speaking, this is stronger than the formulation given above, since we are now talking about the reduction of a whole host of abelian varieties (ones that appear in the family over the Shimura variety) rather than just $A$. This leads me to: -Question: Is anything known about the converse to the original formulation? Suppose that $A$ has potentially good reduction everywhere; then is $MT(A)$ anisotropic-mod-center? - -REPLY [4 votes]: An elliptic curve with integral $j$-invariant has potential good reduction everywhere. If it does not have CM then its Mumford-Tate group is $GL_{2,\mathbb{Q}}$ which is not anisotropic modulo its centre.<|endoftext|> -TITLE: monomials in the universal enveloping of a Lie algebra in terms of the symmetric basis -QUESTION [5 upvotes]: Let $L$ be a finite-dimensional Lie algebra over a field $k$ of characteristic zero and $e_1,\ldots, e_n$ some basis of $L$. The formula $[e_i,e_j] = \sum_k C_{ij}^k e_k$ determines the structure coefficients $C_{ij}^k$. Given any ordered $k$-tuple $I = (i_1,\ldots,i_k)\in \lbrace 1,\ldots,n \rbrace^k$, define $e_I = e_{i_1}\cdots e_{i_k}\subset U(L)$ and -$$ -e^S_I = \frac{1}{n!}\sum_{\sigma\in\Sigma(k)} e_{\sigma(i_1)}\cdots e_{\sigma(i_k)}\in U(L). -$$ -As it is well known, from various forms of a PBW theorem, $e_I$, for all $I$ with $i_1\lt i_2\lt \ldots \lt i_k$, $k \geq 0$, form a basis and also $e^S_I$ (for the same set of $I$-s) form a basis. I need explicit formulas for $e_I$ in the linear basis of $e^S_J$-s where the coefficients are expressed in terms of the structure constants $C_{ij}^k$ (and combinatorial factors). In fact, for my present purposes, it would be enough to know explicitly the deepest, lowest order, linear term (but it is of course the hardest summand in the expansion). For example, for the easiest nontrivial case $k = 2$, -$$e_{(i,j)} = e_i e_j = \frac{1}{2}(e_i e_j + e_j e_i) + \frac{1}{2} \sum_k C_{ij}^k e_k = e_{(i,j)}^S + \frac{1}{2} \sum_k C_{ij}^k e_k,$$ -hence the linear term is $\frac{1}{2} \sum_k C_{ij}^k e_k$. - -REPLY [6 votes]: An explicit formula is given in this paper by L. Solomon. I copy the abstract here: - -Let g be a Lie algebra over a field of characteristic zero. Let T be the tensor algebra of g, let S be the subspace of symmetric tensors and let J be the two-sided ideal of T generated by tensors x⊗y−y⊗x−[x, y]. One formulation of the P-B-W theorem states that T=S⊕J, direct sum. In this paper we give an explicit formula for the projection of T on S defined by this direct decomposition.<|endoftext|> -TITLE: Do Shintani zeta functions satisfy a functional equation? -QUESTION [11 upvotes]: Probably my questions are known or evident to the experts but I'm a bit puzzled. First of all there seem to be two kinds of zeta functions that go under the name of Shintani zeta functions. -First, there are zeta functions $\zeta^{SS}$ associated with so called prehomogenous vector spaces going back to important work by Sato and Shintani (see the original article or this book by Yukie) and then, second, zeta functions $\zeta^S$ that appeared in Shintani's work on special values of Dedekind zeta functions of totally real number fields at negative integers (see Shintani's article or Neukirch's book for example). -1) I'm mainly interested in the question if it is known (or expected) if the latter zeta functions $\zeta^S$ satisfy functional equations. (From what I understand the $\zeta^{SS}$ satisfy functional equations or are expected to satisfy in case it is not proven). -Let me just note that one can write Shintani zeta functions in the following form $$\Gamma(s)^n \zeta^S(s,z,x) = \int_0^\infty \cdots\int_0^\infty \sum_{z_1,\dots , z_n=0}^\infty e^{-\sum_{i=1}^n t_i L_i(z+x)}(t_1\cdots t_n)^{s-1} dt_1\cdots dt_n$$ -where the $L_i(x)$ are linear forms, i.e. essentially we could say that we're looking at multivariable theta-like functions and Mellin transforms thereof. So the question can be rephrased in asking whether these theta-like functions occurring in the above integral satisfy a functional equation/theta inversion formula. (Note that these theta-like functions do in general not come from symplectic structures, i.e. they are not related to abelian varieties (at least as far as I see)). -2) But next to this question I'm also extremely interested in the relationship of these two kinds of zeta functions. In which cases do the two constructions agree? Is there anything known? -Thank you very much in advance! -EDIT: OK, so I could speak a bit with one of the absolute authorities in this field and I learned, that -1) -one shouldn't expect functional equations for single functions $\zeta^S$ but rather for certain finite linear combinations and -2) -one shouldn't expect relations between the two notions of "zeta" functions. -This doesn't destroy the applications I had in mind with my question but I have to rethink the question and will try to give a better and less naive version of it soon. Thank you so far very much for your helpful comments! - -REPLY [2 votes]: there was a talk recently in RIMS Kyota titled something like -"the functional equation of the Shintani zeta function". -You might try looking that up.<|endoftext|> -TITLE: Algebra - Decomposition of a matrix polynomial -QUESTION [5 upvotes]: Dear All, -This is related with a problem that I'm trying to solve on my PhD dissertation in econometrics, and I thought that some mathmatician can know the answer. -What is known about a possible extension, $E$ , of the ring, $ A$ , of all n-by-n matrices with entries in $\mathbb{C}$ such that any non-constant polynomial of $ A[x] $ splits in a product of linear factors in $E[x]$? -$ax = xa$ iff $a$ is in the commutator of $E$. Moreover, $ A$ is unitary. -Thanks a lot, -Federico Carlini - -REPLY [2 votes]: This is true because of following: any matrix over principal ideal domain can be reduced by elementary Gauss transformations to the Smith normal form (see this: http://en.wikipedia.org/wiki/Smith_normal_form). -The ring of polynomials over any field is such a domain. -Smith normal form can be splitted to linear factors. -Elementary Gauss transformations may be of the form $(z, w)\mapsto (z+p(x)w, w)$, where $p(x)$ is not linear. I think that these transformations cannot be splitted in linear factors (I don't know how to prove it - please let me know if you do know this, I'm curious about).<|endoftext|> -TITLE: Nerve of the semi-simplex category -QUESTION [7 upvotes]: The category of simplices $\Delta$ has a terminal object $[0]$, hence its nerve is contractible. What can be said about the nerve of its subcategory $\Delta_{\mathrm{mono}}$ which contains only the coface maps? - -REPLY [8 votes]: Let $C: \Delta_{\mathrm{mono}}\to \Delta_{\mathrm{mono}}$ be the "cone" functor, given on objects by $C([p])=[p+1]$, and on morphisms by $C(\delta)(0)=0$ and $C(\delta)(i) = \delta(i-1)+1$. Then there are natural monomorphisms -$$[p] \to C([p]) \leftarrow [0]$$ -which give a zig-zag of natural transformations relating the identity functor on $\Delta_{\mathrm{mono}}$ to a constant functor. So its got contractible nerve.<|endoftext|> -TITLE: Comparing two measures on trees on $n$ vertices -QUESTION [13 upvotes]: A standard measure on trees on $n$ vertices is the Uniform Spanning Tree (UST) on the complete graph. This is the measure where every tree has equal probability, $1 / n^{n-2}$ by Cayley's formula. -Here is another measure. Take an Erdős–Rényi (i.e. edge independent) random graph $G \in G(n,p)$ with $p$ large enough to ensure that $G$ is asymptotically almost surely connected, and then choose a UST on $G$. -Note that if $p = 1$ these two measures are identical. My guess is that they are close (say in total variation distance), even for much smaller $p$. In particular, suppose that $$p \ge \frac{\log n + \omega}{n},$$ where $\omega \to \infty$ arbitrarily slowly as $n \to \infty$. (This is barely sufficient to guarantee that the probability that $G$ is connected tends to one.) - -Are these actually the same measure in - disguise? If not, can we say that - they are "close" to the same measure, - for example, by putting an upper bound - on the total variation distance - between the two measures that tends to - zero as $n \to \infty$? - -REPLY [4 votes]: It's a nice question. The following is not an answer but too long to put in comment. I think in general you want to ask "how much can you tell of the underlying graph given a sample from the uniform spanning tree". There is a great algorithm due to David Wilson to sample the UST, which consists in growing the tree by successively running loop-erased random walks. See eg http://en.wikipedia.org/wiki/Loop-erased_random_walk. -Here I think you can try to generate the graph at the same time as the loop-erased random walks, which means that for a while you can couple a loop-erased random walk on the complete graph with one on $G(n,p)$. Not sure this method will allow you to go all the way to the connectivity threshsold $p = (1+ \epsilon) \log n / n$, though. Perhaps it would be easier to start with $p=1/2$ !<|endoftext|> -TITLE: Minimal genus, adjunction inequality -QUESTION [5 upvotes]: Let's consider closed simply-connected 4-manifold $M$ and some $a\in H^2(M)$. It is very natural question to estimate minimal $g$ that $a$ can be presented as embedded surface of genus $g$. -As I know there is the adjunction inequality for estimation of minimal genus via Seiberg-Witten theory. -Question 1: Does there exist other methods to estimate minimal genus ? -I heard that there are homeomorphic but not diffeomorphic 4-manifolds $M,N$ such that for some $a\in H^2$, $a$ has different minimal genus in $M,N$. -Question 2: Could you give me such examples? As I understand it should be some manipulations with Seiberg-Witten invariants... - -REPLY [5 votes]: To add a bit to Andrey's answer, you can improve the method of Hsiang and Szczarba (also Kotschick and Matic deserve credit here) a bit using Furuta's approach to Seiberg-Witten theory. The method can give genus bounds on four manifolds where the adjunction formula doesn't apply (for example on $\mathbb{CP}^2 \# \mathbb{CP}^2$). I wrote a short paper on this: Math. Res. Lett. 5 (1998), no. 1-2, 165–183. The arXiv version is here: http://arxiv.org/pdf/dg-ga/9704010v1. Theorem 1.6 gives a typical genus bound from this method. -The basic idea is the following. If your class has certain divisibility properties and it is represented by an embedded surface, then the four manifold obtained by taking a cyclic cover branched along the surface will be a new four manifold with certain properties which can be used to constrain the genus of the embedded surface. Under appropriate hypotheses, the cover will be a spin four manifold with a cyclic group action. Refining Furuta's proof of the 10/8ths theorem for spin four manifolds to include the presence of a cyclic action, one obtains a bound on the size of the signature in terms of the second Betti number which then translates into a genus bound on the original manifold.<|endoftext|> -TITLE: Noteworthy achievements in and around 2010? -QUESTION [24 upvotes]: The goal of this question is to compile a list of noteworthy mathematical achievements from about 2010 (so somewhat but not too far in the past). -In particular, this is meant to include (but not limited to) results considered of significant importance in the respective mathematical subfield, but that might not yet be widely known throughout the community. -Compiling such a list is inevitably a bit subjective, yet this can also be seen as a merit, at least as long as one keeps this implicit subjectivenes in mind. -Thus the specific question to be answered is; - -Which mathematical achievements from about 2010 do you find particularly noteworthy? - -This is perhaps too broad a question. So, a way to proceed could be that people answering focus on their respective field(s) of expertise and document that they did so in the answer (for an example see Mark Sapir's answer). -As my own candidates let me mention two things: -[Note: original version of the question by Alexander Chervov, so these are Alexander Chervov's candidates.] - - -"Polar coding" (Actually it is earlier than 2010, but I asked for "around 2010") - -introduction of "Polar coding" http://arxiv.org/abs/0807.3917 by E.Arikan. -New approach to construct error-correcting codes with very good -properties ("capacity achieving"). -Comparing the other two recent and popular approaches -turbo-codes (http://en.wikipedia.org/wiki/Turbo_code) -and -LDPC codes (http://en.wikipedia.org/wiki/LDPC) -Polar coding promises much simpler decoding procedures, -although currently (as far as I know) they have not yet achieved same -good characteristics -as LDPC and turbo, it might be a matter of time. -It became very hot topic of research in information theory these days - -just in arxiv 436 items found on the key-word "polar codes". - -I was surpised how fast such things can go from theory to practice - -turbo codes were invented in 1993 and adopted in e.g. mobile -communication standards within 10 years. So currently yours -smartphones use it. - - - - -Proof of the Razumov-Stroganov conjecture -http://arxiv.org/abs/arXiv:1003.3376 -So the conjecture lies in between mathematical physics (integrable systems) and combinatorics. There was much interest in it recent years. -Let me quote D. Zeilberger (http://dimacs.rutgers.edu/Events/2010/abstracts/zeil.html): -In 1995, Doron Zeilberger famously proved the alternating sign matrix conjecture (given in 1996, a shorter proof by Greg Kuperberg). In 2001, Razumov and Stroganov made an even more amazing conjecture, that has recently been proved by Luigi Cantini and Andrea Sportiello. I will sketch the amazing conjecture and the even more amazing proof, that is based on brilliant ideas of Ben Wieland. - -REPLY [10 votes]: In differential geometry there have been two old conjectures both solved in spring 2012 within 1 month: - -A proof of the Willmore conjecture by Marques and Neves (arXiv:1202.6036) which states that the Willmore minimizer amongs immersed tori in $S^3$ is the Clifford torus up to Moebius transformation. -A proof of the Lawson conjecture on less than 10 pages by Brendle (arXiv:1203.6597): The only embedded minimal torus in $S^3$ is the Clifford torus. - -One should note that the proofs of the theorems rely on totally different methods.<|endoftext|> -TITLE: Is there always a 'smallest model' $M[x]$ of ZFC? -QUESTION [9 upvotes]: Given $M$ a wellfounded transitive set model of ZFC, and $x$ a set which is not in $M$, is there always a 'smallest wellfounded transitive model' $M[x]$ of ZFC which extends $M$ and contains $x$? -I believe the answer is NO, because there might not be a 'canonical choice-function' for the set $x$ which we want to add. But I'm not sure how to make that precise. -If the above idea can be made precise and the answer is in fact NO, then the natural follow-up question would be: What if we only demand $M[x]$ to be a model of ZF? - -REPLY [13 votes]: It is consistent with ZFC that there is no such model at all. For example, when $M$ is countable, let $x$ be a real coding a relation on $\omega$ revealing that the ordinals of $M$ are countable. Thus, any model $N\models{\rm ZFC}$ containing $M$ and the object $x$ will also view the ordinal height of $M$ as a countable ordinal. Thus, the height of $N$ will be strictly taller than the height of $M$. But it is consistent with ZFC that all countable well-founded models of ZFC have the same height, for example, this is true in the second $L_\alpha$ which is a model of ZFC, since all the transitive models inside that model have the same height as the minimal model of ZFC. -And it doesn't matter if you demand ZFC or just ZF. -But on a postive note, if there is at least one model $N$ of ZF extending $M$ and the object $x$, and $x$ is a set of ordinals in $N$, then there is a least model of ZFC containing $M$ and $x$. This can be seen by closing $M$ and $x$ under the Goedel operations inside $N$.<|endoftext|> -TITLE: Nonnegative additive functions on coherent sheaves -QUESTION [8 upvotes]: Let $(X,\mathcal{O}_X)$ be a Noetherian integral scheme and let $g$ be a (numerical) additive nonnegative function from coherent $\mathcal{O}_X$-modules to $[0,\infty)$. This question may be well known to the expert but I couldn't find a reference: is $g$ a constant multiple of generic rank? If true, do you know of any reference for this? -Notes: - -If $X=\mathrm{Spec}\:R$ is affine with $R$ an integral domain, then a proof can be found in Northcott-Reufel, Theorem 2, p. 303. There are also other proofs. -If $X$ is a projective variety over a field, I think I can prove it, but I don't know any reference for this case. - -I have a feeling this question must have been answered in K-theory. - -REPLY [11 votes]: I suppose that "additive" means that "additive over short exact sequences". If so, this is does not seem too hard, at least if $X$ is separated. -By noetherian induction, you may assume that for all proper integral subscheme $Y$ of $X$, the restriction of $g$ to $Y$ is given by a multiple of the generic rank at $Y$. But every coherent sheaf with support on $Y$ can be given by a successive extension of coherent sheaves of $\mathcal O_Y$-modules; hence the restriction of $g$ to sheaves supported on $Y$ is given by a multiple of the length of the stalk at the generic point of $Y$. On the other hand, for each $n > 0$ denote by $Y_n$ the subscheme defined by the $n^{\mathrm th}$ power of the sheaf of ideals of $Y$; the length of the stalk of $\mathcal O_{Y_n}$ at the generic point of $Y$ is unbounded, but by the positivity of $g$ the value of $g(\mathcal O_{Y_n})$ is bounded by $g(\mathcal O_X)$. Hence this multiple is 0, and $g$ is 0 on all torsion sheaves. In particular, if $F \to G$ is a generic isomorphism of coherent sheaves, $g(F) = g(G)$. -But on the other hand if $F$ is a coherent sheaf of generic rank $r$, there exists homomorphisms $F \to G$ and $\mathcal O_X^r \to G$ of coherent sheaves that are generic isomorphisms; hence $g(F) = rg(\mathcal O_X)$. The conclusion follows.<|endoftext|> -TITLE: Pólya's Random Walk Constants at infinity -QUESTION [9 upvotes]: Let be the probability that a random walk on a d-D lattice returns to the origin. In 1921, Pólya proved that $p(1)=p(2)=1$ -but $p(d)<1$ for $d>2$. -http://mathworld.wolfram.com/PolyasRandomWalkConstants.html -I wonder what we can say about the probability for $d \to \infty$ -In other words, if there is a closed formula or approximation which presents the limit of the probability for very big $d$? -Does the limit exist? -Thank you in advance for any comments or approach to investigate the question. - -REPLY [13 votes]: The table in that Mathworld page suggests that $p(d) \rightarrow 0$ as $d \rightarrow \infty$. -That page also gives a formula for $p(d)$ in terms of a definite integral: -$$ -p(d) = 1 - \left[ \int_0^\infty I_0(t/d)^d e^{-t} dt \right]^{-1}, -$$ -where $I_0$ is a "modified Bessel function" with power series -$$ -I_0(x) = \sum_{n=0}^\infty \frac{(x/2)^{2n}}{n!^2} -= 1 + \frac{x^2}{2^2} + \frac{x^4}{8^2} + \frac{x^6}{48^2} + \cdots . -$$ -[For large $x$ it is known that $I_0(x) \sim (2\pi x)^{-1/2} e^x$, so the integrand decays as a multiple of $x^{d/2}$ for $x \rightarrow \infty$, and the integral is finite iff $d>2$.] Substituting the power series for $I_0(x)$ into the integral, and expanding termwise via $\int_0^\infty t^m e^{-t} dt = m!$, yields -$$ -1 + \frac1{2d} + \frac3{4d^2} + \frac3{2d^3} + \frac{15}{4d^4} + \frac{355}{32d^5} + \cdots -$$ -(the coefficients in powers of $1/2d$ are OEIS sequence A105227). We then compute the asymptotic series -$$ -p(d) \sim \frac1{2d} + \frac1{2d^2} + \frac7{8d^3} + \frac{35}{16d^4} + \frac{215}{32d^5} + \cdots, -$$ -which seems consistent with the Mathworld table. The coefficients in powers of $1/2d$ are OEIS sequence A043546; as often happens, finding a few terms makes it easier to hunt down some of the literature. [Added later: so does posting on Mathoverflow; a comment to this answer by Folkmar Bornemann gives a reference dating back 55+ years - -Montroll, Elliot W: Random walks in multidimensional spaces, especially on periodic lattices, J. Soc. Indust. Appl. Math. 4 (1956), 241–260 (MR0088110) - -— thanks! Here's some gp code for this power series in $w = 1/2d$ and its coefficients, quite similar to Flajolet's Maple code reproduced in the OEIS entry: -N = 20 -I1 = sum(n=0,N,x^n/n!^2,O(x^(N+1))); -Iw = subst(I1,x,w^2*x)^(1/(2*w)); -g = sum(n=0,N,(2*n)!*polcoeff(Iw,n,x)) + O(w^(N+1)); -p = 1 - 1/g -vector(N,n,polcoeff(p,n)) - -This returns -[1, 2, 7, 35, 215, 1501, 11354, 88978, 675569, 4175664, 1725333, -687775083, -19848956619, -438027976068, -8715988203509, -161989586455204, -2784493824166078, -41530410660307610, -406672888265416456, 4420077014249902362] - -and gp readily computes it for $N$ as large as 50, and with some more effort even for $N=100$.] -The form of the asymptotic series can be explained as follows: for each $k=1,2,3,\ldots$, the probability of return to the origin in $2k$ steps is $O(d^{-k})$ as $d \rightarrow\infty$; so the probability of return by the $2k$-step gives $p(d)$ to within $O(d^{-(k+1)})$, and this estimate is a polynomial in $1/2d$ with integer coefficients. For example, the $k=1$ probability is $1/2d$ exactly; for $k=2$, add $2!/(2d)^2 - O(d^{-3})$; "etc.".<|endoftext|> -TITLE: When is the image of a non Lebesgue-measurable set measurable? -QUESTION [5 upvotes]: Hi MathOverflow, -I'm not sure if it makes sense to ask this question in the general setting, but: -Are there any necessary conditions for a function, such that if $N$ is a not Lebesgue measurable, $f(N)$ is Lebesgue measurable? -I am working on a problem, which seems to suggest that there are no 'trivial' conditions on the function (in particular, $f$ can be injective, which is a surprise to me). The problem is a as follows: -Pick a non Lebesgue measurable set $N \subset (0,1) \subset \mathbb{R}$ and write $x \in (0,1)$ in an infinite binary expansion, i.e. $x = 0.x_1x_2...$ with $x_i = 0$ or $1$ and infinitely many $x_i$'s equal to $1$ (this is ok, since $0.1 = 0.0111...$). -Now, take $f(x) = 2 \sum_{i=1}^{\infty} x_i 3^{-i}$. Then $f(N)$ is Lebesgue measurable, since it maps any set to a Cantor-like set (of measure zero) (thanks to Tapio Rajala for the easy solution). -$f$ just takes $x$ to a base $3$ representation with no $1$'s in the expansion, thus is clearly injective. It sort of "spreads out" the elements of set $N$. Also, clearly $f(N) \subset (0,1)$. -The thing that bothers me is that this seems to suggest that this $f$ is able to transform any non-measurable set into a measurable one, without really "loosing information" about it (because it is injective), which just sounds too good to be true. -I tried to look for sources on functions applied on non-Lebesgue measurable sets, but failed to find anything, so if anyone could guide me to some I would highly appreciate it too. -Thanks. - -REPLY [3 votes]: Suppose $A \subset I = [0,1]$ is Lebegue non-measurable, $B \subseteq I$ Lebesgue measurable, and $f: I \to I$ is a measurable function with $A = f^{-1}(B)$. By inner regularity, $B$ is the disjoint union of sets $C$ and $D$ where $C$ is an $F_\sigma$ and $D$ has measure 0. Then $A$ is the disjoint union of $f^{-1}(C)$, which is Lebesgue measurable, and $f^{-1}(D)$. -Thus the only way an injective measurable function can map a nonmeasurable set onto a measurable one is that it maps some nonmeasurable subset to a set of measure 0.<|endoftext|> -TITLE: Induced representations of topological groups -QUESTION [10 upvotes]: Sorry if this is a naive question-- I'm trying to learn this stuff (cross-posted from https://math.stackexchange.com/questions/89248/induced-representations-of-topological-groups) -If $G$ is a group with subgroup $H$, then we have the restriction functor $\operatorname{Res}$ from $G-\operatorname{mod}$ to $H-\operatorname{mod}$. We also have this idea of induction, a functor $\operatorname{Ind}^G_H$ from $H-\operatorname{mod}$ to $G-\operatorname{mod}$. These are adjoints, which means (I think) that $\operatorname{Hom}_G(\operatorname{Ind}^G_H(V), U) \cong \operatorname{Hom}_H(V,\operatorname{Res}(U))$ naturally, for $G$-modules $U$ and $H$-modules $V$. -For locally compact groups, there is a theory worked out by MacKey and others. Actually, I have only read Rieffel's work on the subject (as I come from a functional Analysis background). For a locally compact $G$ and closed subgroup $H$, there is a very satisfactory notion of the functor $\operatorname{Ind}^G_H$ (where we consider "Hermitian modules", i.e. unitary representations on Hilbert spaces). What I don't see is how (or even if) this relates to the restriction functor? - -In the topological setting, are $\operatorname{Ind}^G_H$ and $\operatorname{Res}$ in any sense adjoints? - -A slightly vague rider-- if (as I suspect) the answer is "no", can we be more precise about why the answer is no? - -REPLY [9 votes]: In fact the right relation is -$$ Hom_G ( - , Ind_H^G - ) = Hom_H( Res - , -).$$ -For compact group, it does not really matter by the Peter Weyl theorem, but it is essential as soon as you omit compactness. -I want to add that Frobenius reciprocity usually boils down to tensor-hom adjointness, so essentailly the nuclearity of the module category is necessary. For a positive answer, you have to be more restrictive. -For Mackey's theory of induced representation and its variants of Frobenius reciprocity, have a look at Baruk&Raczka "Theory of group representations and applications" pg. 549. -For the locally profinite groups, there is a really nice treatment in Bushnell-Henniart "Local Langlands on GL(2)" on pg. 18ff. -To add upon Jesse Peterson's example: If $G$ is locally compact, then -$$ Hom_G( L^1 G , 1_G) \cong \mathbb{C} \cong End_H(1_H).$$ -(Hint: The Haar integraal is the only element on the left hand side.) -So the category of Hilbert spaces is not for all purposes the prefered one, you want to work in.<|endoftext|> -TITLE: Simple Random Walk on a Locally Finite Graph - when is it recurrent? -QUESTION [14 upvotes]: I'm giving a talk tomorrow about a result in computer science which I recently proved. It's a recurrence-transience result on a random process which is related in spirit to a simple random walk. My result works on any bounded degree locally finite graph, and I'd like to discuss the analogy to random walks there as well. Unfortunately, I've never studied random walks on graphs other than lattices. It's easy to define a random walk on an arbitrary locally finite graph (if there are $d$ edges out of $v$ then assign each probability $1/d$). Let's assume a very standard hypothesis: there is a fixed $D<\infty$ s.t. the degree of each vertex is less than $D$. Definitions from Doyle-Snell are given at the bottom of this post. -One thing which is known is Doyle and Snell's Theorem: if $G$ can be drawn in a civilized manner in $\mathbb{R}$ or $\mathbb{R}^2$ then the simple random walk on $G$ is recurrent. -Unfortunately, this leaves the question open for graphs which can't be drawn this way. An exercise in Doyle and Snell asks you to find a graph which can't be drawn this way but which is still recurrent. There is some discussion which says if a graph can be drawn in a civilized manner in $\mathbb{R}^3$ and that we can embed $\mathbb{Z}^3$ into a $k$-fuzz of $G$ then $G$ is transient. The rest of their paper gets heavily into the language of flows and I don't really understand it. It doesn't seem to finish the classification. - -What's the current state of knowledge on recurrence vs. transience for random walks on a bounded degree locally finite graph? Has anyone figured this out for graphs other than those studied by the papers mentioned here? - -I also found a paper by Carsten Thomassen which basically says if a graph grows slower than $\mathbb{Z}^2$ then it's recurrent (due to Nash-Williams originally) and if it satisfies an isoperimetric inequality slightly stronger than that of $\mathbb{Z}^2$ then it's transient. I don't understand this paper at all, but I'd be curious to know if it covers a larger class of graphs than Doyle and Snell's results do. -EDIT: -As my audience in this talk is going to be all graph theorists, I'm mostly seeking a purely graph theoretic characterization of recurrent graphs. So for me, "infinite resistance from any point to infinity" is not good enough. I'd rather have a criterion which doesn't require me to choose an embedding and set resistances. Perhaps there is no hope of getting such an answer, but Vincent's answer below makes me believe this problem has been studied outside the context of electrical networks. -Also, based on Vincent's comment, I'm making the definitions clearer: -Doyle-Snell page 105: $G$ can be drawn in a civilized manner if its vertices can be embedded into $\mathbb{R}^d$ so that for some $0 -TITLE: Flatness for infinity functors -QUESTION [8 upvotes]: It is well known that for ordinary categories, if $C$ has finite limits and $D$ is cocomplete, and -$A:C \to D $ is left-exact (i.e. preserves finite limits) then the left-Kan extension of $F$ along the Yoneda embedding $y:C \hookrightarrow Set^{C^{op}}$ is left-exact. I'm pretty sure this is still true for $\left(\infty,1\right)$-categories, once we replace the role of presheaves with that of $\infty$-presheaves, but is this written up somewhere? - -REPLY [2 votes]: For reference, at least when $D$ is an infinity topos, which I believe is probably necessary, this is Proposition 6.1.5.2 in HTT.<|endoftext|> -TITLE: More on completion/compactification of open manifolds -QUESTION [7 upvotes]: This is a follow up to one of my previous questions (81714). Suppose that $M$ is an open manifold, say with a single end. Previously, I was concerned with realizing $M$ as the interior of a compact manifold with boundary $\bar{M}$, essentially adding a boundary to $M$. The responses there were quite helpful, but even after some more reading I hesitate to say that I have a satisfactory understanding of the situation. -Now I would like to consider a similar question in a slightly different direction. I wonder how to realize $M$ as a dense open subset of a compact manifold $\bar{M}$, now without boundary. That is, the extra points $C=\bar{M}\setminus M$ are all interior points of $\bar{M}$. My question is two-fold. -a. Given $M$, what are the possibilities for $C$ and $M$ as topological spaces/manifolds? Is there an analog equivalence relation between possible choices for $C$ like $h$-cobordism for potential boundaries (as in my previous question)? -b. If $C$ and $M$ are somehow fixed in advance (one example of how, though not the exact one I'm most interested in, is to consider $C$ to be a set of points supplied by Cauchy completion with respect to some metric on $M$), What do I need to give $\bar{M}=M\cup C$ manifold structure? In the previous case of adding $C$ as a boundary, I would need a collar neighborhood of the end of $M$ of the form $C\times[0,1)$. But it is not clear to me what I would need to give $C$ the structure of a set of interior points of $\bar{M}$. -Again, I would be interested in answers both in the smooth and topological categories. Specific pointers to the literature are also appreciated. - -REPLY [7 votes]: If $M$, connected and without boundary but not compact, can be embedded as the interior of a smooth compact manifold $M'$ with boundary, then I believe it can be embedded as a dense set in a smooth closed manifold. Can't you just glue two copies of $M'$ together along the boundary and then let $C$ be a "spine" in the second copy of $M'$? -And $\mathbb R^n$ can be embedded densely in any connected closed $n$-manifold, can't it?<|endoftext|> -TITLE: Is there a nice application of category theory to functional/complex/harmonic analysis? -QUESTION [67 upvotes]: [Title changed, and wording of question tweaked, by YC, because the original title asked a question which seems different from the one people want to answer.] - -I've read looked at the examples in most category theory books and it normally has little Analysis. Which, is strange as I've even seen lattice theory be used to motivate a whole book on category theory. -I was wondering is there a nice application of category theory to functional analysis? -It's weird as read that higher category theory is used in Quantum mechanics as it foundation, yet QM has heavy use of Hilbert spaces. -Sorry for cross posting to MSE. However, somebody there suggested I posted it here to get better response (well, more than two replies). I'm surprised that most books on category theory have very little mention of Analysis. Especially since Grothendieck originally studied functional analysis. - -REPLY [10 votes]: I would suggest that the following three applications of category theory to functional analysis can be useful (they have points of contact with some of the earlier answers): they concern three topics (which are related) and at the least provide a unifying thread---in my opinion, they do more---to many themes in abstract analysis (spaces of measures, distributions, analytic functionals, the Riesz representation theorem and Gelfand-Naimark duality). These are: extensions of categories, free topological vector spaces and extended duality. -Extensions of categories. The basic example is the extension of the category of Banach spaces (with continuous linear mapings as morphisms) to the class of locally convex spaces (Wiener) and convex bornological spaces (Buchwalter and Hogbe-Nlend). The first has, of course, long occupied a place in the mainstream of functional analysis, the latter less so. They can be regarded as the categories obtained by "adding inductive, respectively, projective limits. This informal notion has been formalised in the first edition of the book "Saks spaces and applications to functional analysis". Exactly the same process can be applied to the category of metric spaces (where we obtain that of uniform spaces), that of Banach algebras (convex bornological algebras and locally multiplicatively convex algebras). -The examples can be multiplied indefinitely. We mention two further extensions which we shall refer to below---compactological spaces (Buchwalter) (inductive limits of compact spaces) and Saks spaces (adding projective limits to the category of Banach spaces with linear contractions as morphisms---paradoxically, despite the fact that this category {\it has} limits). These example shows that this process often leads to us rediscovering America. However, it does have the advantage over Columbus that we are rediscovering a plethora of continents -with a single expedition, often ones which are known but have hardly been investigated so that a rescrutinising may be worthwhile (after all, Columbus himself had been anticipated by native Americans). -Free locally convex spaces. The basic example here is the following. If we start with the unit interval $I$, then we can consider the free vector space over this set. If we then supply it with the finest locally convex topology which agrees with the original one on $I$ and complete it, we obtain a locally convex space which has the universal property that every continuous function from $I$ into a Banach spaces lifts to a unique continuous linear operator thereon (and is characterised by this property). Not surprisingly, this is just the space of Radon measures on $I$. Our point is that this construction can be carried out in an infinity of analogous situations and leads to a unified aproach to a large array of spaces which are of interest in analysis. We mention -a small sample---compact spaces (universal property for continuous functions), compactologies and completely regular spaces (bounded, continuous functions), metric spaces (bounded Lipschitz-continuous functions), uniform spaces (bounded uniformly continuous functions), compact rectangles in $n$-space ($C^\infty$-functions), -compact manifolds (again $C^\infty$-functions), open subsets of $n$-dimensional complex -space (holomorphic functions). These lead to a long list of interesting spaces (of bounded Radon measures, of uniform measures, of distributions, of analytic functionals) -and many of their basic properties can be deduced from general principles which arise from this method of construction (most important example, duality theorems). Again, many of these spaces are known but the historical path to their discovery was long and stony. It is of advantage to have a natural unified approach to their construction. Again, there -aare further important cases which are indeed known but seem to have passed out of the mainstream despite an obvious demand for them. A particularly important and, in my opinion, unfortunate example (and a perennial favourite for queries in this forum) -is the topic of extensions of the Riesz representation, which we shall now discuss. -Extensions of duality. As we have just mentioned, the classical example is the Riesz representation theorem for compacta. It was initially shown by Buck (for locally compact space and later, by other researchers, to the class of completely regular spaces) that this can be extended to the non-compact case by using the so-called strict topology which can be most succinctly described here as the finest locally convex topology on the space $C^b(S)$ of bounded, continuous fuctions on $S$ which agrees with that of compact convergence on the unit ball for the supremum norm. Again a number of queries on this forum suggest that this topic which barely entered into the mainstream despite the prominence of its proponents (Buck, Beurling and Herz---mainly motivated by questions in harmonic analysis) and, sadly, seems to have vanished without a trace. -There is a simple and natural scheme at work here. If we have a duality between two of the central catogories of analysis, then we can extend it in an obvious way to one between say the category obtiained by adding inductive limits to the first one and projective limits to the second one. This leads almost automatically to the above extension of the Riesz representation to the case of completely regular spaces (or, better, compactological spaces). If we take as our starting point one of the central dualities of abstract analysis (that between a Banach space and its dual as a Banach space, or, if one wants a symmetric duality, as a Waelbroeck space, between a compact space and the Banach space of continuous functions thereon, resp. the same space regarded as a $C^*$-algebra, between a metric space and the space of bounded, Lipschitz functioons, then we can apply this process to obtain a large classes of extended dualities which are useful in abstract analysis. -Why is this useful? Lack of space prevents an elaborate justification of these three methods but I would like to mention the following version of Occam's rasor. One can speculate that one of the reasons for the fact that many of these extended dualities failed to enter into the mainstream of abstract analysis lies in the fact that to analysists accustomed to the Banach space settings, the structures employed here seemed unattractively elaborate and artifical, not to say baroque. (The strict topology is not only not normable, but is not a member of the accepted "nice" classes of locally convex spaces---Frechet, $DF$-, even barrelled or bornological. Of course, any such extension must necessarily be more elaborate than the original duality and the above considerations show that the ones presented here are the simplest that can succeed in the given situations and so are inevitable.<|endoftext|> -TITLE: decidable fragments of first-order logic without the finite countermodel property -QUESTION [7 upvotes]: Say a set of sentences in first-order logic has the finite countermodel property if any sentence in the set that is falsifiable is falsifiable on a finite domain. (Two examples: the set of sentences with only unary predicates; the dual class to the Bernays–Schönfinkel class, i.e., those sentences in prenex form whose universal quantifiers, if any, precede existential quantifiers.) It's obvious that logical validity for any such set is decidable. -But is the converse true? Or are there decidable fragments of first-order logic that lack the finite countermodel property? (I think there ought to be: I'm imagining sentences that may be false only on infinite domains -- i.e. which admit only infinite countermodels -- but which nevertheless permit the computation of an upper bound on the time it takes some semidecision procedure to finish. Though I suppose my question amounts to asking whether that's possible.) - -REPLY [7 votes]: If you restrict attention to the traditional prefix–vocabulary classes, validity in the following fragments is decidable without having the finite model property (note that it is customary in the literature to discuss satisfiability rather than validity, hence you will find there the dual classes): - -Full FO in a language with equality, unary predicates, and a single unary function. -The prefix class $\forall^{*}\exists\forall^{*}$ (i.e., sentences in prenex normal form with only one existential quantifier) in a language with equality, arbitrary predicates, and a single unary function. -Any prefix class with a finite prefix in a fixed language with finitely many relations and no functions. (This is a trivial case: if you further normalize the matrix to CNF, there are only finitely many formulas in the class.) - -A nice survey is in this lecture by Erich Grädel. A comprehensive reference is: E. Börger, E. Grädel, Y. Gurevich, The Classical Decision Problem, Springer, 1997 (reprinted 2001), MR1482227.<|endoftext|> -TITLE: Jacobian Conjecture for unit triangular matrices -QUESTION [8 upvotes]: This question is about the Jacobian conjecture for a special case. I will first explain the Jacobian conjecture (since it is something every mathematician should know about). -Let $k$ be an algebraically closed field. -Consider a map $$F: k^n \rightarrow k^n,$$ -defined by $$F(x_1,\ldots,x_n)=(f_1(x_1,\ldots,x_n),\ldots,f_n(x_1,\ldots,x_n)),$$ -where $f_1,\ldots,f_n$ are polynomials. -The Jacobian of $F$, which I denote $J$, is the determinant of the matrix $dF$ where the $(i,j)$-th entry of $dF$ is $\partial f_i/\partial x_j$. (The matrix $dF$ gives the induced map on the tangent bundle, or maybe it's the cotangent bundle; it doesn't matter for this question.) -Since $F$ is given by polynomials, the entries of $dF$ are polynomials and $J$ is a polynomial. Hence $F$ is a nonsingular map if and only if $J$ is a constant. -The inverse function theorem tells us that $F$ has a smooth inverse map if and only if $J$ is constant. The Jacobian conjecture says that this smooth map is in fact also given by polynomials (in the case where the original map $F$ is given by polynomials). -Question: I would like to know if the Jacobian conjecture is known (or trivial) for the special case where the matrix $dF$ is a triangular matrix with $1$'s on the diagonal. If it is not known, I would like to know if the full Jacobian conjecture is known to be equivalent to this special case. -Motivation: This is a possible strategy for proving that a particular family of maps I have constructed for a particular purpose is in fact invertible within the category of affine algebraic varieties. -EDIT: Clarifying in light of Tom's remark. The inverse function theorem just says that $F$ has a local inverse. The Jacobian conjecture is that $F$ has a global inverse which is given by polynomials. - -REPLY [18 votes]: More simply: -(I'll write this down for the case $n=3$ because writing and reading subscripts makes me tired.) -Let $(u,v,w)=F(x,y,z)$. By hypothesis -$u-x$ has derivative $0$ with respect to $x$, so -$$u=x+P(y,z)$$ -for some $P$. And $v-y$ has derivative $0$ with respect to $x$ and $y$, so -$$v=y+Q(z)$$ -for some $Q$. And $w-z$ has derivative $0$ with respect to all three variables, so -$$w=z+R$$ -for some constant $R$. -Now just write down the inverse: -$$z=w-R$$ -$$y=v-Q(z)=v-Q(w-R)$$ -$$x=u-P(y,z)=u-P(v-Q(w-R),w-R).$$<|endoftext|> -TITLE: Weak operad and deloopings -QUESTION [5 upvotes]: Let $E$ be an operad in topological spaces. $E$ is usually called an $E_{\infty}$-operad, if all the spaces $E_n$ are contracticle. If $E$ acts on a space $X$, then by the recognition principle, $X$ turns out to be an infinite loop space. - -How much of the theory is preserved, if I replace contractible by weakly contractible? Do I still have deloopings, if $E$ acts on a space? - -REPLY [6 votes]: To answer your question and complete Justin's answer, one can pick a cofibrant replacement of $E$ to get a genuine $E_\infty$-operad $F$ acting on the space $X$, and conclude that $X$ has an infinite delooping up to group completion issues. -But the other way round fails: not all infinite loop spaces are acted on by $E$. Example: any group like commutative monoid is an infinite loop space, but not all infinite loop spaces are commutative monoids (otherwise no non-trivial Dyer Lashof operation would exist). -Remark: If $E$ is weakly-contractible, then the constant maps $c: E_n\rightarrow pt$ define an acyclic fibration, in the category of topological operads, from the operad $E$ towards the operad of commutative monoids $C$. -If you have a preferred (cofibrant) $E_\infty$-operad $F$, then you can use the LLP to get an operad weak-equivalence $f: F\xrightarrow{\sim} E$, making $F$ a cofibrant replacement of $E$.<|endoftext|> -TITLE: Voevodsky's counterexample to the existence of a motivic t-structure -QUESTION [24 upvotes]: I have been trying to unravel some of the known relationships between various ideas on mixed motives. I find the literature quite hard to follow -"from experts, for experts". -Voevodsky in "Triangulated categories of motives over a field" [4.3.8] shows that there is no reasonable t-structure on his category $DM^{eff}_{gm}(k)$ (for $k=\mathbb Q$ or most fields, characterized it seems by a certain cohomological dimension condition which flies above my head) but I do not quite understand the meaning of his proof, in particular how he came up with it, though I more or less follow individual steps. Can anyone help me? -Related to this are Bruno Kahn's remarks in his review article in the "Handbook of K-theory" that the etale version of $DM_{gm,et}^{eff}(k)$, with Homsets isomorphic to $DM^{eff}_{gm}(k)$'s after tensoring by $\mathbb Q$ (using some motivic scissoring magic I presume), should have a t-structure whose heart should be (equivalent to) Nori's category. Kahn says this is related to the Hodge conjecture (not "Hodge-type standard"), can anyone flesh this relation out here? -So I wonder what happens when we pass from $gm$ to $gm,et$. Does anybody know? Is it related to (Serre-type) supersingularity-based counterexamples to the existence of a Weil cohomology theory with $\mathbb Q$ or $\mathbb Q_p$ coefficients? -Also, what has been done to relate the cohomological t-structure on the bounded derived category of Nori mixed motives to $DM_{gm,et}^{eff}(k)$? ($DM_{gm}^{eff}(k)$ and $DM_{gm,et}^{eff}(k)$ have a canonical functor to $D^b(NMM(k))$, which should be an equivalence after tensoring with $\mathbb Q$ -Beilinson's "mixed motives" conjecture.) -Are these questions related to CM lifting results of Chai-Conrad-Oort? Do their constructions explain why there is no t-structure for Nisnevich triangulated motives but there is for etale triangulated motives? Should Voevodsky's example be interpreted in that context? -As you see I am starting to divagate, so I would greatly appreciate some enlightenment. -I would also appreciate any comment on the feeling (probably quite ill-informed) that the t-structure on $DM_{gm,et}^{eff}(k)$ should not be too hard to construct but rather a matter of technical mastery of the algebra/arithmetic involved. -And along this line, would the existence of that t-structure yield insight on the Tate and Hodge conjectures, or other conjectures on algebraic cycles? Those seem harder but to confirm (and following Kahn's remark mentioned above): does the Hodge conjecture imply the existence of the motivic t-structure on $DM_{gm,et}^{eff}(k)$? -Finally, I hesitate asking more but... Has anything been tried regarding "bootstrapping" the thorough understanding we have of t-structures on Tate motives to construct t-structures on larger triangulated categories of motives? I think I remember something from Déglise, I have to check... And have those constructions on Tate motives been related to the Nori-Kontsevich tannakian philosophy -e.g. to justify/formalize a hypothetical bootstrapping? - -REPLY [21 votes]: I will try to give some answers. -Voevodsky proved that there could be no 'reasonable' motivic $t$-structure for motives with INTEGRAL coefficients (over a non-algebraically closed field); note that there seems to be no nice $t$-structure already for the (very small!) triangulated subcategory of Artin's motives (generated by motives of varieties of dimension $0$). Most people believe that for the 'rational hull' of this category i.e. for motives with coefficients in $\mathbb{Q}$ the motivic $t$-structure does exist, both for the category of effective geometric motives and for the whole category of geometric motives. Now, for motives with rational coefficients there is no difference between etale and Nisnevich versions of Voevodsky's motives. There is a difference only for motives with integral coefficients; in this setting there is a 'reasonable' $t$-structure for etale Artin motives. Serre's example seems to have nothing to do with this question. -The existence of the motivic t-structure is certainly a very hard conjecture (already for motives with rational coefficients). You need a long list of conjectures on algebraic cycles in order to deduce it. You can find such a list here: http://mrlonline.org/mrl/1999-006-001/1999-006-001-005.pdf -Note that the Hodge conjecture implies all the Grothendieck's standard conjectures over base fields of characteristic 0, but it says nothing about the Murre's conjectures. -Some remarks. -1. Hanamura considers his own category of motives; yet I proved that it is (anti)-equivalent to Voevodsky's motives. -2. There is also the category of Nori's motives. As far as I remember, this is an abelian category. Yet it is certainly a very hard question whether the derived category for it is equivalent to Voevodsky's motives. -3. Another interesting reference is http://arxiv.org/PS_cache/arxiv/pdf/1006/1006.1116v2.pdf<|endoftext|> -TITLE: The main theorems of category theory and their applications -QUESTION [65 upvotes]: This question first arose as I wrote an answer for the question: Is there a nice application of category theory to functional/complex/harmonic analysis?; it can also be regarded as a (hopefully) more focused version of the question Most striking applications of category theory?. - -It seems that category theory began as an organizational tool in topology and algebraic geometry, but by now it has grown into an area of research in its own right with applications all over the place in mathematics. I realized as I was thinking about my answer to the question above that I did not know the statements of the main results in category theory, or even if there are "main results". In a comment, Martin Brandenburg suggested several examples: - -The general adjoint functor theorem -Freyd's representability criterion -Beck's monadacity theorem -Recognition theorems for locally presentable categories -Brown's representability theorem - -He also indicated that these results have numerous unsung applications to other areas of mathematics. This question is essentially an invitation for Martin and anyone else add to this list and explain some of the applications of the items on it. I would not have asked this question on mathoverflow if I believed that such a list already existed; if I am wrong then the question should probably be closed. - -Here is what I have in mind for an answer to this question. It should include the statement of a theorem in pure category theory (ideally using language which is friendly to outsiders) and at least one application to another area of mathematics. The community wiki rule "one theorem per answer" makes sense here, particularly so that others can conveniently add applications to your list. -When I say "theorem in pure category theory", I don't insist that the result be incredibly nontrivial, just that it is a result which is stated and proved in the language of category theory. For example, the statement that $\pi_1$ is a functor belongs to topology, not category theory; on the other hand the Yoneda lemma counts even though it is a "lemma" instead of a "theorem". -When I say "application to another area of mathematics" I am ideally looking for statements which can be formulated without using the language of categories and functors. I want to be clear that I am interested in applications of specific results in category theory, not just results for which categorical thinking is useful (such results are everywhere). - -REPLY [5 votes]: Freyd-Mitchell and Gabriel-Popescu theorems, and also the characterization of co-Grothendieck cats.<|endoftext|> -TITLE: Hyper-complex and quaternionic Kähler Geometry -QUESTION [8 upvotes]: What is the exact relationship hyper-complex and quaternionic Kahler manifolds? From Wikipedia we get that hyper-Kahler manifolds are both hyper-complex and quaternionic Kahler. Thus, the two families have a non-empty intersection. But this is all I can conclude. -Moreover, quaternionic projective space is quaternionic Kahler, but is it also hyper-Kahler? Is it even hyper-complex? - -REPLY [9 votes]: By a recent result of Gauduchon, Moroianu and Semmelmann if a positive quaternion Kahler manifold admits any almost complex structure then it must be the complex Grassmannians $Gr_2(\mathbb C^{n+2})$. The latter is not hypercomlex so there are no hypercomplex manifolds which are also positive Quaternion Kahler. Note that it's even easier if you assume that the hypercomplex structure is compatible with the quaternion Kahler structure which the above argument does not assume but I think the original question did assume. That implies that the twistor bundle is trivial which is well-known not to be possible for positive quaternion Kahler manifolds. I don't know what happens in the negative quaternion Kahler case but I suspect there are no examples there either at least among compact ones. So I would guess that any compact hypercomplex manifold with holonomy in $Sp(n)\cdot Sp(1)$ must be hyper-Kahler.<|endoftext|> -TITLE: cospectral graphs -QUESTION [7 upvotes]: The simple connected graph $G$ has $n$ vertices and we have: -1) $|E(G)|‎\geq‎ \frac{n(n-1)}{3}$ -2) we have the spectrum and degree sequence of $G$ -3) $Spectrum(G)=Spectrum(H)$ -Is $G \cong ‎H$? - -REPLY [2 votes]: Connected strongly regular graphs with parameters $(v,k,\lambda)$ have eigenvalue $k$ appearing once and two other eigenvalues with prescribed multiplicity. In general there are many non-isomorphic graphs for a fixed parameter. There are two graphs with parameters $(16,6,2)$ and 15 with parameters $(25,12,5)$. The number of non isomorphic graphs can grow dramatically.... -All strongly regular graphs with parameter $(v,k,\lambda)$ are regular (so the degree sequence is obvious) and are cospectral. Either the graph or its complement meets your bound on the number of edges. -Andries Brouwer maintains a nice webpage on strongly regular graphs here. (Check out the number of $(36,15,6)$ strongly regular graphs!)<|endoftext|> -TITLE: Topological mean value function. -QUESTION [6 upvotes]: My friend posed a question that how we can generalize the notion of mean value on topological spaces, preferably Poincare complexes, $X$, so that it satisfies cheap properties of mean value on Euclidean spaces. More precisely, we want sequence of continuous maps $ m_n:X^n\rightarrow X$ for every $n$ such that it is symmetric ( or if you like $m_n: SP_n(X)\rightarrow X$ ) and also $ m_n\circ \Delta = id_X$ meaning that mean value of identical points should be the point itself. It seems even these properties are highly restrictive on topology, for example it is easy to show that we cannot define mean value on spheres. I wonder if there exists non-acyclic finite dimensional complex that admits mean value function? - -REPLY [6 votes]: I am not sure whether "acyclic" means connected for you. If yes, then there are obvious counterexamples (finite spaces with the discrete topology). So, assume that the space is connected. If you assume that the space $X$ is a compact, connected polyhedron, then the existence of even one $m_n,$ for $n\geq 2,$ then $X$ is contractible. This is a Theorem of Eckmann (1954). Both assumptions are necessary: The Eilenberg-McLane Space $K(\mathbb{Q}, m)$ admits an $n$-mean for EVERY $n$ (so is an example of the sort you seek), but is not compact. The $n$-solenoid admits an $n$-mean but is not a polyhedron. For, much, much more on this subject see: -MR1490716 (98k:55013) -Hilton, Peter(1-CFL) -A new look at means on topological spaces. (English summary) -Internat. J. Math. Math. Sci. 20 (1997), no. 4, 617–620. - -REPLY [3 votes]: I think there's a mod two homology obstruction. Let's assume $X$ is a $k$-dimensional manifold, compact and without boundary, and let $\mu \in H_k(X, \mathbb Z_2)$ be its fundamental class. Then $\mu$ maps to some element of $H_k(X^n,\mathbb Z_2)$ under $\Delta_*$, and by your symmetry assumption, it has to lie in the diagonal subspace of $H_k(X^n,\mathbb Z_2) \simeq \prod_n H_k(M;\mathbb Z_2)$ (Kunneth), which means that when you further apply $m_{n*}$, the element of $H_k(X,\mathbb Z_2) \simeq \mathbb Z_2$ you get would be represented by a multiple of the integer $n$. So provided $n$ is even, this is a contradiction. If your manifold were orientable it would give you an obstruction for any $n \geq 2$.<|endoftext|> -TITLE: abelian centralizers in almost simple groups -QUESTION [6 upvotes]: Hallo! -I'm looking for a reference. I'm sure that the information I need is already in the literature but I'm having some trouble to find it. Here is the question. -Let $S$ be a non-abelian finite simple group (the only case I'm really interested in is for groups of Lie type) and let $A$ be the automorphism group of $S$. For which primes $p$, every element of order $p$ in $S$ has $C_A(x)$ (the centralizer of $x$ in $A$) abelian? -Typically, one might think that $p$ is a primitive prime divisor of $q^n-1$ (where $q$ is the size of the defining field of $S$ and $n$ is, roughly, its Lie rank). However, already for these elements I'm not able to find any reference for $C_A(x)$ (despite the fact that lots is known on $C_S(x)$). - -REPLY [5 votes]: As comments by other people suggest, this kind of question requires a lot of case-by-study, even for groups of Lie type. In the latter groups, regular semisimple elements certainly get involved when the prime is different from the defining one. The regular unipotent elements in $\mathrm{PSL}_2(\mathbb{F}_p)$ should also be looked at. But in general irregular elements already have non-abelian centralizers in the finite simple group. -Concerning references, there are many kinds of books and papers which provide detail about the classes and centralizers in finite simple groups. Especially when there are no interesting outer automorphisms, these sources should settle your question case-by-case. Among the books is Part 3 of the ongoing AMS series by Gorenstein-Lyons-Solomon on classification of finite simple groups: they give a vast amount of detailed information about known groups, along with references. There is a lot of algebraic group literature which treats the finite groups explicitly, going back to Steinberg's Yale lectures on Chevalley groups and the detailed article by Springer and Steinberg in LN 131 based on the 1968-69 program at IAS. Much other relevant work has gone on over the years, including another AMS book to appear soon by Liebeck and Seitz Unipotent and Nilpotent Classes in Simple Algebraic Groups and Lie Algebras. -I'll try to add more specific comments on your question, but I should emphasize the need to formulate it as narrowly as possible in view of the extra difficulties involved in studying sporadic groups as well as groups of Lie type for bad primes. -(Mentioning some motivation for the question might also be helpful.)<|endoftext|> -TITLE: The second homotopy group of a simple CW-complex -QUESTION [17 upvotes]: Let $X$ be a CW-complex with - -one 0-cell -two 1-cells -three 2-cells -no cells in dimensions 3 or higher. - -Is it always true that $\pi_2(X)\ne 1$? - -REPLY [14 votes]: There are classic examples, coming from small cancellation theory. See -the section of the Wikipedia article on asphericity.<|endoftext|> -TITLE: Smooth four-manifolds with contractible universal cover -QUESTION [11 upvotes]: Let $X$ be a smooth compact four-manifold with definite non-trivial intersection form. Can the universal cover of $X$ be contractible? -It semms to me that the answer is negative when $X$ is simply connected using results of Freedman and Donaldson. Is anything known when $X$ is not simply connected? Donaldson proved that also in this case the intersection form is diagonalizable over $\mathbb Z$. - -REPLY [14 votes]: In algebraic geometry, there are examples of "fake projective planes", which in this context means smooth complex surfaces of general type with the same cohomology ring as the complex projective plane. It is known that the universal cover of such spaces is the complex hyperbolic ball. So the answer to your question is yes. (The first such fake projetive plane was shown to exist by Mumford.)<|endoftext|> -TITLE: Simultaneously minimizing intersections -QUESTION [5 upvotes]: This may be a standard problem in homotopy theory, but I don't know a good reference. -Let $\Sigma$ be a smooth, oriented surface, and let $X_1,X_2$ and $X_3$ be three smoothly embedded curves in $\Sigma$. Can I find smoothly embedded curves $Y_1,Y_2$ and $Y_3$ such that... - -$Y_i$ is homotopic to $X_i$ through smooth embeddings, -each intersection in $Y_1\cup Y_2\cup Y_3$ is a transverse double intersection, and -the number of intersections between $Y_i$ and $Y_j$ is minimal among all pairs homotopic to $X_i$ and $X_j$ through smooth embeddings? - -Roughly speaking, I have three curves, and I want to use homotopy to simultaneously minimize the number of intersections between each pair of curves. - -REPLY [5 votes]: Firstly, "homotopic through smooth embeddings" is known as "isotopic", and there is a well known paper of David Epstein titled "Curves on 2-manifolds and isotopies" where he shows that homotopy is equivalent to isotopy in this setting. -Secondly, if you put a hyperbolic structure on your surface, then the curve shortening flow will isotope your curves into geodesics, which will automatically minimize intersection numbers (this can be seen by going to the hyperbolic plane as in the Klein model, and seeing what intersection numbers mean in terms of lifts and cross-ratios), and will also minimize self-intersection numbers, so will stay embedded if your curves are to begin with. -Thirdly, the geodesics, as above, might have triple intersections, but it is clear (by transversality, if you like) that a small local perturbation will remove the triple points (this would be less obvious if your curves $X_i$ were not embedded, but they are.)<|endoftext|> -TITLE: What is the sandpile torsor? -QUESTION [46 upvotes]: Let G be a finite undirected connected graph. A divisor on G is an element of the free abelian group Div(G) on the vertices of G (or an integer-valued function on the vertices.) Summing over all vertices gives a homomorphism from Div(G) to Z which we call degree. -For each vertex v, let D(v) be the divisor -$d_v v - \sum_{w \sim v} w$ -where $d_v$ is the valence of v and $v \sim w$ means "$v$ and $w$ are adjacent." -Note that D(v) has degree 0. The subgroup of Div(G) generated by the D(v) is called the group of principal divisors. We denote by Pic(G) the quotient of Div(G) by the group generated by the principal divisors, and by Pic^0(G) the kernel of the degree map from Pic(G) to Z. -The notation here suggests that I am really thinking about algebraic curves, not graphs; and that's in some part true! In the work of Matt Baker and his collaborators you can find a really beautiful translation of much of the foundational theory of algebraic curves (Riemann-Roch, Brill-Noether, etc.) into this language. -But that's not really what this question is about. -Lots of people study this abelian group, maybe most notably statistical physicists and probabilists who study dynamical processes on graphs. In those communities, Pic^0(G) is called the sandpile group, because of its relation with the abelian sandpile model. -But that's also not really what this question is about. -What this question is about is the following fact: by the matrix-tree theorem, the number of spanning trees of G is equal to |Pic^0(G)|. -When one encounters a finite set that has the same cardinality as a finite group, but the set does not have any visible natural group structure, one's fancy lightly turns to thoughts of torsors. So: -QUESTION: Is the set S of spanning trees of G naturally a torsor for the sandpile group Pic^0(G)? If so, how can we describe this "sandpile torsor?" -(By "naturally" we mean "functorially" -- in particular, this torsor should be equivariant for the automorphism group of G.) -That question is rather vague, so let me make it more precise, and at the same time try to argue that in at least some cases the question is not ridiculously speculative. The paper "Chip-Firing and Rotor-Routing on Directed Graphs," by (deep breath) Alexander E. Holroyd, Lionel Levine, Karola Meszaros, Yuval Peres, James Propp and David B. Wilson, contains a very interesting construction of a "local" torsor structure for the sandpile group. Suppose G is a planar graph -- or more generally any graph endowed with a cyclic ordering of the edges incident to each vertex. Then the "rotor-router process" described in Holroyd et al gives S the structure of a Pic^0(G)-torsor! (See Def 3.11 - Cor 3.18) This would seem to answer my question; except that the torsor structure they define depends, a priori, on the choice of a vertex of G. A better way to describe their result is as follows: for each v, let S_v be the set of oriented spanning trees of G with v as root. Then the rotor-router model realizes S_v as a torsor for the group Pic(G) / $\mathbf{Z}$ v. -But S_v is naturally identified with S (just forget the orientation) and the natural map Pic^0(G) -> Pic(G) / $\mathbf{Z}$ v is an isomorphism. So for each choice of v, the rotor-router construction endows S with the structure of Pic^0(G)-torsor. Now one can ask: -QUESTION (more precise): Are the torsor structures provided by the rotor-router model in fact independent of v? Do they in fact provide a Pic^0(G)-torsor structure on S which is functorial for maps compatible with the cyclic edge-orderings, and in particular for automorphisms of G as a planar graph? If this is false in general, is there some nice class of graphs G for which it's true? -REMARK: If you are used to thinking about algebraic curves, like me, your first instinct might be "well, surely if the set of spanning trees is a torsor for Pic^0, it must be Pic^d for some d." But I don't think this can be right. Here's an example: let G be a 4-cycle, which we think of as embedded in the plane. Now the stabilizer of a vertex v in the planar automorphism group of the graph is a group of order 2, generated by a reflection of the square across the diagonal containing v. In particular, you can see instantly that no spanning tree in S is fixed by this group; the involution acts as a double flip on the four spanning trees in S. On the other hand, Pic^d(G) is always going to have a fixed point for this action: namely, the divisor d*v. -REMARK 2: Obviously the correct thing to do is to compute a bunch of examples, which might instantly give negative answers to these questions! But it gets a bit tiring to do this by hand; I checked that everything is OK for the complete graph on 3 vertices (in which case the torsor actualy is Pic^1(G)) and then I ran out of steam. But sage has built-in sandpile routines..... - -REPLY [4 votes]: I recently wrote "Rotor-Routing Induces the Only Consistent Sandpile Torsor Algorithm Structure" with Ankan Ganguly which was inspired by this question (and, more directly, a conjecture by Klivans). As one may surmise from the title, we show that the sandpile torsors that comes from rotor-routing on plane graphs are in a sense unique. -We use the term sandpile torsor algorithm to mean a map from the set of plane graphs to the set of sandpile torsor actions (i.e. free transitive actions of the sandpile group on the spanning trees). Rotor-routing actually defines 4 sandpile torsor algorithms which have the same structure: clockwise rotor-routing, counterclockwise rotor-routing, inverse clockwise rotor-routing, and inverse counterclockwise rotor-routing. -The first major challenge in this project was figuring out how to define a consistent sandpile torsor algorithm. In particular, the torsors we obtain on different graphs should be related in some way. With this in mind, we looked at how rotor-routing interacts with contraction and deletion and found a remarkable property. -Let $G$ be a plane graph and suppose $c$ and $s$ are two vertices joined by an edge. Then $c-s$ is a representative for some equivalence class of $Pic^0(G)$. Let $T$ be a spanning tree of $G$ and $T'$ be the spanning tree we get by acting on $T$ by $c-s$ using rotor-routing. - -For an arbitrary $e \in T \cap T'$ (not incident to both $c$ and $s$), consider the plane graph $G/e$. If we act on $T\setminus e$ by $c-s$ using rotor-routing, then we get $T' \setminus e$. -For an arbitrary $e \not\in T \cup T'$, consider the plane graph $G\setminus e$. If we act on $T$ by $c-s$ using rotor-routing, then we get $T'$. - -We use these properties to define consistent sandpile torsor algorithms (along with a third property that lets us restrict to 2-connected plane graphs). The bulk of the paper is spent proving that any sandpile torsor algorithm satisfying the properties of consistency must be equivalent to one of the 4 torsor algorithms coming from rotor-routing. -This was a really fun project and I'd love to talk more about it if anyone has any questions or comments. Thanks to Jordan Ellenberg for posing the question and to everyone who has worked toward answering it!<|endoftext|> -TITLE: Why is the definition of l-adic sheaves so complicated? -QUESTION [11 upvotes]: I find the definition of constructible $\bar{\mathbb Q}_l$-sheaves (or their derived category) on a variety of positive characteristic quite involved and ad Hoc. Roughly it goes as follows: -First one defines constructible sheaves modules over torsion rings in the naive way, then over finite $\mathbb Z_l$ extensions as certain projektive systems of sheaves with bigger and bigger torsion coefficients. To pass from finite $\mathbb Z_l$ extensions to finite $\mathbb Q_l$ extensions one tensors the hom spaces in our category with the quotient field. Finally $\bar{\mathbb Q}_l$ sheaves are certain inductive systems of sheaves over finite $\mathbb Q_l$ extensions. -Is there a good reason a priori to make the definition as it is? Why are more naive definitions flawed? - -REPLY [11 votes]: One reason why one needs a large coefficient group like $\overline{\mathbb{Q}_\ell}$ instead of $\mathbb{Q}$ is that a reasonable cohomology theory should be a functor. Sometimes varieties over finite fields admit morphisms that cannot be captured by those smaller coefficients. -The standard example due to Serre is that of a supersingular elliptic curve $E$ over a finite field. $E$ has an endomorphism ring that is a quaternion order over $\mathbb{Z}$, but elliptic curves over $\mathbb{C}$ have 2-dimensional $H^1$. If you want the cohomology theory to behave like de Rham or singular cohomology, you need an action of the quaternion order on a two-dimensional vector space such that pullback along the multiplication-by-$n$ isogeny acts as multiplication by $n$. The only way to get a quaternion algebra to act on a 2-dimensional vector space is to make sure the quaternion algebra splits over the coefficient field of the vector space, i.e., after taking a tensor product, it becomes isomorphic to a ring of $2 \times 2$ matrices. The splitting is guaranteed over algebraically closed fields, but in this case, it fails to happen over $\mathbb{Q}$, or even $\mathbb{Q}_p$, where $p$ is the characteristic of the field of definition. -As it happens, you can get away with $\mathbb{Q}_\ell$ coefficients much of the time, but if you want to examine eigenvalues of the Frobenius acting on cohomology, it helps to have solutions to polynomial equations on hand. -As some comments on the question explained, one reason why $\mathbb{Q}_\ell$ coefficients are not given by the naïve construction is that the naïve construction will not yield a comparison with singular or de Rham cohomology. Torsion-free coefficients will not see finite covers, and finite covers are the essence of étale site.<|endoftext|> -TITLE: 2-colorings of the reals -QUESTION [12 upvotes]: It's easy to prove that, if $\mathbb{R}$ is well-orderable, then there is a 2-coloring of pairs of reals with no uncountable homogeneous set, i.e., there is an $m: [\mathbb{R}]^2\rightarrow 2$ such that for all uncountable sets $U\subseteq\mathbb{R}$, there are $x, y, z\in U$ such that $m(\lbrace x, y\rbrace)\not=m(\lbrace x, z\rbrace)$. -Proof: Let $<_w$ be a well-ordering of $\mathbb{R}$, and let $m(\lbrace x, y\rbrace)=0$ if $x< y\iff x<_w y$ and $m(\lbrace x, y\rbrace)=1$ if $x< y\iff y<_w x$. Then a homogeneous set for $m$ yields a well-ordered increasing or decreasing sequence of reals with the same cardinality. But there is no uncountable increasing or decreasing sequence of reals, since the reals have a countable dense subset. So we are done. -My question is, What happens if we don't assume that $\mathbb{R}$ is well-ordered? That is: -Question 1: Does there exist a model of $ZF$ in which $\mathbb{R}$ is not well-orderable but there is a 2-coloring of $\mathbb{R}$ with no homogeneous set of the same cardinality as $\mathbb{R}$? -A somewhat related question has to do with the complexity of colorings without nice homogeneous sets: -Question 2: Does every Borel 2-coloring of pairs of reals have a Borel homogeneous set with the same cardinality as $\mathbb{R}$? Does every measurable 2-coloring of pairs of reals have a measurable homogeneous set with the same cardinality as $\mathbb{R}$? - -REPLY [15 votes]: Fred Galvin showed that if $c:[\mathbb{R}]^2\to\lbrace0,1\rbrace$ is such that $c^{-1}(0)$ and $c^{-1}(1)$ both have the Baire property, then there is a perfect set $P \subseteq \mathbb{R}$ which is $c$-homogeneous. (Note that perfect sets have size $2^{\aleph_0}$.) -Since Borel sets have the Baire property and perfect sets are Borel, Galvin's Theorem answers your second question. -Shelah has shown that it is relatively consistent with ZF+DC that every subset of any Polish space (like $[\mathbb{R}]^2$) has the Baire property. Galvin's proof seems to run in ZF+DC, so it looks like it is consistent that every $2$-coloring of $[\mathbb{R}]^2$ has a homogeneous set of size $2^{\aleph_0}$. -I just realized that I misread your first question. The answer to that question is surely yes, but I don't have a handy model to show you right now. (I'll try to find one later.) -By the way, if you want to color triples of reals, you can't always get a perfect homogeneous set. For triples, this is illustrated by the coloring $$c(x,y,z) = \left\{\begin{array}{cc}0 & \mbox{when }2y < x+z \\ 1 & \mbox{when }2y \geq x+z\end{array}\right.$$ where $x < y < z$. But Galvin showed that you can always get a perfect set whose triples assume at most two colors. Blass later showed that for colorings of $n$-tuples, you can always get a perfect set that takes on at most $(n-1)!$ colors and that this is best possible.<|endoftext|> -TITLE: Borel reduction/Wadge hierarchy -QUESTION [5 upvotes]: An equivalence relation $E$ is Borel-reducible to an equivalence relation $F$ if there is a $\Delta^1_1$ function $f$ such that $xEy$ holds iff $f(x)Ff(y)$ holds. A set $A\subset \omega^{\omega}$ is Wadge reducible to a set $B\subset \omega^{\omega}$ if there is a continuous function $g$ such that $A=g^-1[B]$. How far can the analogy between Borel-redution and Wadge-reduction be pushed? In particular is there a notion of the Borel order just like the Wadge order of a set of reals? Also can the Martin-Monk argument showing that the relation of Wadge reduction is wellfounded be replicated for the relation of Borel-reduction? If yes do projective ordinal ($AD$ in general) have any consequences on being Borel-reducible? -My motivation is that I just started reading about the area of classification problems and Borel-reduction and some concepts and definition look very similar to the Wadge hierarchy (this is kind of obvious actually because both notions are about descriptive set theoretic complexity). - -REPLY [5 votes]: Equivalence relations of the same complexity, when considered as sets, need not be mutually continuously reducible. A proof that the quasiorder of Borel equivalence relations up to continuous and Borel reducibility is ill-founded can be found in Louveau and Velickovic: 'A note on Borel equivalence relations' (1994). -Note that Borel equivalence relations with only two classes are well-ordered up to continuous reducibility and the rank of an equivalence relation is the Wadge rank of the equivalence classes. Selivanov and others have results on Borel functions with finitely many values up to continuous reducibility, generalizing continuous reducibility of Borel sets.<|endoftext|> -TITLE: Does the following condition imply the homotopy type of a wedge of spheres? -QUESTION [16 upvotes]: Let me preface this question by saying that I am not an algebraic topologist. -Motivation. I was looking with a colleague at the homotopy type of a family of posets and we were able to show using discrete Morse theory that the order complex of this poset was homotopy equivalent to a space with exactly one-cell in dimensions $0$ to $n$ and no cells of higher dimension. Moreover, we can show that the fundamental group is cyclic. Extensive computer computation shows that it is most likely true that the homology in each dimension from $0$ to $n$ is $\mathbb Z$, Thus it smells like our complex is homotopy equivalent to a wedge of spheres, one from each dimension from $1$ to $n$. Before killing ourselves to prove that the homology is as the computer suggests, we wanted to figure out if this data implies the homotopy type of a wedge. - - -Question. Is is true that if one has an $n$-dimensional CW complex $X$ with exactly one cell in each dimension (from $0$ to $n$) such that $\pi_1(X)=\mathbb Z$ and $H^q(X)=\mathbb Z$ for $0\leq q\leq n$, then $X$ has the homotopy type of a wedge $\bigvee_{q=1}^n S^q$? - -REPLY [25 votes]: No, this is false. For example take $\mathbb{CP}^2\vee S^1\vee S^3$. It admits a cell decomposition and cohomology groups as you describe but clearly has a different homotopy type then a wedge of spheres because the cohomology ring structure is different. - -REPLY [16 votes]: You can take $S^1\vee S^2\vee S^3$ and then use the Hopf fibration from $S^3$ to $S^2$ as the attaching map for a 4-cell onto the 2-cell. This has to have the cohomology (and homotopy) as you described. -(Having now seen Vitali's answer, I guess this is the same as his, but I thought I might as well give this answer anyway.)<|endoftext|> -TITLE: Monge Ampere equations -QUESTION [7 upvotes]: I am a graduate student trying to understand complex Monge-Ampere equations(mostly on complex manifolds with or without boundary, but also in C^n), but I can't put my hand on any monograph/textbook discussing this problem thoroughly. Is there anything out there that could help me? If there isn't, can any of you folks tell me with what articles I should start my reading? -Any piece of information is appreciated. - -REPLY [3 votes]: It's rather too late. However, Vincent Guedj and Ahmed Zeriahi, Degenerate Complex Monge-Ampere Equations seems to be a very carefully written book.<|endoftext|> -TITLE: Motivation behind Panyushev's "constant-averages-along-orbits" conjecture -QUESTION [8 upvotes]: In his article "On orbits of antichains of positive roots" (European Journal of Combinatorics 30 (2009) 586–594, Dmitri Panyushev discusses an interesting self-map on the set of antichains of a finite poset (also discussed earlier by D.G. Fon-der-Flaass in "Orbits of antichains in ranked posets", European J. Combin. 14 (1993) 17–22, and even earlier by A. Brouwer and A. Schrijver in "On the period of an operator, defined on antichains", Math. Centrum Report ZW 24 (1974)). -Given an antichain $A$, we define $X(A)$ as the set of minimal elements of the complement of the order ideal generated by $A$. -One of Panyushev's conjectures, Conjecture 2.1(iii), asserts that for a certain class of posets the average $(1/|O|) \sum_{A \in O} |A|$ is the same for all $X$-orbits $O$, and gives a value for this average. This conjecture was recently proved by D. Armstrong, C. Stump, and H. Thomas in their article "A uniform bijection between nonnesting and noncrossing partitions", http://arxiv.org/abs/1101.1277 . -Does anyone know of any motivation behind Panyushev's conjecture about $(1/|O|) \sum_{A \in O} |A|$? Why would one be interested in this average? It's possible Panyushev may just have noticed the pattern numerically, with no particular theoretical purposes in mind. But I can't help feeling that he saw this conjecture as fitting into a larger story. -As a related question, is anyone in MathOverflow in touch with Panyushev? I tried sending him an email at the address listed in his article (asking the above question) but received no reply. - -REPLY [6 votes]: In "The root poset and its relatives" (https://arxiv.org/abs/math/0502385) Panyushev established (see Corollary 3.4) that the average size of an antichain of the root poset $\Phi^+$ of an irreducible crystallographic root system $\Phi$ is $n/2$, where $n$ is the number of simple roots of $\Phi$. He did this just by observing that the "$\Phi$-Narayana polynomial", i.e., the generating function for antichains of $\Phi^+$ by cardinality, is palindromic, something he in fact observed in his earlier paper "ad-nilpotent ideals of a Borel subalgebra: generators and duality" (https://www.sciencedirect.com/science/article/pii/S0021869303006380) (see Section 6). -In the "ad-nilpotent ideals" paper he is very interested in the fact that the Narayana polynomials are palindromic and uses this palindromicness as evidence in favor of a certain conjectured "duality" for the ad-nilpotent ideals (which correspond bijectively to antichains of the root post). In particular he conjectures (see Conjecture 6.1) the existence of a natural involution on the set of antichains which would send an antichain of cardinality $k$ to an antichain of cardinality $n-k$. As far as I know, this conjecture remains open (see my MO question Panyushev's conjectured duality for root poset antichains). -He couldn't prove the existence of that duality, but in the "On orbits of antichains of positive roots" paper (https://arxiv.org/abs/0711.3353) he accomplished something close using "rowmotion"/"the Fon-der-Flaass operator". Namely, he conjectured a specific way to partition the set of antichains of $\Phi^+$ into "small" sets (of size dividing $2h$ where $h$ is the Coxeter number of $\Phi$) such that in each such set the average size is $n/2$. (So the conjectured involution on the set of antichains would be doing the same thing except with sets of size dividing $2$ instead of $2h$.) -In this way I view Panyushev's homomesy conjecture as an extension of his investigation of the "duality" property for ad-nilpotent ideals, which evolved over the course of the three papers referenced above. -EDIT: Here is some further "context"/speculation: -Let $\Phi$ an irreducible crystallographic root system with Weyl group $W$. As mentioned above, the $\Phi$-Narayana number $N_k(\Phi)$ is the number of antichains of the root poset $\Phi^+$ of cardinality $k$. And the $\Phi$-Narayana polynomial is then $N(\Phi;q) = \sum_{k=0}^{n} N_k(\Phi)q^k$. Note that $N(\Phi;1)=\mathrm{Cat}(\Phi)$ is the $\Phi$-Catalan number. -The previous paragraph gave a "nonnesting" description of $N(\Phi;q)$. There is also a "noncrossing" description. Namely, recall that the lattice of noncrossing partitions of $\Phi$, denoted $NC(\Phi)$, is the induced subposet of the absolute order on $W$ below $c$, where $c$ is any fixed Coxeter element of $W$. Then $NC(\Phi)$ is a graded poset, and the Narayana number $N_k(\Phi)$ is also the number of elements of $NC(\Phi)$ at rank $k$ (so $N(\Phi;q)$ is the rank generating function for $NC(\Phi)$). For these (and other) descriptions of $N(\Phi;q)$, see Theorem 5.9 of Fomin-Reading "Root systems and generalized associahedra" (https://arxiv.org/abs/math/0505518). -Now, as mentioned, Panyushev was very interested in the fact that $N(\Phi;q)$ is palindromic, which is not at all obvious from the nonnesting description. In particular he was looking for a bijective (in fact, involutive) proof of this fact. But there is a nice way to see that $N(\Phi;q)$ is palindromic from the noncrossing description. Namely, there is the Kreweras complementation map $\mathrm{Krew}\colon w \mapsto cw^{-1}$ on $NC(\Phi)$, which takes an element of rank $k$ to an element of rank $n-k$. -Note that while $\mathrm{Krew}$ does provide a bijective proof that $N(\Phi;q)$ is palindromic, $\mathrm{Krew}$ is not an involution (it has order $h$ or $2h$). In fact, here's where the connection to Panyushev's "rowmotion" action comes in: elements of $NC(\Phi)$ under $\mathrm{Krew}$ are in equivariant bijection with antichains of $\Phi^+$ under rowmotion. This was conjectured by Bessis-Reiner (https://arxiv.org/abs/math/0701792) and proved by Armstrong-Stump-Thomas (https://arxiv.org/abs/1101.1277). -As far as I can tell, Panyushev did not know at the time that his action was the "same" as Kreweras complementation. But I do think it's interesting that he came up with his action with the goal of understanding why $N(\Phi;q)$ is palindromic on the nonnesting side (or at least why the average antichain cardinality is $n/2$), while the Kreweras complementation proves this on the noncrossing side.<|endoftext|> -TITLE: Projective modules over quantum groups -QUESTION [14 upvotes]: My question is short: - -How can one calculate $\operatorname{Tor}_{U_q(\mathfrak g)}(k,k)$? - -($k$ is the ground field of characteristic zero). -If we had a regular universal enveloping algebra $U(\mathfrak g)$, then of course this is just $H_\ast(\mathfrak g,k)$. To calculate it, we just take a projective resolution of $k$, for example: -$$\cdots\to U(\mathfrak g)\otimes\mathfrak g^{\wedge 2}\to U(\mathfrak g)\otimes\mathfrak g\to U(\mathfrak g)\xrightarrow\epsilon k$$ -is the standard resolution (the last arrow is the counit). In fact, it is a free resolution! - -Is there a corresponding standard free resolution of $k$ as a $U_q(\mathfrak g)$-module? - -One can of course use the counit for the first map, but after that I'm not sure. - -REPLY [6 votes]: I'll assume that $U_q(\mathfrak{g})$ is the quantum group defined over the rational function field $k=\mathbb{C}(q)$ in the indeterminate $q$. Then Poincare duality is known to hold for $U_q(\mathfrak{g})$; see for example the paper of Chemla (Corollary 3.2.2, J. Algebra 276 (2004), 80-102). This tells us that $\mathrm{Tor}_n^{U_q(\mathfrak{g})}(k,k)$ is isomorphic (as a vector space) to $\mathrm{Ext}_{U_q(\mathfrak{g})}^{d-n}(k,k)$, where $d = \dim \mathfrak{g}$. You can also apply a version of the universal coefficient theorem to show that $\mathrm{Ext}_{U_q(\mathfrak{g})}^n(k,k) \cong \mathrm{Tor}_n^{U_q(\mathfrak{g})}(k,k)^*$ (vector space dual). So in answer to the question of how to compute the Tor group, you compute the Ext group instead. -I am not aware of any general results on explicit projective resolutions for $U_q(\mathfrak{g})$, but in my paper, I was able to compute the Ext groups via an indirect comparison to the ordinary Lie algebra cohomology, and determined that they are all of the same dimension as the corresponding cohomology groups for the Lie algebra $\mathfrak{g}$. -The case when $q$ is not an indeterminate seems more difficult, and I don't know how to handle it at this time.<|endoftext|> -TITLE: Permutation models with a class-sized group -QUESTION [5 upvotes]: I'm working on building a model of ZF where a very weak choice principle fails, in so much as in any permutation model with a set of atoms it is inconsistent that this principle fails. -To get a feel for the sort of results around this sort of construction can someone point me to permutation models in the literature which use a class-sized group? -I'm aware of Blass' paper on SVC, but little else. - -REPLY [7 votes]: Jech's The Axiom of Choice has a couple class permutation models, a review shows theorem 11.2, as well Problems 9.3, 9.4 which you may want to examine (they talk about Injection Principle and Surjection principles from classes to sets, and the failure of them) -Also, while not permutation per se, Monro constructs via symmetric class forcing a Dedekind-finite proper class (that is a class that every function into $\omega$ is bounded). You may find this proof in Independence results concerning Dedekind-finite sets (J. Austral. Math. Soc. 19 (1975), 35–46).<|endoftext|> -TITLE: Vigorous actions on the Cantor set -QUESTION [9 upvotes]: This is related to my previous question here: -Antichains and measure-preserving actions on Boolean algebras -This time I will ask something more precise. -Let $G$ be a group acting by homeomorphisms on the standard Cantor set $X$ (or if you prefer, $G$ is acting on the countable atomless Boolean algebra, which can be realised as the clopen subsets of $X$). Say a non-empty clopen subset $\alpha$ of $X$ is minorising (for $G$) if the following holds: -For every non-empty clopen subset $\beta$ of $X$, there is some $g \in G$ such that $g\beta$ contains $\alpha$. -Now suppose that there is a minorising subset $\alpha$, and that $G\alpha$ (that is the set of $g\alpha$ for $g \in G$) covers $X$. Let $n$ be the smallest size of a subcover of $G\alpha$, that is the smallest number of $G$-translates of $\alpha$ needed to cover $X$. It is clear that $n$ is finite (by compactness) and at least $2$, and that $n$ does not depend on the choice of $\alpha$ (beyond ensuring that $\alpha$ is minorising), so it is an invariant of the action. -The condition $n=2$ is equivalent to the condition that every proper non-empty clopen subset is minorising. (This occurs for instance if $G$ consists of every homeomorphism of $X$.) -Questions: Can $n$ be greater than $2$? What if $G$ is simple, and/or every orbit of $G$ on $X$ is dense? -If $n=2$, can $G$ still have infinitely many orbits on the clopen subsets of $X$? -NB: The existence of a minorising subset means that $G$ destroys most of the extra structure I can think of putting on $X$, such as non-trivial metrics and measures. (Can $G$ act by quasi-isometries?) Effectively $\alpha$ is `as small as possible' up to the action of $G$. - -REPLY [2 votes]: I ran across this question after pondering your more recent question here. -Ramiro has already described examples giving a YES answer to your final question, but here is a large and naturally arising class of examples. -Take $G$ to be the full automorphism group $Aut(T)$ of any finite valence tree $T$ on which $Aut(T)$ acts cocompactly, and take $X$ to be the Cantor set of ends of $T$. Or, you can take $G$ to be any cocompact subgroup of $Aut(T)$, such as the free group of finite rank $n \ge 2$ acting on the universal cover of a rose with $n$ petals, or the fundamental group of any finite graph of groups with finite vertex and edge groups, acting on its Bass-Serre tree. -For such actions, every clopen is minorizing and your number $n$ always equals $2$. The proof uses the fact every orbit of the action of $G$ on $X$ is dense, and that the set of source-sink pairs $x \ne y \in X$ is dense in $X \times X$; these are pairs for which there exists $g \in G$ whose action on $X$ has source--sink dynamics with source $x$ and sink $y$. So, if $C,D$ are any proper clopens then choose a source-sink pair $x,y$ with $x \in D$, $y \not\in D$, then choose a translate of $C$ that contains $x$, then choose $n$ so large that $g^n(C)$ contains $D$. Similarly, if $C$ is any proper clopen in $X$, choose a translate $C_1$ that contains $x$, choose another translate $C_2$ that contains $y$, then choose sufficiently large $n$ so that the pair $g^n(C_1),C_2$ covers $X$. -Also for such actions, there are infinitely many $G$-orbits of clopens. To see this, given any clopen $C$, and letting $D=X-C$, the intersection in $T$ of the convex hulls of $C$ and $D$ is a finite subtree of $T$, the number of vertices of this subtree is an invariant of the orbit of $C$, and this number can be as large as you like.<|endoftext|> -TITLE: the homotopy type of the pointed loop space of a countable cw complex -QUESTION [9 upvotes]: I apologize in advance if this is too elementary for this forum. I have received some help but am still unsure about how to proceed. I am interested in a proof of the following result due to John Milnor: If X is a countable CW complex then the pointed loop space of X (with the compact open topology) has the homotopy type of a CW complex. I have consulted Milnor's "on spaces having the homotopy type of a CW complex" but it references articles I was unable to find after determined searches on google scholar. Does anyone know of a self-contained exposition of this result? - -REPLY [14 votes]: There is a very fine book ``Cellular structures in topology'', by R. Fritsch and R.A. Piccinini -that gives a detailed and self-contained treatment of Milnor's results. It has a wealth of other well-presented material, some of which is little known nowadays.<|endoftext|> -TITLE: Explicit expression for recursively defined functions -QUESTION [5 upvotes]: Consider a function $w(i)$, $i \in \mathbb{N}$, defined recursively by: -$w(0)=w(1)=1$, and -$w(i)={i}^{n}-\sum_{j=1}^{i-1}{i \choose j}w(j)$ for $i>1$. -Is it possible to write $w(i)$ out explicitly as a function of $i$ (i.e., with only $i$ on the right hand side), for $i>1$ ? -In general, given a recursively defined total computable function say, $f(n)$, is it always possible, in principle, to write out $f(n)$ explicitly as a function of $n$, for $n$ large enough? I suspect it's impossible. If so, what is the real reason that prevents us from doing so? -Another well known example that comes to my mind is: -$f(0)=f(1)=c$, and -$f(i)=5f(i-1)(1-f(i-1))$ -where we can choose some value of $c \in (0,1)$ so that the trajectory of $f(i)$ becomes chaotic in $(0,1)$. If we could really express any recursively defined total computable function explicitly, then we could write out the chaotic trajectory in a finite, flat and non-dynamic formula, which seems strange to me. (or can we?) - -Edit: Thank you very much for all the answers. They are really helpful. -@Joel: Is the recursive formation essential when the resulting recursive function doesn't grow faster than it's component functions ${g}_{i}$ (in your notation)? i.e., is it true that given ${g}_{i}$ and recursive function $f$ defined on them, as long as $f$ doesn't grow faster than the fastest of ${g}_{i}$, the recursive formation can be dispensed with? -@Carl: By "...it is not possible to come up with a completely explicit form for every computable function, in some fixed signature" , did you mean to write out the functions symbolically using finite, fixed set of symbols? In @Joel's interpretation, he didn't distinguish between we being able to symbolically write out exponential functions as, say ${2}^{n}$, and just taking the recursive function on multiplication as another "known" basic function ${g}_{i}$. Of course, one may just dismiss ${2}^{n}$ as a "shorthand" for the whole recursive formation, but I do feel they are not quite the same ("${2}^{n}$" is closer to what I had in mind by "explicit". I'd like to know if the distinction makes any sense in any sense). -Another question I have concerns about this "...because then by enumerating all possible forms we would get a numbering ϕ as in the theorem" But just because we can write a completely explicit form for each computable function doesn't mean we can enumerate these forms effectively, does it? If we can't, this does not contradict the theorem. Can you be more specific about this? Thanks! - -REPLY [5 votes]: To follow up on Joel Hamkins' answer, the fundamental obstruction here is totality. Explicit definitions are always going to give total functions. The reason that recursion is so powerful for defining functions is that it is possible to make partial functions, e.g. -$$ -f(0) := 0 \qquad f(2n+2) := f(2n) \qquad f(2n+1) := f(2n+3) -$$ -The following is a standard theorem. - -Theorem. Let $C$ be a countable system of total computable functions $\mathbb{N} \to \mathbb{N}$ with the property - -There is a numbering $\phi_i\colon \mathbb{N} \to C$ such that every $f \in C$ is of the form $\phi_i$ for at least one $i$, and there is a uniform way to compute $\phi_i(j)$ given just $i$ and $j$. - -Then $C$ does not include all total computable functions from $\mathbb{N}$ to $\mathbb{N}$. -Proof. Diagonalization. - -In the context of this question, the philosophical meaning of this theorem is that it is not possible to come up with a completely explicit form for every computable function, in some fixed signature, because then by enumerating all possible forms we would get a numbering $\phi$ as in the theorem. -The closest we can hope for in terms of an explicit form for all total computable functions is something like Kleene normal form, but this still includes an unbounded search. Kleene's normal form theorem says that there is a primitive recursive function $U$ and a primitive recursive relation $T$ such that for every computable function $f$ there is an $e$ such that for all $i$, -$$ -f(i) \simeq U(\mu s . T(e,i,s)). -$$ -where $\mu$ is the unbounded search operator.<|endoftext|> -TITLE: Why tropical geometry? -QUESTION [58 upvotes]: Tropical geometry can be described as "algebraic geometry" over the semifield $\mathbb{T}$ of tropical numbers. As a set, $\mathbb{T}=\mathbb{R}\cup \{ -\infty\}$; this is endowed with addition being given by the (usual) maximum of real numbers and multiplication by the (usual) sum of real numbers. Recently there has been a lot of research on this kind of geometry. The obvious question is now: why are people interested in this? -There are three possible motivations I am aware of: -(1) "Polynomial" equations over $\mathbb{T}$ can be interpreted as linear equalities and inequalities over the classical reals. So people working in linear optimization and similar areas can use tropical geometry as an alternative, more algebraic approach that might lead to new methods and insights. -(2) In quantization jargon, one can interpret $\mathbb{T}$ as a "classical limit" of semifields which are all isomorphic to $\mathbb{R}^{\ge 0}$ with the usual addition and multiplication. More precisely, for any $q>0$ set $\mathbb{T}_q=\mathbb{R}\cup\{ -\infty \}$ and consider the bijection $\log_q:\mathbb{R}^{\ge 0}\to\mathbb{T}_q$ (setting $\log_q(0)=-\infty$). Pushing forward the usual semifield structure on $\mathbb{R}^{\ge 0}$ along $\log_q$, we get a semifield structure on $\mathbb{T}_q$. Then it is easy to check that the semifield $\mathbb{T}$ is, in the obvious sense, the limit of $\mathbb{T}_q$ as $q\to 0$. So if one likes to think in these terms (I do not, but that shall not bother me for now), then real algebraic geometry appears (with a grain of salt) as the quantum version of tropical geometry, which in turn gives tropical geometry an important rôle. -(3) This is only a post hoc justification. Some problems of classical algebraic geometry, mainly in enumerative geometry, have been solved by methods of tropical geometry. The usual strategy is as follows: we have a question in algebraic geometry, to which the answer is supposed to be an integer (say). We then set up an analogous question in tropical geometry, prove that the answers to the two questions agree, and then work on the tropical question, which is usually much simpler to answer. -Besides these three arguments, do you have any other motivations for studying tropical geometry? - -REPLY [12 votes]: Bernd Sturmfels has applied tropical algebraic geometry to phylogenetic trees. IIRC, he says the space of trees whose edges have lengths is a tropical Grassmannian. And the trees used in these models are not rooted because you can't tell from DNA evidence which vertex should be the root.<|endoftext|> -TITLE: Counting Special Graphs -QUESTION [6 upvotes]: Do we have any formula for counting the number of graphs with $n$ vertices, that has exactly $k$ vertices with degree $d$ and the other vertices have different and disjoint degrees? -(Different and disjoint are the same, $d_1$ is different or disjoint rather than $d_2$, iff $d_1\neq‎ d_2$.) -For example, for $n=3, k=2, d=1$, we only have one graph($P_3$) with this property. -Also, for $n=4, k=2, d=2$, we have the only graph with degree sequence $1, 2, 2, 3$. - -REPLY [2 votes]: The two examples you mention have a very nice generalization but the general counting problem seems hopeless. Let me start with the nice result. It is a fun (and easy) exercise to show that in any simple graph there are at least two vertices with the same degree. Define a $g(n,k,d)$ to be a simple unlabeled graph with $n$ vertices and $n-k+1$ distinct degrees, The degree $d$ taken by exactly $k$ vertices and some other $n-k$ degrees taken once each. Also, let $N(n,k,d)$ be the number of $g(n,k,d)$. -You make two interesting remarks. That $N(4,2,2)=1$ the graph having degrees $1,2,2,3$ (this is true) and that $N(3,2,1)=1$ because of the path $P_3.$ However this number is actually $2$ because the complement (an edge and an isolated vertex) is another example. -It turns out that for each $n \ge 2$ there are exactly two $g(n,2,d)$ : a connected one with $d=\lceil (n-1)/2 \rceil$ and the complement which is not connected and has $d=\lfloor (n-1)/2 \rfloor.$ -I'm sure that this is well known but didn't find it in a short bit of looking around. The ingredients of an induction proof are the following operations: - -The complement of a $g(n,k,d)$ is a $g(n,k,n-d-1).$ -If an isolated vertex is added to a connected $g(n,k,d)$ then the result is a disconnected $g(n+1,k,d).$ And if the unique isolated vertex of a $g(n+1,k,d)$ is removed, the result is a $g(n,k,d.)$ -If a new vertex is added to a $g(n,k,d)$ (connected or not) and then an edge is drawn to each old vertex, the result is a $g(n+1,k,d+1).$ And if a $g(n+1,k,d+1)$ has a unique vertex of degree $n$, it may be removed to obtain a $g(n,k,d).$ - -This leaves a few steps to fill in. -Mathscinet gives little information about On graphs having exact two vertices with the same degree. Nor about On graphs having exact three vertices with the same degree. -At the other extreme, the case $k=n$ of graphs regular of degree $d$ is reasonable but not trivial for $d=2.$ (In that it is the number of partitions of a certain type so at least has a name.) For larger $d$ it would be hopeless to get exact numbers for large $n$ . And the general case seems equally difficult. In a few isolated cases an answer might be fairly easy such as $k=n-2$ vertices of degree $1$ (or maybe $n-m$ for small $m$.)<|endoftext|> -TITLE: Dual Lefschetz Operator and Contraction with the Fundamental Form -QUESTION [5 upvotes]: Let $M$ be a Kahler manifold, with metric $g$, fundamental form $\omega$, and dual Lefschetz operator $\Lambda$. Now $\Lambda$, and contraction with $\omega$, both map the two forms $\Omega^2(M)$ to $0$-forms, ie smooth functions. Are they equal? I think this is almost certainly true, but I can't see a clean argument. -Do I need Kahler here? I would guess this works for all complex manifolds. - -REPLY [5 votes]: The question asks whether $\boxed{ \Lambda \alpha = g(\omega,\alpha)}$ for any $\alpha \in \Omega^2(X)$. -This is indeed true, as explained by Gunnar, but here is a simple way of seeing it. Firstly, since we're just using linear operators we can work fiber-wise (hence indeed no closedness of $\omega$ is needed, and whether we work in the context of $TM$ or general vector bundles is irrelevant). Remember that by definition $g(\Lambda \eta, \mu) = g(\eta, L \mu)$ where $L$ is the wedge product with $\omega$. -Using this in our case: let $\alpha \in \Lambda^2 E_x$, then we can rewrite $\boxed{\Lambda \alpha}$ as $g(\Lambda \alpha, 1) = g(\alpha, L \; 1) = \boxed{ g(\alpha, \omega) }$.<|endoftext|> -TITLE: What is the role of equivariance in the Atiyah-Singer index theorem? -QUESTION [18 upvotes]: I'm trying to read through "The index of elliptic operators I," and here's what I understand of the structure of the proof: - -Define (using purely K-theoretic means) a homomorphism $K_G(TX) \to R(G)$ where $G$ is a compact Lie group, $X$ a $G$-manifold, $R(G)$ the representation ring, and $K_G(TX)$ the equivariant (compactly supported) K-theory of the tangent bundle $TX$ of a compact manifold $X$. Call this the topological index. -Show that the topological index is characterized by a collection of properties: any natural transformation $K_G(TX) \to R(G)$ that satisfies an excision-like property, a multiplicative property for fiber bundles, and certain normalization conditions is necessarily the topological index. -Define (using analysis) the analytical index $K_G(TX) \to R(G)$ by taking the index (in the usual sense) of an elliptic (pseudo)differential operator. -Show that the analytical index satisfies the relevant conditions in 2 (which involves a few computations), so it must be the topological index. - -This is all very nice and tidy, but I'm puzzled: what exactly is the role of $G$? Let's say hypothetically that I'm a non-equivariant person and only care about plain differential operators on plain manifolds. It seems (if I understand correctly) that the only place where the $G$-action is relevant is in the multiplicativity condition on the index. -The reason is that the multiplicativity property is used to show that if -$i: X \to Y$ is a closed immersion of manifolds, then the induced map $i_!: K_G(TX) \to K_G(TY)$ (given by the Thom isomorphism and push-forward for an open imbedding) preserves the index. -It seems that the point is to reduce to the case where $X \to Y$ is the imbedding $i: X \to N$ for $ N$ a vector bundle on $X$ and $i$ the zero section. Now in this case there is a principal $O(n)$-bundle $P \to X$ such that $N$ is obtained via $N = P \times_{O(n)} \mathbb{R}^n$, and Atiyah-Singer define a multiplication -$$K(X) \times K_{O(n)} (\mathbb{R}^n) \to K(N)$$ -which, when one takes a specified element of $K_{O(n)}(\mathbb{R}^n)$, is precisely the Thom isomorphism. -If this is the only place in which equivariance enters the proof, it seems strange that the proof would depend on it. Is it possible to phrase the argument in this paper non-equivariantly? (Or, is there a high-concept reason equivariance should be important to the proof?) - -REPLY [15 votes]: As long as you are only interested in Dirac type operators on oriented manifolds, the equivariant $K$-theory can be kicked out of the proof. This is done by Guentner in ''K-homology and the index theorem'', relying on Higson ''On the bordism invariance of the index''. Essentially, the main steps in Guentners argument are: - -Given a manifold $M \subset R^{2n+1}$ with a Dirac bundle, consider the boundary of the disc bundle of the normal bundle $S$, which is a hypersurface. The Dirac bundle on $M$ induces one on $S$, and tensoring with the Clifford bundle on the sphere, you obtain a Dirac operator on $S$. The topological indices of the original bundle and the new one are the same, and the analytical indices likewise. This procedure involves a bit of equivariant considerations, but no equivariant $K$-theory. That some kind of symmetry considerations are important should not bother you, because it is the symmetry that allows you to do computations (of kernels and cokernels of operators). -The hypersurface $S$ is bordant to $S^{2n}$. Both indices are bordism invariant. To be able to use this, the Mayer-Vietoris-sequence in $K$-theory is needed, because the Dirac bundle on $S$ might not be extendable over the bordism. Here the fact that hypersurfaces are $spin^c$ is essential. - -This is all quite similar in spirit to IOE1. Why do Atiyah and Singer need equivariant $K$-theory? The question is why they need $K (\mathbb{R}^n)$, equivariant or nonequivariant, at all. Couldn't they just have taken the fibre product with a specific operator on $\mathbb{R}^n$ of index $1$, which should be possible because the operator is equivariant? This is what they essentially do, but the problem is that the product of two pseudodifferential operators of order $0$ is not uniquely defined, but only up to equivalence. So the twisted product of operators is not well-defined on the operator level. They are forced to consider order $0$ operators because only for order $0$ operators it makes sense to ask that they are equal to the identity outside a compact set. Because Atiyah-Singer consider open manifolds, they need order zero operators, and this why at the passage to the normal bundle they need the $K$-theory of $R^n$, which has to be equivariant if the normal bundle of the manifold is not trivial. -Guentner works with closed manifolds, which avoids these problems. Since he takes differential operators only, the product procedure is straightforward. But to get an operator on the sphere of index $1$, he needs orientability, which is why his proof only applies to oriented manifolds.<|endoftext|> -TITLE: Does Euclidean space have a compact factor? -QUESTION [36 upvotes]: Is $\mathbb{R}^n$ homeomorphic to a product $X \times Y$ with $X$ compact and not a point? -Bing's Dogbone space is a quotient of $\mathbb{R}^3$ with fibers points and arcs, and whose product with $\mathbb{R}$ is $\mathbb{R}^4$, so it doesn't seem to me to big a stretch to think that it may be possible. -Or, is there a notion of dimension which takes care of it swiftly? - -REPLY [5 votes]: Here is a proof which uses only singular homology.$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\To}{\longrightarrow}$$\def\set#1{\lbrace#1\rbrace}$$\newcommand{\Xminusx}{X\setminus\set{x}}$$\newcommand{\Yminusy}{Y\setminus\set{y}}$ -Assume $f:X\times Y\to\RR^n$ is a homeomorphism, and that $X$ is compact. I will prove that $X$ is a singleton by applying repeatedly the Künneth theorem, and using a few basic calculations of singular homology. By default, I use homology with coefficients in $\ZZ$. One can also carry out the exact same proof using homology with coefficients in a field, but the resulting simplifications are fairly inconsequential. -General remarks -The spaces $X$ and $Y$ cannot be empty, and we will fix $x\in X$ and $y\in Y$. Let also $p=f(x,y)\in\RR^n$. Observe that $X$ and $Y$ are Hausdorff, given that $\RR^n$ is Hausdorff. In particular, $\Xminusx$ is open in $X$, and $\Yminusy$ is open in $Y$. Furthermore, as observed in the comments, $X$ and $Y$ are contractible since $\RR^n$ is contractible. In particular, $H_\ast(X)$ is zero in positive degrees, and is $\ZZ$ in degree zero. -Claim 1: $H_n(Y,\Yminusy) \simeq H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr)$ -First of all, consider the pair $X\times(Y,\Yminusy)=\bigl(X\times Y,X\times(\Yminusy)\bigr)$. By the Künneth theorem and the contractibility of $X$, we conclude that -$$ H_n\bigl(X\times Y,X\times (\Yminusy)\bigr) = H_0(X)\otimes H_n(Y,\Yminusy) = H_n(Y,\Yminusy) $$ -The above pair $\bigl(X\times Y,X\times (\Yminusy)\bigr)$ is homeomorphic via $f$ to $\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr)$. The preceding expression thus implies -$$ H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr) \simeq H_n(Y,\Yminusy) $$ -Claim 2: $H_n(Y,\Yminusy)$ has a $\ZZ$ summand -Since $X$ is compact, the image $f(X\times\set{y})$ is compact in $\RR^n$, and thus bounded. Let $R\in\RR^+$ be such that $f(X\times\set{y})$ is contained in the closed ball of radius $R$ centered at $p$, $B_R(p)$. Then we have inclusions -$$ \RR^n\setminus B_R(p) \subset \RR^n\setminus f(X\times\set{y}) \subset \RR^n\setminus\set{p} $$ -which induce homomorphisms on homology: -$$ \ZZ \simeq H_n(\RR^n,\RR^n\setminus\set{p}) \To H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr) \To H_n\bigl(\RR^n,\RR^n\setminus B_R(p)\bigr) \simeq \ZZ $$ -The composition of the two maps is an isomorphism, therefore they exhibit a splitting of the middle group: -$$ H_n(Y,\Yminusy) \simeq H_n\bigl(\RR^n,\RR^n\setminus f(X\times\set{y})\bigr) \simeq \ZZ \oplus A $$ -for some abelian group $A$. -Claim 3: $H_\ast(X,\Xminusx)$ is concentrated in degree zero -Let $i$ be a positive integer. Observe that $f$ gives a homeomorphism between the pairs $\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr)$ and $(\RR^n,\RR^n\setminus\set{p})$. Consequently, -$$ H_{n+i}\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) \simeq H_{n+i}(\RR^n,\RR^n\setminus\set{p}) = 0 $$ -Recall that $\Xminusx$ and $\Yminusy$ are open in $X$ and $Y$, respectively. So we can apply the Künneth theorem to the pair -$$ (X,\Xminusx)\times(Y,\Yminusy) = \bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) $$ -which implies that there is a monomorphism -$$ H_i(X,\Xminusx)\otimes H_n(Y,\Yminusy) \To H_{n+i}\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) = 0 $$ -It follows that $H_i(X,\Xminusx)\otimes H_n(Y,\Yminusy) = 0$. Since $H_n(Y,\Yminusy)$ contains a summand isomorphic to $\ZZ$, we conclude that $H_i(X,\Xminusx)=0$. -Claim 4: $H_0(X,\Xminusx)$ is not zero -We now know that $H_\ast(X,\Xminusx)$ is zero in positive degrees, and it is necessarily a free abelian group in degree zero. Applying once more the Kunneth theorem to $(X,\Xminusx)\times(Y,\Yminusy)$, we obtain an isomorphism -$$\begin{array}{rl} -H_0(X,\Xminusx)\otimes H_n(Y,\Yminusy) \!\!\!\! & = H_n\bigl(X\times Y,(X\times Y)\setminus\set{(x,y)}\bigr) \\ -& \simeq H_n(\RR^n,\RR^n\setminus\set{p}) \\ -& \simeq \ZZ -\end{array}$$ -Consequently, $H_0(X,\Xminusx) \neq 0$. -Conclusion -Since $H_0(X)=\ZZ$, the only way that we can have $H_0(X,\Xminusx) \neq 0$ is if $\Xminusx = \emptyset$. Thus $X=\set{x}$ is a singleton.<|endoftext|> -TITLE: Generalized trigonometric functions $Cos(n) v$ and $Sin(n) v$. -QUESTION [6 upvotes]: I just discovered a paper from 1948, Eine Verallgemeinerung der Kreis-und Hyperbelfunktionen by R. Grammel which introduces functions he calls Cos(n) and Sin(n), representing a parameterization of the curve $x^n + y^n=1$ in $\mathbb{R}^2$ (the unit sphere of the n-norm) (also, I know the notation is pretty bad, if it was my choice I'd probably write something like $\sin_n$). -Grammel then proceeds to prove many identities about these generalizations of the circular sine and cosine that seem to show that they have much in common with the usual trigonometric functions. -Trying to find more information about these functions, I did not succeed in finding anything recent. I wondered if that was perhaps only because the terminology has changed since that paper or if there was some modern sense in which the study of these functions is trivial or uninteresting? - -REPLY [3 votes]: The existing literature on generalized trigonometric functions is scarce and it seems there isn't a comprehensive account of generalized trigonometric functions anywhere. Also there isn't a unified accepted notation and different authors use different notations. -The generalized sine and cosine functions $\sin_{pr}x$, $\ \cos_{pr}x$ are defined by the formulas -$$ -x=\int_0^{\sin_{pr}x}\frac{dt}{\sqrt[p]{1-t^r}},\qquad \cos_{pr}x=\sqrt[r]{1-(\sin_{pr}x)^r}. -$$For generic values of parameters $p,r$ it appears that these functions are not very interesting, but when these parameters have special values, then $\sin_{pr}x$ can be expressed algebraically through Jacobi elliptic functions, and as a consequence one can establish addition theorems analogous to addition theorems for elliptic functions. -Below I give a unified relatively simple discussion of several such cases. The first case $r=2,\ p=2$ is trivial. The second case $r=3,\ p=\frac{3}{2}$ is due to Cayley and Dixon. The third case $r=4,\ p=\frac{4}{3}$ has been considered relatively recently by Edmunds, Gurka, and Lang (Properties of generalized trigonometric functions, J. Approx. Theory 164 (2012), no. 1, 47-56.) I couldn't find the discussion of the fourth case $r=6,\ p=\frac{6}{5}$ in the literature, but it easily follows from the same considerations as in the first three cases. The parameter $r$ is denoted in analogy with Ramanujan's theory of elliptic functions to alternative bases (see "Ramanujan's notebooks, vol. 5" by Bruce Berndt). I don't know whether it is just a coincidence or there is a deeper connection, but $r=2,3,4,6$ are the four signatures for which such alternative theories of elliptic functions have been developed. -Let's consider the case when $\frac{1}{p}=1-\frac{1}{r}$. Then by a series of simple changes of variables one obtains -\begin{align} -\int_0^{u}\frac{dt}{(1-t^r)^{1-1/r}}=\frac{1}{r}\int_0^{u^r}\frac{dt}{(1-t)^{1-1/r}t^{1-1/r}}=\tag{1}\\ -\frac{1}{r}\int_0^{u^r}\frac{dt}{(t-t^2)^{1-1/r}}=\frac{1}{r}\int_0^{u^r}\frac{dt}{\left(\frac{1}{4}-\left(\frac{1}{2}-t\right)^2\right)^{1-1/r}}=\\ -\frac{2^{2-2/r}}{r}\int\limits_{1-2u^r}^{1}\frac{dt}{(1-t^2)^{1-1/r}}=\frac{2^{1-2/r}}{r}\int\limits_{(1-2u^r)^2}^{1}\frac{dt}{\sqrt{t}(1-t)^{1-1/r}}=\\ -\frac{2^{1-2/r}}{r}\int\limits_0^{1-(1-2u^r)^2}\frac{t^{1/r-1}dt}{\sqrt{1-t}}=2^{1-2/r}\cdot\int\limits_0^\sqrt[r]{1-(1-2u^r)^2}\frac{dt}{\sqrt{1-t^r}} -\end{align} -When $r=2,3,4$ the last integral can be inverted in terms of elliptic functions. But the case $r=6$ also can be inverted, since -$$ -\int\frac{dt}{\sqrt{1-t^6}}=-\frac{1}{2}\int\frac{d\left(\frac{1}{t^2}\right)}{\sqrt{\frac{1}{t^6}-1}}. -$$ -As an illustration, according to Edmunds,Gurka, and Lang ($r=4$) one has -$$ -\sqrt{1-\sqrt{\frac{1-\left(\sin_{\frac{4}{3},4}x\right)^2}{1+\left(\sin_{\frac{4}{3},4}x\right)^2}}}=\text{sn}\left(x,\frac{1}{\sqrt{2}}\right). -$$ -Representation in terms of elliptic functions is possible also when $\frac{1}{p}\neq 1-\frac{1}{r}$. This can be seen by substitution $t\to \frac{at+b}{ct+d}$ in the integral (second integral in eq.$(1)$) -$$ -\int\frac{dt}{(1-t)^{1-1/r}t^{1-1/r}}, \quad (r=2,3,4,6), -$$ -and specifying parameters $a,b,c,d$ such that the resulting integral again has he form of an incomplete beta function.<|endoftext|> -TITLE: Automorphic forms and quantum groups -QUESTION [10 upvotes]: The paper Eisenstein series and quantum affine algebras by Kapranov makes contact between automorphic forms and quantum groups. I haven't found even one other paper devoted to this theme. -Have other authors come at this, perhaps from other perspectives? - -REPLY [14 votes]: There are at least two research strands that fit the description - if by automorphic forms you allow me to consider the function field versions. The one closest to your question is in the direct line of Kapranov's very influential paper. The topic of Hall algebras is extremely active (I recommend Schiffmann's beautiful survey). Since Ringel and Lusztig, Hall algebras have been one of the primary means to understand quantum groups. On the other hand, since Kapranov's paper in particular there's been a realization that the entire geometric Langlands program for general linear groups is a subset of the study of the [categorified] theory of Hall algebras - namely the case of Hall algebras for categories of coherent sheaves on curves. The Hecke operators on functions on moduli spaces of bundles are a part of the multiplication action of the Hall algebra of this category on itself, and if we replace functions with sheaves (study Hall categories) we recover the basic questions in the function field version of the theory of automorphic forms. There are innumerable papers this relates to, but I think the most directly relevant to your question are the wonderful papers of Schiffmann and Vasserot on Hall algebras, Macdonald polynomials, geometric Eisenstein series and geometric Langlands (see arXiv search for those two names). -The other major direction is the "quantum geometric Langlands correspondence", which is a q-deformation of the [categorified] theory of automorphic forms on function fields (ie geometric Langlands), which is directly related to the representation theory of quantum groups. (There are various close connections between undeformed, plain old geometric Langlands and quantum groups as well, see e.g. Arkhipov-Bezrukavnikov-Ginzburg and other [amazing] papers of Bezrukavnikov, but this is in a somewhat different direction, though of course everything is related.) Unfortunately not that much is written about quantum geometric Langlands -- its origins are in works of Feigin-Frenkel and Beilinson-Drinfeld, and the idea has been around since the late 90s, but it's hard to find in the literature until recently (though see the withdrawn preprint by Stoyanovsky, still available on arXiv in early versions, for the general idea). In any case a major paper on this topic is Gaitsgory's paper constructing quantum groups directly out of a deformed version of the geometric Satake correspondence. This is the first step in a program of Gaitsgory and Lurie on quantum geometric Langlands which is not publicly documented AFAIK. A recent paper on the topic is Travkin. The subject got a major push from the fact that it arises very naturally from the gauge theory point of view on geometric Langlands due to Kapustin-Witten. -But maybe this response is not the right place to survey what this conjecture is actually about..<|endoftext|> -TITLE: Can every uncountable subset $\mathbb{R}$ be split at some number into two parts of the same cardinality? -QUESTION [13 upvotes]: Is it possible to prove without Continuum Hypothesis that for every uncountable subset $S$ of $\mathbb{R}$ there is a real number $x$ that splits it into two parts of the same cardinality, i.e. $\left|S \cap (-\infty,x)\right|=\left|S \cap (x,\infty)\right|$? -(if the answer to the first question is no) Is this statement equivalent to Continuum Hypothesis? - -REPLY [25 votes]: No, the statement cannot be proven in ZFC without assuming continuum hypothesis or something similar. In fact, it is equivalent to the statement that there are finitely many cardinalities between $\aleph_0$ and $2^{\aleph_0}$, so it is strictly weaker than the continuum hypothesis. -Suppose that there were infinitely many such cardinalities, then you can let $S=\bigcup_{n=1}^\infty S_n$ where $S_n\subseteq(0,1/n)$ has cardinality $\aleph_n$ to obtain a contradition. -On the other hand, if there are only finitely many such cardinalities, then $f(x)=\vert S\cap(-\infty,x)\vert$ must achieve its maximum, say $\aleph_n$ ($n > 0$). If $x_0$ is the infimum of the $x\in\mathbb{R}$ such that $f(x)=\aleph_n$ then $S\cap(x_0,\infty)$ has cardinality $\aleph_n$. Choosing $y_k\in\mathbb{R}$ decreasing to $x_0$, the cardinality of $S\cap(y_k,\infty)$ must be $\aleph_n$ for large enough $k$, otherwise $S\cap(x_0,\infty)=\bigcup_k(S\cap(y_k,\infty))$ is a countable union of sets of cardinality less than $\aleph_n$, so is of cardinality less than $\aleph_n$, giving a contradiction. So, $S\cap(-\infty,y_k)$ and $S\cap(y_k,\infty)$ are both of cardinality $\aleph_n$ for $k$ large enough. - -REPLY [19 votes]: Suppose that the continuum is larger than $\aleph_\omega$. Choose a subset $S_n$ of $(n,n+1)$ of cardinality $\aleph_n$ and let $S=\cup_{n=1}^\infty S_n$. Then for each $x$, $S\cap (-\infty,x)$ has cardinality smaller than $S\cap (x,-\infty)$.<|endoftext|> -TITLE: surjectivity of irreducible representation -QUESTION [6 upvotes]: I don't know how to show the following: -Let $A$ be an associative algebra (not necessary finite-dimensional) and $p\colon A\to End(V)$ be it irreducible finite-dimensional representation. Then $p$ in general is not surjective. -The standard textbooks on representation theory don't contain answer on this and googling doesn't help. -The interesting case for me is an irreducible representation of universal enveloping of semisimple Lie algebra. -Why am I asking: I am reading articles (of G.I. Olshaskiy) on centralizers of Lie subalgebras and trying to understand if $U(gl_{n+m})^{gl_m}$ is a sufficient object to consider or not. -Upd: I've forgot: the ground field is algclosed of char 0. - -REPLY [7 votes]: What's the ground field? Of course if it's $\mathbb{R}$ and $A=\mathbb{R}[t]/(t^2+1)$, then the regular module is irreducible, but the corresponding $p$ is not surjective. Over an algebraically closed field it's true even for infinite-dimensional $A$ though. - -REPLY [5 votes]: Let $D \ne k$ be a central division algebra of finite dimension over a (commutative) field $k$ say with $\dim_k D = n^2$. Now view -$V = D$ as a left module over itself; then $V$ is an irreducible $D$-module. -Since $\dim \operatorname{End}_k(D) = n^4 > n^2$, the image of the mapping $D \to \operatorname{End}_k(V) = \operatorname{End}_k(D)$ is a proper subalgebra.<|endoftext|> -TITLE: Sign conventions for a Chevalley basis of a simple complex Lie algebra -QUESTION [9 upvotes]: Let $R$ be the root system of a simple complex Lie algebra $g$ with respect to some Cartan subalgebra $h$. Chevalley showed there is a basis of $g$ given by the simple coroots {$H_{\alpha_i}=\alpha_i^\vee\in h$} and root vectors $X_\alpha\in g_\alpha$ for each $\alpha\in R$. This basis has the following properties: -$[H_{\alpha_i},H_{\alpha_j}]=0$ -$[H_{\alpha_i},X_\beta]=\beta(H_{\alpha_i})X_\beta$ -$[X_{\alpha},X_{-\alpha}]=H_\alpha=\alpha^\vee\in h$ -($\ast$) $[X_\alpha,X_\beta]=\pm(p+1)X_{\alpha+\beta}$, when $\alpha+\beta\in R$ and $p$ is the greatest positive integer such that $\beta-p\alpha\in R$. Otherwise, if $\alpha+\beta$ is not a root, then the bracket is zero. -References for this can be found in Serre's book on semisimple complex Lie algebras or Humphrey's book or Wikipedia. - -Does anybody know a simple way to determine the sign $\pm$ in the fourth property ($\ast$)? - -I cannot find a reference and my French is not good, so reading the original works by Chevalley and Tits isn't a viable option. In particular, I need to find a sign convention that will work for $g$ of type $F_4$. -Thanks so much. - -REPLY [7 votes]: There is a good discussion of these issues in the paper of A. Cohen, S. Murray and D.E. Taylor, "Computing in groups of Lie type", Math. Comp. 73, Number 247, -1477–1498, (2003), especially section 3 (referring to earlier work, e.g., of Carter). They explain in particular how the signs can be all reduced to so-called "extraspecial pairs", which can be chosen arbitrarily. -In Magma at least, one can see which extraspecial signs have been chosen using the "ExtraspecialSigns" command. For instance, one can see using this that GAP and Magma use (or used, I haven't checked the latest versions...) different constants for E_8.<|endoftext|> -TITLE: Invariance of dynamical system under a transformation -QUESTION [5 upvotes]: I have come across an interesting property of a dynamical system, being transformed by a map, but i haven't been able to figure out why this is happening (for quite some time now actually). Any help is greatly appreciated. Here goes then: -Let M be a n-D manifold and $\dot x=F(x)u_1, F\in \mathbb{R}^{n\times m}, x \in \mathbb{R}^{n}, u_1 \in \mathbb{R}^{m}$ be a control system evolving on M (F is the system matrix i.e. state transition function, and $u_1$ is the input of the system. For all practical purposes $u_1$ is an m-vector from an input space $\mathbb{R}^{m}$). Now let $x=\Psi (y)$ be a coordinate change on M and $u_2=M(y)u_1$ a transformation of the input $u_1$ of the first system. By applying these maps on the system, you get the new equations $\dot y=F(y)u_2$. As you may notice, F is the same in both systems. The problem is why is this happening i.e. for what systems and transformations does this property hold? -A little more elaboration -It is useful to investigate the maps more closely. In the general case one has -$\dot x=D\Psi \dot y$ -$\dot x= F(x)u_1$ -thus -$\dot y=D\Psi ^{-1} F(x)u_1$, (1) -where $D\Psi$ is the Jacobian matrix of $\Psi$. In our case it actually turns out that: -$\dot y=F(y)M(y)u_1$. (2) -You can then consider that $u_2=M(y)u_1$ and get the final system, -$\dot y=F(y)u_2$, -that is, the same system. - By (1),(2) you get, -$D\Psi ^{-1} F(x)u_1=F(y)M(y)u_1 \Rightarrow (D\Psi ^{-1} F(x)-F(y)M(y))u_1=0$. -Since this holds for every $u_1$, you have the condition, -$F(\Psi (y))=D\Psi F(y)M(y)$ -So, what does this condition imply? What systems F and maps $\Psi$ hold this property (of system invariance)? I should note that F is nonlinear and a case study where this actually happens is the kinematic model of a unicycle robot i.e. this. Any ideas? - -REPLY [2 votes]: Let me reply taking M to be the identity (indeed M is somewhat cosmetic to the discussion). -The identity $F\circ Ψ=DΨ\circ F$ is what one considers for example in the Grobman-Hartman theorem, passing from a dynamics to its linearization say at a fixed point. The possibilities for F are endless; $F$ would then be a topological conjugacy, perhaps locally, although maybe not very regular, more precisely at most Hölder continuous in general. Moreover, $F$ need not satisfy any invariance properties, which if I understand correctly is your main concern.<|endoftext|> -TITLE: Explicit way to construct simple complex tori/abelian varieties of dimension at least 2 -QUESTION [7 upvotes]: The following question was motivated by one of the earliest exercises of Complex Abelian Variaties by Birkenhake and Lange during my presentation last year. -It can be shown that any complex torus $X$ $(=V/\Lambda$, where $V$ is a complex vector space and $\Lambda$ is lattice of maximal real dimension in $V)$ admits at most countably many complex subtori. -My question: - -Is there sort of algorithm s.t. one could find simple (not admitting any non-trivial complex subtorus) complex tori of dimension $\geq 2?$ how about simple abelian varieties of dimension $\geq 2?$ - -Note that, $X$ admits a complex subtorus of dimension $g'$ if and only if there exists a subgroup $\Lambda' \subset \Lambda$ of rank $2g'$ s.t. the image of the canonical map $\Lambda' \otimes \mathbb{R} \to V$ is a complex subvector space of $V.$ - -REPLY [9 votes]: Recall that the Néron-Severi group of a complex manifold $X$ is the subgroup of $NS(X)\subset H^2(X, \mathbb Z) $ consisting of first Chern classes of holomorphic line bundles on $X$. -More algebraically, it is the quotient group $PicX/Pic_0X$, as results from the exact sequence -$$ 0\to Pic_0X \to PicX \stackrel {c_1}{\to }NS(X) \to 0 $$ -Since a torus is Kählerian, it will have no divisor at all if its Néron-Severi group is zero. -You can find an explicit calculation of the Picard number $\rho (X)=rank_ {\mathbb Z}NS(X)$ of 2-dimensional tori here, in the Appendix -You will see there for example a calculation of the Picard number of the torus determined by the lattice in $\mathbb C^2$ of matrix -$$\begin{pmatrix} -1&0&ip&ir\\ -0&1&iq & is -\end{pmatrix} -\quad \quad (p,q,r,s \in \mathbb R) $$ -To give a completely explicit example, if $ \; p=1,r=\sqrt 2, q=\sqrt 3, s=\sqrt 5 \;$ then the corresponding (highly non algebraic!) torus has no holomorphic divisor (=curve) whatsoever. - -REPLY [8 votes]: I don't really know what "some sort of algorithm" means, but here is a source of examples of simple abelian varieties. As you probably know, if $L$ is a lattice in $\mathbf{C}$ then $\mathbf{C}/L$ is an abelian variety (of dimension 1). Here are some examples of $L$ coming from arithmetic: take an imaginary quadratic field $K$ living in the complexes, and let $L$ be the ring of integers of $K$. -This construction generalises. Let $K$ be a "CM field", i.e. a totally imaginary quadratic extension of a totally real field $F$. Let $L$ be the integers of $K$. The embeddings $F\to\mathbf{R}$ extend to embeddings $K\to\mathbf{C}$ and the resulting map $L\to\mathbf{C}^n$, $n=\frac{1}{2}[K:\mathbf{Q}]$ has a Riemann form by some basic results in arithmetic. The resulting abelian varieties $\mathbf{C}^n/L$ are typically simple (and have endomorphism ring containing $L$, so quite big). Google for "CM abelian variety" for more information.<|endoftext|> -TITLE: Indecomposable projectives correspond to irreducibles - reference -QUESTION [5 upvotes]: Hello, -We have the following assertion: - -In an abelian category that has enough projectives and in which every object has finite length, indecomposable projectives correspond bijectively to irreducibles (both up to isomorphism). - -I see how to construct a well defined map from irreducibles (up to iso.) to indecomposable projectives, and to show that it is injective. What I do not know is how to show the following: - -Given an indecomposable projective of finite length, how to show that it has a unique irreducible quoteint? - -(I have edited the question; Sorry about the initial imprecision). -Thanks, -Sasha - -REPLY [4 votes]: It's useful for questions like this to go back to the basic literature where some of these ideas are developed in context. (Serganova's lecture notes look helpful but if course rely on older sources.) The applications I'm familiar with arise in various areas of module theory including modular representations of finite groups, but the arguments can be imitated for abelian categories satisfying reasonable finiteness conditions; one could even resort to embedding such categories in module categories (by standard theorems going back to Freyd and Mitchell). -The two-part treatise Methods of Representation Theory (Wiley, 1981+) by Curtis and Reiner provides a lot of general background beyond what is used traditionally for finite groups. In particular, Section 6C of Part I has a clear discussion of projective covers. Most of the interesting results in this direction apply to modules over arbitrary artinian rings. Once one has enough projectives, it's straightforward to show that any finitely generated module has a projective cover (unique up to isomorphism). -These ideas for the BGG category $\mathcal{O}$ of a semisimple Lie algebra (which is artinian) come up in Section 3.9 of my 2008 AMS text but don't involve the special features of that module category. Not only do projective covers of simple modules in the category exist, but one sees immediately from the definition of "essential" map that any indecomposable projective in the category has a unique simple quotient. (Modules here actually have finite length.) To be explicit, the given projective $P$ has at least one simple quotient $L$, which in turn has a projective cover $P_L$. Now you get maps between $P$ and $P_L$, which from the definitions are isomorphisms in both directions. -My main point is that quite a bit of systematic work (for module categories) has been done on these questions, so it's a good idea to be aware of the relevant literature when it's needed for applications.<|endoftext|> -TITLE: Correlations in last-passage percolation -QUESTION [10 upvotes]: Consider the last passage percolation model on $\mathbb{Z}^2$ with, say, geometric weights on each edge. By a landmark result of Johansson (http://arxiv.org/abs/math/9903134), we know that if $T_n(\alpha)$ is the passage time (or distance) between the origin and the point of coordinates $(\alpha n, n)$, then -$$ -\frac{T_n(\alpha) - \omega(\alpha) n}{\sigma(\alpha)n^{1/3}}\to X, -$$ -where $X$ has the Tracy-Widom distribution. Here $\omega(\alpha), \sigma(\alpha)$ are two constants whose value is known and is not important for this question. -I am interested in the following natural question: -Q1) Given $\alpha, \beta$, and letting $\gamma_n(\alpha)$ denote the geodesic between 0 and $(\alpha n, n)$, how many edges do $\gamma_n(\alpha)$ and $\gamma_n(\beta)$ share? -Intuitively, one possible way to approach this question is to first ask -Q2) How big is the covariance between between $T_n(\alpha)$ and $T_n(\beta)$? -The reason why these two questions seem related is that one would expect $\text{cov}(T_n(\alpha), T_n(\beta)) $ to be roughly proportional to the number of edges on $\gamma_n(\alpha)\cap \gamma_n(\beta)$. (At least this is what happens for deterministic paths). -Presumably, Johannson's result tells us that var$(T_n(\alpha))$ is of order $n^{2/3}$ (though it's not a straightforward consequence of that result), so Cauchy-Schwarz implies that the covariance is at most of order $n^{2/3}$. This would suggest that $|\gamma_n(\alpha)\cap \gamma_n(\beta)|$ is at most of order $n^{2/3}$. However, it is hard to believe that this is sharp! -Does anyone know if these questions have been studied ? And what if we only know that $\text{var} (T_n(\alpha)) = o(n)$ (as in this paper, http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aop/1068646373, by Benjamini Kalai and Schramm), does it follows that $|\gamma_n(\alpha) \cap \gamma_n(\beta)| = o(n)$? - -REPLY [7 votes]: EDIT: It was really remiss of me not to mention http://arxiv.org/abs/math-ph/0211040 by Patrik Ferrari and Herbert Spohn, where there are very nice results concerning scalings when the distance between the two directions varies with $n$. The case $\beta=\alpha+O(n^{-1/3})$ is the one where a very nice scaling picture should emerge; maybe http://arxiv.org/abs/1103.3422 (Corwin and Quastel) on the "Airy sheet" is the beginning of this. - -Hi Nathanael. The answer to Q1 will be $O(1)$. For concreteness consider the model with i.i.d. exponential weights at each site (the case with geometric weights at each site should also be fine but the geodesics are not unique so it's slightly messier). The path $\gamma_n(\alpha)$ converges almost surely to a path $\gamma(\alpha)$ as $n\to\infty$ (a "semi-infinite geodesic in direction $\alpha$"). So the question reduces to looking at the intersection of $\gamma(\alpha)$ and $\gamma(\beta)$. (The same behaviour should be seen across a wide class of models but showing that rigorously could be a challenge). -Coupier ( http://arxiv.org/abs/1104.1321 ) is a starting point for results and references concerning existence and uniqueness of such semi-infinite geodesics. -Actually one could certainly say some specific things about the law of the intersection of the two semi-infinite geodesics (in an exactly solvable model like the i.i.d. exponential weight case). The joint law of the geodesics in two different directions from the same point is very closely related to the equilibrium of an appropriately defined two-type particle system. In this case the relevant particle system would be a certain type of process like the Hammersley process, but in discrete time and space and with "fluid" rather than discrete particles. Eric Cator and I have done some work relating joint laws of geodesics to two-type equilibria in this way recently - it is not written up yet but I would be delighted to discuss. The closest thing on paper might be Eric's article with Leandro Pimentel at http://arxiv.org/abs/0901.2450 . The kind of calculations one could do concerning the two-type equilibria might resemble those done for the TASEP by Amir, Angel and Valko in http://arxiv.org/abs/0811.3706 (which make use of a queueing-type representation for the multi-type equilibrium of the kind that Pablo Ferrari and I developed in http://arxiv.org/abs/math/0501291 ).<|endoftext|> -TITLE: What is the relation between a ''homotopy fiber bundle'' and a Serre fibration? -QUESTION [8 upvotes]: Since I got no responses to this question at Stack Exchange, please let me try my luck here. -Call a continuous map $\pi:E\to B$ between CW complexes a homotopy fiber bundle if for any $x$ in the image of $\pi$, there is an open neighbourhood $U\subset B$ of $x$ and homotopy equivalence $\pi^{-1}(U)→U\times F$ over $U$. -I don't know if this has a different name in the literature or even if it is reasonable. Replacing ''homotopy equivalence'' by ''homeomorphism'' should be the definition of an ordinary fiber bundle. - -How relates a ''homotopy fiber bundle'' to the notion of a Serre fibration? - -At least both properties imply that the fibers over connected components are all weakly homotopy equivalent. - -REPLY [10 votes]: There is a nice theorem in geometric topology that a Serre fibration between -actual CW complexes (not just homotopy types thereof) is in fact a Hurewicz fibration. -It's in a paper by Steinberger and West (Covering homotopy properties of maps between -CW complexes or ANRs, Proc AMS92(1984), 573-577), with a correction in a paper by -R. Cauty (Sur les ouverts des CW-complexes et les fibr\'es de Serre, Colloq. Math. 63(1992), 1-7). -Then the rest of your answer applies. But of course the proof of the cited theorem proceeds -by checking homotopical local triviality and then quoting the local to global characterization -of Hurewicz fibrations.<|endoftext|> -TITLE: Operation on Isospectral graphs -QUESTION [5 upvotes]: Suppose $G$ and $H$ are two isospectral connected graphs. Can we say anything about isospectrality of graphs that are obtained by applying a binary operation to $G$ and $H$? -For example, to take one special case, is $G‎\otimes‎G$ (Kronecker product) isospectral? -Which binary operations between $G$ and $H$ preserve the isospectrality? - -REPLY [3 votes]: Chris seems to have forgotten that he and I published a generalization of the NEPS in -C. D. Godsil and B. D. McKay, Constructing cospectral graphs, Aequationes Mathematicae, 25 (1983) 257-268. This contains some general methods of using the tensor product, as well as some other techniques.<|endoftext|> -TITLE: Minimum separating subdivision in Plane -QUESTION [6 upvotes]: Hi -I was thinking about the following problem: -Given a planar Graph embedded in the plane and a set of points $P$ contained in the faces (no face contains more than one point) I want to determine the maximum number of edges I can remove such that in the reduced graph no two points are contained in the same face. -Although I didn't find any publications considering this problem, I guess this must have been studied before. -Any pointers would be helpful to me. -If $P$ only consists of two points $p,q$ it is easy since, we can just compute the shortest cycle containing either $p$ or $q$ but not both in its face. -Thank you -Andy - -REPLY [5 votes]: Your question is equivalent to (a version of) minimum $k$-cut AKA multiway cut. It seems that the general problem is solvable is polynomial time for any fixed $k$, but NP-complete for arbitrary $k$, even restricted to planar graphs. However, Mohammadhossein Bateni, MohammadTaghi Hajiaghayi, Philip Klein, Claire Mathieu give Polynomial time approximation schemes for the planar case.<|endoftext|> -TITLE: Lagrange's theorem for Hopf algebras -QUESTION [15 upvotes]: Under what conditions is a Hopf algebra free over any of its sub-Hopf algebras? -I am reading "Hopf algebras and their actions on rings" by Susan Montgomery, specifically chapter 3. Lagrange's theorem says that if $H$ is a subgroup of $G$ then $\mathbb kG$ is free as a $\mathbb kH$ module. When one tries to extend this to general Hopf algebras, in the finite dimensional setting everything is fine by the Nichols-Zoeller theorem which says that a finite dimensional Hopf algebra $H$ is free over any sub-Hopf algebra $K$. -Unfortunately this is not true in the infinite dimensional setting. A counterexample, which is both commutative and cocommutative, due to Oberst and Schneider, is as follows: Take $F\subset E$ a Galois field extension of degree $2$ with Galois group $G=\lbrace 1,\sigma\rbrace$, and let $\sigma$ act on $\mathbb Z$ by $z\to -z$. Then $G$ acts on $E$ and $\mathbb Z$ so one can take the Hopf algebra $H=(E\mathbb Z)^{G}$. One gets a counterexample showing that $H$ is not free over the sub-Hopf algebra $(E(n\mathbb Z))^G$ when $n$ is even. -The book then mentions that, because of this, perhaps one should look at something weaker than freeness, like being faithfully flat. Here, however, I want to focus on which Hopf algebras are free over their sub-Hopf algebras. It seems that in the literature there are several instances of this kind of result. For example this seems to hold for pointed Hopf algebras. Is there an intuitive explanation of what properties of a Hopf algebra guarantee that it satisfies Lagrange's theorem? -(For example, I do not see how I could have arrived naturally at the example above. A short explanation of that would be great, too.) Is there a reference? - -REPLY [6 votes]: Maybe you find useful Skryabin's paper. -Skryabin, Serge. Projectivity and freeness over comodule algebras. Trans. Amer. Math. Soc. 359 (2007), no. 6, 2597--2623 (electronic). MR2286047 (2008a:16060) -(Check http://arxiv.org/abs/math/0610657 for the preprint (free access).) -EDIT: Other two interesting papers are the following: -Radford, David. Freeness (projectivity) criteria for Hopf algebras over Hopf subalgebras. J. Algebra 45 (2), 1977, 266-273. http://www.sciencedirect.com/science/article/pii/0022404977900354. -Takeuchi, Mitsuhiro. Relative Hopf modules—Equivalences and freeness criteria. -J. Algebra 60 (2), 1979, 452-471 -http://www.sciencedirect.com/science/article/pii/0021869379900930<|endoftext|> -TITLE: Cosets representatives of congruence subgroups -QUESTION [7 upvotes]: I came across Shimura (1971) notes about cosets representatives of the congruence subgroups $ \Gamma_0(N) $. He firstly proves that its index in the modular group $\Gamma$ is -\begin{equation} [\Gamma : \Gamma_0(N)]=N \cdot \prod_{p|N} (1+p^{-1} ) \end{equation} -Then he comes up with a sets of cosets representatives for $\Gamma_0(N)$ in $\Gamma$ made in this way: we first choose pairs $(c,d)$ of positive integers such that -\begin{equation} (c,d)=1, \qquad d|N, \qquad 0 < c \le N/d \end{equation} -then for each pairs we fix integers $a,b$ such that $ad-bc=1$. Our list of cosets representatives is made of the matrices with such entries. -However, let us take for example $N=12$ when we know the index is 24 and thus this is the cardinality of the set of cosets representatives. -Using the rule above, I only find 22 cosets representatives, namely the ones corresponding to the following $(c,d)$ pairs: $$(1,1),(2,1),\dots,(12,1),(1,2),(3,2),(5,2),(1,3),(2,3),(4,3),(1,4),(3,4),(1,6),(1,12).$$ -I also tried to run SAGE and it gives me 24 cosets representatives but they seem redundant, for example $[[1, 0] [2, 1]]$ and $[[1, 2][2, 5]]$ are listed as different cosets representatives, but -$$\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}$$ -and thus it seems to me that these 2 matrices belong in fact to the same coset. -Something is clearly wrong, I hope you can help me. - -REPLY [2 votes]: (2022.03.27, edited after Tireless and hardworking's remark.) -The quotient $\Gamma_0(N) \backslash \mathrm{SL}_2(\mathbf{Z})$ (right cosets) is in bijection with the set of pairs $(c,d)$ where $c$ is a positive divisor of $N$ and $1 \leq d \leq \frac{N}{c}$ satisfies $\operatorname{gcd}(d,c,\frac{N}{c})=1$. -To get the image of a matrix $M = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \in \mathrm{SL}_2(\mathbf{Z})$, first find $u \in (\mathbf{Z}/N\mathbf{Z})^\times$ such that $\gamma u \equiv \operatorname{gcd}(\gamma,N) \bmod{N}$. Then $M$ is mapped to $(c,d):=(\operatorname{gcd}(\gamma,N), \delta u \bmod{\frac{N}{c}})$. It is a good exercice to check that this gives a well-defined and bijective map. -To get an explicit matrix representative associated to $(c,d)$, first find an integer $1 \leq \delta \leq N$ such that $\delta \equiv d \mod{\frac{N}{c}}$ and $\operatorname{gcd}(\delta,c)=1$, then find integers $\alpha,\beta \in \mathbf{Z}$ such that $\alpha \delta- \beta c=1$, and you get your representative $\begin{pmatrix} \alpha & \beta \\ c & \delta \end{pmatrix}$.<|endoftext|> -TITLE: Is the category of quotient of countably based topological spaces cartesian closed ? -QUESTION [8 upvotes]: In "Handbook of categorical algebra Vol 2" from Francis Borceux, the author gives a proof that $Top$ is not cartesian closed. It seems to me that this proof can be adapted to show that the category $\omega$-$QTop$ (the name is not standard) of quotient of countably based topological spaces is not cartesian closed either. -But in the paper 'Comparing cartesian closed categories of (core) compactly generated spaces' from Martín Escardó, Jimmie Lawson, Alex Simpson, -it is stated that the category of quotient of countably based topological spaces, is a cartesian closed category. -So there is probably something that I don't understand well somewhere... -I apologize for the size of my question. But I think this is necessary that I reproduce here my own adaptation of the proof from Francis Borceux's book (almost everything is identical): -Let $[0,1]$ and $\mathbb{Q}$ be topological spaces with their well-known topologies (the both spaces are second countable so there are in $\omega$-$QTop$). Let us suppose that $[0,1]^\mathbb{Q}$ exists in $\omega$-$QTop$. Then the topology on $[0,1]^\mathbb{Q}$ is in particular the coarest countably-based-quotient topology making the evaluation map $ev:[0,1]^\mathbb{Q} \times \mathbb{Q} \rightarrow [0,1]$ continuous. -The proof consists in showing that the fact right above implies that $\mathbb{Q}$ is locally compact, which is a contradiction, since it is not: - -Consider now the constant function $\delta:\mathbb{Q} \rightarrow [0,1]$ mapping everything to 0. -Take any point of $q$ of $\mathbb{Q}$. Since evaluation is continuous we have open neighborhood $V$ of $q$ and $U$ of $\delta$ such that $ev(U \times V) \subseteq [0,\frac{1}{2})$ -We will now show that $\overline{V}$ is compact. -Take an open cover $\bigcup W_i$ of $\overline{V}$. Add $\mathbb{Q} \setminus \overline{V}$ to the open cover. -Then we create of a topology $T$ on $C(\mathbb{Q}, [0,1])$ making the evaluation map continuous: -This is the topology generated by open sets of the form -$$T(A,B)=\{f|\ f(\overline{A} \subseteq B)\}$$ -for $A$ basic open of $\mathbb{Q}$ such that $\overline{A}$ is included in one $W_i$ and $B$ basic open of $[0,1]$. This topology is second coutable and so $(C(\mathbb{Q}, [0,1]), T)$ is in $\omega$-$QTop$. -Sine $[0,1]^\mathbb{Q}$ is structured with the coarest topology making the evaluation map continuous, there is an open $O=T(A_1,B_1) \cap \dots \cap T(A_n,B_n)$ of $T$ such that -$\delta \in O \subseteq U$. - -We now show that $V \subseteq \overline{A_1} \cup \dots \cup \overline{A_n}$. - -Suppose not, then take $x$ such that $x \in V$ and $x \notin \overline{A_1} \cup \dots \cup \overline{A_n}$. Since $\mathbb{Q}$ is completly regular, one can find -a continuous function $f$ such that $f(\overline{A_1} \cup \dots \cup \overline{A_n})=0$ and $f(x)=1$. -Since $f$ coincides with $\delta$ on each $\overline{A_i}$ we have that $f \in O \subseteq U$ but $ev(f,x) \notin [0,\frac{1}{2})$ which is a contradiction. -So $V \subseteq \overline{V} \subseteq \overline{A_1} \cup \dots \cup \overline{A_n} \subseteq W_{l_1} \cup \dots \cup W_{l_n}$. -Thus $V$ is compact. Since we can do the same for any $q \in \mathbb{Q}$ we have that $\mathbb{Q}$ is locally compact. which is a contradiction. -So $\omega$-$QTop$ is not cartesian closed... - -REPLY [6 votes]: Let $C$ be the category of quotients of countably based spaces, and continuous maps between them. The product $X \times Y$ of $X$ and $Y$ in $C$ is computed as the sequentialization of the usual topological product $X \times Y$. This means that we enlarge the topology of $X \times Y$ by all sequentially open subsets (a subset is sequentially open iff every convergent sequence whose limit is in the subset is eventually in the subset). As Todd mentions, there is a problem with your second bullet point. There exists an open neighborhood of $(q,\delta)$ in the space $\mathbb{Q} \times [0,1]^\mathbb{Q}$. But how do you know that the neighborhood may be assumed to be of the form $V \times U$? -The paper Topological and Limit-space Subcategories of Countably-based Equilogical Spaces by Alex Simpson and Matias Menni (Math. Struct. in Comp. Sci., 12:739-770, 2002) contains all the proof and more. I think it will help you see what is going on.<|endoftext|> -TITLE: Getting started: combinatorial optimization for computer scientists -QUESTION [5 upvotes]: I have a background in computer science and I am starting to work on some problems those are basically combinatorial optimization problems. -I have good knowleges of graphs, *-flow algorithms and so on and I took some courses about operations research and similar stuff. -I am looking for one (or two) book to get a uniform and semi-complete view on the topic. - -REPLY [7 votes]: For the short version, Combinatorial Optimization by Papadimitriou and Stieglitz is a good introduction, and at $12, you can't really go wrong. -For the in-depth version, Combinatorial Optimization by Schrijver is pretty encyclopedic.<|endoftext|> -TITLE: Isometric (?) embedding problem. -QUESTION [11 upvotes]: Given a convex surface $S$ in $\mathbb{R}^3,$ you can associate to each point the length of the inward pointing normal contained inside $S.$ Call this quantity $s(x).$ Now, given a positive function $f$ on the sphere $S^2,$ are necessary and sufficient conditions known so that $f$ can be realized as $s$ for some convex realization? There is the obvious uniqueness question, as well. The question can be formulated identically in $\mathbb{R}^n,$ but these sort of questions tend to get harder as $n$ gets larger. On the other hand, it is not obvious what the answer is even for $n=2.$ -EDIT There seem to be confusion about what I mean by "the length of inward pointing normal": Take the point $x.$ There is a line $l$ through $x$ normal to $S.$ Since $S$ is convex, $l$ intersects $S$ twice. The distance between the two intersection points is $s(x).$ - -REPLY [4 votes]: Dear Igor, -Since the answer to your problem is "no" I'm not going to dwell on that, but I feel there is something interesting behind your question that does not admit such a clear cut answer. -Two known facts: -1. Consider the manifold of oriented lines in three-space and, in particular, the open set $U$ formed by the lines that pass through the interior of your convex body. If you use the symplectic structure on the space of oriented lines in Euclidean space, then this open set -is symplectomorphic to the open unit co-disc bundle of your surface. In other words, you can read from $U$ not only the area of your surface (Cauchy-Crofton formula), but also everything related to its geodesic flow. -2. The set of oriented lines normal to your surface is a Lagrangian submanifold that lies inside $U$. -Now consider the John transform (X-ray transform) of the characteristic function of your convex body. Its value at a given line $\ell$ is the length of the segment of $\ell$ that lies in the convex body. In other words, the distance between the entrance point and the exit point. -Your problem can now be written as follows : Restrict the John transform of the characteristic function of a convex body to the Lagrangian submanifold formed by lines normal to its boundary, is this knowledge sufficient to reconstruct the body (i.e., the characteristic function)? This is now an inversion problem in integral geomety. -Generally speaking, one cannot reconstruct the John transform of a function (of three variables) from its restriction to a two-dimensional submanifold on the space of lines, but you can if you take certain three-dimensional submanifolds in the space of lines. Allow me to change the question: -Is there a geometrically-defined three-dimensional submanifold $\Sigma$ of the set of all oriented lines intersecting the interior of a convex body $K$ such that the John transform of $1_K$ can be reconstructed from its restriction to $\Sigma$? -In two dimensions the analogous question has a trivial answer: you need the two-dimensional submanifold of all lines passing through the convex body because of the inversibility of the Radon transform. In higher dimensions, the question looks interesting. -Going back to your original question: one may argue that the characteristic function $1_K$ is a very special function of three variables and that it "feels" like a function of two variables. So the question remains whether one can reconstruct $1_K$ from the knowledge of its John transform on some two-dimensional submanifold (or on some sort of "smallish set") on the space of lines.<|endoftext|> -TITLE: Trivial obstructions and virtual fundamental classes -QUESTION [7 upvotes]: Suppose $X$ is a DM stack, and let $E^\bullet$ be a perfect obstruction theory of $X$ such that the $E^{-1}$ term admits a trivial quotient/sub-bundle. Is it true that the virtual fundamental class $[X, E^\bullet]$ is zero? -If $X$ is smooth, then this is true: In such a case, the virtual fundamental class is the top Chern class of the vector bundle $E^{-1}$, which is zero due to the trivial quotient/sub-bundle and the exact sequence -$$0 \to \mathcal{O} \to E^{-1} \to coker \to 0$$ -together with the multiplicative nature of $c_{top}$. Intuitively, you can use the trivial factor to "move" a section of this bundle away from the zero section. -If $X$ is not smooth, then this argument doesn't work and we must use the intrinsic normal cone of Behrend and Fantechi to compute the virtual fundamental class. Instead of bundles, we obtain a cone $C(E^\bullet)$ contained in $(E^{-1})^\vee$, which we intersect with the zero section of $(E^{-1})^\vee$. In comparison with the smooth case, it seems like we should somehow be able to move the cone out of the zero section using the trivial portion of the bundle, but I don't see how to do this. -Is there an easy argument showing that this class is zero? Are there extra conditions required? -Edit: The method which I have tried to no avail is to use Proposition 5.10 of Behrend-Fantechi. It states loosely that, given two obstruction theories $F$ and $F'$, and certain compatibility data between them, that -$$ -v^![X,F] = [X,F'] -$$ -However, I was not able to find a clear way to have this yield that my desired virtual class is zero. - -REPLY [4 votes]: It turns out that this is true. In the paper "Localizing Virtual Cycles by Cosections" by Kiem and Li, they address the case where one has a surjection $Ob \to \mathcal{O}$. In the case of an injection $\mathcal{O} \to Ob$, one can produce via a diagram chase a corresponding surjection, which yields the claim.<|endoftext|> -TITLE: Gamma-function analogues for Gauss sums -QUESTION [7 upvotes]: I have a Gauss sum, which I have to calculate. I have heard that it has an analogues form with the Gamma function, but couldn't find its formula shape. It would be so nice of you to help me and write the mathematical shape of the link between these two functions. Beforehand thank you. - -REPLY [6 votes]: This is a great question. Suppose $K$ is a locally compact second countable field, and $K^*$ the multiplicitative group of nonzero elements of $K$. Suppose $\phi$ is a homomorphism of the additive group $K$ into $S^1$ and suppose $\chi $ is a homomorphism of the multiplicative group $K^*$ into $S^1$. If the integral -$$\int _{K^*} \phi (x) \chi (x)d^*x $$ makes sense (even as a distribution), then this can be called the "Gamma Function" of $K$. -If we take $K={\mathbb Z}/p{\mathbb Z}$ for a prime $p$, and $\phi(x)=e^{2\pi i x/p}$ , and $\chi (x)$ a character on $K^*$ then we get a Gauss sum. -If we take $K={\mathbb R}$, $\phi (x)=e^{x}$, and $\chi (x)= \mid x \mid ^s$then the above "integral" is the Gamma function. Here the integral , if interpreted as a functional on compactly supported functions on positive reals, makes sense. -In this sense, the Gamma function, or the Gauss sum, is the Mellin transform of an additive character of $K$ with respect to a multiplicative char of the group $K^*$ of non-zero elements.<|endoftext|> -TITLE: primes represented by a binary form -QUESTION [8 upvotes]: Let $D$ be a square free integer. I am looking at primes representable as $x^2+Dy^2$, where $x,y\in\mathbb Z$. I wonder whether it is always true that this set of primes is the union of finitely many arithmetic progressions intersected with the set of all primes? -This looks like a classical questions but I could never find an answer. It is certainly true if $\mathbb Z[\sqrt D]$ is a principal ideal domain. I think I had a proof in the case of class number two but I cannot remember it. It is of course true if one looks at norms of ideals rather than norms of elements but this is not what I want. - -REPLY [5 votes]: EDIT: evidently the Cox book is now print-on-demand, at WILEY -There are two situations where a positive integral form represents all the primes (at least those that do not divide the discriminant) given by arithmetic progressions. On situation, appropriate for your $x^2 + n y^2,$ is when $n$ is an idoneal number, which is when there is only one class per genus. This includes the case of only one genus, but also such thngs as $x^2 + 5 y^2,$ where the other class of that discriminant is in a different genus, $2 x^2 + 2 x y + 3 y^2.$ Note that we also know exactly what primes are represented by the second form. -It is also good enough when a genus of two classes is simply one class and its opposite, as in $3 x^2 \pm 2 x y + 5 y^2.$ Either form represents all primes with certain values modulo 56. Note that this cannot happen with the principal genus. -In general, there are elaborate conditions on what primes are represented by the principal form. Typical for class number 3: (Gauss) a prime $p \equiv 1 \pmod 3$ is represented by $x^2 + 27 y^2 $ if and only if 2 is a cubic residue $\pmod p,$ which is to say that $z^3 -2$ factors into three distinct linear factors $\pmod p.$ A prime $p > 3$ with $(-44 | p) =1$ is represented by $x^2 + 11 y^2$ if and only if $z^3 + z^2 -z+1 $ factors into three distinct linear factors $\pmod p.$ For all the class number three discriminants, see K. S. Williams and R. H. Hudson, Acta Arithmetica, vol. 57 (1991) pages 131-153, Representation of primes by the principal form of the discriminant $-D$ when the classnumber $h(-D)$ is 3 -For classnumber 4: (Gauss) a prime is represented by $x^2 + 64 y^2$ if and only if 2 has four distinct fourth roots $\pmod p,$ which is to say that $z^4 -2$ factors into four distinct linear factors $\pmod p.$ Note that $p \equiv 1 \pmod 8$ anyway. A prime $p > 3$ with $(-56 | p) =1$ is represented by $x^2 + 14 y^2$ if and only if $z^4 + 2z^2 -7 $ factors into four distinct linear factors $\pmod p.$ For all the class number four discriminants,see K. S. Williams and D.Liu, Tamkang Journal of Mathematics, vol. 25 (1994) pages 321-334, Representation of primes by the principal form of negative discriminant $\Delta$ when $h(\Delta)$ is 4 -For a table with all discriminants from $-3$ to $-451,$ all class numbers mixed together, see pages 539-542 in Advanced Topics in Computational Number Theory by Henri Cohen, 1999. The table is called "Discriminants and Hilbert class fields of imaginary quadratic fields." The polynomial given tells you, once a prime is represented by the principal genus of a discriminant, what polynomial must factor completely for the prime to be represented by the principal form itself. This table gives the polynomials $f_n(x)$ promised in Theorem 9.2 on page 180 of Primes of the form $x^2 + n y^2$ by David A. Cox, AMAZON -If you cannot find the two articles or the wonderful Cohen table, email me. My websites are currently down, some of it is usually posted there. The thing about Cohen's table is that it is difficult to find a polynomial that works, it is incredibly difficult to find one with small coefficients, compare some articles by Kaltofen and Yui.<|endoftext|> -TITLE: The composition of derived functors - commutation fails hazardly? -QUESTION [8 upvotes]: Hello, -When we have left exact functors $F: A \to B , G: B \to C$ (between abelian categories), we would like sometimes to state that $D(GF)=D(G)D(F)$ (functors between bounded below derived categories). It holds when $F$ sends some adopted to $F$ class into an adopted to $G$ class. -My question is whether this is only technical issue, or there is something meaningful behind. For example, is there a situation when this does not hold, and somehow thinking that it does hold give wrong answers. Or, is there a more general context in which it always holds. -Sasha - -REPLY [21 votes]: My favorite (counter)example is this: let $A$, $B$, $C$ be the categories of left modules over some rings $R$, $S$, $T$ (respectively), and let $F$ and $G$ be the functors of tensor product with some bimodules, that is $F(M)=K\otimes_RM$ and $G(N)=L\otimes_SN$, where $K$ is an $S$-$R$-bimodule and $L$ is a $T$-$S$-bimodule. Then the left derived functor $\mathbb D (GF)$ is the functor of derived tensor product with the underived tensor product $L\otimes_SK$, while the composition of derived functors $\mathbb D (G)\mathbb D(F)$ is the functor of derived tensor product with the derived tensor product $L\otimes_S^{\mathbb D}K$, that is $\mathbb D(GF)(M)=(L\otimes_SK)\otimes_R^{\mathbb D}M$ and $\mathbb D(G)\mathbb D(F)(M)=L\otimes_S^{\mathbb D}K\otimes_R^{\mathbb D}M$.<|endoftext|> -TITLE: Can a p-adic representation and its twist by a non-crystalline character both have nontrivial $D_{cris}$? -QUESTION [5 upvotes]: For a continuous irreducible representation -$\rho: G_{\mathbb{Q}_p}\rightarrow GL_n(\overline{\mathbb{Q}_p})$, -is it possible for both $D_{cris}(\rho)$ and $D_{cris}(\chi\otimes\rho)$ to be nonzero, where $\chi$ is some non-crystalline character? - -REPLY [7 votes]: Let me use Colmez' article "Representations triangulines" as a reference. Let $V$ be a repn which satisfies your condition. -By proposition 4.3, $V$ is trianguline. By proposition 4.10, the HT weight of $\chi$ has to be an integer. You can then assume that $\chi$ has finite order, and this implies that $V$ is potentially crystalline on an abelian extension of $Q_p$ (aka crystabelline). -Conversely, it seems likely that one can give examples of irreducible crystabelline representations $V$ such that $D_{cris}(V)$ and $D_{cris}(V \otimes \chi)$ are both nonzero, with $\chi$ of finite order (see the examples in 2.4 of Berger-Breuil's "Sur quelques representations potentiellement cristallines de $GL_2(Q_p)$").<|endoftext|> -TITLE: Interesting Applications of the Classical Stokes Theorem? -QUESTION [23 upvotes]: When students learn multivariable calculus they're typically barraged with a collection of examples of the type "given surface X with boundary curve Y, evaluate the line integral of a vector field Y by evaluating the surface integral of the curl of the vector field over the surface X" or vice versa. The trouble is that the vector fields, curves and surfaces are pretty much arbitrary except for being chosen so that one or both of the integrals are computationally tractable. -One more interesting application of the classical Stokes theorem is that it allows one to interpret the curl of a vector field as a measure of swirling about an axis. But aside from that I'm unable to recall any other applications which are especially surprising, deep or interesting. -I would like use Stokes theorem show my multivariable calculus students something that they enjoyable. Any suggestions? - -REPLY [7 votes]: You can tell your students that a clever use of Stokes theorem can give you a Fields medal. Indeed, the proof that the formality map given by M. Kontsevich is a $L_\infty$-morphism, is nothing else than Stokes theorem. A detailed account of this can be found in Deformation quantization of Poisson manifolds. Lett. Math. Phys. 66 (2003), no. 3, 157–216 or with more details in Déformation, quantification, théorie de Lie, 123–164, Panor. Synthèses, 20, Soc. Math. France, Paris, 2005 which is in English, contrary to its title.<|endoftext|> -TITLE: On the uncountability of zero sets -QUESTION [5 upvotes]: If $f$ is any real-valued function, we define its zero set $Z_f = \{ x : f(x) = 0 \}$. Obviously, the zero set of a nice function can be uncountable. e.g., if $f(x) = 0$ on an uncountable domain. -I would like a sufficient condition on functions $f : \mathbb R \to \mathbb R$ for which the following statement holds: $$\mbox{if $Z_f$ is uncountable, then it contains an interval}.$$ -If $X_t$ denotes a Brownian motion, then with probability one, the zero set of $X_t$ is homeomorphic to a Cantor set (hence is uncountable but contains no interval). Since $X_t$ is $\tfrac{1}{2}$-Hölder continuous, this is obviously not sufficient. -Edit: Due to Joel David Hamkin's elegant counterexample below, continuous differentiability is not a sufficient condition for the above statement to hold. Is there a natural sufficient condition? -Edit 2: Thanks, all. I've accepted Joel's answer because it doesn't seem like there is a solution to my problem at this level of generality. The motivation for the question comes from stochastic geometry. I take a realization of a random Riemannian metric $g$ on the Euclidean plane, and consider a certain geodesic $\gamma$. Such a curve is (a.s.) smooth but certainly not analytic. -Given the random environment $g$, the path of the geodesic is determined. I then look at the intersection of the geodesic with a given line segment or circular arc. This intersection could be empty, finite, countably infinite, or uncountable. Under the hypotheses in my model, I have already shown that it cannot be an interval. I was hoping that a general argument would reduce other cases of uncountability to that case, proving that the intersection is countable. I may just have to deal with the possibility it can be uncountable, or find a context-specific argument. - -REPLY [3 votes]: I can't imagine how to sell the following as a natural condition; anyway: Let assume: $f:\mathbb{R}\to\mathbb{R}$ is quasi-analytic at every point but (possibly) countably many exceptional points. So, if $Z_f$ is uncountable, it has an accumulation point $a$ where $f$ is locally quasi-analitic, and since all derivatives of $f$ vanish at $a$, $f$ is locally zero there.<|endoftext|> -TITLE: Recognizing classifying toposes -QUESTION [16 upvotes]: Suppose $\mathbb{T}$ is a geometric theory, $\mathcal{E}$ is a topos, and $M$ is a model of $\mathbb{T}$ in $\mathcal{E}$. Is there any sort of elementary condition on $M$ and $\mathcal{E}$ (or, even better, on the geometric morphism $\mathcal{E}\to \mathbf{Set}$) which would allow us to recognize $\mathcal{E}$ as the classifying topos of $\mathbb{T}$ and $M$ as the generic $\mathbb{T}$-model therein? -I feel like this is a long shot, but I thought I would ask anyway. -Edit: Of course, such a condition could not be expressed in the internal logic of $\mathcal{E}$ (even including non-geometric logic), since then it would be preserved in all slices $\mathcal{E}/X$. This is one reason I feel it's a long shot; but the example of principal bundles mentioned in the comments suggests that it's not an entirely unreasonable question. - -REPLY [4 votes]: This is several years late, but it may be helpful nonetheless. -As alluded to by Buschi, Olivia has given an explicit answer to this in Theorem 2.1.29 of her monograph Theories, Sites and Toposes: - -Let $\mathbb{T}$ be a geometric theory, $\mathcal{E}$ a Grothendieck - topos and $M$ a model of $\mathbb{T}$ in $\mathcal{E}$. Then - $\mathcal{E}$ is a classifying topos of $\mathbb{T}$ and $M$ is a - universal model (i.e. generic model) of $\mathbb{T}$ iff the following - conditions are satisfied: - -The family $F$ of objects which can be built from the interpretations in $M$ of the sorts, function symbols and relation - symbols over the signature of $\mathbb{T}$ by using geometric logic - constructions (i.e. the objects given by the domains of the - interpretations in $M$ of geometric formulae over the signature of - $\mathbb{T}$) is separating for $\mathcal{E}$. -The model $M$ is conservative for $\mathbb{T}$, that is for any geometric sequent $\sigma$ over the signature of $\mathbb{T}$, $\sigma$ is valid in $M$ if and only if it is provable in - $\mathbb{T}$. -Any arrow $k$ in $\mathcal{E}$ between objects $A$ and $B$ in the family $F$ of condition (1) is definable; that is, if $A$ (resp. $B$) - is equal to the interpretation of a geometric formula $\phi(\vec{x})$ - (resp. $\psi(\vec{y})$) over the signature of $\mathbb{T}$, there - exists a $\mathbb{T}$-provably functional formula $\theta$ from -$\phi(\vec{x})$ to $\psi(\vec{y})$ such that the interpretation of - $\theta$ in $M$ is equal to the graph of $k$. - - - -I'm not sure if you were looking necessary and sufficient conditions on $M$ and $\mathcal{E}$, or just merely sufficient conditions, but since this Theorem gives an 'iff' result, one might try and prove the sufficiency of certain (perhaps more intuitive) critiera on $M$ and $\mathcal{E}$ by checking against the conditions listed in this theorem, i.e. by proving results of the flavour: 'If $M$ and $\mathcal{E}$ satisfy condition $X$, then they satisfy the 3 conditions of this theorem.' -Extending this thought, I am curious to see how these conditions relate to the special case mentioned in Dylan's comment. In particular, how does weak contractibility relate to the conditions spelt out by Olivia? This is not obvious to me, but I haven't taken the time to properly work through the details.<|endoftext|> -TITLE: Connectedness of space of ergodic measures -QUESTION [12 upvotes]: Let $X = \Sigma_p^+ = \{1,\dots,p\}^\mathbb{N}$ and let $f=\sigma\colon X\to X$ be the shift map. Let $\mathcal{M}$ be the space of Borel $f$-invariant probability measures on $X$ endowed with the weak* topology. -Now $\mathcal{M}$ is a Choquet simplex, and hence connected. The geometry of its extreme points is a little more subtle. These extreme points are precisely the ergodic measures. Let $\mathcal{M}^e$ denote the collection of ergodic measures in $\mathcal{M}$. Note that $\mathcal{M}^e$ has some nice properties; for instance, there is a natural embedding from the space of Hölder continuous functions into $\mathcal{M}^e$ that takes $\phi$ to its unique equilibrium state $\mu_\phi$. The image of the embedding is the collection of Gibbs measures (for Hölder potentials). -Of course, there are many ergodic measures that do not arise as equilibrium states of Hölder continuous functions, and so I wonder which nice properties of the collection of Gibbs measures extend to $\mathcal{M}^e$. In particular: Is $\mathcal{M}^e$ connected? Path connected? I expect that it is, and that moreover this should happen whenever $X$ is a compact metric space and $f\colon X\to X$ is a continuous map satisfying the specification property, but I don't know a reference and don't yet see how to approach a proof. - -REPLY [6 votes]: I'll flesh out the consequences of Gerald's comment in a (CW-ed) answer. Lindenstrauss, Olsen, and Sternfeld showed in 1978 that if $S_1$ and $S_2$ are compact metrisable simplices such that the extremal points of $S_i$ are dense in $S_i$ for $i=1,2$, then there is an affine homeomorphism from $S_1$ to $S_2$; the unique (up to affine homeomorphism) compact metrisable simplex with the property that its extremal points are dense is called the Poulsen simplex. -In that same paper, it was shown that the Poulsen simplex has the property that its set of extremal points is arc-connected. Since $\mathcal{M}$ is a compact metrisable simplex whenever $X$ is a compact metric space and $f\colon X\to X$ is continuous, and the extremal points of $\mathcal{M}$ are precisely the ergodic measures $\mathcal{M}^e$, it follows that $\mathcal{M}^e$ is arc-connected whenever it is dense in $\mathcal{M}^e$. In particular, the strong specification property introduced by Bowen implies that periodic orbit measures are dense in $\mathcal{M}^e$ (Sigmund 1974), and since such measures are ergodic, this implies that $\mathcal{M}$ is the Poulsen simplex, and hence $\mathcal{M}^e$ is arc-connected, whenever $(X,f)$ has strong specification. -So that's not quite as constructive a proof as the approach following (Sigmund 1977) as suggested in Andrey's answer and the comment following, but it's certainly simpler to write down based on existing results.<|endoftext|> -TITLE: Integers n for which n divides $a^n-b^n$ -QUESTION [5 upvotes]: Hello guys! -Firstly, I'd like to start with a problem: Find all natural numbers n, for which n divides (10^n)-1. Then, I wanna generalize the statement as: Is it possible to find all solutions to the the statement: n divides (a^n)-1, where g.c.d(a,n)=1? Is there any method or algorithm at least, in this case? - -REPLY [4 votes]: I would just add a short sentence to the above answer. Since you are only interested in the case when $b=1$ there is a very simple elementary characterization: $n|a^n-1$ if and only if $\mathrm{ord}_n(a)|n$. That also explains the comment above that $3^m|10^{3^m}-1$. -Unfortunately I think it also means that there is no good `formula' for the answer to your problem.<|endoftext|> -TITLE: Zeros of "exponential" function -QUESTION [5 upvotes]: Define ${f}_{i}(x) = \sum_{j=1}^{i} (-1)^{i-j}{i \choose j}j^x$, where $i=1,2,3,...$ and $x \in \mathbb{R}$. -For integer $x \geq i$, ${f}_{i}(x)$ reduces to ${f}_{i}(x)=i!S(x,i)$, where $S(x,i)$ is Stirling Number of the Second Kind. So in this case, we are able to interpret ${f}_{i}(x)$ using some concrete combinatorial model, say the number of ways one can map a set of $x$ objects onto a set of $i$ objects. -Question one: How can we prove $x=1,2,...,i-1$ are zeros for ${f}_{i}$? Is there a model (as the mapping model for the integer $x \geq i$ case, for example) according to which we can intuitively see that such values must be zeros for ${f}_{i}$? (Those zeros don't seem obvious to me.) -Question two: Are $x=1,2,...,i-1$ the only zeros for ${f}_{i}(x)$, $x \in \mathbb{R}$? - -Edit: For motivation of the function see discussions here Explicit expression for recursively defined functions - -REPLY [18 votes]: [Edited to add sharper bound (number of sign changes) and connection -with "Descartes' Rule of Signs"] -Yes, the zeros at $x=1,2,\ldots,i-1$ are the only real zeros of $f_i$. -We prove that in general an "exponential polynomial" with $d+1$ nonzero -terms, i.e. $A(x) = \sum_{j=1}^{d+1} a_j \exp(\lambda_j x)$ -with distinct $\lambda_j \in {\bf R}$ and each $a_j \in {\bf R}^*$, -can have at most $d$ real roots, counted with multiplicity. -We use induction on $d$, the base case $d=0$ being trivial. -Suppose we've proved the case $d-1$ for some $d>0$. -Now $A_0(x) := e^{-\lambda_1 x} A(x)$ is an exponential polynomial, -with the same number of nonzero terms and the same roots as $A$, -whose $j=1$ term is constant. Hence $A'_0(x) := \frac{d}{dx}(A_0(x))$ is an -exponential polynomial with only $d$ nonzero terms. By the inductive -hypothesis, it thus has at most $d-1$ real roots. But by Rolle's -theorem (easily generalized to allow multiple roots), there's at least -one root of $A'_0(x)$ between each consecutive pair of roots of $A_0$, and -thus of $A$. Therefore $A$ can have no more than $d$ roots with -multiplicity. This completes the induction step and the proof. -In the case at hand, $d=i-1$, each $\lambda_j = \log j$, and each $a_j -= (-1)^{i-j} {i \choose j}$. Having already located $i-1=d$ distinct -real roots of $A$, we deduce that there are no others (and that each of -the known roots is simple). QED -[added later] It is no accident that in our case, where $A(x)$ -actually has $d$ distinct roots, the coefficients $a_j$ alternate -in sign when the $\lambda_j$ are listed in increasing order. -Indeed if we assume without loss of generality that -$\lambda_1 < \lambda_2 < \cdots < \lambda_{d+1}$ -then the number of real roots (counted with multiplicity) is at most -the number, call it $s$, of sign changes in the coefficient sequence -$(a_1,a_2,\ldots,a_{d+1})$; that is, at most the number of -$j \in \lbrace 1,2,\ldots,d \rbrace$ such that $a_j a_{j+1} < 0$. -To see this, we argue by induction as above, noting also that -$A_0(x) = \sum_{j=1}^{d+1} a_j \exp((\lambda_j - \lambda_1) x)$ -has $\lambda_j - \lambda_1 > 0$, so each coefficient -$(\lambda_j - \lambda_1) a_j$ of $A'_0(x)$ has the same sign as $a_j$ -for $j>1$. Thus if $a_1 a_2 < 0$ then $A'_0(x)$ has $s-1$ sign changes, -and we're done as before. If $a_1 a_2 > 0$ then $A'_0(x)$ -has $s$ sign changes. But then $A_0$ is monotone on $x \leq x_0$ -where $x_0$ is the smallest root of $A'_0(x)$. -Since the two leading terms of $A_0$ as $x \rightarrow -\infty$ -have the same sign, it follows that $A_0$ does not vanish on -$x \leq x_0$, and thus by Rolle that $A_0$ has no more real zeros -than its derivative. Therefore in this case too $A_0$ has at most $s$ -real zeros. This completes the inductive step and the proof. QED -This argument may feel familiar, most likely because the same technique -proves Descartes' "rule of signs", which bounds the number of positive -roots of an ordinary real polynomial by its number of sign changes. -Indeed this "rule of signs" is equivalent to the special case -$\lambda_j \in \lbrace 0, 1, 2, \ldots \rbrace$ of our bound, -when $A(x)$ is an ordinary polynomial in $e^x$.<|endoftext|> -TITLE: Residual $p$-finiteness of principal congruence subgroups -QUESTION [7 upvotes]: Let $\Gamma(N)$ be the principal congruence subgroup of level $N$ in $\mathrm{SL}_n(\mathbf{Z})$, where $n\geq 3$. Then $\Gamma(N)$ is residually $p$-finite for all primes $p$ dividing $N$. - -Can $\Gamma(N)$ be residually $p$-finite for any prime $p$ that does not divide $N$ ? - -On a related note: $\Gamma(N)$ is residually $p$-finite for only finitely many primes $p$. The proof I know is somewhat indirect: 1) (Rhemtulla) if a group is residually $p$-finite for infinitely many primes $p$, then it is orderable. 2) (Witte) no finite index subgroup of $\mathrm{SL}_n(\mathbf{Z})$, where $n\geq 3$, is orderable. Is there a more direct / hands-on proof? - -REPLY [12 votes]: No. In fact, I claim that if $G$ is any solvable group and $\phi : \Gamma(N) \rightarrow G$ is a surjection, then $G$ is a finite group and all primes that divide $|G|$ also divide $N$. The key is the following beautiful theorem of Lee and Szczarba. - -Theorem: If $n \geq 3$ and $\Gamma(N)$ is the level $N$ principal congruence subgroup of $\mathrm{SL}_n(\mathbb{Z})$, then $[\Gamma(N),\Gamma(N)] = \Gamma(N^2)$. - -See their paper -MR0422498 (54 #10485) -Lee, Ronnie; Szczarba, R. H. -On the homology and cohomology of congruence subgroups. -Invent. Math. 33 (1976), no. 1, 15–53. DOI link -Anyway, this implies that the derived series of $\Gamma(N)$ is -$$\Gamma(N) > \Gamma(N^2) > \Gamma(N^4) > \cdots.$$ -Any surjection to a solvable group thus contains $\Gamma(N^{2^k})$ in its kernel for some $k$. But it also follows from Lee-Szczarba's work that $\Gamma(M)/\Gamma(M^2)$ is an abelian group all of whose elements have order $M$. This implies that all the primes which divide the order of $\Gamma(N) / \Gamma(N^{2^k})$ also divide $N$. The desired result follows. - -REPLY [10 votes]: No – since $\mathrm{SL}_n(\mathbb{Z})$ has the congruence subgroup property the profinite completion of $\Gamma(N)$ is the same as the congruence subgroup of level $N$ in $\mathrm{SL}_n(\widehat{ \mathbb{Z}})$. -Since $\mathrm{SL}_n(\mathbb{Z}_q)$ does not have any quotient which is a $p$-group, one sees that $\Gamma(N)$ does not have $p$-quotients unless $p$ divides $N$.<|endoftext|> -TITLE: How to write popular mathematics well? -QUESTION [74 upvotes]: Recently, some classmates and I were lamenting the fact that our classmates in other disciplines had almost no conception of what we did, despite the large mathematics population at Waterloo. Instead of giving up in the face of a Very Hard Problem, one of us brought up a column popularizing physics that had a brief run in the school paper, and suggested that we author something similar for mathematics. The column will have some particular constraints that seem challenging to satisfy (self-contained week to week, 500-700 words, try to cover at least some of the current research at UW) but this question is a more general one. -In looking for resources and guidance to help with the writing we have come across several good discussions of topic. We have also found examples of good popular writing and a general discussion of presenting mathematics to a non-mathematical audience. -What we have not found, on MathOverflow or elsewhere, is a popular analogue of the well-answered question "How to write mathematics well?". A lot of the tactical advice of Knuth, Halmos, and others goes out the window when you answer their first question, "Who is your audience?" with "a general university educated public". - -What is your advice for writing good mathematics for a popular audience? What holds for all styles of writing and what is article or book specific? - -REPLY [10 votes]: Work with a concrete example instead of in general or the abstract. Ash & Gross do a wonderful job of this in Elliptic Tales. -Start with a question. (one which normal people might want to know about—not like a pure-number or pure-pattern question.) Tell me why I should care. -Draw pictures. -Tell me what it's like, not just what it is. Plenty of facts that follow logically from definitions are not obvious (or, they're only obvious after X hours / days / years of concentrated thought). As an introduction, you could legitimately describe (for example) a category of smooth compact manifolds as "squishy", or a point-set category as "stippled", without being either dishonest or pedantic. A category of polytopes is "solid" whereas a conformal category "is flexible, but makes things line up". (Being truthful with these adjectives pays too. Many people repeat the phrase "rubber-sheet geometry" for topology, but rubber is elastic, and AT invariants wouldn't be apparent with rubber whereas Hooke's-law type behaviour would be.) Choosing good metaphors, like finding good mathematics, requires taste and skill. -Tell me what it does for you, not what it is. "Compact manifolds help me think about more shapes than I could otherwise, without having to invent too many new tools." "Fourier analysis helps me think about waves." -You can lie a little. (Especially if you follow it up with a remark like "The above isn't exactly true, but I hope it gives you the impression. See [ABC] for the real story.") The Bohr atom is a lie, and not a terribly bad one—especially situated between the plum-pudding model and the spdf model. -Think about the material you're competing with for their interest. Look at the other books in the bookstore or library. Or the magazine rack, or television …. -What is it emotionally that originally grabbed you about [whatever you're writing about]? A mathematical book is necessarily going to be cerebral to some degree, but a person is not a person without feelings. Emotional disingenuity makes prose ugly. -Assume you don't have the reader's full brainpower. Assume they're distracted by other things in their lives and therefore won't pause at the end of a sentence to reflect on its obvious implications. (Here's a place where concision is bad: do some work for the reader; fill in what are the ramifications of what you just said. Ash & Gross again achieve a "Goldilocks" balance with respect to the problem of too many vs too few sentences.) -Assume the reader might skip around, rather than reading sequentially. -Never assign exercises. You're not the boss. -More generally, the reader is the boss. You have to "earn" their time. You have to figure out what they're interested in and meet them there. -Keep it short. More insights per minute = more respect for the reader's time. (I 'll draw the analogy to Edward Tufte's dictum in The Visual Display of Quantitative Information. Focussed, short documents are good. Information density is good. When we define that to be information meaningfully imparted (received and successfully processed) by the reader, per word.) -Variables can be words rather than letters, and equations can be paragraphs. base^exponent is better than m^n (especially three pages later when you refer back to m). And whilst the quadratic equation is simpler as a formula, the Birch/Swinnerton-Dyer conjecture would be easier understood as a paragraph than as . -Jargon is your enemy. Nobody will understand phrases like "complex simple ____". The need to rephrease this is an invitation to think more deeply through what you would normally take for granted. -Omit needless words (like "interesting"), but do include transition sentences and overview sentences. -Read style guides and good non-mathematical authors. Mathematics has a lot, lot, lot of substance to offer the general public. (As opposed to other factoids they might read, which stick to the ribs as much as a Jolly Rancher.) That's a strength you get for free just by the selection of topic, and the work already done by researchers. But the "substance over style" culture has downsides. In particular it's the main reason most people hate mathematics (they never penetrate through the style to suck out the juicy substance). -Pretend that every time you write a sentence, you are also implicitly saying "This is important". If you wouldn't say "This is important", then the idea can be sent to a footnote or appendix, or excised. -Il semble que la perfection soit atteinte non quand il n'y a plus rien à ajouter, mais quand il n'y a plus rien à retrancher. ―St-Exupéry, Terre des Hommes, 1939 -Play around. - -That's a lot of advice, and while I may not be qualified to give it, I think you'll find I've merely collated the advice of people who really do know what they talk about. - -Added: Bill Thurston, Three-dimensional geometry and topology, reader's preface: - -The most efficient logical order for a subject is usually different from the best psychological order in which to learn it. Much mathematical writing is based too closely on the logical order of induction in a subject, with too many definitions before, or without, the examples which motivate them, and too many answers before, or without, the questions they address.<|endoftext|> -TITLE: Short Course Suggestions For High School Students -QUESTION [13 upvotes]: I am planning to teach a course for talented high school students at a summer camp and I need suggestions for possible topics. The students usually have different backgrounds but most of them are familiar with single variable calculus and very basic linear algebra over the reals. The teaching format will be two hours per day, six days per week and two weeks in total. Suggestions for one week courses are also welcome. -There are two things I want about this course. First, it should have a direction and a final goal. So it shouldn't be based on isolated Olympiad-type problems. Second, it should introduce at least one new concept or object which is not a part of the high school curriculum. -For instance, classification of frieze patterns is a good topic. There is a clear goal and one needs to introduce the concept of a group which is new for high school students. Any other suggestions? - -REPLY [3 votes]: You can do Monsky's theorem, that a square cannot be divided into an odd number of equal area triangles. -On the way you will have to do - -p-adic numbers, -Sperner's lemma, -present $\mathbb{R}$ as a vector space over $\mathbb{Q}$. - -In case if you have more time, you could - -use Sperner's lemma to prove Brouwer's fixed point theorem. -use (3) to do Dehn Invariants -and I am sure you can find what to do with p-adic numbers<|endoftext|> -TITLE: A variant of the Stone-Weierstrass theorem? -QUESTION [6 upvotes]: I would like to ask specialists in C*-algebras if the following variant of the Stone-Weierstrass theorem is true. -Suppose $A$ is a C*-algebra and $C$ is its center. Since $C$ is a commutative C*-algebra, there exists a compact space $T$ such that $C$ is isomorphic to the algebra $C(T)$ of continuous functions on $T$. Does this mean that there exists a C*-algebra $B$ such that -1) $A$ is isomorphic to a closed subalgebra in the algebra $C(T,B)$ of continuous mappings $f:T\to B$ with the pointwise algebraic operations (and the topology of uniform convergence on $T$), and -2) this isomorphism turns $C$ into the algebra of scalar mappings, i.e. the mappings of the form $f(x)=\lambda(x)\cdot 1_B$, where $1_B$ is the identity in $B$, and $\lambda(x)$ $\in$ $\mathbb{C}$ for all $x\in T$. -EDIT 21-03-12: All the C*-algebras here are supposed to be unital, excuse me for not mentioning this from the very beginning! - -REPLY [10 votes]: The statement of the Dauns-Hofmann theorem is actually too weak to get bundles of $C^*$-algebras. For completeness, let me state it: -Let $A$ be a $C^*$-algebra. For each $P \in Prim(A)$, let $\pi_P \colon A \to A/P$ be the quotient map. Then there is an isomorphism $\phi$ of $C_b(Prim(A))$ onto the center $ZM(A)$ of the multiplier algebra $M(A)$ such that for all $f \in C_b(Prim(A))$ and $a \in A$ -$$ -\pi_P(\phi(f)a) = f(P)\pi_P(a) -$$ -for every $P \in Prim(A)$. Usually one writes $f \cdot a = \phi(f)a$. -So, the best you could hope for is some kind of sheaf of $C^*$-algebras over the primitive ideal space. Getting local triviality in general is kind of hopeless, I think. A reading recommendation for these matters would be the book "Morita Equivalence and Continuous-Trace $C^*$-algebras" by Raeburn and Williams. -For continuous trace $C^*$-algebras things are quite different. These are all Morita equivalent (or stably isomorphic) to sections in a bundle of compact operators!<|endoftext|> -TITLE: undecidable sentences of first-order arithmetic whose truth values are unknown -QUESTION [25 upvotes]: Godel's undecidable sentences in first-order arithmetic were guaranteed to be true, by construction. But are there examples of specific sentences known to be undecidable in first-order arithmetic whose truth values aren't known? I'm thinking, by contrast, of the situation in set theory: CH is undecidable in ZFC, but its truth value is, in some sense, unknown. -Paris and Harrington showed the strengthened finite Ramsey theorem is true (in the sense of provable in second-order arithmetic) but undecidable in first-order arithmetic. I'm asking for "natural" examples in this general vein -- but whose truth values haven't yet been settled. -EDIT. Let me clarify my interest in the question, which is more philosophical than mathematical. I asked it on the basis of the following passage in Peter Koellner's paper "On the Question of Absolute Undecidability": - -The above statements of analysis [i.e. all projective sets of reals are Lebesgue measurable] and set theory [i.e. CH] differ from the early arithmetical instances of incompleteness in that their independence does not imply their truth. Moreover, it is not immediately clear whether they are settled at any level of the hierarchy. They are much more serious cases of independence. - -What I'm asking is whether there are "much more serious cases" of independence even in first-order arithmetic -- and not in the trivial case of full-on ZFC, like V=L, etc. By a sentence with "unknown truth value," I just mean a sentence that hasn't been proved in a theory stronger than first-order arithmetic. (For example, Paris and Harrington proved the strengthened finite Ramsey theorem in second-order arithmetic.) - -REPLY [21 votes]: Update. I've improved the argument to use only the consistency of $T$. (2/7/12): I corrected some over-statements previously made about Robinson's Q. - -I claim that for every statement $\varphi$, there is a variant way -to express it, $\psi$, which is equivalent to the original -statement $\varphi$, but which is formally independent of any -particular desired consistent theory $T$. -In particular, if $\varphi$ is your favorite natural open question, -whose truth value is unknown, then there is an equivalent -formulation of that question which exhibits formal independence in -the way you had requested. In this sense, every open question is -equivalent to an assertion with the property you have requested. I -take this to reveal certain difficult subtleties with your project. -Theorem. Suppose that $\varphi$ is any sentence and $T$ is any consistent theory containing weak arithmetic. Then there is another sentence $\psi$ such that - -$\text{PA}+\text{Con}(T)$ proves that $\varphi$ and $\psi$ are equivalent. -$T$ does not prove $\psi$. -$T$ does not prove $\neg\psi$. - -Proof. Let $R$ be the Rosser sentence for $T$, the self-referential assertion that for any proof of $R$ in $T$, there is a smaller proof of $\neg R$. The Gödel-Rosser theorem establishes that if $T$ is consistent, then $T$ proves neither $R$ nor $\neg R$. Formalizing the first part of this argument shows that $\text{PA}+\text{Con}(T)$ proves that $R$ is not provable in $T$ and hence that $R$ is vacuously true. Formalizing the second part of this argument shows that $\text{Con}(T)$ implies $\text{Con}(T+R)$, and hence by the incompleteness theorem applied to $T+R$, we deduce that $T+R$ does not prove $\text{Con}(T)$. Thus, $T+R$ is a strictly intermediate theory between $T$ and $T+\text{Con}(T)$. -Now, let $\psi$ be the assertion $R\to (\text{Con}(T)\wedge \varphi)$. Since $\text{PA}+\text{Con}(T)$ proves $R$, it is easy to see by elementary logic that $\text{PA}+\text{Con}(T)$ proves that $\varphi$ and $\psi$ are equivalent. -The statement $\psi$, however, is not provable in $T$, since if it were, then $T+R$ would prove $\text{Con}(T)$, which it does not by our observations above. -Conversely, $\psi$ is not refutable in $T$, since -any such refutation would mean that $T$ proves that the hypothesis -of $\psi$ is true and the conclusion false; in particular, it -would require $T$ to prove the Rosser sentence $R$, which it does not by the Gödel-Rosser theorem. QED -Note that any instance of non-provability from $T$ will require the consistency of $T$, and so one cannot provide a solution to the problem without assuming the theory is consistent. -The observation of the theorem has arisen in some of the philosophical literature you may -have in mind, based on what you said in the question. For example, the claim of the theorem is mentioned in Haim Gaifman's new -paper "On ontology and realism in mathematics," which we read in my course last semester -on the philosophy of set theory; see the discussion on page 24 of Gaifman's paper and specifically footnote 35, where he credits a fixed-point argument to Torkel Franzen, and an independent construction to Harvey Friedman. - -My original argument (see edit history) used the sentence $\text{Con}(T)\to(\text{Con}^2(T)\wedge\varphi)$, where $\text{Con}^2(T)$ is the assertion $\text{Con}(T+\text{Con}(T))$, and worked under the assumption that $\text{Con}^2(T)$ is true, relying on the fact that $T+\text{Con}(T)$ is strictly between $T$ and this stronger theory. The current argument uses the essentially similarly idea that $T+R$ is strictly between $T$ and $T+\text{Con}(T)$, thereby reducing the consistency assumption.<|endoftext|> -TITLE: Find weak equivalences from fibrations and cofibrations -QUESTION [9 upvotes]: Let's suppose I have a category $\mathcal C$ with two weak factorisation systems $(C,F_W)$ and $(C_W,F)$, where $C_W\subset C$ and $F_W\subset F$. -I would like to have a model structure on $\mathcal C$ such that $C$ (resp. $C_W$) are the cofibrations (resp. the acyclic cofibrations) and $F$ (resp. $F_W$) are the fibrations (resp. the acyclic fibrations). -1. Is there always at most one possible choice for the class of weak equivalences? (I've heard that in a model structure, having two classes of arrows determines uniquely the third one, but I don't know how to do this when it's the weak equivalences that I want) -2. Under what conditions does a model structure indeed exists. - -REPLY [5 votes]: Charles and Jonathan have given good answers to (1), but here's another way to recover the weak equivalences from the cofibrations and fibrations, which I think is sometimes convenient. -First of all, the wfss give us notions of "fibrant replacement" (a $(C_W,F)$-factorization of $X\to 1$) and "cofibrant replacement" (a $(C,F_W)$-factorization of $0\to X$). If $F_W\subseteq F$ (equivalently $C_W\subseteq C$), then we have two notions of fibrant-cofibrant replacement. -Now the wfs $(C_W,F)$ defines a notion of "path object", namely a $(C_W,F)$ factorization of the diagonal $X\to P X \to X\times X$. From this we can obtain a notion of "right homotopy": two maps $A\to X$ are right homotopic if the induced map into $X\times X$ factors through some path object for $X$. Dually, $(C,F_W)$ defines a notion of "cylinder object" and thereby "left homotopy". -If the two given wfss underlie a model structure, then the following all characterize the class of weak equivalences: - -$f\in W$ iff some (hence any) fibrant-cofibrant replacement of $f$ is a left (or right) homotopy equivalence (i.e. becomes an isomorphism upon quotienting by homotopy) -$f\in W$ iff some (hence any) cofibrant replacement $f'$ of $f$ has the property that it induces an isomorphism on right homotopy classes of maps into any fibrant object. -Dually, $f\in W$ iff some (hence any) fibrant replacement $f'$ of $f$ has the property that it induces an isomorphism on left homotopy classes of maps out of any cofibrant object. - -In contrast to Charles' and Jonathan's answers, with these definitions it's almost automatic to get 2-out-of-3 (there's probably some fiddlyness with some vs any, but if you have a given functorial realization of your wfss you could just use that factorization to define the replacements). Moreover, $C_W$ automatically has the second property, while $F_W$ automatically has the third. Thus, if you can show that the three definitions agree, then you get $C_W \subseteq C\cap W$ and $F_W \subseteq F\cap W$. The tricky part would now be showing the reverse inclusions (although a standard argument implies that you only need to show one of them).<|endoftext|> -TITLE: The Riemann's Zeta Function represented as a continued fraction and a question of convergence. -QUESTION [12 upvotes]: The Riemann's zeta function can be expressed as a continued fraction as follows -\begin{align*} -\zeta(z)=\newcommand{\bigk}{\mathop{\Huge\vcenter{\hbox{K}}}}\left(1-\bigk_{k=1}^{\infty }\frac{-e^{-2\cdot (coth^{-1}(2k+1))\cdot z}}{1+e^{-2\cdot (coth^{-1}(2k+1))\cdot z}}\right)^{-1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1) -\end{align*} -and its reciprocal as -\begin{align*} -\frac{1}{{\zeta(z)}}=\newcommand{\bigk}{\mathop{\Huge\vcenter{\hbox{K}}}}1-\bigk_{k=1}^{\infty }\frac{-e^{-2\cdot (coth^{-1}(2k+1)) \cdot z}}{1+e^{-2\cdot(coth^{-1}(2k+1))\cdot z}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2) -\end{align*} -Proof: -Note that -\begin{align*} -ln(n)=2 \sum_{m=1}^{n-1} \sum_{k=0}^{\infty}\frac{1}{2k+1}\left( \frac{1}{2\cdot m+1} \right)^{2k+1} -\end{align*} -and that -\begin{align*} -coth^{-1}(2 n +1)=\sum_{k=0}^{\infty}\frac{1}{2k+1}\left( \frac{1}{2\cdot n+1} \right)^{2k+1} -\end{align*} -so -\begin{align*} -ln(n)=2 \sum_{m=1}^{n-1} coth^{-1}(2 m +1) -\end{align*} -so teh Riemann's zeta function is expressed as -\begin{align*} -\zeta(z)=1+\sum_{n=1}^{\infty}\prod_{k=1}^{n-1}e^{-2\cdot(coth^{-1}(2k+1))\cdot z} -\end{align*} -and using Euler's continued fraction formula the result follows. -\begin{equation*} - \zeta(z)= \cfrac{1}{ - 1- \cfrac{e^{-2(coth^{-1}(3))z}}{ - 1+e^{-2(coth^{-1}(3))z}- \cfrac{e^{-2(coth^{-1}(5))z}}{ - 1+e^{-2(coth^{-1}(5))z}- \cfrac{e^{-2(coth^{-1}(7))z}}{ - 1+e^{-2(coth^{-1}(7))z} - \ddots}}}} -\end{equation*} -wich in Gauss' notation is (1) -Now considere -\begin{align*} -f(z)=\newcommand{\bigk}{\mathop{\Huge\vcenter{\hbox{K}}}}\bigk_{k=1}^{\infty }\frac{-e^{-2\cdot (coth^{-1}(2k+1))\cdot z}}{1+e^{-2\cdot(coth^{-1}(2k+1))\cdot z}} -\end{align*} -Using Śleszyński–Pringsheim theorem we can see that $f(z)$ converges for $\Im{z}=0$ and $\Re{z}\geq 0$. This is saying that $1/\zeta(z)$ converges for $x\geq 0$. -My question: can a bigger region of convergence be found using the theory of continued fractions? - -REPLY [13 votes]: (Too long for a comment.) -There's a (somewhat) simpler (Eulerian) continued fraction: -$$\sum_{k=1}^\infty \frac1{k^s}=1+\sum_{k=2}^{\infty} \prod_{j=2}^k \left(1-\frac1{j}\right)^s=\cfrac1{1-\cfrac{\left(1-\frac12\right)^s}{1+\left(1-\frac12\right)^s-\cfrac{\left(1-\frac13\right)^s}{1+\left(1-\frac13\right)^s-\cfrac{\left(1-\frac14\right)^s}{1+\left(1-\frac14\right)^s-\cdots}}}}\;\;\;\;\;\;$$ -but as you can see from comparing successive convergents of this continued fraction and the successive partial sums of the Dirichlet series, it's not terribly useful. -Also, -$$e^{-2z\,\mathrm{arcoth}(2k+1)}=\left(\frac{k}{k+1}\right)^z$$ -so your CF could certainly be simplified a fair bit...<|endoftext|> -TITLE: Do the algebraic integers form a free abelian group? -QUESTION [30 upvotes]: It is a well-known fact, proved in every introductory textbook on algebraic number theory, that if $K$ is an algebraic number field, i.e. a finite extension of $\mathbb{Q}$, then its ring $\mathcal{O}_K$ of integers is a free abelian group. -Does this statement still hold for arbitrary algebraic extensions of $\mathbb{Q}$? In particular, is the underlying abelian group of the ring $\mathcal{O}_{\overline{\mathbb{Q}}}$ of all algebraic integers free abelian? -Should this be true, I am also interested whether anything is known about the dependence of this statement on the axiom of choice, and similar logical questions. - -REPLY [36 votes]: Pontryagin's criterion says that, for a countable, torsion-free, abelian group to be free, it suffices that every finitely many elements lie in a finitely generated pure subgroup. The rings $\mathcal O_K$ for finite extensions $K$ of $\mathbb Q$ show, in view of the result you quoted, that this criterion is satisfied. The axiom of choice is not needed for any of this, because the question is absolute between the full universe $V$ and Gödel's constructible universe $L$.<|endoftext|> -TITLE: Finding Presentations of Groups with GAP -QUESTION [5 upvotes]: Group $A_5$ has presentation $〈 a, b | a^2 = b^3 = (ab)^5 = 1 〉$. Items equal to 1 are relators, so a presentation of $A_5$ as a set of relators could be -$(a^2, b^3, (ab)^5)$ -$Q_{16}$ is SmallGroup(16,9) with $〈 a, b | a^4 = b^2 = abab 〉$. -More groups of size 16 -In GAP, $A_5$ is SmallGroup(60,5). The following code: -RelatorsOfFpGroup(Image(IsomorphismFpGroup(SmallGroup(60,5)))); -will give -[ F1^5*F2^-5, F1^5*F2^-1*F1^-1*F2^-1*F1^-1, F1^-2*F2^2*F1^-2*F2^2 ] -That's not what I'm looking for. How can I get GAP to give me a presentation, or a minimal set of relators? Is there a single line piece of code that will work for most SmallGroup items? - -REPLY [7 votes]: I am not sure what exactly you are looking for (GAP IS giving you some presentation). However, if you are looking for a simpler presentation, check out this discussion: -http://mail.gap-system.org/pipermail/forum/2011/003313.html<|endoftext|> -TITLE: GCH+ Kurepa Families -QUESTION [7 upvotes]: I have a couple of questions about known theorems for GCH+Kurepa families. -Definition first: Let $\kappa$ be a infinite cardinal. A $\kappa^+$ Kurepa family is a family $F$ of subsets of $\kappa^+$ such that $F$ has size $>\kappa^+$ and for every $a<\kappa^+$, the set $\lbrace X\cap \alpha|X\in F\rbrace$ has size $\le\kappa$. -(The definition can be given in terms of tress too). -$KH(\kappa^+)$ is the statement that a $\kappa^+$ Kurepa family exists. -Please correct me, if I am mistaken, but we know that $KH(\kappa^+)$ holds for all infinite $\kappa$ in $L$ (the constructible universe). Also, if $\lambda$ is an inaccessible cardinal and we collapse $\lambda$ to $\aleph_2$, then in the generic extension $KH(\aleph_1)$ fails. -(Look also this On the independence of the Kurepa Hypothesis) -So, my questions are: -1) Do we know any models where GCH holds and $KH(\kappa^+)$ fails for all $\kappa$? -2) If this is not the case, can we at least have GCH+ the failure of $KH(\aleph_{\alpha+1})$, for all $\alpha$ countable ordinals? -3) If (2) is not known either, then fix some $\alpha$ countable ordinal $>0$. Can we have GCH+ the failure of $KH(\aleph_{\alpha+1})$? -4) If the ground model satisfies GCH, after we collapse an inaccessible cardinal to $\aleph_2$ do we still get GCH? -I am sure if I am asking too much. I just want to see what we already know. -PS. What is the right way to pronounce Kurepa? Is it KUrepa (stress on KU), or KuREpa (stress on RE), or KurePA? - -REPLY [7 votes]: Hi Ioannis! I guess you might know the answer by now; if we suppose for 1 that there is a class of inaccessible cardinals in the ground model and force with an Easton support product of Lévy collapses between the inaccessibles, we obtain a model of GCH where $KH(\kappa^+)$ fails for all $\kappa$; the argument is the same as for $\omega_1$. Also the failure of $KH(\kappa^+)$ implies that $\kappa^{++}$ is inaccessible in $L$, so we need the inaccessibles.<|endoftext|> -TITLE: Finite support iterations of $\sigma$-centered forcing notions -QUESTION [13 upvotes]: I am looking for a proof (or better, a reference) of the following fact: -The finite support iteration of $\sigma$-centered forcing notions is again $\sigma$-centered, assuming we iterate less than $(2^{\aleph_0})^+$ steps. -(EDIT: In the first version of the question I forgot to mention that it was Stefan Geschke who suggested that there should be a proof similar to "the product of continuum many separable spaces is still separable") -EDIT: In his 1994 paper "$\sigma$-centred forcing and reflection of (sub)metrizability" in TAMS (MR1179593 94g:54003), Frank Tall writes: - -"It is well known (proved by the same method that proves the product of $\le 2^{\aleph_0}$ separable spaces is separable) that the finite support iteration of $\le 2^{\aleph_0}$ $\sigma$-centred orders is $\sigma$-centred." - -REPLY [7 votes]: I just wanted to point out how Andreas' proof above relates to the separability of products of $2^{\aleph_0}$ separable spaces: -Let $\lambda<(2^{\aleph_0})^+$. -Suppose $\langle X_\alpha:\alpha<\lambda\rangle$ is a family of separable spaces. For each $\alpha<\lambda$ let $\langle x_\alpha^n:n\in\omega\rangle$ enumerate a dense subset of $X_\alpha$. -Let $F:\lambda\times\omega\to\omega$ be as in Andreas' proof. -Now the set of sequences of the form -$\langle x_{\alpha}^{F(\alpha,n)}:{\alpha<\lambda}\rangle$, $n\in\omega$, is dense in $\prod_{\alpha<\lambda}X_\alpha$. -Also note that $\sigma$-centeredness of a forcing notion $\mathbb P$ corresponds to the separability of the Stone space of the completion of $\mathbb P$. -However, I don't know whether it is possible to deduce the $\sigma$-centeredness of short finite support iterations of $\sigma$-centered forcing notions directly from the separability of small product of separable spaces.<|endoftext|> -TITLE: Hopf algebras examples -QUESTION [17 upvotes]: Following Richard Borcherds' questions 34110 and 61315, I'm looking for interesting examples of Hopf algebras for an introductory Hopf algebras graduate course. -Some of the examples I know are well-known: - -Examples related to groups and Lie algebras, -Sweedler and Taft Hopf algebras, -Drinfeld-Jimbo quantum groups. - -(These examples can be found for example in Montgomery's book on Hopf algebras.) -Other interesting examples are Lusztig small quantum groups and the -generalized small quantum groups of Andruskiewitsch and Schneider. -Therefore I'm looking for interesting and non-standard examples maybe related to combinatorics or to other branches of mathematics. - -REPLY [2 votes]: To complement the answer about Hopf algebras in categories other than Vect: in the monoidal categories of stereotype spaces, (Ste,$\circledast$) and (Ste,$\odot$) (which generalize the category (Vect,$\otimes$) of vector spaces) there are lots of examples of Hopf algebras. Some of them are generalizations of the standard constructions, like stereotype group algebras, some others generalize quantum groups, some others appear as envelopes of already constructed Hopf algebras. A detailed explanation can be found here.<|endoftext|> -TITLE: Nonzero digits in n! -QUESTION [10 upvotes]: Can it be shown that a positive fraction of the base-$b$ digits of n! are nonzero (in the limit as $n\to\infty$)? - -REPLY [11 votes]: The bound given by F. Luca and cited in Gjergji Zaimi's answer has been recently improved, although only by a factor $\log \log \log n$. Precisely, in C. Sanna, On the sum of digits of the factorial, J. Number Theory 147 (2015), 836--841, the author proved that -$$\min\{s_b(n!),s_b(\Lambda_n)\} > c_b \log n \log \log \log n , $$ -for all integers $n > e^e$, where $s_b(\cdot)$ is the sum of base-$b$ digits function, $c_b > 0$ is a constant depending only on $b$, and $\Lambda_n$ is the least common multiple of $1, \ldots, n$. Clearly, this inequality also holds with $s_b$ replaced by $t_b$, i.e., the function counting the number of non-zero base-$b$ digits of its argument, since $t_b(m) \geq s_b(m)/b$ for each positive integer $m$.<|endoftext|> -TITLE: Toposes (topoi) as classifying toposes of groupoids -QUESTION [22 upvotes]: A famous theorem of Joyal and Tierney says that each Grothendieck topos is equivalent to the classifying topos of a localic groupoid. I believe that Butz and Moerdijk have shown that if the topos has enough points then one can use a topological groupoid. -I am not a category theorist and I have trouble following the proof in Johnstone's Elephant. I was wondering if someone can explain what is going on for the specific case of presheaves on a monoid. This is a very special kind of topos with enough points and so I am hoping somebody can describe a groupoid explicitly. -If the monoid is cancellative, the topos is an etendue and I know how to get an etale groupoid via inverse semigroup theory so I am interested in the non-cancellative case. -Motivation: I study monoids and so it is interesting to try and understand how a Morita class of groupoids can be associated to a monoid. - -REPLY [10 votes]: Perhaps these slides will be helpful. I'll try to explain what happens in your special case. -Let $M$ be a monoid and let $\mathcal{B} M$ be the topos of right $M$-sets. The points of $\mathcal{B} M$ are the left $M$-sets $P$ that satisfy the following conditions: - -$P$ is inhabited. -Given elements $p_0$ and $p_1$ of $P$, there exist an element $p$ of $P$ and elements $m_0$ and $m_1$ of $M$ such that $m_0 \cdot p = p_0$ and $m_1 \cdot p = p_1$. -Given an element $p$ of $P$ and elements $m_0$ and $m_1$ of $M$ such that $m_0 \cdot p = m_1 \cdot p$, there exist an element $p'$ of $P$ and an element $m'$ of $M$ such that $m' \cdot p' = p$ and $m_0 m' = m_1 m'$. - -For example, the left regular action of $M$ on itself is a point of $\mathcal{B} M$, and as it happens, this point covers all of $\mathcal{B} M$. However, what we need to find is an open cover of $\mathcal{B} M$. The Butz–Moerdijk construction yields such a thing. -Let $K$ be a fixed set of cardinality $\ge \left| M \right|$. An enumeration of $M$ is a partial surjection $K \rightharpoonup M$ with infinite fibres. An isomorphism of enumerations of $M$ is an isomorphism of left $M$-sets making the following diagram commute: -$$\require{AMScd} -\begin{CD} -K @= K \\ -@VVV @VVV \\ -M @>>> M -\end{CD}$$ -Choose a representative in each isomorphism class of enumerations of $M$. We define a groupoid $\mathbb{G}$ as follows: - -The objects are the chosen representatives. -The morphisms $\alpha \to \beta$ are tuples $(\alpha, \beta, m)$ where $m$ is an invertible element of $M$. (We do not require any compatibility with the partial surjections here.) -Composition is given by $(\beta, \gamma, m) \circ (\alpha, \beta, n) = (\alpha, \gamma, m n)$. - -Write $G_0$ (resp. $G_1$) for the set of objects (resp. morphisms) in $\mathbb{G}$. There is a Galois topology on these making $\mathbb{G}$ a topological groupoid: - -The basic open subsets of $G_0$ are the subsets -$$U_{\vec{i}, C} = \{ \alpha \in G_0 : \alpha (\vec{i}) \in C \}$$ -where $\vec{i}$ is an $n$-tuple of elements of $K$ and $C$ is a right $M$-subset of $M^n$. -The basic open subsets of $G_1$ are the subsets -$$W_{\vec{i}, C, \vec{j}, D} = \{ (\alpha, \beta, m) \in G_1 : \alpha (\vec{i}) \in C, \beta (\vec{j}) \in D, \alpha (\vec{i}) \cdot m = \beta (\vec{j}) \}$$ -where $\vec{i}$ and $\vec{j}$ are $n$-tuples of elements of $K$ and $C$ and $D$ are right $M$-subsets of $M^n$. - -The theorem of Butz and Moerdijk is that $\mathcal{B} M$ is equivalent to the topos of equivariant sheaves on this topological groupoid $\mathbb{G}$. Note that the domain and codomain maps $G_1 \to G_0$ are locally connected, hence open a fortiori.<|endoftext|> -TITLE: Lie algebra valued 1-forms and pointed maps to homogeneous spaces -QUESTION [15 upvotes]: Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, and let $(M,p_0)$ be a simply connected pointed smooth manifold. A $\mathfrak{g}$-valued 1-form $\omega$ on $M$ can be seen as a connection form on the trivial principal $G$-bundle $G\times M\to M$. Assume that this connection is flat. Then, parallel transport along a path $\gamma$ in $M$ from $p_0$ to $p$ determines an element in $G$, which actually depends only on the endpoint $p$ since we are assuming that the connection is flat and that $M$ is simply connected. Thus we get a pointed map $\Phi_\omega:(M,p_0)\to (G,e)$, where $e$ is the identity element of $G$. Now, the Lie group $G$ carries a natural flat $\mathfrak{g}$-connection on the trivial princiapl $G$-bundle $G\times G\to G$, namely the one given by the Maurer-Cartan 1-form $\xi_G$. Using $\Phi_\omega$ to pull-back $\xi_G$ on $M$ we get a $\mathfrak{g}$-valued 1-form $\omega$ on $M$ which, no surprise, is $\omega$ itself. So one has a natural bijection between $\{\omega\in \Omega^1(M,\mathfrak{g}) | d\omega+\frac{1}{2}[\omega,\omega]=0\}$ and pointed maps from $(M,p_0)$ to $(G,e)$. -(all this is well known; I'm recalling it only for set up) -If now $\omega$ is not flat but has holonomy group at $p_0$ given by the Lie subgroup $H$ of $G$, then we can verbatim repeat the above construction to get a pointed map $\Phi_\omega:(M,p_0)\to (G/H,[e])$. I therefore suspect by analogy that there should be a natural bijection -$ -\{\omega\in \Omega^1(M,\mathfrak{g}) | \text{some condition}\} \leftrightarrow C^\infty((M,p_0),(G/H,[e])), -$ -but I've been so far unable to see whether this is actually true, nor to make explicit what "some condition" should be (it should be something related to the Ambrose-Singer holonomy theorem and to Narasimhan-Ramanan results on universal connections, but I've not been able to see this neatly, yet). I think that despite my unability to locate a precise statement for the above, this should be well known, so I hope you will be able to address me to a reference. - -REPLY [19 votes]: The question you are asking is a very basic one in the theory of what Élie Cartan called "the method of the moving frame" (in the original French, "la méthode du repère mobile"), so you should be looking that up. Cartan's basic goal was to understand maps of manifolds into homogeneous spaces, say, $f:M\to G/H$, by associating to each such $f$, in a canonical way, a 'lifting' $F:M\to G$ in such a way that the lifting of $\hat f = g\cdot f$ would be $\hat F = gF$ for all $g\in G$. If one could do such a thing, then one could tell whether two maps $f_1,f_2:M\to G/H$ differed by an action of $G$ by checking whether $F_1^*(\gamma) = F_2^*(\gamma)$, where $\gamma$ is the canonical $\frak{g}$-valued left-invariant $1$-form on $G$. -It turns out that it is not always possible to do this in a uniform way for all smooth maps $f:M\to G/H$ (even in the pointed category, which modifies the problem a little bit, but not by much). However, if one restricts attention to the maps satisfying some appropriate open, generic conditions, then there often is a canonical lifting $F$ for those $f$ belonging to this set of mappings, and it can be characterized exactly by requiring that the $1$-form $\omega_F = F^*(\gamma)$ satisfy some conditions. Working out these conditions in specific cases is what is known as the "method of the moving frame". -There's no point in trying to give an exposition of the theory here because it is covered in many texts and articles, but let me just give one specific example that should be very familiar, the Frenet frame for Euclidean space curves. -Here the group $G$ is the group of Euclidean motions (translations and rotations) of $\mathbb{E}^3$ and $H$ is the subgroup that fixes the origin $0\in\mathbb{E}^3$. The elements of $G$ can be thought of as quadruples $(x,e_1,e_2,e_3)$ where $x\in\mathbb{E}^3$ and $e_1,e_2,e_3$ are an orthonormal basis of $\mathbb{E}^3$. -When $f:\mathbb{R}\to\mathbb{E^3}$ is nondegenerate, i.e., $f'(t)\wedge f''(t)$ is nonvanishing, there is a canonical lifting $F:\mathbb{R}\to G$ given by -$$ -F(t) = \bigl(f(t),e_1(t),e_2(t),e_3(t)\bigr) -$$ -that is characterized by conditions on $\omega = F^*(\gamma)$ that are phrased as follows: First, $e_1\cdot df$ is a positive $1$-form -while $e_2\cdot df = e_3\cdot df = 0$, and, second, $e_2\cdot de_1$ is a positive $1$-form while $e_3\cdot de_1 = 0$. -These conditions take the more familiar form -$$ -df = e_1(t)\ v(t)dt,\qquad -de_1 = e_2(t)\ \kappa(t)v(t)dt, -$$ -for some positive functions $v$ and $\kappa$ on $\mathbb{R}$, and they imply -$$ -de_2 = -e_1(t)\ \kappa(t)v(t)dt + e_3(t)\ \tau(t)v(t)dt, -\qquad -de_3 = -e_2(t)\ \tau(t)v(t)dt, -$$ -for some third function $\tau$ on $\mathbb{R}$. -Conversely, any $F:\mathbb{R}\to G$ that satisfies the above conditions on $\omega = F^*(\gamma)$ is the canonical (Frenet) lift of a (unique) nondegenerate $f:\mathbb{R}\to\mathbb{E}^3$. -Without the nondegeneracy condition, the uniqueness fails. Just consider the case in which the image of $f$ is a straight line. -There are similar, but, of course, more elaborate, examples for other homogeneous spaces and higher dimensional $M$, but you should go look at the literature if you are interested in this. -Added in response to request in the comment: There are several excellent sources for the method of the moving frame. I'll just list (alphabetically by author) some of my favorites, which means the ones that I find most felicitous: - -Élie Cartan, "La théorie des groupes finis et continus et la géométrie différentielle traitées par la méthode du repère mobile", Paris: Gauthier-Villars, 1937. (His style takes some getting used to, and so many say that Cartan is unreadable, but, once you get used to the way he writes, there's nothing like Cartan for clarity and concision. I certainly have learned more from reading Cartan than from any other source.) -Shiing-shen Chern, W. H. Chen, and K. S. Lam, "Lectures on Differential Geometry", Series on University Mathematics, World Scientific Publishing Company, 1999. (Chern learned from Cartan himself, and was a master at calculation using the method.) -Jeanne Clelland, 1999 MSRI lectures on Lie groups and the method of moving frames, available at http://math.colorado.edu/~jnc/MSRI.html. (A nice, short elementary introduction.) -Mark Green, "The moving frame, differential invariants and rigidity theorems for curves in homogeneous spaces", Duke Math. J. Volume 45, Number 4 (1978), 735-779. (Points out some of the subtleties in the 'method' and that it sometimes has to be supplemented with other techniques.) -Phillip Griffiths, " "On Cartan’s method of Lie groups and moving frames as applied to uniqueness and existence questions in differential geometry", Duke Math. J. 41 (1974): 775–814. (Lots of good applications and calculations.) -Thomas Ivey and J. M. Landsberg, "Cartan for Beginners: Differential Geometry Via Moving Frames and Exterior Differential Systems", Graduate Studies in Mathematics, AMS, 2003. (Also contains related material on how to solve the various PDE problems that show up in the applications of moving frames.) - -I think that this is enough to go on. I won't try to go into some modern aspects, such as the work of Peter Olver and his coworkers and students, who have had some success in turning Cartan's method into an algorithm under certain circumstances, or the more recent work of Boris Doubrov and his coworkers on applying Tanaka-type ideas to produce new approaches to the moving frame in certain cases.<|endoftext|> -TITLE: Variable-centric logical foundation of calculus -QUESTION [15 upvotes]: Since calculus originated long before our modern function concept, much of our language of calculus still focuses on variables and their interrelationships rather than explicitly on functions. For example, in the assertion "If $y=x^2$ then $\frac{dy}{dx}=2x$," the functions $f$ and $f'$ remain unnamed while the variables $x$ and $y$ take center stage. We interpret this as notational finesse, but there seems to be an important philosophical difference between what we say and what we mean. -I have sometimes wondered: Is there an alternate logical foundation of Calculus in which variables, expressions, and equations are the central ideas, and functions per se are implicit? - -REPLY [14 votes]: Here is another approach, which I believe I first learned from Toby Bartels. Suppose $X$ is an arbitrary differentiable manifold (think of the state space of some physical system), and define a variable (one might also say "observable") to be a smooth real-valued function on $X$. If $x:X\to \mathbb{R}$ is such a "variable", then its differential is, as usual in differential geometry, a smooth function ${\rm d}x:T X \to \mathbb{R}$ on the tangent bundle of $X$. We also have the tangent map $T x : T X \to T\mathbb{R} \cong \mathbb{R}\times\mathbb{R}$, with $T x = (x, {\rm d}x)$. -If $y:X\to \mathbb{R}$ is another such "variable", then it might be related to $x$ by an equation such as $y = x^2$ or $x^2 + y^2 = 4$. Being equalities of real-valued functions, these are pointwise equalities. If $y= x^2$, then we can say that "$y$ is a function of $x$" in the sense that there is a function $f:\mathbb{R}\to\mathbb{R}$ such that $y = f\circ x$, namely $f = \lambda u. u^2$ (see this question). In this case, the chain rule of differential geometry tells us that $T y:T X \to T \mathbb{R}$ is the composite $T X \xrightarrow{T x} T \mathbb{R} \xrightarrow{T f} T \mathbb{R}$. Since $T f (u,v) = (f(u), f'(u) \cdot v)$, this means that (in addition to $y = f\circ x$) we have ${\rm d}y = f'(x) \cdot {\rm d}x$. This is a simple pointwise equality of functions $T X \to \mathbb{R}$, so we can divide by ${\rm d} x$ (at least assuming it is never zero) to get $f'(x) = \frac{{\rm d}y}{{\rm d}x}$, or in this case $\frac{{\rm d}y}{{\rm d}x} = 2x$. -Similarly, if $x^2+y^2=4$, then $y$ is not a function of $x$ in this sense, but $x^2+y^2$ and $4$ are two smooth functions $X\to \mathbb{R}$, where the first is expressed as a composite $$X\xrightarrow{(x,y)} \mathbb{R}\times\mathbb{R} \xrightarrow{\lambda u v. u^2+v^2} \mathbb{R}.$$ Thus the chain rule of differential geometry again gives us $2 x \,{\rm d}x + 2 y \,{\rm d}y = 0$ as a pointwise equality of functions $T X \to \mathbb{R}$, so that we can solve it as usual in elementary calculus to get $\frac{{\rm d}y}{{\rm d}x} = -\frac{x}{y}$.<|endoftext|> -TITLE: When does a site give rise to a Hausdorff topos -QUESTION [8 upvotes]: What conditions on a Grothendieck site $\left(C,J\right),$ are equivalent to the diagonal map $$Sh_J\left(C\right) \to Sh_J\left(C\right) \times Sh_J\left(C\right)$$ being a proper map of topoi? - -REPLY [3 votes]: I think Johnstone's Elephant gives a site characterisation of proper maps between toposes; so I guess the problem reduces to finding the site corresponding to the product $Sh_J (C) \times Sh_J(C)$ - this is surely known?<|endoftext|> -TITLE: Random Functions and Transition Probabilities -QUESTION [7 upvotes]: Let $(S,\mathcal{S})$ and $(T,\mathcal{T})$ be measurable spaces. A transition probability from $S$ to $T$ is a function $\pi:S\times\mathcal{T}\to [0,1]$ such that $\pi(s,\cdot)$ is a probability measure for all $s\in S$ and $\pi(\cdot,B)$ is measurable for all $B\in\mathcal{T}$. -Now let $(\Omega,\Sigma,\mu)$ be a probability space and let $f:\Omega\times S\to T$ be a jointly measurable function. - -Under what conditions is - $\pi_f:S\times\mathcal{T}\to [0,1]$ - given by $\pi_f(s,B)=\mu -> \{\omega:f(s,\omega)\in B\}$ a - transition probability? - -Clearly, the issue is measurability. -Motivation: If $(T,\mathcal{T})$ is standard Borel, there always exists a function such that $\pi=\pi_f$ when $\Omega$ is atomless, this is Proposition 10.7.6. in Bogachev. Moreover, such random functions have been used as a definition of mixed strategies in game theory and I think it would be interesting to understand them in terms of transition probabilities in a general setting. - -REPLY [2 votes]: I've found a rather simple solution. No additional assumptions are necessary. There is a natural method of composing transition probabilities that gives rise to a new transition probability (see for example (2) here). -We can identify probability measures with transition probabilities constant in the first argument. Also, we can identify $f$ with the transition probability $\kappa:\Omega\times S\times\mathcal{T}\to[0,1]$ given by $\kappa(\omega,s,B)=1$ if $f(\omega,s)\in B$ and $\kappa(\omega,s,B)=0$ if $f(\omega,s)\notin B$. -Also, the pointwise product of transition probabilities is a transition probability. -So let $\kappa_\mu:S\times\Sigma\to[0,1]$ the representation of $\mu$ as a tranition probability. Also, let $\kappa_1:S\times\mathcal{S}\to[0,1]$ be the identity transition probability on $S$. Let $\pi:S\times\mathcal{S}\otimes\mathcal{T}\to[0,1]$ be the pointwise product of $\kappa_\mu$ and $\kappa_1$. -Then $\pi_f=\kappa\circ\pi$ and is therefore a transition probability as the composition of transition probabilities.<|endoftext|> -TITLE: Christmas giftgiving -QUESTION [5 upvotes]: Say, that there is a group of $n$ people who decides to share Christmas gifts. -Each person has a budged, he/she will spend at most $m_i \in \mathbb{Q}$ coins on gifts. -Each person must give, exactly $1\leq g\leq n-1$ gifts, -and each person must receive $g$ gifts. Furthermore, the total worth $w$ of gifts received must be -equal for each person. How do one maximize $w$? -Clearly, if $g=n-1$ then each person $i$ will give a gift of value $m_i/(n-1)$ to every other person. Thus, $w$ is the arithmetic mean of the $m_i$:s. -However, if $g=1,$ then every person cannot give more than $\min_i m_i$ -so $w=\min_i m_i$ in this case. -The case when all $m_i$ are equal is also simple, just give $g$ gifts of value $w=m_i/g$ -so that everyone receives $g$ gifts, and $w$ is maximized. -However, what can be said in the general case? Is this equivalent to some known problem, like knapsack or max-flow? -As a graph-theoretical problem, one may view this as a directed 2g-regular graph, on $n$ vertices, where each vertex has out-degree g. Each vertex is a source and a sink, and one wants to maximize the flow so that all sinks receive the same amount. - -REPLY [10 votes]: In the graph theoretic setting, the question is analyzed by N. Megiddo in -Optimal flows in networks with multiple sources and sinks (1973) (google will give you the pdf). Gives an algorithm, does not seem to discuss complexity) -More recently this is discussed (in a more general setting) in the classic Ahuja/Mananti/Orlin, chapter 17. They do discuss complexity, but you have to hunt around a bit.<|endoftext|> -TITLE: Separation condition for higher Deligne-Mumford stacks -QUESTION [8 upvotes]: Let $X$ be a stack of $n$-groupoids on the site of affine schemes over a fixed base, with the etale topology. If $n=1$ then for $X$ to be Deligne-Mumford, aside from having an etale atlas from an algebraic space, one must also impose certain separation conditions on its diagonal. In DGA-V, Lurie remarks that what he defines as higher Deligne-Mumford stacks lacks separation axioms, but that they may be added in by hand later. My question is, what separation axioms should be added? Here, I do not mean "separated" or "quasicompact". What I mean to ask is on which morphisms do I put the appropriate separation conditions? It seems to be that simply imposing them on the diagonal may be too naive, but perhaps I am wrong, hence this question. - -REPLY [4 votes]: If you look at the last lemma here [a link to the stacks project] and -the comment preceding it, you can see that the most appropriate morphisms to impose conditions on are probably the higher diagonals. - -The lemma gives a bunch of conditions for a morphism of algebraic stacks to have various properties. The conditions on the diagonal and second diagonal of the morphism. Precisely, if these two diagonals are universally closed, then the morphism is separated; if they are quasi-compact, then the morphism is quasi-separated; if they are unramified, then the morphism is DM (``Deligne--Mumford''); -if they are monomorphisms, then the morphism if representable by algebraic spaces. -(The point is that for a morphism of algebraic stacks, the second diagonal is always a monomorphism, and hence the diagonals beyond the second are always isomorpisms, and hence satisfy every condition you might impose. But for higher stacks, the second diagonal is presumably not always a monomorphism, and hence a natural thing to do would be to consider morphisms all of whose higher diagonals are universally closed, or quasi-compact, or unramified, or monomorphisms.)<|endoftext|> -TITLE: Any suggestions for a course in Mathematical Logic? -QUESTION [5 upvotes]: I am teaching a topics course for Mathematics majors (at Temple), and am considering Logic as the topic. I was wondering if people (a) have suggestions for an appropriate text and (b) how much might be reasonable to cover in a semester. Is it feasible to end with a little model theory, or would that be a pipe dream? -EDIT -The students are theoretically math majors, and the title of the course is "Senior Problem Solving", so they are seniors with high probability. Which means that they have a couple of courses in real analysis, and one in abstract algebra. - -REPLY [3 votes]: Kenneth Kunen recently wrote a wonderful introduction to Mathematical Logic, called "The Foundations of Mathematics" (ISBN: 978-1-904987-14-7), published in 2009. The book's only prerequisite is the mathematical maturity that an Introduction to Analysis course would provide, so it sounds like your students would be prepared. The book provides a brief introduction to axiomatic set theory, model theory, and computability theory; and it culminates with a proof of Godel's incompleteness theorems and Tarski's theorem on the non-definability of truth. There are also a couple brief discussions of the philosophy of mathematics; these are given from the perspective of the working mathematician, and they are used to motivate the material. And they are very helpful. In fact, the most salient thing about this book is that it is exceptionally clear, well-written, and easy to learn from. (Kunen also wrote "Set Theory: An Introduction To Independence Proofs" which is also exceptionally clear, well-written, and easy to learn from). Your students will be grateful for the fact that this book is available (new) on amazon.com for less than $25.<|endoftext|> -TITLE: Spaces of Finite Subsets -QUESTION [11 upvotes]: $\exp_nX$ is the space whose underlying set is the set of nonempty subsets $S\subseteq X$ with $|S|\le n$. Its topology is the quotient one inherited from the map $X^{\oplus n}\rightarrow\exp_nX$ given by $(x_1,\ldots,x_n)\mapsto\lbrace x_1\rbrace\cup\cdots\lbrace x_n\rbrace$. And $\exp_{m\le n}X$ is canonically embedded in it. -Interestingly, for the case $X=S^1$, we have $\exp_2S^1\approx M\ddot{o}$ (homeomorphic mobius band). Etienne Ghys saw this by considering the mobius band as $\mathbb{R}P^2$ minus the open disk (with $S^1$ as the disk's boundary) and mapping $p\in M\ddot{o}$ to the set of tangency points of the lines tangent to $S^1$ and intersecting $p$. And from this we see that $\exp_1S^1$ is the boundary of the band and not the meridian circle. Now Raul Bott showed that $\exp_3S^1\approx S^3$ (On the Third Symmetric Potency of $S^1$), and someone else showed that $\exp_1S^1\subset S^3$ is the trefoil knot. Furthermore, $\exp_2S^1$ is a Seifert surface of $\exp_1S^1\subset S^3$. -My two curious questions: What happens for $n\ge 3$ and the corresponding embeddings? Are there interesting results for other $X$? - -REPLY [16 votes]: The spaces $\exp_n(S^1)$, as well as the embeddings $\exp_n(S^1) \subset \exp_{n+2}(S^1)$ were studied by Christopher Tuffley in Finite subset spaces of $S^1$, Algebr. Geom. Topol. 2 (2002), 1119–1145, http://dx.doi.org/10.2140/agt.2002.2.1119; MR1998017 (2004f:54008), and, more recently, by Sadok Kallel and Denis Sjerve in Remarks on finite subset spaces, Homology, Homotopy Appl. 11 (2009), no. 2, 229–-250, http://www.intlpress.com/hha/v11/n2/a12/; MR2591920 (2011a:55019). -In particular, based on an argument from Clifford H. Wagner's thesis (Symmetric, cyclic, and permutation products of manifolds, Dissertationes Math. (Rozprawy Mat.) 182 (1980); MR0605369 (82h:55021)), Kallel and Sjerve show that $\exp_n(S^1)$ is a closed manifold if and only if $n=1$ or $n=3$. Furthermore, Tuffley shows that -$$ -\pi_1(\exp_{n+2}(S^1) \setminus \exp_{n}(S^1)) = \langle x, y \mid x^{n+2} = y^{n+1} \rangle. -$$<|endoftext|> -TITLE: Littelmann path operators for an arbitrary positive root -QUESTION [6 upvotes]: Looking at a few of Littelmann's papers, he seems to only apply root operators $f_\alpha$ for $\alpha$ a simple root. However, the definition seems to make perfect sense for $\alpha$ any positive root. (Indeed one can change the hyperplane that divides positive and negative roots and get a new set of simple roots corresponding to that choice.) -My question is: how much have the $e_\alpha$ and $f_\alpha$ been studied or used for nonsimple $\alpha$? -(This question also applies to other path models, like LS paths.) - -REPLY [4 votes]: The question may be somewhat open-ended, but perhaps I can focus some aspects of it. First, I'm not sure which papers by Littelmann you've looked at, but probably the most definitive treatment of his operators occurs in his often-cited paper Paths and root operators in representation theory. Ann. of Math. (2) 142 (1995), no. 3, 499–525. Here and in related papers the objects of study are the finite dimensional irreducible representations of a semisimple Lie algebra over $\mathbb{C}$ or more generally the irreducible integrable representations of a symmetrizable Kac-Moody Lie algebra (for which there is a Weyl-Kac character formula). -In the classical case the representations are classified by their dominant highest weights relative to a fixed choice of positive (or simple) roots. But the computation of the character or weight multiplicities is complicated by the fact that a subweight can usually be reached in many ways from the highest weight by applying root vectors corresponding to negatives of simple roots. So you tend to get a lot of inefficient overcounting and cancellations. Littelmann's formal method is intended to streamline this way of thinking for all related problems by avoiding overcounts. Elegant formulas like Weyl's do involve all positive roots, but it's extremely inefficient in any computational approach to apply the root vectors for all possible negative roots. -While the definitions of Littelmann's root operators don't require simple roots, the applications do require a notion of dominant weight and hence a fixed choice of simple roots. Of course, any root can be "simple" in some set of positive roots, but nothing new is obtained classically. (On the other hand, defining a Kac-Moody algebra in general presupposes a choice of simple roots.)<|endoftext|> -TITLE: How about the Lie algebra over commutative ring? -QUESTION [11 upvotes]: It is just like the linear algebra over commutative ring (maybe advanced linear algebra), that is a nature extension and can make the structure of Lie algebra more algebraic, but I find little book discussing this topic. -I just know Weibel’s book “Homological Algebra” discuss something, but that is just a form, do not pay more attention to the Lie algebra and show some difference between field coefficient and ring coefficient. -Does anybody know something else？ - -REPLY [8 votes]: J.F. Hurley in a series of papers studied Lie algebras obtained by taking the multiplication table (with integer coefficients, due to Chevalley) of simple Lie algebras of classical or exceptional type and considering them over a commutative ring. The results describe center, ideal structure, etc. of such algebras in terms of the underlying ring. See, for example: Ideals in Chevalley algebras, Trans. Amer. Math. Soc. 137 (1969), 245-258; Composition series in Chevalley algebras Pacific J. Math. 32 (1970), 429-434; Centers of Chevalley algebras, J. Math. Soc. Japan 34 (1982), No.2, 219-222. In the joint paper with J. Morita (Affine Chevalley algebras, J. Algebra 72 (1981), N2, 359-373) he does something similar for some Kac--Moody algebras. -Some questions in free Lie algebras were considered over commutative rings, for example: D.Z. Djokovic, On some inner derivations of free Lie algebras over commutative rings, J. Algebra 119 (1988), 233-245, where centralizers of a member of a free generating set are studied. The latter reference is more or less random, probably more can be found in some books (Reutenauer?). - -There are more instances of considering Lie algebras over commutative rings (for example, plenty of papers about automorphisms of some triangular or close to them algebras), but, unlike in the case of Lie algebras over fields, all these are some isolated examples, rather than a coherent theory. The book(s) of Bourbaki recommended by Anatoly Kochubei are, probably, interesting in that regard. Bourbaki tend to state things in the utmost generality, and it is educational to see how quickly they have to give up considering Lie algebras over rings and have to "throw around properties of vector spaces" (quoting Darij Grinberg). -Perhaps the question could be augmented slightly by asking what is the reason for the absence of such a theory for Lie algebras (as opposed, for example, for associative algebras). Perhaps this is related somehow to the fact that classifying some natural (e.g., simple) classes of associative algebras is easier then that of Lie algebras (e.g., root space decomposition technique for Lie algebras which works over algebraically closed fields vs. "idempotent" technique for associative algebras which works over arbitrary fields and even over rings), but I venture into a sheer speculation here.<|endoftext|> -TITLE: On Robin's criterion for RH -QUESTION [15 upvotes]: \begin{equation} -\sigma(n) < e^\gamma n \log \log n -\end{equation} -In 1984 Guy Robin proved that the inequality is true for all n ≥ 5,041 if and only if the Riemann hypothesis is true (Robin 1984). - -I could not get a hold of his paper. I am trying to understand how did he derived the inequality. Can anyone can outline the steps, how Robin derived this criterion? - -REPLY [11 votes]: I have requested a pdf of Robin 1984 from campus scanning service. One highlight of the article that really should be mentioned is this: -For $n \geq 3,$ we have -$$ \sigma(n) \; < \; \; e^\gamma \; n \log \log n \; + \; \frac{ \; 0.64821364942... \; \; n \; }{\log \log n},$$ with the constant in the numerator giving equality for $n=12.$ -see: -Which $n$ maximize $G(n)=\frac{\sigma(n)}{n \log \log n}$? -That, at least, rests on effective bounds of Rosser and Schoenfeld (1962), which can be downloaded from ROSSER -Well, maybe not so directly. R+S do the unconditional bound for $n/\phi(n)$ in Theorem 15, pages 71-72, formulas (3.41) and (3.42). The treatment for $\sigma(n)$ is quite similar in spirit, maybe Robin was the first to write it down. The analogue of the primorials PRIMO and $n^{1-\delta}/\phi(n)$ is the colossally abundant CA numbers and $\sigma(n)/ n^{1 + \delta}.$ -Well, I am not sure where it is written down, but it is easy enough to show that the maximum value, for some $0 < \delta \leq 1, $ of -$$ \frac{ n^{1-\delta}}{\phi(n)} $$ -occurs when the prime factor $p$ of $n$ has exponent -$$ v_p(n) = \left \lfloor \frac{p^{1-\delta}}{p-1} \right \rfloor.$$ -Since, for a fixed $\delta,$ this expression is either 0 or 1 and nonincreasing in $p,$ it turns out that the optima occur at the primorials, the products of the consecutive primes from 2 to something... -From Alaoglu and Erdos, -the maximum value, for some $0 < \delta \leq 1, $ of -$$ \frac{\sigma(n)}{ n^{1+\delta}} $$ -occurs when the prime factor $p$ of $n$ has exponent -$$ v_p(n) = \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1. $$ -This is Theorem 10 on page 455. The results of this construction are the colossally abundant numbers. The construction is originally due to Ramanujan, but the part of his manuscript that dealt with ca numbers was not printed owing to paper shortages at the time. -Hardy and Wright use $d(n)$ for the number of divisors of $n.$ This is in the original paper by Ramanujan. For some $0 < \delta \leq 1, $ the maximum of $$ \frac{d(n)}{ n^{\delta}} $$ occurs when the prime factor $p$ of $n$ has exponent -$$ v_p(n) = \left\lfloor \frac{1}{p^\delta - 1} \right\rfloor. $$ The results are called the superior highly composite numbers SHC. -So, taking all three with $\delta = 1/2,$ we get lemmas -$$ \phi(n) \geq \sqrt{\frac{n}{2}}, \; \; d(n) \leq \sqrt{3n}, \; \; \sigma(n) \leq 3 \left( \frac{n}{2} \right)^{3/2}. $$ -In all three cases, if $\delta$ is such that more than one number $n$ achieves the maximum value of the ratio specified, we are choosing the largest of these $n$'s.<|endoftext|> -TITLE: Quasi-isometry classes of elementary amenable groups -QUESTION [9 upvotes]: Is there any elementary argument showing that there exist uncountably many distinct quasi-isometry classes of elementary amenable groups? How about solvable groups? -For amenable groups it follows from the result of Grigorchuk (proved in the 80's) stating that there are uncountably many groups of intermediate growth with pairwise incomparable growth functions. - -REPLY [4 votes]: The affirmative answer in given by Yves de Cornulier and Romain Tessera in arXiv:1203.4696.<|endoftext|> -TITLE: Extending faithfully flat covers of closed subschemes to open neighborhoods -QUESTION [7 upvotes]: I am curious about the analogue of this question, as stated in the title. Namely, - -If $Z \subset X$ is a closed subscheme and $Y \to Z$ is faithfully flat (let's also say of finite presentation), can we find a map $Y' \to X$ which is flat, whose image contains $Z$ (and is thus a neighborhood of $Z$), and whose restriction to $Z$ is the given map? - -This may be too strong; i.e. just as for étale maps, in the other question, this may be true only Zariski locally on $X$. That's fine too. In that case, in its most basic form it becomes a question of commutative algebra: - -If $R$ is a ring, $I \subset R$ an ideal, and $S$ a faithfully flat $R/I$-algebra, is there a set of elements $f \in R$ such that $\sum (R/I) f = R/I$ and a corresponding set of faithfully flat $R_f$-algebras ${}_f\widetilde{S}$ such that ${}_f \widetilde{S}/I_f \cong S_f$ for each $f$? - -Also, it would be nice to know that, in the event that this is true, it preserves finiteness hypotheses like "finitely presented". -For those who are curious about my motivations: I want to show that "$!$ pushforwards (resp. pullbacks) commute with $*$ pushforwards (resp. pullbacks)" under appropriate circumstances, those being when the $!$'s are along closed immersions and the $*$'s are along faithfully flat covers or vice-versa. At the very least I need to be able to switch the order in which maps of these types appear; that is, rewrite a cover of a closed immersion as a closed immersion into a cover, as my question asks. - -REPLY [6 votes]: (Essentially copied from the comments as requested.) -I think this question is asking for something too strong; here's an example. Let $R = \mathbf{Z}_p$, and let $S_0$ be a finite flat $\mathbf{F}_p$-algebra that does not lift to a finite flat $R$-algebra; an explicit example can be found here. Since $R$ is local, the present question asks: does there exist a faithfully flat $R$-algebra $S$ lifting $S_0$? If there was such an $S$, then the $p$-adic completion $\widehat{S}$ of S would be a finite flat R-algebra lifting $S_0$ (the flatness is clear, and the finiteness comes from Nakayama, completeness, and finiteness modulo $p$), which cannot exist. Hence, there is no such $S$ either.<|endoftext|> -TITLE: p^2 dividing n^8-n^4+1 -QUESTION [16 upvotes]: When trying to see if a number of the form $n^8-n^4+1$ can be divisible by the square of a prime, I found that it can indeed. The first few values for $n$ are -412, 786, 1417, 1818, 2430, 2640, 2809, 2822, 2899 ... -and the first few such primes $p$ (in increasing order) are -73, 97, 193, 241, 313, 337, 409, 433, ... -Interestingly enough, the latter is precisely the beginning of this sequence which lists the primes of the form $x^2+24y^2$. I am quite sure that this cannot be a pure coincidence and that some deep number theory must be involved. The number $24$ is not accidental either, as $n^8-n^4+1=\Phi_{24}(n)$, with $\Phi_k(x)$ being the $k$th cyclotomic polynomial. Maybe there is some relation to the field $\mathbb{Q}(\zeta_{24})$... -So, the question is if the following is true: -Conjecture. A prime $p$ has the form $x^2+24y^2$ if and only if $p^2$ divides $n^8-n^4+1$ for some $n$. - -REPLY [6 votes]: There is another proof which is somewhat easier. -Since the multiplicative group of $\mathbb{Z}/p\mathbb{Z}$ is cyclic there exists -$x$ such that $\Phi_{24}(x) = 0 (mod p)$ iff $24 | p-1$. Notice that in this case -$\Phi_{24}(x)$ has $8$ different roots in $\mathbb{Z}/p\mathbb{Z}$ and the derivative -at each root is not divisible by $p$, which implies that any root if $\Phi_{24}$ -in $\mathbb{Z}/p\mathbb{Z}$ can be lifted to a root in $\mathbb{Z}/p^k\mathbb{Z}$ -(for all $k$) and even in $p$-adics $\mathbb{Z}_p$. -Here there is nothing special about $24$, and one can replace it with any integer.<|endoftext|> -TITLE: Can we represent computable functions by r.e. sets ? -QUESTION [10 upvotes]: As we know, if $f$ is a computable function, then every pre-image of a r.e. set under $f$ is also a r.e. set, i.e. $f^{-1}(X)$ is a r.e.set if $X$ is a r.e. set. So I want to know that if a function satisfies that every pre-image of a r.e. set is also a r.e. set, then can we conclude that $f$ is a computable function? - -REPLY [5 votes]: Some related functions have the following application (D. Myers 2007, unpublished). -Define -$$ -A\le B\quad\Longleftrightarrow\quad A=f^{-1}(B) \quad\text{ for some $f$ that maps every $\Sigma^0_n$ set to a $\Sigma^0_n$ set, for each $n$} -$$ -Theorem: For each $n\ge 1$, the properly $\Sigma^0_n$ sets (the sets in $\Sigma^0_n\backslash \Delta^0_n$) form a single equivalence class under the equivalence relation induced by $\le$. As do the properly $\Pi^0_n$ sets. -Thus $\le$ allows us to look at the arithmetical hierarchy as a degree structure. It is not clear whether this can be done with an arithmetical relation in place of $\le$. (Turing reducibility is $\Sigma^0_3$ hence arithmetical, but the relation $\le$ is on its face $\Sigma^1_1$ only.)<|endoftext|> -TITLE: Why are monads useful? -QUESTION [23 upvotes]: I am a training Algebraic Geometer, and am currently trying to prepare a course on category theory. I would be really thankful if people could tell me about "real-math" applications of theorems about monads which they came across during their career. - -REPLY [27 votes]: Forgetting or ignoring category theory (I hardly knew any in 1971), as Todd says I used monads concretely and constructively to manufacture spaces Y such that $\Omega ^n Y$ is equivalent to $X$, -where $X$ is a space with an action by a suitable operad. Operads encode lots of operations, -and their associated monads coalesce all those operations into a single operation, the -product of the monad. (The portmanteau word operad combines operation and monad, inspired by this -connection and by Lewis Carroll). This concrete coalescence of information is the essence of many, but by no means all, applications of monads. In iterated loop space theory, the combinatorial monads -associated to operads mesh with the more abstract monads $\Omega^n\Sigma^n$ associated to the -adjoint pair of functors $(\Sigma^n,\Omega^n)$.<|endoftext|> -TITLE: Primitive ideals of the universal enveloping algebras of affine Lie algebras -QUESTION [5 upvotes]: Given a finite-dimensional semisimple Lie algebra $\frak g$, take an irreducible representation $V$, and let $ann(V)$ the annihilator of $V$ in $U(\mathfrak g)$. Such ideals are called primitive ideals. Then the variety defined by the associated graded ideal $gr(ann(V))$ of $gr U(\mathfrak{g})=S\mathfrak g$ is known to define the closure of a nilpotent orbit of $\frak g$. See the review by Borho in ICM 1986. -Could you suggest a reference where primitive ideals of the universal enveloping algebras of affine Lie algebras and their associated varieties are studied? What replaces the concept of the nilpotent orbit in that case? - -REPLY [2 votes]: I'm not aware of any reasonable analogue of the nilpotent variety (or related theory of associated varieties) in this infinite dimensional setting. But you may get some inspiration from the work of Joseph, including his book (which has many citations listed on MathSciNet): -Anthony Joseph, Quantum groups and their primitive ideals. Ergebnisse der Mathematik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Areas (3)], 29. Springer-Verlag, Berlin, 1995. -Apart from asking Joseph himself to comment, you might take a look at one of his reviews: -MR2145731 Martino, Maurizio (4-GLAS) The associated variety of a Poisson prime ideal. J. London Math. Soc. (2) 72 (2005), no. 1, 110–120.<|endoftext|> -TITLE: 'Important' applications of p-adic numbers outside of algebra (and number theory). -QUESTION [54 upvotes]: Surely, $\mathbb{Z}_p$ and $\mathbb{Q}_p$ (and their extensions) are very important for algebra and number theory. Do they have any important applications outside of algebra (that I could easily explain to a student)? Here I do not demand the applications to be (purely) 'mathematical'; for example, I wonder whether p-adic numbers have applications to physics (outside of string theory?). Moreover, I am also interested in those applications that are partially 'algebraic', and yet important for some other parts of mathematics. - -REPLY [4 votes]: In my answer to this related MO question, I tried to explain the utility of using $\mathbb{Q}_p$ based toy models in order to better understand questions in mathematical quantum field theory in physics.<|endoftext|> -TITLE: Lovász $\delta$ condition for LLL Algorithm -QUESTION [5 upvotes]: http://en.wikipedia.org/wiki/Lenstra%E2%80%93Lenstra%E2%80%93Lov%C3%A1sz_lattice_basis_reduction_algorithm -What is the importance of the $\delta$ parameter for LLL bases called Lovász condition? -Wiki seems to mention that the higher the $\delta$ the better. -What does it mean when $\delta = 1$? Can we find shortest vector and closest vector easily if $\delta =1$? What are typical error in approximations when $\delta = 1$? -In the SVP and CVP algorithm presented below, is there any connection to $\delta$ on the approximation error. -http://www.cs.tau.ac.il/~odedr/teaching/lattices_fall_2004/ln/cvp.pdf -Finally is there a reference that talks about connections to CVP and minimum distance? - -REPLY [6 votes]: This will answer your questions on $\delta$: -The case $\delta=1$, sometimes referred to as the ideal or optimal LLL-Algorithm, is the most natural way to define the swap condition between vectors after the length reduction step, as it forces to swap $b_k$ and $b_{k+1}$ if and only if its orthogonal 'emergence', i.e. -orthogonal distance of $b_{k+1}$ with respect to the linear span of the $b_1,\ldots,b_{k-1}$ is smaller than the orthogonal emergence of $b_k$. -Some people refer this type of reduction to Hermite, who applies this algorithm in order to prove an upper bound for the Hermite constant in any dimension $d$. -Lovász introduced the relaxation factor $\left(\frac{1}{2}\right)^2< \delta<1$ for the swap condition -$d(b_{k+1},Span(b_1,\ldots,b_{k-1}))<\sqrt{\delta}\cdot d(b_{k},Span(b_1,\ldots,b_{k-1}))$, -in order to be able to prove an upperbound for the time complexity of the algorithm which is polynomial in the magnitude of the initial basis AND the dimension $d$. -Also in the ideal case $\delta=1$ the algorithm is still polynomial with respect to the magnitude of the basis (A. Akhavi, The optimal LLL algorithm is still polynomial in fixed dimension,Theoretical Computer Science 297 (2003),1-3, pp.3–23), but it is still an open problem whether it is also polynomial in the dimension $d$. -The factor $\delta$ can be regarded as a strictness factor: the closer it is to $1$, the more orthogonal the reduced basis will be in general, and therefore both its Hermite factor as well as its approximation factor for the minimal lattice norm $\lambda_1$ are better. -I don't know of any strict theoretical proof of this connection, but extensive experiments as made by Gama, Nguyen and Stehlé (cf. P. Nguyen, D. Stehlé, LLL on average, Algorithmic Number Theory 4076 (2006), pp.1-17 or N. Gama, P. Nguyen, Predicting lattice reduction, Proc. of the EUROCRYPT 2008, Vol. 4965, pp.31-51) show that for LLL with $\delta=0.999$ the average Hermite factor, taking samples from an equidistributed (according to a natural measure on lattices described in D.Goldstein, A.Mayer, On the equidistribution of Hecke points, Forum Mathematicum (2003) 15, 2, pp. 165-189) sequence of integer lattices, increases exponentially by -$ 0.801\cdot 1.0219^d$, -which shows much better performance than predicted by the theoretic upper bound $(4/3)^\frac{d}{4}\approx 1.074^d$. -Using what is called Gaussian heuristics, one can take the Hermite factors to estimate the average approximation factor by -$average \frac{\|b_1\|}{\lambda_1}\approx\sqrt{\frac{\pi\cdot e}{2\cdot d}}\cdot 0.801\cdot 1.0219^d$, -slightly less exponential in $d$. -According to a frequently cited draft of M.Ajtai, Generating random lattices according to the invariant distribution (2006), this heuristics is quite accurate for high dimension lattices. -Unfortunately, I have never found the draft itself or any later on published paper. -However, if you don't want to rely on Gaussian heuristics, the bounds I derived from my own experiments running LLL with $\delta=1$ on integer lattices from dimension $30$ to $100$, choosing random knapsack type bases of magnitude $2^{20d}$, are -$0.3444\cdot 1.0136^d\leq average \frac{\|b_1\|}{\lambda_1} \leq 0.6688\cdot 1.0442^d,$ -the lower bound representing the Gaussian heuristics, again much better than the proven upper bound $(4/3)^{d/2}\approx 1.1547^d$. -For less ideal $\delta$, I don't know any published estimate of the average behaviour. -The experiments I've done by myself seem to verify the thumb rule: the smaller $\delta$, the faster reduction is done, but the more volatile the quality of the results, and moreover the stronger is the exponential growth of the average Hermite and approximation factors.<|endoftext|> -TITLE: Connected graded involutive bialgebras (Hopf algebras) sought (for Dynkin idempotent checking) -QUESTION [11 upvotes]: Again, there is a general and a concrete question: -Concrete question. Let $H$ be a connected graded bialgebra over a commutative ring $k$ with unity (where graded means $\mathbb N$-graded). Then, it is known that $H$ is a Hopf algebra, thus has an antipode $S$. Let $E:H\to H$ be the $k$-linear map which sends every homogeneous element $x\in H_n$ to $nx$ for every $n\in\mathbb N$. (Note that $E$ is both a derivation and a coderivation, i. e., it satisfies $E\left(ab\right)=aE\left(b\right)+E\left(a\right)b$ for all $a$ and $b$, and $\Delta\circ E = \left(E\otimes \mathrm{id} + \mathrm{id}\otimes E\right)\circ\Delta$.) Let $*$ denote convolution of $k$-linear maps $H\to H$. -It is known that $\left(E\ast S\right)\circ \left(E\ast S\right) = E\circ\left(E\ast S\right)$ whenever $H$ is commutative or cocommutative. (The usual way to state this in characteristic $0$ is by saying that the map which sends every homogeneous $x\in H_n$ to $\frac{1}{n}\left(E\ast S\right)\left(x\right)$ (for all $n$) is a projection - the so-called Dynkin idempotent of $H$. For $H$ cocommutative, this projection is the projection on the primitive part of $H$; for its kernel, see Patras, Reutenauer, On Dynkin and Klyachko idempotents in graded bialgebras, Corollary 7. For commutative $H$ the dual assertions hold.) -I have also checked (in the $3$rd degree of the Malvenuto-Reutenauer Hopf algebra) that $\left(E\ast S\right)\circ \left(E\ast S\right) = E\circ\left(E\ast S\right)$ does not hold for general $H$. -The question is whether $\left(E\ast S\right)\circ \left(E\ast S\right) = E\circ\left(E\ast S\right)$ holds for involutive $H$ ("involutive" means that $S^2=\mathrm{id}$). If not generally, at least in characteristic $0$ ? -General question. What are good examples of involutive (this means $S^2=\mathrm{id}$) connected graded Hopf algebras which are neither commutative nor cocommutative? I know that for every Hopf algebra $H$, the ideal $H\cdot \left(\left(S^2-\mathrm{id}\right)\left(H\right)\right)\cdot H$ is a Hopf ideal of $H$ (unless I am mistaken, it is a nice easy exercise), and the quotient Hopf algebra of $H$ by this ideal is the largest involutive quotient Hopf algebra of $H$. Unfortunately it does not seem to have been studied anywhere. I know of only two connected graded Hopf algebras which are neither commutative nor cocommutative: Loday-Ronco and Malvenuto-Reutenauer. The latter is definitely non-involutive, but I can reasonably compute by hand only up to degree $3$, and the involutive quotient in degree $3$ is not a counterexample when $2$ is invertible in $k$. As for the former, I don't have a grasp on it yet; I don't even know whether it is involutive. -EDIT: So it seems that the answer to the Concrete Question is No for the $4$-th graded component of a certain sub-Hopf algebra of the Hopf algebra of ordered (rooted) trees ($\mathbf H_o$ in Loic Foissy's "Ordered forests and parking functions" (arXiv:1007.1547v3). Why am I sure about this? I am not. I have spent hours with the computations, did them a second time writing them up, corrected everything that gave different results on these two tries. But who knows how often I made the same mistake twice? (No, I am not wanting anyone to check this stuff by hand.) -Also, after a total of 2 days wasted on this problem, I am surely not going to do the same for the next natural question: is $\log\operatorname*{id}$ (the logarithm being taken in the convolution algebra) still an idempotent in an involutive connected graded Hopf algebra? (This would be the "Eulerian idempotent" rather than then Dynkin one. These two idempotents seem to be strangely related in some sense, but I have not figured out how exactly. It's as if the Dynkin one tries to be a kind of "logarithmic derivative" of the Eulerian one, but the non-commutativity of the convolution algebra puts a spoke in its wheel.) -Now that the General Question is sufficiently answered with Loic Foissy's website (which is not saying that there can ever be enough combinatorial Hopf algebras), let me ask the Real question: What is the right software to solve such problems in 2011? Sorry for spamming MO with yet another software question, but I cannot believe there hasn't been anything written for such cases yet. - -REPLY [2 votes]: The answers to both the Concrete Question and the analogous question for $\log\operatorname*{id}$ are "No". By that I mean that now I am sufficiently convinced of the correctness of my Sage 5.0 code. I am still surprised that I had to go all the way up to a $5$-th graded component to disprove the $\log\operatorname*{id}$ conjecture (and up to a $4$-th one for Dynkin), but there is no guarantee that my counterexample is minimal. -Correspondence with Marcelo Aguiar gave me a clue towards a different direction: What is the largest involutive quotient of the Malvenuto-Reutenauer Hopf algebra? It clearly surjects onto the abelianization of the Malvenuto-Reutenauer Hopf algebra. In degree $\leq 5$ (at least), this surjection is an isomorphism. What about higher degrees? (This, of course, would explain why I could not find counterexamples in Malvenuto-Reutenauer.) What about the Hopf algebra of ordered forests? (Note that my counterexamples don't work in the Hopf algebra of ordered forests itself; they work in a proper Hopf subalgebra of it.) Both of these Hopf algebras are free and cofree (I think this was proven by Loic Foissy); is there a theorem that the largest involutive quotient of a free-and-cofree connected graded Hopf algebra must equal to its abelianization? -EDIT: The largest involutive quotient of the Malvenuto-Reutenauer Hopf algebra is not equal to its abelianization, and $\log\operatorname*{id}$ is not idempotent on this quotient. The same holds for the quotient of the Malvenuto-Reutenauer Hopf algebra modulo dual equivalence (this is one of the two mutually dual Hopf algebras introduced by Poirier and Reutenauer). I just checked this on Sage, which found a counterexample in degree $7$.<|endoftext|> -TITLE: computing average height-functions for lozenge tilings -QUESTION [9 upvotes]: Can anyone suggest a simple and efficient way (preferably embodied in computer code) to compute the average height function for lozenge tilings of an $a,b,c,a,b,c$ semiregular hexagon? I prefer to scale my height functions so that the height-change along each tile-edge is 1 and so that the lowest possible height of any vertex in any tiling of the region is 0, so that for instance when $a=b=c=1$ the average height function alternates between 1 and 2 on the boundary and is 3/2 in the middle. -When we replace the tiling model by the (dual) dimer model, height becomes associated with faces rather than vertices, and the average height of a face can be written as a linear combination of edge-inclusion probabilities (for the uniform distribution on dimer configurations); each of these probabilities can be written as a ratio of determinants, so I do know one way to compute the average height functions I'm interested in. -But I'm hoping for something simpler, along the lines of the recurrence formulas that apply to domino tilings of Aztec diamonds. Perhaps something like this follows from the recent work of Borodin (or maybe from the earlier work of Kuperberg), or can be obtained from Kuo's graphical condensation method. (And have I mentioned Speyer's paper on the octahedron recurrence? Ah, I just did.) - -REPLY [6 votes]: David Speyer made a blog post a while back saying how to do this: -http://sbseminar.wordpress.com/2009/10/21/rhombus-tilings-and-an-over-constrained-recurrence/ -It wasn't the main point of the blog post, but he does say how to use Kuo's graphical condensation method to compute such things. In short, if $H(a,b,c)$ is the set of perfect matchings on your hexagon, you're supposed to think of Kuo's beautiful technique as a bijection between the sets -$H(a,b,c) \times H(a-1,b-1,c)$ -and -$[H(a-1,b,c) \times H(a,b-1,c)] \cup -[H(a,b,c-1) \times H(a-1,b-1,c+1)] -$ -If you're interested in the expected value of some statistic on $H(a,b,c)$ --- he was talking about edge-placement probabilities, but I don't see why it wouldn't work for the height function at a given face --- you should compute this expectation on both sides of Kuo's bijection. In general that'd be tricky, but for the hexagon, it's quite tractable to do, since you know the sizes of all of these sets explicitly. For instance, MacMahon gave a closed formula for $H(a,b,c)$, which can also be found in Kuo's paper; Speyer even works out the relative sizes in his post for you. This gives you a recursive answer to your question, by dividing through by the term which came from the expected height function of $H(a-1,b-1,c)$. The base case is where one of a,b, or c is zero, which is easy. -That's not code, obviously. Sorry! I'm rather busy travelling over the next few weeks, but I can try to do it in January sometime. If I forget, and nobody else does it, please remind me.<|endoftext|> -TITLE: Every prime number > 19 divides one plus the product of two smaller primes? -QUESTION [57 upvotes]: This is a part of my answer to this question I think it deserves to be treated separately. - Conjecture Let $A$ be the set of all primes from $2$ to $p>19$. Let $q$ be the next prime after $p$. Then $q$ divides $rs+1$ for some $r,s\in A$. -I wonder if this conjecture is already known. I checked it for all $p<692,000$. -A reformulation of the conjecture is this (motivated by Gjergji Zaimi's comment). Let $p > 19$ be prime. Let $A$ be the set of primes $< p$ considered as a subset of the cyclic multiplicative group $\mathbb{Z}/p\mathbb{Z}^*$. Then the product $A\cdot A$ contains $-1$. -It is interesting to know how large $A\cdot A$ is. This seems to be related to Freiman-type results of Green, Tao and others. Also as Timothy Foo pointed out, perhaps Vinogradov's method of trigonometric sums can apply. - -REPLY [11 votes]: Perhaps this might be another perspective on this problem. In an answer to a question I had previously asked on Math Overflow, "A generalized Möbius function?", the following paper of Addison was cited (which I simply copy from that answer): -A Note on the Compositeness of Numbers, A. W. Addison, Proceedings of the American Mathematical Society, Vol. 8, No. 1 (Feb., 1957), pp. 151-154 -In this paper, it is proven that if one divides the integers into $q$ classes according to whether $\Omega(n)$ is $0,1,\dots,q-1 \bmod q$, $q>2$ then the function $C_{q,i}(x)$ which counts integers less than $x$ with $\Omega(n) \equiv i \bmod q$ satisfies -$$ -C_{q,i}(x) - \frac{x}{q} = \Omega_{\pm}\left(\frac{x}{(\log x)^r}\right) -$$ -where $r = 1-\cos(2\pi/q)$. (Here the variables $q$ and $r$ are taken from the paper and mean different things from elsewhere in this page.) -Let's assume that the relative frequencies of $\Omega(n) \bmod q$ should not differ much whether we consider $n\in [1,p^2]$ or $n\in (p\mathbb{Z}-1)\bigcap [1,p^2]$. Then we take $x$ to be about $p^2$ and $q$ to be about -$$ -f(p)=\max_{n \in (p\mathbb{Z}-1)\bigcap [1,p^2]}\Omega(n). -$$ -Then a necessary (but not sufficient) condition for the conjecture is that $C_{f(p),2}(p^2)\geq 1$. I'm not really sure if one can take $x$ and $q$ in this relative range, but if so, then one can't even get a necessary condition from this because, using $1-\cos x \approx x^2/2$ for small $x$, we have -$$ -f(p) \gg (\log p)^{1-\cos(2\pi/f(p))}. -$$<|endoftext|> -TITLE: Computations in $\infty$-categories -QUESTION [15 upvotes]: Direct to the point. -Since now I've looked a lot of presentations of $\infty$-categories, but it seems that the only way to do explicit computations on these objects is via model categories. Is that so, or maybe there are other way of doing computations in $\infty$-categories, and if there such different ways what are they? -Edit: I really want to thank all people that answered this question, unfortunately I can choose just one answer. -Btw because here is yet midnight I wish everybody a happy new year. - -REPLY [20 votes]: (This is an answer to a question below from Akhil Mathew; he wanted examples of -``explicit computations'' since all he knew were classical 1950s calculations and -abstract theory. My answer is too long for a comment and too short to do justice -to the question.) -That is terrible!!! I don't know where to begin, since there are huge masses of explicit calculations in the past half century. You mention infinite loop spaces. There is a theorem that says that the ordinary mod $p$ homology of $\Omega^n\Sigma^n X$ is an explicit functor of the ordinary mod $p$ homology of $X$, with all information (product, Dyer-Lashof operations, coproduct, Steenrod operations) determined. That surely sounds conceptual, but try to find a conceptual proof! We have tons of explicit calculations in the classical and Novikov Adams spectral sequences, without which the solution of the Kervaire invariant problem would be unthinkable. Chromatic theory as a whole is informed by explicit calculations. In unstable homotopy theory, exponent theorems for the homotopy groups of spheres use a remarkable blend of theoretic and calculational techniques. There are tons of explicit calculations in ordinary and generalized cohomology in the past half century. Spin cobordism and its applications to curvature questions is an example of a blend of algebraic topology and differential geometry. I could go on for 100s of pages without pausing for breath, and none of these results use any model category theory, let alone $\infty$ categories. There is a subject out there, with real content.<|endoftext|> -TITLE: Presentation of the Clifford group by generators and relations? -QUESTION [6 upvotes]: The Clifford group $\mathcal{C}_n$ is a matrix group on $\mathbb{C}^{2^n}$ generated by tensor products of the following matrices: -$$ -P = \begin{pmatrix} 1 & 0 \\\\ 0 & i\end{pmatrix} -\quad -H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\\\ 1 & -1\end{pmatrix} -\quad -\wedge\\! X = \begin{pmatrix}1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0\end{pmatrix} -$$ -For my purposes I take equality up to a complex unit, i.e., if $A = e^{i\alpha}B$ then $A = B$. -My question is: - -is there an abstract presentation of this group by generators and relations? - -Or the easier version: - -is there a presentation of $\mathcal{C}_2$ or $\mathcal{C}_3$ by generators and relations? - -I don't care which set of generators is used, although it would be nice if the generators were those given above. - -REPLY [4 votes]: Follow-up, many years later: -In the currently (02/2020) most highly up-voted comment and the discussion below it, I claimed that the Clifford group is a projective representation of the semi-direct product of the phase space translations and the symplectic group. As Ross correctly implies in his response to my answer (and I wrongly deny in my response to him), this property only holds for qudits in odd dimensions, not for the qubit case! For qubits, it is still true that the symplectic group arises as a quotient group of the Clifford group (up to Paulis and multiples of the identity). However, the extension does not split. -[I've since lost my old account and access to the old email address - hence this answer comes from a new account. Apparently, the new account does not have enough "reputation" to directly reply.]<|endoftext|> -TITLE: Economical hard word problem -QUESTION [13 upvotes]: Can anyone give me an example of a very simple word problem, where by "simple" I mean that it has very few generators and relations, that is nevertheless insoluble. To make the question easier, I am prepared to allow "relation schemas" (an example might be that the fifth power of any word is equal to the identity, say), and I'm happy -- in fact, very happy -- to weaken "insoluble" to "insoluble in polynomial time" (in the length of the word). Also, I'm happy to work in a semigroup rather than a group. To make clearer what would count as a good example, let me give the reason behind it. I would like to find a collection of strings (the strings that are equal to the identity in the semigroup) such that recognising membership is difficult, but such that the space of strings in the collection is "interesting to explore", in the sense that one can develop methods for showing quite non-trivially that certain strings belong to the collection, and then build on those methods to get even less trivial examples and so on. -My motivation for that is to have a nice toy model of mathematics itself. So I'd like to be able to develop from the original replacement rules further particularly useful replacement rules that would be like lemmas, and that kind of thing. But I really would like the initial set of generators and relations to be very simple. Does this ring a bell with anyone? -Edit. Many examples of groups with insoluble word problems use sets of integers for which membership is not decidable and encode their membership problems as word problems for suitably constructed groups. I wouldn't consider such examples good ones, because they are simple relative to a set that may be quite complex. I want absolutely simple examples. If that seems like rather a strong demand, remember that I am allowing a considerable weakening of the insolubility condition. - -REPLY [12 votes]: Here is an example that I know of. The following semigroup presentation has a recursively unsolvable word problem. -Generators: a,b,c,d,e -Relations: ac=ca, ad=da, bc=cb, bd=db, abac=abace, eca=ac, edb=be. -OR -Relations: ac=ca, ad=da, bc=cb, bd=db, ce=eca, de=edb, cca=ccae -These examples were published by G.S.Ceitin in 1958 (Trudy Math. Inst. Steklov, Vol. 52, pp. 172--189). Now, whether this suits Timothy's motivation, well,...(?)<|endoftext|> -TITLE: Etale cohomology with coefficients in the integers -QUESTION [22 upvotes]: Here is a basic question. When does $H^1_{et}(X,\mathbb{Z})$ vanish? Using the exact sequence of constant etale sheaves $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}\rightarrow 0$, it is enough to show that $H^1_{et}(X,\mathbb{Q})$ vanishes. It is known, for instance by 2.1 of Deninger's 1988 JPAA paper, that $H^1_{et}(X,\mathbb{Q})$ vanishes when $X$ is normal. -Note: there are two arguments I think are incorrect that claim to show $H^1_{et}(X,\mathbb{Z})$ always vanishes. The first is that $\mathbb{Z}$ is flasque in the etale topology. This is false. For instance, over the function field $\mathbb{C}(x,y)$, the long exact sequence in cohomology for $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}/n\rightarrow 0$ shows that $H^2(\mathbb{C}(x,y),\mathbb{Z})$ is non-zero. So, $\mathbb{Z}$ cannot be flasque. The second argument is that $H^1_{et}(X,\mathbb{Z})=Hom_{cont}(\pi_1^{et}(X),\mathbb{Z})$, where $\pi_1^{et}(X)$ is the etale fundamental group of $X$, which is a profinite group. Since it is profinite, the $Hom$ group above vanishes. But, the claimed equality between $H^1_{et}(X,-)$ and $Hom_{cont}(\pi_1^{et}(X),-)$ only holds for torsion sheaves, as far as I have been able to determine. -I am in fact interested in several things. First, either an example of $X$ such that $H^1_{et}(X,\mathbb{Z})$ is non-zero, or a proof that this always vanishes. Second, the same thing but where we only look at affine $X$. In particular, if it exists, I would love to see an example of a commutative ring $R$ where $H^1_{et}(Spec R,\mathbb{Z})$ is non-zero, if this is possible. - -REPLY [13 votes]: More generally, if $X$ is proper over an algebraically closed field, then $H^1(X,\mathbb Z)$ -is isomorphic to the cocharacter module of the maximal torus of the Picard variety -$Hom(\mathbb G_m,Pic^0)$.<|endoftext|> -TITLE: Dehn surgery on handlebody -QUESTION [6 upvotes]: Assume $V$ is a handlebody and $C$ be a simple closed curve contained in the interior of $V$. -As Sam said, there exists some simple closed curve such that every dehn surgery along it produces a handlebody. So I assume $C$ can not be isotopic to a simple closed curve in $\partial V$. -Obviously, trivial dehn surgery along $C$ produces a handlebody. So My question is: -Is there a different dehn surgery along $C$ which produces a handlebody? -Can we classify all the dehn surgery along $C$ which produce handlebodies? - -REPLY [14 votes]: There's an extensive literature on this and more general questions. -First, let's consider a more precise formulation of the question. -Let $H$ be a handlebody, and let $K\subset H$ be a knot. Let's -assume that $\partial H \subset H-K$ is incompressible. Otherwise, -$H=H_0 \natural H_1$ the boundary connect sum of two handlebodies, -such that $K\subset H_0$, and $\partial H_0 \subset H_0-K$ is -incompressible. Then a surgery on $K$ makes $H$ a handlebody if -and only if the corresponding surgery makes $H_0$ a handlebody. -Thus, one may reduce to considering the case that $K$ is diskbusting. -With this reformulation of the question, the case that $K$ is -a core curve corresponds to when $H$ is a solid torus, and $K$ is -a core curve (since this is the only case in which a core curve is -diskbusting). So assume that $K$ is not a core curve. If $K$ -is isotopic into $\partial H$, then there are infinitely many surgeries -which yield $H$. In this case, there is an annulus going between the -knot and $\partial H$. The surgery slopes which intersect the annulus -slope once give back $H$. Surgery along the annular slope gives -a manifold containing an incompressible surface, by Jaco's lemma. -If the intersection with the annulus slope is $>1$, then the manifold -has incompressible boundary by Theorem 2.4.3 of the cyclic surgery paper. -A result of Wu implies that for knots $K$ which -are not isotopic into the boundary, the distance between boundary-reducible -surgeries is at most one, and therefore there are at most two non-trivial -surgeries which may yield a handlebody. -In the case that $H$ is a solid torus, it was proved by Berge -and Gabai that $K$ must be a 1-bridge braid, and a complete description -was given. There is a famous example (the Berge link) which yields 3 solid torus surgeries, -which shows that Wu's estimate is sharp (there is a 3-fold symmetry permuting -the slopes). -Wu has further results on the case of 1-bridge knots in handlebodies. -Added: Frigerio, Martelli and Petronio show that there are 1-bridge -knots in handlebodies with three handlebody fillings (including the trivial one), -generalizing the example of the Berge link to higher genus boundary, and showing -that Wu's theorem is sharp in general. It seems to be an open question whether -all such examples are 1-bridge (see discussion in another of Wu's papers).<|endoftext|> -TITLE: Complexity of computing expansion of a newform level 18 weight 3 and character [3] - OEIS A116418 -QUESTION [6 upvotes]: I am not familiar with newforms, so this may not make any sense. -OEIS sequence A116418 is Expansion of a newform level 18 weight 3 and character [3] -Numerical evidence suggest that up to $10^5$ -$$ \text{A116418}[n] \equiv \sigma(3n+1) \pmod 3$$ - -What is the complexity of computing A116418[n], possibly assuming $n$ is factored (for modular form coefficients after $n$ is factored the coefficient is efficiently computable). - -Gjergji Zaimi proved a similar congruence involving eta and A116418 is expansion of an eta formula. -Added My main interest is computing $\sigma(3n+1) \mod 3$ and a comment by Dror Speiser suggests the coefficient of the newform is computable in polynomial time assuming $n$ is factored. -The factorization of $n$ is not related related to the factorization of $3n+1$ and for numbers of form $3 \cdot 2^n + 1$ the factorization is trivial. - -Is A116418 really the expansion of the newform or is it a typo in OEIS? -Is the congruence $ \text{A116418}[n] \equiv \sigma(3n+1) \pmod 3$ identity or just coincidence for the the first $10^5$ terms? - -REPLY [8 votes]: [More a comment than an answer, but too long for the comment space] -Call this form -$$ -\varphi := \frac{\eta(q^3)^2 \eta(q^6)^3 \eta(q^9)^2}{\eta(q^{18})} -= q - 2q^4 - 4q^7 + 6q^{10} + 8q^{13} \cdots. -$$ -The listing of coefficients in the OEIS is correct as far as it goes (checked -with copy-and-paste to gp). The form is not CM: the coefficients are -supported on $q^n$ with $n \equiv 1 \bmod 3$ but do not vanish even for $n$ -such as $10$ and $22$ that are $1 \bmod 3$ but not norms from ${\bf -Q}(\sqrt{-3})$. In particular the coefficients aren't multiplicative, so -$\varphi$ isn't quite an eigenform. It seems that the relevant eigenforms are -obtained as follows. Apply $w_{18}$ to get (within a multiplicative factor) -$$ -\phi := \frac{\eta(q^6)^2 \eta(q^3)^3 \eta(q^2)^2}{\eta(q)} = q + q^2 - 2q^4 - -3q^5 - 4q^7 - 2q^8 + 6q^{10} + 12q^{11} + 8q^{13} - 4q^{14} \cdots, -$$ -whose $q^n$ coefficient is 0 if $n \equiv 0 \bmod 3$, and coincides with the -$q^n$ coefficient of $\varphi$ also when $n \equiv 1 \bmod 3$, but need not -vanish for $n \equiv 2 \bmod 3$. Then "experimentally" if $m,n$ are relatively -prime then the $q^{mn}$ coefficient of $\phi$ equals the product of the $q^m$ -and $q^n$ coefficients, unless both $m$ and $n$ are $2 \bmod 3$, when the -$q^{mn}$ coefficient is $-2$ times that product. Hence we obtain an eigenform -by choosing a square root of $-2$ and multiplying the $q^n$ coefficient of -$\phi$ by that square root for each $n \equiv 2 \bmod 3$. -As Dror Speiser notes, the Edixhoven program promises to compute the $q^n$ -coefficient of such a form in time $\log^{O(1)}n$ for $n$ prime, and thus for -all $n$ given the factorization of $n$; but I don't think this has been -implemented yet to the point that one could actually carry out the computation -this way. For specific forms there can be shortcuts that make a $\log^{O(1)}n$ -computation practical (still assuming $n$ is factored), but here I've tried a -few things and not yet(?) found such a shortcut. -[added later] Curiously the images of $\phi$ under the other two $w$ -operators are in the linear span of $\varphi$ and $\phi$: if we write -$$ -\psi = \frac{\eta(q^3)^3 \eta(q^6)^2 \eta(q^{18})^2}{\eta(q^9)} - = q^2 - 3q^5 - 2q^8 + 12q^{11} - 4q^{14} \cdots -$$ -for (a multiple of) the $w_2$ image, then $\phi = \varphi + \psi$, while -$\varphi - 2 \psi$ is the multiple -$$ -\frac{\eta(q)^2 \eta(q^3)^2 \eta(q^6)^3}{\eta(q^2)} -= q - 2q^2 - 2q^4 + 6q^5 - 4q^7 + 4q^8 + 6q^{10} - 24q^{11} + 8q^{13} + 8q^{14} -\ldots -$$ -of the $w_9$ image.<|endoftext|> -TITLE: do spectra have diagonal maps? -QUESTION [9 upvotes]: Topological spaces have diagonal maps $X \rightarrow X \times X$ and $X \rightarrow X \wedge X$, and suspension spectra also have diagonal maps $\Sigma^\infty X \rightarrow \Sigma^\infty(X \wedge X) \cong (\Sigma^\infty X) \wedge (\Sigma^\infty X)$. What about general spectra? (i.e. symmetric spectra, S-modules, or any other convenient definition.) I always assumed you could, but I haven't thought through it carefully. And if not, can we still get a cup product on $E^*(X)$ when $E$ and $X$ are spectra? - -REPLY [13 votes]: The existence of an $E_\infty$-diagonal is an obstruction for equipping a spectrum $E$ with the structure of a suspension spectrum. Conversely, in -Klein, J.R.: Moduli of suspension spectra. Trans. Amer. Math. Soc. 357 (2005), 489–507 -I showed that the existence of a suitably defined notion of $A_\infty$-diagonal on $E$ -is equivalent to equipping $E$ with the structure of a suspension spectrum provided we are in the metastable range. Here "metastable" means $E$ is $r$-connected (for $r \ge 1$) and is weak -equivalent to a cell spectrum of dimension $\le 3r+2$. -There are various elementary ways of defining the notion of $A_\infty$-diagonal, -but they in the end -amount to the existence of a map $\delta: E \to (E\wedge E)^{\Bbb Z_2}$ (for a suitably -defined version of the smash product), which is a homotopy section to the map -$(E\wedge E)^{\Bbb Z_2} \to E$ which is given by passing from categorical to geometric fixed points. The way I do this in the paper is the use the second stage of the Taylor tower -of the functor $E \mapsto \Sigma^\infty \Omega^\infty E$; this second stage turns out to be a model for $(E\wedge E)^{\Bbb Z_2}$.<|endoftext|> -TITLE: Nagata's conjecture, Seshadri constant -QUESTION [5 upvotes]: What is it known now about Nagata's conjecture and Seshadri constant (http://en.wikipedia.org/wiki/Nagata%27s_conjecture_on_curves and http://en.wikipedia.org/wiki/Seshadri_constant) for toric surfaces? It seems that it should be some lower bounds in terms of fans or polytops. Is it true? -Does there exist some simple examples where exact inequalities are proved? - -REPLY [2 votes]: I assume $S$ is a projective smooth toric surface. -If $D_1, \dots, D_n$ are the irreducible toric divisors on $S$, then $-K_S=D_1+\dots+D_n$ is an anticanonical divisor. Thus blowing up any point on any of these divisors one obtains a smooth anticanonical rational surface $\tilde S$; such surfaces are very well known by work of Brian Harbourne. Maybe more simply, by Mori theory, since $-K_{\tilde S}=-\pi^*K_S-E$ is effective on $\tilde S$, the cone of curves is spanned by extremal rays $\mathbb{R}C_i$ with $-K_{\tilde S}\cdot C_i>0$ and components of $-K$; these curves are in our case the exceptional divisor and the birational transforms of a subset of $D_i$'s [EDIT: the previous (italicized) sentence was not correct because new extremal rays do appear with the blowup, but it is still true that these extremal rays can be controlled with Harbourne's results]. Thus to determine nefness on $\tilde S$ and so the Seshadri constant of any ample divisor at the given point is not difficult. In particular these Seshadri constants are rational, and they only depend on the ample class and the $D_i$(s) to which the point belongs. [EDIT: these conclusions are still correct]. -Blowing up at a general point of $S$ may give a surface which is anticanonical (if $-K$ is not fixed on $S$) or non-anticanonical (when $|-K|=\{-K\}$). In the first case, similar considerations would lead to the computation of the Seshadri constant. In the second case, I am afraid the problem can be difficult, and indeed related to the Nagata conjecture. The simplest interesting example would be the following: start with $\mathbb{P}^2$ as a toric surface and blow it up at the three three toric points. Now blow up the resulting surface at its six toric points. The result is a toric surface, the blow up of $\mathbb{P}^2$ at three clusters of three infinitely near points, where the Seshadri constant of your preferred ample divisor $L$ at a general point is presumably unknown. It might be irrational, if $L^2$ is not a square. -An analogon of the Nagata conjecture for toric surfaces (Seshadri constants at sets of $r\gg 0$ general points) can of course be stated, as particular cases of the conjecture stated by Lazarsfeld in 5.1.24 of "Positivity in Algebraic Geometry". I have nothing particularly relevant to say about that, except that it will probably be just as difficult as Nagata's original conjecture.<|endoftext|> -TITLE: Does model-complete in a language with a constant symbol imply EQ? -QUESTION [6 upvotes]: Marker Theorem 3.1.4 says: -Suppose $T$ is a theory in a language with at least one constant symbol. -Then an $L$-formula $\phi(x)$ is $T$-equivalent to a quantifier-free formula iff, whenever $M$ and $N$ are models of $T$, $A \subseteq M$ and $A \subseteq N$, then $M \models \phi(a)$ iff $N \models \phi(a)$ for any $a$ from $A$. --end of theorem -Thus, if we further assume that $T$ is model-complete, then it must follow that $T$ eliminates quantifiers? Where am I going wrong, if it is wrong? - -REPLY [7 votes]: Per JDH's suggestion, I'll turn my earlier comment into an answer. - -Assuming $T$ to be model-complete, then whenever $M$, $N$ and $A$ are all models of $T$, it would certainly follow from $A \subseteq M$ and $A \subseteq N$ that $M \models \phi(a)$ iff $N \models \phi(a)$ for any $a$ from $A$ (as whenever one model of $T$ is a substructure of another, it is in fact an elementary substructure). -But in Marker's condition, $A$ can be any $L$-structure and is not required to be a model of $T$. -Any theory that has elimination of quantifiers is model-complete, but the converse is not true. Note that while Marker's Theorem 3.1.4 is stated for a theory in a language with at least one constant symbol, he notes afterward that the proof can be adapted to cover the case in which $L$ has no constant symbols; so if model-completeness were to have sufficed here, it would've implied that the false converse were true. -Incidentally, one very interesting theory which is model-complete yet does not admit elimination of quantifiers is the theory of the real field with exponentiation. This theory isn't known to be decidable, but MacIntyre and Wilkie showed that its decidability is implied by the real version of Schanuel's conjecture. (This nicely succinct postscript file of Kuhlmann's contains handy references.)<|endoftext|> -TITLE: Computational complexity of integration in two dimensions -QUESTION [6 upvotes]: I have an algorithm for solving a certain problem that requires that I compute two-dimensional integrals as a subroutine, and I'd like to make some kind of statement about its running time. Suppose $S$ is a simply connected region in the plane whose boundary is composed of $n$ line segments and $m$ arcs of some type (say, arcs of a circle, ellipse, or parabola). I'd like to numerically approximate the integral of some function $f(x)$ over $S$. What's the "computational complexity", in big-O notation, of computing this within a given error $\epsilon$? Clearly this depends on lots of factors, such as the curvature of the arcs and the derivatives of the function $f(x)$ but I haven't had much success in pinning down anything concrete. By way of comparison, the wikipedia page for Simpson's rule -http://en.wikipedia.org/wiki/Simpson%27s_rule#Error -says that the error committed by the composite Simpson's rule is bounded (in absolute value) by -$\frac{h^4}{180} (b-a) M$ -where $h$ is the step size and $M = \max_{\xi\in[a,b]}|f^{(4)}(\xi)|$, and therefore the complexity of computing such an integral within precision $\epsilon$ is $M^{1/4}(b-a)^{5/4}\epsilon^{-1/4}$. Can I make any similar statement for a two-dimensional integral over my region $S$? What if I were able to triangulate $S$ into $N$ small pieces? - -REPLY [2 votes]: Since you mentioned triangulation, Quarteroni, Sacco, and Saleri's book gives the following formula in the section on multi-dimensional integration: if your domain of integration is a convex polygon $\Omega$ with a triangulation $\mathcal{T}_h$ consisting of $N_T$ triangles, where $h$ is the maximum edge length in $\mathcal{T}_h$, and your quadrature rule has degree of exactness equal to $n$ with all nonnegative weights, then there exists a positive constant $K_n$, independent of $h$, such that the error $E$ is bounded by -$|E| \leq K_n h^{n+1} |\Omega| M_{n+1}$ -where $M_{n+1}$ is the maximum value of the modules of the derivatives of order $n+1$ and $|\Omega|$ is the area of $\Omega$. The composite midpoint formula and the composite trapezoidal formula have nonnegative weights and degree of exactness $n=1$ and so we have -$|E| \leq K_1 h^2 |\Omega| M_2$ -for some constant $K_1$. It follows that, if your desired error is $\epsilon$, then the maximum length of any edge of your triangulation $h$ must be at most $\epsilon^{1/2}(|\Omega| M_2 K_1)^{-1/2}$, i.e. $O(\epsilon^{1/2}(|\Omega| M_2)^{-1/2})$. If you break $\Omega$ into $N_T$ triangles as I mentioned before, the edge length will be $O(\sqrt{|\Omega|/N_T})$, so you'll need to break $\Omega$ into $O(\frac{|\Omega|^2 M_2}{\epsilon})$ triangles. So, quadratic in the area, linear in the maximum value of the second derivative, and inversely proportional to $\epsilon$. It's not clear to me where convexity enters the picture, so perhaps this will adapt to the general case you've described. The authors state that a proof can be found in Isaacson E. and Keller H. (1966), Analysis of Numerical Methods. Wiley, New York. (This is not my area of expertise so I welcome any corrections) -UPDATE: I checked the Isaacson book and the relevant section, "7.4 Composite Formulae for Multiple Integrals" doesn't make any mention of convexity, so you may be safe in using the $O(\frac{|\Omega|^2 M_2}{\epsilon})$ that I gave before.<|endoftext|> -TITLE: The Norm Map in (group) cohomology via classifying spaces -QUESTION [7 upvotes]: The well-known transfer map in group (co)homology can be defined with only homological algebra, or with algebraic topology via classifying spaces (group cohomology of $G$ is isomorphic to ordinary cohomology of $BG$). Here I fix a group $G$ with finite-index subgroup $H$, and field $k=\mathbb{Z}_p$. Let $p:\tilde{Y}\rightarrow Y$ be a covering map, where $Y$ is a $K(G,1)$-manifold of top dimension $n$ and $\tilde{Y}$ is the cover which corresponds to $H$ (so that it is a $K(H,1)$-manifold); clearly $p$ is a $|G:H|$-sheeted covering map. Now the transfer map $tr^G_H:H_n(G,k)\rightarrow H_n(H,k)$ agrees with the transfer map $H_n(Y;k)\rightarrow H_n(\tilde{Y};k)$, which in terms of chain complexes is induced by $\sigma\mapsto\sum_{\sigma'}\sigma'$ where $\sigma$ is an oriented $n$-cell of $Y$ and $\sigma'$ ranges over the oriented $n$-cells of $\tilde{Y}$ lying over $\sigma$. In cohomology this is $f\mapsto[\sigma\mapsto \sum_{\sigma'}f(\sigma')]$ for cochains. Likewise, the restriction map runs in the opposite direction of the transfer and is induced from the map $BH=EG/H\rightarrow EG/G=BG$. -What is the corresponding classifying space construction in the setting of the norm map defined below? -Let $S_{G/H}$ be the group of permutations of $G/H$ (left coset representatives) and consider $H$ as a subgroup of $S_H$ through left multiplication. The wreath product $S_{G/H}\int H$ is $H^{\oplus|G:H|}\rtimes S_{G/H}$, where $s^{-1}(\prod_{x\in G/H}h_x)s=\prod_{x\in G/H}h_{s(x)}$. From this is the monomial representation $\Phi:G\rightarrow S_{G/H}\int H$ defined by $\Phi(g)=\pi(g)\prod_{x\in G/H}h_{g,t}$ where $gx=x_gh_{g,t}$ (for $x,x_g\in G/H$ and $h_{g,t}\in H$), and $x\mapsto x_g$ induces a permutation $\pi(g)\in S_{G/H}$ (so $\pi$ is the representation of $G$ as a group of permutations of its left coset space $G/H$). -Finally, for $\alpha\in H^*(H,k)$ of even degree, the norm map $\mathcal{N}^G_H:H^{even}(H,k)\rightarrow H^*(G,k)$ is defined by $\mathcal{N}^G_H(\alpha)=\Phi^*(1\int \alpha)$; for convenience I left out the cohomological construction of the element $1\int\alpha$ from $\alpha$. If $\alpha\in H^n(H,k)$ then $\mathcal{N}^G_H(\alpha)\in H^{n|G:H|}(G,k)$. This is a pretty hard construction for me to grasp, but ultimately it is extremely useful in group cohomology (in particular, it was defined by Leonard Evens and used to prove the finite generation result that $H^*(G,\mathbb{Z}_p)$ is Noetherian for any $p$ dividing $|G|$). -The reason this question arose is because $\mathcal{N}^G_H(1+\alpha)=1+tr^G_H(\alpha)+\cdots+\mathcal{N}^G_H(\alpha)$ for $\alpha\in H^n(H,k)$, where the intermediate terms are also transfers, i.e. the norm map is intertwined with the transfer map. Even simpler, $\mathcal{N}^G_H(\alpha+\beta)=\mathcal{N}^G_H(\alpha)+tr^G_H(\mu)+\mathcal{N}^G_H(\beta)$ for some $\mu\in H^*(H,k)$, if $H$ is an index-$p$ normal subgroup ($\alpha,\beta$ are homogeneous elements of even degree). -[[Addendum]] I have actually just stumbled upon a piece of this desired construction, on pg73-75 of Adem & Milgram's Cohomology of Finite Groups. The map given here is $BG\rightarrow (BH)^{|G:H|}\times_{S_{G/H}}ES_{G/H}\simeq B(S_{G/H}\int H)$ and is induced from $\Phi$. I assume from here we can functorially relate cohomology classes of $H$ to that of $S_{G/H}\int H$ and hence obtain $\mathcal{N}^G_H$. - -REPLY [4 votes]: The easiest starting point for equivariant stable homotopy theory is probably the paper -with that title in The Handbook of Algebraic Topology, edited by I.M. James. The paper -modern foundations for stable homotopy theory in the same volume is one of several possible -starting points for spectrum level stuff.<|endoftext|> -TITLE: Irreducibility of Induced Representation over arbitrary field -QUESTION [7 upvotes]: In representation theory of finite groups, there is Mackey test for irreducibility of an induced representation (Serre- Linear Representations of Finite Groups - #7.4). The author has stated it for field $\mathbb{C}$. Is it true for representations over arbitrary field whose characteristic is zero or co-prime to the order of group ? - -REPLY [11 votes]: To reinforce what Vladimir says, I'd replace his "I think so" by "Yes". The basic paper by Mackey in Amer. J. Math. 73 (1951) is retrievable online through JSTOR and might be worth looking at. His initial discussion of finite groups is done over arbitrary fields of characteristic 0, intended to generalize the work of Frobenius while providing background for the study of locally compact groups. It's only a convenience to assume that the ground field is $\mathbb{C}$ (done by Serre in part because his audience for the early lectures consisted of people with applied interests in character theory). -By now there are multiple textbook sources (including Serre, Bump, Isaacs, Dornhoff, Grove, etc.), each with different coverage and emphases. Probably the most comprehensive treatment of induction (along with tensor products and intertwining maps) is found in Part I, Section 10, of the large treatise Methods of Representation Theory by Curtis & Reiner, which goes beyond their pioneering 1962 book. At first they can work over arbitrary commutative rings, while discussing only the general module results; this flexibility is needed later to pass to J.A. Green's work on modular representations. -But in order to get an applicable irreduciblity criterion for induced characters in (10.25), they take a subfield of $\mathbb{C}$ which is a splitting field for the given finite group and all its subgroups. Even this kind of assumption can be weakened, since you only need a field of characteristic 0 containing sufficiently many roots of unity to ensure absolute irreducibility of the characters of the various irreducible representations involved. -If you really want to work with fields of prime characteristic, it's also a standard principle that the character theory works the same way provided only that the characteristic of the field doesn't divide the group order. To deal with absolute irreducibility of characters, you still need the splitting field assumption. - -REPLY [6 votes]: The inclusion ${\rm End}_H (V)\longrightarrow {\rm End}_G ({\rm Ind}_H^G V)$ is much clearer at the level of Hecke algebras. If ${\mathcal H}(G,V)$ denotes the spherical Hecke algebra attached to $V$ (which is known to be isomorphic to $ {\rm End}_G ({\rm Ind}_H^G V)$), then ${\rm End}_H (V)$ naturally identifies with the subalgebra of functions in the Hecke algebra with support $H$ : to $\psi$ you attach the function with support $H$ given by $h\mapsto \pi (h)\circ \psi =\psi \circ \pi (h)$. Then one has just to observe that the hypothesis of Mackey's criterion exactly says that functions in ${\mathcal H}(G,V)$ have support in $H$ ! -For all of this you need to assume that the order of $G$ is prime to the caracteristics of the field of coefficients.<|endoftext|> -TITLE: Results about existence/uniqueness of solution to Euler-Lagrange equations? -QUESTION [11 upvotes]: While studying calculus of variations, there is one question that I feel is missing in the texts I'm reading: -What can we say about the existence and/or uniqueness of solutions to Euler-Lagrange equations? Is there any general result? -I tried to google it, but found nothing. - -REPLY [8 votes]: The so called direct method of the calculus of variations provides one such existence and uniqueness result. -Here is the gist of it. Suppose that $X$ is a reflexive Banach space, e.g. a Hilbert space or a space of the form $L^p(\Omega)$, $p\in (1,\infty)$, $\Omega$ open subset of some Euclidean space. We are given a functional $J$ on $X$, i.e., a function -$$ J : X\to (-\infty, \infty]$$ -and we seek minimizers of such functionals, i.e., points $x_0\in X$ such that -$$J(x_0)=\inf_{x\in X} J(x)$$ -The subset of $X$ where $J$ is finite is called the domain of $J$. It is typically described by various equalities and inequalities called constraints. -Existence Theorem. Suppose that $J$ satisfies the following conditions. -\begin{equation} -\inf_{x\in X} J(x)>-\infty. -\tag{A} -\end{equation} -\begin{equation} - \mbox{The set}\;\;\lbrace -J\leq t\rbrace:=\lbrace x\in X;\;\; J(x)\leq t\rbrace -\;\; \mbox{is convex},\;\;\forall t\in \mathbb{R}. -\tag{B} -\end{equation} -\begin{equation} - \mbox{The set}\;\;\lbrace -J\leq t\rbrace\;\; \mbox{is closed in the norm topology},\;\;\forall t\in \mathbb{R}. -\tag{C} -\end{equation} -\begin{equation} -\lim_{\|x\|\to\infty} J(x)=\infty. -\tag{D} -\end{equation} -Then $J$ admits at least one minimizer. -Remark. I should comment on the four conditions above. Condition (A) states that $J$ is bounded from below. Condition (B) states that $J$ is a convex function in the usual way. Condition (C) states that $J$ is lower semicontinuous in the norm topology. Under the convexity assumption this is equivalent to $J$ being lower semicontinuous with respect to the weak topology. If $J$ happens to be differentiable, then the differential of $J$ at any minimizer $x_0$ is zero. The ensuing equation $dJ(x_0)=0$ translates into the classical Euler-Lagrange equations. The minimizer postulated by the above theorem is unique provided that $J$ is strictly convex. For more about the direct method see Wikipedia and the reference therein. -In general, the objects satisfying the Euler-Lagrange equations are critical points of a functional $J: X\to\mathbb{R}$, i.e., points where the differential of $J$ vanishes. The critical points that are observable and detectable in the real world are stable and these correspond to (local) minimizers of $J$. Sometime, one is interested in not necessarily stable objects, i.e., critical points of $J$ that are not necessarily local minimizers. Morse theory is particularly good at detecting such points. All applications of this theory are based on the following principle. -Suppose that $J: H\to\mathbb{R}$ is a $C^2$ function on a Hilbert space $H$ satisfying some additional compactness assumption (e.g. the Palais-Smale condition). Suppose that there exist real numbers $a < b$ such that the sublevel sets -$$ \lbrace J\leq a\rbrace\;\;\mbox{and}\;\; \lbrace J\leq b\rbrace$$ -are not homeomorphic. Then $J$ admits a critical point $x_0$ such that -$$ J(x_0)\in [a,b]. $$ -For more detail see the booklet by Paul Rabinowitz, Minimax methods in critical point theory with applications to differential equations.<|endoftext|> -TITLE: On the generalized Gauss-Bonnet theorem -QUESTION [20 upvotes]: I am trying to learn about basic characteristic classes and Generalized Gauss-Bonnet Theorem, and my main reference at the moment is From calculus to cohomology by Madsen & Tornehave. I know the statement of the theorem is as follows: - -Let $M$ be an even-dimensional compact, oriented smooth manifold, $F^{∇}$ be the curvature of the connection ∇ on a smooth vector bundle $E$. -$$∫_{M}\mathrm{Pf}\Big(\frac{−F^{∇}}{2\pi}\Big)=χ(M^{2n}).$$ - -My questions are: - -How does this relate to counting (with multiplicities) the number of zeros of generic sections of the vector bundle? -Also, are there other good references for learning this topic? - -REPLY [5 votes]: Mathai and Quillen have a theorem that computes the Euler characteristic as an integral of a form defined using a section of and a connection on the tangent bundle. If one scales the section by a factor $t$, then at $t=0$ one gets the Gauss-Bonnet theorem, and as $t\to\infty$ the integral localizes near the zero set of the section and becomes the Poincar\'e-Hopf theorem.<|endoftext|> -TITLE: Is there a simplicial volume definition of Chern Simons invariants? -QUESTION [9 upvotes]: Suppose we have some compact hyperbolic 3-manifold $M=\Gamma\backslash\mathbb H^3$. Now we know that the hyperbolic volume of $M$ can be defined as (a constant times) the simplicial volume of the fundamental class $[M]\in H_3(M,\mathbb Z)$, which is a homotopy invariant. -Now the hyperbolic volume and Chern-Simons invariant $M$ are connected by the following definition: -$$i(\operatorname{Vol}(M)+i\operatorname{CS}(M))=\frac 12\int_M\operatorname{tr}(A\wedge dA+\frac 32A\wedge A\wedge A)\in\mathbb C/4\pi^2\mathbb Z$$ -where $A$ is any flat connection on the trivial principal $\operatorname{SL}(2,\mathbb C)$-bundle over $M$ whose monodromy is the isomorphism $\pi_1(M)=\Gamma$. This corresponds to a particularly natural homomorphism (based on a dilogarithm) in $H_3(\operatorname{SL}(2,\mathbb C),\mathbb Z)\to\mathbb C/4\pi^2\mathbb Z$ (see work of Neumann and Zickert). -This close connection between the two invariants $\operatorname{Vol}(M)$ and $\operatorname{CS}(M)$ motivates the following question: - -Is there a definition of $\operatorname{CS}(M)$ within the framework of simplicial volume? - -REPLY [8 votes]: If you use eta invariant in place of Chern-Simons invariant, there is almost such a definition, at least in a closely related context. If we restrict to surface bundles over the circle with fiber of fixed genus, then the eta invariant of a fibered 3-manifold can be thought of as a certain kind of class function on the mapping class group. Such eta invariants exist for many different kinds of (unitary) representations (of subgroups of mapping class groups), and under suitable circumstances (see e.g. http://arxiv.org/abs/1003.4977) the functions they define on (subgroups of) mapping class groups are examples of what are known as homogeneous quasimorphisms. -Quasimorphisms arise in the theory of bounded cohomology; there is a duality theory relating them to a (relative) 2-dimensional Gromov norm, called stable commutator length. Gromov norm here is just a synonym for simplicial volume, as in your question.<|endoftext|> -TITLE: gcd of three numbers -QUESTION [15 upvotes]: Let $a, b, n$ be positive integers. Assume that $\gcd(a,b,n)=1$. -It seems that one can prove that there exist two integers -$c$ and $d$ bounded from above by $( \log n )^{O(1)}$ such that -$ \gcd (ac + bd, n) =1$. However the only proof I can see is -by a complicated exclusive-inclusive argument. -I am wondering whether it has been proved somewhere or whether there is -a simpler argument. -Thanks a lot for helping. -Qi - -REPLY [8 votes]: The answer is yes, there exists $c$ and $d$, even with $c = 1$, and -$d \ll (\log(n))^{O(1)}$. This follows from a result of Iwaniec. -It suffices to assume that $(a,b) = 1$. -Suppose that $(n,b) = e$, which implies that $(a,e) = 1$. -Since $b$ is divisible by $e$, it follow that $(a+bk,e) = 1$ -for all integers $k$. Let $m$ denote the largest factor of $n$ such that -$(m,e) = 1$, it clearly satisfies $(m,b) = 1$. -If $(a+bk,m) = 1$, then because $(a+bk,e) = 1$, -one also has $(a+bk,n) = 1$. -Hence the problem becomes: given an integer $m$, and an arithmetic -progression -$$a, a + b, a + 2b, a+3b, a+4b, \ldots $$ -with common difference $b$ prime to $m$, can one find a small integer $d$ such -that $a+db$ is prime to $m$? -Equivalently, let $g \in \mathbf{Z}$ be any multiplicative inverse to $b$ mod $m$, -then does there exist a small $d$ such that -$ag + dbg$ is prime to $m$? Equivalently, does there exist a small $d$ -such that $ag + d$ is prime to $m$? -Recall the definition of the Jacobsthal function: $j(m)$ -is the smallest integer such that any arithmetic progression -of length $j(m)$ (with common difference one) contains an -element which is co-prime to $m$. -If $m|n$, then $j(m) \le j(n)$. -In particular, there exists a $d \le j(m) \le j(n)$ with $c = 1$ such -that $ac+bd$ is prime to $n$. -Finally (the hard part), by a result of Iwaniec (Demonstratio Math. 11 (1978), 225-231 (MR0499895)), if $n$ has -$r$ distinct prime factors, then $j(n) \ll r^2 \log^2 r$, which -implies that -$$j(n) \ll \log^{2}(n) = \log(n)^{O(1)}.$$<|endoftext|> -TITLE: Why is there a unique hyperbolic simplex of largest area? -QUESTION [20 upvotes]: Why is there a unqiue ideal $n$-simplex in $\mathbb H^n$ with largest volume for $n\geq 3$? - -For $n=3$, this is a standard calculation, and for larger dimensions is much harder (see Haagerup and Munkholm); I made the mistake of originally stating that this is easy for all $n$. Anyway, I'm wondering if there is a deeper reason for this fact (one which does not rely on calculation); let me say what type of answer I am looking for. -Recall that this is a crucial step in Gromov's proof of Mostow Rigidity, a (very rough) sketch of which is as follows. If we have two cocompact subgroups $\Gamma_1,\Gamma_2\subseteq\operatorname{Isom}(\mathbb H^n)$ which are isomorphic as abstract groups, then such an isomorphism must extend to a homeomorphism of their boundaries (as hyperbolic groups). In other words, we get a self-homeomorphism of $\partial\mathbb H^n$. But now since simplicial volume is a homotopy invariant, this homeomorphism must preserve the $(n+1)$-tuples of points giving simplices of maximal volume. Then one proves that any self-homeomorphism of $\partial\mathbb H^n$ with this property is in fact induced by an element of $\operatorname{Isom}(\mathbb H^n)$, so $\Gamma_1,\Gamma_2$ are conjugate (via this element) in $\operatorname{Isom}(\mathbb H^n)$. -Is there a high-brow proof of the fact in the title of this question? (that is, one which uses some rigidity results in Lie groups). A first step at answering this question would be to realize that the configuration space of $n+1$ points in $\partial\mathbb H^n$ modulo isometries is nontrivial iff $n\geq 3$. Now volume is some real-analytic function on this moduli space. Is there a nice explanation for why it miraculously has a unique global maximum (in fact, a unique local maximum!) (which thus implies Mostow rigidity)? - -REPLY [19 votes]: Firstly, this is not something you "can easily prove by calculation": The proof (by Haagerup and Munkholm) was published in Acta. -Secondly, in three dimensions, the set of ideal simplices are parametrized by positive triples $\alpha, \beta, \gamma$ such that $\alpha + \beta + \gamma = \pi$ -- these are the dihedral angles. One can then show that the volume is a convex function of the dihedral angles (this is actually true for any convex ideal polyhedron, but the tetrahedron case is at the base of the proof, and is proved differently). This immediately implies that the regular ideal simplex is the one of maximal volume (by symmetry considerations) -- this result (in greater generality) is in a paper of mine in the early nineties: -Euclidean structures on simplicial surfaces and hyperbolic volume -I Rivin -The Annals of Mathematics 139 (3), 553-580 -In higher dimensions the argument does not quite work, but the fact that a simplex is determined by its codimension-two areas, together with the Schlafli differential formula implies that the volume (as the function of dihedral angles) should have the same signature everywhere, and since the volume function is proper on the set of ideal simplices, it should have a maximum, and so should be concave. This, however, does not quite prove Haagerup-Munkholm, since the set of possible dihedral angles is much more complicated in dimensions bigger than three, and is not obviously convex, so the symmetry argument breaks down.<|endoftext|> -TITLE: Lower semicontinuity of Kullback-Leibler divergence -QUESTION [8 upvotes]: The Kullback-Leibler Divergence (KLD) of two PMF's $P(x)$ and $Q(x)$ is $D(P||Q)=\sum_x P(x)\log(P(x)/Q(x))$, with the provisos that $0\cdot \log (0/p)=0$ and $p\cdot \log (p/0)=+\infty$ whenever $p>0$. -It is known that KLD is continuous at $(P,Q)$ if $Q$ is strictly positive over all $x$'s. What can be said otherwise? -To be more specific, assume we are given a sequence of PMF $\{(P_n,Q_n)\}_{n\geq 0}$ s.t. $(P_n,Q_n)\rightarrow (P,Q)$ in the simplex of PFM's (with the topology induced by, say, norm-1 distance). -Is it correct to deduce that -$\lim \inf_{n\rightarrow \infty} D(P_n||Q_n) \geq D(P||Q)$ ? -This would follows if KLD is lower-semicontinuous, right? -Many thanks. - -REPLY [5 votes]: In addition to the conventions you have mentioned, it is also assumed that $0\log(0/0)=0$. -With these conventions, I think, in the finite case, it is always true that $$\lim_{n\to \infty} D(P_n||Q_n)=D(P||Q)$$ As you said, if $Q(x)>0$ for all $x$, its immediate from the Dominated Convergence theorem. -The problem is only when for some $y$, $Q(y)=0$ whereas $P(y)>0$. -In which case $P(y)\log(P(y)/Q(y))=\infty$ and $D(P\|Q)=\infty$ -But since $(P_n,Q_n)\to (P,Q)$, we have $P_n(y)\to P(y)$ and $Q_n(y)\to Q(y)$, whence $$P_n(y)\log(P_n(y)/Q_n(y))\to P(y)\log(P(y)/Q(y))=\infty.$$ Hence $D(P_n||Q_n)\to \infty$ -So in any case we have $\lim_{n\to \infty} D(P_n||Q_n)=D(P||Q)$. -In a general measurable space (i.e., if $P_n, Q_n, P, Q$ are probability measures on some general measure space say $(\mathbb{X}, \mathcal{X})$), I think, we have only lower semicontinuity. -Pardon me, if something is wrong.<|endoftext|> -TITLE: Diffeomorphisms vs homeomorphisms of 3-manifolds -QUESTION [18 upvotes]: For a smooth 3-manifold $M$, is the natural map from the space of diffeomorphisms of $M$ to the space of homeomorphisms of $M$, -$${\sf Diff}(M) \longrightarrow {\sf Top}(M),$$ -a weak homotopy equivalence? Equivalently, is the space of smooth structures on a topological 3-manifold contractible? (This is as opposed to just connected, which is the usual statement of Moise's theorem.) - -REPLY [26 votes]: $\mathsf{Diff}(S^3)$ is homotopy equivalent to $O(4)$, by Hatcher's solution to the Smale Conjecture: Hatcher, Allen E. A proof of the Smale conjecture, Diff(S3)≃O(4). Ann. of Math. (2) 117 (1983), no. 3, 553–607. -Cerf proved that the Smale conjecture implies that $\mathsf{Diff}(M) \to \mathsf{Homeo}(M)$ is a weak equivalence for all $3$--manifolds $M$, in J. Cerf, Sur les difféomorphismes de la sphère de dimension trois ( $\Gamma_4$ = 0 ), Lecture Notes in -Math., vol. 53, Springer-Verlag, Berlin and New York, 1968, I think. See Linearization in 3-Dimensional Topology by Hatcher.<|endoftext|> -TITLE: Are there 'cohomology' functors that respect all Eilenberg-Steenrod axioms except homotopy invariance? -QUESTION [9 upvotes]: What goes wrong in the axiomatic definition of a generalized (co)homology theory if one drops the axiom of homotopy invariance i.e. that homotopic maps should induce the same map in (co)homology? -Or do we have examples? Are there "interesting" or "useful" functors $\mathfrak{h}^{\cdot}:\mathrm{Spaces}\to \mathrm{Ab}$ that respect all Eilenberg-Steenrod axioms except homotopy invariance? Could such an $\mathfrak{h}$ be used to distinguish between two homotopy equivalent non-homeomorphic spaces? -(Take your favourite definition of admissible spaces) - -REPLY [12 votes]: ordinary differential cohomology roughly speaking satisfies the demands of the question -the homotopy axiom fails and is replaced by a variation statement -it is a functor defined for smooth manifolds and smooth maps -the suspension isomorphism is only true in a weaker version -and there are [single space] axioms -Simons& Sullivan first issue of topology -finally it is quite geometric and useful in differential geometry & quantum field theory -it extends to the generalized cohomology context -essentially by dropping the homotopy axioms -thus it pervades a natural class of examples to the spirit of the question -this question is not obviously a good one, but it is and I salute the questioner.<|endoftext|> -TITLE: Why are isometries of Minkowski space necessarily linear? -QUESTION [13 upvotes]: The Mazur-Ulam theorem says that any surjective isometry of normed vector spaces is affine. This argument doesn't seem to apply to Minkowski space (of special relativity) since the metric is indefinite. How would one show that the Poincaré group consists of affine maps? This seems really standard but I can't seem to find it anywhere. - -REPLY [5 votes]: The following general result is described in the answer I posted on math.stackexchange here: if $V_1$ and $V_2$ are finite-dimensional vector spaces of equal dimension over an arbitrary field and they are both equipped with nondegenerate bilinear forms $B_1$ and $B_2$, then a function $\sigma \colon V_1 \rightarrow V_2$ such that $B_1(v,w) = B_2(\sigma(v),\sigma(w))$ for all $v$ and $w$ in $V_1$ must be an isomorphism of vector spaces (linear, injective, and surjective). The proof does not use non-degeneracy of $B_2$ on $V_2$, so that property actually follows from non-degeneracy of $B_1$ on $V_1$. -In the particular case you ask about, where $V_1 = V_2$ and $B_1 = B_2$, the result says: for a finite-dimensional vector space $V$ over an arbitrary field and a nondegenerate bilinear form $B$ on $V$, a function $\sigma \colon V \rightarrow V$ for which $B(v,w) = B(\sigma(v),\sigma(w))$ for all $v$ and $w$ in $V$ must be linear, injective, and surjective.<|endoftext|> -TITLE: "The Galois group of $\pi$ is $\mathbb{Z}$." -QUESTION [19 upvotes]: Last year, in a talk of Michel Waldschmidt's, I remember hearing a statement along the lines of the title of this question, that is, "The Galois group of $\pi$ is $\mathbb{Z}$.". In what sense/framework is this true? What was meant exactly - and can this notion be made precise? - -REPLY [17 votes]: I gather that the idea behind $\mathrm{Gal}(\pi)=\mathbb{Z}\backslash\{0\}$ (not $\mathbb{Z}$, $0$ is not a conjugate of $\pi$!) comes from Euler's formula: -$$\prod_{n\in \mathbb{Z\backslash\{0\}}}\bigg(1-\frac{x}{n\pi}\bigg)=\frac{\mathrm{sin}(x)}{x} \in\mathbb{Q}\{x\} $$ -which can make you think of those $n\pi$ as the conjugates of $\pi$. -But in order for the Galois groups to act transitively on the conjugates you need all the non-zero rationals, so that $\mathrm{Gal}(\pi)=\mathbb{Q^{\times}}$. -For some actual evidence that $\mathrm{Gal}(\pi)=\mathbb{Q^{\times}}$ is the right answer (this is, consistent with the conjectural picture of periods and motives), see sections 3 and 5 of Galois theory, motives and transcendental numbers by Yves Andre, alredy mentioned in the comments.<|endoftext|> -TITLE: Averages of Euler-phi function and similar -QUESTION [7 upvotes]: What are the odds two numbers are relatively prime? This is known to be $\frac{6}{\pi^2}$. The proof involves calculating averages of the Euler phi function. -\[ \phi(1) + \phi(2) + \dotsb + \phi(n) \approx 3 \left(\frac{n}{\pi}\right)^2 + O(n \log n). \] -So even though $\phi$ is rather noisy, its sum is relatively "quiet" behaving like a parabola. Why does all the noise disappear? -I'm wondering how it is possible to compute the exactly coefficient of $n^2$ in this expansion. It seems like a coincidence. -For good measure I have plotted the $\sum \phi$ and $\sum \phi - (\cdot)^2$ as demonstration. - -Also we could consider a related function $\displaystyle \theta(n) = n^2 \prod_{p\mid n} \left( 1 - \frac{1}{p^2}\right)$. Here $\theta(1) + \theta(2) + \dotsb + \theta(n) \approx c \cdot n^3 + \dotsb$. What's the procedure for computing the constant $c$? - -REPLY [8 votes]: This exact question has actually been answered a few times on Math Stack Exchange. -See this for a general approach to finding the mean value of multiplicative functions which are "close" to $n$. Here is the idea: -Heuristic: Notice $f(n)\approx n$, then $\frac{f(n)}{n}\approx 1$. For functions close to one, convolution with the Möbius function will be close to zero, so we can deal with it easily. Lets define $g(n)=(\mu*\frac{f(d)}{d})(n)=\sum_{d|n}\frac{f(d)}{d}\mu\left(\frac{n}{d}\right)$ so that $(1*g)(n)=\frac{f(n)}{n}$. The idea will be to rewrite everything in terms of $g$ since $g(n)$ will be small. -The answer linked above provides the precise computation, and this will cover the Totient function, and the second example you gave above. -For the Totient function in particular, see this answer which also gives history of upper and lower bounds on the error term. In particular, the error term is surprisingly at least $\Omega (x\sqrt{\log \log x}).$ -Hope that helps, - -REPLY [4 votes]: A more analytic way to see this is through Dirichlet series, namely we know that (H&W as mentioned in the other answers is a good reference, but the identity can be seen by the Euler product) -$$ - \sum_{k=1}^\infty \phi(k) k^{-s}= \frac{\zeta(s-1)}{\zeta(s)}. -$$ -Perron's formula gives -$$ \sum_{k=1}^n \phi(k) =\frac 1 {2 \pi i}\int_{c-\infty i}^{c+\infty i} \frac{\zeta(s-1)}{\zeta(s)} \frac{x^s} s ds, \qquad (n < x < n+1)$$ where $ c > 2$ is large enough for the Dirichlet series to be absolutely convergent. From moving $c>2$ to $ 1 < c < 2$ we pick up a residue at $s=2$ coming from the zeta-function's pole at Re$(s)=1$, from where the main term comes. This will be exactly $\frac {x^2} {2 \zeta(2)}=3 x^2/\pi^2$. -The remaining integral can be estimated as an error term (the strong form of the error term as mentioned above will be more difficult to obtain this way however). -The Dirichlet series argument why all the noise disappears (The error term is rather good) is simply that the Dirichlet series in the region $ 1 < $ Re $ (s)<2 $ does not behave too badly (i.e. has no poles, and does not grow too fast when Im$(s)\to \infty$). This method also works for other arithmetical functions (often with worse error terms), for example divisor problems. Cases more difficult to treat with the convolution method, where even more noise remains because the function has poles includes for example estimating sums of the Möbius function or the Von Mangoldt function (the RH gives much better estimates than known unconditional estimates in these cases). - -REPLY [3 votes]: Let's say we want to figure out how many lattice points in a disk of radius $R$ are "visible" (have relatively prime coordinates). Call this function $f(R).$The basic point in proving the first statement is that if you look at lattice points in (say) a disk of radius $R$, that set has a stratification according to to the gcd of the coordinates, so you have a sum like -$\pi R^2 = \sum_{d=1}^\infty f(R/d).$ -If you assume that the probability exists, then we can write the above as: -$\pi R^2 = f(R) \sum_{d=1}^\infty\frac{1}{d^2}= f(R) \zeta(2).$ -While the computation is not rigorous, it gives the right constant, and the right intuition. -By the way, the fact that the limit exists is related to the fact that $SL(2, Z)$ acts ergodically on $\mathbb{R}^2.$ See -Densities in free groups and $\mathbb {Z}^ k $, Visible Points and Test Elements -(I Kapovich, I Rivin, P Schupp, V. Shpilrain) -for group-theoretic applications. -I am quite sure that the $\theta$ computation can be given in the same way if you interpret the sum "geometrically"<|endoftext|> -TITLE: Conjugation between commutative subalgebras of a matrix algebra? -QUESTION [9 upvotes]: Let $K$ be an algebraically closed field and $M_n(K)$ the $K$-algebra of all matrices $n\times n$ over $K$. If $L$ and $M$ are two isomorphic commutative subalgebras of $M_n(K)$, it is true that there exists a regular matrix $S\in M_n(K)$ such that $SLS^{-1}=M$. That is, the isomorphism can be chosen to be inner? - -REPLY [10 votes]: Two isomorphic subalgebra of $M_n(K)$ do not need to be conjugated. The following example is taken from Exercise 161 of my web site http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf -Set $n=p+q$ with $q>p>0$. Then define $\mathcal A$ as the subset of $M_n(k)$ made of the matrices with block form -$$\left(\begin{array}{cc} 0_p & 0_{p\times q} \\\\ A & 0_q \end{array}\right).$$ -Likewise, ${\cal B}$ is made of the matrices -$$\left(\begin{array}{cc} 0_q & 0_{q\times p} \\\\ B & 0_p \end{array}\right).$$ -Both $\cal A$ and $\cal B$ are subalgebras of $M_n(k)$, with dimension $pq$ and the property that $MN=0_n$ for every two elements (of the same algebra). They are obviously isomorphic, because the algebra structure is trivial. But ${\cal A}$ and $\cal B$ are not conjugated in $M_n(k)$. However $\cal B$ is conjugated to ${\cal A}^T$ in $M_n(k)$.<|endoftext|> -TITLE: Ubiquitous Zimin words -QUESTION [10 upvotes]: Let $w$ be a word in letters $x_1,...,x_n$. A value of $w$ is any word of the form $w(u_1,...,u_n)$ where $u_1,...,u_n$ are words. For example, $abaaba$ is a value of $x^2$. A word $u$ is called unavoidable if every infinite word in a finite alphabet contains a value of $u$ as a subword. There is a nice characterization of unavoidable words due to Zimin. A word $u$ in $n$ letters is unavoidable if and only if a value of $u$ is a subword of the $n$th Zimin word $Z_n$ defined by induction: $Z_1=x_1$,...,$Z_n=Z_{n-1}x_nZ_{n-1}$, that is $Z_1=x_1, Z_2=x_1x_2x_1, Z_3=x_1x_2x_1x_3x_1x_2x_1,...$. Zimin words appear very often in algebra. For example, if one lists binary expressions of all numbers $1,2,3,...$ and records the numbers of 0s at the end of the numbers plus 1, one will get $12131214121...$ which is the infinite Zimin word. Values of Zimin words also appeared as $m$-sequences in Levitzki's description of Baer radical (see Jacobson's book "Structure of rings") and in the work of Schutzenberger. The Zimin words have obvious fractal structure, so these words could have appeared in other areas of mathematics as well. - Question. Do Zimin words appear in your area of mathematics? -This might be a "big list" question. But I do not know how big the list is, it may be empty. If it turn out to be big, I will make the question "community wiki". - Update 1. Googling 121312141213121 ($=Z_4$) returns 439 results including a discussion at reddit. - Update 2. The most curious among these links is this link to a US patent. It looks like Zimin words up to $Z_5$ at least have been patented before Zimin introduced them. - Update 3. This is needed for the book with the current title "Words and their meaning" which we are writing together with Mikhail Volkov. There we already have four different applications of Zimin words to different areas of algebra and would like to mention applications outside algebra as well. - -REPLY [5 votes]: I am about to go chair a dissertation defence where the candidate uses an endomorphic image of Zimin words (of his invention) to construct various counterexamples to questions in combinatorics of finite and infinite words, mainly to do with counting palindromes in subwords and scattered subwords of finite and infinite words. The image in question are words on a three-letter alphabet of the form w_0=a, w_1=aba, w_{n+1}=w_nc^nw_n, which the candidate Bojan Basic calls highly potential words. This is obviously a related sequence to Zimin words, and the candidate managed to construct several examples and counterexamples using these, including a counterexample to an already published "theorem" by non-unknown authors. -I suppose it qualifies as an application of Zimin words as highly potential ones are clearly homomorphic images of Zimin words and inspired by them. Don't know if it is too late for your book. Counter-question: has anybody seen these highly potential words before? -Petar Markovic<|endoftext|> -TITLE: Is there a model of Set Theory which thinks it is the Standard Model, i.e. is there a Universe U such that U $\models$ U=V? -QUESTION [6 upvotes]: I asked my friend (a Set Theorist) this question and he said that every model of ZFC thinks it is the Standard Model. But, I'm not sure it is so simple. First, because I don't know how a Universe could test whether or not it is the Standard Model. And second, because I think that various models of ZFC could have different opinions about whether they are the standard model. -For a model which does think it is standard, I'm thinking of a model which has enough evidence to believe it is the standard model, perhaps one obtained after many generic extensions, or one which has "nice" properties, perhaps because PD (projective determinacy) holds there. But, I don't have the knowledge or vocabulary to put this in precise language yet. -I can imagine a universe which does not think that it is the standard model, one which is "close" to its generic extensions, perhaps one where generic filters exist (a universe where MA holds for example). -What do you think? Is there any way to tell, within a given model, whether it has enough evidence to conclude that it is the standard model? Or, is it as simple as my friend suggests and every model thinks it is standard? Or, is it simple because no model thinks it is standard since there are always generic extensions of any given model? - -REPLY [9 votes]: One of the main lessons of set theory is that many of our familiar set-theoretic concepts, such as countability, uncountability, well-orderedness, ill-foundedness, even finiteness, depend on the set-theoretic context in which they are considered. We know that different models of set theory can disagree about whether a set, the same set, is countable or uncountable, about whether a relation is well-founded or ill-founded and even about whether a set is finite or infinite. -First, of course, one can play all kinds of games using the compactness theorem or ultrapowers to make models with vastly different beliefs about the same set. For example, by starting in a universe $V$ and constructing a nonprincipal ultrapower $W=V^\omega/\mu$, one has nonstandard finite sets of natural numbers in $W$ that $W$ thinks are finite, but $V$ thinks have size continuum. -But second, the method of forcing allows us to construct new universes that stay close to the original universe, the ground model, in some ways while differing in others, which we can precisely control. They have the same ordinals, for example, and thus agree on the well-foundedness of the relations they have in common. But meanwhile, we can still make them disagree on other matters, such as the power set operation, since a forcing extension can have additional set of natural numbers and thus additional reals. So the concept of the "set of reals" is not absolute between models of set theory, even when they agree on the ordinals. -The issue of background dependence is sometimes thought not to apply to arithmetic, since we can prove that there is only one standard model of arithmetic. But one can object to this slight-of-hand: the uniqueness of $\mathbb{N}$ is proved in set theory, and so each model of set theory has a unique concept of natural number, but different models of set theory can have different non-isomorphic versions of $\mathbb{N}$, as my ultrapower example above shows. -Because of this pervasive dependency of set-theoretic concepts on the set-theoretic background in which they are considered, some set theorists, perhaps like your friend, are pushed to the view that every model of set theory provides its own concept of standardness, and that the notion of an absolute set-theoretic background is illusory. Every model of set theory look upon itself as the entire universe, a standard world by its own lights. This is a sense in which the notion of standardness itself is relative-to-a-model. -Other set-theorists retain the idea that there is an absolute sense of standardness, that we are living in the real universe of sets, and that it makes sense to judge models as being correct or incorrect about their judgements of well-foundedness. As you know, such questions were precisely the topic of my recent course on the philosophy of set theory, and the debate between these two perspectives is the thrust of my article on the set-theoretic multiverse, along with the other readings we had in the course. -But let me close by mentioning one case often mentioned by set-theorists, where a model of set theory can begin to have an inkling that something is seriously awry with its notions of standardness. Specifically, if ZFC is consistent, then we know that $\text{ZFC}+\neg\text{Con}(\text{ZFC})$ is also consistent. What would it be like to live in a model of this latter theory? In that universe, we would see that all the ZFC axioms are true, but also we would think that the ZFC axioms are formally inconsistent, admitting a proof of a contradiction.<|endoftext|> -TITLE: Is the maximum tree-path length distributed lognormally (in the limit) ? -QUESTION [5 upvotes]: Consider a full binary tree with $k>10$ levels. Let the lengths of individual edges in this tree be i.i.d. random variables with finite moments. Then total lengths of the $2^{k-1}$ source-to-sink paths in this tree are approximately Gaussian by the CLT, regardless of the edge-length distribution. We are interested in the limit distribution ($k\rightarrow\infty$) of the maximum path length in the tree. -Our numerical simulation built 100K independent trees ($k=15$) with $(a)$ uniform and $(b)$ Gaussian edge lengths. The resulting distributions for $(a)$ and $(b)$ did not look qualitatively different and were somewhat skewed to the right. Lognormal distributions provided very close fits --- better than Gumbel and Airy. If lognormal is indeed the limit distribution, we would appreciate references or suggestions on proving this analytically. - -REPLY [3 votes]: Unless I misunderstood your question, this can be entirely rephrased in terms of branching random walks. This goes as follows: at time 0 there is 1 individual at position 0. Each individual gives birth to two descendants, whose position is the position of the parent plus a jump, where all jumps are i.i.d. random variable. You are asking about the maximum position at time $k$, $M_k$. -This is a much studied problem, with deep links to traveling wave partial differential equations such as the Fisher-KPP equation. (Eg, in the space-time continuous case where branching random walk is replaced by branching Brownian motion, the function $u(t,x) = \mathbb{P}(M_t >x)$ solves the KPP equation with initial condition $u(0,x) = 1_{x<0}$.) -See this recent paper http://front.math.ucdavis.edu/1101.1810 by Elie Aidekon, which provides complete answers to your question under minimal assumptions on the jump distribution. The main result is then that $M_k - ck + (3/2) \log k$ converges to a random variable, where $c$ is a constant that is easy to compute. The distribution of the limiting random variable doesn't have to be either Gumbel or lognormal. -In the space-time continuous case where branching random walk is replaced by branching Brownian motion, this result is a famous result originally due to Maury Bramson (1983): Convergence of solutions of the Kolmogorov equation to travelling waves. Mem. Amer. Math. Soc. 44, no. 285.<|endoftext|> -TITLE: unitary irreps of O(p,q) -QUESTION [12 upvotes]: I am interested in the irreducible unitary representations of the orthogonal groups $O(p,q)$. By $O(p,q)$ I mean the real Lie groups which preserve the quadratic form of signature $(p,q)$ in $\mathbb{R}^n$, $n = p+q$ dimensions. Special cases of interest in physics are the conformal group O(4,2), the deSitter group O(4,1) and the anti-deSitter group O(3,2) in dimension 4 = 3+1 (i.e. Minkowski space). I am only interested in the non-compact groups, the compact cases being well-understood. (So I expect that the irreducible unitary representations are all infinite dimensional.) I am not exclusively interested in Lorentzian signature $n = (n-1) + 1$, nor am I exclusively interested in $n=4$. As a theoretical physicist, I am not familiar with the undoubtedly vast literature on representations of non-compact Lie groups, and I would appreciate a few pointers to the most relevant references; those that review the general setting, but especially which address these groups specifically. - -REPLY [14 votes]: The unitary dual of Spin(n,1) is known for all n (Hirai, 1962, see Math Reviews MR0696689). -This gives the unitary dual of the identity component of SO(n,1) (which is a quotient -of Spin(n,1)). The unitary dual of any group can be deduced readily from that of its identity -component. Also SL(2,C)=Spin(3,1), and SL(2,R)=Spin(2,1). -The unitary dual of Sp(4,R) is known (Nzoukoudi, 1983, MR0736241), and Sp(4,R)=Spin(3,2). -The unitary dual of SU(2,2) is known (Knapp/Speh, 1982, MR0645645), and SU(2,2)=Spin(4,2). These -give the unitary duals of SO(3,2) and SO(4,2). -Besides this (and the compact groups) I think the entire unitary dual is not known -for any O(p,q), although for any fixed group a large part of its unitary dual is known. You might look at some papers by Welleda Baldoni and Tony Knapp from the 1980s or so. -Note: much of the literature applies to groups of "Harish-Chandra's class". -This includes all SO(p,q), but O(p,q) only if p+q is odd. So if p+q is even you -have to do a little extra work to get from the unitary dual of SO(p,q) to that -of O(p,q) (for each irreducible unitary representation of SO(p,q) you have to decide -if its induction to O(p,q) has 1 or 2 irreducible summands). -As for www.liegroups.org, we hope to have the complete answer for any group, but -that day has not yet arrived.<|endoftext|> -TITLE: Learning Tropical geometry -QUESTION [21 upvotes]: I'm interested in learning tropical geometry. But my background in algebraic geometry is limited. I know basic facts about varieties in affine and projective space, but nothing about sheaves, schemes etc. -I wanted to know how much algebraic geometry does one need to understand the research literature in tropical geometry. -What are the other subjects which I must know before starting tropical geometry? - -REPLY [13 votes]: Several years ago, I participated in a learning seminar in tropical algebraic geometry and collected several helpful survey articles. (This was before Maclagan and Sturmfels' book was written, which I suspect is excellent.) -Anyway, here were some of the most helpful intro points for me: -Tropical Mathematics, -First Steps in Tropical Geometry, -Tropical Algebraic Geometry, -Introduction to Tropical Geometry, -The Tropical Grassmannian, -The Number of Tropical Plane Curves Through Points in General Position. -Sturmfels, Speyer, and Gathmann all write very well, and Gathmann especially devotes considerable space to giving motivation for the field. Mikhalkin, of course, was the one who pioneered the idea of attacking challenging classical problems (such as counting the number of plane curves of genus $g$ and degree $d$ passing through $3d + g - 1$ points, which had just been solved by Capraso-Harris in the late 90s) using the tropical semifield.<|endoftext|> -TITLE: Theme of Isbell duality -QUESTION [31 upvotes]: Let $C$ be a small category. Isbell duality provides an adjunction $\widehat{C} {{\mathcal{O} \atop \longrightarrow} \atop {\longleftarrow \atop \mathrm{Spec}}}\widehat{C^{\mathrm{op}}}^{\mathrm{op}}$. If I understand correctly, this notation suggests that $\widehat{C}$ consists of "geometric" objects, whereas $\widehat{C^{\mathrm{op}}}$ consists of "algebraic" objects. Then $\mathcal{O}$ associates to some geometric object $X$ the algebraic object of all global functions on $X$, whereas $\mathrm{Spec}$ associates to some algebraic object $A$ the "affine" geometric object associated to $A$. Note both the unit $\eta_X : X \to \mathrm{Spec}(\mathcal{O}(X))$ and the counit $\varepsilon_A : A \to \mathcal{O}(\mathrm{Spec}(A))$ of this adjunction are given by evaluation. As with every adjunction, we get an equivalence of categories between its fixed points, i.e. those $X$ such that $\eta_X$ is an iso, and those $A$ such that $\varepsilon_A$ is an iso (Isbell-dual objects). -In general this formulation does not make sense, but it is well-known in the following special cases which are also alluded in the nlab article. -1) Algebraic geometry: There is an adjunction $\mathrm{Sch} {{\mathcal{O} \atop \longrightarrow} \atop {\longleftarrow \atop \mathrm{Spec}}} \mathrm{Ring}^{\mathrm{op}}$. Unit and counit are just evaluation. It restricts to an antiequivalence of categories between affine schemes and rings. -2) Functional analysis: There is an adjunction $\mathrm{Top} {{\mathcal{O} \atop \longrightarrow} \atop {\longleftarrow \atop \mathrm{Spec}}} {C^*\mathrm{Alg}_1}^{\mathrm{op}}$. Unit and counit are again just evaluation. It restricts to an antiequivalence of categories between compact Hausdorff spaces and commutative unital $C^*$-algebras. -3) Pointless topology: There is an adjunction $\mathrm{Top} {{\mathcal{\Omega} \atop \longrightarrow} \atop {\longleftarrow \atop \mathrm{Spec}}} \mathrm{Frm}^{\mathrm{op}} = \mathrm{Loc}$, where $\Omega$ associates to a topological space the frame of its open subsets, and $\mathrm{Spec}$ associates to every locale the space of principal prime ideals. It restricts to an equivalence between sober spaces and spatial locales and is related to Stone duality. This is very, very similar to 2), we just replace $\mathbb{C}$ with the partial order $2$. -Question. Are these adjunctions really special cases of Isbell duality? If not, how are they related? Is there any more general pattern? -I'm also interested in the case of $(2,1)$-categories $C$. Here $\widehat{C}$ should be the category of pseudo-functors $C^{\mathrm{op}} \to \mathrm{Gpd}$. In this setting there is an adjunction similar to 1) between stacks and cocomplete tensor categories, where $\mathcal{O} = \mathrm{Qcoh}$ associates to every stack $X$ the category of quasi-coherent modules, which may be imagined as categorified global functions on $X$. The fixed points are the tensorial stacks which I study currently. Where does this adjunction arise in the literature? It is similar to an adjunction from derived algebraic geometry (Ben-Zvi, Nadler, Prop. 3.1). -There are related conversations between Jim Dolan and Todd Trimble, which already answer my questions partially. - -REPLY [8 votes]: This seems to be an informal answer: Whenever we have an algebraic object $\mathbb{A}$ of type $A$ in a geometric category $G$, we get an adjunction: $\mathrm{O} := \mathrm{Hom}(-,\mathbb{A}) : G \to A^{op}$ is left adjoint to $\mathrm{Spec} := \mathrm{Hom}(-,\mathbb{A}): A^{op} \to G$ where the latter hom-set is endowed with some kind of Zariski topology. The four adjunctions from above are special cases: -1) Consider the ring object $\mathbb{A}^1 := \mathrm{Spec}(\mathbb{Z}[x])$ in the category of schemes. -2) Consider the $\mathbb{C}$-algebra object $\mathbb{C}$ in the category of locally compact Hausdorff spaces. -3) Consider the Sierpinski space $\{0,1\}$ as a frame. -4) Consider the cocomplete tensor category of quasi-coherent modules as a stack.<|endoftext|> -TITLE: Can $L^{2}$ be represented as a space of functions (not equivalence classes)? -QUESTION [25 upvotes]: Let $X$ be the vector space of all Lebesgue-measurable functions $f:\left[a,b\right]\rightarrowℝ$ such that $\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx<\infty$ (Lebesgue integral). Then we can define an equivalence relation on $X$ as follows: $f \cong g$ if $f(x)=g(x)$ almost everywhere on $\left[a,b\right]$. Then we construct equivalence classes $\tilde{f}=\{g\in X:f\cong g\}$, and the vector space of these equivalence classes is $L^{2}[a,b]$, on which we define the norm $||\tilde{f}||_{1}=\sqrt{\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx}$ (Lebesgue integral). Now some of these equivalences classes are rather special: they contain a continuous function in them, so this is the natural choice for a representative of the equivalence class. Let -$D\subseteq L^{2}[a,b]$ be the subspace containing these special equivalence classes. My basic question is, if we assign the equivalence classes in $D$ their continuous representatives, what are the natural representatives of the other equivalence classes? -We can make this more precise. Let $C[a,b]$ be the vector space of continuous functions $f:\left[a,b\right]\rightarrowℝ$, endowed with a norm $||f||_{2}=\sqrt{\int^{b}_{a}\left|f\left(x\right)\right|^{2}dx}$ (Riemann or Lebesgue integral). Then the norm-completion of this space is in fact $L^{2}[a,b]$. The upshot of all this is that $D$ is dense in $L^{2}[a,b]$, and we have a norm-respecting isomorphism $T:(D,||\cdot||_{1})\rightarrow(C[a,b] , ||\cdot||_{2})$ defined by $T(\tilde{f})\in \tilde{f}$ (assigning each element of $D$ its continuous representative). So now the question becomes, does there exist a continuous linear extension $S$ of $T$ defined on all of $L^{2}[a,b]$ such that $S|_{D}=T$ and $S(\tilde{f})\in \tilde{f}$ ? Well, $T$ is a bounded linear transformation (with operator norm 1) defined on a dense subspace, so it meets all the conditions of the BLT theorem other than the fact that its codomain is not a Banach space. Thus we have to expand $C[a,b]$ to a larger subspace of $X$, so that the codomain of $T$ becomes complete. -There are two potential ways to do this, depending on whether we define the norm $||\cdot||_{2}$ in terms of Riemann or Lebesgue integrals. If we use Riemann integrals, we would need a subspace of $X$ consisting of Riemann-integrable functions, so we would have to answer the following in order to establish completeness: if $f_{n}\rightarrow f$ with respect to the the $||\cdot||_{2}$ (where $f$ need not be continuous), is $f$ necessarily Riemann integrable? (My first instinct is no, because Riemann-integrability requires boundedness, and you can have a sequence of continuous functions with ever-increasing bounds, so that the limit is unbounded). If we use Lebesgue integrals, we would need to ensure that two distinct elements of the subspace cannot have zero distance, so we would have to answer the following: if $f_{n}\rightarrow f$ and $g_{n}\rightarrow g$ with respect to the $||\cdot||_{2}$ norm (where $f$ and $g$ need not be continuous) and $f(x)=g(x)$ almost everywhere on $[a,b]$, then are $f$ and $g$ necessarily the same function? (Again I fear the answer is no, because perhaps you can have a sequence of continuous functions that converges to a function with a removable discontinuity). -I know I've included a lot of convoluted detail, but my fundamental question is relatively simple: can we replace the equivalence classes in $L^{2}[a,b]$ with natural representative functions, using continuous representatives where possible? Or to put it another way: does there exist a subspace $Y$ of $X$ containing $C[a,b]$, on which we can define a norm which will make it isomorphic to $L^{2}[a,b]$? -EDIT: As Gerald has pointed out, a simpler way to phrase my question is that I want a lifting of $L^{2}[a,b]$ or more generally $L^{2}(ℝ^{3})$. -Any help would be greatly appreciated. -Thank You in Advance. - -REPLY [4 votes]: Well, there is a general sense in which your question can be answered in the affirmative. X = L^2 is a Banach space, and every Banach space X can be represented linearly and isometrically as a subspace of the continuous functions on a compact Hausdorff space K. The points of K are the continuous linear functionals on X. You deal with point functopns, not equivalence classes, but you have greatly extended the space of points.<|endoftext|> -TITLE: Does the equation $1 + 2 + 3 + \dots = -\frac{1}{12}$ have a natural $p$-adic interpretation? -QUESTION [31 upvotes]: Consider the equation -$$1 + 2 + 3 + 4 + \cdots = - \frac{1}{12},$$ -"proved" by Ramanujan Euler. One correct way to interpret this is that $\zeta(-1) = - \frac{1}{12},$ where $\zeta(s) = \sum_{n = 1}^{\infty} n^{-s}$ for $\Re(s) > 1$, and $\zeta(s)$ is defined by analytic continuation elsewhere. -I seem to remember being told once that this equation was true in the $p$-adic integers. However, on a moment's reflection this is clearly false; the infinite series does not converge in any $\mathbb{Q}_p$. (I must be misremembering what I was told.) -Is there any argument that an amended version of Euler's statement is true $p$-adically, which does not imitate the usual arguments for $\mathbb{R}$? Is it "obvious" that the denominator should only be divisible by the primes 2 and 3? - -REPLY [21 votes]: For the primes in the denominator, there is an amusing heuristic based on the fact that $n \equiv n^{-1}\pmod p$ holds for all $n\geq 1$ (coprime to $p$) only for $p=2$ and $p=3$. So for these primes the series is like the series $1+1/2+1/3+\cdots$ for $\zeta(1)$, which has a true pole... - -REPLY [12 votes]: First, it seems that this equation was actually first ``proved'' by Euler. In the preface to the book "Elementary theory of $L$-functions and Eisenstein series" by Haruzo Hida, the author gives a beautiful exposition of Euler's manipulations leading to this formula. -The connection to $p$-adic zeta functions seems to be via the Kummer congruences: looking at Euler's formulae, Kummer was apparently lead to the congruences that bear his name, and an appropriate interpretation of the latter by Kubota and Leopoldt half century ago gave rise to the first construction of the $p$-adic analogue of the Riemann zeta function -- the so-called Kubota-Leopoldt $p$-adic zeta function. -Regarding your questions: for the first, I can only say that the zeta values in the Kummer congruences need to be amended to give rise to a continuous function of a $p$-adic variable $s$ (essentially, one needs to remove the corresponding values of the Euler factor $1-p^{-s}$); for the second, I do not see any a priori ``obvious'' reason why $2$ and $3$ should be the only primes in the denominator in Euler's formula. -The first four pages of Pierre Colmez's notes http://www.math.jussieu.fr/~colmez/Kubota-Leopodt.pdf are an excellent reference for the mathematical facts referred to in the last two paragraphs.<|endoftext|> -TITLE: $\mathcal{I}$-functors and infinite loop spaces -QUESTION [12 upvotes]: By the Barratt-Priddy-Quillen theorem, the space $B \Sigma_\infty^+$ is the infinite loop space $\Omega^\infty \Sigma^\infty S^0$. I'm curious about a "high-concept" reason that $B \Sigma_\infty^+$ (and "more generally" $BGL_\infty^+(R)$ for a ring $R$) should be infinite loop spaces (well, almost that). For instance, I read about a theorem as follows: -Define an $\mathcal{I}$-functor $F$ as the following data: $F$ is a functor from the category of finite-dimensional inner product spaces over $\mathbb{R}$ and isometric linear imbeddings to the category of pointed spaces and closed imbeddings, and there are functorial maps $F(V) \wedge F(W) \to F(V \oplus W)$ for each $V, W$ satisfying associativity and commutativity relations. Then it is a theorem that the connected component $F(\mathbb{R}^\infty) = \varinjlim F(\mathbb{R}^n)$ is an infinite loop space. This is because the functorial maps given above lead to an action on $F(\mathbb{R}^\infty)$ of the linear isometries operad, which is an $E_\infty$-operad, and one can invoke May's recognition principle. This provides additional explanation why the infinite orthogonal group is an infinite loop space. -The data above is sort of analogous to the theory of orthogonal spectra. Namely, it states that $F$ is a commutative algebra object in a certain monoidal functor category from inner product spaces to pointed topological spaces; the functor $F(V) :=S ^V$ is another one and orthogonal spectra are precisely modules over this functor. -I'm curious whether there is an analog for symmetric spectra. Namely, let's say we have a functor $G$ from finite sets to pointed spaces, together with maps $G(A) \wedge G(B) \to G(A \sqcup B)$ satisfying associativity and commutativity conditions. I am curious whether the connected component of $G(\mathbb{N}) = \bigcup G([n])$ is an infinite loop space. -By similar reasoning, $G(\mathbb{N})$ admits an action of the "injection operad," which is however only a set-valued operad. So if this is true, perhaps Segal's delooping machinery might be relevant (from his paper "Categories and cohomology theories"). Can it be proved that $G(\mathbb{N})$ is a $\Gamma$-space, in his terminology? - -REPLY [9 votes]: You are looking at the telescope of maps $BG_n\longrightarrow BG_{n+1}$ where the coproduct -of the $G_n$ (thought of as categories) has a structure of permutative category. The group -completion property of infinite loop space machines defined on permutative categories shows -easily that there is a canonical map from $BG_{\infty}$ to the component of the unit (for the permutative structure) in the zeroth space of the associated spectrum that induces an -isomorphism on homology (with any ring of coefficients). Passing to the plus construction, -it becomes an equivalence. -In the kind of examples you refer to with the linear isometries operad, the input looks -quite a lot different. The fact that you get equivalent spectra when both constructions apply is non-trivial. See my paper -The spectra associated to $\mathcal{I}$-monoids. -Math. Proc. Camb. Phil. Soc. 84(1978), 313--322.<|endoftext|> -TITLE: The Classifying Space of the Discrete Heisenberg Group -QUESTION [5 upvotes]: What is the Classifying Space of the Discrete Heisenberg Group? Which paper/book contains a detailed proof? -Thank you for your time. - -REPLY [11 votes]: As was said by Andy, the classifying space of the discrete Heisenberg group $\Gamma$ is $B\Gamma=G/\Gamma$, where $G$ is the 3-dimensional Heisenberg group over the reals. Due to the central extension -$$0\rightarrow\mathbb{Z}\rightarrow\Gamma\rightarrow\mathbb{Z}^2\rightarrow 0$$ -you may view $B\Gamma$ as a circle bundle over the 2-torus. Alternatively, viewing $\Gamma$ as the semi-direct product -$\Gamma=\mathbb{Z}^2\rtimes\mathbb{Z}$, where $\mathbb{Z}$ acts by (powers of) -$\left(\begin{array}{cc} -1 & 1 \\ -0 & 1 -\end{array}\right)$, you can view $B\Gamma$ as the mapping torus of this automorphism of $B\mathbb{Z}^2=\mathbb{T}^2$, i.e. as a 2-torus bundle over the circle.<|endoftext|> -TITLE: algebraization theorems -QUESTION [19 upvotes]: One of the fundamental properties that distinguishes schemes among all contravariant functors $\mathrm{Sch}^\circ \rightarrow \mathrm{Sets}$ is algebraization: a functor $F$ satisfies algebraization if, whenever $S$ is the spectrum of a complete noetherian local ring and $S_n$ are the infinitesimal neighborhoods of the central point in $S$, -$F(S) = \varprojlim_n F(S_n)$. -I only know of two basic algebraization results: (1) Grothendieck's existence theorem gives algebraization when $F$ is the stack of coherent sheaves on a proper scheme, and (2) SGA3.IX.7.1 gives algebraization for maps from tori into affine group schemes. -It is possible to deduce algebraization for many other functors from these. My question is: are there any other basic algebraization results (that don't eventually reduce to one of these) out there? - -REPLY [5 votes]: Bhargav Bhatt has recently proved a remarkable algebraization theorem: -http://arxiv.org/abs/1404.7483 -It implies that formal maps into quasi-compact, quasi-separated schemes may be algebraized. Another version of this result (with slightly different hypotheses) has been proved by Hall and Rydh: -http://arxiv.org/abs/1405.7680<|endoftext|> -TITLE: Best reading on "writing on dirty paper" ? -QUESTION [6 upvotes]: Can one suggest good introduction(s) to "dirty paper coding" ? (Other comments, suggestions etc. are highly welcome). - -Some background. -http://en.wikipedia.org/wiki/Dirty_paper_coding -"Writing on dirty paper" - is the name of the quite famous paper in information theory: -M. Costa (May 1983). "Writing on dirty paper". IEEE Trans. Information Theory 29 (3): 439–441. doi:10.1109/TIT.1983.1056659. -The analogy (which is not 100% correct, but inspiring) - assume that you given a sheet of paper with many dirty places - you need to write a message to another person, such that he will be able to understand it. -The main surprise of the theorem is the following - you can do it the same successfully as if the paper would be clean. -Currently this topic is interesting for industry since "dirty paper" is interference coming from the other users of wireless networks. - -REPLY [2 votes]: What about - -D. Tse, Fundamental of Wireless Communication, Cambridge University Press, 2005, Chap. 10, available free here; -C.B. Peel, On "Dirty Paper Coding", Signal Proc. Mag., May, 2003, avaliable here; - -Also, if your Institution have access I suggest IEEE Xplore -ADDED: - -J. Liu, N. Elia, Writing on Dirty Paper with Feedback, Comm. Inform. Systems, International Press, Vol. 5, No. 4, pp. 401-422, 2005<|endoftext|> -TITLE: Are the Drinfeld compact quantum groups simply connected ? -QUESTION [8 upvotes]: To fix notations : let G be simply connected simple compact group, and $U_q(\mathcal{G})$ the Drinfeld-Jimbo universal algebra quantization of its complexified algebra defined as usual, with q not root of unity. Recall that it's a *-algebra, and I denote $G_q$ the unital $C^*$-algebra dual to it as usual (thus completion of matrix coefficient of representations...). -My question is very naive and loose, so all suggestions are welcome. Is there a general notion of fundamental group (e.g. using like étale), such that $G_q$ has trivial fundamental group? - -REPLY [2 votes]: This is a negative answer to your question, or at least a partially negative one. I don't know if you've thought about things in this way, but there is a naive way to formulate the idea of the fundamental group of a space in terms of its $C^*$-algebra of continuous functions. The problem is that, at least according to my quick calculations, the noncommutative space associated to the quantized function algebra $\mathcal{O}_q(SU_2)$ does not have trivial fundamental group. This may mean that this notion of fundamental group doesn't make much sense for noncommutative spaces, or that quantum phenomena arise which just make these quantum groups not simply connected, or something else. My suspicion is that the first is true, my reasoning being that if there was a good notion of fundamental group for noncommutative spaces, we would have heard of it already. Perhaps somebody can chime in on that? -Anyway, let me tell you what I've been thinking about. The fundamental group of a space $X$ is defined as homotopy classes of based continuous maps $S^1 \to X$. Continuous maps $S^1 \to X$ are in bijection with unital $*$-homomorphisms $C(X) \to C(S^1)$, at least for $X$ compact. So for a compact noncommutative space with corresponding unital $C^*$-algebra $A$, loops should be defined as morphisms (of unital $C^*$-algebras) $A \to C(S^1)$. -I believe that the correct notion of homotopy of morphisms is the following: two morphisms $\Phi_0, \Phi_1 : A \to C(S^1)$ are homotopic if there is a morphism $\Psi : A \to C([0,1]) \otimes C(S^1)$ such that $\mathrm{ev}_0 \circ \Psi = \Phi_0$ and $\mathrm{ev}_1 \circ \Psi = \Phi_1$. This is spelled out in Chapter 1 of Elements of Noncommutative Geometry, by Varilly, Gracia-Bondia, and Figuera. -To say that a loop takes basepoint to basepoint is a little trickier for noncommutative spaces. However, for the quantum groups this is ok because there is still one classical point, namely the identity element, which corresponds to the counit $\varepsilon : \mathcal{O}_q(G) \to \mathbb{C}$. For a loop/morphism $\Phi : \mathcal{O}_q(G) \to C(S^1)$, then, we can say that $\Phi$ preserves basepoints if $\Phi(x)(1) = \varepsilon(x)$ for all $x \in \mathcal{O}_q(G)$. Or in more fancy terms we could say that $\mathrm{ev}_1 \circ \Phi = \varepsilon$. -Finally, we have to define the correct analogue of the constant loop. Again this may be difficult in general, but for the quantum groups we have the counit, so the constant loop is the morphism $x \mapsto \varepsilon(x)1$ of $\mathcal{O}_q(G)$ to $C(S^1)$. -So this gives us most of the elements to define the fundamental group of a noncommutative space with associated $C^*$-algebra $A$: we take based morphisms $A \to C(S^1)$ modulo homotopy, and say that a loop is trivial if it is homotopic (through based morphisms) to the constant loop. I haven't said how you're supposed to compose loops, but this at least defines the fundamental group as a set. - -Now the bad news: for $\mathcal{O}_q(SU_2)$, this doesn't give the right answer. For the sake of completeness, $\mathcal{O}_q(SU_2)$ (for $0 < q < 1$, say) is the unital $C^*$-algebra generated by elements $a,b$ with relations -$$ a^*a + b^*b =1, \quad aa^* + q^2 bb^* = 1, $$ -$$ bb^* = b^*b, \quad ab = qba, \quad ab^* = qb^*a. $$ -The first two relations show that if $\Phi : \mathcal{O}_q(SU_2) \to A$ is a morphism to a commutative $C^*$-algebra $A$, then $(1-q^2)\Phi(b^*b) = 0$, so $\Phi(b) = 0$. (Thus we really need $q \neq \pm 1$ for this to work.) -Now $C(S^1)$ is the universal $C^*$-algebra generated by a single unitary element $u$. This is a quantum subgroup of $\mathcal{O}_q(SU_2)$ via the map $\pi : \mathcal{O}_q(SU_2) \to C(S^1)$ defined by -$$ \pi(a) = u, \quad \pi(b) = 0. $$ -Combining this with the preceding statement, any morphism $\Phi : \mathcal{O}_q(SU_2) \to C(S^1)$ must factor through the quotient map $\pi$, so homotopy classes of morphisms $\mathcal{O}_q(SU_2) \to C(S^1)$ are in bijection with homotopy classes of morphisms $C(S^1) \to C(S^1)$. These clases correspond to homotopy classes of maps $S^1 \to S^1$, so this means that the fundamental group is $\mathbb{Z}$, rather than the trivial group. -Again, my guess is that this notion of fundamental group is just too rigid to make sense for all noncommutative spaces. One needs at least one classical point in order for this to make sense, which excludes any simple algebras, such as noncommutative tori. The question is interesting, though, and I would be glad to know if anybody else has thought about other notions of a fundamental group for noncommutative spaces.<|endoftext|> -TITLE: Is the moduli space of curves defined over the field with one element? -QUESTION [32 upvotes]: There are various frameworks around which enlarge the category of rings to include more exotic objects such as the 'field with one element,' $\mathbb{F}_1$. While these frameworks differ in their details, there are certain things this should be true of any object that deserves to be called $\mathbb{F}_1$. For example, The algebraic K-theory of $\mathbb{F}_1$ should be sphere spectrum, and the theory of toric varieties should be defined over $\mathbb{F}_1$. -Question 1: Is there a moral reason why the moduli space of curves should (or should not) be defined over Spec $\mathbb{F}_1$? -EDIT: For anyone who would like to be more concrete, I'm happy to take the Toen-Vaquie definition of schemes over $\mathbb{F}_1$. (see arXiv:math/0509684). In this setup (and most of the other frameworks I know) an affine scheme over $\mathbb{F}_1$ is just a commutative monoid $M$. After base change to $\mathbb{Z}$ this becomes the monoid ring $\mathbb{Z}[M]$. So here is a more precise question: -Question 2: Does the moduli space of curves $\mathcal{M}_{g,n}$ (over $\mathbb{Z}$, say) admit a covering by affine charts of the form spec $\mathbb{Z}[M_i]$ for commutative monoids $M_i$? If so, can this covering be chosen so that (as in the case of toric varieties) the gluing is entirely determined by maps of monoids? - -REPLY [18 votes]: Here a few remarks from the $\Lambda$ point of view on $\mathbf{F}_1$. (There we say a scheme is defined over $\mathbf{F}_1$ if it admits a $\Lambda$-structure. If the scheme is flat over $\mathbf{Z}$, a $\Lambda$-structure is equivalent to a commuting family of endomorphisms $\psi_p$, one for each prime number $p$, such that each $\psi_p$ agrees with the Frobenius map modulo $p$.) -I prove in http://arxiv.org/abs/0906.3146 that if a scheme of finite type over $\mathbf{Z}$ is defined over $\mathbf{F}_1$, then its motive is pure Tate (or rather becomes so after base change to some cyclotomic field - it's actually false in general without this qualification). So as in Dan Petersen's answer, $M_{g,n}$ won't admit a $\Lambda$-structure unless the pair $(g,n)$ is sufficiently small. -I haven't really thought about the small cases, except just a bit when $(g,n)=(1,1)$. Then $M_{g,n}$ is just the affine line, so it has many $\Lambda$-structures. As far as I know, you can't really say that they have any meaning, but I suspect there is still something interesting to say. -1) There is a $\Lambda$-structure on the completion of $\bar{M}_{1,1}$ at the point at infinity which does have a meaning. The completion is $\mathbf{Z}[[q-1]]$, and the $\Lambda$-structure is defined by $\psi_p(q)=q^p$, for all $p$. The interest in this is that this $\Lambda$-structure prolongs to a $\Lambda$-structure on the universal generalized elliptic curve over $\mathbf{Z}[[q-1]]$ (i.e. the Tate curve). From what I remember, the idea is that a point $z\in\mathbf{C}^*/q^{\mathbf{Z}}$ should be mapped to its image $z^p\in\mathbf{C}^*/q^{p\mathbf{Z}}$. Making this rigorous is not hard, though of course you have to have an actual definition of the Tate curve over $\mathbf{Z}[[q-1]]$, a point that is often glided over. -What I think could be interesting is to study the extent to which these $\Lambda$-structures extend to the whole moduli space. Here is one possible interpretation. A $\Lambda$-structure on a scheme $X$ is equivalent to a section of the canonical projection $W_*(X)\to X$ satisfying a certain associativity property. ($W_*$ is the right adjoint to the Witt vector functor $W^*$. It is an arithmetic analogue of a jet space.) Now consider the map $T\to X$, where $T$ is the Tate curve above and $X$ is the total space of the universal elliptic curve. (This is a stack. One would have to check that all these $\Lambda$ and $W$ concepts make sense for stacks.) Then the projection $W_*(X)\to X$ has a natural section over $T$. Then we can ask: what is the Zariski closure $Z$ of this section in $W_*(X)$? The extent to which $Z$ fails to be a section of $W_*(X)\to X$ should be some measure of the extent to which the $\Lambda$-structure on $T$ fails to extend to one on $X$. -2) One could try to do something similar at CM points, instead of at the boundary of the moduli space. CM elliptic curves admit a certain generalized version of a $\Lambda$-structure. (See the end of my paper cited above.) Do these generalized $\Lambda$-structures extend to the formal neighborhood of the elliptic curve in the universal elliptic curve? If so, one could look at closures of formal sections (as above) to see the extent to which they fail to extend to generalized $\Lambda$-structures on the entire universal elliptic curve. For all this, you'd have to fix an imaginary quadratic field, so all this is probably less fundamental than in 1). -3) Thomas Scanlon tried to convince me a few years ago that the Hecke correspondences have some kind of $\Lambda$ nature. From what I remember, there were certain $\Lambda$-schemes closely related to modular varieties which are not of finite type but which are still finite-dimensional in some model-theoretic sense. I never got around to understanding what he meant (though I always intended to). Perhaps it is related to what I wrote above. -So there are a few pieces of evidence that something is going on with $\Lambda$-structures and modular curves, though it remains to be seen if it's really an identifiable phenomenon and, if so, how interesting it is.<|endoftext|> -TITLE: How do various notions of natural transformation relate to various notions of homotopy in $2Cat$? -QUESTION [8 upvotes]: In what follows, $2$-categories will be strict, and "$2$-functor" will mean "strict $2$-functor". (Please mention which terminological conventions you are using when answering.) I guess that the answer to my question is rather standard for people working daily with $2Cat$ as a $2$- or $3$-category, and I think I may know some people who could answer my question if I ask them privately. However, it seems to me it could be useful to have an answer to that question on MO for the record (and so as to allow those knowledgeable people to earn much sought-after MO reputation and badges). -It is standard to view natural transformations between functors as categorical homotopies. There has been a question about this viewpoint on MO before. What I would like to know is the relationship between higher natural transformations and homotopies. -The starting point is that, given two categories $A$ and $B$ and two functors $F$ and $G$ from $A$ to $B$, a natural transformation from $F$ to $G$ is the same thing as a functor $\Delta_{1} \times A \to B$ which makes the obvious diagram commutative, where $\Delta_{1}$ is the category associated to the naturally ordered set $\{0,1\}$. -Now, if we replace $Cat$ by $2Cat$, we have a whole bunch of possible variants. For instance, if $\mathcal{A}$ and $\mathcal{B}$ are two $2$-categories and $F$ and $G$ are two $2$-functors from $\mathcal{A}$ to $\mathcal{B}$, then a lax natural transformation from $F$ to $G$ gives a lax $2$-functor $\Delta_{1} \times \mathcal{A} \to \mathcal{B}$ which makes the obvious diagram commutative. But we could ask whether a lax transformation between lax functors gives such a lax functor too, and whether these two notions are equivalent in this setting, as in the case of $Cat$ (I have just stated one implication only, and just for strict $2$-functors). The question which arise could therefore phrased as: - -What are the objects in $2Cat$ analogous to $\Delta_{1} \times A$ in the case of $Cat$, with respect to strict transformations between strict $2$-functors, lax transformations between strict $2$-functors, lax transformations between lax $2$-functors? - -I guess this is related to the Gray tensor product, but Gray's style is, woe is me, undecipherable to me. The related $nLab$ page seems to help but not to answer exactly this question (I may well be mistaken). -A related question is the one relating the $2$-categorical viewpoint and the $1$-categorical viewpoint: - -What kind of transformations in $2Cat$ gives what kind of homotopies in $2Cat$? (Here, I primarily think to "homotopy" as an arrow from the product of $\Delta_{1}$ with something, but I do not want to restrict to this case if it turns out this is not the right generalization.) - -This last question does not ask for the universal objects, but rather how various maps (the "various" refering to the degree of laxness) with various universal objects as domain relate to various (same remark) notions of transformations. I have routinely worked with $2Cat$ from the $1$-categorical viewpoint, but much less from the $2$-categorical viewpoint (let alone the dreaded $3$-categorical viewpoint). -If some people feel comfortable with the general setting of $nCat$, I do not have any objection, although I would mind if it should obfuscate what happens in the special case of $2Cat$. -Many thanks in advance! - -REPLY [7 votes]: For strict transformations between strict 2-functors, you can just use the cartesian product $\Delta_1\times A$. -For lax transformations between strict 2-functors, this is what the lax version of the Gray tensor product does. The nLab page is mostly about the pseudo version (which corresponds to pseudo natural transformations), but if you replace "pseudo" with "lax" then the defining isosmorphism -$$2Cat(B\otimes_l C, D) \cong 2Cat(B, Lax(C,D))$$ -gives you what you want if you take $B=\Delta_1$, where $Lax(C,D)$ denotes the 2-category of strict 2-functors and lax natural transformations. -For lax transformations between lax 2-functors, we can invoke the fact that for any 2-category $C$, there is a 2-category $Q_l C$ such that strict 2-functors $Q_l C \to D$ are in bijection with lax 2-functors $C \to D$. Then $\Delta_1\otimes_l Q_l C$ will have the property that strict 2-functors out of it are in bijection with lax transformations between pairs of lax 2-functors out of $C$. I doubt there is a way to get this to work if you want to look at lax functors out of the cylinder object, though.<|endoftext|> -TITLE: How is the Julia set of $fg$ related to the Julia set of $gf$? -QUESTION [32 upvotes]: Let $f$ and $g$ be complex rational functions (of degree $\geq 2$ if that helps). What can be said about the relationship between $J(fg)$ and $J(gf)$, the Julia sets of the composite functions $f \circ g$ and $g \circ f$? -If I'm not mistaken, $f$ restricts to a map $J(gf) \to J(fg)$, and $g$ restricts to a map $J(fg) \to J(gf)$. So $J(fg)$ and $J(gf)$ surject onto each other in a particular way (and, indeed, in a way that commutes with the actions of $fg$ on $J(fg)$ and $gf$ on $J(gf)$). Since Julia sets are completely invariant, the restricted map $f: J(gf) \to J(fg)$ is $deg(f)$-to-one, and similarly the other way round. -So there's some kind of relationship between the two sets. -If $f$ or $g$ has degree one then $J(fg)$ and $J(gf)$ are "isomorphic", in the sense that there's a Möbius transformation carrying one onto the other. Thus, the simplest nontrivial example would be to take $f$ and $g$ to be of degree 2. I don't know a way of computing, say, the example $f(z) = z^2$ and $g(z) = z^2 + 1$. That would mean computing the Julia sets of $gf(z) = z^4 + 1$ and $fg(z) = z^4 + 2z^2 + 1$. -My question isn't completely precise, I'm afraid. But here are some of the things that I would value in an answer: theorems on what $J(fg)$ and $J(gf)$ have in common, examples showing how different they can be, pictures of $J(fg)$ and $J(gf)$ for particular functions $f$ and $g$, and references to where I can find out more (especially those accessible to a non-specialist). Thanks. - -REPLY [13 votes]: Answer. $J(fg)$ is the full $g$-preimage of $J(gf)$. (And vice versa with interchange of -$f$ and $g$). -Proof. Let $A=fg$ and $B=gf$. Then we have a semi-conjugacy $gA=Bg$. -Now it is a general fact, that whenever you have such a semi-conjugacy (of rational functions) -the Julia set of $A$ is the $g$-preimage of the Julia set of $B$. -Proof. The semi-conjugacy can be iterated: $gA^n=B^ng$. -Now, $z\in J(A)$, iff the family $gA^n$ is not normal, iff $B^ng$ is not normal, that is -$z\in g^{-1}(J(B))$. -Added on 8.6.12. By the way, this demonstrates an amazing fact: for every $f$ and $g$, -there exist sets $F$ and $G$ such that $G=f^{-1}(F)$ and $F=g^{-1}(G)$. -This looks surprising to me. Finite sets $F$ and $G$ of cardinality greater than $2$ -with such properties cannot exist, as a simple count shows. -Are there other examples of such $F$ and $G$ ? -Added on 8.7.12. Let $f$ and $g$ be two rational functions. Let $F$ be a closed set, containing -more than 2 points, and such that $(g^{-1}f^{-1}(F))=F$, then $F$ contains the Julia set of $fg$. -And $J(fg)$ does have this property. -(I am writing compositions $fg=f(g)$.) Trivial, but funny.<|endoftext|> -TITLE: Reduction rules for inductive types -QUESTION [8 upvotes]: (I'm not sure if I should post this here rather than at Theoretical Computer Science, I've found a lot of type theory related questions on MathOverflow) -I'm working in Martin-Löf type theory with inductive types. Everything I say below for booleans should be understood with the type of booleans replaced by any inductive type, but for simplicity I'll do it in the case of boolean with match with written as if then else. -My question is about two reduction rules that I've never seen studied anywhere but that seems rather natural to me. I'd like to know if those rules have an "official" name and where I could read about them. -The first one (that I'm calling "lazy match") has the form - -b : bool $\vdash$ if b then u else u $\rightarrow$ u : A - -where A is a type, b : bool $\vdash$ u : A and "$\rightarrow$" is reduction (definitional equality, if you prefer). -I know there is a problem with this rule if b does not terminate, but I'm interested here in type theory as a logical system, so everything is supposed to terminate. -The second one (that I'm calling "deep match") is about exchanging two match and has the form - -b : bool $\vdash$ if (if b then s else t) then u else v $\rightarrow$ if b then (if s then u else v) else (if t then u else v) : A - -where A is a type, b : bool $\vdash$ s, t : bool and b : bool $\vdash$ u, v : A -Intuitively, when you match a match expression, the outer match can be distributed in the branches of the inner match. -Where can I read about these rules? Or are there obvious problems that I haven't seen? -(there are perhaps problems with inductive predicates (as opposed to inductive types), but you can restrict it to inductive types) - -REPLY [4 votes]: I think both rules are an instance of a more general relation: -b : bool ⊢ f (if b then x else y) ≡ if b then f x else f y : A - -Your first rule is with f := const u, but going backwards. The second rule is with -f a := if a then u else v - -I've used ≡ instead of an → since I'm not sure which is side more reduced. -I vaguely recall this being called either χ-conversion or ξ-conversion, but I forget. Perhaps someone else can help more.<|endoftext|> -TITLE: irreducible polynomial with repect to prime number -QUESTION [8 upvotes]: Let $n$ and $n$ are positive integers, $b>1$. -Express $n$ in $b$-basis -$$n = a_kb^k + \cdots + a_1b + a_0.$$ -We consider the polynomial -$$f_{b,n}(X) = a_kX^k + \cdots + a_1X + a_0 \in \mathbb{Z}[X].$$ -Question 1: Let $p$ is a prime number. Then, is it true that $f_{2,p}(X)$ is an irreducible polynomial? -I have checked this question for all $p < 10^6$, and some cases. -Question 2: Is it true that: p is prime number iff $f_{b, p}(X)$ is an irreducible polynomial for all $b>1$? -It is clear that if $n$ is a composite number and $b$ is the least prime divisor of $n$, then $f_{b, n}(X)$ is reducible. - -REPLY [12 votes]: The answer is yes to both questions: it is a theorem of Brillhart, Filaseta, and Odlyzko -(see corollary 2, p.1058): -http://cms.math.ca/10.4153/CJM-1981-080-0<|endoftext|> -TITLE: Topological examples of profinite groups -QUESTION [9 upvotes]: I am preparing a course on profinite groups, to be delievered to early graduate students. The first part of the course will discuss the equivalent characterizations of profinite groups. I will first define a profinite group as a Hausdorff, compact topological group such that the open subgroups form a base for the neighbourhoods of the identity. -I would like to give my students a few ‘natural’ examples. I am looking for examples of profinite groups that give the students something they can wrap their heads around without knowing about products of finite groups and inverse limits. Ideally, I am looking for examples which have more of a topological emphasis. -As an example of what I am looking for: The $p$-adic integers $\mathbb{Z}_p$ can be constructed as the completion of $\mathbb{Z}$ with the metric space structure induced from the $p$-adic absolute value. -I would particularly like to see an example of a group acting on some object which induces some topology on the group. I would like to exclude Galois groups: these will be covered in another part of the course. I would also like to exclude the fundamental groups encountered in algebraic geometry: I do not expect my students be familiar with this material. - -REPLY [3 votes]: One idea would be to introduce the profinite completion. If the students have already met the connection of the fundamental group with covering spaces, why not look at finite covering spaces. (This is essentially doing the SGA1 algebraic fundamental group idea without mentioning the links with alg. geom nor with Galois theory.) You are looking at a group $G$ and the category of finite $G$-sets. As all the permutation groups of the objects concerned are finite, the group acts via its profinite completion. There is a nice master's thesis by Robalo at the IST in Lisbon which looked at various aspects of this situation and may provide a useful reference document for the students as it is written at only slightly above the level at which they will be working.<|endoftext|> -TITLE: 4-regular graphs with every edge in a triangle -QUESTION [12 upvotes]: I am interested in regular graphs in which every edge lies in a triangle. -For 3-regular graphs, only the complete graph $K_4$ has this property, so there's not much to see here. -For 4-regular graphs, there are more graphs, including some infinite families, but few enough and slowly-growing enough that I have some wild hope of a characterisation. -The numbers on 5, 6, ..., 17 vertices are -1, 1, 2, 2, 3, 3, 4, 8, 11, 18, 35, 57 and 106, -which is tiny compared to the total number of 4-regular graphs. -For smaller numbers of vertices there are likely to be numbers of graphs that "just happen" to have every edge in a triangle, but at some stage I am hoping they will fall into some infinite families for which one might get at least a qualitative description. -I have searched "the literature" (i.e. tried MathSciNet and Google) and cannot find any results along these lines which surprised me a bit. But perhaps it is just impossible and those numbers just keep growing... -So my question is: -Does anyone know any results or have any ideas pertaining to regular graphs, in particular 4-regular graphs, with every edge lying in a triangle? -EDIT: I think that Florian Pfender (see comment below) may have basically found the solution. All of the graphs that I have computed (on up to 17 vertices) are either: -(a) The Cayley graph Cay$(Z_n; \{ \pm 1, - \pm 2 \})$ ("the square of a cycle") -(b) The linegraph of a cubic graph, or -(c) have an edge $xy$ such that the closed neighbourhood of $x$ is equal to the closed neighbourhood of $y$ -This latter condition means that the two vertices $x$, $y$ have exactly three common neighbours, say $a$, $b$ and $c$. So we can reduce to a smaller graph by deleting $x$ and $y$ and adding a triangle on $a$, $b$, $c$ perhaps at the expense of introducing multiple edges. -Last thing that remains is to show that when this "reduction process" is completed, the resulting graph is just the line-multigraph of a cubic-multigraph, which is what Florian implied. -I am hopeful of a genuine result here. -(Then on to 5-regular graphs....) -FINAL EDIT: August 2013 - This question has now generated a paper containing the genuine result that I was hopeful for. See http://arxiv.org/abs/1308.0081 and https://symomega.wordpress.com/2013/08/02/regular-graphs-triangles-and-mathoverflow/. -Thanks MathOverflow! - -REPLY [7 votes]: yes, it feels like one could proof my characterization along the following lines: - -If the graph does not contain $K_4^-$, then it is a line graph of a 3-regular graph. -Now continue with a copy of $K_4^-$. The two vertices of degree three must each be incident to another edge. If these two edges meet, then we can reduce the whole structure to a triangle like you describe (my $K_{3,1,1}$). -If the two edges do not meet, then they each must lie in a triangle, so there need to be edges from their endpoints to the vertices of degree 2 in the $K_4^-$. If both these edges meet in the same vertex, we again have a structure we can reduce to a triangle (a triangle in the line graph original, which we can contract keeping 3-regularity). -So we are left with the case where the two edges go to different sides of the $K_4^-$, and again we have two vertices of degree 3 to work with. If these two vertices are connected, then we have a $K_4$ with two triangles attached to it which we can get from the multi-line-graph of edge-doubleedge-edge by $K_{3,1,1}$ing one of the triangles (or alternatively, we can reduce the whole graph by contracting the whole structure to one vertex). -If these two vertices are not connected, then each must lie in a triangle with the neighboring vertex of degree 2, and this process will go on until the two sides meet in a cycle, in which case you got your square of a cycle.<|endoftext|> -TITLE: Can the difference of two distinct Fibonacci numbers be a square infinitely often? -QUESTION [32 upvotes]: Can the difference of two distinct Fibonacci numbers be a square infinitely often? - -There are few solutions with indices $<10^{4}$ the largest two being $F_{14}-F_{13}=12^2$ and $F_{13}-F_{11}=12^2$ -Probably this means there are no identities between near neighbours. -Since Fibonacci numbers are the only integral points on some genus 0 curves the problem is equivalent to finding integral points on one of few varieties. Fixing $F_j$ leads to finding integral points on a quartic model of an elliptic curve. - -Are there other solutions besides the small ones? - -[Added later] Here is a link to elliptic curves per François Brunault's comment. -According to DIOPHANTINE EQUATIONS, FIBONACCI HYPERBOLAS,. AND QUADRATIC FORMS. Keith Brandt and John Koelzer. -Fibonacci numbers with consecutive odd indices are the only solutions to -$$ x^2-3xy+y^2 = -1 \qquad (1)$$ -Fibonacci numbers with consecutive even indices are the only solutions to -$$ x^2-3xy+y^2 = 1 \qquad (2)$$ -Given $F_n$ and $F_{n+2}$ one can compute $F_{n+k}$ using the linear Fibonacci recurrence and $F_{n+k}$ will be a linear combination $l(x,y)$ of $x,y$. Adding $l(x,y)-x=z^2$ to (1) or (2) gives a genus 1 curve. (Or just solve $l(x,y)-x=z^2$ and substitute in (1) or (2) to get a genus 1 quartic). -The closed form of $l(x,y)$ might be of interest, can't find the identity at the moment. -Probably a genus 0 curve with integral points $F_{2n},F_{2n+1}$ will be better. -Added much later - -Does some generalization of the $abc$ conjecture predict something? - -For 3 Fibonacci numbers identities are much easier: -$$ F_{4n+1}+F_{4n+3}-F_3 = L_{2n+1}^2$$ - -REPLY [7 votes]: This is a slight elaboration on joro's comment; I was hoping that someone else would write a better version of this. -The integer points on $x^2-xy-y^2 = 1$ are precisely the pairs $(F_{2n+1}, F_{2n})$. So looking for solutions of the form $F_{2n+1} - F_{2m+1} = z^2$ is looking for integer points on $$ x_1^2 - x_1 y_1 - y_1^2 = x_2^2 - x_2 y_2 - y_2^2 = 1,\ x_1 -x_2 = z^2.$$ -The other three possibilities give similar equations. -Each of these is a $K3$ surface. Here is where a better answer would review the major results on integer points on $K3$ surfaces. But I don't know them, so I'm going to stop here and hope someone else fills it in.<|endoftext|> -TITLE: Analogues of the dihedral group -QUESTION [8 upvotes]: A virtually-$\mathbb{Z}$ group $G$ admits either a epimorphism onto $\mathbb{Z}$ or a epimorphism onto $D_\infty$. -So what happens if one replaces $\mathbb{Z}$ by another group $F$ (like the free group or $\mathbb{Z}^n$). My first guess would be that any virtually-$F$ group $G$ maps surjectively onto one of the groups -$F\rtimes H$, where $H\le $Aut$(F)$ is any finite subgroup. -This is clear in the case where $G$ is semidirect product of $F$ and a finite group $K$; one can simply take $H$ to be the image of $K\rightarrow $ Aut $ (F) $. -So my questions are: -1) Is it still true that there is an epimorphism even in the non-split case? -2) Is it also true that two groups $F\rtimes H_1$ and $F\rtimes H_2$ (with $H_i\subset $ Aut $(F)$ finite) cannot surject onto each other unless $H_1$ and $H_2$ are conjugated (in which case the groups are isomorphic) ? -I doubt that this is true for all groups $F$ but maybe one can find sufficient conditions that guarantee this. - -REPLY [4 votes]: Let me add another class of examples: virtual hyperbolic surface groups, by which I mean groups that have some $\pi_1(S_g)$ as a finite index subgroup, where $S_g$ is the closed, oriented surface of genus $g \ge 2$. Each virtual hyperbolic surface group $\Gamma$ surjects, with finite kernel, onto the fundamental group of a closed hyperbolic 2-orbifold (predecessors to Haefliger's orbispaces mentioned above by Agol). And if you fix $g$ then there are only finite many such orbifold groups which are the targets of a virtual $\pi_1(S_g)$ group. -This class of examples comes from quasi-isometric rigidity of the class of hyperbolic surface groups, a theorem of Gabai and of Cannon and Jungreis, which says that any finitely generated group quasi-isometric to the hyperbolic plane (which includes virtual hyperbolic surface groups) has a surjection as described above for $\Gamma$. -One can mine the theory of quasi-isometric rigidity for other classes of examples.<|endoftext|> -TITLE: Polarizations of K3 surfaces over finite fields -QUESTION [7 upvotes]: Suppose that $X$ is a (projective) K3 surface over a field $k$. A polarization of $X$ is an element $\lambda\in Pic_X(k)$ that is represented over an algebraic closure $\overline{k}$ by an ample line bundle $L$ over $X_{\overline{k}}$. Given such a $\lambda$, we can consider its self-intersection number $(\lambda,\lambda)\in 2\mathbb{Z}$. Suppose that $k$ has finite characteristic $p>2$. Is it possible to find a finite extension $k'/k$ and a polarization $\lambda$ of $X_{k'}$ such that $(\lambda,\lambda)$ is co-prime to $p$? -This question has a negative answer for polarizations of abelian varieties (see BCnrd's answer here), so I'm not optimistic that things are any better for K3 surfaces. On the other hand, one has an affirmative answer to the analogue for polarizations of abelian varieties, if one is allowed to modify the abelian variety up to isogeny. -So here's my backup question: Is there an analogous notion of an 'isogeny' of K3 surfaces that might help here? At the least, two K3 surfaces that are `isogenous' should have isometric Neron-Severi lattices (up to tensoring with $\mathbb{Q}$) and also isomorphic $l$-adic realizations, compatible with the cycle class map from the Neron-Severi lattices. - -REPLY [3 votes]: Let $S$ be a smooth projective surface and $m:=\gcd(\deg \lambda)$ where $\lambda$ runs through the ample cone. Let $m'$ be the gcd of the entries of the intersection matrix of $S$ -Then $m$ equals either $m'$ or $2m'$. The fact that $m'$ divides $m$ is obvious. To prove that $m$ divides $2m'$, note that there exist an effective divisor $D$ on $m$ such that at least one of $gcd(D^2,H^2)=m$ or $gcd(D.H,H^2)=m$ holds. -In particular, there exist infinitely many positive $n$ such that -$\gcd(H^2, (nH+D)^2)=\gcd(H^2,2n(H.D)+D^2)\in \{m,2m\}$. -For $n$ sufficiently large one has that $D+nH$ is ample. -For a $K3$ surface over a finite field it is believed that the geometric Picard number is always even. -So if you want to produce a counterexample you might try to construct a complete intersection of degree 3,2 with geometric Picard number 2, such that the second generator of the Picard group has genus 1 modulo 3, and such that the intersection number of this curve with $O(1)$ is divisible by 3.<|endoftext|> -TITLE: Topological conditions forcing continuity -QUESTION [9 upvotes]: Let $X$, $Y$, and $Z$ be topological spaces. Let $f:X \rightarrow Y$. Further assume that for every continuous function $g:Y \rightarrow Z$, $g \circ f$ is continuous. -Question: Under what conditions on the topology of $Y$ and/or $Z$ can we conclude that $f$ must then be continuous? -This is easily achieved if $Y$ is (essentially) homeomorphic with $Z$, but this seems like drastic overkill. -To me, this seems like some kind of 'universal property' for the continuity of $f$; or maybe some kind of generalized "uniformly continuous" condition? Note: original question due to Nick James, but he's not a big web user, so I am asking on his behalf. - -REPLY [11 votes]: It's possible to find a sufficient condition without putting any restraint on the spaces $X$ or $Y$. -Let $Z$ be the Sierpinski space, that is $Z$ has $\{0,1\}$ as a base set, and $\tau = \{\emptyset, \{0\}, \{0,1\}\}$ as a topology. -If $f: X\to Y$ is not continuous, then there is $V\subseteq Y$ open such that $f^{-1}(V)$ is not open in $X$. Defining $g_V: Y\to Z$ as mapping $V$ to $0$ and $Y\setminus V$ to $1$ we immediately see that $g_V\circ f: X \to Z$ is not continuous. -An interesting (and probably solved) question would be how to characterise the spaces $Z$ that have this universal property for all spaces $X, Y$.<|endoftext|> -TITLE: an operation on binary strings -QUESTION [24 upvotes]: Recently, as part of some joint research, Tom Roby was led to a curious operation on strings of L's and R's which he calls "bounce-reading": We start by reading the string at the left. When the symbol we have just read is an L, the next symbol we read is the leftmost unread symbol; but when the symbol we have just read is an R, the next symbol we read is the rightmost unread symbol. For instance, if our string of L's and R's is LLRRLRLR, we read the positions in the order 1,2,3,8,7,4,6,5, obtaining the bounce-reading LLRRLRRL. -It may be more natural to think of this as an action on circular words; when we read a letter, we delete it, close up the circle, and move either to the left or the right of the deletion-point. -Has this operation been studied before? - -REPLY [2 votes]: I suspect that it has not been studied, but it should be! I base this on the fact that various statistics one might expect to be associated with it do not appear in the OEIS (even with an appeal to superseeker@oeis.org). For example $$2, 3, 6, 7, 19, 27, 36, 79, 130, 384, 473, 710, 2903, 4197$$ -gives the maximum size orbit for strings of lengths up to $16.$ -It turns our that $RRRRRLRLLLLLL$ has an orbit of size $473$ and that is the longest orbit for any string of length 13.<|endoftext|> -TITLE: Finite groups with automorphism mapping $a/b$ of the elements of $G$ to their own inverses? Case $a/b=3/4$? -QUESTION [10 upvotes]: I was helping a friend prepare for his intro abstract final and he mentioned the professor had once asked the question: name a group and an automorphism that takes $3/4$ of the elements of the group to their own inverses (for instance, the dihedral group $D_4$ of order $8$, with identity automorphism). I tried to figure out how to approach this question in general but can't see how. - -1) Can we construct all such groups? - -(it is asserted in comment to the answer below that these are precisely those finite groups whose center has index 4) - -2) Given a rational number $a/b\in [0,1]$ does there exist a finite group $G$ and an automorphism $f$ such that $f$ maps exactly $a/b$ elements of $G$ to their own inverses? - -($a/b=1$ is achieved precisely for the inversion map on an abelian group; otherwise $a/b\le 3/4$ according to the answer below) - -3) Also, can these questions make sense in infinite groups? - -REPLY [18 votes]: This may be a well-known chestnut? (well-known to those that know it well, that is) -The fraction can never be between 3/4 and 1. To prove this, suppose $\phi\colon G\to G$ is an automorphism of $G$ that sends more than 3/4 of the elements of $G$ to their inverses. Let $S=\lbrace g\in G\colon \phi(g)=g^{-1}\rbrace$. -Notice that if $g$, $h$ and $gh$ all lie in $S$ then on the one hand, $\phi(gh)=(gh)^{-1}=h^{-1}g^{-1}$. On the other hand, $\phi(gh)=\phi(g)\phi(h)=g^{-1}h^{-1}$, so that $g^{-1}$ and $h^{-1}$ commute. It follows that $g$ and $h$ commute. -Fix $g\in S$ and consider $A=\lbrace h\colon h\in S\text{ and } gh\in S\rbrace$. There are less than $|G|/4$ $h$'s for which the first condition fails and less than $|G|/4$ $h$'s for which the second condition fails, so that $|A|>|G|/2$. By the above, it follows that the centralizer of $g$ (i.e. the set of $h$'s that commute with $g$) is a superset of $A$. Since the centralizer is a subgroup, by Lagrange's theorem the centralizer of $g$ must be all of $G$. That is $g$ lies in the center of the group. Now we have that more than $1/2$ of the group lies in the center (which is again a subgroup), so that $G$ is Abelian. -Now since $S$ is more than half of the group, the subgroup generated by $S$ must be all of $G$, so that every element of $G$ is a product of elements of $S$. Now $g\in G$, write $g=s_1\ldots s_n$. Then $\phi(g)=\phi(s_1)\ldots\phi(s_n)=s_1^{-1}\ldots s_n^{-1}=s_n^{-1}\ldots s_1^{-1}=g^{-1}$, so that $\phi(g)=g^{-1}$ for all $g\in G$.<|endoftext|> -TITLE: volume of compact simple Lie groups under the natural Euclidean embedding -QUESTION [10 upvotes]: I am looking for a quick reference for the volume formula for all the compact simple Lie groups embedded as matrix groups in the natural way. The one I care most for are the real orthogonal groups. I don't see how to deduce them from the Weyl integration formula. I believe they are of the order diameter raised to the power of the dimension. So for instance for $SO(n)$ the volume should be of order $n^{\Theta(n^2)}$. - -REPLY [5 votes]: As far as I know, the most standard volume formula was obtained by I.G. Macdonald in a very short 1980 paper here. Since the volume depends on the choice of Haar measure, Macdonald starts with the (complexified) Lie algebra of the compact Lie group, fixing a Lebesgue measure there along with a fixed lattice such as a Chevalley $\mathbb{Z}$-form. The second ingredient in his formula comes from the standard invariants in the cohomology calculation for the group. -There is another approach to Macdonald's formula in a later paper by Y. Hashimoto: -On Macdonald’s formula for the volume of a compact Lie group., Comment. Math. Helv. 72 (1997), no. 4, 660–662.<|endoftext|> -TITLE: Equitable Allocation of Individuals to Positions -QUESTION [8 upvotes]: I'm not a mathematician but I working on a problem that feels like it an example of a more general kind of problem and I'm hoping that someone might be able to point me in the right direction. -The problem is trying to find a convex combination of different ways of ranking n items that satisfy certain constraints. Let me be more concrete: Suppose we have $n$ individuals with endowments $b_1 > b_2 > b_3 \ldots b_n$. There are positions $1 \dots n$, each of which gives a value of $v_1 > v_2 > v_3 \ldots v_n \ge 0$ that the individuals can be assigned to. -I want to create a non-deterministic algorithm for assigning individuals to positions that is in some sense "equitable." so that for each individual, their expected proportional value from their positions is equal to their proportional endowment. E.g., suppose we have some method that assigns individual $i$ to position with 1 with probability $p_{i1}$, to position 2 with probability $p_{i2}$ and so on, I want it so that the algorithm generates an allocation such that: -$\forall i, \frac{\sum_{j=1}^n p_{ij} v_j}{\sum_{j=1}^n v_j} = \frac{b_i}{\sum_{j=1}^n b_j} $ -Some observations / thoughts: -For this to work even in the n=2 case, we need to constrain $b_1$ to that it isn't proportionally larger than $v_1$ (otherwise even always placing individual 1 at position 1 wouldn't be enough). -I originally thought this could be framed as a linear programming problem, where the goal is to find weights for each of the $n!$ possible orderings. Maybe this would work, but it would be computationally infeasible. -A particularly nice approach might be one that sequentially assigns positions by having each remaining individual "buy" probability shares with their budget and then draw a winner. Unfortunately, this doesn't have the equitable property above, but I was thinking that perhaps if we thought of the allocation as happening repeatedly, we could give the "losers" (payoff from position too small) a bigger endowment, taken from the "winners" in such a way that expected payoff converges to the equitable outcome. -Anyway, thanks for reading this far and I appreciate any comments, suggestions, answers etc. - -REPLY [5 votes]: This is not a complete answer but too long for a comment. You wrote, - -I originally thought this could be framed as a linear programming problem, where the goal is to find weights for each of the n! possible orderings. Maybe this would work, but it would be computationally infeasible. - -Linear programs with this many variables are solved all the time using column generation: -http://www.columbia.edu/~cs2035/courses/ieor4600.S07/columngeneration.pdf -In your problem, let variable $q_k$ denote the probability of using permutation $\pi_k$, and let variable $p_{ij}$ denote the marginal probability that agent $i$ is assigned to $j$, so you have a total of $n! + n^2$ variables. Let $\pi_k(i,j) = 1$ if agent $i$ is assigned to position $j$ in permutation $k$ and zero otherwise. We can relate the $p_{ij}$'s and $q_k$'s via $n^2$ constraints of the form -$\sum_k q_k \pi_k(i,j) = p_{ij} \forall i,j$ -Now, assume as Kevin did that $\sum_i v_i = \sum_j b_j = 1$. Your equity criterion is just $\sum_j p_{ij} v_j = b_i \forall i$. We obviously must require that $q_k \geq 0$ for all $k$. Thus the feasible set of these permutations can be written as: -$\sum_k q_k = 1$ -$p_{ij} = \sum_k q_k \pi_k (i,j) \forall i,j$ -$q_k \geq 0 \forall k$ -$\sum_j p_{ij} v_j = b_i \forall i$ -There are a total of $1 + n^2 + n$ equality constraints and $n!$ inequality constraints, and you have $n! + n^2$ variables. Any basic feasible solution of this set (i.e. a corner point) therefore must have $n! - (n+1)$ of the inequality constraints active, which means that the corner points have only $n+1$ nonzero $q_k$'s. Thus, if your problem is feasible at all for particular choices of $v$ and $b$, it's possible to find such a set of assignments using at most $n+1$ possible permutations. -You may gain further insight by looking at the dual of this LP -- the primal problem above has $n^2 + n!$ variables, $1 + n^2 + n$ equality constraints, and $n!$ inequality constraints, so its dual will have $n^2 + n!$ constraints, but only $1 + n^2 + n$ variables. Those constraints should have a nice structure to them (something like, one constraint for each permutation, and that constraint sums over all nonzero $\pi_k(i,j)$'s or something), which would give you a polynomial-time separation oracle for solving the dual LP. There may likely be a way to recover a primal solution from the dual using complementary slackness.<|endoftext|> -TITLE: Cyclic cubic numbers as rational linear combinations of roots of unity -QUESTION [8 upvotes]: In the written version of a talk Barry Mazur gave to Friends of the Harvard Mathematics Department on May 5, 2009, there is an interesting question in Footnote 5 (page 8). -He recalls how Gauss wrote $\sqrt p$ (where $p$ is an odd prime) as an explicit rational linear combination of roots of unity (using Gauss sums) and says that he doesn't know any such explicit expression for the roots $\alpha$ of an irreducible cubic polynomial $T^3+bT+c\in\mathbf{Q}[T]$ whose discriminant is a square (so that $\mathbf{Q}(\alpha)$ is a cyclic extension of $\mathbf{Q}$, and hence contained in $\mathbf{Q}(\zeta)$ for some root of unity $\zeta$). -Question. Does anyone know such an explicit expression for the roots of irreducible cubic polynomials whose discriminant is a square ? - -REPLY [4 votes]: For the simplest cubic $x^3-ax^2-(a+3)x-1$ which is cyclic and real -with discriminant $p^2$ where $p=a^2+3a+9$, and when $p$ is prime, -the roots $\theta_j,j=0,1,2$ are translates of the Gauss's cubic -periods $\eta_j$. Explicitly $\theta_j=\eta_j+(L-1)/6$, where -$4p=L^2+27$ and $\eta_j$ are the Gauss's periods. Since -$\sum_{j=0}^2 \eta_j=-1$, one can certainly replace $(L-1)/6$ by -linear combinations of $\eta_j$. This relation appeared (and also -the quartic and sextic case) in E. Lehmer, "Connections between -Gaussian periods and cyclotomic units", Maths Comp. Vol 50, No. 182 -(1988) 535-541.<|endoftext|> -TITLE: Wiener Tauberian Theorem for nonunimodular group -QUESTION [5 upvotes]: Is there a nonunimodular group for which Wiener's Tauberian theorem is true? -Is a locally compact topological group whose volume grows polynomially with radius always unimodular? - -REPLY [7 votes]: To the first is yes also. The example is already given above: ax+b group. It is the semidirect product $\mathbb R^\times \ltimes \mathbb R$. Leptin has proved in Leptin, H., Ideal theory in group algebras of locally compact groups. Invent.Math. 31 (1975/76), no. 3, 259-278 that every semidirect product of separable abelian l.c. groups has the Wiener property (in the sense that every proper closed two-sided ideal is annihilated by a nonzero *-representation).<|endoftext|> -TITLE: Multiple Hodge integrals and integrability -QUESTION [8 upvotes]: It is known that a generating function of the linear Hodge integrals is a tau function of the KP hierarchy, namely a one-parameter deformation of the Kontsevich-Witten tau-function (see Kazarian). Are there known any generating functions of the multiple Hodge integrals (integrals with multiple $\lambda$-classes) with nice integrable properties? - -REPLY [6 votes]: There's been a lot of interest on triple Hodge integrals in the last 10 years, mostly after the work of Mariño and Vafa on framed knots (see here). The question of integrability was first studied, as far as I am aware, by Jian Zhou in his paper on Hodge Integrals and Integrable Hierarchies, where two subcases (1-partition and 2-partition triple Hodge integrals) are found to be related to KP and 2-Toda respectively. The general case is considered by Aganagic, Dijkgraaf, Klemm, Mariño and Vafa in their Topological Strings and Integrable Hierarchies paper; the relevant integrable hierarchy is the 3-KP hierarchy. See also Mariño's book on Chern-Simons, Matrix models and Topstrings about it. -Slightly switching to autobiography, a somewhat different interpretation of the multi-Hodge integrals generating functions as dispersive deformations of the Witten-Kontsevich tau funcion is alluded to here. This is closer in spirit to the Dubrovin-Zhang study of 1+1 integrable hierarchies of topological type.<|endoftext|> -TITLE: Cayley graphs and its subgraphs -QUESTION [7 upvotes]: I have two questions about Cayley graphs. Any answers will be appreciate. -1) Do we have any Cayley graph that has Petersen graph as its induced subgraph? -2) Suppose $Cay(G,S)$ be a Cayley graph that $G$ is a finite group. Can we characterize any induced subgraphs of $Cay(G,S)$? -Thanks for any answer and guidance. - -REPLY [2 votes]: To answer your first question, take a look at [P. Erdos and A. B. Evans. Representations of graphs and orthogonal Latin square graphs. J. Graph Theory 13 (1989), no. 5, 593-595.] -Actually, It was shown that every graph $G$ is the induced subgraph of a circulant graph (a cayley graph on a cyclic group).<|endoftext|> -TITLE: rigidity of eigenvalues of circular ensemble -QUESTION [5 upvotes]: Given a circular unitary ensemble, with the following joint density: -$p(\theta_1,\ldots, \theta_n) = Z_n \prod_{j < k} |e^{i \theta_j} - e^{i \theta_k}|^2$, -is the following statement true? With high probability the eigenvalues are within distance $\mathcal{O}(1)$ from the evenly spaced set of $n$ points $(0,2\pi/n, 4\pi/n, \ldots, 2(n-1)\pi/n)$, rotated by some angle $\theta$. More precisely, is it true that -$ P[\int_\alpha d(\{\theta_1, \ldots, \theta_n\}, \{0, 2\pi/n, \ldots, 2(n-1)\pi/n\} + \alpha (\text{mod }2\pi)) < C] \to 1$ -for some sufficiently large constant $C$? -Here the distance is the induced Riemannian distance on $\mathbb{T}^n/ S_n$, where the action of $S_n$ on $\mathbb{T}^n$ is permutation of the coordinates. -I know Erdos Schlein Yau have proved a rigidity theorem for Wigner ensembles, but their result is slightly weaker than what I need. It seems natural to investigate this question for the exactly solvable case. - -REPLY [2 votes]: If I understand correctly, what you are looking for is Lemma 10 (when m=1) in http://arxiv.org/pdf/1210.2681v3.pdf by Elisabeth and Mark Meckes.<|endoftext|> -TITLE: Fibration of Batanin/Leinster $\omega$-groupoids -QUESTION [7 upvotes]: Is there (defined somewhere) a notion of fibration between two weak $\omega$-groupoids in the sense of Batanin/Leinster? -I tried to search on Google and in Higher Operads, Higher Categories of Tom Leinster, but I haven't found anything. -This would probably be very useful for interpreting Martin-Löf type theory in the category of Batanin/Leinster weak $\omega$-groupoids. - -REPLY [6 votes]: There isn't one existing. There is something on the completely strict omega-categorical case in Michael Warren's article "The strict ω-groupoid interpretation of type theory" (available from his web page at IAS). -However we have a PhD student here at Macquarie working on higher fibrations and all that so hopefully there will be something on the weak case in due course!<|endoftext|> -TITLE: Conjecture on Markov-Hurwitz Diophantine equation -QUESTION [14 upvotes]: Somebody asked about the Markov Diophantine equation recently - https://math.stackexchange.com/questions/94394/diophantine-equation-in-positive-integers -and I was reminded of some unfinished business. In 1907, A. Hurwitz wrote -Uber eine Aufgabe der unbestimmten Analysis, in Archiv der Mathematik und Physik, pages 185-196. I am going to abbreviate...with integers $x_1, \ldots, x_n,$ define -$$ T = x_1^2 + \cdots + x_n^2 $$ and let -$$ P = x_1 \ldots x_n.$$ -For some positive integer $a,$ consider the Markov-Hurwitz equation in positive integers, -$$ T = a P.$$ -For any solution to M-H, we can create a new solution with some $x_j$ replaced by -$$ x_j' = \frac{aP}{x_j} - x_j.$$ -So as with the Markov tree, any solution can be transformed into a "fundamental solution" (Grundlosung) by a finite sequence of such transformations. That is, eventually, no $x_j$ can be reduced by this process, which is what happens when all -$$ 2 x_j^2 \leq aP. $$ -The stuff that happens next is easier to type if we order the entries, so an OFS, or Ordered Fundamental Solution, is -$$ x_1 \geq x_2 \geq x_3 \geq \cdots \geq x_n, \; T = a P, \; 2 x_1^2 \leq aP.$$ -Hurwitz shows many things. In an OFS, we have $$ a x_3 \cdots x_n \leq n. $$ -We get $$ a \leq n.$$ Also, most entries are 1, indeed if -$ k > 2 + \left\lfloor \log_2 n \right\rfloor$ then $x_k = 1.$ -Whenever $a=n,$ the only OFS has all $x_j = 1.$ Hurwitz gives a table of all OFS for $n \leq 10.$ There can be more than one value of $a$ for each $n.$ -Kap and I investigated a bit for larger $n.$ For example, the first time there is more than one OFS for a single $(n,a)$ pair is -$n = 14, a = 1.$ One OFS is $(3,3,2,2,1,1,\ldots,1).$ There is another, $(6,4,3,1,1,\ldots,1).$ -We are finally getting to my conjecture. Whenever $n = 5 w^2 - 6,$ there is an OFS with $$a=1, \; x_1 = 3 w, \; x_2 = 2 w, \; x_3 = 3, \; x_4 = x_5 = \cdots = x_n = 1.$$ The common value of $T=aP$ is $18 w^2.$ Considerable computation suggests that this is the largest that $x_1$ ever gets, and the only occurrence of equality, at least once $n \geq 3.$ So we have -CONJECTURE: In an OFS with $n \geq 3,$ we always have $$ x_1 \leq \sqrt{\frac{9(n+6)}{5}}$$ and equality occurs only when -$ n = 5 w^2 - 6.$ -So that is the question. I can make no guarantee that any advanced mathematics is invloved in any way, only that I put a ton of effort into proving this and never got it. Let me know if more information would be helpful, I've got tons. Pantloads. -ADDED: as far as I know the second best OFS is related to the best, -$$n = 5 w^2 - 7, \; \; a=1, \; x_1 = 3 w -1, \; x_2 = 2 w, \; x_3 = 3, \; x_4 = x_5 = \cdots = x_n = 1.$$ The common value of $T=aP$ is $18 w^2 -6w.$ -SEPTEMBER 2013: there have been two recent articles on Apollonian Circle Packing. Motion between quadruples of integers in an ACP is accomplished by exactly the same trick as Hurwitz and Markov before him, which is called Vieta Jumping in high-school competitions. If we have the integer equation $$ x_1^2 + \cdots + x_n^2 = F(x_1, x_2, \cdots, x_n), $$ where $F$ is a symmetric polynomial with each term "squarefree," each exp[onent of a variable is either $0$ or $1,$ then we may "jump" to another $n$-tuple by fixing $n-1$ variables and switching just one. That is, write the relation as $x_j^2 - B(\mbox{other}) x_j + C(\mbox{other}) = 0.$ One root is the current value of $x_j,$ the other is also an integer $x_j' =B(\mbox{other}) - x_j. $ The ACP equation is $$ w^2 + x^2 + y^2 + z^2 = 2 wx + 2 w y + 2 w z + 2 x y + 2 x z + 2 y z. $$ It turns out that there are infinitely many distinct orbits of the Apollonian (Vieta) Group acting on integer quadruples solving the equation. - -REPLY [6 votes]: Do the Apollonian and Markoff-Hurwtiz problems share the same "jumping" movement? That would depend on one's interpretation of "share." There is a link; one that my research followed. -I was first introduced to the Hurwitz equations via the Markoff equation; and the fractal nature of the exponent of growth lead me to the work of Boyd on the Apollonian Packing Problem. I later heard about a certain class of K3 surfaces ... those surfaces that are smooth $(2,2,2)$ forms in $P\times P\times P$. (Note that $P\times P\times P$ is not the same as $P^3$.) These are surfaces in three projective variables whose defining equation is quadratic in each variable. They therefore have the same "jumping" action as the Markoff equation. These surfaces have the advantage of being smooth, so the jumps end up being automorphisms of the surface. The automorphisms act on the group of divisors on the surface (the Picard group), and this action shares a lot of similarities with the Apollonian packing. -The Apollonian packing can be thought of this way: The bilinear form that you mention in your note is a Lorentz product. That is, we can write it as $X^tJX=0$, where $X=(w,x,y,z)$ and $J$ is a symmetric matrix with signature $(3,1)$. (That is, $J$ has $3$ positive eigenvalues and one negative eigenvalue.) Thus, a surface of the form $X^tJX=-1$ is a hyperboloid of two sheets, and one sheet is a model of 3D hyperbolic geometry, with distance defined by $\cosh(|XY|)=-X^tJY$. Three dimensional hyperbolic space has the Poincare upper half space model, and planes in this model are (Euclidean) hemi-spheres perpendicular to the boundary. A plane is therefore represented by a circle. The "jumps" used in the Apollonian packing is a subgroup of the group of all such jumps, and is a subgroup of isometries in the hyperbolic space. The packing is an orbit of one plane under the action of this group. Because the fundamental domain for the group of "jumps" has infinite volume, the packing has an associated fractal. -For the K3 surfaces, the Lorentz product is the intersection pairing, which has signature $(1,3)$ (if the Picard group has dimension $4$). The group of automorphisms on the K3 surface (analogs of the jumps for the Markoff equation) acts as isometries (analogs of jumps for the Apollonian packing) on the underlying hyperbolic space. The orbit of a circle for these cases are not always tangent, as they are in the Apollonian case, but are sometimes perpendicular. An example is pictured on my homepage (http://faculty.unlv.edu/baragar/). Note that one of the jumps in the Picard space came from an automorphism that isn't exactly a "jump" on the surface, though it has a very interesting interpretation. -There is no K3 surface, though, that is associated to the Apollonian packing.<|endoftext|> -TITLE: Exponential and Logarithm Mapping on Stiefel Manifold -QUESTION [9 upvotes]: The Stiefel Manifold is defined as -$$ -\mathrm{St}(p,n):= \{ X\in \mathbb{R}^{n\times p} :\ X^T X = I_p \}. -$$ -Recall that the tangent space at a point $X\in \mathrm{St}(p,n)$ is given by -$$ -T_X{\mathrm{St}(p,n)} = \{\xi\in \mathbb{R}^{n\times p}:\ X^T\xi + \xi^T X = 0 \}. -$$ -Given a point $X\in \mathrm{St}(p,n)$ and a tangent vector $\xi \in T_X{\mathrm{St}(p,n)}$, -it is possible to express the exponential map $\exp_X(\xi)$ using the matrix exponential function. A formula is given in the paper www.mit.edu/~wingated/introductions/stiefel-mfld.pdf . -My question is whether the inverse $\log_X(\cdot): \mathrm{St}(p,n) \to T_X\mathrm{St}(p,n)$ can also be expressed using the matrix logarithm. -A related question is the following: Instead of the exponential function one may define other retractions such as -$$ - R_X(\xi) = (X+\xi)(I_p + \xi^T\xi)^{-1/2} -$$ -or the closest point projection -$$ - R_X(\xi) = \pi(X+\xi), -$$ -where $\pi$ maps a matrix $A$ to the closest element in the Stiefel manifold. This projection can be easily computed using SVD. -Again, the question is wheter one can find simple formulas for the inverses of these retractions. -Any helpful comments would be greatly appreciated. -edit: at this point I do not really care which metric is used (the Euclidean metric or the one inherited from the orthogonal group). - -REPLY [3 votes]: If the question is still relevant to you, have a look at the paper: - -R. Zimmermann: A Matrix-Algebraic Algorithm for the Riemannian Logarithm on the Stiefel Manifold under the Canonical Metric, SIAM. J. Matrix Anal. & Appl. 38-2 (2017), pp. 322-342, arXiv:1604.05054, doi: 10.1137/16M1074485<|endoftext|> -TITLE: infinite dimensional CAT(0) groups -QUESTION [15 upvotes]: Usually a CAT(0) group is defined to be a group acting properly isometrically and cocompactly on a CAT(0) space, but I would like to consider only those groups that act properly, isometrically and cocompactly on a finite-dimensional CAT(0) space. -So is there a group I have to leave out? -Not every CAT(0) space with a proper isometric cocompact group action is finite-dimensional. For example the trivial group acts on the compact CAT(0)-space $[0;1]^\mathbb{N}$. - -REPLY [8 votes]: First, I suppose that by proper action you mean the one in the sense of Bridson and Haefliger, otherwise you would have to regard ${\mathbb R}$ as a $CAT(0)$ groups. Now, it follows from Eric Swenson's paper "A cut point theorem for CAT(0) groups" (Journal of Diff. Geometry, 1999) that the ideal boundary of the $CAT(0)$ space $X$ (on which a group $G$ acts geometrically) is finite-dimensional. This suffices for many practical purposes. For instance, it follows (from Bestvina's work) that $G$ has finite cohomological dimension over ${\mathbb Q}$ and, if you consider torsion-free groups, over ${\mathbb Z}$ as well. (This immediately excludes Thompson's group, etc.) In particular, geometric dimension of $G$ is finite, $G$ has finite type, etc. From this you can make pretty much the same algebraic conclusions about $G$ as in the case when $G$ acts geometrically on a finite-dimensional $CAT(0)$ space. Thus, in the torsion-free case, I do not think you are missing (or gaining) much by restricting to finite-dimensional $CAT(0)$ spaces. (For instance, I do not see how assuming finite dimension of $X$ would help with proving that $G$ has finite asymptotic dimension.) -I am not sure what happens in the case of groups with torsion: It is conjectured by Swenson that a $CAT(0)$ group $G$ cannot contain infinite torsion subgroups. Maybe it would be easier to exclude some infinite torsion subgroups (say, the infinite permutation group) -using the assumption that $G$ acts geometrically on a finite-dimensional $CAT(0)$ space, but I do not see how. -Swenson's work had a follow-up paper by Geoghegan and Ontaneda -http://arxiv.org/abs/math/0407506 where they weaken some of his assumptions and strengthen some of his conclusions. -Note: In view of Swenson's result it is tempting to say: Take the closed convex hull (in $X$) of the ideal boundary of the $CAT(0)$ space $X$ and show that it is finite-dimensional. It might work, but, in general, convex hulls in $CAT(0)$ spaces tend to be much bigger than expected.<|endoftext|> -TITLE: Is $ \sum\limits_{n=0}^\infty x^n / \sqrt{n!} $ positive? -QUESTION [99 upvotes]: Is $$ \sum_{n=0}^\infty {x^n \over \sqrt{n!}} > 0 $$ for all real $x$? -(I think it is.) If so, how would one prove this? (To confirm: This is the power -series for $e^x$, except with the denominator replaced by $\sqrt{n!}$.) - -REPLY [4 votes]: Although the following does not provide another proof (perhaps it is possible to attempt one on this basis) I found it nice to see the following pictures. -Let's take from the series $f(x) = \sum_{k=0}^\infty {x^k \over \sqrt{k!}}$ the following variants in the same spirit as we have the hyperbolic and trigonometric series from the exponential-series: -$$\begin{array}{} - \small \exp_{\tiny \sqrt{\,}}(x) &=& f(x) \\ - \small \cosh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k} \over \sqrt{(2k)!}} \\ - \small \sinh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k+1} \over \sqrt{(2k+1)!}} \\ -\small \tanh_{\tiny \sqrt{\,}}(x) &=& { \sinh_{\tiny \sqrt{\,}}(x)\over \cosh_{\tiny \sqrt{\,}}(x) } \\ -\small \cos_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (-1)^k {x^{2k} \over \sqrt{(2k)!}} \\ -\small \sin_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (-1)^k {x^{2k+1} \over \sqrt{(2k+1)!}} \\ -\end{array}$$ -The answer to your question is equivalent to say, that always (="for real $x$") - -$\small \cosh_{\tiny \sqrt{\,}}(x)$ is larger than $\small \sinh_{\tiny \sqrt{\,}}(x) $ $\qquad \qquad$ or that -$\small \mid \tanh_{\tiny \sqrt{\,}}(x) \mid \lt 1$ - - -To illustrate this I've plotted the $\sinh_{\tiny \sqrt{\,}}$ and $\cosh_{\tiny \sqrt{\,}}$-curves: - -This gives surely an extremely familiar impression... -The $\tanh_{\tiny \sqrt{\,}}$-curve looks completely familiar too: - -and the image suggests, that indeed the absolute value of $\small \tanh_{\tiny \sqrt{\,}}(x) $ very likely is smaller than $1$ for all real $x$. - -However, things are different for the $\sin_{\tiny \sqrt{\,}}$ and $\cos_{\tiny \sqrt{\,}}$ curves - they deviate strongly from the nicely periodic common trigonometric functions: - -and combined they do not give a circle, but some ugly thing, strongly distorted (y-axis by $\small \cos_{\tiny \sqrt{ \,} }(\phi)$, x-axis by $\small \sin_{\tiny \sqrt{ \,} }(\phi)$, $\phi$ from $-5$ to $+5$) :<|endoftext|> -TITLE: Graph of dependencies from a Latex file -QUESTION [12 upvotes]: This question has been "manually migrated" to TeX-SX: https://tex.stackexchange.com/q/40200/86 - -Apologies if the question is not very appropiate for Mathoverflow. It seems to me more appropiate here than in the other 'exchange' sites. -Is there an IT tool to create a graph of dependencies from a Latex file? The sense is the following: -It just occurred to me that if everyone creates propositions with proofs (usually) afterwards and these proofs use \eqref, \ref \cite to call to other results it should be feasible to create a graph of dependencies of results, given a paper written in Latex. -I think such a thing would be useful for any mathematician (check dependencies, recursive arguments, that there are lemmas which have a need, possibility of suggesting equivalences, writing well-ordered documents...) so I would be surprised if this does not exist yet, but I couldn't find it anywhere. - -REPLY [4 votes]: As a rule, you cannot depend upon math papers making every dependency explicit, meaning you cannot extract nearly so much information from this directed graph as you imagine. In addition, there isn't any reason this graph should be acyclic since forward references frequently get used in outlines and motivational text. -That said, there are use cases like identifying all the backwards references. For example, you could find all backwards \ref commands using this perl script I call earlyref.pl : -#!/usr/bin/perl -n -BEGIN { %labels = (); } -while (/\\(label|ref)\{([A-Za-z0-9_]+)\}/g) { - if ($1 eq "label") { $labels{$2} = 1; next; } - next if ($labels{$2}); - print "Early \\ref{$2} on line $. in $ARGV" -} -close if eof(ARGV); - -You'll find this handles multiple filename arguments correctly because the last line resets the line number $.` when appropriate. -There is considerably more you can do using scripting languages with built in regular expressions. I've written a chcite script which changes all your \cite commands for switching between different co-authors .bib files, for example.<|endoftext|> -TITLE: Permission to use Online Notes -QUESTION [16 upvotes]: I am a new professor in Mathematics and I am running an independent study on Diophantine equations with a student of mine. Online I have found a wealth of very helpful expository notes written by other professors, and I would like to use them for guided reading. I am wondering whether it is customary to ask permission of the author before using his or her online notes for my own reading course. -Also, if anyone has suggestions for good sources Diophantine Equations please feel free to enlighten me. - -REPLY [3 votes]: The Stanford Copyright & Fair Use Overview is pretty clear about what constitutes fair use in the classroom and what doesn't: - -Rules for Reproducing Text Materials for Use in Class -The guidelines permit a teacher to make one copy of any of the following: a chapter from a book; an article from a periodical or newspaper; a short story, short essay, or short poem; a chart, graph, diagram, drawing, cartoon, or picture from a book, periodical, or newspaper. -Teachers may photocopy articles to hand out in class, but the guidelines impose restrictions. Classroom copying cannot be used to replace texts or workbooks used in the classroom. Pupils cannot be charged more than the actual cost of photocopying. The number of copies cannot exceed more than one copy per pupil. And a notice of copyright must be affixed to each copy. -Examples of what can be copied and distributed in class include: - -a complete poem if less than 250 words or an excerpt of not more than 250 words from a longer poem -a complete article, story, or essay if less than 2,500 words or an excerpt from any prose work of not more than 1,000 words or 10% of the work, whichever is less; or - one chart, graph, diagram, drawing, cartoon, or picture per book or per periodical issue. - -Not more than one short poem, article, story, essay, or two excerpts may be copied from the same author, nor more than three from the same collective work or periodical volume (for example, a magazine or newspaper) during one class term. As a general rule, a teacher has more freedom to copy from newspapers or other periodicals if the copying is related to current events. -The idea to make the copies must come from the teacher, not from school administrators or other higher authority. Only nine instances of such copying for one course during one school term are permitted. In addition, the idea to make copies and their actual classroom use must be so close together in time that it would be unreasonable to expect a timely reply to a permission request. For example, the instructor finds a newsweekly article on capital punishment two days before presenting a lecture on the subject. -Teachers may not photocopy workbooks, texts, standardized tests, or other materials that were created for educational use. The guidelines were not intended to allow teachers to usurp the profits of educational publishers. In other words, educational publishers do not consider it a fair use if the copying provides replacements or substitutes for the purchase of books, reprints, periodicals, tests, workbooks, anthologies, compilations, or collective works. - -Note that this does not apply if the work is licensed appropriately. For example, the Stanford Copyright & Fair Use Overview quoted above is licensed CC BY-NC and this answer is licensed CC BY-SA. For course notes, it makes sense to use such a license since that is probably the intended use of the work. Contact the author(s) and recommend that they license their work in a manner they find appropriate. - -The Stanford Copyright & Fair Use Overview also has a clear notice regarding online material: - -If You Want to Use Material on the Internet -Each day, people post vast quantities of creative material on the Internet -- material that is available for downloading by anyone who has the right computer equipment. Because the information is stored somewhere on an Internet server, it is fixed in a tangible medium and potentially qualifies for copyright protection. Whether it does, in fact, qualify depends on other factors that you would have no way of knowing about, such as when the work was first published (which affects the need for a copyright notice), whether the copyright in the work has been renewed (for works published before 1978), whether the work is a work made for hire (which affects the length of the copyright) and whether the copyright owner intends to dedicate the work to the public domain. If you want to download the material for use in your own work, you should be cautious. It's best to track down the author of the material and ask for permission. Generally, you can claim a fair use right for using a very small portion of text for commentary, scholarship or smilar purposes. - -The thing to take away from this is that there is no substitute for proper licensing. If you intend for something to be used in some way or another: make that clear!<|endoftext|> -TITLE: Upper bounds on the difference of consecutive zeta zeros -QUESTION [8 upvotes]: There are many results on the spacing of the gaps between nontrivial zeros of the $\zeta$ function, from trivial (average value is $\frac{2\pi}{\log\gamma_n}$) to difficult (bounds on max and min values of the normalized gap). Are any reasonable upper bounds known? I'd like to have something that says, given any $\varepsilon>0,$ there is some N beyond which the gaps $\gamma_{n+1}-\gamma_n$ is at most $\varepsilon.$ This seems a weak request given the asymptotic behavior but I haven't found anything along these lines. -Any ideas? -I asked the question on math.se but did not get an answer. - -REPLY [2 votes]: Edward Charles Titchmarsh, The theory of the Riemann zeta-function, IX. The general distribution of the zeros pages 191 to 193. -Oxford at the Clarendon press 1951. - -9.12. We shall now obtain a more precise result of the same kind. $\dagger$ -THEOREM 9.12. For every large positive T, $\zeta(s)$ has a zero $\beta+i \gamma$ satisfying -$$|\gamma-T|<\frac{A}{\log\log\log\;T}.$$ -This was first proved by Littlewood by a detailed study of the conformal representation used in the previous proof. This involves rather complicated calculations with elliptic functions. We shall give here two proofs which avoid these calculations. -In the first, we replace the rectangles by a succession of circles. Let $T$ be a large positive number, and suppose that $\zeta(s)$ has no zero $\beta+i \gamma $ such that $T-\delta \leq \gamma \leq T+\delta$, where $\delta < \frac{1}{2}$. Then the function -$$f(s)=\log\zeta(s),$$ -where the logarithm has its principal value for $\sigma > 2$, is regular in the rectangle -$$-2 \leq \sigma \leq 3, \;\;\;\;\;\;\; T - \delta \leq t \leq T + \delta$$ -$\dagger$ Littlewood (3); proofs given here by Titchmarsh (13), Kramaschke (1). -Let $c_v, C_v, \textbf{C}_v, \Gamma_v$ be four concentric circles, with centre $2-\frac{1}{4}v\delta+iT,$ and radii $\frac{1}{4}\delta,\frac{1}{2}\delta,\frac{3}{4}\delta,\delta$ respectively. Consider these sets of circles for $v=0,1,...,n,$ where $n=[12/\delta]+1,$ so that $2-\frac{1}{4}n\delta \leq -1,$ i.e. the centre of the last circle, or to the left of, $\sigma = -1$. Let $m_v, M_v,$ and $\textbf{M}_v$ denote the maxima of $|f(s)|$ on $c_v, C_v,$ and $\textbf{C}_v$ respectively. -Let $A_1,A_2,...$ denote the constants (it is convenient to preserve their identity throughout the proof). We have $\textbf{R} \{ f(s) \} < A_1\log T$ on all the circles, and $|f(2+iT)|t^{A_5}$ for $\sigma \leq -1, \;\; t>t_0,$ so that $M_n>A_5 \log T.$ Hence -$$A_5 -TITLE: Alternating sum of square roots of binomial coefficients -QUESTION [43 upvotes]: Let -$$ c_n = \sum_{r=0}^n (-1)^r \sqrt{\binom{n}{r}}. $$ -It is clear that $c_n = 0$ if $n$ is odd. Remarkably, it appears that despite the huge positive and negative contributions in the sum defining $c_{2m}$, the sequence $(c_{2m})$ may be very well behaved. - -Is $c_n > 0$ for all even $n$? - -An affirmative answer will imply that the function $F(x) = \sum_{n=0}^\infty x^n/\sqrt{n!}$ is always strictly positive, thereby answering this earlier question. -Numerical computation using Magma shows that $c_n > 0$ if $n$ is even and $n \le 2000$. To give some illustrative values, $c_{100} = 0.077737 \ldots$, $c_{1000} = 0.019880 \ldots $ and $c_{2000} = 0.013317 \ldots$. -A comment by Mark Sapir on the earlier question suggests a stronger result might hold. - -Is $c_{n} > c_{n+2} > 0$ for all even $n$? - -I have checked that this is the case for all even $n \le 2000$. -It is very natural to ask what happens if we replace $\sqrt{\binom{n}{r}}$ with $\binom{n}{r}^\alpha$ for $\alpha \in (0,1)$. For $n\le 250$ the generalized version of the conjecture continues to hold if $\alpha = k/10$ where $k \in \mathbf{N}$ and $k \le 9$. Of course when $\alpha = 1$ we have $c_n = 0$ for all $n$, so, as David Speyer remarked in a comment on the earlier question, there is a good reason for the cancellation in this case. - -REPLY [46 votes]: Here's a proof of the positivity of -$$ -c_n(\alpha) := \sum_{r=0}^n (-1)^r {n\choose r}^\alpha -$$ -for all even $n$ and real $\alpha < 1$. It follows -(via M.Wildon's clever $F(x) F(-x)$ trick at mo.84958) that -$\sum_{n=0}^\infty \phantom. x^n / n!^{\alpha} > 0$ for all $x \in\bf R$. -[EDIT fedja has meanwhile provided a very nice direct proof of -the positivity of $\sum_{n=0}^\infty \phantom. x^n / n!^{\alpha}$.] -The key is to write $c_n(\alpha)$ as a finite difference -$$ -\sum_{r=0}^n \phantom. (-1)^r {n\choose r} \cdot {n\choose r}^{\alpha - 1} -$$ -and show that the Gamma interpolation -$$ -\bigl(\Gamma(r+1)\Gamma(n-r+1) / n!\bigr)^{1-\alpha} -= n!^{\alpha-1} \exp\bigl((1-\alpha) (\log\Gamma(r+1) + \Gamma(n-r+1)\bigr) -$$ -of ${n\choose r}^{\alpha - 1}$ has a positive $n$-th derivative -for all $r \in [0,n]$. -This in turn follows from the fact that the expansion of -$\log\Gamma(r+1) + \log\Gamma(n-r+1)$ in a Taylor series about $r = n/2$ -has positive $(r - (n/2))^k$ coefficient for each $k=2,4,6,\ldots$. -[The coefficient vanishes for odd $k$ because -$\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is an even function of $r-(n/2)$.] -Indeed the well-known formula -$$ -\log \Gamma(x) = -\gamma x - \log x + \sum_{j=1}^\infty - \left[ \frac{x}{j} - \log \left( 1 + \frac{x}{j} \right) \right] -$$ -shows that the $k$-th derivative of $\log\Gamma(x)$ is positive -for all $x>0$ and $k=2,4,6,\ldots$, because this is true for -$-\gamma x - \log x$ and for each term in the sum; explicitly -the derivative is $k! \phantom. \sum_{j=0}^\infty (x+j)^{-k}$ which is -positive termwise. Therefore in the Taylor expansion -$$ -\log \Gamma(r+1) = \log(n/2)! + \sum_{k=1}^\infty \phantom. g_k (r-(n/2))^k -$$ -each of $g_2,g_4,g_6,\ldots$ is even. -Since $\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is -$$ -2\log(n/2)! - + 2 \Bigl( g_2 (r-(n/2))^2 + g_4 (r-(n/2))^4 + g_6 (r-(n/2))^6 + \cdots\Bigr), -$$ -the claim follows. [EDIT David Speyer notes that the convergence -of the Taylor series on $|r-(n/2)| \leq n/2$ requires justification, -and that the justification is easy because the $\Gamma(z)$ has no zeros -and poles only at $0,-1,-2,\ldots$ so the radius of convergence is $(n/2)+1$.] -Multiplying by $1 - \alpha$ and substituting into the exponential series, -we deduce that $(\Gamma(r+1) \Gamma(n-r+1))^{1-\alpha}$, too, -is a positive combination of even powers of $r-(n/2)$. -Now if a function $g$ has positive $n$-th derivative, then its -first finite difference -$$ -g(x+1) - g(x) = \int_x^{x+1} g'(y) dy -$$ -has positive $(n-1)$-st derivative; repeating this argument $n$ times, -we find that the $n$-th finite difference is positive, and we're done.<|endoftext|> -TITLE: (Closures of sets of) operations in topological groups. -QUESTION [12 upvotes]: Let $G$ be a topological group. For each $n \in \mathbb{Z}$, consider the continuous functions $f_{n} \colon G \to G : x \mapsto x^{n}$, and set $F := \{f_{n} \mid n \in \mathbb{Z}\}$. -Is there a nice/easy description of the closure of the set $F$ in $C(G,G)$ with respect to the topology of pointwise convergence? And what is the induced subspace topology on this closure? -Example: If $G$ is the (topological) circle group, then $F = Hom(G,G)$ is closed in $C(G,G)$ with respect to the topology of pointwise convergence. However, I do not know how the induced subspace topology on $F$ can be described. (It is known from Pontrjagin Duality that compactness of $G$ implies that the compact-open topology on $F$ coincides with the discrete topology, but this does not tell anything about the topology of pointwise convergence.) -Ideally, I would like to have a general statement, e.g, for compact Hausdorff groups. Does anybody know a suitable reference? - -Edit: -I originally stated the example differently, namely -Example: If $G$ is the group associated to one-dimensional sphere, then $F$ is closed in $C(G,G)$ and the topology of pointwise convergence on $F$ coincides with the discrete topology. -However, in this example, the topology of pointwise convergence on $F$ does not coincide with the discrete topology. The proof that I had mind requires this topology to be frist-countable on $F$, but I cannot find an argument for this. In general, topological spaces with countable underlying set need not be first-countable (see Arens-Fort Space), so the second statement of the example is just not true. - -REPLY [2 votes]: Here is what I think happens in the category of compact (Hausdorff) groups. I know it is true in the category of profinite groups and I assume the argument carries over. First of all I believe the closure in the compact-open topology and the pointwise convergence topology are the same. The closure should be described this way. -Let $C$ be the free compact group on 1-generator $a$. To each element $\nu$ of $C$ and compact group $G$, we get a cts map $\nu_G\colon G\to G$ as follows. Let $x\in G$. By the universal property of $C$ there is a unique cts map $C\to G$ sending $a$ to $x$. Define $\nu_G(x)$ to be the image of $\nu$ under this map. This gives a cts mapping $C\to Cts(G,G)$ (natural in $G$) in the compact open topology. The closure of the family $F$ of mappings in the -OP's question is the image of $C$ in $Cts(G,G)$. -To make this clearer, the mapping on $G$ associated to $a^n$ with $n\in \mathbb Z$ is $x\mapsto x^n$. - - -Edit. My intuition above was from the profinite case. In the profinite setting the family of mappings $x\to x^n$ with $n\geq 1$ is equicontinuous. Indeed if $G$ is profinite, then it has a fundamental system of entourages for its uniformity consisting of partitions into cosets of an open normal subgroup. Now if $x,y$ are in the same coset of an open normal subgroup $N$ then $x^nN=y^nN$ for all $n\geq 1$. This gives equicontinuity in the sense of uniform spaces (in fact all these maps are contractive in the uniform sense). It follows that for profinite groups the closure in the compact-open topology is the pointwise closure and hence compact. The discussion above using free pro-cyclic groups works then in the profinite setting. -In the general setting of compact groups I was probably naive. Doubling on a circle is chaotic and not contractive nor are its powers equicontinuous. So most likely what I wrote above doesn't work in general. -However do note that all maps of raising to a negative power are pointwise limits of positive powers. So the OP's statement for the circle case about the topology being discrete is incorrect.<|endoftext|> -TITLE: A Linear Algebra Problem -QUESTION [7 upvotes]: Given a matrix $A\in \mathbb{R}^{n\times n}$, I am looking for -a symmetric matrix $S\in\mathbb{R}^{n\times n}$ such that -$$ -S A + A^T S = I -$$ -$A$ can be assumed to be regular (with positive determinant, if this is of any help). -The difficulty is of course that $S$ must be symmetric, otherwise one could simply take $2S = A^{-T}$. In principle this is a linear equation with $\frac{n(n+1)}{2}$ unknowns and this can be solved for $S$. -Is there a nicer way to find $S$ such as a closed solution formula using some factorization? Has this problem been studied anywhere? - -REPLY [11 votes]: These matrix equations are called Lyapunov equations and are extensively studied in control theory. -For instance, if $A$ is Hurwitz (all eigenvalues in the left half-plane), then the unique symmetric solution of $A^TX+XA+Q$ is -$$ -X=\int_0^\infty e^{A^T t } Q e^{At} dt. -$$<|endoftext|> -TITLE: Symplectic groups $Sp_{2m}(2)$ as $2$-transitive permutation (i.e. Galois) groups -QUESTION [7 upvotes]: I am looking for information about the symplectic groups $Sp_{2m}(2)$ as permutation group acting on quadratic forms. -Consider the block matrices -$$e=\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad -f=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=e-e^T$$ -on the vector space $(\mathbb{F}_2)^{2d}$ equipped with the standard basis. Consider the symmetric bilinear form $\phi(u,v)=ufv^T$, and let $\Omega$ be the set of all quadratic forms $\theta(u)$ such that -$$\phi(u,v)=\theta(u+v)-\theta(u)-\theta(v).$$ -In particular the quadratic form $\theta_0(u)=ueu^T$ is $\in\Omega$, and any other element of $\Omega$ can be shown to be of the form -$$\theta_a = \theta_0(u)+\phi(u,a).$$ -Now $Sp_{2m}(2)$ acts on $\Omega$, and it turns out that the action splits in two distinct orbits -$$\Omega^+=\{\theta_a|\theta_0(a)=0\},\qquad \Omega^-=\{\theta_a|\theta_0(a)=1\},$$ -of size respectively $2^{m-1}(2^m+1)$ and $2^{m-1}(2^m-1)$. The group $Sp_{2m}(2)$ acts $2$-transitively on each of these orbits, see Chap. 7 of Permutation Groups (Dixon, Mortimer) for more details. -Question: what can be said of the action of these two sets? Here are two more specific questions: is the stabilizer of one point acting imprimitevely, for some block structure? What are the orbits of a $2$-point stabilizer? -Motivation: I am studing the Galois groups of polynomials (trinomials) over function fields in characteristic $p$, which can be proven to be $2$-transitive. This was done by Abhyankar, Galois theory on the line in nonzero characteristic (1996), which computed the Galois group of many trinomials, and I think that his results can be extended to cover more cases. If I'm wrong I will happen to have learned something about $2$-transitive groups. -The $2$-transitive permutations groups are classified (affine groups, alternating/symmetric, projective, symplectic $Sp_{2m}(2)$, unitary $PGU_3(q^3)$, Suzuki $Sz(q)$ and Ree $R(q)$, plus a few sporadic groups). Computing the local Galois group at a ramified place it is possible to describe the action of a subgroup, the inertia subgroup, as a permutation group on the roots. This allows to rule out certain familes of $2$-transitive groups, and sometimes it is possible to determine completely the Galois group. And the symplectic group at the moment is the family that I find more difficult to understand. - -REPLY [13 votes]: Both of these actions are 2-primitive, so the 1-point stabilizer acts primitively on the remaining points. -The 1-point stabilizers in the two actions are the orthogonal groups ${\rm SO}^{\pm}_{2m}(q)$, -and the 2-point stabilizers are the maximal parabolic subgroups of these orthogonal groups with structure $2^{2m-2}.{\rm SO}^{\pm}_{2m-2}(q)$. They are the stabilizers of isotropic points in the actions of the orthogonal groups on their natural modules. -The 2-point stabilizers have 4 orbits, two of length 1. The other two are the sets of points that are perpendicular to or not perpendicular to the fixed isotropic point. They apparently have lengths $2^{2m-2}$ and $2^{2m-2} \pm 2^{m-1}$. (That last claim is based on computations with $m \le 6$, but I expect I could work it out if obliged to!) -Maurizio: To answer your query, I did all of the calculations by hand, except for the lengths of the orbits of the 2-point stabilizers, but since they are just the numbers of isotropic vectors that are or are not orthogonal to a given isotropic vector, this should be a routine calculation. -I will try and find a reference for information about these 2-transitive actions of ${\rm Sp}_{2m}(2)$. The easiest way to construct them on a computer is as the action on the cosets of their orthogonal maximal subgroups. This was straightforward in Magma, because the standard functions constructing these groups resulted in the orthogonal groups already being subgroups of the symplectic groups. Irritatingly, in GAP, the standard functions produced groups with different invariant bilinear forms, although one could be conjugated to the other by a permutation matrix. -Here is a GAP calculation to construct the 136-degree 2-transitive representation of ${\rm Sp}_8(2)$. -gap> G:=Sp(8,2);; -gap> H:=SpecialOrthogonalGroup(1,8,2);; -gap> Display(InvariantBilinearForm(G).matrix); -. . . . . . . 1 -. . . . . . 1 . -. . . . . 1 . . -. . . . 1 . . . -. . . 1 . . . . -. . 1 . . . . . -. 1 . . . . . . -1 . . . . . . . -gap> Display(InvariantBilinearForm(H).matrix); -. 1 . . . . . . -1 . . . . . . . -. . . 1 . . . . -. . 1 . . . . . -. . . . . 1 . . -. . . . 1 . . . -. . . . . . . 1 -. . . . . . 1 . -gap> P:=PermutationMat((2,8)(4,6),8,GF(2));; -gap> HP:=H^P;; -gap> IsSubgroup(G,HP); -true -gap> I := Image(FactorCosetAction(G,HP));; -gap> LargestMovedPoint(I); -136 -gap> Orbits(Stabilizer(I,1)); -[ [ 2, 17, 40, 25, 45, 21, 68, 36, 41, 14, 29, 75, 96, 27, 24, 130, 102, 70, - 66, 73, 88, 23, 49, 43, 38, 11, 20, 56, 92, 42, 13, 31, 30, 133, 93, - 136, 72, 8, 100, 131, 115, 129, 123, 9, 47, 52, 86, 28, 19, 89, 116, - 51, 74, 105, 67, 22, 34, 6, 76, 54, 81, 7, 64, 44, 37, 3, 26, 135, 85, - 108, 132, 78, 58, 61, 134, 94, 107, 113, 125, 84, 118, 104, 127, 5, 33, - 98, 50, 32, 46, 114, 90, 77, 10, 48, 122, 69, 35, 18, 65, 57, 12, 99, - 117, 83, 4, 55, 106, 60, 82, 59, 128, 87, 126, 109, 95, 62, 79, 111, - 121, 53, 120, 80, 97, 101, 91, 71, 16, 112, 39, 15, 103, 119, 63, 124, - 110 ] ] -gap> Orbits(Stabilizer(I,[1,2],OnTuples)); -[ [ 3, 17, 54, 22, 50, 15, 21, 7, 42, 56, 51, 12, 18, 62, 52, 25, 23, 8, 11, - 71, 44, 53, 46, 69, 49, 36, 20, 16, 68, 65, 58, 70, 38, 29, 24, 9, 19, - 4, 55, 35, 13, 28, 63, 41, 48, 72, 67, 61, 33, 40, 5, 14, 26, 64, 37, - 39, 60, 59, 34, 31, 10, 27, 6, 45, 43, 66, 32, 30, 47, 57 ], - [ 73, 98, 117, 77, 122, 93, 101, 104, 133, 75, 110, 119, 81, 88, 78, 103, - 127, 126, 107, 90, 89, 113, 102, 95, 99, 114, 123, 94, 136, 132, 131, - 79, 135, 80, 108, 112, 92, 76, 83, 84, 86, 87, 106, 97, 134, 85, 116, - 100, 124, 118, 111, 115, 105, 121, 96, 128, 91, 120, 129, 130, 125, 74, - 82, 109 ] ]<|endoftext|> -TITLE: Subcategories which still give a Yoneda embedding -QUESTION [10 upvotes]: If $\mathbf{C}$ is a category, then the Yoneda functor which sends $a$ to $Hom_\mathbf{C}(-,a)$ is a fully faithful embedding of categories -$$ \mathbf{C}\rightarrow \mathbf{Func}(\mathbf{C}^{op},\mathbf{Set})$$ -Given any subcategory $\mathbf{B}\subseteq \mathbf{C}$, there is a similar functor -$$ \mathbf{C}\rightarrow \mathbf{Func}(\mathbf{B}^{op},\mathbf{Set})$$ -which restricts $Hom_\mathbf{C}(-,a)$ to arguments in $\mathbf{B}$. -This functor need not be an embedding in general, but there are many examples where it is, and where its an interesting statement that it is. Examples: - -The category of finite sets, inside the category of sets; or more generally, any subcategory of set containing a non-empty set will work. -The category of sets and maps which factor through a finite set, inside the category of sets (so $\mathbf{B}$ need not be full). -The category of affine schemes (ie, $\mathbf{Comm}^{op}$), inside the category of schemes. -The category of open subsets of $\mathbb{R}^n$ and smooth maps between them, inside the category of smooth $n$-dimensional manifolds. -The category of abelian groups, inside the category of groups. - -My question is, is there a name or a nice characterization of this subcategories? I'm writing up some notes for a class this semester, and I want to make a remark to this effect. In the absence of a good name, I was going to call them `Yoneda subcategories'. - -REPLY [14 votes]: Let $B$ be a full subcategory of $C$. The condition that $C \to \mathrm{Set}^{C^{op}} \to \mathrm{Set}^{B^{op}}$ is fully faithful is easily seen to be equivalent to the condition that for every $c \in C$ the set of all morphisms from objects in $B$ to $c$ is a colimit diagram. In other words, $c$ is the colimit of the canonical diagram $(B \downarrow c) \to B \to C$. Therefore one then calls $B$ a dense subcategory of $C$. If you just require that every $c$ is the colimit of some diagram which factors over $B$, then $B$ is called colimit-dense. This is a weaker condition: $\{R\}$ is not dense in $\mathrm{Mod}(R)$, but it is colimit dense. Note that already $\{R \oplus R\}$ is dense in $\mathrm{Mod}(R)$. -You mention that the open subsets of $\mathbb{R}^n$ constitute a dense subcategory of the smooth $n$-manifolds. But actually already $\mathbb{R}^n$ (together with all its endomorphisms) suffices since this is also the local model for the open subsets of $\mathbb{R}^n$. Consequently, we get a fully faithful embedding from the category of smooth $n$-manifolds into the category of right $M$-sets, where $M$ is the monoid of all continuous maps $\mathbb{R}^n \to \mathbb{R}^n$. I don't know if this is of use at all, but I think this is an interesting point of view. It is a sort of algebraic representation from a geometric category of interest, very similar to the functor of points approach in algebraic geometry (where we take affine schemes as a dense subcategory). -You can find more about dense subcategories at the nlab and in "Abstract and Concrete Categories - The Joy of Cats" (Examples 2.11 + Exercise 12.D, online). The definition also plays a central role in the definition of locally presentable categories; see the book by Adamek and Rosicky. -Yoneda's Lemma now just asserts that $C$ is dense in $C$, what else should we expect? There are other formal similarities with the topological notion of the same name: Let $B \subseteq D \subseteq C$ be full subcategories. If $B \subseteq C$ is dense, then $D \subseteq C$ is dense. [If $B \subseteq D$ and $D \subseteq C$ are dense, then $B \subseteq C$ is dense. -- This is not correct, see comment.]<|endoftext|> -TITLE: $\ell$-adic Weil cohomology theory -QUESTION [8 upvotes]: I have a reference or counterexample request. Suppose $k$ is a field and $\ell\neq char(k)$. There are several common references that show that $H^i_{et}(-, \mathbb{Q}_\ell )$ is a Weil cohomology theory when $k$ is separably closed. -I've spent several hours skimming through Milne's Etale Cohomology, the 1994 Motives volume, SGA articles, online searches, etc and I can't seem to determine whether or not $\ell$-adic cohomology forms a Weil cohomology theory when you don't assume you are in some "geometric" situation by making assumptions on the field. -Is there a reference that proves this is still a Weil cohomology theory or is it just false in this case? Thanks. -(This might be in SGA somewhere, but my skimming of French is rather slow and any specific related statement I find tends to throw in being over an algebraically closed field.) -Edit/Update: Everyone is commenting on the non-finitely generatedness, so I'll be more specific. That isn't really the interesting thing to me. Do you still have some sort of cycle class map that behaves nicely (functorially)? For instance, that paper Timo listed seems to imply that as long as finiteness is satisfied when you plug in a particular variety, everything else seems to be fine, but I haven't had time to seriously look at it yet. - -REPLY [7 votes]: See e.g. the first page of Uwe Jannsen, Continuous étale cohomology: http://www.springerlink.com/content/u526616k42051570/ (For varieties over number fields, the cohomology groups in question will in general not be finite dimensional.)<|endoftext|> -TITLE: Alternating sums of GCDs -QUESTION [6 upvotes]: The exciting question on alternating sums of binomial coefficients triggered me to ask the following much simpler question (sorry if it is too simple, but I'm not a number theorist, so I must be overlooking something obvious). -Consider the sum: -$$s_n := \sum_{i=1}^n (-1)^{i-1}\gcd(n+1,i).$$ -Simple numerical experiments suggest the following: - -If $n=2p-1$ for a prime $p$, then $s_n = 1$ [Edit] as Kevin observed, this one follows immediately, because the gcd is either $1$, $2$, or $p$, and it is easy to see which term contributes which part (even, odd, etc.)) -If $n$ is an odd number [edit: not of the form $2p-1$, prime $p$] (see A166257), then $s_n < 0$ -otherwise, of course, $s_n=0$. - -How should I prove this? Observe that the $\alpha$-generalization of $s_n$ (i.e., summing $\mbox{gcd}^\alpha$ terms instead) does not satisfy the above observations. - -REPLY [9 votes]: One can compute this sum explicitly. Let $n+1=2^a \prod_i p_i^{\alpha_i}$ with $a\geq 1$, then we have: -$$\sum_{i=1}^n (-1)^{i-1}\mbox{gcd}(n+1,i)=(-1)^{n-1}(n+1)-a2^{a-1}\prod_i\left((\alpha_i+1)p_i^{\alpha_i}-\alpha_ip_i^{\alpha_i-1}\right)$$ -as was proved in - -Laszlo Toth, "Weighted Gcd-Sum Functions", Journal of Integer Sequences, Vol. 14 (2011) - -So $s _n = (n+1) \left( (-1)^{n-1} -\frac{a}{2}\prod _i (1+\frac{\alpha _i(p _i-1)}{p _i})\right)$. It is easy to see from here that if $n$ is odd, then $s_n$ is positive only if $n=2p-1$. - -REPLY [5 votes]: Actually both (2) and (3) are true, and there is a relatively short proof: -First of all we have that -$$s_n=\sum_{d|n+1, d< n+1}d\sum_{i=1, (i, (n+1)/d)=1}^{(n+1)/d-1}(-1)^{id-1}.$$ -Sorry for the bad typesetting, the compiler here doesn't seem to like \substack. Ok to be sure this is clear, the inner sum is over $i$ running from $1$ to $(n+1)/d-1,$ which are coprime to $(n+1)/d$. -First notice for later reference that when $d$ is even the inner sum over $i$ is negative. On the other hand if $d$ is odd then the inner sum is equal to -$$\sum_{i=1, (i, (n+1)/d)=1}^{(n+1)/d-1}(-1)^{i-1}.$$ -Furthermore if $(n+1)/d$ is odd then the map $i\mapsto (n+1)/d-i$ is a bijection between integers coprime to $(n+1)/d$, and it reverses parity. Therefore when both $d$ and $(n+1)/d$ are odd we have that -$$\sum_{i=1, (i, (n+1)/d)=1}^{(n+1)/d-1}(-1)^{i-1}=0.$$ -Plugging this in above proves (3) right away. -To prove (2) we write $n+1=2^km$ with $m$ odd. Suppose that $d|(n+1),~d<(n+1)/2,$ and $d$ is odd. Then $2^k d|(n+1)$ and $2^k d < n+1,$ and pairing the contribution in the above sum from $d$ with that from $2^kd$ gives -$$d\sum_{i=1, (i, (n+1)/d)=1}^{(n+1)/d-1}(-1)^{i-1}-2^kd\sum_{i=1, (i, (n+1)/2^kd)=1}^{(n+1)/2^kd-1}1$$ -$$\le d\varphi \left( 2^k\cdot\frac{n+1}{2^kd}\right)-2^kd\varphi \left(\frac{n+1}{2^kd}\right)$$ -$$=-2^{k-1}d\varphi \left(\frac{n+1}{2^kd}\right)<0.$$ -As we mentioned before, all of the even divisors $d$ contribute a negative amount to $s_n$, so everything is accounted for except for the case when $d=(n+1)/2$ is odd. In this case we have that $k=1$, and the term in the top sum corresponding to $d=(n+1)/2$ is -$$ \frac{n+1}{2}\sum_{i=1, (i,2)=1}^1 (-1)^{i-1} = \frac{n+1}{2} . $$ -However using our previous calculation the sum of contributions from all of the other odd divisors $e<(n+1)/2$, paired with the divisors $2e$, is -$$\le -\sum_{e|(n+1)/2,e<(n+2)/2}e\varphi\left(\frac{n+1}{2e}\right)$$ -$$=-\frac{n+1}{2}\cdot\sum_{e|(n+1)/2,e>1}\frac{\varphi (e)}{e}$$ -$$=-\frac{n+1}{2}\left(\prod_{p^a||(n+1)/2}\left(1+a(1-1/p)\right)-1\right).$$ -Now if $(n+1)/2$ is composite then either $a>1$ for one of the primes above, or there is more than one prime in the product above (also if you are trying to follow this calculation then note that all primes in the product above must be odd). In either case it is trivial to check that the quantity above is strictly less than $-(n+1)/2$, therefore $s_n<0$. -So to summarize, the even divisors contribute a negative amount to the overall sum, we pair every odd divisor $d$ with the contribution from $2^kd$ to get a negative quantity, and we combine these quantities with the divisor $(n+1)/2$ if it is odd. That finishes the proof.<|endoftext|> -TITLE: Relative commutants of abelian von Neumann algebras -QUESTION [22 upvotes]: This question arose from the discussion over at the question Centralizers in $C^*$-algebras. -Which von Neumann algebras $N$ satisfy the property that $A' \cap N = B' \cap N \implies A = B$, for all commutative von Neumann subalgebras $A, B \subset N$? -Note that $N = \mathcal B(\mathcal H)$ has this property by von Neumann's double commutant theorem, and perhaps this property characterizes $\mathcal B(\mathcal H)$. It is clear that $N$ must be a factor by considering $A = \mathbb C$ and $B = \mathcal Z(N)$. If $\mathbb F_2 = \langle a, b \rangle$ is the free group on two generators, then by considering the Fourier expansion of elements in $L\mathbb F_2$ it is not hard to see that for $A = L\langle a \rangle$ and $B = L\langle a^2 \rangle$ we have $A' \cap L\mathbb F_2 = B' \cap L\mathbb F_2$ thus $L\mathbb F_2$ does not have this property. -Also note that if we were to consider the case when $A$ and $B$ are allowed to be non-commutative then relevant is Corollary 4.1 in Popa's paper On a Problem of R.V. Kadison on Maximal Abelian $*$-Subalgebras in Factors which shows that every type $II$ factor $N$ contains a hyperfinite subfactor $R$ such that $R' \cap N = \mathbb C$. - -REPLY [3 votes]: I learned recently that (see here) Popa proved that every separable II$_1$ factor $M$ contains (an embedding of) the hyperfinite von Neumann algebra $R$ such that $L^2M\ominus L^2R\cong _RL^2(R\overline{\otimes}R^{op})^{\oplus \infty}_R$. -It is a standard fact that this implies $A'\cap M\subseteq R$ for every diffuse subalgebra $A\subseteq R$. -Then, as in $R$, there are diffuse abelian von Neumann subalgebras $A\neq B$ such that $A'\cap R=B'\cap R$. Therefore, by the above result, we also have $A'\cap M=B'\cap M$. -For example, write $R=L(\mathbb{Z}\wr \mathbb{Z})=L(\langle t\rangle \wr \langle s\rangle)$, $A=L(\langle s\rangle)$ and $B=L(\langle s^2\rangle)$. -Or write $R=L((\mathbb{Z}/2\mathbb{Z})G)\rtimes G$, $A=L((\mathbb{Z}/2\mathbb{Z})G)$ and $B=L(\{x\in (\mathbb{Z}/2\mathbb{Z})G: x_{e_G}=\bar{0}\})$. -In both examples, we have that $A'\cap R=B'\cap R=A\neq B$. - -A few comments: -(1) Following Kadison's paper, a von Neumann subalgebra $A\subseteq M$ is called normal if $(A'\cap M)'\cap M=A$. There are several old papers (see e.g. Anastasio's paper) giving concrete examples of the most extreme case of non-normal abelian subalgebras, i.e. thick subalgebras following Bures's book here. Recall $A\subseteq M$ is thick if $A'\cap M$ is a masa in $M$; equivalent characterizations can be found in Lemma 10.1 of this book. -(2) If $A$ is abelian, then since $(A'\cap M)'\cap M=\cap_BB$, where $B$ is a masa in $M$ containing $A$, we know that the above question is the same as asking whether there is an abelian von Neumann subalgebra $A\subset M$ such that $A$ is not normal, e.g. $A\neq A'\cap M$ and $A$ is thick. -(3) Let $B\subseteq M$ be a masa. Then every diffuse proper von Neumann subalgebra of $B$ is NOT normal iff $B$ satisfies the disjointness property, i.e. if $C\subset M$ is a masa with $B\cap C$ being diffuse, then $B=C$. -(4) Popa's result implies every separable II$_1$ factor $M$ contains a mixing masa, as we can take it to be a mixing masa in $R$, where $R\hookrightarrow M$ is the above mixing inclusion. Inside a mixing masa, every proper (diffuse) von Neumann subalgebra is thick. It seems not clear whether one can always find proper thick subalgebras inside every singular, equivalently weakly mixing, masa in a separable II$_1$ factor.<|endoftext|> -TITLE: Atiyah-Patodi-Singer Eta invariant and Chern-Simons form -QUESTION [18 upvotes]: I am trying to understand the Atiyah-Patodi-Singer index theorem in the case of Dirac operators in four dimensions. I have three questions about the eta invariant: -1) Is eta a topological invariant (or geometric invariant)? -2) Which is its relation with the three dimensional Chern-Simons form? -3) In how many non-trivial cases the eta invariant is explicitly calculable? - -REPLY [9 votes]: Ad (3). Computing $\eta$-invariants on the nose is notoriously difficult. However, sometimes one needs $\eta$-invariants as ingredients of other differential topological invariants ($\rho$-invariants as in Pauls answer are an example. The Eells-Kuiper invariant is another one of a slightly different flavour). In this case, one can often compute $\eta$-invariants up to some correction terms, which are more accessible. Let me sketch a few things that work sometimes. Most, but not all of the following is explained in this overview article. Most of the following is applicable to 3-manifolds. - -Direct computation from the spectrum of the operator. This works in very few cases where one does not expect $\eta(D)=0$ from the very beginning. The first problem is that one needs to know the spectrum. Hitchin did this for Berger spheres in his phd thesis. Somewhat related are the papers by Millson and Moscovici-Stanton that relate $\eta$-invariants of locally symmetric spaces (including hyperbolic 3-manifolds) to $\zeta$-functions associated with the geodesic flow. -Using the Atiyah-Patodi-Singer theorem. Try to find an explicit cobordism to a manifold where the $\eta$-invariant is known, and apply APS. Kruggel used this method to compute Kreck-Stolz invariants of Eschenburg spaces. -If $M$ admits a finite normal covering $M=\tilde M/\Gamma$ and the $\Gamma$-equivariant $\eta$-invariant of $\tilde M$ is known, one can recover $\eta$- and $\rho$-invariants of $M$. The computation of $\eta$-invariants for lense spaces is a combination of (2) and (3). -Find a similar operator $\tilde D$ on the same vector bundle, such that $\eta(\tilde D)$ is computable, then try to access the difference. This has been done by Deninger and Singhof for Heisenberg manifolds, or here for the cubical Dirac operator on quotients of compact Lie groups. -If the manifold $M$ happens to be a fibre bundle over a base $B$, one can use the adiabatic limit technique of Bismut-Cheeger and Dai. This also works for Seifert fibrations, see also Liviu's answer. However, one needs to know the $\eta$-forms of a family of fibrewise operators, and one has to assume that their kernels form a vector bundle over $B$. -If $M=M_-\cup_{M_0}M_+$ with $M_0$ totally geodesic, one can use a gluing theorem, see Kirk-Lesch. Now one needs to know the $\eta$-invariants of the pieces with respect to suitable boundary conditions, and a contribution from $M_0$ relating those boundary conditions. This technique has been applied here. - -I am happy to add to this list, just leave a comment.<|endoftext|> -TITLE: why we need rigid geometry? -QUESTION [16 upvotes]: Hello, everyone. -I want to ask some questions about rigid geometry. -1.what is the motivation of rigid geometry? -2.what is the applications of rigid geometry for solving arithmetic problems, especially for studying the fundamental groups of algebraic curves? what the beautiful theorems which were first proved by rigid geometry method? -thank you very much - -REPLY [3 votes]: It is known that both the Jacquet-Langlands correspondence and the local Langlands correspondence for GLn can be realized in the étale cohomology of a Lubin-Tate tower (or, more precisely, in the étale cohomology of the Berkovich space attached to the rigid analytic space which is the generic fibre of the Lubin-Tate tower).<|endoftext|> -TITLE: Rearrangement-style inequality with lots of terms and little evidence -QUESTION [7 upvotes]: This is another of the problems I designed for contests but wasn't able to solve on my own for years. Maybe AoPS is a better place for it, but let me try it here. -[UPDATE: I have streamlined the exposition after zeb's wonderful proof of my conjectures. Everything stated below as a "conjecture" is true. Note that some comments, as well as fedja's and Suvrit's answers below, refer to an older version of these conjectures, which was false.] -Let $n\in\mathbb N$ be $\geq 2$, and let $a_1$, $a_2$, ..., $a_n$ and $b_1$, $b_2$, ..., $b_n$ be nonnegative reals such that $a_1\geq a_2\geq ...\geq a_n$ and $b_1\geq b_2\geq ...\geq b_n$. Let $A_n$ denote the $n$-th alternating group, and $S_n$ denote the $n$-th symmetric group. We use the symbol $-$ for set-theoretical complement (since the backslash means something else in group theory). -Product-sum conjecture. Then, -$\left(-1\right)^{\binom{n}{2}}\left(\prod\limits_{\pi\in A_n} \left(\sum\limits_{k=1}^n a_kb_{\pi\left(k\right)}\right) - \prod\limits_{\pi\in S_n - A_n} \left(\sum\limits_{k=1}^n a_kb_{\pi\left(k\right)}\right) \right) \leq 0$. -Sum-maximum conjecture. We have -$\left(-1\right)^{\binom{n}{2}}\left(\sum\limits_{\pi\in A_n} \max\left\lbrace a_k + b_{\pi\left(k\right)} \mid k\in\left\lbrace 1,2,...,n\right\rbrace \right\rbrace - \sum\limits_{\pi\in S_n - A_n} \max\left\lbrace a_k + b_{\pi\left(k\right)} \mid k\in\left\lbrace 1,2,...,n\right\rbrace \right\rbrace \right) \leq 0$. -zeb has proven both of these conjectures (I am still interested in an analysis-free proof, but rather convinced that zeb's is the natural one). Here are some easy observations: - -If the Product-sum conjecture holds, then so does the Sum-maximum one, by the standard "tropical geometry" trick (replace $a_k$ by $x^{a_k}$, replace $b_k$ by $x^{b_k}$, and watch the asymptotics of the sides while $x$ goes to $\infty$). -Both conjectures hold for $n\leq 3$ for very simple reasons. -By the same argument as in the proof of Cauchy's and Vandermonde's determinants, the difference -$\prod\limits_{\pi\in A_n} \left(\sum\limits_{k=1}^n a_kb_{\pi\left(k\right)}\right) - \prod\limits_{\pi\in S_n - A_n} \left(\sum\limits_{k=1}^n a_kb_{\pi\left(k\right)}\right) $ -is divisible (as a polynomial) by $\prod\limits_{1\leq i < j\leq n}\left(a_i-a_j\right) \prod\limits_{1\leq i < j\leq n}\left(b_i-b_j\right)$. The question is whether the quotient has the same sign as $\left(-1\right)^{\binom{n}{2}}$. I wouldn't be surprised if it is even a polynomial with all coefficients having that same sign, and maybe even Schur-positive times $\left(-1\right)^{\binom{n}{2}}$, whatever this means for symmetric polynomials in two sets of indeterminates. (For $n\leq 3$, this quotient is $1$.) - -REPLY [6 votes]: Ok, I have a functional generalization of your Product-Sum conjecture using a very simple method. -Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be any function with a nonnegative $\binom{n}{2}$th derivative. I claim that we have the following functional inequality: -$\sum_{\pi\in S_n} (-1)^{\sigma(\pi)}f(\sum_i a_ib_{\pi(i)}) \ge 0$. -Plugging in $f(x) = -(-1)^{\binom{n}{2}}\log(x)$, we see that your inequality holds as long as $\sum_i a_ib_{n+1-i} \ge 0$. -To prove the functional inequality, it clearly suffices to prove it in the case that the $b_i$s are all distinct positive integers, so assume from now on that this is the case. Let $x_i = e^{a_i}$. Note first that in the special case in which $f(x) = e^x$, we get that -$\sum_{\pi\in S_n} (-1)^{\sigma(\pi)}\prod_ix_i^{b_{\pi(i)}} = \det((x_i^{b_j})_{i,j})$ -which is $\prod_{i\lt j}(x_i-x_j)$ times the Schur polynomial $s_{(b_1-n+1,...,b_n)}(x_1,...,x_n)$, and as is well known Schur polynomials have all of their coefficients nonnegative. For every monomial $x^m = \prod_i x_i^{m_i}$, let $c_m$ be its coefficient in the Schur polynomial $s_{(b_1-n+1,...,b_n)}(x_1,...,x_n)$. -Next, let $S_a$ be the shift operator - i.e., let $S_a(f)(x) = f(x+a)$. Then it is easy to check that we have -$\sum_{\pi\in S_n} (-1)^{\sigma(\pi)}f(\sum_i a_ib_{\pi(i)}) =\sum_mc_m(\prod_{i\lt j}(S_{a_i}-S_{a_j}))(f)(\sum_ia_im_i)$, -which is nonnegative since we have $(\prod_{i\lt j}(S_{a_i}-S_{a_j}))(f)(x) \ge 0$ for any $x$ and any function $f$ with nonnegative $\binom{n}{2}$th derivative.<|endoftext|> -TITLE: References to SGA 8 and descent theory -QUESTION [9 upvotes]: In Geometric Invariant Theory, by Mumford, Fogarty, and Kirwan, if there is a mention of descent theory, it almost always comes along with a reference to SGA 8, Theorem 5.2 (see the end of the proof of prop 6.9 on page 119, for example). Now I know SGA 8 was never made, but I was wondering: - -Does anyone have a good guess as to what this theorem should say? -Does anyone have a good reference for a quick and "hands off" introduction to descent theory? I am really just looking to understand the "gist" of it. - -REPLY [5 votes]: For question 1, see the comment above. -Collecting the answers to question 2: - -Grothendieck's original FGA, starting with TDTE I -Vistoli's chapter in FGA explained, for the connection with stacks -What is descent theory? for a very short overview -Bosch-Lütkebohmert-Raynaud, Néron Models (recommended by BCnrd in the above-mentioned thread) -Waterhouse, Introduction to Affine Group Schemes, containing a 20-page introduction primarily concerned with the affine case - -"Community wiki" post, feel free to modify.<|endoftext|> -TITLE: Surjective Maps onto $\aleph$-numbers -QUESTION [6 upvotes]: We denote by $\frak p\le q$ the abbreviation that there is $f:\frak p\to q$ which is injective, and by $\frak p\le^\ast q$ we abbreviate that there is a surjection from $\frak q$ onto $\frak p$. -If $X$ is a set in a universe of ZF, denote by $H(X)=\min\lbrace\alpha\mid\alpha\nleq X\rbrace$ known as The Hartog number of $X$. It is clear that $H(X)\nleq X$. -Under the axiom of choice $X\leq Y\iff X\leq^\ast Y$, however without the axiom of choice the $\leq^\ast$ order can behave quite strangely. For example, it is possible that $\omega_1\nleq2^\omega$, but it is always true that $\omega_1\leq^\ast2^\omega$. This may occur when the continuum is a countable union of countable sets, or if we are in a Solovay-like model. -Monro showed in [1] that it is consistent to have for every $\kappa$ an infinite set $p$ such $H(p)=\omega$ and $\kappa\leq^\ast p$. -In those two instances we have examples where the axiom of choice fails, and $H(x)\leq^\ast x$. If we do not wish to add large cardinal into the mix both the examples above are such that the axiom of countable choice already fails in. -Questions: - -How much choice can we have while still finding sets with the property $H(x)\leq^\ast -x$? -How consistent is this property with the different type of weak choice principles (e.g. if it is consistent with DC/AC/W for some $\kappa$, would it also be consistent with BPI and its relatives)? - - -Bibliography: - -G.P. Monro, Independence results concerning Dedekind finite sets. Journal of the Australian Mathematical Society (Series A) (1975), 19 : pp 35-46. - -REPLY [4 votes]: I originally posted this question in hope that someone else knew of a reference for an answer, however it seemed to me that indeed the best way is to solve this on my own. I tried to imitate Monro's proof, to a certain extent, and I believe that I have succeeded: -$\renewcommand{\Dom}{\operatorname{Dom}}\renewcommand{\Htg}{\operatorname{Htg}}$ -Let $M$ be a model of ZFC, $\kappa$ is an uncountable regular. Consider the notion of forcing $(P,\le)$ where $p\in P$ is a function from a subset of $\kappa\times\kappa$ into $2$, and $|\Dom(p)|<\kappa$. -Let $M[G]$ be a generic extension, $G_i = \lbrace\alpha<\kappa\mid\exists p\in G: p(i,\alpha)=1\rbrace$ and $H=\lbrace G_i\mid i<\kappa\rbrace$. We give them "canonical" names, $\dot G_i = \lbrace\langle p,\check n\rangle \mid p(i,n)=1\rbrace$ and $\dot H=\lbrace\langle 1, \dot G_i\rangle\mid i < \kappa\rbrace$. -We now proceed to define the symmetric model. If $\pi$ is a permutation of $\kappa$ we can think of $\pi$ as an automorphism of $P$ by: $$\pi p =\lbrace \langle\pi i,\alpha,j\rangle\mid \langle i,\alpha,j\rangle\in p\rbrace$$ -Let $\mathscr G$ be the permutation group of these automorphisms. Let the subgroups filter be generated by countable support, that is $\dot f$ is a symmetric name if and only if there is a countable set of conditions such that whenever they are fixed pointwise, $\dot f$ is fixed. -We note two things: - -For $\pi$ fixes (pointwise) countably many conditions then only $<\kappa$ many points must be fixed by $\pi$. -For $i<\kappa$ we have $\pi\dot G_i = \dot G_{\pi i}$, so all the $\dot G_i$ are symmetric and trivially hereditarily symmetric. From this follows that $\dot H$ is also hereditarily symmetric. - -Let $N$ be the symmetric model, we will show the following: - -$H\in N$, which is obvious since $\pi(\dot G_i)=\dot G_{\pi i}$ so $\pi(\dot H)=\dot H$ for every $\pi\in\mathscr G$. -$N\models DC$: -Let $S\in N$ be a non-empty set and $R\in N$ a binary relation defined on all elements of $S$. In $M[G]$ we have a function which witnesses $DC$. For every $n\in\omega$ we have $n,f(n)\in N$ therefore $\langle n,f(n)\rangle\in N$. Each ordered pair has a countable support, and the union of this countable collection is countable (in $M$) therefore supports $\dot f$ as a symmetric name, as wanted. -$N\models\Htg(H)=\aleph_1$: -Assume towards contradiction that $\dot f$ is a symmetric name such that for some $p_0\in G$ we have $p_0\Vdash\dot f:\check\omega_1\to H\text{ injective}$. Let $E$ be a support for $\dot f$. There exists $i<\kappa$ that for every $\alpha$ and $p\in E$ the pair $\langle i,\alpha\rangle\notin\Dom(p)$. -Let $p\le p_0$ be an extension of $p$ such that for some $\alpha<\omega_1$ we have: $$p\Vdash \dot f(\check\alpha)=\dot G_i$$ -We choose now $j<\kappa$ which for every $\alpha$ and $q\in E\cup\lbrace p\rbrace$ we have $\langle j,\alpha\rangle\notin\Dom(q)$, and define $\pi$ to be induced by the 2-cycle: $(i\ j)$. -We have that $\pi$ fixes $E$ pointwise, since neither $i$ nor $j$ appear in relevant coordinates in the conditions of $E$; we have that $\pi p$ and $p$ are compatible since if $x$ is in the shared domains then $i$ and $j$ do not appear in relevant positions and so the value given to $x$ by the two conditions is similar. Lastly $\pi p\Vdash (\pi\dot f)(\pi\check\alpha)=\pi(G_i)$, since $\pi$ fixes $\dot f$ and $\check\alpha$ we have: $$\pi p\Vdash\dot f(\check\alpha)=\dot G_{\pi i}$$ -Thus we arrive the wanted contradiction since $\pi p\cup p$ extends $p_0$, but it contradicts $p_0$ by forcing that $\dot f$ is not a function. -$N\models\kappa\leq^\ast H$: -We define from $f:H\to\kappa$ by $f(G_i) = \min\{\beta\mid\beta\notin G_i\}$. Since $G_i\in N$ for all $i$ and $H\in N$ we have that indeed $f\in N$. Now we only have to show that it is onto $\kappa$, this is a simple genericity argument: -Given $\beta<\kappa$, and a condition $p\in P$ there exists $i<\kappa$ such that $\langle i,\alpha\rangle\notin\Dom(p)$ for all $\alpha$. Therefore $p\cup\lbrace\langle i,\alpha,1\rangle\mid\alpha<\beta\rbrace\cup\lbrace\langle i,\beta,0\rangle\rbrace\le p$ as wanted. By genericity there is some $i<\kappa$ such that $f(G_i)=\beta$. - - -In this model $N$ described above we have $DC$ therefore we have no infinite Dedekind finite sets, however we have a non-well orderable set which can be mapped onto an ordinal (possibly) much higher than its Hartog number. I suspect that this easily generalized to higher cardinalities as well.<|endoftext|> -TITLE: Terminology of "covariant derivative" and various "connections" -QUESTION [5 upvotes]: I apologize for the long question. My ultimate aim is to understand the existing terminology and find appropriate terminology for covariant derivatives on vector bundles which generalizes to "covariant" derivatives on fiber bundles (a linear Ehresmann connection is to a [linear] covariant derivative as a nonlinear Ehresmann connection is to [fill in the blank]). Or if such terminology already exists, to find out what it is. -A covariant derivative on a vector/tensor bundle $E \to M$ is an $\mathbb{R}$-linear map of the form $\nabla \colon \Gamma(E) \to \Gamma(E \otimes T^*M)$. As I understand it, the "covariant" part of this comes from the fact that the $T^*M$ component changes covariantly under coordinate changes and not how the $E$ component changes. Is this correct? A pedantic followup question is prompted by some inconsistency I've found in various literature: Must $E$ be a tensor bundle induced from tangent/cotangent vector bundles for the "covariant" qualifier to still apply? -The motivation for the qualifier "covariant" seems to ultimately stem from coordinate-based definitions and considerations. I've also seen uses of "covariant" to mean "independent of coordinate choice" in a more abstract setting. Is this also valid? Does the notion and terminology of a Koszul connection supplant the coordinate-minded covariant derivative? I'm guessing that the $\nabla$ formulation as a map producing a section of a tensor bundle came from Koszul. -Linear Ehresmann connections are in one-to-one correspondence with covariant derivatives/Koszul connections, and there is a notion of a nonlinear Ehresmann connection on a fiber bundle. I have come up with a corresponding definition for a "nonlinear covariant derivative/Koszul connection" on a fiber bundle, which has as a natural example in the fiber bundle $\Pr_2^{S\times M} \colon S \times M \to M$, where $M$ and $S$ are smooth manifolds (noting that $\Gamma(\Pr_2^{S\times M})$ can be identified naturally with $C^\infty(M,S)$). Under certain natural identifications, if $\nabla$ denotes this nonlinear operator (definition left out to avoid clutter) and $\phi \in C^\infty(M,S)$, then $\nabla \phi = T \phi$, i.e. this connection/derivative is the tangent map operator. This suggests the question: Can the tangent map operator be considered a covariant derivative? It is certainly independent of any coordinate choice. -Apart from the questions interspersed throughout the text above, my main question is: Is there already terminology for this "nonlinear covariant derivative". If not, would "nonlinear covariant derivative" or "nonlinear Koszul connection" be appropriate for such? The proliferation of "connections" (e.g. Levi-Civita, Cartan, Ehresmann, Koszul, affine, etc.) suggests that "connection" is preferred over "derivative", is this correct? - -REPLY [4 votes]: As I understand it, the "covariant" - part of this comes from the fact that - the T∗M component changes covariantly - under coordinate changes and not how - the E component changes. Is this - correct? - -Yes. - -The motivation for the qualifier - "covariant" seems to ultimately stem - from coordinate-based definitions and - considerations. I've also seen uses of - "covariant" to mean "independent of - coordinate choice" in a more abstract - setting. Is this also valid? - -Yes on this one as well. - -Is there already terminology for this - "nonlinear covariant derivative". If - not, would "nonlinear covariant - derivative" or "nonlinear Koszul - connection" be appropriate for such? - -There is a notion of nonlinear connection, even for the tangent bundle. Recall that the auto-parallel curves of the Levi-Civita connection of a metric are the geodesic. Recall also that these curves have a variational characterization, the local minimizers of length. The geodesic equations are the Euler-Lagrange equations of the length functional. The metric is used to define the lagrangian of this functional. More generally, given a lagrangian on a manifold, i.e., a function on the total space of the tangent bundle, the extremals of the resulting functional can be interpreted as the auto-parallel curves of a nonlinear connection on the tangent bundle. It is called nonlinear because the parallel transport defined by this connection is a nonlinear map. Finsler geometry is a special incarnation of this philosophy. The lagrangean in this case is a function whose restriction to each tangent space is a norm on that tangent space.<|endoftext|> -TITLE: Stochastic process describing long-term fluctuations -QUESTION [5 upvotes]: I need to model a process that has large, smooth and mean-reverting long-term fluctuations and some small short term wiggles, a sample path looks like this: - -My first idea was to model it as an integral of an Ornstein-Uhlenbeck process (because the graph of the differences of the process look nicely mean-reverting), but this would not capture the long term fluctuations. The mean-reversion level of the differences should somehow incorporate this information, but I don't know how to achieve it. Any ideas which process would be reasonable? - -REPLY [3 votes]: There are two ways this is commonly handled, both still based on O-U: - -Deterministic long-term fluctuations. -Separate timescale processes - -Let's say your real time series of interest is $Y(t)$. In the first case, often used by econometricians to handle seasonality, one sets up a deterministic process $F(t)$, and then run Ornstein-Uhlenbeck as a variation on that process. That is to say, we define the stochastic variable $X(t)=Y(t)-F(t)$ and then specify -$$ -dX_t = -\alpha_t X_t + \sigma_t dW -$$ -The second case is more physical, and one uses a two-dimensional OU process -$$ -dY^{(1)}_t = -\alpha_t^{(1)} (Y^{(1)}_t - \bar{Y})+ \sigma^{(1)}_t dW_1 -$$ -$$ -dY^{(2)}_t = -\alpha_t^{(2)} Y^{(2)}_t + \sigma^{(2)}_t dW_2 -$$ -with $\alpha_1 \ll \alpha_2$, after which we define $Y=Y^{(1)}+Y^{(2)}$. -The math is all still quite tractable, particularly so with constant $\alpha,\sigma$. -It's worth noting that, with respect to calibration, you can treat the problem as separable to the extent that $\alpha_1 \ll \alpha_2$ is really true. -That is to say, you get a far better fit using a Fourier transform. Use a low-pass filter to obtain data for finding $\sigma^{(1)}$ and $\alpha^{(1)}$. Then, you use a high-pass filter to get $\sigma^{(2)}$ and $\alpha^{(2)}$. -For example assume $\alpha^{(1)}, \sigma^{(1)}$ are roughly daily and $\alpha^{(2)}, \sigma^{(2)}$ are roughly annual. We define -$$ -f(\omega)={\cal{F}}(Y) -$$ -and let -$$ -f^{(1)}(\omega) = f(\omega) \times \mathbb{1} \lbrace \omega<(30d)^{-1} \rbrace -$$ -and -$$ -f^{(2)}(\omega) = f(\omega) \times \mathbb{1} \lbrace \omega\geq(30d)^{-1} \rbrace, -$$ -taking due care of the degeneracies in fourier transforms for real data sets. Now set -$$ -X^{(1)} = \text{Re} \left[ \cal{F}^{-1}\left( f^{(1)} \right) \right] -$$ -and -$$ -X^{(2)} = \text{Re} \left[ \cal{F}^{-1}\left( f^{(2)} \right) \right]. -$$ -Running a standard maximum-likelihood estimate on $X^{(2)}$ will find $\sigma^{(2)}$ and $\alpha^{(2)}$. Subsampling $X^{(2)}$ on, say, the same 30 day interval as the cutoff will give a time series appropriate for estimating $\sigma^{(1)}$ and $\alpha^{(1)}$. -Running these two separate two-dimensional calibrations (say as 2-dimensional maximum-likelihood estimates) is far more stable than a single four-dimensional calibration. You still have to be careful. For example if $\sigma^{(2)}$ is too small you have no hope of finding it in all the noise of the lower-frequency process.<|endoftext|> -TITLE: Self-tightening knot -QUESTION [10 upvotes]: Is there a way, for some finite $L>1$, to tie two length $L$ pieces of rope together, such that any finite force is not enough to pull them apart? -The type of rope I have in mind is something like cylindrical with radius 1, unbreakable, unstretchable, perfectly flexible, non-self-intersecting and has length $L$, but I am open to other models. -Suppose that rope 1 has ends $A$ and $B$, rope 2 has ends $C$ and $D$. Tie $B$ and $C$ together. Pull $A$ and $D$. Is there a knot that holds for every coefficient of friction $e>0$ and every force $F>0$ applied to the two ends? - -REPLY [3 votes]: The answer to your question is obviousely "NO", since you want $L$ to be fixed. -So let me consider the following question instead: - -Given the coefficient of friction $e>0$, is there a knot that holds any force? - -I would bet that the answer to this question is "YES". -Obviously we should have $L\to\infty$ as $e\to 0$. -A right way to proceed would be to take the Ashley's big book of knots suggested by Matt and look for a knot which admits a sequence of iterations of some kind. -Even if you made right guess for iterated knot, actual proof that it holds any force might be difficult. -Now let me explain why I would bet for "YES". -Assume in addition you are allowed to make any metal ring with zero coefficient of friction with the rope. Then this - -(source: psu.edu) -would solve your problem. -The metal ring is marked by black, the green and red ropes alternate; -i.e., they go under and over the black ring in turn. -(The width of the ring should be about 1.<|endoftext|> -TITLE: Presenting Lawvere theories? -QUESTION [6 upvotes]: (Re)Reading Lawvere theories versus classical universal algebra, I was reminded of a question I have had for quite some time: - -What is the best way to present Lawvere theories? - -By present, I mean give a (mathematical) description of the theory, by means of syntax. The Lawvere theory will be the denotation of that syntax. -I believe I understand the technical advantages that they provide over classical universal algebra (basically in terms of proving theorems over the models). My 'problem' is that I am trying to mechanize (parts of) mathematics. And to do that, one needs examples, which means constructing particular theories; more precisely, that means writing something down, in finite terms which denotes a theory. -Lawvere theories seems to largely abstract this away, unlike classical universal algebra, which gives quite explicit tools for writing things down explicitly. With Lawvere theories, I am reduced to figuring out how to present a small category with finite products such that every object is isomorphic to a finite product of copies of a distinguished object x. I have read quite a bit of CT, and I have yet to find a book which gives me tools as convenient as those of universal algebra for that purpose. - -REPLY [4 votes]: Developing @AndrejBauer's suggestion to present them as single-sorted equational theories, you could write something like: - -Let $\mathsf{Grp}$ denote the Lawvere theory with the following - presentation. -Generators: - -$c : 2 \rightarrow 1$ -$e : 0 \rightarrow 1$ -$i : 1 \rightarrow 1$ - -Relations: - -$c(c(x,y),z) \equiv (x,c(y,z))$ -$c(e,x) \equiv x,\;\; c(x,e) \equiv x$ -$c(i(x),x) \equiv e, \;\; c(x,i(x)) \equiv e$ - - -But, perhaps you were looking for something a little deeper. - -REPLY [3 votes]: One way to present a Lawvere theory would be via a finite-product sketch. A finite-product sketch $(\mathcal{A}, \mathbb{L})$ is a small category $\mathcal{A}$ equipped with a collection $\mathbb{L}$ of cones over finite discrete diagrams in $\mathcal{A}$; a model in a category $\mathcal{E}$ is a functor $\mathcal{A} \to \mathcal{E}$ which sends the cones in $\mathbb{L}$ to product cones. A Lawvere theory $\mathcal{T}$ can naturally be regarded as a sketch $(\mathcal{T},\mathbb{L})$ where $\mathbb{L}$ consists of all the finite product cones; the sketch $(\mathcal{T},\mathbb{L})$ has the same models as the Lawvere theory $\mathcal{T}$. -Conversely, every sketch $(\mathcal{A},\mathbb{L})$ ``freely generates" a Lawvere theory $\mathcal{T}$ with the same models. One way to put it is this. The sketches form a 2-category $Sketch$ where a 1-cell $(\mathcal{A},\mathbb{L}) \to (\mathcal{A}',\mathbb{L}')$ is a functor $\mathcal{A} \to \mathcal{A}'$ which sends cones in $\mathbb{L}$ to cones in $\mathbb{L}'$, and 2-cells are natural transformations. Lawvere theories also form a 2-category $Law$ where 1-cells are product-preserving functors and 2-cells are natural transformations. By regarding a Lawvere theory as a sketch, we obtain a fully-faithful 2-functor $Law \to Sketch$, and this 2-functor has a left 2-adjoint. -A single-sorted sketch is a sketch $(\mathcal{A},\mathbb{L})$ equipped with a designated object $A_0$ such that for every object $A \in \mathcal{A}$, there is a cone in $\mathbb{L}$ with vertex $A$ and all of its legs equal to $A_0$. Modifying the notion of 1-cell in $Sketch$ accordingly yields a similar relation to 1-sorted Lawvere theories. -For example, a sketch for the category of groups might consist of the opposite of the category of free groups on $\leq 3$ generators, with all product diagrams indicated. -We could get even more syntactic and, following Barr and Wells (see Chapter 4 - a great resource on sketches) define a sketch to consist merely of a reflexive graph $\mathcal{G}$ eqipped with certain diagrams $\mathbb{D}$ in the free category on $\mathcal{G}$ and certain (finite-product) cones $\mathbb{L}$ in the free category on $\mathcal{G}$; a model in a category $\mathcal{E}$ is a morphism of reflexive graphs from $\mathcal{G}$ to the underlying reflexive graph of $\mathcal{E}$ which sends the diagrams of $\mathbb{D}$ to commutative diagrams and the cones of $\mathbb{L}$ to limiting cones. We can view Barr-Wells sketches as containing $Sketch$ and $Law$ as reflective subcategories just as before. -If we use a Barr-Wells sketch, we can give a truly finite presentation of the theory of groups. It is a reflexive graph with objects $G^0,G^1,G^2,G^3$ with arrows $e:G^0 \to G^1$, $m:G^2 \to G^1$, $i: G^1 \to G^1$, various projection arrows, cones which are designated by $\mathbb{L}$ to make $G^i$ into a product of $i$ copies of $G^1$, and one diagram each for associativity, left and right unitality, and left and right inverse laws.<|endoftext|> -TITLE: Distribution of big component of set partitions -QUESTION [8 upvotes]: Consider the set $S_n = \{1, \dotsc, n\},$ and consider the set $P(n, k)$ of partitions of $S_n$ into $k$ parts (the cardinality of $P(n, k)$ is the Stirling number of the second kind $S(n, k).$ Define a function $M$ on $P(n, k),$ where $M(x)$ is the size of the biggest piece of the partition. Is there anything known about the distribution of the values of $M$ (as $n, k$ become large)? I assume that the answer is "yes", but am having trouble finding references. - -REPLY [4 votes]: I haven't managed to find the answer to precisely your question but here are a couple of references that might be useful. -Vershik and Yakubovich have a paper on The limit shape and fluctuations of random partitions of naturals with a fixed number of summands. It addresses partitions of $n$ with around $\sqrt{n}$ summands, but doesn't seem to have exactly the result you're asking about. -If you haven't already looked at it, Chapter 1 of Pitman's Combinatorial stochastic processes seems quite relevant to your question. In particular he states something which he calls "Kolchin's representation of Gibbs partitions". For the special case of uniformly random partitions, this can be stated as follows, I think. Fix a positive parameter $\xi$ and let $X_1,X_2,\ldots$ be iid with distribution Poisson$(\xi)$ (Added on edit: the $X_i$ should be conditioned to be strictly positive). Also, let $K$ be Poisson$(e^{\xi}-1)$ and independent of the $X_i$. -Then for any $n$, conditional on the event that $X_1+\ldots+X_K=n$, the vector $(X_1,\ldots,X_K)$ is distributed as the vector of sizes of the parts of a uniformly random partition of $\{1,\ldots,n\}$, listed in exchangeable random order. -You could then try conditioning both on $X_1+\ldots+X_K=n$ and on $K=k$, and playing with the parameter $\xi$, to read off information about partitions of $\{1,\ldots,n\}$ into $k$ parts.<|endoftext|> -TITLE: Status of Hadamard matrix conjecture -QUESTION [11 upvotes]: I would like to know if any progress has been made on Hadamard conjecture : - -Hadamard matrix of order $4k$ exists for every positive integer $k$. - -REPLY [5 votes]: With all due respect to Colin McLarty, here is (in my not so humble opinion) a better answer. -The conjecture (For every positive integer $k,$ there is a square matrix $H$ of order $4k$ such that $H$ is binary with entries being $1$ or $-1$, with $HH^t = 4k~I$) is not resolved, but there is much work going on in the area. - -The conjecture is generalized to the Hadamard maximum determinant problem (with binary matrices of all orders, not just $4k$), which in turn is generalized to the determinant spectrum problem (range of determinant function over binary matrices of a given order, a personal favorite of mine). Will Orrick helps maintain a website http://indiana.edu/~maxdet presently for these problems. My motivation for working on the determinant spectrum problem came from attempting an oblique approach to HMC. -when extended to complex entries of modulus 1, such a matrix exists for every order. Such matrices are studied for their own interest, but many who peruse this part of the literature and say they are doing this without a thought to the (real number version of the) Hadamard matrix conjecture, well, let me stop short of name-calling: I don't believe that statement for a moment. -other extensions concern combinatorial designs, partial Hadamard matrices, equivalence under a variety of relations, and ways of generating representatives. - -If one takes off the blinders and asks what progress on and around the Hadamard Matrix Conjecture is being made, the Wikipedia and Mathworld articles mentioned in other posts are a good start. In addition to the maxdet website and similar websites, the ArXiv has at least 10 articles from the last few years on recent work. -Gerhard "Progress Isn't Just A Number" Paseman, 2017.11.02. - -REPLY [3 votes]: According to the current version on the MathWorld website by Wolfram, -http://mathworld.wolfram.com/HadamardMatrix.html : -"… the smallest unknown order [$of\ a\ possible\ Hadamard\ matrix$] is 668." -I suppose this is what you wanted to know. In addition, the website provides important references on this subject.<|endoftext|> -TITLE: Ring of Witt vectors and p-adics -QUESTION [10 upvotes]: This is probably an easy question, but I'm not able to figure it out. -Are the following the same: - -Field of fractions of the ring of Witt vectors over the algebraic closure of $\mathbb{F}_p$ -Algebraic closure of the field of p-adics (which is the field of fractions of the ring of Witt vectors over $\mathbb{F}_p$) - -In other words, does the operation of taking the field of fractions of the ring of Witt vectors commute with the operation of taking the algebraic closure? - -REPLY [15 votes]: no: the Witt ring of $\bar{F_p}$ is a complete DVR and so its field of fractions will be a complete local field; but the algebraic closure of $Q_p$ is not complete. -However, take the maximal unramified extension of $Q_p$; this is a non-complete field. Its completion $F$ is the fraction field of the Witt ring of $\bar{F_p}$ and the Witt ring itself is the ring of integers in $F$. -(Serre's Local Fields contains all of this and much more!!)<|endoftext|> -TITLE: Has Sid Sackson's "Hold That Line" been analyzed? -QUESTION [9 upvotes]: In Sid Sackson's classic book A Gamut of Games, he introduces a game that he calls "Hold That Line." Briefly, it is an impartial pencil-and-paper game played on a finite grid of dots. The first player connects two dots with a horizontal, vertical, or diagonal line; thereafter, each player extends the given piecewise-linear curve at one end, making sure to keep it self-avoiding. -Analyzing this game looks like a nice project in combinatorial game theory, suitable for undergraduates, but has it already been studied? A quick search only turned up a brief discussion in Alan Lipp's book The Play's the Thing. -EDIT added March 2019: This question (among others) was raised by Jim Henle in his Spring 2019 Mathematical Intelligencer article, "Mathematical Treasures from Sid Sackson," but without any analysis (other than the observation that the normal form version has a Tweedledum-Tweedledee winning strategy for the first player). Henle has a webpage for recording comments about his article, but as of this writing, there are no comments about Hold That Line. - -REPLY [6 votes]: Jindřich Michalik of Charles University has just proved that the player going first has a winning strategy for the 4 x 4 game. He also analyzes 2 x n and 3 x 3 games in a paper, "A Winning Strategy for Hold That Line" that will appear in The Mathematical Intelligencer later this year or early in 2021.<|endoftext|> -TITLE: Why "Categorify"? Relating to link/knot homologies... -QUESTION [12 upvotes]: Hey Everyone! -So I am new blood in the topic of Khovanov Homology and related topics. According to my basic reading the idea is to get the Jones polynomial as the Euler Characteristic of a certain homology theory. My question is, why do this? I mean this in the sense that why is it important/interesting to have this "quantum invariants" $\rightarrow$ "homology theories" transformation? If this question is perhaps to basic and /or unsuited for the site I would appreciate some bibliography on the matter as Khovanov's "A categorification of the Jones polynomial" article doesn't seem to answer this question for me. -Thanks! - -REPLY [9 votes]: As suggested by Budney and Agol's answers, one short answer to the question "why categorify [3-manifold invariants]" is "to get 4-manifold invariants." -This is directly related to the standard answer to the question "why is homology better than the Euler characteristic" which is: "functoriality." Functoriality in the case of Khovanov homology leads to maps coming from 4-dimensional cobordisms between knots, which one hopes will give interesting information about knot cobordisms (this has been realized in the work of Jake Rasmussen). -More specifically, the work of Donaldson/Floer/Witten et al in the 80's and early 90's led to an invariant for 3 and 4 manifolds which you might call "Donaldson-Floer theory" and which (more or less) fit together into what came to be called a "3+1 dimensional topological quantum field theory." D-F theory gave many exciting results about the smooth structure of 4-manifolds, but it was defined using complicated non-linear analysis and was very hard to compute. -Around the same time, there appeared another TQFT, the Witten-Reshetikhin-Turaev (WRT) theory. This was a 2+1 dimensional TQFT, and the work of Reshetikhin-Turaev showed that it had a purely combinatorial definition. Moreover, it gave invariants of knots which were directly related to the Jones polynomial. -According to the introduction of Khovanov's paper, his original motivation goes back to ideas of Crane and Frenkel on lifting the WRT theory from a 2+1 to a 3+1 theory, in the hopes of getting a combinatorially-defined invariant of smooth 4-manifolds with the same power as D-F theory. To understand, from their point of view, why this is related to "categorification" in the algebraic sense (a term which they may have helped define? I'm not sure), see their paper "Four dimensional topological quantum field theory, Hopf categories, and the canonical bases." There, they say - -"the second idea motivating our - construction is that replacing an - algebraic structure with a similar - categorical algebraic structure lifts - the dimension of the corresponding - TQFT by 1" - -Currently, Khovanov homology is not defined for all knots and knot cobordisms in all 3 and 4 manifolds, and it gives limited (albeit very interesting!) 4 dimensional information. It remains a fundamental problem to get a complete combinatorial 4-manifold invariant generalizing Khovanov homology and fit it into a TQFT, and to understand its relationship with (categorified) representation theory and smooth 4-manifold topology. -I should add that there is a different point of view on this question, which takes the "categorification of quantum groups" as a fundamental goal and interesting problem in its own right. But this should probably be addressed by someone who knows more about it than I do!<|endoftext|> -TITLE: Periodic lightray paths trapped between two nested mirror circles -QUESTION [12 upvotes]: I wonder if the periodic paths of a lightray trapped between two nonconcentric circles, -each perfectly reflecting, are known. -The behavior of such rays seems chaotically complicated. For example, left below the highlighted -initial ray has slope $\frac{1}{12}$, while that to the right has slope $\frac{1}{9}$. -(The green circle has radius $\frac{3}{4}$.) After 500 reflections, neither ray has become periodic. - -    - -I suspect the combination of dispersive and focusing reflection -(from the inner and outer circles respectively) -leads to this complex behavior. -But perhaps the periodic paths are known? -I only know the 2-cycles from rays collinear with the circle centers, -and those that Noam Elkies kindly identified in his comment: -"regular $n$-gons (and $(n/k)$-gons, i.e. stars) that stay so close to the outer circle that they never hit the inner one." -(Related MO question: "Trapped rays bouncing between two convex bodies.") - -Update 1. I found one! :-) No doubt among the "simple short cycles" that Noam had in mind: - -                 - - -Update 2. -With the benefit of the search terms helpfully provided by Ian Agol and Igor Riven, -I found a useful Physical Review paper by -G. Gouesbet, S. Meunier-Guttin-Cluzel, and G. Grehan, -"Periodic orbits in Hamiltonian chaos of the annular billiard," -Phys. Rev. E 65, 016212 (2001). Their approach is more experimental than theoretical: - -Periodic orbits embedded in the phase space are - systematically investigated, with a focus on inclusion-touching periodic orbits, up to symmetrical orbits of - period 6. Candidates for periodic orbits are detected by investigating grayscale distance charts and, afterward, - each candidate is validated (or rejected) by using analytical and/or numerical methods. - -The (unstable) 3-cycle I found they label "2(1)1" in their -(barely discernable) Fig.5 inventory below: - -REPLY [8 votes]: I don't know how to answer your question, but I'll make a reformulation of the question. -Consider the unit tangent bundle to the outer circle, and consider -the subset consisting of vectors pointing into the circle. This is -an annulus, parameterized by $(\theta,\varphi)\in A=[0,2\pi]\times [0,\pi]/\{ (0,\varphi)\sim (2\pi,\varphi)\}$, where $\theta$ parameterizes the point on the outer circle, and $\varphi$ gives the angle that the unit vector makes with the tangent vector. One may consider the first-return map under the geodesic flow $F: A\to A$ (flow in the direction of the vector until you hit the outer circle again, then reflect). This is a piecewise smooth map. For each $\theta$, there are angles $0< f_1(\theta) < f_2(\theta) < \pi$ such that $F(\theta,\varphi)= (\theta+2\varphi, \varphi)$ for $0\leq \varphi \leq f_1(\theta)$ and $f_2(\theta)\leq \varphi \leq \pi$ (in particular, $F$ is the identity on the boundary of $A$). For fixed $\theta$ and $f_1(\theta) < \varphi < f_2(\theta)$, $F(\theta,\varphi)$ is a more complicated trigonometric function depending on how the geodesic reflects off the inner circle and bounces back to the outer circle, but it has the property that $\varphi$ is increasing, and $\theta$ is decreasing. There is a natural measure on geodesic flow in the plane, the Liouville measure, which restricts to a measure on $A$. Clearly $F$ preserves this measure. I haven't computed the measure, but it is invariant under rotation, so is independent of $\theta$, and is invariant under reflection $(\theta,\varphi)\to (\theta,\pi-\varphi)$. One could reparameterize the $\varphi$ coordinate in terms of the Liouville measure to get an area-preserving homeomorphism of the annulus. So I would suggest you could do a literature search for results on periodic points for area-preserving homeomorphisms of an annulus. - -REPLY [5 votes]: Following up on @Ian's comments: the magic words are "monotone twist maps of the annulus" and "Aubry-Mather theory". googling either of the above (or looking in Katok-Hasselblad, who have a whole chapter on the subject) is your best bet.<|endoftext|> -TITLE: On prime numbers -QUESTION [6 upvotes]: Let $p_n$ be the n-th prime number and $c_n$ be the n-th composite number. We have -$$ -\lim_{n \to \infty}\frac{1}{n} \sum_{r=1}^{n}\frac{p_n^2}{p_n^2 + p_r^2} -= \lim_{n \to \infty}\frac{1}{n} \sum_{r=1}^{n}\frac{c_n^2}{c_n^2 + c_r^2} = \frac{\pi}{4}. -$$ -The beauty of the above result is that the first limit is a series over prime and the other is a series over composites. Similar results hold if the sequence of primes (or composites) are replaced by the sequence of natural number. This is a specific example of a general family of results of this kind. -The question is understand why such a relationships hold. I have been working on this and wanted to share with and learn from other number theorists. Has anyone seen similar results in mathematics literature? Any reference would be helpful. - -REPLY [22 votes]: The sums above are Riemann sums over different partitions for the function $\arctan(x)$ between $x=0$ and $x=1$, so the limit will equal the integral which is $\frac{\pi}{4}$. Both of these converge to the same value because they are not too weirdly distributed among $[0,1]$. -Remark: We need to use the fact that there exists $\theta<1$ with $p_n-p_{n-1}\ll p_n^\theta$. (we can take $\theta=7/12$) For the primes, we know that this tells us that if $j\geq n^{7/12+\epsilon}$, then -$$p_{n+j}-p_{n}\sim j\log n. $$ -Edit: I added why $p_{n+j}-p_{n}\sim j n^{7/12}$ for $j\geq n^{7/12+\epsilon}$ is important after reading some of the comments. It tells us/(or actually comes from) how things will look in short intervals for primes. It is not true that for general sequences with $\alpha_{i}-\alpha_{i-1}\ll n^{-\delta}$ the Riemann sum works out, rather for sequences where sums over short intervals is very close to the identity function. -Edit 2: This is more of a remark because I have a feeling someone will wonder about this. The reason why we need it to be close to the identity on short intervals is because we are weighting with the identity, $\frac{1}{n}$, rather then $x_i-x_{i-1}$ which is what is used in the definition of the Riemann integral. Summation tricks to move to these short intervals allows us to make the desired conclusion. Note that the limit will hold for any bounded monotonic integrable $f$, and any sequence satisfying the condition.<|endoftext|> -TITLE: Examples of "Monster" groups -QUESTION [12 upvotes]: I am planning a talk for a general graduate student audience. The topic is exotic examples of countable discrete groups ("monsters"). Some examples of properties that I'm interested in are: -1.) Non-amenable groups without free subgroups -2.) Groups such that $x^n=e \hspace10pt\forall x$ -3.) Groups with all proper subgroups cyclic -4.) Groups such that every proper subgroup is finite and cyclic of a given order -5.) Groups such that every elements has roots of all orders -The main source I have been using so far is a survey by Mark Sapir http://arxiv.org/abs/0704.2899. -I would like additional sources. Additional properties to the ones above would also be great. Also examples that arise "naturally" (say as a group of symmetries of some nice space rather than a combinatorial construction would be great.) - -REPLY [2 votes]: This is really a long comment, but points to a number of things that answer this question. -Mark Sapir has made some useful comments in answers to a couple of my questions that may be helpful to you. One reference that comes to mind is his Lacunary hyperbolic groups with Olshanskii and Osin. Additionally, have a look at Olshanskii's book Geometry of defining relations in groups. In this book many such examples are generated that solve various problems.<|endoftext|> -TITLE: Good Surface,Bad Surface-Surface classification -QUESTION [6 upvotes]: Maybe this question be very simple, but I don't know why it is hard for me. Thanks for any guide and help. -We say a surface $S$ (2-dimensional metric(compact) Riemannian surface) is good (denote by $GS$), if every $2n$, $n\geq1$, points on surface can be separate by some geodesic to two distinct subsets $V_1$ and $V_2$, where $|V_1|=|V_2|=n$. Also, if a surface $S$ is not good, we say it is bad an denote it by $BS$. -For example, it is not difficult to show that plane is a $GS$. Also, a sphere is $GS$. -1) Do we have some $BS$ examples(class of examples)? -2) Can we characterize the $GS$ and $BS$ surfaces? -I can't find any $BS$ examples and also I can't prove that they are $GS$. -For example, is Klein Bottle $GS$ or $BS$? -Is there any related works and questions about this post? - -REPLY [11 votes]: Assuming that "geodesic" in this question means "simple closed geodesic", then every complete hyperbolic surface $S$ of finite area is "bad": You cannot even separate an arbitrary pair of points. The reason is that the union of simple closed geodesics on $S$ is nowhere dense (even more, its closure has Hausdorff dimension 1) by the result of Birman and Series, "Geodesics with bounded intersection numbers on surfaces are sparsely distributed", Topology 24 (1985). The paper is available at: -http://www.math.columbia.edu/~jb/bdd-int.no-sparce.pdf -In view of this theorem, there exists an open disk $D\subset S$ which is disjoint from all simple closed geodesics in $S$. Now, take two points from this disk. I did not check it, but it is quite likely that Birman-Series result also holds in the case of negatively pinched variable curvature. -Hyperbolic surfaces are probably still "bad" if you allow non-simple closed geodesics, but pairs of points no longer suffice; one could try to use Hausdorff dimension arguments in the products of hyperbolic surface by itself to get a contradiction.<|endoftext|> -TITLE: Non-computational software useful to mathematicians -QUESTION [38 upvotes]: The MathOverflow question Open source mathematical software contains a list of programs that are useful to perform various computational tasks, such as computer algebra systems. -However, evaluating complicated formulas is not all that a professional mathematician needs to do. For instance, another important part of it is communicating results, producing papers and slides. There was a Mathoverflow question devoted specifically to -tools for collaborative paper writing. -It would be useful to identify and gather on MO a list of areas of activities where research mathematicians can use software as part of their professional activity. -I think it would also be useful to gather on MO a list of software that does not strictly do computations, but is nevertheless useful to those who research and teach mathematics. -The first example that springs to mind is of course $\LaTeX$, but there is much more: - -citation and literature management software, such as Jabref, Zotero, Mendeley -conference management software, such as Open Conference Systems (never actually used it, but it seems interesting) -reference tools such as the Online Encyclopedia of Integer Sequences and Plouffe's Inverter -Euclidean geometry software such as Geogebra -diff'ing and merging tools, such as latexdiff, kdiff3 -specific LaTeX packages such as Beamer for presentations or several drawing packages (TikZ, Eukleides). - -I find some of those real gems, and I'd like to find out more examples. So, the question is: - -Can you provide (other) examples of programs that are useful to professional mathematicians in their job, while not being strictly speaking "software that does complicated computations"? - -The question is a little broad and perhaps if there is much software relevant to a specific activity it will be wise to ask, based on input given to the question, a more specific question. Also please do not interpret this question too broadly (see discussion below). -Also check these MO questions: Tools for collaborative paper-writing (mainly regarding revision control software), Most helpful math resources on the web (mainly regarding online databases of something). - -REPLY [2 votes]: When collaborating in person, I frequently work at the chalkboard with my collaborator, periodically taking pictures of the board with a smartphone, and then posting the pictures in a shared Dropbox folder. That way, we don't spend our time together transferring notes from the board onto sheets of paper. -Most of my collaborators are employed elsewhere, so I wanted to replicate this experience for remote collaboration. Specifically, we wanted access to a common virtual whiteboard. Such technology is available under a freemium model with MIRO. When working together, we access MIRO via iPads while simultaneously chatting face-to-face on Google Hangouts. When the wifi is good, MIRO's virtual whiteboard exhibits near-zero time lag, which makes for a seamless working experience. To save our work, we periodically export the whiteboards as pdf files and save them in a shared Dropbox folder. -(As an aside, I've found that over time, I'll forget the mathematical context of various pictures in the Dropbox folder. It's extremely helpful to maintain a README.txt file in the same folder that provides a short description of the contents of each picture.)<|endoftext|> -TITLE: Is the Hochschild chain complex $C_*(A, A)$ a $B_\infty$-module over the Hochschild cochain complex $C^*(A, A)$? -QUESTION [5 upvotes]: Let $A$ be a dg-algebra, or more generally an $A_\infty$-algebra. Then it is well known that the Hochschild cochain complex $C^*(A, A)$ computing Hochschild cohomology is a $B_\infty$-algebra, see for example, the paper of Bernhard Keller "Derived invariance of higher structures on the Hochschild complex" available on his pageweb. -I would like to know whether the Hochschild chain complex $C_*(A, A)$ (which computes Hochschild homology) is a $B_\infty$-module over the Hochschild cochain complex $C^*(A, A)$? -Can someone give me the precise defintion, or a precise reference, of the action of $C^*(A,A)$ -on $C_*(A, A)$ if the answer is Yes? - -REPLY [6 votes]: This is the subject of Section 2 in that paper (sorry for self-promotion). -Chains actually have two $B_\infty$-module structures (over cochains). Those two module structures are moreover compatible (see Theorem 2.4 of the above paper for a precise statement).<|endoftext|> -TITLE: Presenting work in progress -QUESTION [26 upvotes]: At some point in the evolution of a project, one has to decide when to start talking about it in seminars and at conferences. I am interested in hearing about how other mathematicians make this decision. Of course the decision to give a talk on work in progress may depend on the state of the project, on the venue and the audience, and on other factors. -My specific questions are: - -At what stage of a project do you consider it to be ready to talk about? -How does the venue (departmental seminar/colloquium/conference/other) affect your decision? -What other factors do you consider when deciding whether to talk about work in progress? - -I would be interested in hearing about relevant experiences that have informed your decision process, but please keep anecdotes anonymous. Mainly I want to hear about people's criteria for evaluating when their work is ready to be publicly presented. - -REPLY [4 votes]: (1) If I've thought carefully enough about a project that I have something to teach my audience by presenting what I know, I'll talk about it (qualifying the parts that I'm unsure of). I prefer to give talks about what I'm most excited by at the moment; this isn't always what is most complete. -(2) The venue doesn't matter, but if there are fancy-pants mathematicians in the audience I will be more nervous about presenting partial results -(3) If I have something that is complete, that I'm excited about, and that will be new to the audience I'm talking to I'll usually choose that over something that is in progress. -Maybe I should be more worried about being scooped, but (a) I'm not presenting a proof of the Riemann Hypothesis (if I were it would be wrong) (b) Its hard to scoop someone based on a talk that they gave (c) Publish-or-perish pressure aside, most of the mathematicians that I've met are people of integrity (d) if someone uses my ideas to do something I might have done but haven't yet done this is a credit to my work (and moreso if they cite me).<|endoftext|> -TITLE: Does the image of a section lie in the regular part -QUESTION [5 upvotes]: Say $f:X\to C$ is a family of curves. More precisely, $C$ is a smooth projective irreducible curve over a field, $f$ is a flat morphism of schemes and $X$ is a normal projective irreducible surface. -Say I take a section $P:C\to X$. Does the image of $P$ lie in the nonsingular part of $X$? -What conditions (weaker than regularity) on $X$ do I need? Does it help if every Weil divisor on $X$ is $\mathbf{Q}$-Cartier? - -REPLY [10 votes]: Example: -Let $Y$ be the projective cone over a conic, so for instance, let $X$ be defined by $xz=y^2$ in the projective 3-space with coordinates $[x:y:z:w]$ and consider the projection to the $[x:w]$-axis: -$$ g:Y\dashrightarrow \mathbb P^1$$ -$$ [x:y:z:w]\mapsto [x:w]\quad\qquad $$ -This is defined everywhere except at $[0:0:1:0]$, so let $X$ be the surface one obtains by resolving the indeterminacies of $g$ and $f:X\to \mathbb P^1$ the induced morphism. -This $X$ is a normal, projective, irreducible surface and $f$ is a dominant morphism onto $C=\mathbb P^1$. It follows that $f$ is flat. -$X$ is also non-singular everywhere except at $[0:0:0:1]$ where it has an $A_1$-singularity which is one of the simplest surface singularities. In particular, it is $\mathbb Q$-factorial, i.e., every Weil divisor is $\mathbb Q$-Cartier. -There is also a section that goes through the singular point: -$\sigma: \mathbb P^1\to X$ defined by $[x:w]\mapsto [xw^2:x^2w:x^3:w^3]$ which originally maps to $Y$, but lifts to $X$ by taking the strict transform of the image. Notice that the image of this section is $\mathbb Q$-Cartier. -Analysis: The above example shows that unfortunately what you want does not necessarily hold, even under pretty strict conditions. The only condition short of non-singularity I could imagine is to assume that $X$ is factorial, that is, every Weil divisor (or at least the section you are considering) is Cartier. This is of course very easy: the section is non-singular since it is isomorphic to $C$ and if a variety contains a smooth Cartier divisor, then it is smooth along that divisor. -One could prove this last statement in a little more complicated way that shows why this fails in the example above: -The intersection of the section and any fiber has to be $1$, because the projection is an isomorphism. If the section is Cartier, this means that the fiber cannot be singular at the intersection point, so not only $X$ is non-singular, but even $f$ is smooth along the section. The problem with this argument in the case when the section is only $\mathbb Q$-Cartier is that the intersection number of the section with a fiber could be $1$ and the fiber still be singular. -In the example, the section is a smooth curve, but it is not Cartier (this actually follows from the above, but it is usually proved by computing the self-intersection, which is $\frac 12$), but the fiber through the singular point of $X$ is singular and the section goes through that point. Twice the section is Cartier and its intersection with the singular fiber is $2$, so the intersection of the section with the fiber is $1$ and we get no contradiction.<|endoftext|> -TITLE: The graph of congruences between modular forms -QUESTION [17 upvotes]: Let $S$ be the (countable) set of holomorphic cuspidal new eigenforms of weight $\geq 2$. Any $f\in S$ has a level $N_f$ and a canonically normalized Fourier expansion $f(z)=\sum_{n=1}^{\infty}a_f(n)e^{2\pi i nz}$ with $a_f(1)=1$ and $a_f(n) \in \overline{\mathbf{Z}}$. -Form a graph $\mathcal{G}$ as follows: Take the vertex set of $\mathcal{G}$ to be the set $S$, and join $f$ and $g$ by an edge if the algebraic integers $(a_f(n)-a_g(n))_{n \nmid N_f N_g}$ generate a nontrivial ideal in $\overline{\mathbf{Z}}$; in other words, $f$ and $g$ are joined if their Fourier coefficients are related by a nontrivial congruence. -Is $\mathcal{G}$ connected? - -REPLY [11 votes]: Suppose that $f$ has level $N$, and suppose that $N$ is divisible by $p$. -Then it is well known that $f$ is congruent modulo (some prime above) $p$ to a form -$g$ of level $M$ dividing $N$ (and high weight), where $M$ is prime to $p$. -In particular, by induction, all forms $f$ are connected to a form $g$ of level $1$ in -at most $d$ steps, where $d$ is the number of distinct prime divisors of $N$. Yet all -level one forms are congruent to $\Delta$ modulo $2$. This is actually related to ideas behind the proof of Serre's conjecture: -http://en.wikipedia.org/wiki/Serre%27s_modularity_conjecture<|endoftext|> -TITLE: Orbifold fundamental group and configuration space -QUESTION [11 upvotes]: I'm not very familiar with (even simple examples of) orbifolds, so my first question is: - -Let $C_2$ be $\mathbb{C}$ with one cone singularity at 0 of index 2. What is the fundamental group of $C_2$ minus $k$ points ? - -My naive answer is: take $\mathbb{C}^*$ minus the same $k$ points. Its fundamental group is freely generated by the $k+1$ loops around the punctures. Now decide that you don't have a "hole" in 0 anymore, but a cone singularity, meaning that the generators corresponding to a loop around 0 is now of order 2. Then I would say that the fundamental group of $C_2$ minus $k$ points is $\langle a_0,\dots,a_k | a_0^2=1\rangle$, ie $Z_2\ltimes F_{2k}$, where $F_{2k}$ is generated by {$a_i,a_0a_ia_0, i\geq 1$}. -Now recall the following construction: take the pure braid group $P_n$ with its standard generators $x_{i,j}, 1\leq i < j\leq n$ given by taking the $j$th strand, letting it go behind all other strand, loop around the $i$th one and going back. Then it's quite easy to see that the subgroup generated by the $x_{i,n}$ is free: it is the subgroup of pure braids for which all but the last strand are fixed straight lines. In fact, it leads to a semidirect product decomposition $P_n=P_{n-1}\ltimes F_{n-1}$. This decomposition is actually a so called "almost direct" product, which is quite an important fact. -This construction has a nice geometric interpretation: let $X_n$ be the configuration space of $n$ points in $\mathbb{C}$, and recall that $P_n=\pi_1(X_n)$. Then the map $X_n \rightarrow X_{n-1}$ which forget the last coordinate is a locally trivial fibration with fiber $\mathbb{C}$ minus $n-1$ points. Then it induces a (split) short exact sequence of fundamental groups -$$1\rightarrow F_{n-1} \rightarrow P_n \rightarrow P_{n-1}\rightarrow 1$$ -Let's try to do something similar with the "orbifold braid group" of $C_2$, that is the fundamental group $P_n(C_2)$ of $O_n=${$z_1,\dots,z_n \in C_2, z_i \neq z_j$}. -It seems to me that $P_n(C_2)=P_{n+1}/ \langle x_{1,i}^2=1,i=2 \dots n+1 \rangle$. -The above construction seems to work "at the algebraic level": let $G_n$ be the subgroup of $P_n(C_2)$ generated by (the images of) $x_{i,n+1}$. What is stated in this paper (in a slighty different form) is that $P_n(C_2)=P_{n-1}(C_2) \ltimes G_n$, and that it is an almost direct product too. -But $G_n$ satisfies some relations, for example $x_{i,n+1}$ and $x_{0,n+1}x_{i,n+1}x_{0,n+1}$ commute for a given $i$, hence it is not isomorphic to the fundamental group of $C_2$ minus $n-1$ points (at least if my first naive try is not wrong). While this construction strongly looks like to and shares many algebraic properties with the construction for $P_n$, it does not seems to come from a natural geometric construction. So my real question is: - -Am I wrong somewhere ? Is there a natural interpretation of $G_n$ ? - -Edit: Here is roughly what happen: assuming that $n=2$ for the sake of simplicity, it doesn't make sense to "freeze" the first strand (and its negative) and to make the second one loop around because the following relation holds: - -now pushing the red loop (seen as a loop in the 2-punctured plane) to the bottom plane, we see that it has to be identified with its conjugate by a loop around the two strands at once, ie by the product of the generators of $F_2$. Therefore, this product has to be central, leading to the relation holding in $G_n$ above. So one can ask: - -Is there a topological space modelled on this situation, i.e. which looks like to the "complementary in $\mathbb{C}\times[0,1]$ of two strands modulo homotopy". Or at least, is there a way to prove that there are no other relations than declaring that the big loop is central ? - -REPLY [5 votes]: For the first question, see: -MR0285619 (44 #2837) -Hoare, A. Howard M.; Karrass, Abraham; Solitar, Donald -Subgroups of finite index of Fuchsian groups. -Math. Z. 120 1971 289–298. -On the first page they talk about what orbifold groups look like (and refer back to Fricke-Klein...)<|endoftext|> -TITLE: Matrix inequality $(A-B)^2 \leq c (A+B)^2$ ? -QUESTION [12 upvotes]: Let A and B be positive semidefinite matrices. It is not hard to see that $(A-B)^2 \leq 2A^2 + 2B^2$. In fact, $2A^2 + 2B^2 - (A-B)^2 = (A+B)^2$ is positive semidefinite. -My question is: Is there a constant c (independent of A and B and the dimension) such that -$$(A-B)^2 \leq c (A+B)^2?$$ -Thanks. - -REPLY [3 votes]: A weaker version is also not true, that is, there is no constant $c$ such that -$$\lambda_j(A-B)^2 \leq c \lambda_j(A+B)^2, ~~ j=1, 2,\ldots$$ -where $\lambda_j(A)$ means the $j$-th largest eigenvalue of $A$. -Nevertheless, it is trivial to see that -$$trace(A-B)^2 \leq trace(A+B)^2.$$<|endoftext|> -TITLE: Homotopy type of tensors of Moore spectra -QUESTION [17 upvotes]: I would like to hear what's known about the homotopy type of smash products of mod-$p^j$ Moore spectra, for $p$ an odd prime. -First, here is what I'm specifically interested in: there is a short exact sequence $$0 \to \mathbb{Z} \xrightarrow{p^j} \mathbb{Z} \to \mathbb{Z}/p^j \to 0.$$ Tensoring this short exact sequence against your favorite group $G$ yields an exact sequence $$\cdots \to G \xrightarrow{p^j} G \to G \otimes \mathbb{Z}/p^j \to 0,$$ which exhibits $G \otimes \mathbb{Z}/p^j \cong G / p^j G$ as $G$ with its $p^j$-divisible part stripped out. -Moore spectra play a related role in homotopy theory: they are defined by the cofiber sequence $$S \xrightarrow{p^j} S \to M(p^j).$$ Smashing through with any spectrum $X$ gives the new cofiber sequence $$X \xrightarrow{p^j} X \to X \wedge M(p^j),$$ and chasing this around shows that the homotopy group $\pi_n X \wedge M(p^j)$ is a mix of $\pi_n X / p^j(\pi_n X)$, as one would expect, together with the $p^j$-torsion of $\pi_{n-1} X$, which is new and different. So, though $X \wedge M(p^j)$ is sometimes written $X / p^j$, and though this notation suggests a useful analogy, this isn't exactly true, and we have to be careful about things we expect to follow from the algebraic setting. -The specific algebraic fact I'm interested in is that the composition of the tensor functors $- \otimes \mathbb{Z}/p^j$ and $- \otimes \mathbb{Z}/p^i$ for $j > i$ has a reduction: $$- \otimes \mathbb{Z}/p^j \otimes \mathbb{Z}/p^i \cong - \otimes \mathbb{Z}/p^i.$$ The exact translation of this statement to Moore spectra and the smash product is not true --- one can, for instance, compute the reduced integral homology of $M(p^i) \wedge M(p^j)$ to see that there are too many cells around for it to be equivalent to $M(p^i)$ alone. However, the same homology calculation suggests something related: there is an abstract isomorphism between the reduced homology groups of $M(p^j) \wedge M(p^i)$ and those of $M(p^i) \wedge M(p^i)$. This is, of course, also true for groups; it is indeed the case that $\mathbb{Z}/p^i \otimes \mathbb{Z}/p^j \cong \mathbb{Z}/p^i \otimes \mathbb{Z}/p^i$. This is what I want to know: - -For $j > i$, is $M(p^j) \wedge M(p^i)$ homotopy equivalent to $M(p^i) \wedge M(p^i)$? - -If this is not true, I'm willing to throw in some extra qualifiers. For instance, is the situation improved if we work $K(n)$-locally? Is it true only when $j \gg i$? What if we additionally restrict to $j \gg i \gg 0$? -This specific question aside, I am also interested in any and all known features of the homotopy type of $M(p^i) \wedge M(p^j)$ --- any favorite fact you have that would help me get a grip on them. I'm also specifically interested in variants of the above question for generalized Moore spectra: can anything similar be said about those? - -REPLY [4 votes]: You might find this paper useful: -\bib{MR760188}{article}{ - author={Oka, Shichir{\^o}}, - title={Multiplications on the Moore spectrum}, - journal={Mem. Fac. Sci. Kyushu Univ. Ser. A}, - volume={38}, - date={1984}, - number={2}, - pages={257--276}, - issn={0373-6385}, - review={\MR{760188 (85j:55019)}}, - doi={10.2206/kyushumfs.38.257}, -}<|endoftext|> -TITLE: Can anyone analyze this misere game? -QUESTION [11 upvotes]: Problem -Let $* = \{0\}$ be the one matchstick nim game, let $*2 = \{0,*\}$ be the two matchstick nim game, let $*3 = \{0,*,*2\} = *2+*$ be the three matchstick nim game, let $g = \{0, *2+*3, *2+*2+*2\} = *2_{22}2_30$, and finally let $h = \{*3, g+*, g+g\}$. - -Question: For which numbers $n$ does the misere game consisting of $n$ copies of $h$ have the same outcome as the misere game consisting of $n+1$ copies of $h$? - -The $n$s below $4000$ with this property are $1, 2, 3, 12, 14, 17, 38, 40, 56, 74, 101, 227, 319, 464, 692, 1025, 1189,$ and $1781$. - -Second question: Is the set of such $n$s infinite, with density $0$? - - -Motivation -There is a recent conjecture due to Ezra Miller and Alan Guo (see Conjecture $4.5$ in this paper) which implies that the set of multiples of any misere game that are wins for the previous player to move is eventually periodic. -I found the game I am calling $h$ while I was searching for a counterexample to their conjecture. It seems likely that it can be completely analyzed. If the answer to the second question is yes, then $h$ is clearly a counterexample to their conjecture. - -Computation -To calculate the winning positions quickly, it is useful to think of playing a sum of such games as moving around on a four dimensional lattice, where the coordinates correspond to the numbers of copies of $*, *2, g,$ and $h$ that are currently in play. A move corresponds to decreasing one of the positive coordinates while increasing some of the other coordinates, and the position $(0,0,0,0)$ is a win for the next player to move. Thinking this way, you can easily calculate the set of winning positions inductively. -Since $*+* = 0$ even as misere games, we can further simplify by imagining that we are playing a game on $\mathbb{Z}/2\times \mathbb{N}^3$, which gives a computational speedup. In theory one could probably rephrase this again as a fairly simple (directional) two-dimensional cellular automata (by taking two dimensional diagonal slices of the three dimensional lattice), so some version of the hashlife algorithm should give an exponential computational speedup, but I haven't found a way to make it work yet. - -REPLY [4 votes]: I recently managed to convert this problem into a cellular automata, and the answer to the second question appears to be no, making this question uninteresting. -However, I think some people might appreciate the details of the computations, so I'll put them here (and then close the question? I'm not sure how I should deal with answering my own question.) -First, the results of the computations. Let $N(x)$ denote the number of values $n$ below $x$ such that $n$ copies of $h$ and $n+1$ copies of $h$ have the same outcome. Then -$N(10^3) = 15, N(10^6) = 86, N(10^9) = 148, N(10^{12}) = 196,N(10^{15}) = 205,$ and $N(10^{54}) = 205$. The last value of $n$ below $10^{54}$ is (if I didn't make any mistakes) $3,814,460,171,075$. -Next, the details of the computation. I start by visualizing the situation as a coloring of $\mathbb{N}^3$ in the colors red, green, and blue. The coloring of $(i,j,k)$ corresponds to the outcome of the game formed by $i$ copies of $h$, $j$ copies of $g$, and $k$ copies of $*2$ as follows: green corresponds to a lost game, red corresponds to a game that loses when you add $*$, and blue corresponds to a game that is a first player win and remains a first player win when you add $*$ to it. In order to calculate the color of $(i,j,k)$, one needs to know the colors of $(i,j,k-1)$, $(i,j-1,k)$, $(i,j-1,k+2)$, $(i,j-1,k+3)$, $(i-1,j,k+1)$, $(i-1,j+1,k)$, and $(i-1,j+2,k)$. -In order to convert this to a cellular automata, I take "diagonal cross sections" of $\mathbb{N}^3$. I lay out one such diagonal cross section as follows: -(3,0,0) (3,0,1) (3,0,2) (3,0,3) (3,0,4) (3,0,5) (3,0,6) (2,0,7) (2,0,8) ... - (3,1,0) (3,1,1) (2,1,2) (2,1,3) (2,1,4) (2,1,5) (2,1,6) (2,1,7) ... - (2,2,0) (2,2,1) (2,2,2) (2,2,3) (2,2,4) (2,2,5) (2,2,6) ... - (2,3,0) (2,3,1) (2,3,2) (2,3,3) (1,3,4) (1,3,5) ... - (1,4,0) (1,4,1) (1,4,2) (1,4,3) (1,4,4) ... - ... - -To go from this cross section to the next higher cross section, we just need to replace the cell marked $(2,0,7)$ with $(3,0,7)$, the cell marked $(2,1,2)$ with $(3,1,2)$, the cell marked $(1,3,4)$ with $(2,3,4)$, etc. With the cells laid out like this, it is easy to see that these updates only require the knowledge of the colors of cells within two steps of the cell you would like to update. Unfortunately, most cellular automata only allow the transition rules to depend on the values of the cells immediately adjacent, so I had to emulate this cellular automata with a slightly more complicated one. -The end result is this rule table, with these colors which you can load in a program such as Golly and experiment with. The automata needs a single white square to get started, and then proceeds to compute winning positions automatically. When it detects an $n$ such that the outcome of $(n,0,0)$ matches the outcome of $(n+1,0,0)$, it fires a cyan signal upwards, so you can calculate $N(x)$ by simulating up to generation $12x+14$ and counting the number of cyan signals produced. Here is a screenshot of a computation in progress:<|endoftext|> -TITLE: Topos associated to a category -QUESTION [23 upvotes]: For each topos $\mathbb E$ let $\mathcal O(\mathbb E)$ be the locally presentable category of objects in $\mathbb E$. We can make $\mathcal O$ into a contravariant functor to the category of locally presentable categories (with morphisms being cocontinuous) by assigning to each geometric morphism $(f^\*, f_\*)$ the functor $f^\*$. By [Mac Lane, Moerdijk: Sheaves in Geometry and Logic] this functor is representable, that is there is a topos $\mathbb A$, called the object classifier, such that there is a natural equivalence -$$ -\mathrm{Hom}(\mathbb E, \mathbb A) \to \mathcal O(\mathbb E). -$$ -Now I wonder whether $\mathcal O$ has a right adjoint, which I want to call $\operatorname{Spec}$ due to the analogy with algebraic geometry, that is whether there exists a contravariant functor $\operatorname{Spec}$ from the category of locally presentable categories to the category of topoi (with geometric morphisms) such that there is a natural equivalence -$$ -\mathrm{Hom}(\mathbb E, \operatorname{Spec}\mathcal C) \to \mathrm{Hom}(\mathcal C, \mathcal O(\mathbb E)) -$$ -of categories. -(Here, topos shall mean Grothendieck topos.) - -REPLY [19 votes]: This is described in the paper - -Bunge and Carboni, The symmetric topos, Journal of Pure and Applied Algebra 105:233-249, 1995. - -which describes the sense in which the construction you call Spec is analogous to the symmetric algebra construction. -Bunge and Carboni give a biadjunction between the bicategory R, which is the opposite of the bicategory of Grothendieck toposes, and the bicategory A of locally presentable categories and cocontinuous functors (equivalently, left adjoints).<|endoftext|> -TITLE: bound for zeros of a polynomial with bounded integer coefficients -QUESTION [6 upvotes]: Let $f$ be a monic polynomial with bounded integer coefficients and such that all zeros are (in absolute value) greater than $1$. How close can the zeros of $f$ reach $1$ (in absolute value)? - -More precisely, let -$$ -f = a_0 + a_1 X + \cdots + a_{n-1}X^{n-1} + X^n -$$ -with $a_i \in \mathbb{Z}$ and $\lvert a_i \rvert < M$. Suppose $f(z)=0$ implies $\lvert z \rvert > 1$ (so all zeros have absolute value greater than $1$). - -What is an (explicit) positive function $B(n,M)$ with - $$\lvert z \rvert - 1 \geq B(n,M)$$ - for any zero $z$ of $f$? - -If it is easier, then $M=2^n$ can be assumed. Further, I am only interested in the behavior for large $n$, i.e., I want something like -$$ -\frac{1}{\lvert z \rvert - 1} = O(B(n,2^n)) -$$ -for $n\to\infty$ and for each zero $z$ of any monic polynomial $f$ with degree at most $n$ and integer coefficients bounded by $2^n$. - -REPLY [4 votes]: Just a brief remark that if $M=2$ and the constant term is $\pm2$, then these are called Garsia numbers. It is known that $z=1$ is a limit point for this set (and some computational results as well). Perhaps, you'll find the following recent paper useful as far as the techniques are concerned. - -REPLY [3 votes]: $\def\conj#1{\overline{#1}}\DeclareMathOperator\Res{Res}$If $z$ is a zero of $f$, then $|z|^2-1=z\conj z-1$ is a zero of the resolvent $g(w)=\Res_z(\conj f(z),z^nf((w+1)/z))$. You can extract a bound on the (integer) coefficients of $g(w)$ from the definition, and then e.g. Cauchy's bound will give you a lower bound on $|z|^2-1$, which in turn implies a lower bound on $|z|-1$. I don’t feel like working it out myself now, but you should get something of the form $B(n,M)\ge M^{-O(n)}$.<|endoftext|> -TITLE: Hovey's unit axiom in monoidal model categories -QUESTION [10 upvotes]: Let $\mathcal{C}$ be a monoidal model category in the sense of Hovey's book. He assumes the following unit axiom not considered in other references (e.g. Schwede-Shipley): given a cofibrant replacement of the monoidal unit $q\colon QI\stackrel{\sim}\rightarrow I$ and a cofibrant object $X$, the morphisms $q\otimes X$ and $X\otimes q$ are weak equivalences. -This axiom obviously holds if $I$ is cofibrant. Moreover, if I'm not mistaken, a stronger version holds in all examples I know: $q\otimes X$ and $X\otimes q$ are weak equivalences for any $X$. -Does any of you know any example where this stronger version is false? Is it always true under some 'standard' assumptions on $\mathcal{C}$? - -REPLY [4 votes]: I can only answer half of your question: namely, a standard condition under which the more general unit axiom holds. I don't know of any examples where Hovey's unit axiom holds but this more general one does not. The hypothesis is that cofibrant objects are flat, i.e. smashing with cofibrant objects preserves weak equivalences. This hypothesis comes up all the time in Hovey's work, and also appears in the paper of Schwede-Shipley where the Monoid Axiom is first introduced. There you need cofibrant objects to be flat in order to conclude that if $R\rightarrow S$ is a weak equivalence of ring objects, then $Ho(R-mod)\cong Ho(S-mod)$. -Suppose that cofibrant objects are flat. Then we know Hovey's unit axiom automatically, since $QI\rightarrow I$ is a weak equivalence and so for any cofibrant $Y$, $Y$ smashed with this map is still a weak equivalence. To see that the more general unit axiom holds, let $X$ be any object (not necessarily cofibrant). Then we have the following commutative diagram: -$$ \begin{array}{rrrr} -QI\otimes QX & \rightarrow & QX & \rightarrow & X\\\ -\downarrow & & & & \downarrow \\\ -QI \otimes X & & \rightarrow & & X -\end{array} -$$ -Here the top maps are weak equivalences by Hovey's unit axiom and by the definition of $QX$. The left vertical map is a weak equivalence because cofibrant objects are flat and $QI$ is cofibrant. The right vertical is clearly a weak equivalence because it's the identity. Thus, by the 2-out-of-3 property, the bottom horizontal is a weak equivalence. -EDIT: It should go without saying, but in a non-symmetric setting "cofibrant objects are flat" means both $K\otimes f$ and $f\otimes K$ are weak equivalences. So the argument above works if you apply the twist natural transformation to everything, and we also have $X\otimes QI \rightarrow X$ is a weak equivalence.<|endoftext|> -TITLE: References for classical Yang-Mills theory -QUESTION [13 upvotes]: I am looking for a reference to study classical (i.e., not quantized) Yang-Mills theory. -Most of the sources I find focus on mathematical aspects of the theory, like Bleecker's book Gauge theory and variational principles, or Baez & Muniain's Gauge fields, knots and gravity. -But I am more interested something similar to the standard development of electromagnetism, as can be found, for example, in Landau & Lifchitz's course on theoretical physics. To be more exact, I would like to learn about the field equations (Yang-Mills and Einstein equations), but also about the corresponding Lorentz Law, the energy of the Yang-Mills field,... and the analogous concepts of what is made for the electromagnetism. -That is, I look for a rigorous exposition where, at the same time, I could learn whether it is possible to prove, at the classical level, the quick decrease of the strong interaction, and things like that. - -REPLY [2 votes]: My personal suggestion is 'Differential Geometry, Gauge Theories, and Gravity' by M. Gockeler and T. Schucker. However, it assumes a fairly high degree of mathematical sophistication (it's one of the texts in the 'Cambridge Monographs on Mathematical Physics). If you do get it, it is really worth the effort to master the topics inside, and the range of topics covered is fascinating.<|endoftext|> -TITLE: Stone-Weierstrass for monotone functions -QUESTION [5 upvotes]: Let $\; f : [0,1] \to \mathbb{R} \;$ be continuous and non-decreasing. $\;\;$ Let $\epsilon$ be a real number such that $\; 0 < \epsilon \;$. - -Does it follow that that there exists a real polynomial $p$ such that $p$ is non-decreasing on - -$\;$ $[0,1]$ - -$\;$ all of $\mathbb{R}$ - -and for all members $x$ of $[0,1]$, $\; |f(x)+(-(p(x)))| < \epsilon \;\;$? - -REPLY [5 votes]: In fact, the Bernstein polynomials approximating $f$ are non-decreasing on $[0,1]$. A cute way to see this is via coupling (I learned this from Lindvall's book Lectures on the Coupling Method): -The $n$th Bernstein polynomial $p_n(x)$ can be written as $\mathbf{E}\Big[f\big(\frac{\sum_{i=1}^n Z^x_i}{n}\big)\Big]$, where $Z^x_i$ are Bernoulli random variables with parameter $x$. If $0\leq x\leq y\leq 1$, then we can define the variables $Z^x_i$ and $Z^y_i$ on the same probability space, such that $Z^x_i\leq Z^y_i$, which immediately gives $p_n(x)\leq p_n(y)$.<|endoftext|> -TITLE: Decomposition of positive definite matrices. -QUESTION [12 upvotes]: It is known that a $n^2 \times n^2$ positive semidefinite matrix $A$ cannot always be written as a finite sum -$$ -A=\sum_{j} B_j \otimes C_j -$$ -with $B_j$ and $C_j$ positive semidefinite matrices (of size $n \times n$). For example, it can be seen that the matrix -$$ -\begin{pmatrix} -1 & 0 & 0 & 1 \\\ -0 & 0 & 0 & 0 \\\ -0 & 0 & 0 & 0 \\\ -1 & 0 & 0 & 1 -\end{pmatrix} -$$ -is not the finite sum of Kronecker products of positive semidefinite $2 \times 2$ matrices. -Is the statement true if $A$ is positive definite? (i.e., $A$ is invertible?). -Edit: this question is a slight variant of a previous question. - -REPLY [11 votes]: The following is just a minor variation of Martin Argerami's proof of the old question. I am even copying his equations and some of his text. If you are +1ing this post, please also +1 his one (if not already done). -Here is a counterexample for $n=2$. Let $\varepsilon $ be a positive real $< -\dfrac{1}{2}$. The matrix -$ -a= \begin{bmatrix} -1 & 0 & 0 & 1-\varepsilon \\ -0 & \varepsilon & 0 & 0 \\ -0 & 0 & \varepsilon & 0 \\ -1-\varepsilon & 0 & 0 & 1 -\end{bmatrix} -\in \mathrm{M}_{4}\left( \mathbb{C}\right) -$ -is positive-definite, but it cannot be written as a sum of tensor products -of nonnegative-semidefinite $2\times 2$-matrices. Here is why: -Assume the contrary. Thus, $a$ is written in the form -$a=\sum_{j} \left[ \begin{matrix} -\alpha _{j} & \overline{\gamma _{j}} \\ -\gamma _{j} & \beta _{j} -\end{matrix} \right] -\otimes \left[ -\begin{matrix} -\alpha _{j}^{\prime } & \overline{\gamma _{j}^{\prime }} \\ -\gamma _{j}^{\prime } & \beta _{j}^{\prime } -\end{matrix} \right] -=\left[ -\begin{matrix} -\sum_{j}\alpha _{j}^{\prime }\alpha _{j} & \sum_{j}\alpha _{j}^{\prime } -\overline{\gamma _{j}} & \sum_{j}\overline{\gamma _{j}^{\prime }}\alpha _{j} -& \sum_{j}\overline{\gamma _{j}^{\prime }}\gamma _{j} \\ -\sum_{j}\alpha _{j}^{\prime }\gamma _{j} & \sum_{j}\alpha _{j}^{\prime -}\beta _{j} & \ast & \ast \\ -\ast & \ast & \sum_{j}\beta _{j}^{\prime }\alpha _{j} & \ast \\ -\ast & \ast & \ast & \ast -\end{matrix}\right] $, -where $j$ ranges from $1$ to some positive integer $N$. Since each $ -\begin{bmatrix} -\alpha _{j} & \overline{\gamma _{j}} \\ -\gamma _{j} & \beta _{j} -\end{bmatrix} -$ is nonnegative-semidefinite, we have $\alpha _{j}\geq 0$, $\beta _{j}\geq 0$, -and $\alpha _{j}\beta _{j}\geq \left\vert \gamma _{j}\right\vert ^{2}$ for -all $j$. Similarly, $\alpha _{j}^{\prime }\geq 0$, $\beta _{j}^{\prime }\geq -0$, and $\alpha _{j}^{\prime }\beta _{j}^{\prime }\geq \left\vert \gamma -_{j}^{\prime }\right\vert ^{2}$ for all $j$. -Now, comparing the entries of $a$ in this equation, we get $\varepsilon -=\sum_{j}\alpha _{j}^{\prime }\beta _{j}$ (from the $\left( 2,2\right) $-th -entry) and $\varepsilon =\sum_{j}\beta _{j}^{\prime }\alpha _{j}$ (from the $ -\left( 3,3\right) $-th entry). Taking the arithmetic mean of these two -equations, we get -$\varepsilon =\dfrac{1}{2}\left( \sum_{j}\alpha -_{j}^{\prime }\beta _{j}+\sum_{j}\beta _{j}^{\prime }\alpha _{j}\right) -=\sum_{j}\dfrac{\alpha _{j}^{\prime }\beta _{j}+\beta _{j}^{\prime }\alpha -_{j}}{2}\geq \sum_{j}\sqrt{\alpha _{j}^{\prime }\beta _{j}\beta _{j}^{\prime -}\alpha _{j}}$ -(by AM-GM, since we are dealing with nonnegative reals). But -for every $j$, we have -$\sqrt{\alpha _{j}^{\prime }\beta _{j}\beta -_{j}^{\prime }\alpha _{j}}=\sqrt{\alpha _{j}\beta _{j}}\sqrt{\alpha -_{j}^{\prime }\beta _{j}^{\prime }}\geq \left\vert \gamma _{j}\right\vert -\left\vert \gamma _{j}^{\prime }\right\vert $ -(since $\alpha _{j}\beta_{j}\geq \left\vert \gamma _{j}\right\vert ^{2}$ and $\alpha _{j}^{\prime }\beta _{j}^{\prime }\geq \left\vert \gamma _{j}^{\prime }\right\vert ^{2}$ -), so this becomes -$\varepsilon \geq \sum_{j}\underbrace{\left\vert \gamma -_{j}\right\vert \left\vert \gamma _{j}^{\prime }\right\vert }_{=\left\vert -\overline{\gamma _{j}^{\prime }}\gamma _{j}\right\vert } = -\sum_{j}\left\vert \overline{\gamma _{j}^{\prime }}\gamma _{j}\right\vert -\geq \left\vert \sum_{j}\overline{\gamma _{j}^{\prime }}\gamma -_{j}\right\vert $ -(by the triangle inequality). But since $1-\varepsilon -=\sum_{j}\overline{\gamma _{j}^{\prime }}\gamma _{j}$ (by comparing the $ -\left( 1,4\right) $-th entry of the matrices in the above equation), this -becomes $\varepsilon \geq \left\vert 1-\varepsilon \right\vert $, what -contradicts the definition of $\varepsilon $.<|endoftext|> -TITLE: Irreducibility of Induced Representation -QUESTION [7 upvotes]: Mackey's test for irreducibility of induced representation over $\mathbb{C}$ is: -Let $G$ be a finite group, $H\leq G$, $W$ be a representation of $H$, and $W^x$ be conjugate representation of $H^x=xHx^{-1}$. Then following are equivalent: -(i) $Ind^G_H(W)$ is irreducible. -(ii) $W$ is irreducible and for each $x\in G\setminus H$, $W$ and $W^x$ have no common irreducible component, when restricted to $H\cap H^x$. -There is question posted, about changing the ground field, and the answer posted is (are) "Yes". -But, $(ii)\Rightarrow (i)$ is true for any field of characteristic zero or prime to $|G|$; how does $(i)\Rightarrow (ii)$ for such fields? - -REPLY [7 votes]: No. Over $\mathbb R$ let $G$ be the quaternion group of order $8$, $H$ the subgroup of order $2$, $W$ the nontrivial one-dimensional representation. -EDIT For an even simpler example, see Kevin Ventullo's comment!<|endoftext|> -TITLE: Any example of a non-strong monad? -QUESTION [23 upvotes]: Looking for an example of a monad that is not strong. -The reason being, a strong monad (wrt cartesian product) is an "applicative functor" (in functional programming); an example of a non-strong monad would be useful to see what's breaking in its "applicativity". - -REPLY [2 votes]: Any partially ordered set $(X,\le)$ is a category, and on such a category, the product $\times$ boils down to meets (greatest lower bounds) and monads boil down to 'closure operators'. -For example, the well-known transitive (reflexive) closure is such a closure operator where the underlying category is the category $\mathcal{C}$ of graphs for a fixed vertex-set $V$, i.e. the objects of this category are relations $E \subseteq V\times V$ and $E_1\le E_2$ iff $E_1\subseteq E_2$. Then, it's easy to see that taking the transitive closure $TE=E^*$ of a relation is a closure operator, i.e. a monad on this category $T\colon \mathcal{C}\to\mathcal{C}$. -In order to see that this monad is non-strong in general, consider some $V$ with at least three (distinct) elements, $x,y,z\in V$. Let $E = \{(x,y), (y,z)\}$ and $R=\{(x,z)\}$. Then, -$$ -R\times E = R\cap E = \emptyset -$$ -and so $T(R\times E) = T\emptyset = \{(v,v) \mid v\in V\}$. But on the other hand, -$$ -R\times TE = \{(x,z)\}. -$$ -and so $R\times TE \not\le T(R\times E)$, which shows that transitive closure is a non-strong monad.<|endoftext|> -TITLE: Symbolic computations with differential operators (universal envelopings i.e. non-commutative variables) ? -QUESTION [5 upvotes]: Please give suggestions about soft to make symbolic computations with NON-commutative variables. -Typical examples I am interesting - Capelli identities -http://en.wikipedia.org/wiki/Capelli's_identity -For example let -2x2 matrix X be defined: -$(x_{11}~~~ x_{12})$ -$(x_{21}~~~ x_{22})$ -and D is defined: -$(\partial_{x_{11}}~~~ \partial_{x_{12}})$ -$(\partial_{x_{21}}~~~ \partial_{x_{22}})$ -Then there is identity: -$det^{column}(XD^t+diag(1,0)) = det(X)det(D)$ -Question Is there soft which can easily check it ? -The ideal would be if some one can provide example of code checking this thing. -(I mean it is true, of course, just to understand how code such things). -More general things I would like to do - some computations in universal enveloping -of Lie algebras - like check two expressions commute. -e.g. check that ef+h^2+h is Casimir for sl(2). - -I am familiar with MatLab and Mathematica - but it seems they cannot do this. -May be I am wrong ? -I know that MatLab can differentiate Diff( p,x) - will give symbolic derivative of symbolic function "p" in x. But it seems MatLab cannot do things like d*x-x*d =1... -I have heard that Macaulay2 can do such things - but once I had trouble just installing it, -so I am quite afraid of it... May be I am wrong ? Does it have some graphic interface ? - -REPLY [2 votes]: There is a boat-load of mathematica packages for Lie Algebra computations. Some examples are: -SuperLie -Quantum Mathematica.<|endoftext|> -TITLE: Has anyone seen this sort of graph property used before? -QUESTION [9 upvotes]: Consider the following property of a graph $G$: -The graph $G$ has no independent cutset of vertices, $S$, such that the number of components of $G-S$ is more than $|S|$ (the size of $S$). -(That is, cannot delete 1 vertex and leave 2+ components, cannot delete 2 independent vertices and leave 3+ components etc.) -For some as-yet-unexplained reason, this property has arisen in a couple of questions relating to chromatic roots; needing a name we called this property $\alpha$-1-tough, which uses the notation from graph toughness plus the adjective $\alpha$ to indicate "independent". -Basically we believe that $\alpha$-1-tough graphs are well-behaved with respect to chromatic polynomials; the evidence is that various small graphs that violate certain reasonably well-founded and natural conjectures are very clearly NOT $\alpha$-1-tough. -Having failed miserably at all attempts to prove anything sensible using this property, I wondered if anyone anywhere has seen this, or a similar, graph property appear anywhere. -(I have posted a longer article about this on my (shared) blog, but am not sure of the policy about posting links to your own stuff so I won't do so just in case.) -Edit: The blog entry is http://symomega.wordpress.com/2012/01/06/chromatic-roots-the-multiplicity-of-2/ - -REPLY [3 votes]: A more relaxed notion of independent (or stable) cutsets -- in which the number of remaining components is not relevant -- was studied in relation to the chromatic number in a 1983 paper by Tucker, see -http://dx.doi.org/10.1016/0095-8956(83)90039-4 -More recently, Brandstädt et al. proved that it is NP-complete to recognize whether a graph has a stable cutset even for restricted graph classes, see -http://dx.doi.org/10.1016/S0166-218X(00)00197-9<|endoftext|> -TITLE: Problem about expectation of maximum partial sum -QUESTION [7 upvotes]: Given a number $m$, a random composition (strong) of this number into $n$ positive parts so that we can get $n$ random variable $X_1, X_2,\dots, X_n$ with $$X_1+X_2+\cdots+X_n=m$$ -Note that all compositions of $m$ have the same probability. -Let $Y_i = X_i - \mathbb{E}(X_i)$, $S_i = Y_1 + \cdots + Y_i$ -I want to calculate $\mathbb{E}(\max_{1 \le i \le n} S_i)$. - -REPLY [2 votes]: I can't comment , but based on Noam D. Elkies's numbers and OEIS A000435 it looks as if -$$c_n = \frac{(n-1)!}{2 n^n} \sum_{k=0}^{n-2} \frac{n^k}{k!}.$$<|endoftext|> -TITLE: Is there an inclusion of $L_\infty(G)$ into $C_0(G)^{**}$? -QUESTION [6 upvotes]: This is probably a silly question with which I am stuck however. Let $G$ be a locally compact group. It seems to me that there is a canonical map of $L_\infty(G)$ into $M=C_0(G)^{**}$ (the latter is the enveloping von Neumann algebra of $C_0(G)$). I would reason as follows: - -Let $I$ be the annihilator of $L_1(G)$ in $M$. Then a) $I$ is weakly closed (obvious); b) $I$ is an ideal. To prove the second thing, consider the quotient map $Q: M\to L_1(G)^*$. If we identify $L_1(G)^*$ with $L_\infty(G)$, then $Q$ becomes a homomorphism (it is a homomorphism on $C_0(G)$, and is weakly continuous). So the kernel of $Q$ is an ideal. -For every weakly closed ideal in a von Neumann algebra, there is a complementing ideal $J$ such that $M=I\oplus J$. (it is known that there is a central projection $p\in M$ such that $I=Mp=pM$, then put $J=(1-p)M$). -Now $M/I\simeq J\simeq L_\infty(G)$, so this isomorphism gives an inclusion in the title. - -Why this worries me: since $C_0(G)^{**}$ is the dual space of $M(G)$, the space of Radon measures, then such an inclusion $T$ would mean that we can associate to every $f\in L_\infty(G)$ its value at every point of $G$, by $f(t)=Tf(\delta_t)$, where $\delta_t$ is the probability measure concentrated at $t$. -Am I doing a silly mistake? Thank you in advance! - -REPLY [6 votes]: Since you are in the commutative setting, you can present the construction more simply. $M(G)=L_1(G)\oplus_1 S(G)$, where $S(G)$ is the space of complex measures that are singular with respect to Haar measure, and hence $M(G)^∗=L_\infty (G)\oplus_\infty S(G)^∗$.<|endoftext|> -TITLE: Minimum Number of Partial Views to Cover a Smooth Shape -QUESTION [5 upvotes]: From a viewpoint outside a circle in the plane, only part of the circle is visible, where a point on the circle is visible from the viewpoint if the line segment from the viewpoint to the point on the circle meets no other point of the circle. Call the set of points of the circle that are visible from a given viewpoint the partial view from the viewpoint of the circle. The partial views of enough viewpoints cover the circle (by compactness). What is the minimum number of viewpoints whose partial views cover the circle? Three. Jack up the dimension, and ask the same question about a sphere. The minimum number of viewpoints whose partial views cover the sphere is five. Why? Say there is a viewpoint over the North pole. Then the equator is not seen, no matter how far away is the the viewpoint. Likewise, from beneath the South pole a viewpoint cannot see the equator. So, from those two viewpoints the partial views see all of the sphere except for a (possibly very narrow) band around the equator. By the result for the circle, three more viewpoints are necessary to cover the band. -My question is, what is there in the mathematical literature that addresses the general question of determining the minimum number of viewpoints required to cover a given smooth shape in three-dimensional Euclidean space? For example, what is that minimum number for a torus? Six. - -REPLY [5 votes]: To expand on my comment about the non-smooth equivalent problem, the problem is discussed in Brass, Moser, Pach, Research Problems in Discrete Geometry, Sec. 3.3, under the title -"Levi-Hadwiger covering problem and illumination." They say this about the smooth version [p.136]: - -Levy [Le55] had already observed that if $C$ has a smooth boundary - (i.e., no two supporting hyperplanes pass through the same boundary point of $C$), - then $H(c)=d+1$. - -       - -Here $C$ is a convex body, and $H(C)$ is "the least integer $H$ such that $C$ can be covered -by $H$ smaller homothetic copies of itself." -Boltyanski proved that an illumination version of this problem (which may be the same -as what the OP posed above [but see below]) has the identical answer for any $C$. So -I think that Vitali Kapovitch's answer is the answer for arbitrary smooth $C$. -Another source for the non-smooth case is the Wikipedia article -"Hadwiger conjecture (combinatorial geometry)." -Cubes/hypercubes or parallelepipeds seem to be the worstcase; in fact, that is Hadwiger's conjecture. -In $\mathbb{R}^3$, 16 lights are known to suffice, but it is seems likely that no more than 8 are needed. -But please note the specific definition of illumination/visbility: - -For the purposes of this problem, a body is only considered to be illuminated if for each point of the boundary of the body, there is at least one floodlight that is separated from the body by all of the tangent planes intersecting the body on this point; thus, although the faces of a cube may be lit by only two floodlights, the planes tangent to its vertices and edges cause it to need many more lights in order for it to be fully illuminated.<|endoftext|> -TITLE: Minimal Model Program for surfaces over algebraically closed fields of characteristic p -QUESTION [5 upvotes]: Let $k$ be an algebraically closed field of characteristic $p>0$. -I have been trying to find out unsuccessfully if there is a mmp for algebraic surfaces over $k$. I know minimal surfaces are classified thanks to Zariski, Mumford, others in this setting and that is not my question. I want to 'run' LMMP of pairs not to find minimal model but as a tool to prove several stuff. -In particular I would like to know if there is a Cone Theorem (i.e. giving me that extremal rays have non-positive self-intersection, I suspect the answer is yes) and if there is a contraction theorem (I suspect not yet). -By contraction theorem I mean that if given $(X,D)$ where $X$ is a surface $D$ is an effective divisor and the pair is klt, if $K_X+D$ is not nef I can find a curve $E$ with negative self-intersection such that $E$ can be contracted. -I am aware I am being vague with the formulation but I do not want to constrain your imagination. -Now, if someone also knows if flips and termination of flips are possible, please share :) - -REPLY [8 votes]: In the surface case, MMP in char p is known. See Koll'ar-Kov'ac's preprint on Koll'ar's webpage. -In dimensional 3, the existence of divisorial contractions and flipping contractions is known as EWM (so the target is only known as a algebraic space). See Keel's paper BASEPOINT FREENESS FOR NEF AND BIG LINE BUNDLES. - I'm not sure about the termination of flips. The existence of flips is certainly not known (at this moment). -In higher dimensions, I think almost nothing is known.<|endoftext|> -TITLE: When is a coadjoint orbit an integrable system (in a weak sense explained below)? -QUESTION [21 upvotes]: Let $X$ be an affine holomorphic symplectic variety of dimension $2n$, with the associated Poisson bracket { , }. Let's say it's an integrable system when there are $n$ algebraically independent holomorphic functions $I_i$ ($i=1,\ldots,n$) on $X$ such that they Poisson-commute: $\{I_i,I_j\}=0$. -Pick a simple Lie algebra $\mathfrak{g}$, and pick a coadjoint orbit $O\subset \mathfrak{g}^*$. As is well-known, this is naturally a holomorphic symplectic variety: $x,y\in \mathfrak{g}$ gives a function on $O$, and its Poisson bracket is then given by the Lie bracket: - -$\{x,y\}_{PB}=[x,y]$. - -I wonder which coadjoint orbit is integrable, in the sense given above. -For example, any regular orbit $O$ of $\mathfrak{g}=\mathfrak{gl}(n)$ is integrable: let $X:\mathfrak{g}^*\to\mathfrak{g}$ be the identification via the invariant inner product, choose a generic element $a\in \mathfrak{g}$, and consider a function on $O$ given by - -$P_k(t)= \mathrm{tr} (X+ta)^k$ - -where $t$ is a complex number. (I'm sorry for a slightly confusing notation, but $a$ here is a constant function on $O$ taking value in $\mathfrak{g}$.) -After some manipulation, $P_k(t)$ and $P_{k'}(t')$ are seen to Poisson-commute. Therefore, coefficients $P_{k,i}$ of $t^i$ of $P_k(t)$ all Poisson-commute. Note that $P_{k,k}$ is a constant function on $O$, because $O$ is a coadjoint orbit. Then there are in total $1+2+\cdots+(n-1) = (\dim \mathfrak{gl}(n)-\mathrm{rank} \mathfrak{gl}(n))/2$ independent commuting operators. -Let me conjecture that all coadjoint orbits are integrable. -Another large class of affine holomorphic symplectic varieties are Nakajima's quiver varieties, which include all the coadjoint orbits of $\mathfrak{gl}(n)$. A similar question can be asked: which quiver variety is integrable in this sense? - -Update -Thanks to the answers so far, I could track down references, see e.g. the end of Sec. 4 of this paper, showing that any coadjoint orbit of real compact Lie algebras is integrable. I guess the proof should carry over to the semisimple orbits (and presumably nilpotent Richardson orbits) of complex semisimple Lie algebras, in the holomorphic sense. So the question now is: how about nilpotent orbits in $\mathfrak{g}_{\mathbb{C}}$? -Update 2 -Indeed, A. Joseph says in this article that he essentially proved that Richardson orbits are integrable in this article (in the setting of enveloping algebras, not the associated graded.) It would be interesting e.g. non-special orbits are not integrable... but I have no idea. - -REPLY [11 votes]: Here is answer (YES) from Alexey Bolsinov who is one the main experts in these questions. -"The answer is YES -There is a very general construction allowing to construct an integrable system on more or less any coadjoint orbit for an arbitrary Lie algebra (non necessarily semisimple). This is a recent paper by Vinberg and Yakimova available in arxiv -http://arxiv.org/abs/math/0511498 -Complete families of commuting functions for coisotropic Hamiltonian actions -In the particular case you are talking about (SEMI-SIMPLE g) the positive answer follows from 2 results: -1) the so-called shifts of polynomial invariants give a completely integrable system on a singular adjoint orbit O(b) in a semi simple Lie algebra G if and only if -the index of the centralizer of b coincides with the index of G -(my paper in Izvestija AN SSSR, 1991 and Acta Appl. Math. 1991), both available on my home page -Bolsinov A.V. Commutative families of functions related to consistent Poisson brackets// Acta Appl. Math., 24(1991), pp. 253-274. -I also conjectured that -this condition ind Cent (b) = ind G, in fact, holds true for all singular elements b\in G and checked it for G=sl(n) (in particular for all nilpotent) -2) This conjecture (widely known as Elashvili conjecture) has been proved for an arbitrary semi simple Lie algebra and for all elements (in fact the proof is easily reduced to nilpotent elements) -First, Elashvili did it by, in some sense, straightforward computation which in the most difficult case of e_8 involved some computer program (unpublished) -Recently a conceptual proof has been done by Jean-Yves Charbonnel (IMJ), Anne Moreau (available in arxiv) -http://arxiv.org/abs/1005.0831 -The index of centralizers of elements of reductive Lie algebras -To the best of my knowledge, this is the only known universal way to construct an integrable system on an arbitrary orbit. -Remark: I am talking about classical integrable systems, not quantum. These systems can be quantized too, but this is another story. -"<|endoftext|> -TITLE: Analogue of van der Corput sequence for prime numbers -QUESTION [6 upvotes]: A van der Corput sequence is a low-discrepancy sequence over the unit interval first published in 1935 by the Dutch mathematician J. G. van der Corput. It is constructed by placing a decimal point and writing the base $n$ representation of the sequence of natural numbers in the reverse order. Read more. -This question is motivated by curiosity to study the analogue of van der Corput sequence for prime numbers. Consider the sequence $v_p$ formed by placing a decimal point and writing the digits of the sequence of prime numbers $p$ in the reverse order (in base 10). The first few terms of this sequence are -$$ -0.2, 0.3, 0.5, 0.7, 0.11, 0.31, 0.71, 0.91, 0.32, 0.92, \ldots -$$ -Clearly $v_p$ is not equidistributed in the unit interval and therefore it is not a low discrepancy sequence. I am curious to know if $v_p$ has anything interesting property in it. -Q1. What is the mean value of the sequence $v_p$? In other words does the following limit exist? -$$ -\lim_{x \rightarrow \infty}\frac{1}{x}\sum_{p \le x}v_p. -$$ -Edit (Adding Timothy's guess as a question) -Q2. What would be the mean value of base $b$? -Q3. Edit (@ one more variation for this problem) -Find a closed form representation of -$$ -\lim_{x\to\infty}\frac{1}{x}\sum_{p\le x}f(v_p) -$$ -where $f$ is any function Riemann integrable in $(0,1)$? -Considering the analogy with the equidistribution theorem, I think this should be possible. - -REPLY [11 votes]: While $\lbrace v_p \rbrace$ is clearly not equidistributed in $(0,1)$, it is equidistributed in -$$ -\Pi_{10} := [.1,.2) \cup [.3,.4) \cup [.7,.8) \cup [.9,1) -$$ -by the prime number theorem (PNT) for arithmetic progressions modulo powers of $10$. In particular, the average tends to $0.55$, the average of the midpoints $0.15$, $0.35$, $0.75$, and $0.95$ of these four intervals. -[I see that Frank Thorne wrote much the same thing in a comment as I was editing this...] -[added in reply to further inquiries:] Replacing $10$ by an arbitrary base $b>1$, we likewise use the PNT for arithmetic progressions (APs) modulo powers of $b$ to show that the sequence is equidistributed in -$$ -\Pi_b := \lbrace x \in [0,1) : \gcd(\lfloor bx \rfloor, b) = 1 \rbrace. -$$ -(I made the intervals half-open in $\Pi_{10}$ for consistency with this definition, though it doesn't change the distribution.) That's a union of $\varphi(b)$ intervals of length $1/b$ whose set of left endpoints is symmetric about $1/2$, so the average is $1/2 + 1/(2b) = (b+1)/2b$ as Timothy Foo surmised. -On further thought, not only does equidistribution in $\Pi_b$ follow from the PNT for APs modulo every power of $b$, but the two statements are readily seen to be equivalent.<|endoftext|> -TITLE: The first eigenvalue of the laplacian for complex projective space -QUESTION [15 upvotes]: What is the exact value of the first eigenvalue of the laplacian for complex projective space viewed as $SU(n+1)/S(U(1)\times U(n))$? - -REPLY [18 votes]: The spectrum of the Laplacian of $\mathbb C P^n$ with the Fubini-Study metric is - -$$Spec(\Delta_{\mathbb C P^n})=\{4k(n+k):k\in\mathbb N\} \quad\quad(*)$$ - -So, the first non-zero eigenvalue of $\mathbb C P^n$ is $\lambda_1=4n+4$. -Note this matches with the fact that $\mathbb C P^1$, with the FS metric, is isometric to the $2$-sphere of radius $1/2$, whose first non-zero eigenvalue is $\lambda_1=8$. - -Let me quote a brief justification of (*) that I had written here: -Using the classic spherical harmonics theory, one obtains the $k$-th eigenvalue of the $n$-dimensional round sphere $S^n$ to be $k(k+n-1)$, and its multiplicity is $\binom{n+k}{k}-\binom{n+k-2}{k-2}$. By looking at eigenfunctions of the Laplacian on $S^n$,$S^{2n+1}$ and $S^{4n+3}$ (note they are the unit spheres of $\mathbb R^{n+1}$, $\mathbb C^{n+1}$ and $\mathbb H^{n+1}$) that are respectively invariant under the natural actions of $\mathbb Z_2$, $S^1$ and $S^3$, one can obtain the eigenfunctions hence the $k$-th eigenvalue of the projective spaces $\mathbb R P^n$, $\mathbb C P^n$ and $\mathbb H P^n$, respectively. These are, respectively, $2k(n+2k-1)$, $4k(n+k)$ and $4k(k+2n+1)$. -If you can understand some French, you will find a thorough explanation of the above in the book by Berger, Gauduchon, Mazet, "Le spectre d'une variete Riemannienne", Lecture Notes in Math, Springer, vol 194; but beware of a small typo [pointed out to me by G. Wei] on the multiplicity of the $k$th eigenvalue of $S^n$ (cf. above). -More generally, it is possible to compute the Laplace spectrum of a compact homogeneous space using some representation theory, as explained e.g. in Nolan Wallach's "Minimal immersions of symmetric spaces into spheres" or in more detail on his book "Harmonic analysis on homogeneous spaces" (Pure and applied mathematics, 19).<|endoftext|> -TITLE: History of the four-colour problem -QUESTION [43 upvotes]: It is stated in many places that the first published reference to the four-colour problem (aka the four-color problem) was an anonymous article in The Athenæum of April 14, 1860, attributed to de Morgan. -I was poking around in earlier issues of The Athenæum and found this on page 726, June 10, 1854: -Tinting maps.—In tinting maps, it is desirable for the sake of - distinctness to use as few colours as possible, and at the same - time no two conterminous divisions ought to be tinted the same. - Now, I have found by experience that four colours are necessary - and sufficient for this purpose,—but I cannot prove that this is - the case, unless the whole number of divisions does not exceed five. - I should like to see (or know where I can find) a general proof of - this apparently simple proposition, which I am surprised never - to have met with in any mathematical work.   F.G. -I cannot find any mention of this item anywhere, so my question is whether this information is new. -As far as I can tell, "F.G." is not identified by the magazine. Two obvious candidates are Francis Guthrie and his brother Frederic Guthrie, who discussed the question starting in 1852. An outside possibility is Francis Galton, who was involved in the problem at a later date (see Crilly, Notes and Records of the Royal Society of London, Vol. 59, No. 3 (Sep. 22, 2005), pp.285-304). So my second question is "who was F.G."? -Added: http://arxiv.org/abs/1201.2852 Suggestions for improvement welcome. - -REPLY [10 votes]: First, congratulations to Brendan for a remarkable discovery. I think it -opens up the possibility of more documentary evidence on the early history of the 4CC. We know that in addition to the famous letter to Hamilton in 1852, DeM -also wrote to Whewell (9/12/53) and Ellis (24/6/54) about it. I discussed these -letters in a short paper in the Archive for History of Exact Sciences in 1983 -(pp163-170). The fact that the letter to Ellis is very soon after the the FG -Athenaeum piece raises some obvious questions. Did Ellis write to DeM about -the 4CC? Why is the DeM to Ellis letter in the Whewell papers at Trinity? -Was there other correspondence stemming from the FG piece?<|endoftext|> -TITLE: Implicit function theorem at a singular point? -QUESTION [11 upvotes]: Let $F:\mathbb{R}^2 \rightarrow \mathbb{R}$ be three times continuously differentiable in some open neighborhood $\mathcal{U}$ of $(0,0)$. Suppose that $F(0,0) = F_x(0,0) = F_y(0,0) = F_{xy}(0,0) = 0$ and that $F_{yy} \not = 0$ and $F_{xx}(0,0)/F_{yy}(0,0) < 0$. -Obviously I cannot apply the implicit function theorem. Under which circumstances can I still express y as a function of x locally aoround (0,0)? It seems as though it should be $y'(0) = \pm \sqrt{-\frac{F_{xx}(0,0)}{F_{yy}(0,0)}}$... -EDIT: Fixed typo. - -REPLY [8 votes]: Scketch of an elementary proof. Assume $F\in C^2(\mathbb{R}^2, \mathbb{R})$ and -(changing sign to $F$ if needed) -$$F(0,0)=F_x(0,0)=F_y(0,0)=F_{xy}(0,0)=0\, ,$$ -$$F_{xx}(0,0) <0\, ,\qquad F_{yy}(0,0) > 0\, .$$ -Therefore there exist $\eta > 0$ and $\epsilon > 0$ such that -$F_{yy}(x,y) > 0$ for $|x| \le \eta$ and $|y|\le\epsilon$. So for all $|x|\le \eta$ -the function $y\mapsto F(x,y)$ is strictly convex on the interval -$[-\epsilon,\epsilon]$. In particular $F(0,\pm\epsilon) >0$ because -$F(0,0)=F_y(0,0)=0$. Since $F(0,\pm\epsilon) >0$ and $F_{xx}(0,0) < 0$, we also -have by continuity $F(x,\pm\epsilon) >0$ and $F_{xx}(x,0) < 0$, for all -$|x|\le\delta$ for some $0 < \delta\le\eta$; thus $F(x,0) < 0$ for $0 < |x|\le -\delta$. Now, since for all $|x|\le\delta$ the function $y\mapsto F(x,y)$ is -strictly convex on the interval $[-\epsilon,\epsilon]$, positive at $y=\pm\epsilon$ -and negative at $y=0$, for any $0 < |x|\le\delta$ we have $F(x,y)=0$ exactly for one -$0 < y < \epsilon$ and one $-\epsilon < y < 0$, always with $F _ y (x,y)\neq0$, -while $F(0,y)=0$ exactly for $y=0$ if $|y|\le\epsilon$. This proves that the trace -of the zero-set of $F$ on $[-\delta,\delta]\times [-\epsilon,\epsilon]$ is the union -of the graphs of two functions, $y_+: [-\delta,\delta]\to [-\epsilon,\epsilon]$ and -$y_-: [-\delta,\delta]\to [-\epsilon,\epsilon]$ defined so that $\operatorname{sgn} -y _ + (x)=\operatorname{sgn} x$ and $\operatorname{sgn}y _ - (x)=-\operatorname{sgn} -x$. Note that the fact that $\epsilon$ is arbitrary immediately implies that $y_+$ -and $y_-$ are continuous at $x=0$ and vanish there. Actually, if we locate the zero-set of $F$ with a bit more care we also have that $y _ \pm (x) $ -is derivable at $x=0$: this follows from the fact that $F$ satisfies an inequality locally at the origin: -$$\big(F_{xx}(0,0)+o(1) \big)x^2/2 +\big( F_{yy}(0,0)+o(1) \big)y^2/2 \le F(x,y) $$ -$$\le \big(F _ {xx} (0,0) +o(1) \big) x^2/2 + \big( F_{yy} (0,0) +o(1) \big)y^2/2$$ -so that $(x,y _ \pm (x))$ belongs to a very thin cone around the line $y= \pm F_{xx}(0,0)/F_{yy}(0,0)x$, meaning that $y _ \pm (x)$ is derivable at $x=0$ and $ y ' _ \pm (0)= -\sqrt {- \frac { F _ {xx}(0,0) } { F_ {yy} (0,0) } }\, . $ - Moreover, since for all -$ (x,y) \in \{ F = 0 \} \cap [ -\delta, \delta] \times [-\epsilon,\epsilon] -\setminus\{ (0,0)\}$ we have $F _ y(x,y)\neq 0$ the standard Implicit Function -Theorem ensures that $y_+$ and $y_-$ are $C^0([ -\delta, \delta])\cap C^1([ -\delta, \delta]\setminus \{ 0 -\}) $, with $$F_x(x,y_\pm(x))+F_y(x,y_\pm(x))\dot y_\pm(x)=0\quad , \quad \forall x\neq0 .$$ To prove that they are $C^1([ -\delta, \delta])$ note that for $|x|+|y|\to 0$ -$$F_x(x,y)=F_{xx}(0,0)x+o(|x|+|y|)$$ -$$F_y(x,y)=F_{yy}(0,0)y+o(|x|+|y|)\, .$$ -Therefore we have -$$F _ {xx}(0,0)+y' _ \pm (x)\frac{y _ \pm(x)}{x}F _ {yy}(0,0)=\Big(1+|y' _ \pm(x)|\Big)\Big(1+\big|\frac{y _ \pm(x)}{x}\big|\Big)o(1)\, .$$ -Since $\frac{y_\pm(x)}{x}=y_\pm'(0)+o(1)$, and $\pm y'_ \pm(x)\ge0$ this implies that -$y^\prime _ \pm (x) \to y^\prime _ \pm (0) $ for $ x \to 0 $ , so $ y^\prime _ \pm $ is continuous and $ y _ \pm \in C^1 ([ - \delta , \delta ] ) $ .<|endoftext|> -TITLE: Intersection of complemented subspaces of a Banach space -QUESTION [14 upvotes]: The following seems a very basic question in the theory of complemented subspaces of Banach spaces, but I was not able to find a reference, so I wish to ask it here. - -Question. Let $X$ be a Banach space, and let - $V$ and $W$ be complemented subspaces - of $X$. Is it true that $V \cap W$ - is a complemented subspace? If not, is it true under (nontrivial) additional assumptions? - -In the case of a Hilbert space $X$, where the answer is of course yes, the orthogonal projector onto $V \cap W$ may be found as a strong limit of operators $P_{V\cap W}=\lim_{n\to \infty}(P_V P_W)^n $ . Is there a similar procedure to obtain a linear projector onto $V\cap W$ in the general case of a Banach space $X$? - -REPLY [16 votes]: The answer to the first question is "no". You can see this with specific examples, but here is a more conceptual approach: Take $Y$ an uncomplemented subspace of $X$ and in $Z:= X\oplus_1 X$ identify $Y\oplus 0$ with $0 \oplus Y$ in the obvious way; that is, mod out from $Z$ the subspace $\{(y,-y) | y \in Y\}$. $X\oplus 0$ and $0 \oplus X$ are norm one complemented in the resulting quotient space of $Z$ but their intersection $Y \oplus 0 = 0 \oplus Y$ is not complemented. (This is just a categorical push out construction specialized to the appropriate category of Banach spaces.) -The answer is yes if the subspaces are norm one complemented and the space $X$ is uniformly convex. This is intuitive, because if $P$ is a norm one projection on a uniformly convex space and $x$ is not in the range of $P$, then $\|Px\| < \|x\|$, since otherwise all vectors on the line segment from $x$ to $Px$ would have norm $\|x\|$. Hence one guesses that playing ping pong with two norm one projections $P$ and $Q$ will produce a norm one projection onto $PX\cap QX$. To see that this works without doing any computations or calculating rates of convergence (at the risk of making experts cringe), set $P_1=P$, $P_{2n} = QP_{2n-1}$, $P_{2n+1}=PP_{2n}$. Let $x\in X$ and let $a=a(x)$ be the limit of the nonincreasing sequence $\|P_n x\|$. I claim that $\|P_{n+1}x - P_{n}x\| \to 0$. Indeed, (1/2)$\|P_{n+1}x + P_{n}x\|$ also converges to $a$, so the claim follows from the uniform convexity of $X$. Let $V$ be a limit in the weak operator topology of some subnet of $P_{2n}$. By the claim, the corresponding subnet of $P_{2n+1}$ also converges to $V$ in the weak operator topology. From this it is evident that $V$ is a norm one projection onto $PX\cap QX$. -ADDED 13 Jan. 2012: Notice that in the first construction $X$ can be uniformly convex, in which case $Z$ (and therefore also every quotient of $Z$) is isomorphic to a uniformly convex space.<|endoftext|> -TITLE: shape of random q-weighted lattice path -QUESTION [9 upvotes]: Where can I find a detailed write-up of the asymptotic shape of a $q$-weighted Young diagram inside an $a$-by-$b$ box, especially one that uses a variational approach? -Equivalently, we can look at probability distribution on binary sequences of length $a+b$ consisting of $a$ 0's and $b$ 1's, where the probability of a particular sequence is proportional to $q$ to the power of the number of inversions (here an inversion is a pair of indices $i \lt j$ such that the $i$th bit is 1 and the $j$th bit is 0); equivalently, the probability of a particular sequence is proportional to $q$ to the power of the sum of the indices $i$ such that the $i$th bit is 0. -What I mean more precisely is that $a+b$ goes to infinity while $a/b$ converges to some finite non-zero number, and we rescale space by a factor of $a+b$. I looked at this about twenty years ago, and as I recall, if $q$ goes to 1 at the right rate, the picture stabilizes and one gets an asymptotic shape theorem, but I never worked out all the details of the argument. I could try to reconstruct and patch it, but even then I was pretty sure that the result must be "well known to those who know it", and I'm even more confident of that now, given all the work that people are doing on harder problems of a similar flavor. So this must be in the literature, but where? (Maybe in the exclusion-model literature?) - -REPLY [4 votes]: This question is treated as well as the fluctuation problem in a paper I wrote with Dan Beltoft and Cédric Boutillier -http://arxiv.org/abs/1008.0846 -to appear in Moscow Math Journal. -The approach is based on the use of $q$-Gauss polynomial and the proof is a $q$-analog of the -Moivre Laplace proof of the CLT. Fluctuations are given by the bridge of a Ornstein-Uhlenbeck process (tightness is the most delicate point of the paper). -I am sure the variational approach is efficient for the more elementary limit shape problem since the functional you have to minimize is the same as in Vershik problem of the limit shape of Tableaux diagrams having a prescribed (large) surface (leading to "Vershik's curve" e^(-x)+e^(-y)=1 ). The only difference comes from the boundary conditions. As a result, the limit shape is nothing but the restriction of Vershik's infinite curve to some interval of R (there is only one possible choice of interval which will match with the prescribed boundary conditions). -I don't know if I was completely clear, but there is a discussion of this kind at the end of our paper as well as some recent reference of F. Petrov considering the limit shape problem.<|endoftext|> -TITLE: center of fundamental group of finite volume-hyperbolic orbifold -QUESTION [5 upvotes]: Is center of a fundamental group of finite volume-hyperbolic n-orbifold trivial? -Is there a good reference that the proof is wriiten? - -REPLY [6 votes]: Let $G$ be the fundamental group of your orbifold, and recall that $G$ is a lattice in the group of isometries of hyperbolic space. -An element in the center of $G$ has to fix every fixed point of any loxodromic and any parabolic element of $G$. The fixed points of such elements are dense in the boundary at infinity of hyperbolic space (this is a consequence of the fact that the orbifold has finite volume, I think that a proof of this fact can be found in Ratcliffe's book, for example). -This implies that any element in the center of $G$ acts as the identity on the boundary of hyperbolic space, so the center of $G$ is trivial.<|endoftext|> -TITLE: Module category equivalent to graded module category? -QUESTION [9 upvotes]: Main Question -Let $R$ be a graded ring, graded by the nonnegative integers. Denote by $\mathrm{gr}R-\mathrm{Mod}$ the category of $\mathbb{Z}$-graded left $R$-modules with morphisms that preserve the grading. Is there a ring $S$ such that the category $S-\mathrm{Mod}$ of left $S$-modules is equivalent to $\mathrm{gr}R-\mathrm{Mod}$? -As the category of graded $R$-modules is abelian, the Freyd-Mitchell theorem guarantees an exact embedding into a module category, but this is not necessarily an equivalence of categories, right? - -Motivation -My motivation for the question is an offhand remark made to me indicating that for a given ring $A$, there is a ring $S$ such that the category of complexes of $A$-modules is equivalent to the category of $S$-modules. If you define the graded ring $R = A[t]/(t^2)$, graded by powers of $t$, then (I think) complexes of $A$-modules are equivalent to graded $R$-modules, so the question is reduced to the main question stated above. -Of course, it could be the case that the answer to the original question I asked is negative, and yet the offhand remark is still true, in which case I would be interested in hearing about why that is. - -Edit -I neglected to mention that I was hoping for a unital ring $S$ with unital modules. As several people have pointed out in the comments, this is not possible. I thought I would put up an argument to show why this is in case people come looking at this post in the future. -Theorem 1 of Chapter 4, Section 11 of the book Categories and Functors, by Bodo Pareigis, gives a complete characterization of when an abelian category $\mathcal{C}$ is a module category. The criteria are that $\mathcal{C}$ must contain a progenerator (i.e. a finite, projective generator; I had to look that up) and it must contain arbitrary coproducts of that generator. -Now let's see that $\mathrm{gr}R-\mathrm{Mod}$ cannot contain a finite progenerator. Take any finite (hence finitely generated) projective module $P = \bigoplus_{n \in \mathbb{Z}} P_n$. Since $P$ is finitely generated, there is some index $k_0$ such that $P_n = 0$ for $n -TITLE: Original proof of Pappus' Hexagon Theorem -QUESTION [5 upvotes]: Does anyone know where I can find an English translation, preferably online or in a book the library of a small liberal arts college would be likely to have, of the original proof of Pappus' hexagon theorem from projective geometry? - -REPLY [8 votes]: It is in book 7 of Pappus's Collection. There is an English translation published by Springer. It's an expensive book, so I'll reproduce the proof & figure here: - - -It relies on earlier lemmas; there's an explanation that includes them on Wikipedia.<|endoftext|> -TITLE: Reference request: Equivariant Topology -QUESTION [18 upvotes]: I am teaching a graduate seminar in equivariant topology. The format of the course is that I will give 2-3 weeks of background lectures, then each week a student will present a topic. The students have all taken a basic course in algebraic topology (they know homology/cohomology and fundamental groups), but some may not know much more topology than that. Topics will likely include equivariant cohomology, (equivariant) bundles and characteristic classes, equivariant K-theory, and important classes of examples interspersed, including toric varieties, homogeneous spaces, and the Hilbert scheme of points in $\mathbb{C}^2$. My personal goal is to learn a bit about Bredon cohomology for compact, connected Lie groups (I'm happy to restrict that a bit, but probably not to finite groups). -Some references that I already have in mind include those listed in David Speyer's question and answer about equivariant K-theory. For Bredon cohomology, there are two books: Equivariant Cohomology Theories by G. Bredon, and Equivariant Homotopy and Cohomology Theory by J.P. May (with many other contributors). - -Reference request: What are classic papers in equivariant topology that every student should read? - -REPLY [19 votes]: If I may be so bold, I would actually strongly suggest you start with finite groups, rather than compact Lie. While many of the results in equivariant homotopy are true in both cases, the formulations for finite groups are often easier to understand. Additionally, there are twists that show up in the compact Lie case which just make exposition (and I find comprehension) a good bit trickier. -For a finite group, it is very easy to carry out computations with Bredon homology and cohomology. In fact, it's easy to write down chain complexes of Mackey functors which do everything for you. For compact Lie, you can of course, do the same thing; I personally find it substantially harder and less intuitive.<|endoftext|> -TITLE: The historical development of automorphic geometry -QUESTION [5 upvotes]: Background: - Today the notion of automorphic geometry is often framed in the context of the Langlands program, in particular what is sometimes called the Langlands reciprocity conjecture. This is often phrased as the expectation that all motives are automorphic. The historical development of this notion is fairly extensive, and can certainly be traced back more than a century to the work of Felix Klein and his student Adolf Hurwitz. One historical thread that therefore could be identified is the "elliptic lineage", with important contributions from Klein, Hurwitz, Hecke, Artin, Eichler, Taniyama, Shimura, Weil, Langlands, Deligne, Serre, Wiles, Taylor, and probably others. -I'm mainly interested in whether the notion of automorphic geometry (motives) has roots that go back farther than Klein and Hurwitz, but I'd also like to use this historical framework to ask what other "automorphic lineages" might be identified, possibly with overlaps with the elliptic lineage above. -Questions: -1) Is there work on automorphic geometry that predates that of Klein and Hurwitz? -2) Are there other lineages in the automorphic genealogy? - -REPLY [2 votes]: A common answer to question 1 is to mention the entries of Gauss's diary from 1814, including famously (but not restricted to) the last one, in which he studies some properties of biquadratic reciprocity and their links with the division of the lemniscate. Gauss understood these properties to derive from properties of Punktgitter (lattices of points) in the complex planes and his formulation is surprisingly close to the statement that (some) elliptic curves with CM are automorphic objects. -Though I am not sure that this is the best way to think about the work of Gauss from a historical point of view (he himself seems to have been unsatisfied with this work and to have considered it of relatively low value compared to the latter approach of Einsenstein) and though these findings had no impact on mathematics at the time for the simple reason that they were kept private for over 80 years, it certainly counts as work in "automorphic geometry" done in the early XIX° century.<|endoftext|> -TITLE: Extreme points of a set of probability measures -QUESTION [11 upvotes]: Consider the set of Borel-measurable probability measures over the interval $[0,1]$ with a given mean, say 1/2. To be precise, I'm talking about the following set $$M=\left(\mu\in \Delta([0,1]):\int x\mu(dx)=1/2\right)$$ -This is a convex set, with convex combinations defined as $\mu=\alpha \eta +(1-\alpha)\xi$ when $\mu(E)=\alpha \eta(E)+(1-\alpha)\xi(E)$ for every Borel set $E\subset[0,1]$. -My question is: what is the set of extreme points of this convex set? -My conjecture is that it is the subset of probability measures with support consisting of at most two points. I'm fairly convinced that this is true, but so far I haven't been able to come up with an elegant argument to show this and I thought that perhaps this is a well-known result. It is not hard to show for the case of atomless measures or for the case of discrete measures, but I'd like to have the general case. - -REPLY [14 votes]: This question (where you prescribe a set of moments, on an arbitrary measure space) is completely answered in this very cool paper. (G. Winkler, Extremal points of moment sets).<|endoftext|> -TITLE: Realizing groups as commutator subgroups -QUESTION [23 upvotes]: What are the groups $X$ for which there exists a group $G$ such that $G' \cong X$? -My considerations: - -$\bullet$ If $X$ is perfect we are happy with $G=X$. -$\bullet$ If $X$ is abelian then $G := X \wr C_2$ verifies $G'=\{(x,x^{-1}): x \in X\} \cong X$. -$\bullet$ If $X$ satisfies the following properties: -(1) $X \neq X'$, (2) The conjugation action $X \to \text{Aut}(X)$ is an isomorphism, -then there is no $G$ such that $G' = X$ (consider the composition $G \to \text{Aut}(X) \cong X \to X/X'$, it is surjective so its kernel contains $X$, contradiction). For instance, the symmetric group $S_n$ verifies (1) and (2) if $n \neq 2,6$. - -I have been looking for this problem on the web but I didn't find anything. -Do you have any reference and/or suggestion on how to solve this problem? - -REPLY [7 votes]: Let me quote some well-known results and perhaps related problems which may be illuminating! -Let $G$ be a non-abelian finite $p$-group having cyclic center. Then, there is no finite $p$-group $H$ such that $G$ is isomorphic to a normal subgroup of the derived subgroup $[H,H]$ of $H$. In particular, $G$ cannot be isomorphic to the derived subgoup of some $p$-group $H$. The latter is a famous result due to Burnside. The former is a slight generalization of problem. See H. Heineken, On normal embedding of subgroups. -Geom. Dedicata 83, No.1-3, 211-216 (2000). -Related problem: Let $V$ be a non-empty set of words in the free group on the countable set -$\{x_1,x_2,\dots\}$. We call a group $G$ is {\bf integrable with respect to $V$}, whenever there is a group $H$ such that $G\cong V(H)$, where $V(H)$ is the verbal subgroup of $H$ generated by $V$, i.e., $$V(H)=\langle v(h_1,\dots,h_n) | v\in V, h_i \in H \rangle,$$ -(the subgroup of $H$ generated by the values of words of $V$ on the elements of $H$) -For example, if one takes $V=\{[x_1,x_2]=x_1^{-1}x_2^{-1}x_1 x_2\}$, then $V(H)$ is the derived subgroup of $H$ for any group $H$ and the problem is the same as it proposed. -One may write (maybe for some propaganda) -$$\int G \; dV=H \Longleftrightarrow G=V(H).$$ -In the case, $V=\{ [x_1,x_2]\}$, $$\int G=H \Longleftrightarrow G=H'$$ and so -$$\int G=H \Longleftrightarrow \int G=H \times A$$ for any abelian group $A$. (This may remind the constant term in the integral of a function!) -I have used the latter notation which has no benefit but only perhaps inspiring in -[A. Abdollahi, Integral of a group, 29th Iranian International Conference on Mathematics, Amirkabir University of Technology (Tehran Polytechnic), Iran, March 28-31, 1998.] -I would like to say that the problem given a group $G$, find groups $H$ such that $G=[H,H]$" have been studied in a more general contex which may be found with key wordsnormal embedding of subgroups" and the above-mentioned paper of Heineken is a good start. -Also it may be worth-mentioning that by a result of Allenby -R.B.J.T. Allenby, Normal subgroups contained in Frattini subgroups are Frattini subgroups, Proc. Amer. Math. Soc, Vol. 78, No. 3, 1980, 318- -if $N$ is a normal subgroup of a finite group $G$ which is contained in the Frattini subgroup of $G$, then $N=\Phi(U)$ for some finite group $U$. -Of course, for the class of finite $p$-groups the Frattini subgroup is the verbal subgroup -generated by the words $x_1^p, [x_1,x_2]$.<|endoftext|> -TITLE: Extensions of the Koebe–Andreev–Thurston theorem to sphere packing? -QUESTION [12 upvotes]: The Koebe–Andreev–Thurston theorem states that any planar graph can be represented -"in such a way that its vertices correspond to disjoint disks, which touch if and only if -the corresponding vertices are adjacent" (to quote Günter Ziegler, Lectures on Polytopes, Springer, 1995 p.117. -(See also the Wikipedia article, "Circle packing theorem.") -          - -(source: uci.edu) - -(Image due to David Eppstein, here.) - -What is the corresponding statement for spheres in $\mathbb{R}^3$? - Every graph $G$ satisfying property $X$(?) can be represented by touching spheres. - -This is surely known—Thanks for pointers! - -REPLY [13 votes]: I don't know of any results on this. There are some reformulations in terms of Gram matrices, but I don't know if they help. -For $K_6$, there is no realization by touching spheres. Suppose you had such a realization. The property of having tangent spheres realizing a graph is invariant under Mobius transformations. In particular, one may perform a Mobius transformation sending the tangency between two spheres to infinity in $\mathbb{R}^3$. The two spheres are sent to parallel planes, and the other 4 spheres are tangent to both of these planes and to each other. In particular, the midplane between these two planes intersects the other 4 spheres in 4 tangent circles of equal radius. But 4 equal radius circles in the plane cannot be simultaneously tangent. So $K_6$ cannot be realized by tangent spheres. - -REPLY [8 votes]: According to corollary 4.6 of "Representing Graphs by Disks and Balls (a survey of recognition-complexity results)" these graphs are NP hard to recognize. - -REPLY [7 votes]: Yes, certain restrictions are well known. One reference is Kuperberg & Schramm here ("Average kissing numbers for non-congruent sphere packings", 1994) which says that such graphs would have to have average degree <15. A more recent reference is Benjamini & Schramm here ("Lack of Sphere Packing of Graphs via Non-Linear Potential Theory", 2009) which shows that certain low degree infinite graphs are not realizable this way.<|endoftext|> -TITLE: Are nilpotent orbits degenerations of semi-simple orbits ? -QUESTION [7 upvotes]: "Examples first:" -Consider so(3,C). (Co)Adjoint Orbits can be described by equations -x^2+y^2+z^2 = R. -R=0 - is nilpotent cone - algebraic closure of the orbit of nilpotent element. (It is union of two orbits - {0} and {Cone w/o {0}} ). -R$\ne$ 0 are orbits of semi-simple elements. -So we have degeneration R->0 - semi-simple orbit degenerates to nilpotent. -Question Is there similar description for the other nilpotent orbits in higher dimensions e.g. for gl(n,c) ? I mean can we write some equations depending on parameters F_t(g)=0, -such that for general "t" we get semi-simple orbits, but for specific values we have nilpotent orbit (more precisely their closures)? (Here "t" can be vector and F is vector-valued algebraic function). -Of course this can be done the biggest orbit - for nilpotent cone itself. -Consider matrices "M" which satisfy the condition, that -their characterestic polynom is fixed with values eigs $a_i$: -$det(M-x) = (x-a_1)(x-a_2)...(x-a_n)$ -For $a_i$ generic - this is semisimple orbit, but if $a_i = 0$ we get nilpotent cone. -Question Reformulated Is it possible to do the same for smaller dimensional orbits ? - -As far as I heard nilpotent orbits can be described by the equations on their rank and -$M^l=0$, however this does not seems to answer the question. - -Part of motivation for asking is related to the following questions: -On an affine analogue of the fact $\mathbb{C}[\mathfrak{g}]^G$ is a polynomial ring -Primitive ideals of the universal enveloping algebras of affine Lie algebras -In particular if the answer would be YES - then probably we can do the same in the -"affine case" so answering the question "What replaces the concept of the nilpotent orbit in that case?" - -REPLY [7 votes]: The answer is often "yes". Here is the sketch of how to obtain a nilpotent orbit as a degeneration of semisimple orbits in the $GL_n$ case. Let $d$ be a partition of $n$ with $k$ parts and $\overline{\mathcal{O}}_{d'}$ be the closure of the conjugacy class of nilpotent $n\times n$ matrices with Jordan blocks sizes given by the dual partition $d'.$ Denote by $\mathcal{O}_d(t_1,\ldots, t_k)$ the conjugacy class of the block diagonal matrix with scalar diagonal blocks $t_i I_{d_i}.$ Then -$$\lim_{t\to 0}\ \mathcal{O}_d(t_1,\ldots, t_k)=\overline{\mathcal{O}}_{d'}.$$ -This is manifested on the level of defining equations using Oshima's approach from - -A quantization of conjugacy classes of matrices. Adv. Math. 196 (2005), no. 1, 124–146. - -For a general $\mathfrak{g},$ this amounts to the induction of (zero-dimensional) orbits and to the correspondence between semisimple and regular orbits. In particular, every Richardson nilpotent orbit can be obtained as a degeneration in the same way. However, the defining equations are not known to the same degree of explicitness. -On the other hand, if $\mathfrak{g}$ is simple of type other than "A" then the minimal -nilpotent orbit is rigid, meaning that it cannot be deformed within the family of adjoint orbits. Existence of rigid orbits makes quantization of orbits a non-trivial task, since a very natural prescription for quantization of semisimple orbits needs to be supplemented by ad hoc quantizations of rigid orbits (several papers of Joseph addressed this question). Rigid orbits have been completely classified: if my memory serves, the answer is in Collingwood-McGovern.<|endoftext|> -TITLE: Reference request: Roger Howe's Schur lectures -QUESTION [13 upvotes]: I have spent a lot of time trying to track down the following without any luck: -Perspectives on invariant theory: Schur duality, multiplicity-free actions and beyond, -The Schur lectures (1992) (Tel Aviv), 1995, pp. 1–182.MR1321638 (96e:13006) -by Roger Howe. -Does anyone know where I could find a copy? - -REPLY [14 votes]: This appears to be a (complete?) scanned version: - Chapters 1-4 - Chapter 5 -Appendix -References<|endoftext|> -TITLE: Is the cardinality of occuring torsion subgroups in cofinite lattices in SL(2,R) bounded? -QUESTION [5 upvotes]: Let $\Gamma$ be a cofinite lattice in $PSL(2,\mathbb{R})$ with torsion subgroup $H$. -Is the a uniform bound on the cardinality of $H$? - -REPLY [6 votes]: There are triangle groups $(2, 3, n)$ for any $n>6,$ so I would say the answer is NO<|endoftext|> -TITLE: When is the adjoint of a hypoelliptic operator also hypoelliptic? -QUESTION [8 upvotes]: Suppose that $M$ is a smooth manifold with a measure $\mu$ and let $L^2(M, \mu)$ be a space of all square-integrable functions on $M$. -Recall that $L$ is a hypoelliptic differential operator if for every $f \in \mathcal{D}(L)$, if $Lf$ is in $C^\infty(M)$ then $f$ is also in $C^\infty(M)$. -Could anyone give a reference regarding the example when $L$ is hypoelliptic but its adjoint w.r.t to $\mu$ is not hypoelliptic? Could this happen to the Hörmander operators, when $L$ is defined as -$$ -L = \sum_i X_i^2 + X_0 -$$ -and the Lie algebra generated by $\{X_i\}$ spans the entire tangent space? -I am mostly interested in the case when $\mu$ is induced by the Riemannian metric and $M$ is complete. -Thanks! - -REPLY [6 votes]: Hormander's operator $L=X_0+\sum_{1\le j\le k} X_j^2$, where the $X_j$ are real smooth vector fields with the Lie algebra of $\{(X_j)\}_{0\le j\le k}$ generating the tangent space is hypoelliptic as well as its adjoint since the Lie algebra condition does not change by taking adjoints. -On the other hand, $\frac{\partial}{\partial t}+t\Delta_x$ is hypoelliptic whereas its adjoint $-\frac{\partial}{\partial t}+t\Delta_x$ is not hypoelliptic, -Bazin.<|endoftext|> -TITLE: Decompose tensor product of type $G_2$ Lie algebras. -QUESTION [15 upvotes]: Let $G$ be a semisimple Lie algebra over $\mathbb{C}$. Let $V(\lambda)$ be the irreducible highest weight module for $G$ with highest weight $\lambda$. If $G$ is of type A, we can decompose $V(\lambda)\otimes V(\mu)$ using Littlewood-Richardson rule. In other types, if $\lambda$ and $\mu$ are given explicitly, we can use Weyl character formula to compute the decomposition. My question is can we have some rules similar as Littlewood-Richardson rule for other types (especially for type $G_2$)? Does Littelmann's path model work for this? - -REPLY [16 votes]: Since the decomposition of tensor products has a complicated history, it's worth adding some comments to ARupinski's answer. -1) Though Weyl's character formula is fundamental for finite dimensional representation theory, it doesn't lead immediately to a method for decomposing tensor products. However, Steinberg derived in 1961 such an elegant method (which however involves an impractical double summation over the Weyl group) here. -2) As early as 1937, Brauer wrote down the explicit recipe referred to here (and often in the applied literature directed toward physicists) as Klimyk's rule: this was a short article in C.R. Acad. Sci. Paris 204, more easily located in Brauer's collected papers. The selling point of this rule is that it requires only the knowledge of one highest weight along with the full character of the second module (given for example by Weyl or by Kostant's equivalent later method). These matters are discussed in Section 24 of my -1972 Springer graduate text, where a proof of Brauer's formula is formulated in the exercises. -3) Klimyk's early work appeared around 1967 in Russian, with an English translation soon after in an AMS volume: -MR0237712 (38 #5993) 22.60, Klimyk, A.U. -Multiplicities of weights of representations and multiplicities of representations of -semisimple Lie algebras. (Russian) -Dokl. Akad. Nauk SSSR 177 (1967) 1001–1004. -4) It's important to realize that Klimyk's work has involved not just tensor products of finite dimensional representations, but also tensor products in which one factor is allowed to be infinite dimensional of various special types. This is part of a much broader program for Lie group representations involving Kostant and others. -5) From a purely computational point of view, getting explicit results for type $G_2$ is not at all easy because dimensions grow so fast. There used to be some explicit printed tables, which always stop when the going gets tough. The older methods of Brauer and Klimyk are anyway inherently inefficient, requiring a huge amount of cancellation as indicated by ARupinski. Combinatorists have found Littelmann's approach (generalizing Littlewood-Richardson for type A) more appealing, but I'm unaware of literature illustrating the method explicitly for type $G_2$. -6) It's also worth mentioning that interesting special features of tensor products have been studied using algebraic geometry by Kumar and others in response to the old "PRV Conjecture", but here the concern is about the occurrence of specific summands in a tensor product decomposition and not the entire picture. (To get some perspective on the range of "geometric" literature about tensor product decompositions, take a look at Kumar's 2010 ICM report. This is on his homepage but apparently not on arXiv: http://math.unc.edu/Faculty/kumar/papers/kumar60.pdf)<|endoftext|> -TITLE: Relation between combinatorial manifolds and PL manifolds -QUESTION [11 upvotes]: In -W. W. Boone, W. Haken, and V. Poenaru, On Recursively Unsolvable Problems in Topology and Their Classification, Contributions to Mathematical Logic (H. Arnold Schmidt, K. Schütte, -and H. J. Thiele, eds.), North-Holland, Amsterdam, 1968. -a combinatorial manifold is defined as a simplicial complex with the property that the star of every vertex is combinatorially equivalent to the standard $n$-simplex. -(two simplicial complexes are combinatorially equivalent if they possess linear subdivisions, the associated abstract simplicial complexes of which are isomorphic) -This is equivalent to the condition that the link of every vertex be a combinatorial $(n-1)$-sphere (=boundary of the standard $n$-simplex) if the underlying manifold has no boudnary. -However, -in -A. Ranicki (ed.), The Hauptvermutung Book, K-Monographs in -Mathematics, vol. 1, Kluwer Academic Publishers, Dordrecht, -Boston, London, 2010. -on page 4, this condition is used to define the term -``combinatorial manifold (or PL manifold)''. -I find this very weird; -a PL manifold should be defined as a topological manifold with a maximal atlas of homeomorphisms with PL coordinate changes (and I know a lot of authors who use this definition). -The obvious question now is: -Is a simplicial complex, the vertices of which have $S^{n-1}$ as link, -the same as a topological manifold with a maximal atlas of homeomorphisms piecewise linear coordinate changes? -Of course, Ranickis nomenclature implies that it does. -Obviously, -the condition on the links can be used to construct such an atlas. -However, the converse puzzles me, -as it seems to be equivalent to the question if every manifold with a maximal PL atlas admits a triangulation. -If anyone could point me to an article where this problem is addressed, -I would be thrilled. -Best regards, -Malte - -REPLY [5 votes]: It is claimed here that "A PL manifold is easily shown to be PL homeomorphic to a simplicial complex that is a so-called combinatorial manifold [37]", [37] being Hudson's Piecewise Linear Topology. I think the whole thing is worked out in Chapter 3 of that book.<|endoftext|> -TITLE: What do $\Gamma$-sets classify? -QUESTION [16 upvotes]: The category $\Gamma^{\mathrm{op}}$ is defined to be a skeleton of the category of finite pointed sets (see also this question). Then $\Gamma$-spaces, meaning space-valued presheaves $\Gamma^{\mathrm{op}}\to \mathrm{Spaces}$, can be used to present spaces with commutative and associative multiplication up to all higher homotopies. This is similar to how the category $\Delta^{\mathrm{op}}$ can be defined as a skeleton of the category of finite total orders with distinct endpoints, and presheaves $\Delta^{\mathrm{op}}\to \mathrm{Spaces}$ (simplicial spaces) can be used to present spaces with associative multiplication up to all higher homotopies. -It is known that the topos of simplicial sets is the classifying topos for total orders with distinct endpoints. Does the topos of set-valued presheaves on $\Gamma$ have a similar interpretation? - -REPLY [2 votes]: Of course the above answer of @AndreasBlass exhausts it all, but it may be still amusing to compute directly the statement in the previous comment by @CharlesRezk. -One additional piece of information is this: for an object $X$ in a topos $T$, giving a geometric morphism from a topos $E$ to $T/X$ is equivalent to giving a geometric morphism $f:E\to T$ together with a point $1\to f^*(X)$. -Applying this to $T=Psh(fSet^{op})$ and $X=$ the generic object, which is the embedding $X:(fSet^{op})^{op}\to Set$ gives that the category of pointed objects of $E$ is equivalent to the category of gm's from $E$ to $Psh(fSet^{op})/X$. Now the latter is equivalent to presheaves over the Grothendieck construction of $X$ which in turn is equivalent to the category of finite pointed sets (objects are pairs $(n,i)$ where $n$ is a finite set and $i$ is an element of $X(n)$, i.~e. of $n$, etc....), as a particular case of the equivalence -$$ -Psh(C^{op})/\hom_C(c,-)\simeq Psh((c/C)^{op}) -$$<|endoftext|> -TITLE: Density in van der Waerden's theorem -QUESTION [12 upvotes]: Color the positive integers using just two colors. By van der Waerden's theorem, we can find a $k$-term arithmetic progression as long as we consider a long interval. -I imagine it is possible to find a $k$-term arithmetic progression so that the terms in the progression have minimal gaps by possibly taking an even longer interval (of some fixed size depending only on $k$). If so, how do these minimal gaps behave as a function of $k$? - -REPLY [6 votes]: They grow very fast. Denote by N(l) the largest integer such that we can color the numbers with two colors from 1 to N(l) without an l-long arithmetic progression. Now if you use this coloring for (k-1)/2-long sequences and put such colorings after each other, then any k-long arithmetic progression will be longer than N((k-1)/2), which means that even its difference will be at least N((k-1)/2)/k, which is HUGE.<|endoftext|> -TITLE: Galois actions in towers and class field theory -QUESTION [6 upvotes]: I'm trying to understand how this works in terms of Galois theory and local class field theory. Assume we have an extension of local fields $E/L/K$ s.t. $E/L$ and $L/K$ are abelian. I'm interested in recognizing when $E/K$ is Galois. Clearly, $E/K$ is Galois if and only if $E$ is always fixed by an extension of an $L/K$ automorphism to $E$, but this is tricky to compute. -I overheard a brief conversation that this can be done through Galois groups and some group actions that occur in the tower, but I haven't found anything explicit through google. We should be able to see from how $Gal(L/K)$ acts on something whether or not the extension is Galois. I'm having trouble seeing what the action should be. I hope someone who knows what I'm talking about could write it down explicitly. Since the Galois groups should correspond through local class field theory to very concrete objects which are quotients of $E^\times$, $L^\times$ and $K^\times$. I was wondering how this action on the Galois side is expressed on the local field side? -I'm interested in this since it clearly would provide a tool for constructing some solvable extensions of e.g. $\mathbb{Q}_p$. I apologize for being fuzzy, but I don't know how to be more explicit. - -REPLY [2 votes]: If $L/K$ is abelian and $K/k$ cyclic, then $L/k$ is going to be normal if the Galois group at the gottom acts on the class group attached to $L/K$ via class field theory, and it will be abelian if action fixes the classes. This is due to Hasse and follows easily from the standard results in class field theory. This is Prop. 1.2.8 in my survey on class field towers (I apologize for the outdated content. All of this needs to be rewritten). -There are similar constructions in Kummer theory going back to Kummer; this can be used in several proofs of the Kronecker-Weber theorem and should be e.g. in Washington's book; see also Prop. 2.2.1 in the survey -If the base extension is abelian and not cyclic, stuff happens. There are many articles investigating this problem, but nothing as simple as in the cyclic case.<|endoftext|> -TITLE: Fitting a mesh to a density function -QUESTION [10 upvotes]: Suppose I have a probability density function defined on a region in the plane (in my case, the pdf is of the form $f(x) = \alpha\|x\|^{-\beta}$, and the region is the unit disk). For large $N$, is it possible to place $N$ points $X_1,\dots,X_N$ in the region so that the points $X_i$ are distributed according to $f(x)$, and also form a mesh of (approximately) equilateral triangles? This is clearly trivial when $f(x)$ is uniform (just put the $X_i$ in a uniform triangular lattice). -For the non-uniform case, obviously some triangles will be larger than others, but I want each individual triangle to be approximately equilateral (e.g. maximum side length and minimum side length are within 1% of each other, etc.). One possibility for the non-uniform case would be to sample $N$ points independently at random from $f(x)$ and then take their Delaunay triangulation, but I don't think there is a guarantee that the triangles will be roughly equilateral (i.e. some will be long and skinny) as $N$ becomes large. -The picture below is along these lines, if you ignore the big ugly hole in the center; each triangle is roughly equilateral, but points are not uniformly distributed. -     (source: Wayback Machine) - -REPLY [7 votes]: Here is one possible interpretation of your question. - -Assume a probability density function $f$ is given. - Is there a sequence of triangulations $T_n$ with $\varepsilon_n$-equilateral triangles such that counting probability measure on nodes converges to $f$ and $\varepsilon_n\to 0$ as $n\to\infty$. - -(Say a triangle is $\varepsilon_n$-equilateral if the ratio of maximum side length and minimum side length is $\le 1+\varepsilon$.) -I am almost sure that the answer is "YES" if and only if $f$ is conformal factor of a flat metric; -i.e., if and only if $f=e^{2{\cdot}\phi}$ and $\Delta \phi\equiv 0$.<|endoftext|> -TITLE: Any more generalization of Fermat's Little Theorem? -QUESTION [15 upvotes]: Fermat's Little Theorem: If $p$ is a prime and $\gcd(a,p)=1$ then $a^{p-1} \equiv1\pmod p$. -Over the years, Fermat's Little Theorem have been generalized in several ways. I am aware of four different generalizations as given below. -1. Euler: If $\gcd(a,n)=1$ then $a^{\phi(n)} \equiv1 \pmod n$. -2. Ramachandra: $\sum_{d|n}\mu(d)a^{n/d} \equiv 0\pmod n$. (Fermat's Little Theorem follows when $n=p$ is a prime and has only two divisors 1 and $p$. -3. Let $d$ be a divisor of $\phi(n)$. There are exactly $d$ distinct positive integers $r_k, (k=1,2, \ldots d)$ such if $\gcd(x,n)=1$ then $x^{\phi(n)/d} \equiv r_k \pmod n$ for some $(k=1,2, \ldots d)$ (Euler's generalization itself is a special case of this result when $d=1$.) -4. Florentin Smarandache: $a^{\phi(n_s)+s} \equiv a^s \pmod n$ where $s$ and $n_s$ are defined in Smarandache's paper -I would like to know if there is any other generalization of Fermat's Little Theorem. - -REPLY [11 votes]: Let $A$ be a square integer matrix. Then -$$\sum_{d | n} \mu(d) \text{tr}(A^{n/d}) \equiv 0 \bmod n.$$ -After assuming WLOG that $A$ has non-negative entries, the clearest proof I know of this result proceeds by relating the above expression to aperiodic walks on a graph with adjacency matrix $A$; see these two blog posts.<|endoftext|> -TITLE: Nonstandard Methods ( or Model Theory ) and Arithmetic Geometry -QUESTION [6 upvotes]: I hear that the nonstandard methods are applied to many problems in various fields of mathematics such as functional analysis, topology, probability theory and so on. -So, I have become interested in using nonstandard methods to my research areas, which are in and around arithmetic geometry. -Questions: - -What kind of useful applications of nonstandard methods to arithmetic geometry exist? -Is there any recommendation of introductory textbook or PDF file to study nonstandard methods in arithmetic geometry? (I heve studied the "nonstandard analysis" to a certain extent: construction of ultraproducts, the transfer principle etc. But I have few knowledge of nonstandard methods for algebra or algebraic geometry.) -Is there any relationship between the transfer principle and Hasse principle? - -Please give me any advice. - -REPLY [3 votes]: Since you also mention algebraic geometry, you may want to take a look at "Model Theory and Algebraic Goemetry" edited by Elizabeth Bouscaren (Springer LNM 1696). Its primary purpose is to introduce Hrushovki's proof of Mordell-Lang for function fields. Except for some elementary model theory, it is self-contained, and shows that deep results in model theory can be used to prove nontrivial statements. Although this might be more geometric than arithmetic, perhaps some of the techniques could still be useful to you.<|endoftext|> -TITLE: Growth of smallest closed geodesic in congruence subgroups? -QUESTION [11 upvotes]: Let $\Gamma$ be one of the classical congruence subgroups $\Gamma_0(N)$, $\Gamma_1(N)$ and $\Gamma(N)$ of $SL(2, \mathbb{Z})$. -How does the lower bound for the length of primitive geodesics on $\Gamma \backslash \mathbb{H}$ depending on $N \rightarrow \infty$? -Any suggestions? - -REPLY [11 votes]: For a hyperbolic element $A\in SL(2,\mathbb{Z})$, we have the length of the closed geodesic is given by $\ln[(tr^2(A)-2+\sqrt{tr^4(A)-4tr^2(A)})/4]$, and this is monotonic in $|tr(A)|$ for $|tr(A)|>2$. For $A\in \Gamma(N),\Gamma_1(N)$, we have $tr(A)\equiv 2 (\mod N)$, and $tr(A)\neq \pm 2$, so the smallest that $|tr(A)|$ can be is when $|tr(A)|=N-2$ (I suppose for $N>4$). This gives a lower bound on the shortest geodesic for $\Gamma_1(N)$. This is realized by the matrix -$$A=\left[\begin{matrix} 1-N & 1 \\ -N & 1 \end{matrix}\right]$$ -For $\Gamma(N)$, one can obtain a better lower bound. Consider the matrix $$A=\left[\begin{matrix} 1+aN & bN \\ cN & 1+dN \end{matrix}\right]$$ with $\det(A)=1$. Then we have $(1+aN)(1+dN)-bcN^2=1$, so $a+d+(ad-bc)N=0$. This implies that $a+d\equiv 0(\mod N)$, so $tr(A)=2+(a+d)N \equiv 2 (\mod N^2)$. Thus, we get a lower bound of $|tr(A)|\geq N^2-2$. This is realized by the matrix -$$A=\left[\begin{matrix} 1-N^2 & N \\ -N & 1 \end{matrix}\right]$$ -Edit: (I'm modifying the answer to address Vitali's question in the comments below). -For a matrix $A\in \Gamma_0(N)$, we have -$$A=\left[\begin{matrix} a & b \\ cN & d \end{matrix}\right]$$ with $\det(A)=1$. - Then $ad-bcN=1$ implies $ad \equiv 1(\mod N)$. We want to minimize $tr(A)=a+d$ subject to the constraint $ad\equiv 1(\mod N)$. Conversely, if $ad=1+kN$ for some $k$, then the matrix -$$A=\left[\begin{matrix} a & k \\ N & d \end{matrix}\right]\in \Gamma_0(N)$$ -has trace $a+d$. So the minimal trace of a hyperbolic element in $\Gamma_0(N)$ is given by $\min \{ a+d >2 | ad\equiv 1 (\mod N)\}$. -Let's reformulate this problem. -$ad\equiv 1(\mod N)$ is equivalent to the characteristic polynomial $\lambda^2-tr(A)\lambda+1\equiv(\lambda-a)(\lambda-d) (\mod N)$, i.e. the characteristic polynomial of $A$ reduces $(\mod N)$. So we want to minimize -$\min \{ t > 2 | \lambda^2-t\lambda+1 \equiv 0 (\mod N), some \lambda \}$. -If $t$ is even, then we complete the square to get $(\lambda-t/2)^2 \equiv t^2/4-1 (\mod N)$, that is $t^2/4-1$ is a quadratic residue $(\mod N)$. If $t$ is odd, then $N$ must be odd if $\lambda^2-t\lambda+1\equiv 0 (\mod N)$, so multiplying by $4$, this is equivalent to $(2\lambda-t)^2\equiv t^2-4 (\mod N)$. -Thus, the minimal trace is given by -$\min \{ t>2 | t^2-4$ is a quadratic residue $(\mod N)$, $N$ odd, or $t^2/4-1$ is a quadratic residue $(\mod N)$, $N$ even $\}$. -Thus, since there are infinitely many $N$ such that $3^2-4=5$ is a quadratic residue $(\mod N)$ (e.g. the sequence $N=a^2-3a+1$), we have that the systole does not approach $\infty$. -Also, the systoles are unbounded from above. To see this, note that if $j$ is not a quadratic residue $(\mod N)$, then it is not a quadratic residue $(\mod kN)$ for any $k$. For $t>2$, choose $n(t)$ such that $t^2-4$ is not a quadratic residue $(\mod n(t))$. Then the number $N(T)=n(3)n(4)\cdots n(T)$ has the property that the minimal trace of $\Gamma_0(N(T))$ is bigger than $T$. In particular, the systole of $\Gamma_0(N!)$ $\to \infty$. - -REPLY [3 votes]: For more general results and extensive references, see Katz, Schaps, Vishne, Logarithmic growth of systole of arithmetic riemann surfaces along congruence subgroups (JDG, 2007)<|endoftext|> -TITLE: Structure of $[S(\mathfrak{g})\otimes S(\mathfrak{g})]^G$ for semisimple $\mathfrak{g}$ -QUESTION [5 upvotes]: Let $\mathfrak{g}$ be a semisimple Lie algebra over $\mathbb{C}$. $S(\mathfrak{g})^G$ is a polynomial algebra with rank $\mathfrak{g}$ generators. Call them $c_i(x)$, where $x\in \mathfrak{g}$ and $i=1,\ldots,\mathrm{rank}\ g$. -Now, consider $G$ acting diagonally on $\mathfrak{g}\oplus\mathfrak{g}$. What is the ring of invariants $R=[S(\mathfrak{g})\otimes S(\mathfrak{g})]^G$ ? (I guess this is a textbook material; sorry for not doing the homework of looking up the books, but it's a Saturday today and it's easier to ask here than to go to the library at my institute in a cold weather...) -For $x\oplus y \in \mathfrak{g}\oplus\mathfrak{g}$, I can easily see $R\ni c_i(x + t y)$ for a scalar $t$. Do they generate $R$ (polynomially)? - -REPLY [7 votes]: Most of what is known in the case of $GL(n,K)$ acting diagonally by conjugation on $r$-tuples of $n \times n$ matrices can be found in Chapters 8-10 of my CBMS notes "The Polynomial Identities and Invariants of $n \times n$ matrices". There $K$ is a field of characteristic zero, the fixed ring is called the ring of invariants of -$n \times n$ matrices, and it is denoted $C(n,r)$. -First Fundamental theorem (H. Weyl): $C(n,r)$ is generated by traces of monomials. -Second Fundamental Theorem (B. Konstant, C. Procesi, Y. P. Razmyslov): Let $S_r$ be the symmetric group on $r$ letters, and let $J(n,r)$ be the two-sided ideal of the group ring -$KS_r$ corresponding to Young diagrams with $\geq n+1$ rows. Then there is an easily defined -map from $KS_r$ onto the multilinear elements of degree $r$ in $C(n,r)$ whose kernel is $J(n,r)$. -Nagata-Higman Theorem: If an algebra $A$ (noncommutative, without a unit) satisfies -$x^n = 0$ for every $x \in A$, then there is an integer $N = N(n)$ such that -$x_1 \cdots x_N = 0$ for all $x_1, \dots ,x_N \in A$. Procesi made the brilliant -observation that the least such $N$ valid for all $A$ is also the least integer such that $C(n,r)$ is generated by traces of monomials of length $\leq N$. Using the Second -Fundamental Theorem Procesi and Razmyslov showed that $N(n) \leq n^2$. Studying the -Nagata-Higman Theorem, E. N. Kuzmin showed that $N(n) \geq n(n+1)/2$ and conjectured that -$N(n) = n(n+1)/2$. His conjecture has only been verified for $n \leq 4$. -General theorems of M. Hochster - J. L. Roberts and M. P. Murthy imply that $C(n,r)$ is a -unique factorization domain and Gorenstein, and M. van den Bergh proved that it is -Cohen-Macaulay. -Results of several authors show that $C(n,r)$ is a polynomial ring if and only if $n = 1$, -or $r = 1$, or $(n,r) = (2,2)$, and is a complete intersection if and only it is a polynomial ring or $(n,r) = (2,3)$, or $(n,r) = (3,2)$.<|endoftext|> -TITLE: Euler's divergent series sum n!*(-1)^n: what is known about the resulting constant? -QUESTION [14 upvotes]: Much of the theory of continued fractions has been developped by Euler in the 18th century. The little survey "Euler: continued fractions and divergent series (and Nicholas Bernoulli)", mentions towards the end the continued fraction $$f(x)=\dfrac{1}{1+\dfrac{x}{1+\dfrac{x}{1+\dfrac{2x}{1+\dfrac{2x}{1+\dfrac{3x}{1+\dots}}}}}}$$ which Euler has "derived" from the divergent power series $1-x+2x^2-6x^3+24x^4-+...$ . -For $x=1$, the continued fraction converges to a limit $f(1)\approx 0.5963475922$. I was wondering: what is known about the values $f(x)$, in particular $f(1)$? Are they known to be transcendental for $x\in\mathbb N$? Can they be expressed by other known constants? - -REPLY [2 votes]: Regarding the expression of $f(1)$ -in terms of other known constants: -$$f(1) = -e\left(\gamma + \sum_{n\geq 1} (-1)^n \frac1{n\cdot n!} \right) -\qquad\quad \tag{$*$} \label{id}$$ -$$\qquad\qquad\qquad= -e \left(\gamma - 1 + \frac1{4} - \frac1{18} + \frac1{96} - \frac1{600} + \cdots \right)$$ -where $e = \sum_{n\geq0} \frac1{n!} \approx 2.718$ and -$\gamma \approx 0.577$ is the Euler-Mascheroni constant. -This infinite sum expression -follows from the identity $f(1) = -e {\rm Ei}(-1)$ -and is on the Wikipedia page for the Gompertz constant. -I recommend also taking a look at the survey Euler's constant: Euler's work and modern developments by Jeffrey Lagarias, -where it appears as (2.5.11). - -Since -$f(x) ``=" \sum_{n\geq 0} (-1)^n n! x^n$ -and -$e^{-1} = \sum_{n\geq 0} (-1)^n \frac{1}{n!}$, -the identity ($*$) can be expressed more symmetrically as -$$ - \left(\sum_{n\geq 0} (-1)^n n! \right) \left(\sum_{m\geq 0} (-1)^m \frac{1}{m!} \right) -\quad ``=" \quad \gamma + \sum_{n\geq 1} (-1)^n \frac1{n\cdot n!}.$$ -This formula suggests an argument for evaluating $f(1)$ which avoids differential equations, if one is willing to make some convenient cancellations of divergent series after rearranging the two-index summation on the left-hand side. - -For arbitrary nonzero $x$, the relation ($*$) generalizes to $f(x) = - \frac1{x} e^{1/x} {\rm Ei}(-\frac1{x})$ -where ${\rm Ei}(x)$ denotes the exponential integral -$${\rm Ei}(x) = \int_{-\infty}^x \frac{e^t}{t} dt. $$ -Using the Taylor expansion of ${\rm Ei}(x)$, -$$f(x) = - \frac1{x} e^{1/x} \left(\gamma + \ln x + \sum_{n\geq 1} \frac{(-1)^n}{n \cdot n!} \frac1{ x^n} \right).$$<|endoftext|> -TITLE: From Sato grassmannian to spectral curve -QUESTION [5 upvotes]: Assume a tau-function of the KP integrable hierarchy is fixed by the point of the Sato grassmannian (that is by a semi-infinite set of Laurant series $\varphi_k(z)=\sum_{m>0}a_{km}z^{-k+m}$). Can one restore a spectral curve that corresponds to this solution? - -REPLY [8 votes]: This is explained in Segal-Wilson. Essentially, realize your point in the Grassmannian as as a space W of functions w(z), then look for all functions g(z) such that g(z)W is included in W. These functions form a commutative algebra, and Spec of this algebra is your spectral curve. Of course for most points of the Grassmannian the commutative algebra is just C. -But this is all beautifully explained in Segal-Wilson.<|endoftext|> -TITLE: Behavior of sectional curvature under metric deformations -QUESTION [16 upvotes]: Metric deformation: -Let $(M,g_0)$ be a Riemannian manifold and consider a (sufficiently smooth) deformation of $g_0$, $$g_t=g_0+th+O(t^2), \quad 0< t<\varepsilon $$ where $h$ is some symmetric (0,2)-tensor. A natural (and important) question is how the sectional curvatures of $g_0$ change under this deformation, e.g., what is the infinitesimal change in terms of $h$. More precisely, given two $g_0$-orthonormal vectors $X$ and $Y$ in $T_pM$, define the (unnormalized) sectional curvature $$k(t)=g_t(R_t(X,Y)Y,X),$$ where we are using the appropriate sign convention on $R$. - -Q: What is the explicit formula for $k'(0)=\frac{d}{dt}k(t)\big|_{t=0}$? - - -Possible (but different?) answers: -I have found a few papers with an answer, but (understandably) none provide the complete argument. Unfortunately, it seems like some of them are really different, and it would be very helpful if someone could point out if they coincide for some (possibly silly) reason I am not seeing. - -Berger'66 (Trois remarques sur les variétés riemanniennes à courbure positive)/Bourguignon, Deschamps, Sentenac'72 (Conjecture de H. Hopf sur les produits de variétés): -$$k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)$$ - -Strake'87 (Curvature increasing metric variations): -$$k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+h(R_0(X,Y)Y,X)$$ - - -$$-k(0)(h(X,X)+h(Y,Y))$$ -3. Topping'06 (Lectures on Ricci Flow): $$k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+\tfrac12h(R_0(X,Y)Y,X)$$ -$$-\tfrac{1}{2}h(R_0(X,Y)X,Y)$$ -EDIT: I should point out that, although formula 3) as I wrote above is NOT a correct expression for $k'(0)$, I (embarrassingly) misinterpreted it from Prop. 2.3.5 in Topping's lecture notes -- which actually contains the correct formula (matching Vitali's answer below). He gives a general expression for $\frac{d}{dt}g_t(R^t(X,Y)Z,W)\big|_{t=0}$, and by using the Ricci identity it becomes clear that his formula is indeed the same as Vitali's. I sincerely apologize for the confusion. -Note that all answers above coincide if $(M,g_0)$ has non-negative sectional curvature and $X$ and $Y$ span a plane of zero $g_0$-curvature. As Strake remarks, the self-adjoint endomorphism $A_X Z=R(Z,X)X$ is positive-semidefinite and $g(A_X Y,Y)=0$. (Nevertheless, to the best of my understanding, THESE HYPOTHESES ARE NOT ASSUMED in the references in 1). Also, it seems to me that answers 2 and 3 DO NOT COINCIDE. - -REPLY [16 votes]: Formula 2) is the correct one in general except it's the derivative of the sectional curvature i.e of $\frac{k_t(X,Y)}{|X\wedge Y|^2_t}$ (and not just of $k_t(X,Y)$) for an orthonormal frame $X,Y$ with respect to the original metric. This accounts for the last term in formula 2. -For $k'(0)$ itself the correct formula is -$$ -k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+h(R(X,Y)Y,X) -$$ -As Deane said, this is a straightforward calculation. But since the OP seems to be struggling with it I'll supply some details. Let $\nabla^t$ be the Levi-Civita connection for $g_t$. Then $\nabla^t_XY=\nabla_XY+tS(X,Y)+O(t^2)$ where $S$ is a (2,1)-tensor and $\nabla=\nabla^0$. -Let's first compute $k'(0)$ in terms of $S$. As usual let's work in normal coordinates around $p$ with $X,Y$ coordinate fields. -We have $$\langle R^t(X,Y)Y,X\rangle_t=\langle R^t(X,Y)Y,X\rangle_0+t\cdot h(R(X,Y)Y,X)+O(t^2)=$$ -$$=\langle \nabla^t_X\nabla^t_YY,X\rangle_0-\langle \nabla^t_Y\nabla^t_XY,X\rangle_0+t\cdot h(R(X,Y)Y,X)+O(t^2)$$ - Next we expand the first term -$$\langle \nabla^t_X\nabla^t_YY,X\rangle_0=\langle \nabla^t_X(\nabla_YY+tS(Y,Y)),X\rangle_0+O(t^2)=$$ -$$\langle \nabla_X\nabla_YY,X\rangle_0+t\langle S(X,\nabla_YY)+\nabla_XS(Y,Y), X\rangle_0+O(t^2)=$$ -$$ \langle \nabla_X\nabla_YY,X\rangle_0+t\langle \nabla_XS(Y,Y), X\rangle_0+O(t^2)$$ -where in the last equality we used that $\nabla_YY(p)=0$. -After a similar computation for $\langle \nabla^t_Y\nabla^t_XY,X\rangle_0$ we get that -$$k'(0)=\langle\nabla_XS(Y,Y)-\nabla_YS(X,Y), X\rangle_0+h(R(X,Y)Y,X)$$ -$$=\nabla_X S(Y,Y,X)-\nabla_YS(X,Y,X)+h(R(X,Y)Y,X)$$ -where we lowered the index and turned $S$ into a $(3,0)$-tensor $S(X,Y,Z)=\langle S(X,Y),Z\rangle_0$ -Lastly, recall that -$\langle \nabla_XY,Z\rangle=\frac{1}{2}[X\langle Y,Z\rangle+Y\langle X,Z\rangle -Z\langle X, Y\rangle]$ for coordinate fileds. This easily gives -$S(Y,Y,X)=\frac{1}{2}[Yh(X,Y)+Yh(X,Y)-Xh(Y,Y)]=\nabla_Yh(X,Y)-\frac{1}{2}\nabla_Xh(Y,Y)$ and -$S(X,Y,X)=\frac{1}{2}\nabla_Yh(X,X)$ written invariantly as tensors. -Altogether this gives -$$ -k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+h(R(X,Y)Y,X) -$$ as promised. -Formulas 1) and 3) are not true in general but might be true in the specific circumstances where they are applied in the papers in question.<|endoftext|> -TITLE: Extracting countable chains from linear orders -QUESTION [7 upvotes]: There is a well-known fact in infinite combinatorics asserting that for each infinite linear order $P$ there is a countable subset $R\subseteq P$ of order type either $\omega$ or $\omega^{*}$ -(by $\omega^{*}$ I mean set of natural number with reversed order). It seems to be a non-trivial result - for example, one can derive it from the Baumgartner-Hajnal theorem but this is, in my taste, too heavy machinery. -Do you know who iss responsible for this result? Are there any cheaper ways (than the BH-theorem) to obtain it? - -REPLY [9 votes]: It seems to me that a natural solution is to use Ramsey's theorem $\aleph_0 \to (\aleph_0)^2_2$: enumerate a countable subset, and color two points depending on whether the enumeration agrees with the given order. -This proof seems "cheaper" to me: Wlog the linear order $P$ is a subset of the rationals. Find a limit point $r$. Wlog $r$ is a limit point of the points from $P$ below $r$ -- so you can find an increasing sequence converging to $r$. - -REPLY [8 votes]: You can derive it straight from Ramsey's Theorem. We may assume the linear order is countably infinite, say $x_0,x_1,x_2,\ldots$ enumerates it. Define the coloring $c:[\omega]^2\to2$ by $c(i,j) = 0$ iff $i \lt j$ and $x_i \lt_P x_j$ (and color 1 if the two orderings disagree). If $H$ is a homogeneous set for $c$, then the subsequence $\langle x_i \rangle_{i \in H}$ is either increasing or decreasing with respect to ${\lt_P}$. -This result is actually not quite as strong as Ramsey's Theorem (for pairs and two colors). See Hirschfeldt & Shore, Combinatorial principles weaker than Ramsey's theorem for pairs, JSL 72 (2007), 171-206. - -REPLY [8 votes]: Here's another "cheap" proof. If your linear ordering doesn't have a decreasing $\omega$-sequence, then it's well-ordered, and therefore order-isomorphic to an ordinal. Since it's infinite, that ordinal is at least $\omega$, and so your ordering not only has a subset of order-type $\omega$ but has an initial segment of order-type $\omega$. (I've used the axiom of choice, or at least dependent choice, to infer "well-ordered" from the non-existence of a decreasing $\omega$-sequence, but some choice is needed in any proof. Without choice, there can be infinite linearly ordered sets with no countably infinite subsets.)<|endoftext|> -TITLE: Two implicit references in Serre's *Groupes de Galois : le cas abélien* -QUESTION [7 upvotes]: In his exposé at the Galois bicentenary conference, Serre makes two references which are not quite explicit. -The first reference occurs (at 22:30 in the video) when he is talking about Dedekind's XIth supplement to Dirichlet's Vorlesungen and says that a certain French graduate text on number fields is essentially based on that supplement. -Question Which book is he referring to ? -I think it is Pierre Samuel's Théorie algébrique des nombres, but would like to hear other people's opinions. -The second reference occurs (at 47:00 in the video) when he is discussing Artin's general reciprocity law and says that its formulation was so simple that apparently his contemporaries did not quite believe it. -Question Which arithmeticians does Serre have in mind ? -I believe he is thinking of Hasse in particular. Artin is reported to have said that Hasse told him that the conjectural general reciprocity law couldn't possibly be true. -Which other mathematicians of the time expressed their disbelief (before Artin actually proved his own conjecture) ? - -REPLY [5 votes]: I certainly don't claim that I can answer your questions authoritatively , but here are two small remarks. -1) Samuel's book is certainly an excellent guess: it is actually the only textbook in French I can think of entirely devoted to elementary algebraic number theory. -2) There was a preliminary draft of a text by Bourbaki to be inserted in His Commutative Algebra, called Contre-Rédaction de la Différente, rédaction n°410. -It was written by Samuel, and Bourbaki's étiquette was that if one of His collaborators had written a text that was not to be integrated in the corresponding volume in a foreseeable future, then that person was free to publish it on his own. -Precedents include Lang's book on the cohomology of groups, and Godement's classic on sheaf theory. -Acknowledgment would be given in some coded way like Godement's amusing: -Il est bien évident que ce livre n'aurait jamais vu le jour sans l'aide précieuse et les encouragements enthousiastes (quoique partiellement intéressés) que nous ont prodigué [sic] certains géomètres et tout spécialement N.Bourbaki ... -This applies to the case at hand: Samuel's book is quite similar to Bourbaki's draft, with obvious changes stemming from the fact that of course Samuel couldn't use sophisticated tools, like étale algebras, used in the draft. - Samuel writes ...I want particularly to thank the master of my generation,N.Bourbaki,who has had the kindness to show me his unpublished manuscripts...<|endoftext|> -TITLE: Segal's Original Definition of a Topological Category -QUESTION [6 upvotes]: Nowadays we can associate to a topological space $X$ a category called the fundamental (or Poincare) $\infty$-groupoid given by taking $Sing(X)$. -There are many different categories that one can associate to a space $X$. For example, one could build the small category whose object set is the set of points with only the identity morphisms from a point to itself. It is claimed that the classifying space of this category returns the space: $BX=X$ -The inspiration for these examples comes from three primary sources: Graeme Segal's famous 1968 paper Classifying Spaces and Spectral Sequences, Raoul Bott's Mexico notes (taken by Lawrence Conlon) Lectures on characteristic classes and foliations, and a 1995 pre-print called Morse Theory and Classifying Spaces by Ralph Cohen, G. Segal and John Jones. -In each of these papers there is a notion of a topological category. It is not just a category enriched in Top, since the set of objects can have non-discrete topology. Here is the definition that I can gleam from these articles: -A topological category consists of a pair of spaces $(Obj,Mor)$ with four continuous structure maps: - -$i:Obj\to Mor$, which sends an object to the identity morphism -$s:Mor\to Obj$, which gives the source of an arrow -$t:Mor\to Obj$, which gives the target of an arrow -$\circ:Mor\times_{t,s}Mor\to Mor$, which is composition. - -Were $i$ is a section of both $s$ and $t$, and all the axioms of a small category hold. -Is the appropriate modern terminology to describe this a Segal Space? What would Lurie call it? Based on reading Chris Schommer-Pries MO post and elsewhere this seems to be true. Would the modern definition of the above be a Segal Space where the Segal maps are identities? Also, why do we demand that the topology on objects be discrete for Segal Categories? Is there something wrong with allowing the object sets to have topologies? - -REPLY [4 votes]: Topological categories were invented by Charles Ehresmann in the late 1950s, and can be seen in his 1959 paper I think called Catégories topologique et catégories differentiable. The usage 'topological category' for a Top-category is much newer.<|endoftext|> -TITLE: Is there a closed formula for the generating function of some trinomial coefficients? -QUESTION [6 upvotes]: We learn in calculus how to obtain a sum of binomial coefficients $\frac{(2d)!}{(d!)^2}$ in terms of a generating function -$\sum_{d \geq 0} \frac{(2d)!}{(d!)^2} x^d$ -by the Taylor series of $(1-4x)^{-1/2}$ at $x=0$. -My question: is there a way to do this for trinomial coefficients? In particular what is -$\sum_{d \geq 1} \frac{(3d-1)!}{(d!)^3} x^d =?$ -I can't imagine this not being studied before, but can not find a specific answer after a few futile hours of searching. - -REPLY [6 votes]: What is "a way"? Of course, your question (in even more general form) was asked -centuries ago and gave rise to hypergeometric series, series of the form $\sum_n c_n$ -with ratio $c_{n+1}/c_n$ being a rational function of index $n$. The most convenient -form is therefore the hypergeometric $_4F_3$ series expression given in Robert's answer. -Note that the hypergeometric function is not algebraic (i.e., transcendental) in -this case, so that no formula like $(1-4z)^{-1/2}$ can be given; -all algebraic hypergeometric instances are now tabulated thanks to -a fantastic result of Beukers and Heckman. -Dyson's famous 1962 -paper Statistical theory of the energy levels of complex systems -originated the study of constant term identities. In particular, -Dyson's ex-conjecture states that for $a_1,\dots,a_n$ nonnegative integers -$$ -\text{constant term} -\prod_{1\leq i\neq j\leq n} \biggl(1-\frac{x_i}{x_j}\biggr)^{a_i} -=\frac{(a_1+a_2+\cdots+a_n)!}{a_1!a_2!\cdots a_n!}\,. -$$ -This could serve a different basis for other type generating functions.<|endoftext|> -TITLE: A trick or a general technique? (Probabilistic Method) -QUESTION [11 upvotes]: Suppose we have some positive quantites $P$ and $Q$ which depend on some choices that we make, and we want to show that some choice makes the quotient $P/Q$ fall below some cool bound. -One idea is to make our choices randomly in some way and show that $P/Q$ is small on average. That is, we can use the trivial bound -$$ -\min P/Q \leq \mathbf{E}[P/Q]. -$$ -But this inequality is unlikely to be of much use, because we still have to compute $P/Q$ for a random choice. A much more useful inequality arises as follows. Observe that -$$ -\mathbf{E}[P - Q\mathbf{E}[P]/\mathbf{E}[Q]] = 0, -$$ -whence -$$ -\min P/Q \leq \mathbf{E}[P]/\mathbf{E}[Q]. -$$ -This inequality is much more likely to be useful because now we can compute expectations first and then take the quotient. Moreover, in some cases this will even give a better bound than the other inequality. -I'm not looking so much for a detailed explanation of what's going on in this specific inequality, but rather for general intuition. Is this just a trick? How can other tricks like this be anticipated? -(Setting: In proving the discrete Cheeger inequality, $P$ is the number of edges coming out of a subset of a graph and $Q$ is the minimum of the size of the subset and the size of its complement, but this question is about general technique and not this specific problem.) - -REPLY [11 votes]: One way to see this technique is as a way of dealing with certain bad cases. $E[P/Q]$ can be unhelpfully dragged up by the inclusion of certain cases where $Q$ is small and $P$ is medium. $E[P]/E[Q]$ is not nearly so distorted. In particular, take $Q=0$, $P>0$. The first inequality becomes totally unhelpful, as $E[P/Q]=\infty$. But $E[P]$ and $E[Q]$ are still finite. -Any time a naive probabilistic bound is poor due to some very very bad cases, you should look for a trick to exclude these cases or dull their impact on the bound. In particular, you should try to change the probability measure, reducing it on those cases. -Here, since we want to avoid small $Q$, we multiply the probability measure by $Q/E[Q]$. It will still be a probability measure after this, and the new expectation will be: -$E'[P/Q]=E[PQ/QE[Q]]=E[P/E[Q]]=E[P]/E[Q]$<|endoftext|> -TITLE: Representations by positive definite binary quadratic forms -QUESTION [10 upvotes]: It's known that the number of representations of an integer $k$ by sum of two squares is -$$ -4\;\sum_{d|k}\left(\frac{-4}{d}\right) -$$ -or -$$ -4\sum_{d|k,\; d \textrm{ odd}} (-1)^{\frac{d-1}{2}}= 4(d_1(k)-d_3(k)) -$$ -where $d_1(k)$ and $d_3(k)$ are the numbers of the divisors of $k$ of the forms $4m+1$ or $4m+3$ respectively. -I would like to have references about similar formulas for the number of representations of an integer by the quadratic form -$$ -x^2+N\;y^2 -$$ -with $N$ a positive integer, or more generally, any positive definite binary quadratic form. Thanks.-. - -REPLY [11 votes]: There are going to be very few of these that come out so cleanly. There is not going to be anything clean unless there is only one class per genus "idoneal" (your discriminant $-4N$), and I would not really be confident unless there is only one class entirely for that discriminant. In the latter case, we have your $N=1,2,3,4,7.$ -For possibilities with more general types of expressions, see BERKOVICH -Here we go, Leonard Eugene Dickson, Introduction to the Theory of Numbers (1929), exercises on pages 80-81, for section 51. If $m$ is positive and odd, the number of representations of $2^k m$ by $x^2 + 2 y^2$ is double the excess of the number of divisors $\equiv 1 \; \mbox{or} \; 3 \pmod 8$ of $m$ over the number of divisors $\equiv 5 \; \mbox{or} \; 7 \pmod 8$ of $m.$ -The number of representations of any positive $n$ by $x^2 + x y + y^2$ is $6E(n),$ where -$E(n)$ is excess of the number of divisors $\equiv 1 \pmod 3$ of $n$ over the number of divisors $\equiv 2 \pmod 3$ of $n.$ Also, if $n=2^k m$ with $m$ odd, then $E(n)=0$ when $k$ is odd, while $E(n) = E(m)$ when $k$ is even. -Using the same $E(n)$ as above, the number of representations of $2^k m$ with $m$ odd by $x^2 + 3 y^2$ is $0$ if $k$ is odd, $2E(m)$ if $k=0,$ and $6 E(m) $ if $k$ is even and nonzero. -Dickson sticks with one class per discriminant on pages 80-81, then allows one class per genus on pages 84-86. -Note that we cannot expect such simple formulas when there are more than one classes per genus. If there were pretty formulas, we would have a simple way to decide whether an individual prime were represented. However, for example, a prime $p \equiv 1 \pmod 3$ is represented by $x^2 + 27 y^2$ if and only if 2 is a cubic residue $\pmod p.$ If $p \neq 7,2$ and $(-14|p) = 1,$ then $p = u^2 + 14 v^2$ is possible in integers if and only if there is an integer solution to $(z^2 + 1)^2 \equiv 8 \pmod p.$ -Here is a good one. If $m$ is positive and odd, the number of representations of $m$ by $x^2 + xy + 3 y^2$ is double the excess of the number of divisors $\equiv 1, \; 3, \; 4, \; 5, \; \mbox{or} \; 9 \pmod {11}$ of $m$ over the number of divisors $\equiv 2, \; 6, \; 7, \; 8, \; \mbox{or} \; 10 \pmod {11}$ of $m.$ Now, a subset of these numbers is represented by $x^2 + 11 y^2,$ by a simple formula. However, for a single prime, if $p > 11$ has -$ (-44 | p) = 1,$ then we have an integral representation $p = u^2 + 11 v^2$ if and only if $z^3 + z^2 - z + 1 \equiv 0 \pmod p$ has an integral solution. But there is always a representation $ p = x^2 + x y + 3 y^2$ as soon as $ (-44 | p) = 1.$ -In case of interest, the primes up to 500 represented by $x^2 + 11 y^2$ are -11 - 47 - 53 -103 -163 -199 -257 -269 -311 -397 -401 -419 -421 -499 -while the primes up to 500 represented by $3 x^2 \pm 2 x y + 4 y^2$ are -3 - 5 - 23 - 31 - 37 - 59 - 67 - 71 - 89 - 97 -113 -137 -157 -179 -181 -191 -223 -229 -251 -313 -317 -331 -353 -367 -379 -383 -389 -433 -443 -449 -463 -467 -487<|endoftext|> -TITLE: Coherent Sheaves on Noetherian schemes -QUESTION [7 upvotes]: Let $X$ be a Noetherian scheme (in particular, we assume that it has only finitely many irreducible components). Is it true that for any open set $U$, the ring $\Gamma(U, \mathscr{O}_X)$ is a Noetherian ring. Let $\mathscr{F}$ be a coherent sheaf on $X$. Is it true that for any open set $U$, $\Gamma(U,\mathscr{F})$ is a finitely generated $\Gamma(U,\mathscr{O}_X)$ module. - -REPLY [10 votes]: There exists a noetherian scheme , which is even a variety over a field $k$, such that -$\Gamma(X, \mathscr{O}_X)$ is not a Noetherian ring. - It is given as Exercise 21.9. D. in Ravi Vakil's wonderful online book. -Ravi takes for $X$ the total space of the vector bundle associated to a locally free sheaf $\mathcal E$ of rank 2 on an elliptic curve $E$. -The locally free sheaf is the direct sum $\mathcal E=\mathcal P\oplus \mathcal N$ of a an invertible sheaf of positive degree $\mathcal P$ and of a non-torsion invertible sheaf $\mathcal N$ of degree $0$ on $E$. -(Full disclosure: I have just discovered that exercise and I have not yet checked it and the ones leading to it in detail. But to say that I trust Ravi is a vast understatement...)<|endoftext|> -TITLE: Extending braidings to tensor powers -QUESTION [5 upvotes]: Given a braiding $\Psi: X \otimes Y \to Y \otimes X$ for two objects $X,Y$ in a monoidal category, it seems reasonable to assume that $\Psi$ extends uniquely to a braiding $X^k \otimes Y^l \to Y^l \otimes X^k$, for all $k,l \in {\mathbb N}$. The proof would surely be based upon the Yang--Baxter property of $\Psi$ and the fact that one can express any permutation as a product of transpositions. However, I can't seem to write it down exactly. - -REPLY [5 votes]: $\newcommand{\id}{\mathrm{id}} \newcommand{\ot}{\otimes}$ -Your assumption is correct. $\Psi$ does extend uniquely to the map that you want. By drawing string diagrams and playing with them, you can intuitively see what to do, namely: every time you see an $X$ strand to the left of a $Y$ strand, use $\Psi$ to braid $X$ over $Y$. However, there may be many ways to do this. For example if $k=l=2$, you can do -$$ (\id_Y \ot \Psi \ot \id_X) \circ (\id_Y \ot \id_X \ot \Psi) \circ (\Psi \ot \id_X \ot \id_Y) \circ (\id_X \ot \Psi \ot \id_Y), $$ -or you can do -$$ (\id_Y \ot \Psi \ot \id_X) \circ (\Psi \ot \id_Y \ot \id_X) \circ (\id_X \ot \id_Y \ot \Psi) \circ (\id_X \ot \Psi \ot \id_Y), $$ -and the braid relations show that those are the same map. Obviously this is easier to see if you draw a picture. -The challenge is how to efficiently prove that, given any $k$ and $l$ and any possible choice of way you build your braiding, that you get the same map. This is the sort of thing that is known as a "coherence theorem." These are often stated in the fashion: Every diagram in (some category) commutes, where the category is a sort of "free category" whose morphisms are the structural ones that you are talking about. -There are coherence theorems for the associativity morphisms in a monoidal category (see MacLane's book, Chapter VII, Section 2) which is what allows you to drop parentheses when doing iterated tensor products. And similarly there is a coherence theorem for morphisms built from the braidings in a braided monoidal category. This is spelled out in detail in Joyal and Street's 1993 article Braided Tensor Categories.<|endoftext|> -TITLE: morita equivalence for categories -QUESTION [21 upvotes]: Suppose we have two categories such that presheaves on them are equivalent. What can be said in this situation? - -REPLY [3 votes]: In short: it happens exactly when there is a bigger category and disjoint full embeddings of the two categories $\mathcal A,\mathcal B$ into it, such that every arrow is a composition of arrows of the form $a\to b$ or $b\to a$. - -It's easy to see that the Eilenberg-Watts theorem holds for monoid actions, moreover, it also holds for categories, presheaves and profunctors in place of rings, modules and bimodules: -Theorem. -Let $\mathcal A$ and $\mathcal B$ be categories, and $F:\mathcal{Set}^{\mathcal A}\to\mathcal{Set}^{\mathcal B}$ has a right adjoint, then there is a profunctor $U:\mathcal A\not\to\mathcal B$, such that $F\simeq (H\mapsto H\otimes_{\mathcal A}U)$. -In other words, in the realm of modules (even over categories), the only adjunctions are the tensor-hom adjunctions. -[As a note, $U$ can be obtained from $F$ by setting $U(a,b)=F(\hom_{\mathcal A}(a,-))\,(b)\ $.] -If $F$ is an equivalence of categories, then its inverse $G$ is both its left and right adjoint, so by the above theorem, $F\simeq(-\otimes U)$ and $G\simeq(-\otimes V)$ for some profunctors $U:\mathcal A\not\to\mathcal B$ and $V:\mathcal B\not\to\mathcal A$. -And, by $F$ and $G$ being inverses to each other, we obtain $U\otimes V\simeq \hom_{\mathcal A}$ and $V\otimes U\simeq\hom_{\mathcal B}$, as the hom functors are the identites for profunctor composition $\otimes$, meaning that $U$ and $V$ form an equivalence in the bicategory of categories and profunctors. -As for such, there exist an adjoint equivalence with arrows $U$ and $V$ in $\mathcal{Prof}$, that is, profunctor isomorphisms $\eta:\hom_{\mathcal A}\to U\otimes V$ and $\varepsilon:V\otimes U\to\hom_{\mathcal B}$ satisfying the zigzag identities. -Writing up these identities for the inverse of $\eta$, we receive a structure called Morita context: -$$ 1_U\otimes\varepsilon=\eta^{-1}\otimes 1_U \quad -\text{ and }\quad\varepsilon\otimes 1_V=1_V\otimes\eta^{-1}\quad\ (*)\,,$$ -which can be very neatly interpreted as follows: -First, the elements of the sets $U(a,b)$ for a profunctor $U:\mathcal A^{op}\times\mathcal B\to\mathcal{Set}$ can be regarded as heteromorphisms $a\to b$ which we can add to the disjoint union $\mathcal A+\mathcal B$, and their compositions is described by the action of $U$ on the morphisms. -Thus, we obtain a bigger category $\mathcal U$ containing both $\mathcal A$ and $\mathcal B$ and no arrows of the form $b\to a$. -Composition ($\otimes$) of profunctors can be viewed as having formal compositions $\langle u,v\rangle$ of these heteromorphisms $u\in U(a,b),\ v\in V(b,a')$, quotienting out by $\langle u\beta,\,v\rangle\sim\langle u,\,\beta v\rangle$ for arrows $\beta\in\mathcal B$. -Thus, the above maps $\varepsilon:U\otimes V\to\hom_{\mathcal A}$ and $\eta^{-1}:V\otimes U\to\hom_{\mathcal B}$ extend the composition operations to $\mathcal U\cup\mathcal V$, assigning certain arrow $a\to a'\,\in\mathcal A$ to every $\langle u,v\rangle\in(U\otimes V)(a,a')$ and similarly for compositions $b\to b'$ in $\mathcal B$, and the above two equations $(*)$ state exactly the altogether associativity of these operations. -In other words, we obtain a bigger category, which is called a bridge $\mathcal A\rightleftharpoons\mathcal B$, that is a category containing $\mathcal A$ and $\mathcal B$ as disjoint full subcategories and no more objects (thus, it is a symmetric version of a profunctor, in a sense). -One direction of my initial statement then follows, since the composition maps $\varepsilon$ and $\eta^{-1}$ are in particular surjective. -For the other direction, their surjectivity in the presence of equations $(*)$ implies invertibility. - -If $\mathcal M:\mathcal A\rightleftharpoons\mathcal B$ is a bridge with surjective zig-zag compositions, then in particular, for every object $a\in Ob\mathcal A$, its identity $1_a$ also factors as $uv$ for some $u:a\to b,\ v:b\to a,\ b\in Ob\mathcal B$, but then $vu:b\to b$ is an idempotent, -so if we take the idempotent completion $\mathcal M^{id}$ of $\mathcal M$, that will be a bridge $\mathcal A^{id}\rightleftharpoons\mathcal B^{id}$, in which every object of $\mathcal A$, and thus also each of its idempotents, is isomorphic to some object of $\mathcal B^{id}$ and vice-versa. -The existence of such a bridge $\mathcal C\rightleftharpoons\mathcal D$ is actually equivalent to the equivalence of $\mathcal C$ and $\mathcal D$ (at least in the presence of choice): fixing isomorphism for each $a$ yields an equivalence functor $\mathcal C\to\mathcal D$, and the other direction can be proved by a similar argument as above. -So we obtain that $\mathcal A^{id}\simeq\mathcal B^{id}$. - -For more on the subject, refer to this paper.<|endoftext|> -TITLE: Max Noether's residual intersection theorem (Fundamentalsatz): importance and applications -QUESTION [5 upvotes]: I would like to ask a couple of naive question about the following theorem of Max Noether: -http://en.wikipedia.org/wiki/AF%2BBG_theorem -In the book of Fulton, page 60 -http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf -this theorem is called Max Noether's fundamental theorem. -Question 1. Are there some nice applications of this theorem apart from the simple ones that a given in the book of Fulton? -Question 2. Why this theorem is called fundamental? Is this for some historical reason? -Is this theorem a part or a starting point of some modern theory? -(Unfortunately I was not able to get answers to these questions by simple googling) - -REPLY [7 votes]: Perhaps one of most famous consequences of Noether "AF+BG Theorem" is Cayley-Bacharach Theorem, that I state below. - -Theorem (Cayley-Bacharach). Let $X_1, X_2 \subset \mathbf{P}^2$ be two plane curves of degree $d$ and $e$, respectively, meeting in a collection of $d \cdot e$ distinct points $\Gamma$. If $C \subset \mathbf{P}^2$ is any plane curve of degree $d+e-3$ containing all but one point of $\Gamma$, then $C$ contains all of $\Gamma$. - -When $d=e=3$ one has Chasles Theorem: if $\Gamma$ is a collection of $9$ points in $\mathbf{P}^2$ which are complete intersection of two cubics, then any cubic $C$ passing through $8$ of the points of $\Gamma$ contains the remaining point as well (this is essentially Proposition 3, page 63 in Fulton's book). -For a nice discussion of these results and their relation with Noether's theorem see the paper by Eisenbud, Green and Harris Cayley-Bacharach Theorems and Conjectures, Bull. Amer. Math. Soc. 33 (1996).<|endoftext|> -TITLE: 2/3 power law in the plane -QUESTION [13 upvotes]: I've recently come across a particular problem whose solution turns out to be a probability distribution given by $f(x) = \alpha \|x\|^{-2/3}$ in the unit disk in $\mathbb{R}^2$ and zero elsewhere (I alluded to this in a previous question -Fitting a mesh to a density function -which was very helpfully answered by Anton Petrunin). Does this distribution appear in any other contexts? I've seen a $2/3$ power law in reference to metabolic rates of animals: -http://www.ncbi.nlm.nih.gov/pubmed/19906667 -and in kinematics: -http://www.ncbi.nlm.nih.gov/pubmed/9844558 -but both of the preceding cases appear to be looking at rather one-dimensional quantities (and they're positive powers rather than negative in my case, not an important distinction); they have $f(t) = \alpha t^{2/3}$, where $t$ represents mass in the first case and angular velocity of the tip of a pen in the second. This seems different from the situation that I'm describing. To put it succinctly, -"Are there natural quantities that are proportional to the distance to some point, raised to the $-2/3$ power?" -(This may be more appropriate for another forum; if so, I welcome any suggestions) - -REPLY [3 votes]: It is a theorem of Renyi and Soulanke that the cardinality of the boundary of a convex hull of a uniformly distributed random point set of cardinality $N$ in a smooth convex set grows like $N^{1/3},$ so in particular, if you take a point set in a disk of radius $R,$ so that the density is $1,$ then the cardinality of said convex hull boundary grows like $R^{2/3}.$ A similar statement was shown by Baranyi et al for lattice points in that same disk, see the recent question: Convex hull of lattice points in a circle<|endoftext|> -TITLE: Question about Thurston's paper "A norm for the homology of 3-manifolds" -QUESTION [13 upvotes]: I have a question about the proof of Theorem 5 in Thurston's paper "A norm for the homology of 3-manifolds". It's the theorem that asserts that the fibered faces are, well, fibered. More precisely, fix a compact 3-manifold $M$, and let $B_x$ be the Thurston polytope. Then the theorem asserts that the set of elements of $H^1(M;\mathbb{R})$ which are representable by non-singular closed $1$-forms is a union of cones on open top-dimensional faces of $B_x$ (minus the origin). -Here's the part that I am having trouble with. Assume that $M$ is closed, and let $\alpha$ be any non-singular closed $1$-form. Thurston already proved that $\alpha$ lies in the cone on an open top-dimensional face of $B_x$. If $S$ is a surface in $M$, then denote by $[S] \in H^1(M;\mathbb{R})$ the Poincare dual cohomology class. Consider an incompressible surface $S$ in $M$ such that $[S]$ lies in the cone on the same open top-dimensional face of $B_x$ as $\alpha$. Most of the work of the proof goes into showing that linear combinations $t[S] + u \alpha$ with $t \geq 0$ and $u > 0$ are representable by nonsingular closed $1$-forms. That I have no problem with. However, Thurston then asserts that $[S]$ can be so represented, and the only justification he gives is that you can get it by "iterating this construction". I cannot figure out what he means here. -Can anyone help me? Thanks! - -REPLY [8 votes]: It's been a very long time since I've read this paper, and I haven't been able to find a copy online, so my apologies in advance if what I'm about to write is nonsense. -Think in terms of projective coordinates (i.e. I won't say "cone" any more). I think the lemma you describe also holds for surfaces $S$ on the boundary of the face in question. (If this is not the case, then never mind, I'll delete this answer.) That is, we can take $[S]$ to be a vertex of the face, and we know that $t[S] + u\alpha$ is representable by a nonsingular 1-form whenever $\alpha$ is, for $t \ge 0$ and $u>0$. In other words, starting at a representable point in the interior of the face, we can move a arbitrary distance toward any of the corners of the boundary of the face, so long as we stay in the interior. Since the interior of the face is the open convex hull of the corners, we can reach any point in the interior this way by letting $S$ run through the corners of the face. -(EDIT: As Agol points out in a comment below, instead of "corners" I should say rational points in the boundary of the face.)<|endoftext|> -TITLE: Reference for a dual isoperimetric problem and solution -QUESTION [5 upvotes]: I am trying to track down the first published solution to the following problem: -What curves within the unit disc in the plane and endpoints on the unit circle, minimize their length (within the ball) while dividing the area of the ball into two regions of a given ratio? -The solution is, of course, circular arcs orthogonal to the unit circle. - -The only complete solution and proof that I've been able to find is embedded in a much stronger theorem within "Stability for Hypersurfaces of Constant Mean Curvature with Free Boundary" by A. Ros and E. Vergasta, Geometriae Dedicata, volume 56, 1995. But I suspect that there is an earlier proof for this specific case. -If anyone can point me in the right direction, I would very much appreciate it. -Edited for clarity of problem. - -REPLY [3 votes]: I think that a straight variational approach has a good chance of yielding the desired conclusion. Fix two points $z_0,z_1$ on the boundary of the unit disk. Denote by $C$ the positively oriented arc of the circle that runs from $z_0$ to $z_1$. -Let $\mathcal{P}_{z_0,z_1}$ the set of paths $[0,1]\to \mathbb{C}$ with endpoints $z_0$ $z_1$. $\mathcal{P}_{z_0,z_1}$ is an affine space modeled on the vector space of maps $[0,1]\to\mathbb{C}$ that vanish at endpoints. -If $\gamma\in \mathcal{P}_{z_0,z_1}$ is an embedded path inside the disk, then the area between $\gamma$ and $C$ is given by the integral -$$\int_C xdy -\int_\gamma xdy $$ -Your constraint can now be given a simpler form -$$\int_\gamma xdy =const. $$ -Now solve the constrained variational problem -$$ \min\left\lbrace \int_0^1 |\dot{\gamma}(t)| dt;\;\;\gamma\in \mathcal{P}_{z_0,z_1},\;\;\int_\gamma xdy =const\right\rbrace. $$ -If we write $\gamma(t)= x(t) + i y(t)$ then the constraint equation can be rewritten as -$$ \int_0^1 x(t) \dot{y}(t) dt =const. $$ -This defines a quadratic hypersurface in the affine space $\mathcal{P}_{z_0,z_1}$. -If we define -$$L, F: \mathcal{P}_{z_0,z_1}\to \mathbb{R}, $$ -$$ L(\gamma)= \int_0^1 |\dot{\gamma}(t)| dt,\;\; F(\gamma)= \int_\gamma xdy$$ -then the Euler-Lagrange equations for the above variational problem have the form -$$ dL +\lambda dG=0 \tag{$EL_\lambda$}$$ -where $\lambda\in\mathbb{R}$ is a Lagrange multiplier and $dL$ and $dF$ are the differentials of $L$ resp. $F$. The differential of $F$ is a linear function so ($EL_\lambda$) can be viewed as a nonlinear eigenvalue problem. It can be written very explicitly and I believe that playing with it will yield the desired conclusion.<|endoftext|> -TITLE: A Variance-Tail Description for Continuous Probability Distributions -QUESTION [8 upvotes]: Start with a continuous probability distribution given by a density function f(x). Let X be a real random variable whose distribution is given by the probability distribution. -I would like to ask about the following specific way to describe the distribution in terms of the variance of tails. -Given a number t between 0 and 1 let s be such that the probability that $X\ge s$ equals t. (In other words, $\int_s^{\infty}f(x)=t$.) Let $W_X(t)$ be the variance of $X$ conditioned on $X\ge s$. -The function $W_X(t)$ seems an interesting way to describe the original probability distribution, and its tail behavior. I wonder if there is an explicit useful way to move from $W$ back to $f$; Can you give a dictionary between the tail behavior of $f$ and of that of $W$; And an explicit description of $W$ for some "famous" probability distributions. (E.g. the normal distribution, and the Tracy Widom distribution.) And of course if this kind of "transform" is considered in the literature. -UPDATE: Many thanks to Robert and Brendan to their answers. Robert's asymptotic expansion seem amazing. I will be happy to understand the asymptotics as t goes to 0. -Looking at various distribution one sees that when the distribution is exponential, W is constant; when the distribution is normal and thus decays doubly exponentially, W(t) behaves like $\log^{-1}(1/t)$. I am curious how the asymptotic decay of $W(t)$ translates to the tail behavior of the probability distribution. For the tracy Widom distribution when you ask what is the probability that $X \ge M+t \sigma$ you get an expression like $e^{-t^{3/2}}$ and for the probability that $X \le M-t \sigma$ you get something like $e^{-t^3}$. I wonder how these tails behavior can be described in terms of $W_X(t)$ and $W'_X(t)$ respectively, where $W'_X(t)$ deals with conditioning on $X \le s$ rather than on $X \ge s$. So let me form a simple question: -Question: If $W_X(t)$ behaves like $\log^a (1/t)$, for some real $a>0$, what does it say about the tail behavior of $X$. (More generally, how the decay of $W$ translates to the decay of the distribution.) - -REPLY [2 votes]: Expression for the quantile function (the inverse CDF) for the case $E(X) = 0$: -Let $\overline{F}(s) = 1 - F(s)$. Then, $t = \overline{F}(s)$ and -$$\frac{d}{ds} E(X^2|X \geq s) = \frac{d}{ds} \Big( \frac{\int_s^{+\infty}x^2f(x)dx}{\int_s^{+\infty}f(x)dx} \Big) =$$ -$$ = \frac{-s^2 f(s) \int_s^{+\infty}f(x)dx+f(s) \int_s^{+\infty}x^2f(x)dx}{\big(\int_s^{+\infty}f(x)dx\big)^2} = $$ -$$ = \frac{-s^2 f(s) + f(s) W(\overline{F}(s))}{\overline{F}(s)}$$ -On the other hand, -$$ \frac{d}{ds} E(X^2|X \geq s) = \frac{d}{ds} W(\overline{F}(s)) = -W'(\overline{F}(s)) f(s)$$ -where $W'$ is the derivative of $W$. Therefore, -$$ \frac{s^2 - W(\overline{F}(s))}{\overline{F}(s)} = W'(\overline{F}(s))$$ -or, using $t$, -$$ \frac{(\overline{F}^{-1}(t))^2 - W(t)}{t} = W'(t) $$ -and -$$ \overline{F}^{-1}(t) = \sqrt{tW'(t) + W(t)}$$ -($\overline{F}^{-1}(t)$ should be positive in some neighborhood of $t = 0$). -Then, since $\overline{F}^{-1}(t) = F^{-1}(1-t)$, -$$ F^{-1}(x) = \sqrt{(1-x)W'(1-x) + W(1-x)} $$ -I hope that it's correct and that it can tell something about the decay of the distribution when $t$ tends to $+0$. Maybe it can also be extended to the case $E(X) \neq 0$.<|endoftext|> -TITLE: Can we put a probability measure on every $\sigma$-algebra? -QUESTION [16 upvotes]: The following question has puzzled me for some time: - -Let $(\Omega,\Sigma)$ be a nonempty, - measurable space. Does there - necessarily exist a probability - measure $\mu:\Sigma\to[0,1]$? - -If there exists a nonempty measurable set $A$ such that no nonempty subset of $A$ is measurable (an atom), we can simply let $\mu(B)=1$ if $A\subseteq B$ and $\mu(B)=0$ otherwise. So the problem is only interesting if the $\sigma$-algebra has not atoms. This rules out every countably generated $\sigma$-algebra. An example of a $\sigma$-algebra that has no atoms but supports a probability measure is $\{0,1\}^\kappa$ for $\kappa$ uncountable, which we can endow with the coin-flipping probability measure. - -REPLY [6 votes]: The following is a (corollary of a) theorem of Sierpinskii from 1933: -If $\mu:\mathcal{P}(\Omega) \to [0,1]$ is a probability measure and $|\Omega|$ is smaller than the first weakly-inaccessible cardinal, then there must be a countable $A \subseteq \Omega$ such that $\mu(A)=1$.<|endoftext|> -TITLE: Linearization of a fact about functions between finite sets -QUESTION [10 upvotes]: Suppose $X$, $Y$, and $Z$ are finite sets. If we have a function -$$f : X \longrightarrow Y$$ -and another -$$g : Y \longrightarrow Z$$ -then the composite function $g \circ f$ has the property that -$$ |\mbox{Im} ( g \circ f )| \leq |Y|. $$ -If we replace $f$ by a formal linear combination of functions $X \longrightarrow Y$, and $g$ by a formal linear combination of functions $Y \longrightarrow Z$, then we may "compose" these formal combinations (enforcing the distributive law) to obtain a linear combination of functions $X \longrightarrow Z$, none of which have image exceeding $|Y|$ in cardinality. -My question is the converse: can any such linear combination be obtained? -I posted this question at math.stackexchange, but there are no responses after a week: -https://math.stackexchange.com/q/97513/22621 - -REPLY [3 votes]: No. Consider e.g. the case $X = Z = \{1,2,3\}$, $Y = \{1,2\}$, and the formal linear combination $h_1 + h_2$ where $h_1(1) = h_1(2)=1$, $h_1(3)=2$, $h_2(1)=h_2(3)=2$, $h_2(2)=3$. -If $h_1 = g_1 \circ f_1$, we must have $f_1(1) = f_1(2) \ne f_3(3)$ and -$\text{Im}(g_1) = \{1,2\}$. Similarly if $h_2 = g_2 \circ f_2$ we must have $f_2(1)=f_2(3) \ne f(2(2)$ and $\text{Im}(g_2) = \{2,3\}$. But $h_{12} = g_2 \circ f_1$, which would have -$h_{12}(1) = h_{12}(2)$ and $\text{Im}(h_{12}) = \{2,3\}$, and similarly $h_{21} = g_1 \circ f_2$, are not part of this formal linear combination.<|endoftext|> -TITLE: Three consecutive quadratic residues problem -QUESTION [8 upvotes]: Prove that doesn't exist $N\in\mathbb{N}$ with property: for all primes $p>N$ exist $n\in\{3, 4,\ldots, N\}$ such that $n, n-1, n-2$ are quadratic residues modulo $p$. - -REPLY [7 votes]: This is Theorem 2 in D. Lehmer & E. Lehmer, On runs of residues, -Proceedings of the American Mathematical Society, Vol. 13, No. 1 (Feb., 1962), 102-106. -The proof there uses quadratic reciprocity and Dirichlet's Theorem on primes in arithmetic progression. They also deal with similar problems for k-th powers.<|endoftext|> -TITLE: Large geodesically convex subsets of tori -QUESTION [6 upvotes]: Let $X=\mathbb T^d=\mathbb R^d/\mathbb Z^d$ and let $E$ be a proper open subset of $X$. We say $E$ is geodesically convex if for any $x,y\in E$ the shortest geodesic connecting $x$ and $y$ lies in $E$. -Question. How large can the Haar/Lebesgue measure of $E$ can be? -For example, is $d=2$, then it seems that this cannot exceed $1/2$. Say, $[0,1)\times [0,s)$ is geodesically convex if and only if $s\leq1/2$. (If $s>1/2$, then $[x,x+\delta]$ is not the shortest geodesic for any $\delta\in(1/2,s)$ and any $x\in(0,1)$.) -Is it true for any $d\ge2$ that the measure of such an $E$ cannot exceed $1/2$? - -REPLY [4 votes]: (This is a new answer; my original answer was completely wrong.) -Assume $\mathop{\rm vol}E>\tfrac12$. -Then it contains two opposite points say $x$ and $x'=x+(\tfrac12,\tfrac12,\dots,\tfrac12)$. -WLOG we can assume that $x=0$. -Taking minimizing geodesics form $(\tfrac12,\tfrac12,\dots,\tfrac12)$ to $y\approx 0$, we get that all main diagonal of unit cube -$$\square^n=(0,1)\times(0,1)\times\dots\times(0,1)$$ -lie in $E$. -Then apply the following lemma: -Trivial Lemma. Let $\square^n$ be open unit cube in $\mathbb R^n$ and $E\subset \square^n$ be a locally convex open set which contains all main diagonals of $\square^n$ then $E=\square^n$. -To prove the lemma, note that local convexity + conectedness in $\mathbb R^n$ implies convexity.<|endoftext|> -TITLE: What vector space does the Kauffman bracket skein algebra of FxI act on? -QUESTION [6 upvotes]: The Kauffman bracket skein module $K_t(F\times I)$ (where $t$ is an indeterminant and $F$ is a closed surface) is an associative algebra (the operation being "stacking" links in the $I$ direction). It is a quantum deformation of the coordinate ring of the character variety $\operatorname{Hom}(\pi_1(F),\operatorname{SL}(2))//\operatorname{SL}(2)$. This skein algebra naturally acts on the vector space $\mathcal H_F$, the vector space associated to $F$ by the WRT TQFT. - -Is there a direct construction of this vector space (e.g. in terms of generators and relations) which makes the action of $K_t(F\times I)$ manifest? - -I know only two constructions of the vector space $\mathcal H_F$, and both are unsatisfactory for the purposes of this question. First, we could pick a complex structure on $F$, and let $\mathcal H_F$ be the global holomorphic sections of a particular line bundle $\mathcal L$ over the $\operatorname{SU}(2)$ character variety of $\pi_1(F)$. In this construction, I don't know of any explicit way to realize the action of $K_t(F\times I)$. Second, we could say the vector space is generated by all $3$-manfolds with boundary $F$, and quotient by the subspace which has zero WFT invariant when paired with all three-manifolds with boundary $-F$. This is unsatisfactory because the relations aren't "local", and they require understanding the WRT invariants for all three-manifolds. - -I am looking for a definition of the vector space in the spirit of the definition of $K_t(F\times I)$, i.e. some generators modulo some local "picture" relations. - -REPLY [6 votes]: The answer is $K_t(H)$, where $H$ is a handlebody with boundary $F$. If $t$ is a root of 1 and we are taking the usual semisimple quotient, then $K_t(F\times I)$ is isomorphic to a matrix algebra and $K_t(H)$ isomorphic to the standard representation. Also, in this case we can let $H$ be any 3-manifold with boundary $F$ -- it doesn't even need to be connected.<|endoftext|> -TITLE: an example of a Morse-Bott function -QUESTION [8 upvotes]: Hi, -I want to find an example of a Morse-Bott function such that for at least one of the critical submanifolds, the orientation sheaf O is nontrivial. - -REPLY [6 votes]: You can get a large class of examples if you look at the real projective space $\mathbb{RP}^n$. Each point $L\in \mathbb{RP}^n$ is a one dimensional subspace in $\mathbb{R}^{n+1}$ and we denote by $P_L$ the orthogonal projection onto $L$. Fix a unit vector $v\in \mathbb{R}^{n+1}$ and denote by $U$ the orthogonal complement in $\mathbb{R}^{n+1}$ of the line $L_v$ spanned by $v$. Consider the function -$$f_v: \mathbb{RP}^n\to\mathbb{R}, \;\; f_v(L)=(P_Lv,v). $$ -Note that -$$ 0 \leq f_v(L)\leq 1,\;\;\forall L.$$ -This is a Morse-Bott function with precisely two critical submanifolds: the locus of minima where $f=0$ and consisting of the real projective space $\mathbb{P}(U)\subset \mathbb{RP}^n $ of lines in $U$, and a unique maximum point, $L_v\in\mathbb{RP}^n$ where $f_v(L_v)=1$. -The normal bundle of $\mathbb{P}(U)\subset \mathbb{RP}^n $ is the tautological real line bundle over $\mathbb{P}(U)$ which is nonorientable.<|endoftext|> -TITLE: Reference for embedding an infinite direct product of matrix algebras into the hyperfinite $II_1$ factor -QUESTION [5 upvotes]: In some calculations I am writing up, -$\newcommand{\cR}{{\mathcal R}}$ - I want to add - as a fairly throwaway remark - that any countable product (= $\ell^\infty$-direct sum) of matrix algebras can be embedded inside $\cR$, the hyperfinite ${\rm II}_1$. I think I have managed to hack out an explicit embedding, by realizing $\cR$ as an infinite tensor product of matrix algebras into which I embed successive blocks of my original product algebra, but the inductive construction seems both tedious and wasteful. -Assuming that I have not made a mistake, and that such product algebras do embed as von Neumann subalgebras of $\cR$, does anyone know of a reference for this, so that -(a) I don't have to tax the reader's patience by slogging through something tedious and routine -(b) I can refer the reader to a better construction than the one I currently have? -Moreover, my guess is that something slightly more general should hold: a finite Type ${\rm I}$ with separable predual should embed into $\cR$. Again, does anyone know of a reference, or a counter-example if I have made a mistake? - -REPLY [11 votes]: Murray and von Neumann showed that if $p \in \mathcal R$ is a non-zero projection then $p\mathcal R p \cong \mathcal R$, i.e., the fundamental group of $\mathcal R$ is all positive reals (You should be able to find this in most books that discuss the hyperfinite II$_1$ factor. Also, note that this is easy to see if $p$ has rational trace by viewing $\mathcal R$ as an infinite tensor product of matrices). Thus, if $\{ p_n \}_{n \in \mathbb N}$ is a partition of $1$ by non-zero projections then we obtain an embedding -$$ -\oplus_{n \in \mathbb N} \mathbb M_n(\mathbb C) \subset \oplus_{n \in \mathbb N} \mathcal R \cong \oplus_{n \in \mathbb N} p_n \mathcal R p_n \subset \mathcal R. -$$ -Note also that since $L^\infty([0, 1], \lambda)$ embeds into $\mathcal R$ (e.g., as tensor products of diagonal matrices), it follows that $\mathbb M_n(\mathbb C) \overline \otimes L^\infty([0, 1], \lambda) \subset \mathbb M_n(\mathbb C) \overline \otimes \mathcal R \cong \mathcal R$. Thus, since any separable abelian von Neumann algebra $A_n$ is a von Neumann subalgebra of $L^\infty([0, 1], \lambda)$ it follows from above that $\oplus_{n \in \mathbb N}( A_n \overline \otimes \mathbb M_n(\mathbb C) )$ embeds into $\mathcal R$. All separable finite type I von Neumann algebras are of this form where some of the $A_n \overline \otimes \mathbb M_n(\mathbb C)$ terms may be omitted (this can be found for instance in Takesaki, vol. 1, chapter 1). Thus all separable finite type I von Neumann algebras embed into $\mathcal R$.<|endoftext|> -TITLE: Can a Lagrangian submanifold of ${\mathbb R}^{2n}$ be dense ($n>1$)? -QUESTION [19 upvotes]: I'm betting `yes, sure!', but don't see it. Could someone please point me toward, -or construct for me, a Lagrangian submanifold immersed in -standard symplectic ${\mathbb R}^{2n}$ for $n > 1$, whose closure is all of ${\mathbb R}^{2n}$? -(For an $n =1$ example, one can use the leaves arising from -this modification by Panov of irrational flow on the two-torus.) -Strong preference given to analytic immersions of ${\mathbb R}^n$. -Holomorphically immersed complex lines which are dense in complex 2-space -- i.e. dense ${\mathbb C}$'s in ${\mathbb C}^2$ -- are well-known. Ilyashenko in 1968 showed that the typical solution of the typical polynomial ODE (in complex time) yields such a curve. Following his line -of thought, it might be easier to construct an entire singular Lagrangian foliation of ${\mathbb R}^{2n}$ whose typical leaf is dense, rather than the one submanifold. -Motivation: I have a certain unstable manifold related to a Hamiltonian system. -It is Lagrangian. I would like to be ``as dense as can be'', so I'd like to know -how dense can that be. - -REPLY [22 votes]: Your question already has the answer in it for $n=2$. Take a connected complex curve $L\subset\mathbb{C}^2$ that is dense in $\mathbb{C}^2$. Then $L$ is Lagrangian for the real part of the holomorphic $2$-form $\Upsilon = dz^1\wedge dz^2$. This real part of $\Upsilon$ is equivalent to the standard symplectic structure on $\mathbb{R}^4$ by a linear change of variables. -Added comment about injectivity: Note, by the way, that one can easily arrange for such an $L$ to be a submanifold, not just the image of an immersion (i.e., the immersion is injective). One simple explicit way to do this is to select constants $\lambda_1,\ldots,\lambda_k$ such that the subgroup in $\mathbb{C}^\times$ generated by the numbers $\mathrm{e}^{2\pi i\lambda_1},\ldots,\mathrm{e}^{2\pi i\lambda_k}$ is dense in $\mathbb{C}^\times$ and consider the linear differential equation -$$ -\frac{dy}{dx} = \left(\frac{\lambda_1}{x-x_1}+\cdots + \frac{\lambda_k}{x-x_k}\right)\ y -$$ -where $x_1,\ldots,x_k\in \mathbb{C}$ are distinct. The graph of any nonzero multi-valued solution $y(x)$ over $\mathbb{C}\setminus\{x_1,\ldots,x_k\}$ will then be dense in $\mathbb{C}^2$. (Consider the holonomy around the punctures $x_j$.) Of course, these graphs are the Riemann surfaces associated to the multivalued functions -$$ -y = y_0 (x{-}x_1)^{\lambda_1}\cdots(x{-}x_k)^{\lambda_k} -$$ -(when $y_0\not=0$). These are obviously integral curves (leaves) of the polynomial $1$-form -$$ -\omega = (x{-}x_1)\cdots(x{-}x_k)\ dy - q(x) y\ dx -$$ -for some polynomial $q$ of degree at most $k{-}1$ in $x$. Aside from the obvious closed leaves $x-x_j=0$ and $y=0$, the rest of the leaves are dense submanifolds. (This just gives a simple, explicit example of the general theorem that Richard quoted.) -Dense analytic curves in $\mathbb{R}^2$: It is not hard to construct dense, connected analytic curves in $\mathbb{R}^2$: There exist analytic metrics on the $2$-sphere that have geodesics that wander densely over the surface. Now take such a geodesic and remove a point from $S^2$ through which the geodesic doesn't pass. What's left is a dense analytic curve in $\mathbb{R}^2$. If you are willing to use Finsler metrics, you can even do this with a rotationally invariant real analytic Finsler metric on the $2$-sphere (eg. Katok's examples), so that you can write down the dense analytic curve very explicitly. -I'll think about the case $n>2$. I don't see it yet either, but maybe it's not too hard.<|endoftext|> -TITLE: Stability condition for vector bundles -QUESTION [5 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{C}$, and fix $A$ an ample divisor as the polarization. We say a vector bundle $E$ to be (semi-)stable, if for any proper subsheaf $F$ of $E$, $\mu(F)<\mu(E)$ (resp. $\leq$). -I guess it is not sufficient to check just subbundles $F$ of $E$. But is there a counterexample? Or more precisely, is there an example of $(X, A, E, F)$ satisfying the following conditions? -(1)$X,A,E$ are as above, and $F$ is a proper subsheaf of $E$ breaking the stability condition, i.e., $\mu(F)\geq \mu(E)$. -(2)There exists no vector bundle which break the stability condition, i.e., $\mu(F')<\mu(E)$ for any subbundle $F'$ of $E$. - -REPLY [8 votes]: There are many examples of unstable bundles on a projective surface that have no non-trivial subbundles. For example, if $k$ is an integer with $k < 3$ and $I$ is the sheaf of ideal of $m$ distinct points in $\mathbb P^2$, with $m > 0$, there exists an extension -$$ -0 \longrightarrow \mathcal O \longrightarrow E \longrightarrow I(k) \longrightarrow 0 -$$ -on $\mathbb P^2$ in which $E$ is locally free. Furthermore, the Chern classes of $E$ are $c_1(E) = k$ and $c_2(E) = m$ (for this, see page 103 of "Vector bundles on complex projective spaces", by Okonek, Schneider and Spindler). If $k < 0$, this vector bundle is clearly unstable; but for most values of $k$ and $m$ it can not split as a direct sum of line bundles, hence it cannot contain a line subbundle (since every extension of line bundles on $\mathbb P^2$ splits).<|endoftext|> -TITLE: Products with compactly generated spaces -QUESTION [5 upvotes]: It is well known that if $X$ and $Y$ are topological spaces with $X$ locally compact Hausdorff and $Y$ compactly generated, then $X \times Y$ (with the ordinary product topology) is compactly generated. Does this fail if $X$ is compact, compactly generated, but not Hausdorff? - -REPLY [3 votes]: I think the space $Z=\mathbb{Q}^* \times \mathbb{Q}$ works for your Question. -(Here $\mathbb{Q}^*$ is the one point compactification of the rational numbers) -At first let me recall some properties of $\mathbb{Q}^*$ which you could find Them in the Monthly Article (Between $T_1$ and $T_2$) http://www.jstor.org/discover/10.2307/2316017?uid=3738280&uid=2&uid=4&sid=47699073521117. - -$\mathbb{Q}^*$ is a $KC$ space.(i.e. every compact subset of this space is closed) -It's easy to show that $\mathbb{Q}^*$ and $\mathbb{Q}$ are compactly generated. - -Now I bring Two theorems from the Article "On KC and K-spaces" which are important in the sequel to show that $Z$ is the space which you needed. you could find this Article from: -http://texedores.matem.unam.mx/publicaciones/index.php?option=com_remository&Itemid=57&func=startdown&id=304 -Theorem 1: Let $X , Y$ be topological spaces.If $X$ is $KC$ and $Y$ is hausdorff; Then $X\times Y$ is $KC$. -Theorem 2: In the topological space $X$ the following are equivalent: - -$X$ is locally compact and Hausdorff. -$X^*\times X$ is $KC$ and compactly generated. - -For the sake of Theorem 1, you could see that $\mathbb{Q}^* \times \mathbb{Q}$ is $KC$. Also from part one of theorem 2 and because $\mathbb{Q}$ is not locally compact, We can conclude that $\mathbb{Q}^* \times \mathbb{Q}$ is not compactly generated.<|endoftext|> -TITLE: Maximal ideals in a polynomial ring over the real numbers. -QUESTION [8 upvotes]: Let $\mathbf{R}$ be the field of real numbers. What are the generators of the maximal ideals of the polynomial ring $\mathbf{R}[x_1, ... , x_n]$? If instead of $\mathbf{R}$ one considers the field $\mathbf{C}$ of complex numbers, then Hilbert's Nullstellensatz implies that each maximal ideal $\mathfrak{m}$ of $\mathbf{C}[x_1, ... , x_n]$ is generated by $n$ generators of the form $x_i-a_i$. - -REPLY [21 votes]: There are two kind of maximal ideals in $\mathbf{R}[x_1, \ldots, x_n]$: the ideals corresponding to real points of $\mathbf{A}^n_{\mathbb{R}}$, i.e. of the form $$(x_1-a_1, \ldots, x_n-a_n), \quad a_i \in \mathbf{R}$$ -and the ideals corresponding to pairs of complex-conjugated points, that after a real change of coordinates can be put in the form $$(x_1^2+a_1, x_2-a_2, \ldots, x_n-a_n), \quad a_i \in \mathbf{R}, \quad a_1 >0.$$ -This follows from the generalized weak Nullstellensatz, saying that if $\mathbf{K}$ is any field and $\mathfrak{m} \subset \mathbf{K}[x_1, \ldots, x_n]$ is a maximal ideal, then the field $\mathbf{K}[x_1, \ldots, x_n]/ \mathfrak{m}$ is a finite extension of $\mathbf{K}$. In particular, when $\mathbf{K}=\mathbf{R}$ this extension must have degree at most $2$. -For a reference, see Arrondo's notes A geometric introduction to commutative algebra, in particular Example 0.6 and Corollary 5.14.<|endoftext|> -TITLE: 3-manifolds with solvable fundamental group -QUESTION [17 upvotes]: Is there a nice reference for the classification of closed 3-manifolds with solvable (nilpotent, abelian, etc.) fundamental group, assuming the Geometrization Conjecture? - -REPLY [5 votes]: Although I am late to answer this question, I wanted to put in a sales pitch for the flow charts at the end of Thurston's book (Figures 4.22 and 4.23). They do a great job for dealing with your questions for oriented manifolds. -In the first chart, Thurston uses "almost" to mean virtually, i.e. has a finite index subgroup with some property. Considering compact manifolds with abelian fundamental group, we have the following classification. (The references are to Thurston's book although older references exist for nearly all of these statements.) If $\pi_1(M)$ is finite, then $M$ is a lens space with finite cyclic fundamental group (see Theorem 4.4.14). If $\pi_1(M)$ is not finite and abelian, then $\pi_1(M)= Z, Z\times Z\times Z$ . The first case implies that $M\cong S^1\times S^2$ (see Exercise 4.7.1) and the second implies that $M \cong T^3$ (see Theorem 4.3.4). -As Richard Kent points out the paper of Peter Scott is a wonderful reference and these last two facts are discussed in greater detail there. -Finally to finish our classification for oriented manifolds, we also have to worry about the cusped manifolds. In general, the second chart has this information. The one caveat is that the unknot and Hopf link complements ($S^1\times D^2$ and $T^2 \times I$) might not neatly fit on this list, because they do not admit a cofinite $H^3$, $\widetilde{PSL(2,R)}$ or $H^2 \times R$ structures. However, unknot and Hopf link complements should be added to our list of manifolds having abelian fundamental groups as well since their fundamental groups are $Z$ and $Z\times Z$, respectively. I believe they are not considered in this chart because they would not show up in a decomposition prescribed by the Geometrization. For example, if you attach a Hopf link complement to the boundary of geometric piece via a boundary identification the resulting manifold remains geometric. -The non-orientable case, includes $S^1 \times P^2$, which has fundamental group $Z \times Z/2Z$ (see Epstein's paper Theorem 9.1).<|endoftext|> -TITLE: Number of integer combinations x_1 < ... < x_n ? -QUESTION [5 upvotes]: I asked this question earlier on math.stackexchange.com but didn't get an answer: - -Let $0 < a_1 < ... < a_n$ be integers. Is there a closed formula (or some other result) for the number $N(a_1,...,a_n)$ of integer combinations $0 < x_1 < ... < x_n$ such that $x_i \le a_i$ $(i=1,...,n)$ ? - -Of course, if $a_i = a_n - n + i$ for all $i$, then $N = \binom{a_n}{n}$. -I considered the following model: Let $B_1 \subseteq ... \subseteq B_n$ be nested boxes. $B_i$ contains $a_i$ balls that are labeled by $1,...,a_i$. Choose one ball from each box (without repetition) and afterwards sort the balls. Then $N$ equals the number of different combinations that can be obtained in this way. -Example: $n=3$ -For the chosen balls $b_i \in B_i$ there are the following possibilities: -1) $b_3 \in B_3 \setminus B_2$, $b_2 \in B_2 \setminus B_1$, $b_1 \in B_1$. The balls -are already sorted and there are $(a_3-a_2)(a_2-a_1)a_1$ possibilities. -2) $b_3,b_2 \in B_2 \setminus B_1$, $b_1 \in B_1$. After sorting $b_2,b_3$ there are -$\frac{(a_2-a_1)(a_2-a_1-1)}{2!}\cdot a_1$ possibilities. -3) $b_3,b_2,b_1 \in B_1$. After sorting there are $\frac{a_1 (a_1-1)(a_1-2)}{3!}$ possibilities. -4) $b_3 \in B_3 \setminus B_2$, $b_2,b_1 \in B_1$. After sorting $b_1, b_2$ there are $(a_3-a_2) \cdot \frac{a_1 (a_1-1)}{2!}$ pssibilities. -5) $b_3 \in B_2 \setminus B_1$, $b_1,b_2 \in B_1$. After sorting $b_1, b_2$ there are $(a_2-a_1) \cdot \frac{a_1 (a_1-1)}{2!}$ pssibilities. -Generalizing this pattern yields the formula - -$$N(a_1,...,a_n) = \sum_{\nu} \prod_{i=1}^n \binom{a_i-a_{i-1}}{\nu_i!}$$ - -$(a_0 := 0)$ where $\nu_i$ is the number of balls choosen from $B_i \setminus B_{i-1}$. The sum is taken over all $\nu=(\nu_1,...,\nu_n)$ such that $0 \le \nu_i \le i$ and $\nu_1 + ... + \nu_n = n$. -But this is far from a closed formula. I do not even know the exact number of summands. -Also note that there is a recursion formula - -$$N(a_1,...,a_n) = N(a_1,...,a_{n-1}) + N(a_1,...,a_{n-1},a_n -1)$$ - -but I wasn't able to guess a closed form thereof. - -Edit: Thank you all very much for your answers. Each one deserves to be accepted. Unfortunately this isn't possible in MO. I therefore accepted William's since Proctor's formula in the linear case seems to be most helpful in the application I have in mind. - -REPLY [12 votes]: The simplest formula is the determinant -$$ -\left| \binom{a_i+j-i}{ j-i+1}\right|_{i,j=1,\dots,n}. -$$ -For example, when $n=3$ this is -$$ -\begin{vmatrix} -\displaystyle \binom{a_1}{1}&\displaystyle \binom{a_1+1}{2}&\displaystyle \binom{a_1+2}{3}\\ -1 &\displaystyle \binom{a_2}{1}&\displaystyle \binom{a_2+1}{2}\\ -0 & 1 &\displaystyle \binom{a_3}{1} -\end{vmatrix} -$$ -In general there are 1's in the diagonal below the main diagonal and 0's below that, so when expanded there are $2^{n-1}$ terms. -It is unlikely that there is a simpler formula. -This formula is most easily proved by inclusion-exclusion. We start with the set of positive integer sequences $(x_1,\dots, x_n)$ satisfying $1\le x_i\le a_i$ for each $i$ and use inclusion-exclusion to count those sequences satisfying none of the conditions $x_i\ge x_{i+1}$, using the fact that it's easy to count the sequences satisfying any subset of them. For example, with $n=3$ the determinant expands to -$$a_1a_2a_3 - a_1\binom{a_2+1}{2} -\binom{a_1+2}{2} a_3 - + \binom{a_1+2}{3}.$$ -Here the term $a_1\binom{a_2+1}{2}$, for example, counts sequences $(x_1,x_2,x_3)$ satisfying $a_1\ge x_1\ge 1$ and $a_2\ge x_2\ge x_3\ge 1$. The crucial fact that makes this work is that since $a_2 < a_3$, the condition $a_2\ge x_2\ge x_3\ge 1$ implies that $a_3\ge x_3$. -As Vladimir noted, an equivalent formula replaces strong with weak inequalities. This formula (in a more general form) was given by B. R. Handa and S. G. Mohanty, ``On $q$-binomial coefficients and some statistical applications," SIAM J. Math. Anal. 11 (1980), 1027--1035. A recent proof of their formula, with further references, is in my paper with Nicholas Loehr, -Note on enumeration of partitions contained in a given shape, Linear Algebra Appl. 432 (2010), 583--585. A preprint version can be found on my home page, -http://people.brandeis.edu/~gessel/homepage/papers/. -While I'm plugging my own papers, I'll note a paper of mine closely related to the paper of Pemantle and Wilf that William mentioned: A probabilistic method for lattice path enumeration, J. Statist. Plann. Inference 14 (1986), 49--58, also available on my home page.<|endoftext|> -TITLE: Is $\omega$ absolute in set theory without foundation? -QUESTION [5 upvotes]: Let $\text{ZF}^-$ be the set theory without powerset, choice, and foundation. Consider the following notions: - -Wellfounded sets -$$WF(c) \Leftrightarrow (\forall x \subseteq TC(c)) \left[x \neq \emptyset \rightarrow (\exists y \in x) (\forall t \in x) [t \notin y]\right]$$ -Ordinals -$$ON(c) \Leftrightarrow WF(c) \wedge \text{Transitive}(c) \wedge (\forall x,y \in c)[x = y \vee x \in y \vee y \in x]$$ -$\omega$ -$$x = \omega \Leftrightarrow ON(x) \wedge (\forall y \in x)[y = \emptyset \vee (\exists z)[y = z \cup \{z\}]] $$ - -Are those $\Delta_1$ notions? Are those notions absolute between transitive models of $\text{ZF}^-$? More precisely, is there a model $V$ of $\text{ZF}^-$ with transitive classes $N,M \models \text{ZF}^-$ s.t. those notions are not absolute between $N$ and $M$? Does the situation change if we add powerset or choice to our theory? -Those notions are absolute once we have foundation, moreover they are definable by a $\Delta_0$ formula. I was wondering if they are still absolute without assuming foundation, but that requires more work to show; or if they cease being absolute at all. If so, is there an easy way to construct a counterexample? - -REPLY [10 votes]: None of these three notions is absolute, even if you retain powerset and the axiom of choice. -In Boffa’s set theory (which contains ZFC without foundation, and is conservative over ZFC with respect to the well-founded kernel), every extensional set-like binary relation is isomorphic to a transitive class with $\in$. In particular, you can take the ultrapower of the universe over a nonprincipal ultrafilter on a countable set, and let $M$ be its transitive collapse. Then there are nonstandard integers in $M$, i.e., $\omega^M\ne\omega$. -References for Boffa’s axiom: - -Maurice Boffa, Forcing et négation de l’axiome de Fondement, Académie Royale de Belgique, Mémoires, Classe des Sciences, Collection $8^o$, II. Série 40, No. 7 (1972). -David Ballard and Karel Hrbáček, Standard Foundations for Nonstandard Analysis, Journal of Symbolic Logic 57 (1992), No. 2, 741–748.<|endoftext|> -TITLE: Orthogonal subgroups of dual group -QUESTION [6 upvotes]: This question arises from this one. -Let $G$ be a finite abelian group and $H$ a subgroup of $G$. Let $\widehat{G}$ be group of all characters of $G$ and let $H^\perp = \{\chi \in \widehat{G} : \chi = 1 \text{ on } H\}$. Does anyone know a good reference proving the identity $(H^\perp)^\perp = H$? -I'm interested in a reference (instead of a proof) because I only need to use this fact in a non representation theory paper and won't have the space needed to introduce and prove what is needed. - -REPLY [4 votes]: It is Proposition 3.4 in the book "Washington: Introduction to Cyclotomic Fields". - -REPLY [3 votes]: There are very pedagogical proofs in [1] section 3.5 and in [2] section 6.1. In both references, the most typical properties of orthogonal subgroups of finite-abelian groups are proven, including the one you are looking for. I have read both texts and used both for preparing seminars, and I find the exposition quite clear. I have cited both of these references in one of my papers: I think the first of them [1] has never appeared in a Journal, but it is quite a trusted paper in my field (quantum computation), so you could cite the arXiv; the second reference [2] has recently been published in QIC Vol. 13. No.11&12, 1007 (2013). -In fact, this property is one of many consequences of the Pontryagin-Van Kampen duality, and holds for all locally compact Abelian groups (finite Abelian groups are a particular case of those; the integers, the real numbers, the torus are also locally compact Abelian; see also this related question of mine in MathOverflow). If you would like to see a proof of your statement in the more general case, I recommend you to take a look at the online notes of the course Introduction to Topological Groups, by Dikran Dikranjan, University of Udine. The relevant sections for you are: - -Section 11 Pontryagin-van Kampen duality, in particular 11.4.; -the annihilator $A_{\widehat{G}}(H)$, defined at the beginning of section 11.4.2, is precisely the subgroup of the characters of $G$ that you are looking at; -theorem 11.5.4. proves your statement. - -As far as I know, Dikranjan's notes have never been published; still, they are (in my opinion) a very useful resource. Some parts of them (but sadly not the ones you need) have appeared in An elementary approach to Haar integration and Pontryagin duality in locally compact abelian groups, Dikranjan and Stoyanov, Topology and its Applications, Vol. 158, no. 15, 2011 Elsevier Science.<|endoftext|> -TITLE: Density of strictly ergodic measures in the d-bar topology -QUESTION [8 upvotes]: I am currently studying a problem which deals with cocycles of highly -noncompact operators on Hilbert space, with the base transformation -being the full shift on two symbols. In my particular situation it -turns out that the top Lyapunov exponent of a fixed cocycle, -considered as a function on the space of invariant measures, is discontinuous with respect to the weak-* topology but Lipschitz continuous with respect to Ornstein's $\overline{d}$-metric. -However I do not have very much intuition for the topology on the -invariant measures induced by $\overline{d}$, and the resources on -this seem relatively limited (the best I have found so far is Glasner's book Ergodic theory via joinings). I am currently trying to understand the generic features of the set of ergodic measures with respect to the $\overline{d}$-metric. -With regard to the space of shift-invariant measures under the weak-* topology, the following facts have been well-known for many years, mostly dating back to Parthasarathy's 1961 paper On the category of ergodic measures: - -The space of measures under the weak-* topology is a compact metrisable space (elementary, as these things go). -Measures supported on a single periodic orbit are dense. In particular, ergodic measures are dense, and strictly ergodic measures (i.e. those whose support is uniquely ergodic) are dense (Parthasarathy 1961). -Weak-mixing measures are a dense residual set. (Density follows from the density of strong-mixing measures; the $G_\delta$ property follows from a modification of Parthasarathy's work.) -Strong-mixing measures are a dense meagre set. (Meagreness is due to Parthasarathy. Density of strong-mixing measures follows from density of Gibbs measures.) -Fully-supported measures are a dense residual set. (Fairly elementary.) -Zero-entropy measures are a dense residual set. (Follows from density of periodic orbits and semicontinuity of entropy.) - -Most of these statements have known analogues in the $\overline{d}$-metric, namely: - -The space of measures under the $\overline{d}$-metric is complete but not separable. -Measures supported on periodic orbits are not dense. -The space of ergodic measures is closed, as are the spaces of strong-mixing and Bernoulli measures. -Fully-supported measures are a dense residual subset of the ergodic measures. -Entropy is continuous. - -However, it is not clear to me whether or not strictly ergodic measures are dense in the ergodic measures in the $\overline{d}$-metric. Does anyone know whether or not this is the case? - -REPLY [2 votes]: The strictly ergodic measures are dense in the ergodic measures in the $\bar d$-metric. The proof (actually, a sketch of the proof) is given in Single orbit dynamics by Weiss (Theorem 4.4', page 46).<|endoftext|> -TITLE: Symmetric polynoms are Hopf algebra ? What for one needs co-product ? -QUESTION [10 upvotes]: From many sources I am hearing that symmetric polynomials not only an algebra, but also Hopf algebra. -Would someone be so kind to explain where the coproduct (antipode) comes from ? -And what it is useful for ? -It is mentioned e.g. here: -Avatars of the ring of symmetric polynomials -Let me pay some attention to details of the question. -Is it correct that there is some "natural" Hopf structure OR different authors writes about different and they are interesting each one in each specific task ? -Is it important to go to the limit n->infinity (which I heard many times, but not really like/understand) or we can stay in the more custom setup of $C[x_1,...,x_n]^{S_n}$ ? -What is puzzling for me that $C^n$ can be seen as an additive group and so $C[x_1...x_n]$ is Hopf algebra, but this does not seem to survive on $C[x_1...x_n]^{S_n}$. -E.g. $\Delta(xy)=xy\otimes 1+x\otimes y +y\otimes x + 1\otimes xy$ which is not in -$C[x,y]^{S_2} \otimes C[x,y]^{S_2}$. - -REPLY [6 votes]: In the great answers given so far, I didn't find a simple direct description of the natural comultiplication operation on symmetric functions. So I'll just copy and paste my comment from this question which tries to explain it in as few characters as are allowed in a comment. It shows why infinitely many variables are needed: we need the Hilbert hotel (twice the number of variables must equal the number of variables). OK, since I see copy-and-paste from comments doesn't work without manual repair anyway, I'll add a few words of clarification as well. -Symmetric functions are (degree-bounded power series) in infinitely many variables, and order doesn't matter (due to the "symmetric" part). Now rename the variables $x_0,y_0,x_1,y_1,x_2,\ldots$ and decompose a symmetric function $s=\sum_i u_iv_i$ as a sum of products of a symmetric function $u_i$ in the $x$'s and one $v_i$ in the $y$'s; then $\Delta(s)=\sum_i u_i\otimes v_i$. For instance for elementary symmetric function one has $\Delta(e_k)=\sum_{i+j=k}e_i\otimes e_j$ since the monomials can be arbitrarily spread across the $x$'s and $y$'s (and similarly for complete homogeneous symmetric functions), while for power sums $\Delta(p_k)=p_k\otimes 1+1\otimes p_k$, since the monomials involve a single $x_i$ or a single $y_i$, but the two cannot mix in a power sum. -I may add that multiplication and comultiplication are dual in operations in this case; Zelevinsky calls it a (positive) self-adjoint Hopf-algebra.<|endoftext|> -TITLE: Non-bimeromorphic compactifications -QUESTION [10 upvotes]: Let $X$ be a (smooth, non-compact) complex space. By a compactification of $X$ we mean a compact complex space $\overline X$ which contains a dense open subset biholomorphic to $X$ (we shall identify this subset with $X$) and such that $\overline X\setminus X$ is a proper analytic subset (with no conditions on its codimension). -In particular, given two (different) compactifications of $X$, they always contain biholomorphic dense open subsets. -My question is: given two compactifications of $X$, are they necessarily bimeromorphic? -More precisely, does the biholomorphism between the two dense open subsets always extend to a global bimeromorphic map? -I guess the answer is no, but after a moment of reflection I cannot find any counterexample. -Perhaps, it would suffice to look at compactifications of $\mathbb C^2$... - -REPLY [7 votes]: Another example, different but diffeomorphic to Polizzi's, is given by -$E \times \mathbb C$, where $E$ is a fixed elliptic curve say $ \mathbb C^* / \lbrace z \mapsto 2z \rbrace$. -It can be compactified as the projective surface $E \times \mathbb P^1$ or as the (non-Kähler) Hopf surface $\mathbb C^2-\{0\} / \lbrace (z,w) \mapsto (2z,2w) \rbrace$. -Notice that the two compactifications have fields of meromorphic functions of different transcendence degree over $\mathbb C$, i.e., they have different algebraic dimensions.<|endoftext|> -TITLE: When is there a deformation of a given singularity to a normal singularity -QUESTION [11 upvotes]: Question: Given a variety $X_0$ with a singularity (say Cohen-Macaulay), when does this exist as a special fiber of a flat family $X \to C$ mapping to a smooth curve $C$, such that the generic fiber is normal? -My one thought is that perhaps this reduces to checking whether some 1-dimensional thing is smoothable via some Bertini argument perhaps (using the R1 + S2 characterization of normality). -Of course, maybe there is some easy reference on deformations of singularities which answers this? (Hopefully) -Example: -I actually have a very specific example in mind whose presentation is given below: -$$\frac{\mathbb{F}_2[[a,b,c,d]]}{(bc, d^2 + ab^2, dc)}.$$ -Feel free to take the algebraic closure of the field by the way. I can also discuss how it comes about if it is relevant and I can also describe how one can do an integral domain version if it helps. -This example is Cohen-Macaulay, has two components, and is 2-dimensional (checked via Macaulay2), so I can't use positive characteristic analogs of the sort of results for Du Bois singularities saying that various non-CM things can't be deformed to Cohen-Macaulay things as in Kollar-Kovacs (such theorems do hold in positive characteristic but are not published). -This example is F-injective (the characteristic 2 analog of Du Bois) but it is not F-pure (the characteristic 2 analog of semi log canonical). - -REPLY [5 votes]: Your $X_0$ is Cohen-Macaulay of codimension $2$ in affine space, so determinantal (Hilbert-Burch). When also $\dim X_0\le 3$ it is smoothable; see Schaps' paper in Am. J. Math., vol. $99$, for all this.<|endoftext|> -TITLE: Software package to manipulate representations -QUESTION [8 upvotes]: Hi, -I am looking for a software package that will allow me to experiment with the irreducible representations of lie groups (for example, $SL(2,p)$) over the complex field and over finite fields. That is, I would like to get the corresponding matrices for group elements. -Thanks, -Shachar - -REPLY [10 votes]: For representations over the complex field, I know that GAP does a good job. (I'm not sure if it can do modular representations as well, but I wouldn't be surprised.) -Here is some example code to get you started: -G:=SL(2,3);; -reps:=IrreducibleRepresentations(G);; -Elements(G); -List(G,g->g^reps[5]); - -This prints the elements of the group $SL(2,3)$: -[ [ [ 0*Z(3), Z(3)^0 ], [ Z(3), 0*Z(3) ] ], - [ [ 0*Z(3), Z(3)^0 ], [ Z(3), Z(3)^0 ] ], - [ [ 0*Z(3), Z(3)^0 ], [ Z(3), Z(3) ] ], - [ [ 0*Z(3), Z(3) ], [ Z(3)^0, 0*Z(3) ] ], - [ [ 0*Z(3), Z(3) ], [ Z(3)^0, Z(3)^0 ] ], - [ [ 0*Z(3), Z(3) ], [ Z(3)^0, Z(3) ] ], - [ [ Z(3)^0, 0*Z(3) ], [ 0*Z(3), Z(3)^0 ] ], - [ [ Z(3)^0, 0*Z(3) ], [ Z(3)^0, Z(3)^0 ] ], - [ [ Z(3)^0, 0*Z(3) ], [ Z(3), Z(3)^0 ] ], - [ [ Z(3)^0, Z(3)^0 ], [ 0*Z(3), Z(3)^0 ] ], - [ [ Z(3)^0, Z(3)^0 ], [ Z(3)^0, Z(3) ] ], - [ [ Z(3)^0, Z(3)^0 ], [ Z(3), 0*Z(3) ] ], - [ [ Z(3)^0, Z(3) ], [ 0*Z(3), Z(3)^0 ] ], - [ [ Z(3)^0, Z(3) ], [ Z(3)^0, 0*Z(3) ] ], - [ [ Z(3)^0, Z(3) ], [ Z(3), Z(3) ] ], - [ [ Z(3), 0*Z(3) ], [ 0*Z(3), Z(3) ] ], - [ [ Z(3), 0*Z(3) ], [ Z(3)^0, Z(3) ] ], - [ [ Z(3), 0*Z(3) ], [ Z(3), Z(3) ] ], - [ [ Z(3), Z(3)^0 ], [ 0*Z(3), Z(3) ] ], - [ [ Z(3), Z(3)^0 ], [ Z(3)^0, Z(3)^0 ] ], - [ [ Z(3), Z(3)^0 ], [ Z(3), 0*Z(3) ] ], - [ [ Z(3), Z(3) ], [ 0*Z(3), Z(3) ] ], - [ [ Z(3), Z(3) ], [ Z(3)^0, 0*Z(3) ] ], - [ [ Z(3), Z(3) ], [ Z(3), Z(3)^0 ] ] - ] - -followed by complex matrices representing each element: -[ [ [ 0, -1 ], [ 1, 0 ] ], - [ [ E(3)^2, -E(3) ], [ 1, 0 ] ], - [ [ -E(3), -E(3)^2 ], [ 1, 0 ] ], - [ [ 0, 1 ], [ -1, 0 ] ], - [ [ E(3), E(3)^2 ], [ -1, 0 ] ], - [ [ -E(3)^2, E(3) ], [ -1, 0 ] ], - [ [ 1, 0 ], [ 0, 1 ] ], - [ [ E(3), E(3)^2 ], [ 0, 1 ] ], - [ [ E(3)^2, -E(3) ], [ 0, 1 ] ], - [ [ 1, 0 ], [ E(3), E(3)^2 ] ], - [ [ -E(3)^2, E(3) ], [ E(3), E(3)^2 ] ], - [ [ 0, -1 ], [ E(3), E(3)^2 ] ], - [ [ 1, 0 ], [ -E(3)^2, E(3) ] ], - [ [ 0, 1 ], [ -E(3)^2, E(3) ] ], - [ [ -E(3), -E(3)^2 ], [ -E(3)^2, E(3) ] ], - [ [ -1, 0 ], [ 0, -1 ] ], - [ [ -E(3)^2, E(3) ], [ 0, -1 ] ], - [ [ -E(3), -E(3)^2 ], [ 0, -1 ] ], - [ [ -1, 0 ], [ E(3)^2, -E(3) ] ], - [ [ E(3), E(3)^2 ], [ E(3)^2, -E(3) ] ], - [ [ 0, -1 ], [ E(3)^2, -E(3) ] ], - [ [ -1, 0 ], [ -E(3), -E(3)^2 ] ], - [ [ 0, 1 ], [ -E(3), -E(3)^2 ] ], - [ [ E(3)^2, -E(3) ], [ -E(3), -E(3)^2 ] ] -] - -Change "5" to other numbers to see different representations. Also, note that the GAP symbol -Z(p) - -denotes a generator of the multiplicative group of the finite field $\mathbb{F}_p$. Similarly, -E(k) - -denotes a primitive $k^{\mbox{th}}$ root of unity.<|endoftext|> -TITLE: How many vertices/edges/faces at most for a convex polyhedron that tiles space? -QUESTION [13 upvotes]: I wonder if this problem has already been examined before: -Consider a convex polyhedron that tiles $\mathbb R^3$. What is the maximum of vertices/edges/faces that such a polyhedron can have? -Intuitively, it seems that the truncated octahedron is best possible for edges (36) as well as for vertices (24) -Its packing is also known as "bitruncated cubic honeycomb". -For faces, we can do better than 14, as there is a polyhedron with 16 faces that can be obtained as follows: Take a truncated tetrahedron and add on each triangular face a small pyramid that is a quarter of a tetrahedron. The tessellation of it is known as the quarter cubic honeycomb, with each small tetrahedron "distributed" among its four neighbors. -Questions: -Are these best possible? -What about the corresponding problem in higher dimensions? -In $\mathbb R^4$, it looks like the polytope yielded by the equivalent of the "Quarter cubic honeycomb" tiles it. This one, based on the truncated 5-cell has 25 cells, 60 faces, 60 edges, and 25 vertices, and so for cells and vertices, it does again (slightly) better than the 24-cell with its 96 faces, 96 edges, and 24 vertices. - -REPLY [11 votes]: In $\mathbb{R}^3$ this is a famous problem.See this nice reference. (Danzer, Grunbaum, Shepard -- Does every type of polyhedron tile three-space). The best example at the time of the writing had 38 faces (an example of Engel). For a lattice (periodic) tiling, the problem was solved (I think) by Delone (AKA Delaunay). - -REPLY [10 votes]: To supplement Igor's citation of Engel's polyhedron, here it is: -      - - -The figure above is from the paper by -Branko Grünbaum and G. C. Shephard, -"Tilings with congruent tiles", -Bull. Amer. Math. Soc. (N.S.), Volume 3, Number 3 (1980), 951-973 -and is copied from Peter Engel's paper "Über Wirkungsbereichsteilungen von kubischer Symmetrie", Zeitschrift für Kristallographie, volume 154 (1981), number 3–4, 199–215. -It is certainly remarkable that this polyhedron tiles space! -Addendum (10Oct13): -Now that Günter brought this back to the front page with -the precise citation (thanks!), I'll add -that a student Tiffany Liu and I 3D-printed copies of the -18-sided plesiohedra found by Löckenhoff and Hellner that tile space -so we could see how they fit together. Here they are inside the 3D printer:<|endoftext|> -TITLE: Verifying claims in the proof of the Rigidity Lemma (Mumford, GIT) -QUESTION [6 upvotes]: In Chapter 6 of Mumford's Geometric invariant theory, during the proof of the rigidity lemma, there are two statements I'm not sure how to verify. The general setup is: -$p : X \rightarrow S$ is flat, $S$ connected, and $H^0(X_s, o_{X_s}) \cong k(s)$ for all points $s \in S$. - -In the first part, we're assuming $\epsilon : S \rightarrow X$ is a section, and that $S$ consists of one point. Mumford says: "One checks that $p_*(o_X) \cong o_S$." I found a proof when $p$ is projective (and even proper, I think), which works because this is going to be used on projective abelian schemes, but the general case is still bothering me. -In the second part, $X$ still has the section $\epsilon$, but $S$ is now general (i.e. not just a point), and $p$ is a closed map. During the proof, $Z$ is a closed subscheme of $X$. Mumford claims the statement: - - -If $p^{-1}(t) \subset Z$ (set-theoretically), for any $t \in S$, then for all artin subschemes $T \subset S$ concentrated at $t$, $Z$ contains $p^{-1}(T)$ as a subscheme. - -implies that $Z$ contains an open neighborhood of $p^{-1}(t)$. Intuitively, I think of the artin subscheme as a thickening of the point, and so if I contain an entire fiber then I get "a little bit extra", making $Z$ contain an open neighborhood. I'm wondering how I should do this more formally. -Thanks for any help, it is much appreciated! - -REPLY [2 votes]: Regarding 1): Suppose that $A$ is an Artin local ring with maximal ideal $\mathfrak{m}$. By the local criterion for flatness, we have isomorphisms of $\mathcal{O}_X$-modules $\mathfrak{m}^i/\mathfrak{m}^{i+1}\otimes_A \mathcal{O}_X\simeq \mathfrak{m}^i \mathcal{O}_X/\mathfrak{m}^{i+1}\mathcal{O}_X$, in particular the latter is free and $\Gamma(\mathfrak{m}^i \mathcal{O}_X/\mathfrak{m}^{i+1}\mathcal{O}_X)\simeq \mathfrak{m}^i/\mathfrak{m}^{i+1}$ via the natural adjunction isomorphism. -We argue that the natural morphism $\mathfrak{m}^k\to \Gamma(\mathfrak{m}^k\mathcal{O}_X)$ is an isomorphism by descending induction on $k$. For large enough $k$, both sides are zero, so this is true. For the induction step, use the diagram -$\require{AMScd}$ -\begin{CD} -0@>>>\mathfrak{m}^{k+1} @>>> \mathfrak{m}^{k} @>>>\mathfrak{m}^{k}/\mathfrak{m}^{k+1} @>>>0\\ -@. @VVV @VVV @VVV @.\\ -0@>>>\Gamma(\mathfrak{m}^{k+1}\mathcal{O}_X)@>>> \Gamma(\mathfrak{m}^{k}\mathcal{O}_X)@>>> \Gamma(\mathfrak{m}^k\mathcal{O}_X/\mathfrak{m}^{k+1}\mathcal{O}_X)@>>>R^1\Gamma(\mathfrak{m}^{k+1}) -\end{CD} -Here, the third vertical arrow is the isomorphism by flatness and the projection formula that I claimed above. It follows that the second vertical arrow is an isomorphism by the five lemma.<|endoftext|> -TITLE: Schemes and meaning of "geometric intuition" -QUESTION [19 upvotes]: I recently started studying algebraic geometry together with a couple of friends and especially in discussions online we keep reading about developing geometric intuition. There are some questions on this website about developing geometric intuition, but none of them really ask for what geometric intuition is. It doesn't seem obvious what the whole concept of geometric intuition means in the context of modern algebraic geometry, so it seems hard to judge when one has started developing it. Especially, when problems in number theory can often be related to algebraic geometry and then solve by using this geometric intuition. -Let me give an example from elementary analysis, where this is completely obvious. Take the squeeze theorem. Anyone can visualize two graphs in their head and "see" that anything between them must get pushed to the same point. Then the proof just corresponds to having learned how to translate a picture to a formal epsilon-delta argument. My question is then that is an algebraic geometer or arithmetic algebraic geometer working actually seeing nice pictures of lines, surfaces and curves in their head? Or is it just a matter of having developed experience with how different algebraic objects behave? The latter wouldn't seem any different from having developed intuition about, say, field theory through experience and this intuition could hardly be called "geometric" by anyone. -Feel free to close if this question is considered inappropriate for this website, I certainly understand. The reason for posting here instead of math.stackexchange is that graduate students in my department don't seem to have the experience themselves to answer it. I'm sure that this intuition keeps developing for a long time after finishing graduate studies. Hence, I was hoping for an experienced audience hopefully willing to answer. - -REPLY [27 votes]: Vote this answer up if you consider yourself an algebraic geometer, and (in the course of your work) actually see nice pictures of lines, surfaces, and curves in your head. - -REPLY [2 votes]: Vote this answer up if you consider yourself an algebraic geometer, and (in the course of your work) do not see nice pictures of lines, surfaces, and curves in your head.<|endoftext|> -TITLE: Differential Equation Examples for Calculus Students -QUESTION [9 upvotes]: I've been teaching calculus courses for a while now, and something always bothers me each time I teach it. Students always seem to have trouble connecting with the differential equation material for the following reason: I always tell them that they are one of the most important topics for applications of calculus (this is mainly a course for students in the sciences) and that all sorts of fields use them...and then all I have to tell them are things that are to a certain extent quite dull: exponential growth, Newton's law of cooling, the logistic equation, and a few other of the classics. While each of these is quite important and do have broad applications, I've never seen anyone be shocked to learn that populations of rabbits breeding in the wild grow approximately exponentially. -My knowledge of applied fields isn't terrible, but I'm still at a loss as to what plausible models I could teach them about where the global results are not immediately obvious, so I ask: what are some simple differential equations, simple enough for a freshman calculus class, which occur in the sciences and have behavior interesting enough the catch peoples' interest? - -REPLY [3 votes]: I am not really familiar with the US system, so this might be too advanced. -Maybe it would be interesting to discuss partial differential equations that can be reduced to ordinary differential equations, if one uses symmetries. Take the hydrogen atom for example: You get a decomposition in a radial and spherical part, the second can be solved by separation of variables. So you get three ordinary differential equations and their solution will give you a description of the hydrogen atom in non relativistic quantum mechanics. -There are of course a lot of other examples from physics, where similiar reasoning is applied.<|endoftext|> -TITLE: How many Tutte polynomials of complete graphs are known? -QUESTION [9 upvotes]: I would like to compute the Tutte polynomial of the complete graph $K_n$ for n as large as possible. Using a program by Björklund, Husfeldt, Kaski, Koivisto (here), I managed to compute up to n=18 on my home computer (in serial) in less than a day. Overall, I've been very impressed by this program. -I'm likely to include the results of these computations in an upcoming paper, so, for comparison, I would also like to mention what people have done previously in the area. (Also, I'd like to check that the computations are consistent with one another.) -Question: How far have others computed the Tutte polynomial of the complete graph? - -REPLY [13 votes]: There is a simple formula for the generating function of $T_{K_n}(x,y)$, which is more cleanly expressed in terms of the (equivalent) "coboundary polynomial" $X_M(q,t) = (t-1)^{r(M)} T_M(1+\frac{q}{t-1}, t)$: -$$ -1+q\sum_{n \geq 1} X_{K_n}(q,t) \frac{x^n}{n!} = \left( \sum_{n \geq 0} t^{n \choose 2}\frac{x^n}{n!}\right)^q. -$$ -This is essentially due to Tutte in the 50s; see this paper and the references in it. -For what it's worth, using similar methods one easily obtains formulas for other similar families such as complete bipartite graphs (for graphs) and classical root systems (for matroids).<|endoftext|> -TITLE: Two Definitions of "Character" of topological groups -QUESTION [11 upvotes]: When I first met the concept of "characters" of topological groups in Pontryagin's book "Topological groups", it was defined as follows: - -Let $G$ be a topological group. A character of $G$ is a homomorphism of topological groups from $G$ to the torus $T=\mathbb{R}/\mathbb{Z}$. Here, the torus $T$ has the induced topology from the usual topology of the real line $\mathbb{R}$. - -I found the same definition on Wikipedia. So, I think this is a standard definition of character. But, in a lot of modern articles, it seems to me that characters are defined as follows: - -Let $G$ be a topological group. A character of $G$ is a homomorphism of topological groups from $G$ to the discrete additive group $\mathbb{Q}/\mathbb{Z}$. - -Clearly, these two definitions cannot be the same for all topological groups. However, if $G$ is a (discrete) finite group, then the two definitions agree. -Questions: - -What kind of conditions on a topological group $G$ one needs in order to identify the two definitions of character? - -If $G$ is a "profinite group", then do the two definitions agree? If the answer is yes, then how can one prove it? - - -Please give me any advice. -Later -I found a way to answer the second question. One only needs to show the following: for any profinite group $G$ and any continuous homomorphism $f:G \to T$, the image of $f$ is finite. This statement can be shown as follows. The torus $T$ has an open neighborhood $U$ of $0$ which contains no non-trivial subgroup of $T$. Since $G$ is profinite, there exists an open normal subgroup $H$ of $G$ satisfying $H \subset f^{-1}(U)$. This implies $H \subset \ker(f)$. So, the map $f$ factors through the finite group $G/H$. This implies that the image of $f$ is finite. -Hence the two definitions of character agree for any profinite group. - -REPLY [7 votes]: It is easy to miss the point that in the second definition $\mathbb Q/\mathbb Z$ is required to be discrete in Hiro's question. -Hence even if a continuous morphism $f:G\to T$ has image in $\mathbb Q/\mathbb Z$ it cannot automatically be considered as a continuous map $f_0:G\to \mathbb Q/\mathbb Z$ and so might not be a character in the second sense. -An example for this failure is to take $G=T_{tors} =\mathbb Q/\mathbb Z\subset T$ , the torsion subgroup of $T$ with its induced topology from the circle and for the character $f $ (in the first sense) the inclusion $f:G\hookrightarrow T$. -Even though $G$ is torsion, the corestricted morphism $f_0:G\to \mathbb Q/\mathbb Z$ is not a character in the second sense, since it is not continuous. -However if a character on a compact group $G$ happens to have values in $\mathbb Q/\mathbb Z$,so that both definitions can be compared, then its image in the circle is finite and the two concepts coincide. -Xandi explains in his answer that this is always the case for profinite groups.<|endoftext|> -TITLE: Quasi-nilpotent trace class operators as limits of nilpotents -QUESTION [11 upvotes]: In as yet unwritten work with T. Figiel and A. Szankowski we make an observation in a Banach space context that for Hilbert spaces reduces to: -If $T$ is a quasi-nilpotent (i.e., has only $0$ in its spectrum) trace class operator on a Hilbert space, then $T$ is the limit, in the trace class norm, of finite rank nilpotent operators. -Is this a known result? The proof is simple but employs tools that I think are not generally used when studying nilpotents, so the observation might have been overlooked. A Google search produced nothing promising. My colleagues R. Douglas, C. Foias, and C. Pearcy, as well as G. Weiss, did not know the result, but that is not conclusive as the first three are even older than I and Gary never worked on quasi-nilpotent operators. -The general topic of quasi-nilpotents as limits of nilpotents has been studied a lot. For a start, click -here. - -REPLY [7 votes]: This question appears as the third exercise in the third chapter of Kenneth Davidson's textbook Nest Algebras. Based on the material presented in the third chapter, here is the intended proof (with the numbers referencing the text). -Let $Q$ be a quasinilpotent trace class operator. Since $Q$ is compact, there exists a maximal nest $\mathcal{N}$ of invariant subspaces of $Q$ (Corollary 3.2). Let $\mathbb{A}$ be the collection of (one-dimensional) atoms of $\mathcal{N}$. If $A \in \mathbb{A}$, the map $T \mapsto P(A)T|_A$ is an algebra homomorphism of the nest algebra $\mathcal{T}(\mathcal{N})$ into $\mathbb{C}$ and thus $P(A) Q|_A$ is the zero scalar for all $A \in \mathbb{A}$ as $Q$ is quasinilpotent (see 3.3). -By the Erdos Density Theorem (Theorem 3.11) there exists a net $(R_\lambda)$ of finite-rank contractions in $\mathcal{T}(\mathcal{N})$ that converge to the identity in the strong-$\ast$ topology. Since $Q^\ast$ is also a trace class operator, it is easy to verify that $R^\ast_\lambda Q^\ast$ converges to $Q^\ast$ in the trace class norm (Proposition 1.18). Thus $QR_\lambda$ converges to $Q$ in the trace class norm. -Since each $QR_\lambda$ is a finite-rank operator, it suffices to show that each $QR_\lambda$ is nilpotent. Since each $R_\lambda$ is a finite-rank element of $\mathcal{T}(\mathcal{N})$, each $R_\lambda$ is a finite sum of rank one operators of the form $xy^*$ where $y \in N^\bot$ and $x \in N_+$ for some element $N \in \mathcal{N}$ ($N_+$ being the successor of $N$) (see 3.7 and 3.8). If we multiply $R_\lambda$ by $Q$, we obtain a similar decomposition of $QR_\lambda$ as a sum of such operators (where, if $x \in N_+$, $Qx \in N_+$ as $Q \in \mathcal{T}(\mathcal{N})$). However, if $N_+ \neq N$, $N_+ \ominus N$ is an atom so if $x \in N_+$, $Qx \in N_+$ and thus $Qx \in N$ as $Q$ is the zero scalar on all atoms. Hence each $QR_\lambda$ is a finite sum of rank one operators of the form $xy^*$ where $y \in N^\bot$ and $x \in N$ for some element $N \in \mathcal{N}$. Since $\mathcal{N}$ is a nest, it is then easy to see that each $QR_\lambda$ is a nilpotent operator.<|endoftext|> -TITLE: Tensor product of coherent modules -QUESTION [6 upvotes]: Let $X$ be a ringed space. Recall that an $\mathcal{O}_X$-module $M$ is called coherent if it is of finite presentation and for every open $U \subseteq X$ and any integer $n \ge 1$, the kernel of every morphism of $\mathcal{O}_U$-modules $\mathcal O_U^{\oplus n} \to M|_U$ is of finite type. Coherent modules constitute an abelian category (in contrast to modules of finite presentation or just of finite type). See the Stacks project, modules, section 12. In general $\mathcal{O}_X$ might be not coherent. - -Question. If $M,N$ are coherent $O_X$-modules, is it true that $M \otimes_{\mathcal{O}_X} N$ is coherent? - -I guess that this will be false in this generality. So let us restrict to schemes, w.l.o.g. affine schemes. Here a module $M$ over a ring $A$ is coherent if it is of finite presentation and every submodule of finite type is of finite presentation. - -REPLY [8 votes]: In the analytic category this is indeed true: if $\mathscr{F}$, $\mathscr{G}$ are coherent analytic sheaves on a complex space $X$, then $\mathscr{F} \otimes_{\mathscr{O}_X} \mathscr{G}$ is also coherent. -For a reference, look at [Grauert-Remmert, Coherent Analytic Sheaves], Proposition at the bottom of page 240. -It seems to me that their proof works also in the algebraic category, maybe you should check.<|endoftext|> -TITLE: Weak Vector Bundles -QUESTION [8 upvotes]: The following notion has arisen in a paper I'm writing. -Definition. A map $p: E\to B$ of spaces -is said to be weak vector bundle if for all compact subspaces $K \subset B$ -the restriction of $p$ to -$K$, i.e., $p_{|K}: E_{|K} \to K$, has the structure of a vector bundle. -Of course, when $B$ is compact this $p$ is just a vector bundle, so it's the non-compact case which is relevant. -Questions: Does this concept appear in the literature? If so, can I get a reference? -Have characteristic classes been developed in this setting? -Added Later: Francesco's remark suggests that I should add the question: Are there any examples? - -REPLY [8 votes]: A (rank $n$) vector bundle over a space $X$ is the "same thing as" a principal $Gl_n$ bundle over $X$. Now, consider the functor which assigns to each space $X$ the (discrete) group $Hom\left(X,Gl_n\right).$ We may regard this in fact to be a functor into groupoids, which happens to land in groups. This functor is not a stack, but we can take its stack completion in a universal way (i.e. stackify it) and the resulting functor $BGl_n$ assigns a space $X$ the groupoid of principal $GL_n$-bundles over $X$. Concretely this functor is calculated by taking the (weak) colimit over all (open) covers $U$ of $X$ of $Hom\left(X_U,Gl_n\right),$ where the latter is the groupoid, whose objects are continuous homomorphisms from $X_U$- the associated Cech groupoid- to $Gl_n$ (regarded as a topological groupoid with one object), and whose arrows are given by continuous natural transformations. It's straight forward to verify that for each cover $U$ this is simply a collection of continuous maps $$g_{ij}:U_i \cap U_j \to GL_n,$$ satisfying a cocycle condition, i.e. cocycle data for a principal $Gl_n$-bundle over the cover $U$. This is how to see that the resulting stack actually assigns the groupoid of all principal $Gl_n$-bundles. Now all of this was done for stacks with respect to the "open cover Grothendieck topology". But, there is another Grothendieck topology on compactly generated Hausdorff spaces, which is in may ways more natural. I call it the "compactly generated Grothendieck topology" here: http://arxiv.org/abs/0907.3925 . It is a Grothendieck topology such that every open cover is still a cover, but in general, there are MORE covers. Basically, a non-open covering of a space becomes a cover for this Grothendieck topology if and only if its restriction to each compact subset can be refined by an open covering. If we instead take the stack completion of $$X \mapsto Hom\left(X,Gl_n\right)$$ with respect to the compactly generated Grothendieck topology, what one gets is the concept of principal $Gl_n$-bundles which are only locally trivial when restricted to compact subsets, and using the corresponding cocycles, one can build a "weak vector bundle" in your sense. In particular, this collapses to the definition of an ordinary vector bundle when the base is locally compact Hausdorff. -In fact, such a "weak" principal $Gl_n$ bundle is the same as a continuous homomorphism $$X_V \to Gl_n$$ where $X_V$ is the Cech groupoid associated to such a generalized cover (the concept of a Cech groupoid makes sense for non-open covers). This induces a continuous map between (fat) geometric realizations $$||X_V|| \to BGl_n,$$ and I prove in the same paper I referenced above that $$||X_V||$$ is always weakly homotopy equivalent to $X$. So this should allow you to do characteristic classes as well.<|endoftext|> -TITLE: Non-enumerative proof that there are many derangements? -QUESTION [47 upvotes]: Recall that a derangement is a permutation $\pi: \{1,\ldots,n\} \to \{1,\ldots,n\}$ with no fixed points: $\pi(j) \neq j$ for all $j$. A classical application of the inclusion-exclusion principle tells us that out of all the $n!$ permutations, a proportion $1/e + o(1)$ of them will be derangements. Indeed, by computing moments (or factorial moments) or using generating function methods, one can establish the stronger result that the number of fixed points in a random permutation is asymptotically distributed according to a Poisson process of intensity 1. -In particular, we have: -Corollary: the proportion of permutations that are derangements is bounded away from zero in the limit $n \to \infty$. -My (somewhat vague) question is whether there is a "non-enumerative" proof of this corollary that does not rely so much on exact combinatorial formulae. For instance, a proof using the Lovasz Local Lemma would qualify, although after playing with that lemma for a while I concluded that there was not quite enough independence in the problem to make that lemma useful for this problem. -Ideally, the non-enumerative proof should have a robust, "analytic" nature to it, so that it would be applicable to other situations in which one wants to lower bound the probability that a large number of weakly correlated, individually unlikely events do not happen (much in the spirit of the local lemma). My original motivation, actually, was to find a non-enumerative proof of a strengthening of the above corollary, namely that given $l$ permutations $\pi_1,\ldots,\pi_l: \{1,\ldots,n\} \to \{1,\ldots,n\}$ chosen uniformly and independently at random, where $l$ is fixed and $n$ is large, the probability -that these $l$ permutations form a $2l$-regular graph is bounded away from zero in the limit $n \to \infty$. There is a standard argument (which I found in Bollobas's book) that establishes this fact by the moment method (basically, showing that the number of repeated edges or loops is distributed according to a Poisson process), but I consider this an enumerative proof as it requires a precise computation of the main term in the moment expansion. - -REPLY [17 votes]: L. Lu and L. A. Szekely have successfully applied the Lopsided (i.e. Negative Dependency Graph) Lovasz Local Lemma to this problem. -A negative dependency graph is as a dependency graph except that independence is replaced with the inequality -$$ -\Pr\left(A_k \middle\vert \bigwedge_{i \in S} A_i\right) \leq \Pr(A_k) -$$ -for any fixed event $A_k$ and collection $S$ of non-neighbors of $A_k$. Erdos and Spencer showed the Lovasz Local Lemma holds with negative dependency graphs in place of dependency graphs. -Let $\Omega$ be the probability space of all perfect matchings of $K_{n,n}$ equipped with the uniform distribution. For a partial matching $M$ of $K_{n,n}$, define the event -$$ -A_M = \{M^\prime \in \Omega \mid M \subseteq M^\prime\} -$$ -(all perfect matching that contain the partial matching $M$). -Given a collection $\mathcal{M}$ of partial matchings of $K_{n,n}$, construct a graph with vertex set $\{A_M \mid M \in \mathcal{M}\}$ and set two matchings adjacent if their union is not again a matching. L. Lu and L. A. Szekely showed this graph is a negative dependency graph. -Finally we can address the problem at hand. Let the partite sets of $K_{n,n}$ be $\{1, \dots, n\}$ and $\{1^\prime, \dots, n^\prime\}$. For $1 \leq i \leq n$, let $M_i$ be the one-edge matching $ii^\prime$. Viewing perfect matchings of $K_{n,n}$ as permutations of an $n$-element set, the event $\bigwedge_{i=1}^n \overline{A_{M_i}}$ contains precisely those permutations not having a fixed point. Choosing $x_i = \frac{1}{n}$ for the purposes of the Lopsided Lovasz Local Lemma, we get -$$ -\Pr\left( \bigwedge_{i=1}^n \overline{A_{M_i}} \right) \geq \left(1 - \frac{1}{n}\right)^n, -$$ -which converges to $\frac{1}{e}$ as $n \rightarrow \infty$.<|endoftext|> -TITLE: Reshetikhin-Turaev as a 3-2-1-theory -QUESTION [15 upvotes]: I keep reading that the Reshetikhin-Turaev construction actually yields a 3-2-1 tqft. I know the construction that associates to a suitably decorated surface a vector space built up from a hom-space in a modular tensor category and to a decorated 3-manifold a linear map between these vector spaces as described in the book by Bakalov and Kirillov or in Turaev's blue book. -But to obtain a 3-2-1 theory, I would need a 2-category first. I suppose in this case this would be the category of all small linear categories. What I don't see is, how the Reshetikhin-Turaev construction associates a functor to surfaces and a natural transformation to 3-cobordisms. Or is this the wrong way to think about this? -I have the feeling that this was probably already asked and answered here, but I could not find it. If so, sorry. - -REPLY [4 votes]: I really like Hiro Lee Tanaka's answer. This is just a long comment. -The Witten-Reshitkhin-Turaev TQFT needs more data than a three-manifold, or a surface. You can decorate the three-manifolds with a framing, or a $p_1$ structure, and then decorate surfaces with a Lagrangian subspace of their first homology and an integer, and put colors on one manifolds, and the whole thing works... in theory. Actual computations are a bear. -The most successful modes of computation descend from coloring surgery diagrams to define morphisms between reduced skein modules as can be found in the seminal paper of Habegger, Masbaum and Vogel on TQFT derived from the Kauffman bracket. -Blanchet, C.; Habegger, N.; Masbaum, G.; Vogel, P. Topological quantum field theories derived from the Kauffman bracket. Topology 34 (1995), no. 4, 883–927. -In practice all the extra data can be accounted for quite elegantly by accepting that the TQFT underlying WRT is four dimensional. I think this idea is due to Kevin Walker. -The four dimensional theory is so trivial, that you can account for it with homological data, and hence decategorify the theory to get a less elegant $3$-dimensional TQFT. The ambiguities you are eliminating are in fact just scalar multiplication by a power of the central charge, which is a root of unity. -Here is what I think the extension looks like from top down. -Choose a level $r$ -The Crane-Yetter invariant is assigned to four manifolds. -The reduced Kauffman bracket skein module at level $r$ is assigned to three-manifolds. -The surface $F$ is assigned the category of left modules over the reduced Kauffman bracket skein algebra of the surface. The morphisms in the category correspond to four dimensional cobordisms. -A one manifold is assigned a $2$-category. The objects are relative reduced skein modules of three manifolds, so that there is a disk in the boundary of the three-manifold for each circle in the one manifold, inside of which is an oriented arc that is the boundary of relative skeins. The morphisms come from gluing three manifolds along the complement of the disk, and are induced by including relative skein modules. The two morphisms are given by morphisms between the reduced skein modules induced by surgery diagrams corresponding to the four manifolds. -Pretty sketchy. -Practically speaking, current interest in quantum invariants is focused on their asymptotic behavior as the level goes to infinity. Generally people are focused on the magnitude of the invariants not the phase, so all the fancy book keeping to keep track of phase to get a three-dimensional theory is wasted effort.<|endoftext|> -TITLE: Example of cone of numerically effective curves which is not polyhedral -QUESTION [5 upvotes]: I think I have seen more than one reference in which the cone of numerically effective curves can be 'not polyhedral', i.e. with an infinite number of extremal rays -I cannot remember where I read that the usual example is the blow-up of the plane in the $9$ points of intersection of $2$ general cubics. This should give an infinite number of $-1$-curves, but I don't manage to see why it should be infinite! As far as I see the strict transform $C$ of any curve in the pencil would give me $C^2=\pi*(C)+ \sum E_i^2=9-9=0$ where $E_i$ are the exceptional divisors. -What am I missing? - -REPLY [6 votes]: Yes, this is the standard example of a variety whose cone of curves has infinitely many extremal rays. (A reference is Koll\'ar--Mori, p.22). -To see why there are infinitely many (-1)-curves: each of the 9 points you blow up gives a (-1)-curve E_i, as you know. But the E_i are also sections of the elliptic fibration determined by the pencil. So one can use the group structure of the generic fibre (which is an elliptic curve over the function field k(P^1)) to translate any of the E_i to any other section of the fibration; since the action is by automorphisms of the surface, the other sections must be (-1)-curves also. As long as there are infinitely many sections (which is guaranteed by the assumption that the cubics are general) one gets infinitely many (-1)-curves this way. -Edit: It's worth mentioning that there are other examples of surfaces with non-finitely generated cone of curves, which differ from the example above in an interesting way. For example, suppose X is a K3 surface which has Picard rank at least 3 and no (-2) curves. Then the (closed) cone of curves of X consists of all classes in N^1(X) satisfying x^2 ≥ 0 and x.H ≥ 0 for any fixed ample class H. This is the standard "round cone" in N^1, with uncountably many extremal rays. (A similar example is obtained by taking X to be an abelian surface with Picard rank at least 3.) -As mentioned in the comments, Volume 1 of Lazarsfeld's great book Positivity in Algebraic Geometry is the best reference for these kinds of questions.<|endoftext|> -TITLE: Should I write to the referee? -QUESTION [6 upvotes]: I have submitted a paper to a good journal about 3 months ago. Now I have realized that there is a mistake in the paper. Now I have completely corrected it. It is not a very serious error, but I think it could create some problems in reading the paper. In particular there is a definition that should be changed, but not in a trivial way. The main results of the paper remain essentially unchanged. -So my question is: should I write to the editors of the journal in order to inform the referee about this problem or should I wait some comments by the referee? -Thank's! - -REPLY [6 votes]: If the correction will help the referee understand your paper better, you should definitely write it up and send it to the editors, who will pass it on to the referee. Having served as a referee I can honestly say that I would always appreciate receiving clarifications, and would not hold that against anyone. You might also send a very short note, explaining the changes.<|endoftext|> -TITLE: A question about local connectedness in metric spaces -QUESTION [6 upvotes]: Must every compact and connected metric space be locally connected at at least one -of its points? - -REPLY [5 votes]: One more picture: Brouwer--Janiszewski--Knaster continuum:<|endoftext|> -TITLE: Which trigonometric identities involve trigonometric functions? -QUESTION [11 upvotes]: Another question that's getting no answers on stackexchange: -Once upon a time, when Wikipedia was only three-and-a-half years old and most people didn't know what it was, the article titled functional equation gave the identity -$$ -\sin^2\theta+\cos^2\theta = 1 -$$ -as an example of a functional equation. In this edit in July 2004, my summary said "I think the example recently put here is really lousy, because it's essentially just an algebraic equation in two variables." (Then some subsequent edits I did the same day brought the article to this state, and much further development of the article has happened since then.) -The fact that it's really only an algebraic equation in two variables, $x^2+y^2=1$, makes it a lousy example of a functional equation. It doesn't really involve $x$ and $y$ as functions of $\theta$, since any other parametrization of the circle would have satisfied the same equation. In a sense, that explains why someone like Norman Wildberger can do all sorts of elaborate things with trigonometry without ever using trigonometric functions. -But some trigonometric identities do involve trigonometric functions, e.g. -$$ -\sin(\theta_1+\theta_2)=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2 -$$ -$$ -\sec(\theta_1+\cdots+\theta_n) = \frac{\sec\theta_1\cdots\sec\theta_n}{e_0-e_2+e_4-e_6+\cdots} -$$ -where $e_k$ is the $k$th-degree elementary symmetric polynomial in $\tan\theta_1,\ldots,\tan\theta_n$. These are good examples of satisfaction of functional equations. -So at this point I wonder whether all trigonometric identities that do seem to depend on which parametrization of the circle is chosen involve adding or subtracting the arguments and no other operations. In some cases the addition or subtraction is written as a condition on which the identity depends, e.g. -\begin{align} -\text{If } & x+y+z=\pi \\ \text{then } & \tan x+\tan y+\tan z = \tan x\tan y\tan z. -\end{align} -QUESTION: Do all trigonometric identities that do involve trigonometric functions, in the sense that they are good examples of satisfaction of functional equations by trigonometric functions, get their non-triviality as such examples only from the addition or subtraction of arguments? Or is there some other kind? And if there is no other kind, can that be proved? -In comments below the stackexchange posting, Gerry Myerson mentioned the identities -$$ -\cos \frac x2=\sqrt{\frac{1+\cos x}{2}} -$$ -and -$$ -\prod_{k=1}^\infty \cos\left(\frac{x}{2^k}\right)= \frac{\sin x}{x} -$$ -The latter is somewhat like the one involving tangents above: One can say that if $x_n = x_{n+1}+x_{n+1}$ for $n=1,2,3,\ldots$ then -$$ -\prod_{k=1}^\infty \cos \left(\frac{x_k}{2^k}\right) = \frac{\sin x_1}{x_1}. -$$ -A similar but simpler thing applies to the half-angle formula. -Postscript: Wikipedia's list of trigonometic identities is more interesting reading than you might think. It has not only the routine stuff that you learned in 10th grade, but also some exotic things that probably most mathematicians don't know about. It was initially created in September 2001 by Axel Boldt, who was for more than a year the principal author of nearly all of Wikipedia's mathematics articles---several hundred of them. - -REPLY [6 votes]: Most of these identities have a different explanation, via holonomic functions, aka solutions of linear differential equations with polynomial coefficients. And, in the end, it is all a matter of linear algebra and bounding dimensions. This is very well explained in slides by Bruno Salvy (where trigonometrics are explicitly used for the simple examples, before moving on to the 'fun' stuff). -To a certain extent, there is still a lot of geometry still going on in the above, but the geometry of the parameter space, rather than looking directly at the traces of the functions. -Many details of the method can be found in the paper A non-holonomic systems approach to special function identities by Chyzak, Kauers and Salvy.<|endoftext|> -TITLE: Minimum dimension for sphere packing a graph in Euclidean space -QUESTION [6 upvotes]: Igor Pak suggested I ask this as a separate question. In Extensions of the Koebe–Andreev–Thurston theorem to sphere packing? it was asked whether there were simple conditions to decide whether a finite graph could be expressed by a bunch of spheres in $\mathbb R^3,$ two spheres touching if and only if the relevant vertices shared an edge. -Scott Carnahan and I are of the opinion that any graph on $n$ vertices can be placed in $\mathbb R^{n-1}$ in the manner described. It is proved in Igor's book that the complete graph can be placed in $\mathbb R^{n-2}$ and no smaller dimension, one regular simplex with unit radii and then one extra sphere in the center. Of course, the complete graph can also be placed as a regular simplex with all unit spheres in $\mathbb R^{n-1}.$ But the varying radius question seems more fortunate, we get the same answer, if it works, in $\mathbb H^{n-1}$ and $\mathbb S^{n-1}.$ -So, that is the initial question, can anyone prove that any graph on $n$ vertices, however many or few edges, can be placed in $\mathbb R^{n-1}$ as a set of spheres, if we allow varying radius? -Secondarily, and I have not the slightest idea, is there any sort of expected value of the minimum dimension, or, at least, some sort of "normal behavior" for this, meaning that "most" graphs on $n$ vertices need a minimum dimension of about ____? - -REPLY [7 votes]: Start with a regular simplex with unit length edges in ${\mathbf{R}}^{n-1}$, representing $K_n$. In any non-degenerate simplex, one can increase or decrease the length of any edge by a sufficiently small amount, leaving all other edge lengths fixed and flexing the dihedral angle opposite to the edge. Do this to increase the length of every edge in $K_n$ that is not present in your given graph $G$, one edge at a time. Finally, place radius-1/2 balls at the vertices of the resulting simplex. The result is a sphere packing representing $G$.<|endoftext|> -TITLE: Using Vinogradov's theorem for finding prime solutions to a linear equation (an exercise from Vaughan's book) -QUESTION [5 upvotes]: I'm trying to solve an exercise from Vaughan's book, "The Hardy-Littlewood Method" (ex. 3 in chapter 3: Goldbach's problems, p.36), because I want to use the result stated in it. It is a variation of Vinogradov's theorem on every large enough odd integer being the sum of 3 primes. -Here I need to show that there are "many" triples of prime numbers less than or equal to $N$ that solve the equation with integer coefficients: -$b_1p_1+b_2p_2+b_3p_3-b_4=0$. -More specifically, I need to show that if -$R(N)=\sum_{(p_1,p_2,p_3)\mid p_i\leq n, b_1p_1+b_2p_2+b_3p_3-b_4=0} (\log p_1)(\log p_2)(\log p_3)$ - then $R(N)=J(N)\mathfrak S +O(N^2/\log ^AN)$ where $J(N)$ is the number of integer solutions to $b_1m_1+b_2m_2+b_3m_3-b_4=0$ satisfying $m_i\leq N$. -Following the Proof as given in Vaughan and Nathanzon, I have defined $F_i(x)=\sum _{p\leq N}\log p\cdot e(b_ipx)$ so that $R(N)=\int _0 ^1 F_1(x)F_2(x)F_3(x)e(-b_4x)dx$. -When integrating over the major arcs $\mathcal M$, I guess I should approximate $F_i(x)$ by $G_i(x)=\frac{c_q(b_i)}{\phi (q)}u_i \left(x-\frac{a}{q}\right)$ where $u_i(y)=\sum _{m\leq N}e(mb_iy)$ and evaluate the integral $\int _\mathcal M G_1(x)G_2(x)G_3(x)e(-b_4x)dx$. -So, I start by integrating $G_1(x)G_2(x)G_3(x)e(-b_4x)$ over a singel major arc $(\frac {a}{q}-\frac {Q}{N}, \frac {a}{q}+\frac {Q}{N})$. After a change of variables I end up with the integral $\int _\frac {-Q}{N}^\frac {Q}{N}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$, and I want to bound the difference between the last integral to: $J(N)=\int _\frac {-1}{2}^\frac {1}{2}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$. I guess this is how I get the $J(N)$ in the main term of the desired expression for $R(N)$. -If so, I want to bound $\int _\frac {-1}{2}^\frac {-Q}{N}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$ and $\int _\frac {Q}{N}^\frac {1}{2}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$. In the proof of Vinogradov's Theorem, this difference is $O(Q^2/N^2)$ where $Q=\log ^BN$. But now, in the case dealt with in the exercise, this difference seems to be huge: $O(N^3)$. Since, when $b_iy$ is an integer $e(b_iy)=1$ which gives $u_i(y)=N$ and $|u_1(y)u_2(y)u_3(y)|=O(N^3)$. -To summarize, my questions are: -How do I bound $\int _\frac {Q}{N}^\frac {1}{2}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$? Does anyone know of a place where this claim is proved? -Thanks! - -REPLY [7 votes]: It's a long time since this was posted. I don't know why you didn't ask me directly. There is an oversight in exercise 2. The main term on the RHS in question 2 should be $\gcd(a_1,a_2,a_3)^{-1}J(n)\frak S$. One has to use intervals of length $1/\gcd(a_1,a_2,a_3)$.<|endoftext|> -TITLE: If self-composition of an analytic function on the disk gives a linear fractional map, is the original function linear fractional? -QUESTION [5 upvotes]: Suppose $f: \mathbb{D} \rightarrow \mathbb{D}$ is analytic. Furthermore, suppose $f \circ f = g$, and $g$ is a linear fractional map. Does this guarantee that $f$ is linear fractional? I know it would be true if the domain was the sphere instead of the disk, or if $g$ was onto the disk (as we already have that $f$ is injective because $g$ is). But I want this to be true in general. Any insight? - -REPLY [7 votes]: Yes. Consider the sequence of open disks -$$\mathbb D\subset g^{-1}(\mathbb D)\subset g^{-2}(\mathbb D)\subset \dots . -$$ -Let $\mathbb D'$ be the union. This is either an open disk or the complement of a point in the Riemann sphere. You can extend $f$ to $\mathbb D'$ by defining $f(z)=g^{-n}(f(g^n(z)))$ for sufficently large $n$. Now $f$ maps $\mathbb D'$ to itself in a one to one and onto fashion, so it is a linear fractional map.<|endoftext|> -TITLE: compactness on Weinstein fixed point thm of isometries of positively curved spaces -QUESTION [6 upvotes]: Theorem (Weinstein - Synge) Let $(M,g)$ be a compact oriented positively curved $n$-dimensional Riemannian manifold. Supose $f$ is an isometry that preserves orientation if $n$ is even or reverses it if $n$ is odd. Then $f$ has a fixed point. -This result is also true for conformal diffeomorphisms and implies classical Synge's theorem for compact manifolds. Using the celebrated Soul Theorem we see that the compactness hypotesis in Synge's theorem can be replaced by completness. -Question Can we do the same on Weinstein's theorem? -Thanks! -PS: the even-dimensional version of Synge's theorem is that a compact positively curved oriented even-dimensional Riemannian manifold be simply connected. - -REPLY [8 votes]: The isometry group of a complete open connected manifold of positive sectional curvature has a fixed point. This is stated in Corollary 6.3 of the paper by Cheeger-Gromoll where they prove the soul theorem [Ann. of Math, 1972].<|endoftext|> -TITLE: Bertini's theorem in char p for base point free linear system -QUESTION [12 upvotes]: I always believed the following statement: if $X$ is a smooth variety over an algebraically closed field of positive characteristic, assuming we know that the general member of a base point free linear system $|L|$ is reduced, then indeed a general member is smooth. -However, I realize this is not obvious, though all the examples I know which fail this Bertini theorem has non-reduced fibers. -So I was wondering whether indeed this statement is true. Or counterexamples are known. -[REEDIT:] After Laurent Moret-Bailly's nice counterexample. I was then wondering whether a counterexample exists when the linear system induces a birational morphism to its image. The assumption now is closer to the one in Bertini's theorem which says the statement is true when we assume the morphism is an embedding. -This sort of counterexamples will always help birational geometers to think what's going on in characteristic $p$. - -REPLY [17 votes]: In characteristic 3, consider the surface $V\subset \mathbb{P}^2\times\mathbb{A}^1$ with equation $y^2z=x^3-tz^3$ (where $t$ is the coordinate on $\mathbb{A}^1$). It is easily seen to be smooth. The general fiber of the projection on $\mathbb{A}^1$ is a plane cubic which is reduced but not smooth. Taking a suitable projective completion $V\hookrightarrow X$ you get a morphism $X\to\mathbb{P}^1$ which is a counterexample.<|endoftext|> -TITLE: Morita equivalence of acyclic categories -QUESTION [5 upvotes]: (Crossposted from math.SE.) -Call a category acyclic if only the identity morphisms are invertible and the endomorphism monoid of every object is trivial. Let $C, D$ be two finite acyclic categories. Suppose that they are Morita equivalent in the sense that the abelian categories $\text{Fun}(C, \text{Vect})$ and $\text{Fun}(D, \text{Vect})$ are equivalent (where $\text{Vect}$ is the category of vector spaces over a field $k$, say algebraically closed of characteristic zero). Does it then follow that $C, D$ are equivalent? (If so, can we drop the finiteness condition?) -Without the acyclic condition this is false; for example, if $G$ is a finite group regarded as a one-object category, $\text{Fun}(G, \text{Vect})$ is completely determined by the number of conjugacy classes of $G$, and it is easy to write down pairs of nonisomorphic finite groups with the same number of conjugacy classes (take, for example, any nonabelian group $G$ with $n < |G|$ conjugacy classes and $\mathbb{Z}/n\mathbb{Z}$). -On the other hand, Mariano's answer on math.SE shows that the result is true if $C, D$ are either both taken to be free categories on graphs or both taken to be posets by basic results in the representation theory of quivers. (This and idle curiosity are the main motivation for the question.) - -REPLY [2 votes]: The answer to your question is "no". A counterexample is given in this paper by Leroux in Example 1.6.<|endoftext|> -TITLE: Fiber functors to derived categories -QUESTION [12 upvotes]: Suppose that $G$ is an algebraic group over a field $k$. Then for any $k$-algebra $R$, a fiber functor from $\text{Rep}_k(G)$ to the category of projective modules over $R$ is the same as a $G$-torsor on $\text{Spec}(R)$, by Theorem 3.2 of Deligne-Milne. -Now suppose that instead of a $k$-linear faithful exact tensor functor into the category of projective modules, I have a functor (taking exact sequences of representations to distinguished triangles, and $\otimes$ to $\otimes^L$, say) to the bounded derived category of coherent modules over $R$. Do I get some sort of "derived $G$-bundle" on $\text{Spec}(R)$? What should that mean? - -REPLY [3 votes]: I think you want your functor to be a delta-functor from the abelian category of representations to D^b_{Coh}(R) which is moreover compatible with tensor products. By the way. I suggest we replace D^b_{Coh}(R) by the derived category of perfect complexes, just to be in vogue. -A special case is where R = k. Then D^b_{Coh}(k) is the category of finite dimensional graded k-vector spaces endowed with a suitable structure of derived category. Suppose we only look at those functors where delta is always zero. (By the way, if the group scheme is linearly reductive, this will always be the case.) Then we obtain simply a k-linear faithful exact tensor functor into the category of graded vector spaces. If we forget the grading, then we obtain a G-torsor over k as you say. Suppose that this torsor is trivial, e.g., if k is algebraically closed. Then what's left over is a grading on each representation of G, compatible with tensor product. Such a thing is given by a cocharacter, i.e., a group scheme homomorphism G_m ---> G. -Thus, in the very special case just discussed, there seems to be a little bit more structure, than for usual tensor functors. Before trying to answer your very interesting question in full, we should try to answer it in the case where G is the additive group over k. My feeling would be that delta still has to be zero...<|endoftext|> -TITLE: Image of a Galois representation -QUESTION [6 upvotes]: Notation: - -$E$ is a non-CM Elliptic curve over $\mathbb{Q}$. -$p$ is an ordinary prime. -$f$ - cuspidal eigenform of weight $k$ = 2 attached with $E$. -$\rho_f$ - the global 2-dimensional $p$-adic Galois representation attached with $f$. -$\rho_f$ : $G_S$ $\rightarrow$ $\mathrm{GL}_2({\mathbb{Z}}_p)$. -$G_S:= \mathrm{Gal}(\mathbb{Q}_S/{\mathbb{Q}})$, where $\mathbb{Q}_S$ - maximal unramified extension outside the set -$S=\{\text{ bad primes of } E \} \cup\{ p, \infty \}$. - -Assume that the residual representation $\overline{\rho}_f$ is $p$-split. -The prime $p$ is an ordinary prime. So the the image of $\rho_f$ restricted to the decomposition group $G_p:=\mathrm{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$ will be of the form $\rho_f$ $\mid$ $G_p$ $\sim$ -$\begin{pmatrix} a & * \\ 0 -& d \end{pmatrix}$. The residual representation $\overline{\rho}_f$ is $p$-split. So, $\overline{\rho}_f$ $\sim$ -$\begin{pmatrix} \omega \lambda_p^{-1}(\overline{a}_p) & 0 \\ 0 -& \lambda_p(\overline{a}_p) -\end{pmatrix}$, where $\lambda_p$ is an unramified character which sends $\mathrm{Frob}_p$ to $\overline{a}_p$, $\overline{a}_p \in \mathbb{F}_p$ is the mod $p$ reduction of the -$p$-th coefficent $a_p$ of $f$, and $\omega$ -is the $p$-adic cyclotomic character. - -Question: What is the image of the representation $\rho_f: G_p \rightarrow \mathrm{GL}_2(\mathbb{Z}/p^n \mathbb{Z})$, - where $(\mathbb{Z}/p^n\mathbb{Z})^2 \simeq (E[p^n])$, the $p^n$-torsion points of $E$, - for some fixed $n \geq 2$? Is it possible to compute it (or atleast it's order) by using MAGMA/SAGE/PARI? - -REPLY [2 votes]: One can say something about the image of $\rho_f|G_p$ by checking if $f$ has a companion form mod $p^n$. This can be explicitly done because one knows what the weight of this companion (if it exists) should be ($p^{n-1}(p-1)$, since $k=2$) and the conguences mod $p^n$ that the form's Fourier coefficients must satisfy vis a vis the $a_p$'s. One need only check that these congruences are satisfied up to the Sturm bound to conclude that the companion exists. If a companion mod $p^n$ exists then $\rho_f|G_p$ mod $p^n$ splits and $p^n$ won't divide the order of its image, else it will.<|endoftext|> -TITLE: Measures on general topological groups -QUESTION [8 upvotes]: I am interested in the group algebras of non-locally compact groups. What references can you advise? -This is a wide question, so I list more concretely what I would like to see: - -Here X can be even any Tychonoff space. Is there a description of the dual space of C0(X) as measures, or some close theorem? I have found a theorem in [H. König, Measure and integration] which describes measures as the dual of some space of semicontinuous functions. It is interesting, but I want the opposite: let the functions be "good", what measures will we get? -Are there any algebras, which consist more or less of measures, and can be associated to every topological group, or at lest to a Polish group? I've seen treatments of (left) uniformly continuous functions on groups and their dual space which is a Banach algebra. Is there a good survey of the topic? Are there other algebras? - -Thank you, and obviously I would greet more than one answer. -EDIT: Question 1 is answered below, and for question 2 I have found meanwhile a paper and a bunch of references in it: Anthony To-Ming Lau, J. Ludwig: Fourier–Stieltjes algebra of a topological group, Advances in Mathematics 229, Issue 3 (2012), 2000–2023. For my purposes, this is enogh at the moment, so I close the question. - -REPLY [5 votes]: Since a topological group has a natural uniform structure, the appropriate framework is, I think, the space of uniform measures. This was studied by the Czechoslovakian school. In functional analytical terms, it is the dual of the space of bounded, uniformly continuous functions on a uniform space---not with the uniform norm, of course (that would produce measures on the Samuel compactification). The rather more elaborate structure used is described in the article "Measures as functionals on uniformly continuous functions" by Pachl. A strong hint that this is the right concept is provided by the fact that the resulting space of measures has the natural universal property for bounded, uniformly continuous functions from the uniform space into complete, locally convex spaces. Pachl has also considered measures on topological groups in this context---see his papers "Uniform measures on topological groups" and "Uniform measures and convolutions on topological groups"---all readily available on internet.<|endoftext|> -TITLE: Combinatorial Morse functions and random permutations -QUESTION [28 upvotes]: This question has its origin in combinatorial topology. In the 90s R. Forman proposed a discrete counterpart of Morse theory. In his case, a Morse function on a triangulated space is a function that assigns a number to each face and satisfying certain conditions. -The theory has found beautiful applications, but it has one limitation: discrete Morse functions are hard to find, unlike the smooth case where smooth Morse functions are a dime a dozen. That made me think that maybe there aren't too many such discrete Morse functions. So naturally one can ask, really, how many Morse functions are out there on a given triangulated space. -The present question deals with the simplest triangulated space, namely a line segment divided into $n$-subintervals. The problem of counting the combinatorial Morse functions on this triangulated space reduces to the following purely combinatorial problem. -Consider the group $S_{2n+1}$ of permutations of the set -$$ V_n:=\lbrace 0,1,\dotsc,2n\rbrace. $$ -A point $i\in V_n$ is called an interior point if $i\neq 0,2n$. An interior point $i\in V_n$ is a local minimum of a permutation $\phi\in S_{2n+1}$ if -$$ \phi(i-1)> \phi(i) <\phi(i+1). $$ -A local maximum is defined in a similar fashion. Here is now the question. - -Denote by $p_n$ the probability that - a random permutation of $S_{2n+1}$ has - the property that all its interior - local minima (if any) are even and all its - interior local maxima (if any) are odd. Is it - true (as I believe) that - $p_n\to 0$ as $n\to\infty$? Can - one be more precise about the - behavior of $p_n$ as $n\to\infty$? - -For example, when $n=1$ the permutations of $\lbrace0,1,2\rbrace$ satisfying the above constraints are -$$ (0,1,2), (2,1,0), (0,2,1), (1,2,0). $$ -Hence $p_1=\frac{2}{3}$. -Thanks. - -REPLY [14 votes]: This is a sketch of a solution by my student Yan Zhuang and me. A paper with all the details (and further results) can be found in Counting permutations by alternating descents, Electronic J. Combin. 21 (4) (2014), Paper #P4.23. -We find the exact exponential generating function for counting these permutations and derive the asymptotics from it. -First, there is no reason from the enumerative point of view to consider only permutations of odd length. So let $u_n$ be the number of permutations of $\{1,2,\dots, n\}$ in which every peak is even and every valley is odd, where a peak of $\pi\in S_n$ is an $i$, with $1\pi(i+1)$, and valleys are defined similarly. The first few values of $u_n$ are $u_0=1$, $u_1=1$, $u_2=2$,$u_3=4$, $u_4=13$, $u_5=50$, $u_6=229$. -Let -\begin{equation} U(x) = \sum_{n=0}^\infty u_n \frac{x^n}{n!}.\end{equation} -The nicest formula for $U(x)$ is -\begin{equation} U(x) = \left( 1-E_1x +E_3 \frac{x^3}{3!}-E_4\frac{x^4}{4!}+E_6\frac{x^6}{6!}-E_7\frac{x^7}{7!} +\cdots\right)^{-1}, -\tag{1}\end{equation} -where -$$\sum_{n=0}^\infty E_n \frac{x^n}{n!} = \sec x + \tan x.$$ -A formula equivalent to (1) which is more useful for asymptotics is -\begin{equation}U(x) = \frac{3\sin\frac x 2 + 3\cosh \frac {\sqrt3}{2} x} -{3\cos \frac x 2 -\sqrt 3\sinh \frac{\sqrt3}{2} x}. -\tag{2}\end{equation} -Since $U(x)$ is meromorphic with a single simple pole on its circle of convergence, at $x=\alpha:=1.299828316\cdots$, we find by standard techniques that $u_n/n!$ is asymptotic to $2\beta^{n+1}$, where $\beta:=\alpha^{-1}=.7693323708\cdots$. -Formula (2) can be proved by finding a recurrence for $u_n$, converting it to a differential equation, and solving it. (We need to use an auxiliary sequence, and consider even and odd $n$ separately, so we actually get a system of four differential equations.) -We can give a more conceptual proof of (1). We first note that the similar-looking exponential generating function -\begin{equation}\left( 1-x + \frac{x^3}{3!}-\frac{x^4}{4!}+\frac{x^6}{6!}-\frac{x^7}{7!} +\cdots\right)^{-1}\tag{3}\end{equation} -(OEIS sequence A049774) -counts permutations with no increasing run of length greater than 2. -We may define an alternating run of a permutation $\pi$ to be a maximal subsequence of the form $\pi(2i)<\pi(2i+1)>\pi(2i+2)<\cdots\mathrel{<\atop >}\pi(j)$ or $\pi(2i+1)>\pi(2i+2)<\pi(2i+3)>\cdots \mathrel{<\atop >}\pi(j)$. Then the permutations counted by $u_n$ are permutations with no alternating run of length greater than 2, and (1) can be proved in a way that is analogous to the proof of (3). -Formulas involving the numbers $E_n$ and “alternating descents”, using the same basic idea, have been proved by Denis Chebikin, -Variations on descents and inversions in permutations, -Electronic J. Combin. 15 (2008), Research Paper R132.<|endoftext|> -TITLE: Is a bialgebra with all group-like elements invertible a Hopf algebra? -QUESTION [14 upvotes]: We know that in a Hopf algebra all group-like elements are invertible. Is the converse also true? Here is the precise formulation of my question : -Let $B$ be a bialgebra and $GLE$ = { $g \in B ~|~ g \neq 0, \Delta (g) = g \otimes g$ } the set of group-like elements. We know that this set is a monoid and that if B has an antipode, namely if $B$ is a Hopf algebra, then $GLE$ is a group. -Now, suppose that in the bialgebra $B$, every group-like element is invertible, does $B$ then have an antipode? (it would be easy to define an antipode $S: B \rightarrow B$ on $GLE$ by $S(g) = g^{-1}$, but what about the other elements?) -If there is a known counterexample, then what would be the extra-condition required to assure the existence of the antipode? - -REPLY [5 votes]: Sufficient conditions are known for the existence of an antipode on a bialgebra $A$ over a Commutative ring $R$ that do not assume that $A$ is either commutative or cocommutative. Assume that $A$ -is graded, $A_i = 0$ for $i<0$, and $A_0$ is $R$-free. Let $\epsilon\colon A\longrightarrow R$ be the counit (or augmentation). Assume that $\epsilon^{-1}(1)$ is a group under the multiplication of $A$. For each grouplike element $g$, let $A_g = Rg \oplus \bar{A_g}$, where the set $\bar{A_g}$ of positive -degree elements of $A_g$ is the set of all elements $x\in A$ such that -$$\psi(x) = x\otimes g + \sum x'\otimes x'' + g\otimes x$$ -where the $x'$ and $x''$ are both of positive degree. Assume that $A$ is the direct sum over $g\in GLE$ of the $A_g$. Then $A$ has an antipode $\chi$. If $A$ is commutative or cocommutative, -then $\chi^2$ is the identity, but not in general otherwise. The conditions are motivated by thinking -about the homology of an $H$-space $X$ with coefficients in $R$.<|endoftext|> -TITLE: The relation of nilpotent orbits and simple singularities, for orbits smaller than subregular ones -QUESTION [7 upvotes]: Pick a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$. There is a partial ordering among nilpotent orbits defined by $O\geq O'$ iff $\bar O\supset O'$. -The unique maximal element under this partial order is the regular nilpotent orbit, and the unique sub-maximal element is the subregular nilpotent orbit. Denote the former by $O$ and the latter by $O'$. Then $\dim O-\dim O'=2$; Brieskorn and Slodowy showed that the transversal slice to $O'$ inside $O$ is the simple singularity of type $\mathfrak{g}$. -Now, there are many pairs of nilpotent orbits $O$ and $O'$ such that $O\geq O'$ and $\dim O-\dim O'=2$. The transversal slice to $O'$ inside $O$ should be a simple simgularity of some type. -How can I determine the type, given $\mathfrak{g}$, $O$ and $O'$? - -REPLY [6 votes]: I just saw this question. See -Baohua Fu, Daniel Juteau, Paul Levy, Eric Sommers. -"Generic singularities of nilpotent orbit closures" - arXiv:1502.05770 -where we do the same job as Kraft and Procesi, but for exceptional types. We posted the preprint only this year, but we already had most of the results when the question was asked (and gave talks about this). -So most of the time, it is indeed a simple singularity, but sometimes there are several branches, and there is also a nonnormal singularity with smooth normalization which occurs in all exceptional types. -Let me give some information about the methods. Baohua Fu had studied in previous work Q-factorial terminalizations of nilpotent orbit closures. It turns out that they are all "generalized Springer maps", or just the normalization. Restricting to a slice, they provide a resolution of the surface singularity (a Q-factorial terminal singularity is smooth in codimension 2), and actually it is the minimal resolution. If the terminalization corresponds to inducing some nilpotent orbit from a smaller Lie algebra ("generalized Springer map"), then one can understand the exceptional fiber using a formula of Borho and MacPherson. This is sufficient in most cases, but for the singularities between subregular and subsubregular, we also had to use orbital varieties... See our preprint for more details.<|endoftext|> -TITLE: Fibre cardinality of an unramified morphism -QUESTION [7 upvotes]: Let $\varphi: X \to Y$ be a finite, dominant, unramified morphism of varieties over an algebraically closed field. If necessary, we can assume $X$ and $Y$ to be nonsingular. I am trying to prove that -$$\mathrm{deg}(\varphi):=[K(Y):K(X)]=|\varphi^{-1}(P)|$$ -for every point $P\in Y$. The statement is very easy to prove for curves. However, I am completely stuck trying to prove it for higher dimensions. I cannot find this statement or a similar one in literature, it would also be a great help if someone could point me there. - -REPLY [4 votes]: After I wrote the comments above, I found the following reference : -Formula (12.6.2), p. 329 in Görtz-Wedhorn, Algebraic Geometry I, Viehweg & Teubner Verlag -for (a generalisation of) the equality you are looking for, when $\phi$ is assumed flat (which is true if you assume that $X$ and $Y$ are non-singular, as pointed out in the comments of K. M. Pera and S. Kovacs).<|endoftext|> -TITLE: Primitive elements of a tensor product of bialgebras -QUESTION [12 upvotes]: Given a field $k$ of characteristic $0$. For every $k$-bialgebra $A$, let $\mathrm{Prim} A$ denote the $k$-vector subspace of $A$ consisting of all primitive elements of $A$. -What conditions can we put on two $k$-bialgebras $A$ and $B$ to ensure that $\mathrm{Prim}\left(A\otimes B\right) = k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right)\otimes k$ ? -I haven't given this much thought, but I am not good at constructing counterexamples and it seems pointless to try proving anything here before having an "upper bound" on how far we can go. The only results I know about is that $k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right) \otimes k \subseteq \mathrm{Prim}\left(A\otimes B\right)$ always holds (for trivial reasons), and that if $A$ and $B$ are two connected graded cocommutative bialgebras, then $\mathrm{Prim}\left(A\otimes B\right) = k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right)\otimes k$ (as a consequence of Cartier-Milnor-Moore and Poincaré-Birkhoff-Witt). -It sounds rather natural to assume $A$ and $B$ to be cocommutative (after all, $\mathrm{Prim} A$ is always $=\mathrm{Prim}\left(A^c\right)$, where $A^c$ the greatest cocommutative sub-bialgebra of $A$), but I am not sure whether we can WLOG assume this to be so (maybe $\left(A\otimes B\right)^c$ is greater than $A^c\otimes B^c$ ?). - -REPLY [3 votes]: There is a short proof in Proposition 2.12 of my paper http://arxiv.org/pdf/1502.02150v1 which works over an arbitrary base ring.<|endoftext|> -TITLE: What is the Generating Function for Skew Young Diagrams? -QUESTION [21 upvotes]: The Problem -This strikes me as a very natural problem which should have been asked (and solved?) already. -For each positive integer k, find a nice expression for the following generating function in the variable x: -$$ -\sum_{\lambda/\mu} x^{|\lambda|}. -$$ -Here \lambda ranges over all partitions and \mu over those partitions contained in \lambda for which the skew Young diagram \lambda/\mu has k nodes i.e. each partition of n is weighted by the number of partitions of n-k it contains. -Examples: k=1, the function is $\frac{x}{(1-x)}P(x) $, where $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function. So in this case I'm just enumerating partitions by the number of removable nodes. The formula is equivalent to the well-known fact that every partition has one more addable than removable node. -I've computed the cases k=2,3,4 also (k=4 was painful - I broke it into 14 possible types of skew-diagrams). For k=2 the generating function is $\frac{ x^2(2-x)}{(1-x)(1-x^2)}P(x).$ -It seems plausible that there is a polynomial F_k(x) of degree at most k(k-1)/2 (with leading coefficient \pm 1) so that the power series is -$$ -\frac{ x^k F_k(x)}{(1-x)(1-x^2)..(1-x^k)}P(x). -$$ -If $F_k(x)$ exists, it's easy to see that it must have lowest terms -$p_k+p_{k+1}x+2(p_{k+2}-1)x^2+...$, where p_n=number of partitions of n, after which the terms depend on congruences for k. This suggests that its complicated. Perhaps there is no nice expression for $F_k(x)$. Even knowing whether $F_k(x)$ exists is of interest to me. Maybe there is a neater way of expressing the entire generating function? -Motivation -The coefficient of $x^n$ in the generating function is the dimension of the centre of a certain subalgebra of the complex group algebra of the symmetric group of degree n. This is ${\mathbb C}S_n^{S_{n-k}}$, the centralizer of the subgroup $S_{n-k}$ in ${\mathbb C}S_n$. It is easy to see that this has as ${\mathbb C}$-basis the $S_{n-k}$-orbit sums in $S_n$. The centre is indexed by pairs $(\chi,\phi)$, where $\chi$ is an irreducible character of $S_n$, and $\phi$ is an irreducible character of $S_{n-k}$ occuring in the restriction of $\chi$. The formulation above is then an easy consequence of the parametrization of irreducible characters of $S_n$, and the classic branching rule. -Literature -Yoshiaki Ueno, On the Generating Functions of the Young Lattice, J. Algebra 116 (1988) 261--270. -This gives a generating polynomial for the partitions contained in a given partition \lambda in terms of a determinant involving Gaussian coefficients. It's a beautiful result, but it did not give me any insight into my problem. - -REPLY [21 votes]: This problem is a special case of Exercise 3.150(a) of Enumerative -Combinatorics, vol. 1 (second ed.). The polynomial $A_{\lbrace -k\rbrace}(x)$ of this exercise is the $F_k(x)$ of the present -question. The solution to this exercise gives a recipe for computing -$F_k(x)$ which can probably be used to compute quite a few values and -to prove some properties such as its degree. In the solution to part -(b) there is given the formula $F_3(x)=3+2x-x^2-x^3$.<|endoftext|> -TITLE: chromatic number of the hyperbolic plane -QUESTION [33 upvotes]: A notorious problem in combinatorics is the following: -If we color $\mathbb{R}^2$ so that no pair of points at unit distance get the same color, what is the fewest number of colors required? -This number is sometimes called "the chromatic number of the plane" $\chi$ (since it is really a graph coloring problem on an infinite graph), and it is easily see that $ 4 \le \chi \le 7$ but these bounds have stood for quite a while. -It is natural to ask about coloring other metric spaces, and many people have. In particular, there are bounds known for the chromatic number of $\mathbb{R}^d$ for $d \ge 3$ (and some asymptotics as $d \to \infty$). Also there has been some work on coloring the two-dimensional sphere of radius $r$. (A nice reference for this kind of problem is Soifer's "The mathematical coloring book.") -My question is whether anyone has looked at the chromatic number of the hyperbolic plane. As with the sphere, there is a free parameter --- one could either take fixed curvature $-1$ and let the distance vary, or fix unit distance and let the constant negative curvature vary. -We might not expect to be able to solve this in general, since determining the chromatic number of the plane seems difficult, and now we have an infinite family of such problems. But we should be able to put bounds, and I am wondering what is known. -In particular, the following two questions come to mind. I would appreciate insights or pointers to references if these things have been previously studied. - -(1) Is there a $5$-chromatic unit distance graph in the hyperbolic plane (for some constant negative curvature)? -(2) Is there an absolute upper bound - on the chromatic number of the hyperbolic - plane, that holds for all constant - negative curvatures? - -One might guess that the chromatic number of the hyperbolic plane is increasing as the constant negative curvature decreases, but if so does it grow without bound? - -REPLY [6 votes]: A new paper on the topic has appeared, and while it does not answer your question I find the result interesting enough to be mentioned in a separate answer. -DeCorte and Golubev prove there that for $r$ large enough ($r\ge 12$ suffices), the distance-$r$ graph of the hyperbolic plane with curvature $\kappa=-1$ has measurable chromatic number at least $6$ (here "measurable" means that the maps sending a point to its color is Lebesgue Measurable). It is known for long that the measurable chromatic point of the plane is at least $5$ (Falconer 1981), so the interesting point here is that the lower bound gets better for large distances (i.e. for very negative curvature, if one prefers to fix $r=1$ and let $\kappa$ run over $(-\infty,0)$). -Note that under the Solovay axioms, the measurable hypothesis can be dropped since all sets are Lebesgue measurable (but then we don't have the full axiom of choice, and thus no DeBruin-Erdös theorem: we cannot deduce that under the Solovay axioms there is a finite unit-distance graph whose chromatic number is at least $6$).<|endoftext|> -TITLE: convex polytopes with many faces and edges but few cells and vertices -QUESTION [6 upvotes]: For a convex polytope $P$ in $\mathbb R^4$, denote by $N_0,N_1,N_2,N_3$ respectively the number of vertices, edges, faces, cells. By Euler's formula, we know $N_0+N_2=N_1+N_3$, which means there is a sort of equilibrum among the $N_i$. But I wonder if there exist upper and/or lower bounds for $f(P):=\frac{N_1+N_2}{N_0+N_3}$ where $P\subset\mathbb R^4$ is any convex polytope. -E.g. for the regular one called 24-cell, we have $f(P)=4$. - -REPLY [3 votes]: Each cell is a convex polytope in 3-space and so has at least 4 faces, whereas each face should have only 2 cells, or $N_3\geq 2N_4$. The same argument for the dual polytope gives $N_1\geq 2N_0$, which gives a lower bound of $f\geq 2$, attained by the $5$-simplex. -This is analogous to the polyhedral bound of $2E \geq 3F,3V$.<|endoftext|> -TITLE: Volume inequality between projections of a convex symmetric set in $\mathbb R^3$ -QUESTION [6 upvotes]: Let $K$ be a centrally symmetric convex body in $\mathbb R^3$ with volume ${\rm vol}(K)=1$. For any subset $F \subset \lbrace1,2,3\rbrace$, let $K_F$ be the projection of $K$ in $\mathbb R^F$. - -Question: What is the best constant $C$, such that -$${\rm vol}(K_{\lbrace 1 \rbrace}) \leq C \cdot {\rm vol}(K_{\lbrace 1,2 \rbrace}) \cdot {\rm vol}(K_{\lbrace 1,3 \rbrace}).$$ - -Here, the volume of $K_F$ is computed in $\mathbb R^F$. With some work one can prove that $C=2$ is good enough. This is elementary and follows for example from two applications of Lemma 3.1 in -J. Bourgain and V.D. Milman, New volume ratio properties for convex symmetric bodies in $\mathbb R^n$, Invent. Math. 1987 vol. 88 (2) pp. 319-340. -My guess is that maybe $C=1$ works, but I am not sure. - -REPLY [5 votes]: Picking up from anton's answer, notice that by Cauchy-Schwartz: -$\int_0^1 u \cdot v \cdot dx \leq \sqrt{ \int_0^1 u^2 \cdot dx \cdot \int_0^1 v^2 \cdot dx}$ -So a bound in the case $u=v$ is sufficient. We wish to show: -$\int_0^1 u^2 dx \leq C \left( \int_0^1 u dx\right)^2$ -Fix a triangle with the same average value of $u$, and that attains its maximum over the same $x$-coordinate. If $u$ is equal to that triangle, a trivial calculation gives a value of $C$ of $4/3$. We want to show that it not being a triangle makes $\int u^2$ no bigger. In fact, letting $t$ be the triangle, we have $\int u^2 \leq \int ut\leq \int t^2$. -This is because $(t-u)$ is positive closer to the maximum value, and negative further, so $u$ and $t$ are both higher where $(t-u)$ is positive than where it is negative, so $\int u(t-u)\geq 0$, $\int t(t-u)\geq 0$. -Therefore $C=4/3$ is a bound, obtained by setting $u$ and $v$ to triangles. One can check that this does indeed come from a polytope in $\mathbb R^3$ - an octahedron, for instance. -Edit: This proof isn't quite complete, because the area on one side of the triangle is not necessarily the area on one side of the function. We can remedy this by choosing an unbalanced "triangle" with a discontinuity where the maximum should be, which still satisfies $C=4/3$.<|endoftext|> -TITLE: What is the advantage of the approach of valuations to the Riemann-Roch Theorem for curves (a la Chevalley)? AKA theory of algebraic functions in one variable -QUESTION [9 upvotes]: Hello, -Some books and courses take the approach to Riemann-Roch for curves considering only the algebraic viewpoint of algebraic extensions of a field k(t) without going into any algebraic geometry (no varieties, no topology, no dimension, no sheafs etc') divisors are defined using equivalence classes of valuations of the field, differentials have some 'wierd' definition as well. and riemann roch is proved only in that language, with no regard to concepts such as bundles. -This can be seen in books such as Chevalley, Introduction to the Theory of Algebraic Functions of One Variable. -Or even in Neukirch Algebraic number theory. -Since I know of treatments of Riemann-Roch in the general setting for bundles over algebraic curves in books such as Kempf, Algebraic Varieties, or even in Vakil's notes: -http://math.stanford.edu/~vakil/725/bagsrr.pdf -My question is, what is the advantage of the approach using only algebra and the language of valuations instead of using concepts from algebraic geometry (such as sheaf) -Thanks - -REPLY [6 votes]: Hello Blade. -Reading your question one is tempted to dive into the -interesting and diverse history of the various approaches -to the Riemann Roch Theorem. -However, an answer to your question from the point of view -of a mathematician of today in my opinion depends on the -person who is asking. -The amount of concepts and theory you have to learn before -being able to fully understand the proof of Riemann Roch -is smaller for the valuation theoretic approach compared to -the one taught in Algebraic Geometry. -Of course this is not an advantage for somebody who is -interested in geometry and wants to focus future activity -in that area anyway. -On the other hand if one is tending more towards Algebraic -Number Theory, then it is a demand to learn valuation -theoretic concepts, even for non-discrete valuations. -Finally if one wants to study Arithmetic Geometry or -Non-archimedian Analysis geometric and valuation theoretic -concepts appear almost on an equal level. - -REPLY [5 votes]: You didn't mention Weil, Basic Number Theory, where the case of a finite field of constants is handled, really only using Pontryagin duality. There is an elegant theory of John Tate that seems somewhat magical. I think the right question would be, "what is the trade-off?" rather than "what is the advantage?". Technically simpler methods require fewer prerequisites; and life is finite. -There certainly is some advantage, if you are doing number theory, to have a version of Riemann-Roch that does not require you to work over an algebraically closed field. Other than that, I don't really know. Riemann-Roch has been understood to work for higher dimensions, in some form, for about a century now, I think. If you are interested in the generalisations, there is a good case for starting from sheaves because the theory is slick. (Probably also true for R-R for vector bundles on curves, which Weil himself handled with bare hands c.1938, but makes more sense in Hirzebruch's book). These days, I suppose, valuation theory looks like a technical tool for singularity theory. -For some people, we won't really be comfortable saying we understand R-R until the Riemann Hypothesis is sorted out. I.e. the function field-number field analogy is still potent, and the view from arithmetic is that the circle of ideas is not yet complete. For that reason it is hard to rule out any approach, in terms of heuristic value anyway.<|endoftext|> -TITLE: Asymptotic number of certain functions without fixed points -QUESTION [16 upvotes]: Terry Tao's question Non-enumerative proof that there are many derangements? suggests -the following. What classes $C_n$ of functions $f\colon \lbrace -1,2,\dots,n\rbrace \to \lbrace 1,2,\dots,n\rbrace$ -have the property that if $a(n)$ is the number of functions in $C_n$ -and $b(n)$ is the number without fixed points, then -$\lim_{n\to\infty} a(n)/b(n) = e$? Examples include all functions, -permutations, and alternating permutations. (I don't know a simple -proof for alternating permutations.) Rather than lots of examples, a -general theorem that includes all (or most) of these examples would be -more interesting. - -REPLY [4 votes]: Here is a generalization of the derangements case. -Take a Latin rectangle with $n$ columns and $k\le n^{1-\epsilon}$ rows, such that, for each $i$, the value $i$ does not appear in the $i$-th column. Now consider the set of all permutations that extend this rectangle to $k+1$ rows; i.e., the permutations which are derangements of each of the given $k$ rows. Then this set of permutations has the fraction $e^{-1}(1+o(1))$ of derangements in it. -This follows from Theorem 6.2 of my paper on Latin rectangles with Chris Godsil.<|endoftext|> -TITLE: Help me find good math questions for my students -QUESTION [7 upvotes]: I am a teacher at 西铁一中。 I teach mathematics in English for students going abroad. -Now this is my problem, there are few mathematics books written in English that are at the level of high school, most are geared towards graduate school. -The book for the school is very dry and basic. I would like to be able to present better problems for my classes. I need help finding good problems, ones with some historical context or from real life events. Example: using a closed Leontief input-output model for linear algebra. -Or something along this line: -The quality of oxygen that can dissolve in water depends on the temperature of the water. (So thermal pollution influences the oxygen content of water.) The graph shows how oxygen solubility S varies as a function of the water temperature T. -(a) What is the meaning of the derivatives S'(T)? What are its units? -(b) Estimate the value of S'(16) and interpret it. -with graph - -From Calculus (6th Edition) by James Stewart. - -My other problem is that I've got a group of 6 students that are übersmart. I gave them the AMC 10 & 12 problems, they solve them in class, then I gave them the USAMO problems, and they solve 4 problems in an hour (I gave them both days). Should I just give them the Putnam exams? Now, English textbooks cost a lot and I have to buy them out of my own pocket. Could you help me find online resources which are at the level of a high school student that is just learning calculus and which help him/her understand the more advance mathematical topics? Most I have found are just beyond their mathematical understanding. -The other 24 students are of lower ability than those 6 students. -In summary: -1) Where to find more realistic and historical problems for calculus and prob/stats classes? To help engage my students and improve their English. The graphs and data for the problems, don't worry I can solve it for the answers. Problems that bring in the other subjects : history, economics, etc. -2) What to do about the übersmart students? How to structure the class so the übersmart students don't get bored, but I don't leave the rest of the students behind? -3) Are there any books that are really graduate analysis, number theory, etc. however, are written for high school students? -4) What's your favorite word problem? - -REPLY [8 votes]: At the risk of self-promotion (is this an MO no-no? if so, I apologize), my book A Friendly Introduction to Number Theory is quite well suited for bright high school students, while at the same time giving them a taste of real mathematics. I know that the book is used in this country at some high schools, because I've had teachers write to me and tell me that they're using it. -Even if you or your students don't want to purchase the book (which is fine), the first six chapters are available for free at -http://www.math.brown.edu/~jhs/frint.html -So at least your students can use these 6 chapters to get started and to see if they enjoy studying number theory.<|endoftext|> -TITLE: first chern class and spin structures -QUESTION [5 upvotes]: Let M be a compact complex manifold. Then is it true that if the first Chern class of M is even, then M admits a spin structure? - -REPLY [12 votes]: Yes. An oriented real vector bundle is spin if and only if its second Stiefel-Whitney class vanishes. If $E_\mathbb{C}$ is a complex vector bundle and $E_\mathbb{R}$ is the underlying real bundle then the second Stiefel-Whitney class is given by $w_2(E_\mathbb{R}) = c_1(E_\mathbb{C})$ mod 2. The details appear somewhere in chapter 2 of Spin Geometry by Lawson and Michelsohn.<|endoftext|> -TITLE: Exactness of completed tensor product of nuclear spaces -QUESTION [8 upvotes]: Let $0 \to V \to W \to L \to 0$ be a strict short exact sequence -of (complete) nuclear spaces, i.e. it is a short exact sequence of -(complete) nuclear spaces, all the maps are continuous, the map $V -\to W$ is a closed embeding, the topology on $V$ is induced from -$W$ and the map $W \to L $ is open. Let $U$ be a (complete) nuclear -space. -Is it true that the sequence obtained by completed tensor product with $U$ -(i.e. $0 \to V \hat{\otimes} U \to W \hat{\otimes} U \to L -\hat{\otimes} U \to 0$) -is also strict short exact sequence? -We know that this is true if all the spaces are Frechet or if all -the spaces are dual Frechet, but is this true in general? - -REPLY [6 votes]: Unfortunately, this is not true in general. -Let $E$, $F$, $G$ and $H$ be complete nuclear lcs such that -$$ 0 \longrightarrow E \stackrel{S}{\longrightarrow} F \stackrel{T}{\longrightarrow} G \longrightarrow 0 $$ -is a strict short exact sequence. Note that the sequence -$$ 0 \longrightarrow E \mathbin{\hat\otimes} H \stackrel{S \mathbin{\hat\otimes} \text{id}_H}{\longrightarrow} F \mathbin{\hat\otimes} H \stackrel{T \mathbin{\hat\otimes} \text{id}_H}{\longrightarrow} G \mathbin{\hat\otimes} H \longrightarrow 0 $$ -is a strict short exact sequence if and only if $T \mathbin{\hat\otimes} \text{id}_H : F \mathbin{\hat\otimes} H \to G \mathbin{\hat\otimes} H$ is surjective; everything else is automatic by the mapping properties of the injective resp. projective topology. -Let $w$ and $w*$ denote the weak and weak-$*$ topologies, and let $T' : G_{w*}' \to F_{w*}'$ denote the adjoint of $T$. -It follows from [Sch99, §IV.9.4] that $F \mathbin{\hat\otimes} H \cong \mathfrak{B}_e(F_{w*} \times H_{w*})$. The latter is linearly isomorphic with $\mathfrak{L}(F_{w*}' , H_w)$, and the following diagram commutes: -$$\require{AMScd} -\begin{CD} -F \mathbin{\hat\otimes} H @>T \mathbin{\hat\otimes} \text{id}_H >> G \mathbin{\hat\otimes} H \\ -@VV V @VV V \\ -\mathfrak{L}(F_{w*}' , H_w) @> R \mapsto RT' >> \mathfrak{L}(G_{w*}' , H_w) \\ -\end{CD}$$ -Since $G \cong F/E$, we have $G' = E^\perp$, and the weak-$*$ topology on $G'$ coincides with the relative $F_{w*}'$ topology of $E^\perp$. Thus, we see that $T \mathbin{\hat\otimes} \text{id}_H$ is surjective if and only if every weak-$*$-to-weak continuous operator $E^\perp \to H$ can be extended to $F'$. This is not always the case: - -Counterexample. Let $E$ be a closed, non-complemented subspace of a nuclear Fréchet space $F$.¹ Furthermore, let $G := F/E$, and let $H := G_\beta'$ be the strong dual of $G$. Then $E$, $F$ and $G$ are nuclear Fréchet spaces (so in particular reflexive), and $H$ is a complete, nuclear (DF)-space. -In a Fréchet space, all weakly complemented subspaces are complemented, so $E$ is not weakly complemented in $F$. By duality, $E^\perp = G'$ is not weak-$*$ complemented in $F'$. Since $G$ is reflexive, the weak and weak-$*$ topologies on $H = G_\beta'$ coincide, so we have $\text{id}_H \in \mathfrak{L}(G_{w*}' , H_w)$. Since $G'$ is not weak-$*$ complemented, the map $\text{id}_H \in \mathfrak{L}(G_{w*}' , H_w)$ has no extension in $\mathfrak{L}(F_{w*}' , H_w)$. - -Convesely, if $E$ is (weakly) complemented, then $T \mathbin{\hat\otimes} \text{id}_H$ is always surjective. -¹: Concrete examples of this kind are given in this answer, or in [MV97, Exercise 31.4], or in [DM76], among others. -References. -[DM76] Plamen Djakov, Boris Mitiagin, Modified construction of nuclear Fréchet spaces without basis, Journal of Functional Analysis, vol. 23 (1976), issue 4, pp. 415–433. DOI: 10.1016/0022-1236(76)90066-5. -[MV97] Reinhold Meise, Dietmar Vogt, Introduction to Functional Analysis (1997), Oxford Graduate Texts in Mathematics, Clarendon Press, Oxford. -[Pie72] Albrecht Pietsch, Nuclear Locally Convex Spaces (1972), Ergebnisse der Mathematik und ihrer Grenzgebiete, Springer, Berlin. -[Sch99] H.H. Schaefer, M.P. Wolff (translator), Topological Vector Spaces, Second Edition (1999), Springer Graduate Texts in Mathematics, Springer, New York.<|endoftext|> -TITLE: Löwner-John Ellipsoid: incribed and circumscribed -QUESTION [6 upvotes]: I have two questions about the -Löwner-John ellipsoid, one just terminology, the other -more substantive. -Let $K$ be a convex body in $\mathbb{R}^d$. - -Q1. - Is "the - Löwner-John ellipsoid" - the unique ellipsoid of maximal volume contained in $K$, - or the unique ellipsoid of minimal volume containing $K$? - I have seen it used in both senses. -Q2. - Let $E^+$ be the containing/circumscribing ellipsoid - and $E^-$ the contained/inscribed ellipsoid of min and max - volume respectively, for the same $K$. - (a) Are there bounds known on - $\mathrm{vol}(E^+)/\mathrm{vol}(E^-)$? - (b) Any other interesting relationships known between $E^+$ and $E^-$, e.g., alignment of axes? - -Thanks for pointers! - -REPLY [10 votes]: Q1: Most often it is the maximal volume ellipsoid contained in $K$. -Q2: (a) John's theorem implies that $E^+ \subseteq d E^-$ in general, and $E^+ \subseteq \sqrt{d} E^-$ if $K$ is centrally symmetric, and both these inclusions are sharp (consider a simplex or a cube, respectively), giving of course $d^d$ and $d^{d/2}$ for the best possible upper bounds on the ratios of volumes in the nonsymmetric and symmetric cases respectively. -(b) Not that I'm aware of, but there certainly may be some. -The friendliest reference I can think of is K. Ball, "An elementary introduction to modern convex geometry".<|endoftext|> -TITLE: Schwartz Kernel theorem for tempred functions -QUESTION [5 upvotes]: Let $T(R)$ denote the space of tempered functions on the line, -i.e. the smooth functions that give Schwartz function after a -multiplication by any Schwartz function, equipped with the natural -nuclear topology. e.g. the topology induced from the strong -(convergence on bounded sets) topology on the endomorphism space of -the space of Schwartz functions. -Is it true that tempered functions on the plane is the completed tensor square -of tempered functions on the line, i.e. $T(R) \hat{\otimes} T(R) -=T(R^2)$? - -REPLY [4 votes]: This is proved in 4.1 of: -Michel Dubois-Violette, Andreas Kriegl, Yoshiaki Maeda, Peter W. Michor: Smooth *-algebras. Progress of Theoretical Physics Supplement 144 (2001), 54-78. arXiv:math.QA/0106150. -pdf -According to L. Schwartz, there are two kind of tempered spaces, $\mathcal O_M$, and $\mathcal O_C$. See the paper, where your question is proved for both of them.<|endoftext|> -TITLE: compression of a Turing machine run sequence -QUESTION [7 upvotes]: consider a Turing machine with a set of states $s_n$ and alphabet symbols $a_n$. now consider a "run sequence" generated from a starting input in the following sense. the run sequence is defined as the sequence of state-symbol pairs that ensue in the computation. call the $i$th ensuing state $s'_i$ and the $i$th ensuing symbol $a'_i$. -then the run sequence is the sequence $[(s'_1,a'_1),(s_2,a'_2),(s'_3,a'_3),...]$. or equivalently of composite symbol-pairs $[s'_1a'_1,s'_2a'_2,s'_3a'_3,...]$. if the computation terminates after $z$ steps then $i\leq z$. (note the run sequence is also defined for sequences that do not terminate, in that case its an infinitely long sequence but a (long?) finite initial subsequence alone can be considered.) -now consider large run sequences for large inputs and a large $z$. ie consider the question for recursive machines only. - -question: what can be said about the compressibility of the Turing machine run sequence? - -"compressible" means: is there an algorithm that can be used in the sense of data compression. -my current suspicion (without proof so far) is that if the machine is recursive and runs in some time $O(f(m))$ or (space $O(h(m))$ resp.) where $f(m)$ is some "std" function of input length $m$ say logarithmic, polynomial, exponential etc. then the run sequence is indeed compressible in some sense. -does anyone know a similar formulation of this problem that can be found in the literature or is studied somewhere? thx! - -motivation: a possible useful formulation for understanding separations between complexity classes. think this problem might help bridge relationships between time and space complexity separations, tradeoffs etc. - -REPLY [4 votes]: A run sequence $r = [s'_1a'_1,s'_2a'_2,s'_3a'_3,...]$, of a Turing machine $M$ on input $x$ is a highly compressible string: -$$K(r) = K(\langle M, x, z \rangle) + c \leq |\langle M \rangle| + | x | + \log z + c' $$ -where $z$ is the number of steps of the run sequence.<|endoftext|> -TITLE: Random Unfoldings of the Cube -QUESTION [9 upvotes]: Motivated by unfoldings of the dodecahedron in How To Fold It -- -How many (labeled or unlabeled) unfoldings of the 1 x 1 x n stack of unit cubes are there? - -JORourke (4Nov16): John's original image is now lost: -http://www.freeimagehosting.net/newuploads/f3dkg.gif -JDM (11/15/16): The original image is certainly lost from 4 years ago. Here is a quick example with a 1 × 1 × 1 cube - - -The dissections of the cube are related to each other by "hinge" moves (which I've tried to draw). The graph of unfolding of the cube seems connected by these hinge moves. - -It's equivalent to consider cutting the cube along the edges. Then we're looking for uniform spanning trees on the surface of the cube. -JDM (11/15/16) I've added some figures of the shapes I was trying to unfold. - -For my 4 (unlabeled) (incomplete list of) shapes, how often do they appear among (uniformly random) unfoldings of the labeled cube?? JDM (11/15/16) This part of the question is also ambiguous. I had also been concerned in the 1 × 1 × n case the diagrams are not planar. David Speyer's answer looks possibly correct. The graph of unfoldings connected by hinge moves looks non-trivial even in the 1 × 1 × 1 case.) - -REPLY [10 votes]: Let $a_n$ be defined by the recursion -$$a_{n+2} + 4 a_{n+1} + a_n =0,\ a_1=-4,\ a_2=15.$$ -Let $b_n$ be defined by the recursion -$$b_{n+2} + 6 b_{n+1} + b_n =0,\ b_1 =-6,\ b_2=35.$$ -The number of unfoldings of the labeled $1 \times 1 \times n$ cube is $-4 a_n^2 b_n$. (I might have a sign error here.) -We count spanning trees using the matrix tree theorem. Let $G$ be the dual graph to the $1 \times 1 \times n$ cube (so $G$ has $4n+2$ vertices). Let $V$ be the vector space of functions on the vertices of $G$. Let $\Lambda: V \to V$ be the Laplacian operator. Then $\Lambda$ has a one dimensional kernel -- the constant function -- and we are supposed to compute the coefficient of $t$ in $\det(\Lambda - t \mathrm{Id})$. -The cyclic group of order $4$ acts on the $1 \times 1 \times n$ cube by rotation around the long axis. This action commutes with $\Lambda$. So we can compute $\Lambda$ separately on each of the four eigenspaces of this rotation. The corresponding matrices are -$$\begin{pmatrix} --4 & 1 & 0 & 0 & 0 & \cdots & 0 \\ - 1 &-4 & 1 & 0 & 0 & \cdots & 0 \\ - 0 & 1 &-4 & 1 & 0 & \cdots & 0 \\ - 0 & 0 & 1 &-4 & 1 & \cdots & 0 \\ - 0 & 0 & 0 & 1 &-4 & \cdots & 0 \\ - & & & & & \ddots & \\ - 0 & 0 & 0 & 0 & 0 & \cdots & -4 \end{pmatrix} \ \mathrm{on\ the} \pm i \ \mathrm{eigenspace}$$ -$$\begin{pmatrix} --6 & 1 & 0 & 0 & 0 & \cdots & 0 \\ - 1 &-6 & 1 & 0 & 0 & \cdots & 0 \\ - 0 & 1 &-6 & 1 & 0 & \cdots & 0 \\ - 0 & 0 & 1 &-6 & 1 & \cdots & 0 \\ - 0 & 0 & 0 & 1 &-6 & \cdots & 0 \\ - & & & & & \ddots & \\ - 0 & 0 & 0 & 0 & 0 & \cdots & -6 \end{pmatrix} \ \mathrm{on\ the} -1 \ \mathrm{eigenspace}$$ -$$\begin{pmatrix} --4 & 4 & 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ -1 & -2 & 1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0\\ -0 & 1 &-2 & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\ -0 & 0 & 1 &-2 & 1 & 0 & \cdots & 0 & 0 & 0\\ -0 & 0 & 0 & 1 &-2 & 1 & \cdots & 0 & 0 & 0\\ -0 & 0 & 0 & 0 & 1 &-2 & \cdots & 0 & 0 & 0\\ - & & & & & \ddots & \\ -0 & 0 & 0 & 0 & 0 & 0 & \cdots & 1 & -2 & 1 \\ -0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 4 & -4 \end{pmatrix} \ \mathrm{on\ the}\ 1 \ \mathrm{eigenspace}$$ -The first two matrices are invertible and obey the recursions $a_n$ and $b_n$ respectfully. (See the recursive formula displayed in this article.) -The last matrix has a one dimension kernel. Taking the cofactor by deleting the last row and column, one gets a matrix whose determinant is $-4$, as can be checked from the same recursion as above. -Putting it all together, the number of spanning trees is $-4 a_n^2 b_n$. -I don't understand the statistical question you want to ask about the unfolded shapes. - -Remark: Using the cyclic symmetry broght this computation down into the range of reasonable computation. But, even without that, I immediately knew that the answer would obey some linear recursion. Consider building a spanning tree of $G$ one layer at a time as we travel along the long axis of the cube. At the $k$-th partial stage, there are $4$ vertices of $G$ on the exposed boundary; call them $u_k$, $v_w$, $w_v$, $x_k$. The graph so far is a forest, each connected component of which contains at least one of the four exposed vertices. There are $15$ ways to group these $4$ vertices into connected components, one of which is actually impossible for planarity reasons. So there is some $14 \times 14$ matrix which records, for example, if the vertices on one level are grouped as $(\{ u_k, v_k \}, \{ w_k \}, \{ x_k \})$, how many ways are there to extend the forest to one where the vertices on the next level are grouped as $( \{ u_{k+1} \}, \{ v_{k+1}, w_{k+1}, x_{k+1} \})$. Taking powers of this $14 \times 14$ matrix and then pairing with some row and column vectors to handle end effects, you get this count. -This is what I learned to call the "transfer matrix method" (although Wikipedia seems to call something else by that name) and it solves almost all "linear" combinatorics problems.<|endoftext|> -TITLE: C*-algebras with bizzarre structure of projections -QUESTION [13 upvotes]: This is probably well-known to the experts but I could not find any answer neither in my head nor in the literature: Is there a (unital) C*-algebra such that its projections do not form a lattice (under the usual ordering)? Certainly, this cannot be a von Neumann algebra. - -REPLY [16 votes]: Here is perhaps the simplest example. Let $A$ be the C*-algebra of all sequences of $2 \times 2$ matrices converging to a scalar multiple of diag(1,0). Let $p$ be the constant sequence diag(1,0), and $q$ a sequence of rank 1 projections converging to diag(1,0) but never exactly equal. Then $p$ and $q$ have no upper bound at all. This example can be tweaked to make it unital by allowing any limit matrix at infinity and taking $q$ to alternate diag(1,0) and nearby but unequal projections. Then $p$ and $q$ have no least upper bound.<|endoftext|> -TITLE: Does the class category of ZF-algebras satisfy the Multiverse axioms? -QUESTION [8 upvotes]: I read with interest both Hamkins Multiverse Axioms and Joyal and Moerdijk's algebraic set theory. Both of these perspectives takes a set-theory universe as an object, and consider collections of set-theory universes. The former lives in the setting of first-order logic, while the later lives in the setting of category theory (specifically monads and algebras). -Has there been an attempt to compare these theories? In particular (I hope the following is a well-founded question), does Hamkins' Multiverse axiomatize (the set-theory part, i.e. ignoring the intuitionistic and topos theoretic part) the class category of ZF-algebras? -I'm not too familiar with both these theories, but I have a great interest in Foundations. I would be thankful for more detailed comparisons beyond the particular question that I ask. - -REPLY [2 votes]: I think the two theories should be regarderd as existing at different levels. A category of classes, in the sense of algebraic set theory, is not a collection of models of set theory, but (an abstraction of) the collection of all classes relative to one model of set theory. In particular, if $V$ is a model of set theory, then the collection of all classes in V is a category of classes. -One can then, if one wants, define a notion of "internal model of set theory" inside a category of classes, and in the "canonical" example this would just be the notion of a class-model of set theory relative to $V$. The model $V$ itself is recoverable as the "initial" such internal model. Joel's excellent answer explains why such a collection of "internal models" will not generally satisfy the multiverse axioms.<|endoftext|> -TITLE: Algebraically closed fields with proper maximal subfields -QUESTION [7 upvotes]: Is there a classification of the algebraically closed fields that have maximal proper subfields ? -And if an algebraically closed field has a maximal proper subfield, is that subfield unique ? -Summarizing the answers, an algebraically closed field has a maximal subfield if and only if its characteristic is zero and such a maximal subfield is never unique. - -REPLY [15 votes]: If $F$ is a maximal proper subfield of a field $K$, then $K=F(x)$ for any $x\in K\setminus F$. Next, $x$ must be algebraic over $F$ (otherwise $F\subsetneq F(x^2)\subsetneq F(x)\subset K$). So $K$ is finite over $F$, and if $K$ is algebraically closed it is well known (cf. KConrad's comment) that $F$ is a real closed field and $K=F(\sqrt{-1})$. - -REPLY [8 votes]: This occurs iff the field has characteristic 0. By KConrad's comment, being characteristic 0 is certainly a necessary condition. Conversely, given an algebraically closed field K of characteristic 0, we can use Zorn's Lemma to find a maximal ordered subfield F. Since K is algebraically closed, F must be real closed. But also K must be algebraic over F or else we could pick a transcendental element and adjoin it to F (make it infinitely larger than all elements of F, i.e., use lexicographic ordering). Hence K must be of degree 2 over F and thus F is a maximal proper subfield. -Also by KConrad's comment, this is never unique; just apply an automorphism of K that takes, e.g., $\sqrt[3]{2}$ to $\omega \sqrt[3]{2}$, where $\omega$ is a primitive cube root of unity.<|endoftext|> -TITLE: Aschbacher classes and $\mathbb{F}_p$-subspace stabilizers in classical linear groups -QUESTION [5 upvotes]: I am reading the Kleidman-Liebeck book ("The subgroup structure of the finite classical groups") which is about the Aschbacher classification of maximal subgroups of the classical almost simple groups. But thinking just about natural actions of classical groups I got stuck in the following problem. -Consider the group $GL(n,q)$ of invertible $\mathbb{F}_q$-linear maps $\mathbb{F}_q^n \to \mathbb{F}_q^n$, where $n$ is a positive integer and $q=p^k$ is a prime power. This group acts on $V := \mathbb{F}_q^n$ in the natural way, and this action is $\mathbb{F}_q$-linear, in particular it is $\mathbb{F}_p$-linear. $\mathbb{F}_p$-linearity yields an injective homomorphism $GL(n,q) \to GL(nk,p)$, and I will identify $GL(n,q)$ with a subgroup of $GL(nk,p)$ by means of this homomorphism. Let $W$ be a proper $\mathbb{F}_p$-subspace of $V$, and let $H$ be the stabilizer of $W$ in $GL(nk,p)$, i.e. the set of invertible $\mathbb{F}_p$-linear transformations $f$ of $V$ such that $f(W)=W$. -Since $GL(n,q)$ acts transitively on non-zero vectors, $L := H \cap GL(n,q)$ is a proper subgroup of $GL(n,q)$, so it is contained in a maximal subgroup of $GL(n,q)$ which belongs to one of the Aschbacher classes. Which one? -If $W$ spans a proper $\mathbb{F}_q$-subspace of $V$ then $L$ is of parabolic type (class $\mathcal{C}_1$). -What happens if $\langle W \rangle_{\mathbb{F}_q} = V$? My guess is that in this case $L$ falls in class $\mathcal{C}_5$ (subfield stabilizers), but unfortunately I do not know enough about the linear groups to know how to think about this or where to look at. -I have been thinking about this and I asked people, but the question does not seem easy. This is just to say that I hope it is not trivial for experts, in which case I apologize. - -REPLY [3 votes]: I believe that the stabilizer $H$ of a subspace must be either reducible as a subgroup of ${\rm GL}(n,q)$ or it can be written over a proper subfield of ${\mathbb F}_q$. -Suppose that it is irreducible, so we have an irreducible representation $\rho$ of $H$ over ${\mathbb F}_q$. If I am remembering this correctly then, if you regard this $\rho$ first as a representation of degree $kn$ over ${\mathbb F}_p$ and then think of that as a representation of degree $kn$ over ${\mathbb F}_q$, then the resulting representation is $\rho + \rho^\phi + \cdots + \rho^{\phi^{k-1}}$, where $\phi$ is the Galois automorphism of order $k$ of ${\mathbb F}_q$. -If $H$ is not writeable over a proper subfield of ${\mathbb F}_q$, then the representations $\rho, \rho^\phi, \ldots, \rho^{\phi^{k-1}}$ are all inequivalent. But since $H$ stabilizes a proper subspace over ${\mathbb F}_p$, the restriction to that subspace as a ${\mathbb F}_q$ representation would be equivalent to a sum of a proper subset of $\{\rho, \rho^\phi, \ldots, \rho^{\phi^{k-1}}\}$. But then, since that sum is writeable over ${\mathbb F}_p$, it must be stabilized by $\phi$, which is not possible is $\rho, \rho^\phi, \ldots, \rho^{\phi^{k-1}}$ are all inequivalent. -I expect an expert in representation theory would be able to explain it better than I have, but I think this is essentially correct!<|endoftext|> -TITLE: SO(p,q) and Howe Duality -QUESTION [10 upvotes]: I recently learned of a relationship between the representations of the groups $SO(p,q)$ and $SL(2,\mathbb{R})$ which is part of an apparently much larger set of ideas known as Howe Duality. My question is a bit open ended, but can someone point me to a good entry point (review articles, lectures) for learning more about Howe duality and in particular the $SO(p,q)$-$SL(2,\mathbb{R})$ duality? Thanks. - -REPLY [6 votes]: A good introduction is "Non-Abelian Harmonic Analysis: Applications of SL(2,R)" -by Roger Howe and Eng Chye Tan, especially Chapter III, Section 2.<|endoftext|> -TITLE: Delaunay triangulations and convex hulls -QUESTION [6 upvotes]: This is a reference request. -I have the impression that those who work in computational geometry are accustomed to the following. You have some locally finite set of sites in $\mathbb{R}^n$ and you want to think about their Delaunay triangulation, and then you look at the map -$$ -(x_1,\ldots,x_n) \mapsto (x_1,\ldots,x_n,x_1^2+\cdots+x_n^2)\in\mathbb{R}^{n+1}, -$$ -and the Delaunay triangulation then corresponds to the "lower convex hull". For suitably bounded sets of sites (e.g. all lying in some particular half-plane, or---more drastic still---actually bounded in the sense of having finite diameter) there's also an "upper convex hull" that sometimes gets some attention. -But there's a simpler correspondence that might be less well-known, if vague impressions from talking to a few people are right. Suppose you have a finite set of sites on the sphere $(x_1^2+\cdots+x_n^2=1)$ (not $\le1$) and you measure distances between sites along great circles. Voronoi cells are parts of the surface bounded by great circles; the Delaunay triangulation connects sites whose Voronoi cells touch each other. The statement is this: -The facets of the convex hull of the set of sites are precisely the Delaunay triangles. (Or Delaunay simplices if you want to go to higher dimensions.) -There's no occasion to distinguish between "upper" and "lower". I think this might be counterintuitive to those accustomed to "upper" and "lower". Suppose two sites in a bounded set of sites in the plane are connected by a line that's on the boundary of the "upper" convex hull. Then we suddenly realize that the world is not flat; what we thought was a flat plane is a small region of an immense sphere. Those two Voronoi cells that we had thought were unbounded and not neighbors actually touch each other in an antipodal region. Hence no upper/lower distinction, and we have a simpler proposition than one that involves those. -Here's the question: Is the proposition in bold above found somewhere in the literature in this simple form, with neither any distinction between "upper" and "lower" nor between "nearest"- and "farthest"-point diagrams? -In an earlier thread at n-dimensional voronoi diagram, Joseph O'Rourke pointed out a paper by Kevin Q. Brown that works with spheres rather than paraboloids and deals with the convex hull, but still works with "nearest" and "farthest" because he's ultimately concerned with the plane, not with the sphere. - -REPLY [4 votes]: I have nothing to add to the literature question, but I thought I would include an image of -the Voronoi Diagram on a sphere to illustrate that -"Voronoi cells are parts of the surface bounded by great circles": -           - -           - -Mathematica image by Maxim Rytin. - -REPLY [3 votes]: This paper by Bobenko and Izmestiev "Alexandrov's Theorem, Weighted Delaunay Triangulations, and Mixed Volumes" is based upon a generalization of this observation. -One reason that this convex-hull characterization may seem obscure is that it really only applies to points on the sphere. The paragraph in the OP that begins "There is no reason ..." is misleading because it implies that the planar Delaunay triangulation can be characterized in this way. The lower envelope of the convex hull of the traditional lifting (to a paraboloid) gives exactly the combinatorics of the planar DT. However, going to a sphere, by stereographic projection, say, is only going to work if the radius of the sphere is "big enough". For a fixed number of points n, there will be no upper bound on the radius that is big enough for all point sets of size n. -The Delaunay triangulation is brittle, and unless unusual demands on the vertex configurations are made, the combinatorial structure can change with an arbitrarily small perturbation of the metric. I.e., the point set can be arbitrarily close to being degenerate.<|endoftext|> -TITLE: A more efficient way to generate random graphs with a given degree sequence? -QUESTION [5 upvotes]: In graph data mining it is often useful to generate random (simple) graphs with a given degree sequence (e.g. in searching for network motifs), and ideally these would be uniformly at random. -[For this question, we'll consider the undirected case (the directed case can also be treated similarly, but it's a bit more messy).] -We could use the configuration model to generate these graphs (each node is given stubs corresponding to its degree, and these stubs are connected uniformly at random; repeat until a simple graph is obtained). But, unfortunately, often this results in too many restarts, and is impractical. This is worse for larger graphs, or graphs with unfavourable degree distributions. [When it does work, however, the configuration model is surprisingly quick.] -In cases when the configuration model is impractical, a switching method is often used, where two edges {a,b} and {c,d} are replaced by {a,d} and {b,c}, provided no clashes arise (loops or parallel edges). If we perform this operation a zillion times, we'll get something empirically fairly close to the uniform distribution (it is not actually uniform, but is plenty good enough for most real-world studies, where other errors dominate). - -Let me propose another scheme: We select some induced subgraph H, and use the configuration model on H alone. Repeat this a zillion times. -Question: Has this scheme, or a similar scheme, been considered previously? If so, would it result in a distribution that is "empirically close" to uniform? Could it possibly be more efficient than the "switching pairs of edges" method? - -When I say, "more efficient" I don't mean O(...) efficiency. E.g. simply being twice as fast would be great. -[NB. In the above, I'm (for the time being) putting aside the process of choosing the subgraph H, which is a problem in itself.] - -REPLY [3 votes]: There is the algorithm of Blitzstein and Diaconis -- they claim very good practical performance.<|endoftext|> -TITLE: About the category of von neumann algebras -QUESTION [23 upvotes]: I am looking for one (or more) reference about properties of the category of von Neumann algebra. -More precisely, in an answer of a previous question, Dmitri Pavlov mentions -that the $W^*$ category is complete and cocomplete. I would be happy to have a reference for these facts. -On a related note, at the page 84 of the volume 3 of his treatise "Theory of -operator algebras", Takesaki says: -"Namely, -if one insists to have the universal property for the inductive limit, then one has -to treat a non-separable von Neumann algebra as the inductive limit of a sequence -of separable von Neumann algebras. Therefore, we shall not consider the general -theory of inductive limits of von Neumann algebras." -This got me curious. Where can I find a reference for an example of an inductive system -of separable von Neumann algebras with non separable limit ? Is it possible for this example -to be a countable inductive system ? -And this makes me wonder : why is nonseparability such a problem in the theory of von Neumann algebra, as Takesaki suggests ? - -REPLY [6 votes]: I agree with Dmitri Pavlov that separability is not so important in the modern theory of von Neumann algebras, and this answers the second question. However, an example answering the first question has not yet been given, so here is one. -We take $\mathbf{vNA}$ to be the category of von Neumann algebras and normal *-homomorphisms, and the category $\mathbf{C}^*\mathbf{Alg}$ to be the category of unital C$^*$-algebras and *-homomorphisms. I will be using the fact that the forgetful functor $U : \mathbf{vNA} \rightarrow \mathbf{C}^*\mathbf{Alg}$ has a left adjoint $\mbox{-}^{**}$, the enveloping von Neumann algebra (the double dual of a C$^*$-algebra). -First, observe that $C(2^\omega)$ is the colimit of the diagram -$$ -C(2) \rightarrow C(2^2) \rightarrow C(2^3) \rightarrow \cdots -$$ -in $\mathbf{C}^*\mathbf{Alg}$. (The easiest way I know to prove this is to define the colimiting map on the indicator functions of clopen subsets of $2^\omega$, and then use Stone-Weierstrass to extend this to all of $C(2^\omega)$.) -Each of the C*-algebras $C(2^n)$ is finite-dimensional, and therefore isomorphic to its own enveloping von Neumann algebra. Therefore the diagram above is also a diagram in $\mathbf{vNA}$, consisting of separable von Neumann algebras. Because left adjoints preserve colimits, we have that $C(2^\omega)^{**}$ is the colimit of this diagram in $\mathbf{vNA}$. -The Dirac delta functions $\{\delta_x\}_{x \in 2^\omega}$ form an uncountable norm-discrete subset of $C(2^\omega)^*$. Therefore $C(2^\omega)^*$ is not norm-separable, so $C(2^\omega)^{**}$ does not have a separable predual, i.e. it is not a separable von Neumann algebra. But as we saw above, it is an inductive limit of finite-dimensional (and therefore separable) von Neumann algebras.<|endoftext|> -TITLE: Is there any relationship between Bourbaki's Epsilon Calculus and Lambda Calculus? Is $\lambda x$ the same as $\tau_x$? -QUESTION [5 upvotes]: Is there any relationship between Bourbaki's Epsilon Calculus and Lambda Calculus ? Whether $\lambda {x}$ is same as $\tau _{x}$ ? Are the rules of Meta-Mathematics (Criteria of Substitution, Formative Constructions) described in Bourbaki equally applicable to Lambda Calculus ? - -REPLY [17 votes]: Bourbaki's tau-box notation is somewhat insane (e.g., see Adrian Mathias's A Term of Length 4,523,659,424,929), so I'll eventually answer in terms of Hilbert's epsilon-calculus. -But first, the laws of variable binding are identical for all reasonable classical and intuitionistic calculi (matters are a little different for substructural and modal logics, but not in a fundamental way). The basic idea is always the same: certain terms introduce bound variables, and within the scope of the binder (a) you can refer to the introduced variable, and (b) variables of the same name introduced outside of it are shadowed. Furthermore, renaming bound variables does not change the meaning of a term (this is called alpha-equivalence), and when substituting a term for a variable, bound variables need to be renamed in order to avoid capturing the free variables of the substituted term (this is called alpha-conversion). -Frege was the first person to really figure out how variable binding works, and this was a sufficiently important discovery that he is a great logician despite the small matter of the inconsistency of his foundational system. (And I do not mean that ironically: it really is a small matter compared to the magnitude of the achievement of understanding binding.) -The similarity you see comes from the fact that $\forall x.\;A$, $\exists x.\;A$, $\epsilon x.\;A$, and $\lambda x.\;e$ are all binding forms, and must all respect the same laws of variable binding. However, the difference between them is best understood in terms of their different types[*], taking $\iota$ to range over individuals, and $o$ to range over propositions. -$$ -\begin{array}{lcl} -\forall & : & (\iota \to o) \to o \\\ -\exists & : & (\iota \to o) \to o \\\ -\epsilon & : & (\iota \to o) \to \iota \\\ -\lambda & : & (A \to B) \to (A \Rightarrow B) \\\ -\end{array} -$$ -What $\forall$ and $\exists$ do is to take a term of type proposition, that has a free variable of type individual, and construct a proposition from it. $\epsilon$, on the other hand, is a choice operation. It says that if you give it a predicate (i.e., a term of type proposition with a free variable of type individual), it will give you a canonical individual satisfying that proposition. $\lambda$ is syntax for function-abstraction. If you give it a term of type $B$ with a free variable of type $A$, it will give you back a function value transporting $A$s to $B$s. -Now, note that unlike $\forall$, $\exists$, and $\epsilon$, lambda-abstraction works at all types. This is why Church introduced the lambda calculus: he had the idea that you could model variable binding once, with lambda, and then introduce $\forall$, $\exists$, and $\epsilon$ as additional constants in the lambda calculus. -If you really have your heart set on Bourbaki's notation, since it eliminates all variables from the calculus, then I recommend you look at de Bruijn indices instead. The intuition there is also link-based, but it is a radically more efficient representation than Bourbaki's. -[*] I am playing a bit fast and loose here, since the right way to understand these things is really in terms of categorical algebra and structural proof theory: I should be using homs and contexts, but no matter.<|endoftext|> -TITLE: slick-proof-of-trick-for-counting-domino-tilings -QUESTION [14 upvotes]: The trick for rewriting the number of domino tilings of a simply-connected finite lattice region as the absolute value of the determinant of a matrix (due I believe to Kasteleyn and Percus, but if Fisher deserves credit for this too please set me straight!) hinges on a combinatorial lemma relating the number of vertical dominos in a tiling to the parity of the permutation-matrix associated with the tiling. -What's the most direct way to see this? -My favorite approach is to set up a distributive lattice structure on the set of tilings, and use it to show that every tiling may be obtained from every other by rotating 2-by-2 blocks, and then use this last fact to prove the lemma, but this is a bit indirect. I've seen a way to do it with Pick's Theorem, but I suspect that there's a simpler way. Right now I'm thinking about an approach that involves superimposing matchings to get unions of cycles, and arguing that the retiling associated with such a cycle can always be achieved by a succession of retilings of 2-by-2 blocks, but maybe there's an even simpler way. -The trick applies more generally to all bipartite planar graphs, and Kasteleyn showed that you could handle arbitrary planar graphs by this method (using Pfaffians), but for right now I'll be satisfied with an argument that applies in the case of dominos on a square grid. (As you might guess, the motivation behind my question is pedagogical; I'll be teaching a class on this material tomorrow.) - -REPLY [9 votes]: $\def\RR{\mathbb{R}}$This isn't the right answer for your class, because it works for a level of generality you don't need, but last summer I found what I think is the right way to prove this result for a general bipartite planar graph or, working harder, for any planar graph and I thought I'd record it here. I'm sure this isn't original, but I didn't find it in a quick literature search. -Let $G$ be a bipartite graph. We'll define an immersion of $G$ to $\RR^2$ to be a map $\phi: G \to \RR^2$ which is injective except that $\phi$ is allowed to make edges cross transversely in their interiors; $\phi$ is not permitted to create triple crossings. For example, we can send the vertices of $G$ to generic points of $\RR^2$ and join them by line segments. -Assign a weight $x_e$ to each edge of $G$. For a perfect matching $M$ of $G$, define $\mathrm{cross}(M, \phi)$ to be the number of places where edges of $M$ cross. Define the matching polynomial of $(G, \phi)$ to be -$$\mu(G, \phi) = \sum_M (-1)^{\mathrm{cross}(M, \phi)} \prod_{e \in M} x_e$$ -where the sum is over all perfect matchings of $G$. -We claim that there is a way to take the adjacency matrix of $G$ and replace the entry corresponding to $e$ by $\pm x_e$ to make a matrix $A$ with $\mu(G, \phi) = \det A$. In particular, if $G$ is a planar graph, we can get the generating function for matchings with no signs. -Proof First, suppose that all of the black vertices lie on the line $y=0$ and all the white vertices lie on the line $y=1$, in the same order as the rows and columns of the matrix, and joined by line segments. Then we can put $x_e$ in all positions of $A$, and $(-1)^{\mathrm{cross}(\phi, M)}$ is the sign of the permutation $M$. -Now, consider varying $\phi$ while preserving $G$ and we claim that the result stays true. If we vary $\phi$ generically, we can get from any immersion to any other while passing through singularities of the following sorts: - -A triple crossing of edges. The same signs work on both sides of the triple crossing. -Two edges become tangent. On one side of the singularity, they cross $m$ times, on the other side they cross $m+2$ times. The same signs work on both sides of the tangency. -An edge $e$ forms a cusp. On one side of the cusp, it passes through itself, on the other side it doesn't. Negate the sign of $x_e$. -$\phi$ sends a vertex into the interior of an edge $e$. Negate the sign of $x_e$. - -So in every case, as the topology of $\phi$ changes, we can change the signs in the matrix. $\square$ -The analogous argument works with general graphs and Pfaffians, starting with the vertices all lying on a circle.<|endoftext|> -TITLE: Calculating the Riemann Curvature tensor out of sectional curvature -QUESTION [8 upvotes]: Hi everybody, -my question is the following: Let $(M^n,g)$ be a Riemannian manifold and $e_1,\ldots,e_n$ be an orthonomal frame in a point. Assume, that we now the sectional curvatures of all planes, spanned by these vectors, i.e. we know the components $R_{ijij}$ of the curvature tensor. -Is it then possible to calculate all other components $R_{ijkl}$? -The problem is, that if I want, e.g. to calculate $R_{ijik}$ by the polarization identity, I need to know the sectional curvature spanned by $e_1,e_j+e_k$ which I assume to be unknown. -Is it then possible to calculate, or at least to estimate the sectional curvature from above by $R_{ijij}$, $R_{jkjk}$ and $R_{ikik}$? - -REPLY [7 votes]: There is an explicit, but complicated, formula for going from the sectional curvature back to the curvature tensor in equation (1.10) on page 16 of Cheeger and Ebin's book.<|endoftext|> -TITLE: The "grassmannian" of a simplicial complex -QUESTION [10 upvotes]: This question is mainly a reference request – I have a construction which seems natural, so I am quite convinced it should be standard, but I don't know what it is called. -Take an $n$ dimensional simplicial complex $X$ and define a new simplicial complex $X_{k}$ in the following way – the vertices of $X_{k}$ will be $k$ dimensional simplices of $X$ and two such vertices will be connected by an edge if the two corresponding $k$ simplices in $X$ are contained in a $k+1$ simplex in $X$. This is the skeleton of $X_{k}$ and complete it by gluing higher dimensional simplices whenever they are formed (so you get an $n-k$ simplicial complex). -My questions – what is the standard name for this and where can I find a reference on? Moreover, is it true that whenever $X$ is contractible then $X_k$ is simply connected (and even that the homologies of order < $n-k$ vanish)? - -REPLY [3 votes]: What you describe is similar to something well-known that does satisfy your condition (ie, has the same homotopy as $X$ in dimensions less than $n-k$). I'm speaking of the dual complex to the dual $(n-k)$-skeleton of your simplicial complex. Let me call this $X^{\ast (n-k)\ast}$ (which is short for $((X^\ast)^{(n-k)})^\ast$). This is not a simplicial complex in general, only a cone complex. It has one vertex for each $k$-simplex; vertices corresponding to those $k$-simplices that belong to the same $(k+1)$-simplex are connected by an "edge" (that is, a cone over those vertices - notice that there can be more than two of them); and so on. You can find the discussion of dual complexes in standard texts on PL topology, but cone complexes as such are not clearly exposed there; see references on cone complexes in my answer here.<|endoftext|> -TITLE: Ihara zeta function -QUESTION [6 upvotes]: Is there a natural connection between the Ihara zeta function of a graph, -and (for instance) the Riemann zeta function of certain varieties over finite fields ? - Thanks. - -REPLY [6 votes]: RH for the Ihara zeta function will correspond to the graph being Ramanujan (if the graph is (q+1)-regular). -The zeta function for varieties over finite fields is more related to Ruelle's zeta function, but you can see Ihara zeta function as a special instance of it, using symbolic dynamics representation of a walk in your graph as a dynamical system. -A nice reference for this material is Audrey Terras' book - "Zeta Functions of Graphs: A Stroll through the Garden"<|endoftext|> -TITLE: How many solutions are there to $\sum_{i=1}^3 x_i^2+x_iy_i+y_i^2=k$? -QUESTION [7 upvotes]: Let $k$ be a positive integer. Let -$$Q= -\begin{pmatrix} - 1 &1/2& & & & \\ -1/2& 1 & & & & \\ - & & 1 &1/2& & \\ - & &1/2& 1 & & \\ - & & & & 1 &1/2\\ - & & & &1/2& 1 -\end{pmatrix}. -$$ - -How many solution $x\in\mathbb Z^6$ are there to $\quad x^tQx=k$? - -This is equivalent to: - -How many solution $x\in\mathbb Z^6$ are there to - $$x_1^2+x_1x_2+x_2^2+ x_3^2+x_3x_4+x_4^2+ x_5^2+x_5x_6+x_6^2=k?$$ - -or to - -How many solution $x\in \mathbb Z\left[\omega \right]^3$ are there to $\quad x^* I_3 x=k$? - -where $I_3$ is the $3\times3$-identity matrix and $\omega=\frac{1+\sqrt{-3}}{2}$. -I know that there is a formula for this number (there is only one class in its genus), but I don't know it. -This question is related to - -which integers take the form x^2 + xy + y^2 ? -https://math.stackexchange.com/questions/44139/how-many-solutions-are-there-to-fn-m-n2nmm2-q/ - -but they don't answer my question. - -REPLY [7 votes]: This is a supplement to Noam Elkies' nice answer. The coefficients $s(k)$ can be expressed as -$$ s(k)=27\sum_{d\mid k}\chi(k/d)d^2-9\sum_{d\mid k}\chi(d)d^2, $$ -hence the function $\varphi$ is a linear combination of -$$E_1:=\sum_{k=1}^\infty\sum_{d\mid k}\chi(k/d)d^2q^k -\quad\text{and}\quad -E_2:=1-9\sum_{k=1}^\infty\sum_{d\mid k}\chi(d)d^2q^k.$$ -The latter functions are proportional to the standard Eisenstein series -$$ E_1':=\sum'_{m,n\in\mathbb{Z}}\chi(m)(mz+n)^{-3} -\quad\text{and}\quad -E_2':=\sum'_{m,n\in\mathbb{Z}}\chi(n)(mz+n)^{-3},$$ -which form a basis of the space of modular forms $M_3(\Gamma_0(3),\chi)$, hence indeed $\varphi$ lies in this space. For more details see Section 7.1 in Miyake: Modular Forms, especially Lemma 7.1.1 and Theorem 7.1.3.<|endoftext|> -TITLE: A variational principle for amenable groups -QUESTION [7 upvotes]: Update: If somebody is interested, in Sec. 3, Theorem 3.5, of http://arxiv.org/abs/1203.2301 the variational principle for amenable groups such that every conjugacy class is finite is proved. -Let $G$ be a countable amenable group, $f\in \ell^\infty(G)$. Denote by $L,R$ respectively the set of all right-, left-invariant means. Denote by $P$ the set of all finitely additive probability measures on $G$. -There is some evidence coming from Game Theory that the following conjecture holds: - -Conjecture: For all $\lambda\in L$, the functional $\mu\in P\rightarrow\int\int f(xy)d\mu(x)d\lambda(y)$ attains its maximum in some $\rho\in R$. - -Notice that this functional is not weak*-continuous, so it is not clear even that the maximum should exist. -Thanks in advance for any help, -Valerio - -REPLY [2 votes]: Here I propose a proof showing that the conjecture above holds at least for one left-invariant mean. -Denote by $L$ and $R$ respectively the set of all left-, right-invariant means on $G$. Denote by $P$ the set of all finitely additive probability measures on $G$. -Theorem: There is $\lambda\in L$ such that the functional $\mu\in P\rightarrow\int\int f(xy)d\mu(x)d\lambda(y)$ attains its maximum in some $\rho\in R$. -Proof: By Theorem 3 (and discussion below) of Heath-Sudderth, On a theorem of De Finetti, oddsmaking and game theory, Ann. Math. Stat. (1972), vol. 43, No. 6, 2072-2077, the following equality holds: -$$ -\inf_{\nu\in P}\sup_{\mu\in P}\int\int f(xy)d\mu(x)d\nu(y)=\sup_{\mu\in P}\inf_{\nu\in P}\int\int f(xy)d\mu(x)d\nu(y) -$$ -Denote by $L$ this common value. Fix $\epsilon_1,\epsilon_2>0$. The RHS tells us that -(1) there is $\mu\in P$ such that $\int f(xy)d\mu(x)\geq L-\epsilon_1$, for all $y\in G$. -The LHS tells us that -(2) there is $\nu\in P$ such that $\sup_{\mu\in P}\int\int f(xy)d\mu(x)d\nu(y)\leq L+\epsilon_2$. -Let $F_n$ be a right-Folner sequence for $G$ and define $\nu_n(A)=\frac{1}{|F_n|}\sum_{g\in F_n}\nu(gA)$. It is straightforward to prove that $\nu_n$ still verifies property 2 and so does any weak* limit, which is a left-invariant mean. Denote by $\nu_{\epsilon_2}$ this left-invariant mean. Now, let $\epsilon_2$ go to zero and let $\lambda$ be any weak*-limit of the net $\nu_{\epsilon_2}$. Hence $\lambda\in L$ and it verifies the following property -(2 bis) $\int\int f(xy)d\mu(x)d\lambda(y)\leq L$, for all $\mu\in P$. -Now, let us get back to property 1, and uniformize $\mu$ as above but using a left-Folner sequence. In this way, any weak*-limit is right-invariant and still verifies property 1. So, we have gotten the following property: -(1 bis) There is $\rho_{\epsilon_1}\in R$ such that $\int f(xy)d\rho_\epsilon(x)\geq L-\varepsilon_1$ (contantly in $y$). -Now, the set of real numbers of the shape $\int f(xy)d\rho(x)$, for some $\rho\in R$, is compact and so, by 1 bis, there is $\rho\in R$ such that $\int f(xy)d\rho(x)=L$. Now, using 2 bis, it follows that our functional, for that particular $\lambda\in L$, attains indeed its maximum in $\rho\in R$, as claimed.<|endoftext|> -TITLE: Can every finite poset be realized as divisors of an algebraic curve? -QUESTION [13 upvotes]: Let $D_1$, ... , $D_n$ be a finite set of divisor classes on a nonsingular projective irreducible algebraic curve. We say that $D_1\geq D_n$ if the line bundle defined by $D_1-D_n$ has a section. This obviously satisfies the axioms of a partial order. -Suppose $\{x_1,....,x_n\}$ is a finite partially ordered set. Does there exist a (projective, nonsingular) algebraic curve of sufficiently high genus, and a set of divisors on it, that are isomorphic as a partially ordered set to $\{x_1,...,x_n\}$? - -REPLY [14 votes]: Choosing a general set of $n$ points on a curve of genus at least $n$, you can assume that the divisor they define has a unique global section. Let $P$ denote the poset generated by all the possible sums with coefficients $0,1$ of these $n$ points. This poset is the same as the poset of subsets of an $n$-element set. -Since every finite poset seems to be a subposet of the poset of subsets of a finite set [EDIT: this is true-see the comments below], just embed your poset in a "power set poset" and remove the unwanted divisors, to deduce that what you want is true.<|endoftext|> -TITLE: Quantum coordinate ring at root of unity -QUESTION [7 upvotes]: Noah Snyder gave a great answer to this question about different versions of a quantum group $U_q(\mathfrak g)$ when $q$ is a root of unity. I want to ask about forms of the deformed coordinate ring $\mathcal O_q(G)$ when $q$ is a root of unity. -Let's focus on $SL(2)$. Recall that the "small quantum group" (see Noah's answer) is obtained by dividing $U_q(\mathfrak s\mathfrak l_2)$ by the ideal generated by $E^e,F^e,K^e-1$ (which is central since $q^e=1$). The quotient is a finite-dimensional Hopf algebra $\overline{U_q}(\mathfrak s\mathfrak l_2)$. Since $\mathcal O_q(SL(2))$ and $U_q(\mathfrak s\mathfrak l_2)$ are in duality, one expects that there is a corresponding finite-dimensional subalgebra of $\mathcal O_q(SL(2))$ (the set of elements annihilated by $E^e,F^e,K^e-1$). What is it, and are there references about it? Is there a good reason why it is hard to work with compared to the Hopf algebra perspective? - -REPLY [8 votes]: For what concerns De Concini-like integer form the Sl_2 case (and more) is treated in quite some detail in -"Quantum function algebra at roots of 1" De Concini-Lyubashenko, Adv. Math. 108, 205-262 (1994). -The powers of usual $a,b,c,d$ generators form a commutative Hopf subalgebra and the duality relation is explained in detail (Proposition 1.4 of ref. cit.) -A number of general algebraic properties are contained in two papers by Benjamin Enriquez -"Le centre des algèbres de coordonnèes des group quantiques aux racines $p^\alpha$-ièmes de l'unité" Bull. Soc. Math. Fr. 122, 443-485 (1994) -"Integrity, Integral closedness and finiteness over their centers of the coordinate algebras of quantum groups at $p^\nu$-th roots of unity" Ann. Sci. Math. Quebec 19, 21-47 (1995). -The multiparameter cas was considered by -Costantini-Varagnolo -"Multiparameter quantum function algebras at roots of 1" Math. Ann. 306, 759-780 (1996). -More recently, also, -Costantini "On the quantum function algebra at roots of 1" Comm. Algebra 32, 2377-2383 (2004).<|endoftext|> -TITLE: Generalizations of the Rayleigh(-Beatty) theorem -QUESTION [5 upvotes]: For a given irrational number $\alpha>0$ and a real number $\beta$, -the inhomogeneous Beatty sequence -sequence $S_{\alpha,\beta}$ is the set $\lbrace\lfloor n\alpha+\beta\rfloor:n=1,2,\dots\rbrace$ -(the case $\beta=0$ corresponds to a homogeneous Beatty sequence). -If $\beta=0$, the two homogeneous Beatty sequences -$S_{\alpha_1,0}$ and $S_{\alpha_2,0}$ partition the set of positive integers -iff $1/\alpha_1+1/\alpha_2=1$. There is also a similar result for inhomogeneous -$S_{\alpha_1,\beta_1}$ and $S_{\alpha_2,\beta_2}$: assuming that neither -$n\alpha_1+\beta_1$ nor $n\alpha_2+\beta_2$ is an integer for $n=1,2,\dots$, -the sequences partition $\mathbb Z_{>0}$ iff $1/\alpha_1+1/\alpha_2=1$ and -$\beta_1/\alpha_1+\beta_2/\alpha_2=0$. -Question. -For a given $k\ge3$, what are the conditions on $\alpha_1,\dots,\alpha_k$ -(and on $\beta_1,\dots,\beta_k$ in the inhomogeneous case) to ensure that -the sets $S_{\alpha_i,\beta_i}$, $i=1,\dots,k$, partition the positive integers. -It looks like the book -Old and new problems and results in combinatorial number theory -by P. Erdős and R.L. Graham (which I do not have) mentions a version -of the problem, but I am interested in some (possibly very recent) progress -in the direction. My interest is motivated by the study of functional -equations of the Mahler-type generating functions of the Beatty sequences. - -REPLY [5 votes]: In 1973, Fraenkel showed that, for fixed $k \geq 3$, if $\alpha_i = (2^k - 1)/2^{i-1}$ and $\beta_i = -2^{k-i} + 1$ for $i = 1, 2, \ldots k$, then the $k$ Beatty sequences $S_{\alpha_i,\beta_i} := \lbrace{\lfloor n\alpha_i + \beta_i\rfloor\rbrace}_{n\geq 1}$ partition the positive integers. Many other cases have been proved by Simpson (1991). -Fraenkel also conjectured that any partition of the positive integers into $k \geq 3$ Beatty sequences $S_{\alpha_i,\beta_i}$, with $\alpha_i$, $\beta_i$ real and $0 < \alpha_1 < \alpha_2 < \cdots < \alpha_k$, satisfies $\alpha_i = (2^k - 1)/2^{i-1}$ for $i = 1, 2, \ldots, k$. -To date, Fraenkel's conjecture has been proved for up to $k=7$ sequences. I would recommend taking a look at this paper by Tijdeman (2001), who proved the conjecture for $k = 5, 6$ (and for $k = 3$ in an earlier paper). Altman, Gaujal, Hordijk (1997) proved it for $k = 4$, and more recently, Barát and Varjú (2003) verified the conjecture for $k=7$. It's a tantalising open problem, which I have dabbled with recently too (albeit from a different point of view).<|endoftext|> -TITLE: Cohomology $H^*(G,K)$ of wreath products -QUESTION [6 upvotes]: Let $G = Sym(a) \wr Sym(b)$ be a wreath product of symmetric groups - I'm particularly interested in the Weyl group of type $B$, $Sym(2) \wr Sym(n)$. Let $k$ be a field of characteristic $p$. -What is $H^*(G,k)$? -If $i \leq p-3$ and we're in the symmetric group case, then $H^i(Sym(n), k)=0$. -If $i=1$ and $G$ is as above then $H^1(G, k)=0$, for all $a$ and $b$, unless we're in characteristic 2. -Is anything else possible to say? What I REALLY want is that the symmetric group result generalises so that: -If $i \leq p-3$, then $H^i(G, k)=0$ for $i \leq p-3$, for $G$ a wreath product as above. -Any ideas if this is true? It might be that it holds with fewer restrictions on $i$ and $G$ - the proof I know for the symmetric group case uses the Schur functor and tilting modules for $GL_n$. However, the result proved is FAR more general (it concerns all Specht modules) - so maybe this $GL_n$ approach isn't needed. - -REPLY [6 votes]: Set $H^\ast(-) := H^\ast(-,k)$ and $S_a = Sym(a)$. Let $G$ be the wreath product that fits into the extension -$$ 1 \to S_a^b \to G \to S_b \to 1.$$ - -Claim: $H^n(G) = 0$ for $1 \le n \le p-3.$ - -Proof: The LHS spectral sequence corresponding to the extension is -$$E_2^{pq} = H^p(S_b, (H^\ast(S_a)^{\otimes b})^q)$$ -where -$$(H^\ast(S_a)^{\otimes b})^q = \oplus_{i_1 + ... + i_b = q} H^{i_1}(S_a) \otimes \cdots \otimes H^{i_b}(S_a).$$ -Let $1 \le q \le p-3$. Then $i_j \le p-3$ for $j=1,...,b$ and not all $i_j$ can be zero. Hence $H^i(S_a)=0$ for $1 \le i \le p-3$ implies $(H^\ast(S_a)^{\otimes b})^q = 0$. -Thus $E_2^{\ast,q}=0$ for $1 \le q \le p-3$ and $E_2^{p,0} = 0$ for $1 \le p \le p-3$. This shows $H^n(G) = 0$ for $1 \le n \le p-3$. -Remarks: 1) Even more is true for wreath products: -$$H^\ast(G) \cong H^\ast(S_b,(H^\ast(S_a)^{\otimes b})$$ -as graded rings (cf. Nakaoka: Homology of the Infinite Symmetric Group, Ann. of. Math. 73(1961),229-257, Theorem 3.3). -2) Since the extension splits, $H^\ast(S_b)$ is a direct summand of $H^\ast(G)$. Hence the vanishing range for the cohomology of $H^\ast(G)$ stated above cannot be improved.<|endoftext|> -TITLE: Victor Miller basis for higher $N$ // why is this bilinear form perfect? -QUESTION [5 upvotes]: Hello. -I am trying to understand the proof of Thm 9.23 in -http://wstein.org/books/modform/modform/newforms.html#congruences-between-newforms -. Let $S_k(\Gamma)$ be the cusp forms for a subgroup $\Gamma_1(N) -\subseteq \Gamma \subseteq \Gamma_0(N)$. Let $\mathbb{T} := -\mathbb{Z}[T_1, T_2, T_3, ...]$ be the hecke algebra viewed as a subset of -$End(S_k(\Gamma))$ (not as abstract double coset stuff!). Let $S_k(\Gamma, -\mathbb{Z}) := S_k(\Gamma) \cap \mathbb{Z}[[q]]$ then the author claims -that -$ \langle \cdot, \cdot \rangle : \mathbb{T} \times S_k(\Gamma, \mathbb{Z}) -\mapsto \mathbb{Z},$ -$\langle T, f \rangle = a_1(T(f))$ -is a perfect bilinear form and because of finite-rank-arguments i already -know that it suffices to show that the $\mathbb{Z}$-module homomorphism -$T \mapsto \langle T, \cdot \rangle \in Hom_\mathbb{Z}(S_k(\Gamma, -\mathbb{Z}), \mathbb{Z})$ is surjective but i am unable to understand why -this is the case. Only for $N=1$ i have found a workaround (see below). - -This is what i have tried: -As was pointed out to me here -"Hecke algebra" finitely generated? , -the set $S_k(\Gamma, \mathbb{Z})$ is a finitely generated free -$\mathbb{Z}$-module. Let us take a basis $S_k(\Gamma, \mathbb{Z}) = -\mathbb{Z}f_1 \oplus ... \oplus \mathbb{Z}f_n$. The -$Hom_\mathbb{Z}(S_k(\Gamma, \mathbb{Z})) = \mathbb{Z}\phi_1 \oplus ... -\oplus \mathbb{Z} \phi_n$ where $\phi_i(f_j) = \delta_{ij}$. -Let us assume that we select $\Gamma_0(N)$ with trivial character. Then, -the ring $\mathbb{T}$ is nothing else than the $\mathbb{Z}$-module -$\mathbb{T} = \mathbb{Z}T_1 + \mathbb{Z}T_2 + \mathbb{Z}T_3 + ...$ -(because $T_{p^e} = T_p T_{p^{e-1}} - p^{k-1} T_1$ so that every -polynomial in the $T_m$ is actually a sum in the $T_m$). We want to show that there is a $T$ such that $\langle T, f_j \rangle = \phi_1(f_j) = \delta_{1j}$ so that we search -for a $\mathbb{Z}$ linear combination of the $T_m$ mapping to $\phi_1$, say $T = v_1 T_1 + ... + v_l T_l$. Then, $\langle T, f_j \rangle = \phi_1(f_j) = \delta_{1j}$ is equivalent to -$\begin{pmatrix} a_1(f_1) & \cdots & a_l(f_1) \\ \vdots & & \vdots \\ a_l(f_n) -& \cdots & a_l(f_n) \end{pmatrix} \cdot \begin{pmatrix} v_1 \\ \vdots \\ v_l -\end{pmatrix} = e_1$ -Let us denote the matrix on the left hand side by $A$, then the question -is: is $A$ a surjective map seen as a map $A : \mathbb{Z}^l \mapsto -\mathbb{Z}^n$ (where we can select $l$ as big as we want)? Rephrased i ask for the following: - -For a subgroup $\Gamma$ that satisfies some properties (i.e. $\Gamma$ a -congruence subgroup or so) is there a $\mathbb{C}$-basis $f_1, ..., f_n$ of $S_k(\Gamma)$ with fourier coefficients in $\mathbb{Z}$ such that the matrix $A$ becomes surjective -as a map from $\mathbb{Z}^l$ to $\mathbb{Z}^n$ (for example: when -selecting $n$ columns, does the matrix consisting of these columns satisfy -$\det(Matrix) = \pm 1$? I.e. is there something like a Victor Miller basis -for modular forms of higher level)? - -For $\Gamma = SL_2(\mathbb{Z})$, this is easy, because one has a Miller -basis, i.e. a $\mathbb{C}$-basis $f_1, ..., f_n$ of -$S_k(SL_2(\mathbb{Z}))$ with integral fourier coefficients such that $a_i(f_j) = \delta_{ij}$ so that $A = -(Id ~~ ...)$. Note: if $f_1, ..., f_n$ are a $\mathbb{C}$-basis for $S_k(SL_2(\mathbb{Z}))$ with integral fourier coefficients then this does not necessarily imply that $f_1, ..., f_n$ form a $\mathbb{Z}$-basis for $S_k(\Gamma, \mathbb{Z})$ but this suffices to see that the Hecke algebra is finitely generated by the first $r$ Hecke operators since one can do the same argument with $\mathbb{Z}f_1 \oplus ... \oplus \mathbb{Z}f_n$ in place of $S_k(\Gamma, \mathbb{Z})$ since the behavior of an operator in $End(S_k(\Gamma))$ is determined by its actions on $f_1, ..., f_n$. -Note that for $N > 1$ this seems to be true in "many" cases: For example, typing the following lines in magma -M := ModularForms(Gamma1(16),3); -Basis(CuspidalSubspace(M)); - -yields -[ -q - 189*q^9 + 132*q^10 + O(q^12), -q^2 - 136*q^9 + 94*q^10 + O(q^12), -q^3 - 92*q^9 + 66*q^10 + O(q^12), -q^4 - 57*q^9 + 38*q^10 + O(q^12), -q^5 - 33*q^9 + 22*q^10 + O(q^12), -q^6 - 17*q^9 + 9*q^10 + O(q^12), -q^7 - 8*q^9 + 4*q^10 + O(q^12), -q^8 - 3*q^9 + O(q^12), -q^11 + O(q^12) -] - -So here one would select columns 1,2,3,4,5,6,7,8,11. -Also note the following: Using the Sturm bound and selecting $l \geq \max(n, -\operatorname{SturmBound})$ one can show that the matrix $A$ has a full -$\mathbb{Q}$-rank, i.e. there are $v_1, ..., v_l \in \mathbb{Q}$ -satisfying the above relation but can they be somehow found to be in -$\mathbb{Z}$? -Could someone help me out of this mess? -Best regards, -Fabian Werner - -REPLY [5 votes]: You pose your questions for a general $\Gamma$, but I'm not sure that quite makes sense; in general the Hecke algebra won't be commutative and will have a very different structure, and $S_k(\Gamma, \mathbb{Z})$ won't necessarily span $S_k(\Gamma, \mathbb{C})$. So let's assume $\Gamma$ is $\Gamma_0(N)$ or $\Gamma_1(N)$ for some $N$. -Let $\mathcal{S} = S_k(\Gamma, \mathbb{Z})$ and $\mathcal{T} \subseteq End_{\mathbb{Z}}(\mathcal{S})$ the Hecke algebra. Then (as I think you know) it's not too hard to show that $\mathcal{T} \to Hom(\mathcal{S}, \mathbb{Z})$ and $\mathcal{S} \to Hom(\mathcal{T}, \mathbb{Z})$ are injections. This alone is enough to imply that the pairing is nondegenerate, hence becomes perfect after extending scalars to $\mathbb{Q}$. -So it suffices to show either that $\mathcal{T} \to Hom(\mathcal{S},\mathbb{Z})$ is surjective, or that $\mathcal{S} \to Hom(\mathcal{T},\mathbb{Z})$ is surjective (because the cokernels of these maps are finite groups of the same order, equal to the det of the matrix of the pairing with respect to $\mathbb{Z}$-bases of either side). -You seem to be trying to do the first, but the second is (I think) easier. Let $\phi \in Hom(\mathcal{T},\mathbb{Z})$ and consider the formal power series $f = \sum_{n \ge 1} \phi(T_n) q^n$. It's clear that $f \in \mathbb{Z}[[q]]$; but also that $f \in S_k(\Gamma, \mathbb{Q})$, so $f \in \mathcal{S}$. Clearly we have $(T_n, f) = a_n(f) = \phi(T_n)$ and thus we're done. -Now you can construct a "Miller-like" basis as follows. [EDIT: This doesn't actually work, see comments.] The $T_n$'s generate $\mathcal{T}$, so there is some finite subset which is a $\mathbb{Z}$-basis of $\mathcal{T}$; let these be $T_{n_1}, \dots, T_{n_d}$. Then there is a dual basis $f_1, \dots, f_d$ of $\mathcal{S}$ such that $(T_{n_i}, f_j) = \delta_{ij}$, and there are your Miller-like basis. (You can't necessarily take the $n_i$ to be $\{1, \dots, rk(\mathcal{S})\}$ though, as your example above shows.)<|endoftext|> -TITLE: Closest 3D rotation matrix in the Frobenius norm sense -QUESTION [10 upvotes]: Given a 3 by 3 matrix $M$ I would like to find the rotation matrix $R$ minimizing the Frobenius norm: -\begin{equation} - \|R-M\|_F -\end{equation} -Is there a closed form solution for $R$, or is it possible to express $R$ as the solution to a linear system? I would like to avoid gradient descent if possible. - -REPLY [16 votes]: Let $M=U\Sigma V$ be the singular value decomposition of $M$, then $R=UV$. If you want $R$ to be a proper rotation (i.e. $\det R=1$) and $UV$ is not, replace the singular vector $\mathbf{u}_3$ associated with the smallest singular value of $M$ with $-\mathbf{u}_3$ in the $U$ matrix. An appropriate reference for this answer is: -N. J. Higham. Matrix nearness problems and applications. In M. J. C. -Gover and S. Barnett, editors, Applications of Matrix Theory, pages 1–27. Oxford University Press, 1989.<|endoftext|> -TITLE: Schottky locus in genus 2 -QUESTION [10 upvotes]: Let $\phi_g : \mathcal{M}_g \rightarrow \mathcal{A}_g$ be the period mapping from the open moduli space of genus $g$ Riemann surfaces to the moduli space of $g$-dimensional principally polarized abelian varieties over $\mathbb{C}$. Thus for a Riemann surface $S$ the image $\phi_g(S)$ is the Jacobian of $S$. The Schottky problem consists in determining the image of $\phi_g$. -It is classical that $\text{Im}(\phi_2)$ is exactly the set of abelian varieties that are not isomorphic to a product of elliptic curves. This is asserted in many places, but I have not been able to find a nice discussion of it in the literature. Does anyone know one? The more down-to-earth, the better. - -REPLY [2 votes]: According to MR0364265 (51 #520) Oort, Frans; Ueno, Kenji "Principally polarized abelian varieties of dimension two or three are Jacobian varieties" (J. Fac. Sci. Univ. Tokyo Sect. IA Math. 20 (1973), 377–381) covers this (I don't have access to the paper and am going solely by the review).<|endoftext|> -TITLE: Positive but not completely positive? -QUESTION [16 upvotes]: The only example I know of a positive map which is not completely positive is the transpose map on $M_n(\mathbb{C})$. Of course, one can come up with minor perturbations of this (compose it with, or add it to, a completely positive map, etc). Are there other known examples? - -REPLY [6 votes]: A Jordan homomorphism between C*-algebras is always positive, but it is completely positive only if it is a homomorphism. To get some examples, say $C_r^*(G)$ is the reduced C*-algebra of a discrete group or groupoid $G$. Then the map $g\mapsto g^{-1}$ (a generalized transpose) induces a Jordan homomorphism on $C_r^*(G)$ which is positive but not completely positive.<|endoftext|> -TITLE: Non standard Algebraic Topology -QUESTION [20 upvotes]: Let *$\mathbb R$ a field of non-standard real numbers (or any real closed field) equipped with its natural generalized metric $d(x,y)=|x-y|$. Equip *$\mathbb R^2$ and *$\mathbb R^3$ with the $\ell^1$-(generalized)-metric. - -Question: Does there exist an homeomorphism between *$\mathbb R^3$ and *$\mathbb R^2$? - -Well, this is the simplest subquestion of the most general one - -Question: Is there anybody developing non standard Algebraic Topology? If not, is there any particular reason? - -Thanks in advance, -Valerio - -REPLY [10 votes]: As mentioned in the comments, the actual topology on the non-standard extension can be quite nasty. This is illustrated for example in the first set of problems in these notes. A solution is to replace standard topological notions by definable analogues. Then things mostly work in an arbitrary o-minimal structure. This is also explained in the above notes. -More specifically on algebraic topology in the o-minimal settings, there are several papers by Berarducci and by Edmundo<|endoftext|> -TITLE: Geometric treatment of the Ward-Takahashi identity -QUESTION [6 upvotes]: The quantum field theory generalisation of Noether's theorem about symmetries and conservation laws is the Ward-Takahashi identity. -What is a suitable treatment of this in the context of differential geometry and a modern setting? -Thanks - -REPLY [5 votes]: From a path integral point of view, the Ward-Takahashi identity is a straightforward consequence of the fundamental theorem of calculus. -Let $\delta$ be a vector field on the space $\mathcal{F}$ of fields which expresses the action of the group of gauge symmetries. Suppose that $d\phi$ is a measure which is invariant under these gauge symmetries. The fundamental theorem of calculus tells us that the integral of the total derivative -$\int_{\mathcal{F}} \delta(g) d\phi = 0,$ -for any $g$ such that the contribution from the boundary of $\mathcal{F}$ is $0$. -In the special case where $g$ is the product $\mathcal{O}e^{-S}$ of a gauge-invariant observable and the exponential of a gauge-invariant action, we get a constraint on the path integral (expectation with respect to $e^{-S(\phi)}d\phi$): -$0 = \int_{\mathcal{F}} \delta(\mathcal{O}(\phi) e^{-S(\phi)})d\phi = \int_{\mathcal{F}}\bigl[\delta(\mathcal{O}) - \mathcal{O} \delta(S) \bigr]e^{-S(\phi)}d\phi = \langle \delta \mathcal{O}\rangle - \langle \mathcal{O} \delta(S) \rangle.$ -If we use the fact that the variation of the action is the spacetime integral of the divergence of the current $\delta(S) = \int_\Sigma \nabla \dot{} J$, then we get the usual Ward-Takahashi identity: -$\langle \delta \mathcal{O}\rangle = \int_\Sigma \langle \mathcal{O} \nabla \dot{} J\rangle.$ -Everything I've written here is only obviously true for finite-dimensional integrals. However, any path integral should be arbitrarily well approximated by such finite-dimensional integrals, so one can at least hope to transport the final result, even if the steps themselves can not be carried over.<|endoftext|> -TITLE: Decomposability of Hausdorff measure -QUESTION [10 upvotes]: Consider $s$-dimensional Hausdorff measure $\mathcal{H}^s$ on the Borel sets in $\mathbb{R}^n$. -$\mathcal{H}^s$ is not $\sigma$-finite if $s < n$, but it is semifinite (on Borel sets!) -Is it known whether $\mathcal{H}^s$ can be decomposable, i.e. can there be a partition of $\mathbb{R}^n$ into disjoint Borel sets $\{X_i:i\in I\}$ ($I$ necessarily uncountable) such that $\mathcal{H}^s(X_i)<\infty$ for all $i$ and, for every Borel set $E$, $\mathcal{H}^s(E)=\sum\limits_{i\in I}\mathcal{H}^s(E\cap X_i)$? Does the answer depend on any set-theoretic assumptions? - -REPLY [4 votes]: There are cardinality $c$ Borel sets of finite ${\cal H}^s$ measure in ${\mathbb R}^n$. Assuming the Continuum Hypothesis, well-order these by the first uncountable ordinal as $E_\alpha$ for $\alpha \in A$, and define $X_\alpha = E_\alpha \backslash \bigcup_{\beta < \alpha} E_\beta$. Then $X_\alpha$ are disjoint Borel sets of finite ${\cal H}^s$ measure. For any Borel set $E$ of finite ${\cal H}^s$ measure, $E = E_\alpha = \bigcup_{\beta \le \alpha} (E \cap X_\beta)$ for some $\alpha$, -and ${\cal H}^s(E) = \sum_{\beta \le \alpha} {\cal H}^s(E \cap X_\beta)$ with $E \cap X_\beta = \emptyset$ for all $\beta > \alpha$. -Hmm, I'm not sure about the case ${\cal H}^s(E) = \infty$: I think it's true that $E$ has Borel subsets of arbitrarily large finite ${\cal H}^s$ measure, but I don't have a reference at hand and can't think of a proof. If this is the case, then the result is true.<|endoftext|> -TITLE: Bending an elastic, inextensible sheet of paper into a teardrop shape -QUESTION [6 upvotes]: I take an rectangular sheet of paper, of height $H$ and Young's Modulus $E$, and in the absence of gravity, I bend it into a "teardrop" shape so that the edges along the top and bottom touch only along a single line. -What analytic function describes this "teardrop" shape? - -REPLY [8 votes]: All in here (if using Euler Bernoulli beam theory) -http://sci-toys.com/bent_paper_problem.pdf - -(Added by Joseph O'Rourke). Here is Fig.1 from the paper by Antoni Colom, - "Analysis of the shape of a sheet of paper when two opposite edges are joined," PDF link above.<|endoftext|> -TITLE: The consistency of Martin's Axiom -QUESTION [10 upvotes]: In learning about the Consistency of Martin's Axiom through Kunen and Jech with help from other set theorists, I have come to a basic question about marrying these proofs: -What is the connection between the nice names argument in Kunen and the boolean valued proof in Jech regarding a limit on the number of partial order-names for a partial order which satisfies the countable chain condition? -I am more familiar with the nice names argument and trying to understand why every boolean value of an element of a set is determined by a countable anti chain in the partial order. I can't see how values in the boolean algebra are boolean sums of the values of an anti-chain in the partial order. Any help to understanding the link would be appreciated. - -REPLY [3 votes]: I like to think about the matter topologically. -Let $X=MF(P)$ be the set of all maximal filters of $P$. -Give $X$ the topology generated by the sets -$N_p=\{U\in X: p\in U\}$. Declare $B(P)$ to be -the algebra of regular open subsets of $X$ (that is, -sets equal to the interior of their closures). -Given a regular open set $R$, pick a maximal family -$A$ of pairwise disjoint subsets of $R$ of the form -$N_p$ for some $p\in P$. The union of $A$ is dense -in $R$, so $R$ is the smallest regular open set -containing this union. In other words, every element -of $B(P)$ is the sum of an antichain from $P$.<|endoftext|> -TITLE: The difference between a handle decomposition and a CW decomposition -QUESTION [29 upvotes]: Let $M$ be a compact finite-dimensional smooth manifold. I have a question about the relationship between the statements that a Morse function induces a handle decomposition for $M$, and that it induces a CW decomposition for $M$. -A Morse function induces a handle decomposition -Denote by $X(M;f;s)$ the manifold $M$ with an $s$--handle attached by $f\colon\,(\partial D^s)\times D^{n-s}\to M$. - -Theorem: Let f be a $C^\infty$ function on $M$ with no critical points on $f^{-1}[-\epsilon,\epsilon]$ except $k$ nondegenerate ones on $f^{-1}(0)$, all of index $s$. Then $f^{-1}[-\infty,\epsilon]$ is diffeomorphic to $X(f^{-1}[-\infty,-\epsilon];f_1,\ldots,f_k;s)$ (for suitable fi). - -Historical note: This was stated by Smale in 1961, with proof outline. Milnor's Morse Theory, Theorem 3.2 states and proves a weaker, homotopy version of the theorem, where there is only one handle in play. I asked about a proof of this theorem in this MO question, and it turned out that the first complete proof appeared in Palais, simplified lated by Fukui [Math. Sem. Notes Kobe Univ. 3 (1975), no. 1, paper no. X, pp. 1-4]. There's an alternative proof given in Appendix C to Madsen-Tornehave. -Discussion: Roughly, the theorem states that passing a critical point of a Morse function corresponds to attaching a handle. Thus, a Morse function induces a handle decomposition for $M$. -A Morse function induces a CW decomposition -Let $f$ be a Morse function on $M$. Choosing a complete Riemannian metric on $M$ determines a stratification of $M$ into cells $D(p)$ (the unstable (descending) manifold for a critical point $p$ of $f$) in which two points lie in the same stratum if they are on the same unstable manifold. Each $D(p)$ is homeomorphic to an open cell, but the closure $\overline{D(p)}$ can be complicated. - -Theorem: The union of compactified unstable manifolds $\bigcup \overline{D(p)}$ gives a CW decomposition of $M$ that is homeomorphic to $M$. - -Historical note: A nice discussion of this theorem may be found in Bott's excellent Morse Theorem Indomitable, page 104. Milnor's Morse Theory derives a homotopy version of this statement (Theorem 3.5) from the homotopy version of the statement that a Morse function induces a handle decomposition (Theorem 3.2). The theorem seems to have been first proven by Kalmbach, and was recently strengthened to give the explicit characteristic maps by Lizhen Qin (understanding his papers is the motivation for my question). - -The two statements given above look to me as though they should be very similar, especially since the homotopy version of the second follows directly from the homotopy version for the first in Milnor's book. But briefly searching through the literature makes it seem that they are virtually independant- papers proving one aren't even cited in papers proving the other, and the proofs look to me to be entirely unconnected. I don't understand why, probably because I'm having difficulty breaking free from the intuitive picture of the proof in Milnor's book, which works fine up to homotopy. - -Question: Can you give an example, or intuition, for a case in which one of the above theorems is difficult but the other is easy? Is there an example for a compact finite-dimensional manifold with a Morse function such that the handlebody decomposition can be read straight off the Morse function, but the reading off the CW decomposition takes substantial extra work? Or the converse? -Stated differently, where does the "up to homotopy proof" on page 23 of Milnor conceptually collapse when we are working up to diffeomorphism instead of up to homotopy? - -REPLY [3 votes]: Another reference I would like to mention is -Sharko, V.V. Functions on manifolds, Translations of Mathematical - Monographs, Volume 131. American Mathematical Society, Providence, RI (1993). -Algebraic and topological aspects, Translated from the Russian by V. - V. Minachin. -He really does use some aspects of the crossed complex related to the handlebody decomposition, rather than the CW-filtration. We have suggested this as a line of possible development in our book "Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids" (EMS Tract 15, 2011) since we work there with filtered spaces, and you get such from a Morse function on a manifold. I feel there is more to do there, using for example the tensor product technology explained in our book. This tensor product does reflect the usual cell decomposition of the product $E^m \times E^n$, $m,n \geqslant 0$, where $E^m, E^n$ have cell decompositions with 3 cells.<|endoftext|> -TITLE: Convolution algebras for double groupoids? -QUESTION [15 upvotes]: There is a lot of work of course on convolution algebras of measured groupoids, and this gives "Noncommutative geometry". However there is a lot of interest in algebraically structured groupoids, for example groupoids internal to the categories of say groups, or groupoids, or Lie algebras. Thus double groupoids, i.e. groupoids internal to groupoids, can be seen as "more noncommutative" than groupoids. They are also quite difficult to understand in general, though special cases have been studied extensively, e.g. 2-groupoids, and what are called 2-groups (groupoids internal to groups). They all have relations with crossed modules. -How does one find then a jacking up of Noncommutative Geometry to take into account these algebraically structured groupoids? -The usual formula for a convolution of functions $f,g$ on a finite groupoid $G$ given by -$$(f*g)(z) = \sum _{xy=z} f(x)g(y)$$ -can be extended to a kind (or rather many kinds!) of matrix convolution looking at all decompositions of $z$ as a matrix composition (I find it difficult to write this down in this system!) but it all depends on the size $m \times n$ of the matrix, and because one needs the interchange law it all gets complicated and not reducible to the individual compositions in the double groupoid. -If this could be done, it might open up new worlds! - -REPLY [2 votes]: Pedro Resende understands this well. The interchange law in a double algebra (defined by Resende) is not satisfied by a double groupoid convolution algebra but I think that doesn't necessarily mean that a category fibred over the double groupoid is not a double category. So you can drop the interchange law, well at least that is what we considered doing. In the end it seemed that the idea of a weak Hopf algebra by Natale and Andruskiewitsch was the best approach! (So there is already a counterpart of a Hopf algebra for a group coming from coproduct if not a counterpart of a group convolution algebra coming from product.)<|endoftext|> -TITLE: The Continuum Hypothesis and Countable Unions -QUESTION [21 upvotes]: I recently edited an answer of mine on math.SE which discussed the implication of the two assertions: - -$AH(0)$ which is $2^{\aleph_0}=\aleph_1$, and -$CH$ which says that if $A\subseteq 2^{\omega}$ and $\aleph_0<|A|$ then $|A|=2^{\aleph_0}$. - -We know they are indeed equivalent under the axiom of choice (and actually much less). It is also trivial to see that $AH(0)\Rightarrow CH$. However the converse is not true, indeed in Solovay's model (or in models of AD) there are no $\aleph_1$ many reals, but $CH$ holds since every uncountable set of reals has a perfect subset. -While revising my answer I tried to find a reference whether or not in the Feferman-Levy model, in which the continuum is a countable union of countable sets, satisfies the continuum hypothesis (we already know that it does not satisfy $AH(0)$). -To my surprise the answer is negative. There exists a set whose cardinality is strictly between the continuum and $\omega$, the construction is described in A. Miller's paper [1] in which he remarks that in the Feferman-Levy the constructed set cannot be put in bijection with the continuum. -I was wondering whether or not this is always true in models in which the continuum is a countable union of countable sets, or is this just one of the peculiarities of the Feferman-Levy model. - -Questions: - -Let $V$ be a model of $ZF$ in which $2^{\omega}$ can be written as the countable union of countable sets. Does $CH$ fail in $V$? - -Suppose that $V$ is a model of $\omega_1\nleq2^\omega$ and $CH$, does this imply that $\omega_1$ is regular (which means inaccessible in $L$)? - - - - -Bibliography: - -Miller, A. A Dedekind Finite Borel Set. Arch. Math. Logic 50 (2011), no. 1-2, 1--17. - -REPLY [7 votes]: The answer for the second question is no. Truss proved in [1] that if we repeat Solovay's construction from a limit cardinal $\kappa$, we obtain a model in which the following properties: - -Countable unions of countable sets of real numbers are countable; -Every well-orderable set of real numbers is countable; -Every uncountable set of reals has a perfect subset; -DC holds iff $\omega_1$ is regular iff $\kappa$ is inaccessible in the ground model; -Every set of real numbers is Borel. - -This shows that it is possible to have $CH+\aleph_1\nleq2^{\aleph_0}+\operatorname{cf}(\omega_1)=\omega$. However it does not answer the original (first) question. - -Bibliography: - -Truss, John, Models of set theory containing many perfect sets. Ann. Math. Logic 7 (1974), 197–219.<|endoftext|> -TITLE: What is a pointed Hopf algebra? -QUESTION [7 upvotes]: Hi, -I would like to know what pointed Hopf algebras are and why it is that they are important. -Thank you. - -REPLY [15 votes]: While I don't deem the question "what is a pointed Hopf algebra" appropriate, I sympathize with the second one. Back when I was attending a Hopf algebra course, this was exactly my question, and I didn't obtain a good (for me!) answer to it until I studied combinatorial Hopf algebras. -Many Hopf algebras that appear in combinatorics (the tensor and shuffle Hopf algebras of a vector space, as well as the Hopf algebras $\mathbf{Symm}$, $\mathbf{QSymm}$, $\mathbf{NSymm}$, Loday-Ronco, Malvenuto-Reutenauer, trees, ordered trees, ...) are naturally connected graded or at least connected filtered (like the universal enveloping algebra of a Lie algebra). There is a lot to be said about this kind of Hopf algebras. Most importantly, their "connected filtered" property helps proving things about them; it more or less gives us a way to proceed by induction over the degree. -However, at one moment, non-connected graded and filtered Hopf algebras started to appear in combinatorics: e.g., the $\widetilde{\mathcal H}_{\mathcal T}$ in Dominique Manchon's "Hopf algebras, from basics to applications to renormalization". Just knowing that they are graded does not help (we could have the whole Hopf algebra concentrated in degree $0$). So we need some other condition that these Hopf algebras satisfy. One such condition (which is trivially satisfied in the case of $\widetilde{\mathcal H}_{\mathcal T}$) is that the $0$-th part of the grading (or, more generally, of the filtration) is spanned by grouplike elements. In other words, every simple subcoalgebra of the $0$-th part of the filtration is $1$-dimensional. (Unfortunately, this condition is not geometric, i. e., it can change under an algebraic extension of the ground field. But it is a good start, and for the cases that appear in combinatorics, it holds over any field, because the grouplike elements usually belong to the combinatorial basis.) -Now assume you are given just a random Hopf algebra without filtration. Can you use any of these things that you have proven for connected filtered Hopf algebras? Well, you can canonically define a filtration on it (at least over a field!), the so-called coradical filtration. If the coradical $C_0$ (the $0$-th part of the filtration) is $1$-dimensional, then your Hopf algebra becomes connected filtered, and you have won. Hopf algebras like this are said to be irreducible. If the coradical $C_0$ is spanned by grouplike elements, then you are in the second case discussed above, and the Hopf algebra is said to be pointed. -It is easy to see (using Artin-Wedderburn) that cocommutative Hopf algebras over an algebraically closed field are pointed (while clearly not being always irreducible!), so "pointed" (or, more precisely, "pointed after base change to the algebraic closure") is a kind of weakening of "cocommutative", and you can try to generalize everything you know about cocommutative Hopf algebras to the pointed case: e. g., Cartier-Milnor-Moore becomes Cartier-Kostant.<|endoftext|> -TITLE: Knot Invariants from Twisted Quantum Doubles -QUESTION [12 upvotes]: In "Topological Gauge Theories and Group Cohomology", Dijkgraaf and Witten construct a 3-manifold invariant from a finite group $G$ and 3-cocycle $\omega$. I would think there is also an associated knot invariant, coming in particular from the twisted quantum double $D^\omega(G)$, but haven't been able to find a place where it is explicitly described or computed. -If it exists, could someone point me to the right reference? I would be interested in both a general construction and a specific example, for instance computing it in the case of $\mathbb{Z}_n$. - -REPLY [4 votes]: You may be interested in this paper by Chen. -One of the simplest nontrivial knot invariants is the number of "Fox n-colorings", which count homomorphisms of the knot group to a dihedral group $D_{2n}$ sending the meridians to reflections. This count can be interpreted as a value in a Dijkgraaf-Witten theory on the knot complement. Chen's paper computes those invariants in the nontwisted case in terms of the representation category of the quantum double $D(D_{2n})$ and gives a nice formula for the number of colorings of a rational link. -I tried computing values of the twisted invariant, but it turns out that these are equal to the untwisted invariants times a power of $i$ depending on the writhe. -It may be possible that something more intersting happens with a more complicated group: $D_{2n} = \mathbb{Z}_2 \ltimes \mathbb{Z}_n$ is the simplest possible case where you get interesting homomorphism counts, and I'd be curious what happens in the twisted (and untwisted) case for other semidirect products. In particular, the twisting that actually affects the DW invariants seems to mostly involve the $\mathbb{Z}_2$ part, which has few interesting cocycles.<|endoftext|> -TITLE: What sort of structure can amorphous sets support? -QUESTION [7 upvotes]: Assuming the Axiom of Choice, every cardinal is either finite (i.e., an element of $\omega$) or Dedekind-infinite (i.e., in bijection with a proper subset of itself). This dichotomy is not true in ZF, however; in particular, it is consistent with ZF that there exist sets which are non-finite but cannot be partitioned into two non-finite pieces. Such sets are called "amorphous," and models of ZF containing amorphous sets can be constructed by building a permutation model of ZFA (ZF with urelements) and applying the Jech-Sochor embedding theorem. -Naturally, there are lots of types of structure which amorphous sets simply cannot have. For example, no amorphous set can be linearly ordered, since a linear order on a non-finite set allows us to either inject $\omega$ into the set, or partition the set into a "left" and "right" piece, both of which are non-finite. However, amorphous sets can still have some structure. For example, say that a set is "even" if it can be written as a disjoint union of 2-element subsets. Then, again using permutation models, we can construct a model of ZF which contains an even amorphous set (the lack of choice prevents us from using the evenness of the set to partition it into two non-finite pieces). -My question is twofold. First, say that a set is "odd" if it can be written as the disjoint union of a singleton and an even set. -Question 1: Is ZF+"there exists an amorphous set"+"every set is either even or odd" consistent? (It occurs to me that this may depend on whether the above "or" is meant inclusively or exclusively.) -My main issue with this question is that I don't see how such a model would be constructed. For example, we can use a permutation model to create a single amorphous even (or odd) set, but how do we then conclude that every set is either even or odd? I don't know how to do this at all, so this motivates my second question: -Question 2: How does one build models of ZF+"there exists an amorphous set"+"every set is ---," in general? (Where --- denotes some arbitrary niceness property.) - -REPLY [5 votes]: First I'll remark that for the first question, note that the or cannot be exclusive since $\omega$ is both even and odd. -Now to take some definitions from [1] if $A$ is an amorphous set, let $U$ be a partition of $A$ into infinitely many parts. It follows that every $A\in U$ is finite, and all but finitely many have the same size. Let $n$ be that size (Truss calls this gauge of $U$). -We say that $A$ is strictly amorphous (also known as strongly amorphous) if there are no partitions with gauge $>1$; we say that $A$ is bounded amorphous if there is a finite bound on the possible gauges; and unbounded amorphous otherwise. -Some remarks about the first question: -Note that if $A$ is strictly amorphous then it is neither odd nor even. Also note that if $A$ is bounded amorphous (say by gauge $n$) then taking $U$ a partition of $A$ into parts of size $n$ and perhaps discarding a few elements gives us that $A/U$ is strictly amorphous: -Otherwise we could partition $A/U$ into pairs (or more) and the union of each part would give a partition of $A$ whose gauge is $>n$. -Therefore, to answer this question one first has to assure that no bounded amorphous set exists. If no bounded amorphous exists (which is consistent) then every amorphous set is either odd or even (depending on the partitions with gauge $2$). -There are a handful of examples to answer the second question in the paper. -I'll add a final remark that a bounded amorphous cannot have a group structure, but an unbounded amorphous can. In particular it is possible to have vector spaces which are amorphous sets (see my answers here and here). - -Bibliography: - -J.K.Truss, The structure of amorphous sets, Annals of Pure and Applied Logic, 73 (1995), 191-233.<|endoftext|> -TITLE: A "couple" of questions on Gauss's mathematical diary -QUESTION [19 upvotes]: Throughout my upbringing, I encountered the following annotations on Gauss's diary in several so-called accounts of the history of mathematics: - -"... A few of the entries indicate that the diary was a strictly private affair of its author's (sic). Thus for July 10, 1796, there is the entry -ΕΥΡΗΚΑ! num = Δ + Δ + Δ. -Translated , this echoes Archimedes' exultant "Eureka!" and states that every positive integer is the sum of three triangular numbers—such a number is one of the sequence 0, 1, 3, 6, 10, 15, ... where each (after 0) is of the form $\frac{1}{2}n(n+1)$, $n$ being a positive integer. Another way of saying the same thing is that every number of the form $8n+3$ is a sum of three odd squares... It is not easy to prove this from scratch. -Less intelligible is the cryptic entry for October 11, 1796, "Vicimus GEGAN." What dragon had Gauss conquered this time? Or what giant had he overcome on April 8, 1799, when he boxes REV. GALEN up in a neat rectangle? Although the meaning of these is lost forever the remaining 144 are for the most part clear enough." - " - -The preceding paragraphs have been quoted verbatim from J. Newman's The World of MATHEMATICS (Vol. I, pages 304-305) and the questions that I pose today were motivated from my recent spotting of [2]: - -Why is there no mention whatsoever to the REV. GALEN inscription in either Klein's or Gray's work? -What is the reason that E. T. Bell expressed that Gauss had written the Vicimus GEGAN entry on October 11, 1796? According to Klein, Gray, and (even) the Wikipedians it was written on October 21, 1796. As far as I understand, Klein and Gray are just reporting the dates that appear on the original manuscript. Did Bell actually go over it? -Last but not least, is there a compendium out there of all known potential explanations to the Vicimus GEGAN enigma? The only ones whereof I have notice can be found on page 112 of [1]: - - -"... Following a suggestion of Schlesinger [Gauss, Werke, X.1, part 2, 29], Biermann ... proposed that GA stood for Geometricas, Arithmeticas, so reading GEGAN in reverse as Vicimus N[exum] A[rithmetico] G[eometrici cum] E[xspectationibus] G[eneralibus]. Schumann has since proposed other variants; including, for GA, (La) G(rangianae) A(nalysis)..." - -Heartfelt thanks for your comments, reading suggestions, and replies. -References - -J. J. Gray. " A commentary on Gauss's mathematical diary, 1796-1814, with an English translation". Expo. Math. 2 (1984), 97-130. -F. Klein. "Gauß' wissenschaftliches Tagebuch 1796–1814". Math. Ann. 57 (1903), 1–34. -M. Perero. Historia e Historias de Matemáticas. Grupo Editorial Iberoamérica, 1994, pág. 40. - -REPLY [4 votes]: I believe I have a reference for Schuhmann: -Schuhmann E. 1976 Vicimus GEGAN, Interpretationsvarianten zu einer Tagebuchnotiz von C.F. Gauss, Naturwiss. Tech. Medezin. 13.2, 17-20.<|endoftext|> -TITLE: Existence, uniqueness, and regularity for linear parabolic PDE on a complete Riemannian manifold -QUESTION [12 upvotes]: Let $M$ be a smooth manifold with a complete Riemannian metric $g$ and $E$ a smooth vector bundle over $M$ with an inner product and compatible connection $\nabla$. Let $K: E \rightarrow E$ be a smooth bundle map. Then I believe the following is true: -Given any smooth compactly supported section $u_0$ of $E$, there exists a unique solution $u: [0,\infty) \times M \rightarrow E$ of the linear heat equation -$$ -\partial_t u = g^{ij}\nabla_i\nabla_j u + Ku -$$ -such that $u$ is smooth for $t > 0$ and the $L^p$ norm of $u(t,\cdot)$ is bounded for each $t \ge 0$ and $1 \le p \le \infty$. -If I'm wrong, please explain. Otherwise, is there a reference for this somewhere? If not the exact theorem, a proof of a similar theorem that can be adapted to a proof of this? -ADDED: I'm mostly interested in proving the existence statement and preferably using a standard PDE approach. It appears to me that there is a straightforward argument starting by approximating the equation by the standard constant coefficient heat equation on a sufficiently small co-ordinate chart and patching together local solutions to the constant coefficient equation to get a global section. You then use a priori estimates to show that the global section is an approximate solution to the original system. This can then be iterated into an exact solution. But the details are a bit tedious, so it would be nice if I could find it all written down somewhere. And it would be nice to be clear about what, if any, assumptions on the Riemannian metric are needed to make this all work. -ADDED: I'm being a little slow here. Since everything is linear and homogeneous, it suffices to prove existence for initial data supported on a sufficiently small co-ordinate chart. If you start with compactly supported initial data, then you can always write it as a finite sum, where each term is supported in a sufficiently small open set. -ADDED: I took another quick look at a partition of unity argument for reducing the question to solving a parabolic PDE on an open set in $R^n$. It looks like this really does work pretty easily. But it still seems like something that someone should write up carefully and publish in a book somewhere. - -REPLY [2 votes]: I've played with such things 15 years ago. Here's what I remember... -Sure you need some semi-boundedness condition on K. Uniqueness should then follow via semigroup domination (aka Kato's inequality) from scalar case. -I. Shigekawa, L^p Contraction Semigroups for Vector Valued Functions, J. Funct. Analysis 147 (1997), 69-108 -Shigekawa had rediscovered Barry Simon's semigroup domination criterion (i.e. Kato's inequality) and generalized it to the vector valued things. -See also: -R.S. Strichartz, Analysis of the Laplacian on the Complete Riemannian Manifold, J. Funct. Analysis 52 (1983), 48-79. Scalar case: Theorem 3.5 - but has uniqueness only for 1< p<∞. -(Some details in Strichartz' proof(s) for vector valued Laplacians not optimal (classic problems with doing tensor calculus and Stokes...) not even in the follow-up paper "L^p contractive projections and the heat semigroup for differential forms" J. Funct. Analysis, 65 (1985), 348-357) -Perhaps look also for El Maati Ouhabaz ca. 1999. He's written a book, "Analysis of Heat Equations on Domains" (2004) but I never got hold of it.<|endoftext|> -TITLE: Have any publications been made in this area of group theory? -QUESTION [5 upvotes]: For a group $G$ and a tuple $J = (g_1,g_2 ... g_n) \in G^k$ for $k$ some constant, define a parametrized word $w : G^k \rightarrow G$ to be a function which takes $J$ to some product of the elements in $J$. -So $w(J) = g_1g_1g_2$ for $k \geq 2$ would be an example. -The structure of the space of all $w$ for a particular group modulo the equivalence relation of functional equality is not trivial. -For instance, over $\mathbb{Z}_2$, $g_1g_1g_2 = g_2$ for all $J$ ,and for a finite abelian group the space of $w$ is clearly finite. -I don't know whether this topic has been covered before; It seems simple enough that someone might have done work on it, but I cannot find anything. Does anyone know what this area might be called? - -REPLY [3 votes]: From comments it seems you want to know when your group/monoid of functions is finite for all k. These are equivalent and the answer is when G is locally finite of finite exponent. The group you are looking at is the free group of rank k in the variety generated by G. If G has infinite exponent the words $x^n$ are all distinct functions on G. -If G is of exponent n and is not locally finite then it has an infinite k-generated subgroup so the free group in the variety generated by G is infinite. If G is locally finite, then since varieties of groups are closed under direct limits, it follows G belongs to the variety of groups generated by finite groups of exponent n. This variety has finite free objects on finite generating sets by the solution to the restricted Burnside problem. Thus the free objects in the variety generated by G are finite as well. -If G is finite things are trivial since there are finitely many k-ary functions on G. In fact it is a classical result of Birkhoff shows the variety generated by a finite universal algebra is locally finite.<|endoftext|> -TITLE: What is the obstruction for a local set of models of a curve to come from a global model? -QUESTION [5 upvotes]: If $X_{\mathbb{Q}}$ is a curve over $\mathbb{Q}$, we get a curve $X_{\mathbb{Q}_p}$ over $\mathbb{Q}_p$ for every prime $p$. -My question is about the reverse process. Say we are given curves $X_{\mathbb{Q} _ p}$ for every $p$ such that $X_ { \bar {\mathbb{Q}} _ p} \cong X_ {\bar{ \mathbb{Q}} _ q}$ for any two primes $p$ and $q$ (where the isomorphism is as schemes; alternatively, they are isomorphic when base-changed to an algebraically closed field that contains both $\mathbb{Q}_p$ and $\mathbb{Q} _ q$). Then is there a nice way to describe the obstruction for these models to come from a global curve $X_{\mathbb{Q}}$ over $\mathbb{Q}$? - -REPLY [2 votes]: I think the following interpretation of your question is false for cardinality reasons. It occurred to me before I saw Donu's answer but seems to have a similar flavour. -I will say that two curves over a field are twists of each other if they become isomorphic over an algebraic closure. -Interpretation. Fix a curve $C$ over $\mathbf{Q}$. For every place $v$ of $\mathbf{Q}$, let $X_v$ be an arbitrary (random) twist of $C_v$. Does there exist a curve $D$ over $\mathbf{Q}$ such that $D_v$ is isomorphic to $X_v$ for every $v$ ? -In genus $0$, not every such family can come from a conic over $\mathbf{Q}$, as Felipe has remarked. -In genus $g>0$, if we start with a hyperelliptic curve $C$, then we get uncountably many families $(X_v)_v$ by taking random quadratic twists at each place $v$. All these families satisfy your hypotheses, and some cannot come from the countably many genus-$g$ curves over $\mathbf{Q}$. -Examples. Suppose we have an elliptic curve $E$ over $\mathbf{Q}$ which is the only curve in its isogeny class, and assume moreover that Ш$(E)$ is trivial. It follow that if a genus-$1$ curve $C$ is such that its jacobian $J$ is isomorphic to $E$ almost everywhere locally, then $J$ is $\mathbf{Q}$-isogenous to $E$, and hence $\mathbf{Q}$-isomorphic to $E$, and hence $C$ is isomorphic to $E$ (because Ш$(E)$ is trivial). -Construct the family $(X_v)_v$ by taking $X_v=E_v$ for every place $v\neq2$ of $\mathbf{Q}$, and perversely take $X_2$ to be a quadratic twist of $E_2$, so that $X_2$ and $E_2$ are not $\mathbf{Q}_2$-isomorphic. -If there were a genus-$1$ curve $C$ such that $C_v$ is $\mathbf{Q}_v$-isomorphic to $X_v$ at every $v$, then $C$ would have to be $\mathbf{Q}$-isomorphic to $E$ (by the choice of $E$, as explained above), which is impossible because they are not $\mathbf{Q}_2$-isomorphic. -It remains to find such an $E$. I'm sure this can be done by looking at the tables made by Cremona and Stein. Could someone please confirm this hunch ?<|endoftext|> -TITLE: Embedding torus in space such that its 6-fold symmetry extends -QUESTION [13 upvotes]: The following question is Problem 1.1.2.c in Thurston's book "Three-dimensional geometry and topology". I have not managed to solve it despite quite a bit of effort. -One can obtain a 2-dimensional torus $T$ by identifying the sides of a hexagon in an appropriate way (see, for example, here). By rotating this hexagon, we can obtain an order $6$ self-map of $T$. The question is whether we can embed $T$ into either $\mathbb{R}^3$ or $S^3$ such that this self-map extends to an order $6$ self-map of the ambient space. My guess is that the answer is "no", at least for $\mathbb{R}^3$. I'm less sure about $S^3$. -Thanks! - -REPLY [17 votes]: Lurking here on MO, I've noticed that unanswered questions get bumped to the top periodically. Since this question was answered by Ryan Budney in the comments, I've decided to write his answer here (marked "community wiki" so I get no reputation points) to prevent this from happening. -The answer is no for both $\mathbb{R}^3$ and $S^3$. I'll give the details for $\mathbb{R}^3$; the other case is similar. Fix an embedding $T^2 \hookrightarrow \mathbb{R}^3$. The first step is to show that $T^2$ is the boundary of a closed regular neighborhood $N$ of a knot. This is a nontrivial fact; for an exposition, see for example this. The space $N$ is a solid torus, and thus up to homotopy there exists exactly one simple closed curve $\gamma$ in $T^2$ which bounds a disc in $N$. Any homeomorphism $\phi : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ which preserves $T^2$ must take $\gamma$ to a curve on $T^2$ homotopic to $\gamma$. In other words, the restriction of $\phi$ to $T^2$ must fix a nonzero vector in $H_1(T^2;\mathbb{Z})$. But the automorphism in the question fixes no such vector.<|endoftext|> -TITLE: A simple question about the degree of some vector bundle over rational curve. -QUESTION [5 upvotes]: Let $E$ be a holomorphic vector bundle (infact complex vector bundle is enough) over $\mathbb{P}^1$. Let $c: \mathbb{P^1} \rightarrow \mathbb{P^1}$ be the anti-holomorphic involution, $c(z)=\frac{-1}{\bar{z}}$, and after all suppose we have a commutative diagram (I couldn't draw it here, imagine it yourself) -$\pi: E \rightarrow \bar{E} $ -where $\pi$ is an anti-holomorphic (or anti-complex linear) involution covering $c$. In this situation, is it true that $c_1(E)$ should be even! - -REPLY [12 votes]: Complex curves with anti-holomorphic involutions correspond to real algebraic curves. Your involution has no fixed points, so your curve corresponds to a real curve $C$ of genus 0 with no real points (there is only one). Equivariant vector bundles come from vector bundles on $C$, so the first Chern class of your equivariant vector bundle comes from a 0-cycle on $C$, hence it is enough to see check that $C$ only has 0-cycles of even degree. This is easy: suppose that it had a cycle of odd degree. Since it is has a point of degree 2, it would have a cycle of degree 1; by Riemann-Roch, this would correspond to a real point of $C$, which does not exist.<|endoftext|> -TITLE: uniqueness of laplace's equation applied to electrostatics -QUESTION [11 upvotes]: Does any one know when it was first proved that a conducting surface of any shape, not just a sphere with no charges inside it, but arbitrary charges outside it, has no total charge inside it? As late as 1925 one sees vague arguments to this effect, but only much later the argument using uniqueness of solutions of Laplace's equation for given boundary values. - -REPLY [8 votes]: I should apologize for placing this question here, because I've just found the answer. -It was George Green (of Green's functions); it occurs on page 49 of his mathematical papers, originally published in 1828.<|endoftext|> -TITLE: spin structures on full flag manifolds -QUESTION [5 upvotes]: It is known that any full flag manifold $G/T$ is a spin manifold. -For example, we can prove it using that $G/T$ is a complex manifold, -by computing its 1st Chern class as follows: -For full flag manifolds we have that the first Chern class is given by -$c_{1}(G/T)= 2\delta_{G} \cdot $ generator of $H^{2}(G/T, \mathbb{Z})$, where $$\delta_{G}=\frac{1}{2}\sum_{\alpha\in R^{+}}\alpha.$$ -Here $R^{+}$ are the postive roots of $G$. -It is well known that $\delta_{G}=\Lambda_{1}+\cdots+\Lambda_{\ell}$, where $\Lambda_{i}$ are the fundamental weights and $\ell=\dim T= {\rm rank}G$. -Then -$$ c_{1}(G/T)= 2\delta_{G} \cdot g =2(\Lambda_{1}+\cdots+\Lambda_{\ell}) \cdot g,$$ where $g\in H^{2}(G/T, \mathbb{Z})$ is the generator. Therefore -$c_{1}(G/T)$ is even and since for a complex manifold $M$ the second Whitney class $w_{2}(M)$ -is the reduction $\mod 2$ the first Chern class $c_{1}(M)$, we conclude that -$$w_{2}(G/T)=0 \ \ thus \ \ G/T \ \ is \ a \ spin \ manifold.$$ -Question How can we give a proof of this fact by a different way, for example, by using arguments from homology or cohomology theory? - -REPLY [7 votes]: $G/T$ is a co-adjoint orbit in $\mathfrak g^*$. The normal bundle to the inclusion $G/T\rightarrow \mathfrak g^*$ is trivial, so the tangent bundle of $G/T$ is stably trivial. This implies its Stiefel-Whitney and Pontryagin classes vanish. -(Argument stolen from Dan Freed "Flag Manifolds and Infinite Dimensional Geometry" in MSRI proceedings volume "Infinite Dimensional Groups with Applications".)<|endoftext|> -TITLE: Non congruence subgroups containing congruence subgroups. -QUESTION [5 upvotes]: Does there exist Fuchsian groups, which is not conjugated in $SL(2, \mathbb{R})$ to a subgroup of $SL(2, \mathbb{Z})$, but still contains a congruence subgroup? - -REPLY [6 votes]: How about the normalizer of $\Gamma_0(N)$ in $PSL_2(\mathbb{R})$ for $N > 1$, which is the subgroup generated by $\Gamma_0(N)$ and $\begin{pmatrix} 0 & -1 \\ N & 0 \end{pmatrix}$? -(I wouldn't call these "non-congruence subgroups" as you do in your title: that technical term is usually parsed as "(non-congruence) subgroup". This isn't a subgroup of $PSL_2(\mathbb{Z})$ at all, but it certainly contains a principal congruence subgroup; perhaps we should call it a "congruence non-subgroup"?)<|endoftext|> -TITLE: Homology of infinite loop spaces $QX$ -QUESTION [10 upvotes]: Let $X$ be a simply connected space. By $Q$ I denote $\Omega^{\infty}\Sigma^{\infty}$. Then $QX$ is an infinite loop space and the homology $H(QX)$ in $\mathbb{F}_p$ is a Hopf algebra over the Dyer-Lashof algebra. -Now there is a monomorphism $H(X) \rightarrow$ $H(QX)$ induced from $X \rightarrow QX$. -My question is if $H(QX)$ is generated over $H(X)$ in some sense. -Or is there some other relation between these two homologies? - -REPLY [14 votes]: Let $X$ be a connected space, and let $\lbrace x_\lambda\rbrace$ be a homogeneous basis for $H_\ast(X;\mathbb{F}_2)$. Then -$$H_\ast(QX;\mathbb{F}_2) = \mathbb{F}_2 [Q^I x_{\lambda} \mid I\mbox{ admissible of excess }e(I)>\mathrm{dim}\,\, x_\lambda ].$$ -That is, the homology of $QX$ is a polynomial algebra with generators certain iterated Kudo-Araki operations on the basis of the homology of $X$. There is a similar result with coefficients mod $p$, $p$ an odd prime, involving Dyer-Lashof operations and the Bockstein operator. (The situation is reminiscent of Serre's Theorem on the cohomology of Eilenberg-Mac Lane spaces.) -The reference is (Section 5 of) -Dyer, Eldon; Lashof, R. K. -Homology of iterated loop spaces. -Amer. J. Math. 84 1962 35–88. -You will also find a nice discussion in -Eccles, P. J. Characteristic numbers of immersions and self-intersection manifolds. Topology with applications (Szekszárd, 1993), 197–216, Bolyai Soc. Math. Stud., 4, János Bolyai Math. Soc., Budapest, 1995.<|endoftext|> -TITLE: Path integral and harmonic oscillator -QUESTION [8 upvotes]: Maybe this is not a research level question. I post it because I heard that the path integral can be rigorous by Brownian motion. But my knowledge of probability is so limited. -If $$L=\frac{1}{2}(-\frac{d^2}{dx^2}+x^2),$$ -we know that $Sp(L)=\{1/2,3/2,5/2,...\}$. So we get -$$\mathrm{Tr}[e^{-L}]=\frac{1}{2\sinh1/2}.$$ -I would like to recover it by following "method". -If $E_x$ denote the expectation of the Brownian motion $x_.$ start from x. By Feymann-Kac formula, we have -$$e^{-L}f(x)=E_x[e^{-\frac{1}{2}\int_0^1x_s^2ds}f(x_1)].$$ -If $p(x,y)$ denote the kernel of $e^{-L}$, we get -$$p(x,y)=E_x[e^{-\frac{1}{2}\int_0^1x_s^2ds};x_1=y]\frac{e^{-\frac{1}{2}|x-y|^2}}{\sqrt{2\pi}}$$ -where $E_x[...;x_1=y]$ is the conditional expectation. -So we get -$$\mathrm{Tr}e^{-L}=\int_{x\in \mathbb{R}}\frac{dx}{\sqrt{2\pi}}E_x[e^{-\frac{1}{2}\int_0^1x_s^2ds};x_1=x]$$ -All the thing is rigorous until now. But in some physics book, it follows that the right side is -$$\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\dot{x}_s^2+x^2_sds}\mathcal{D}x -=\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\langle-\Delta +1x,x\rangle} \mathcal{D}x$$ -where $$\mathcal{D}x=\mathrm{det}^{1/2}(-\Delta)\frac{dx}{(2\pi)^{\infty/2}}.$$ -As the finite cas, -$$\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\langle-\Delta +1x,x\rangle} \mathcal{D}x=\frac{\mathrm{det}^{1/2}(-\Delta )}{\mathrm{det}^{1/2}(-\Delta +1)}=\mathrm{det}^{-1/2}(1+(-\Delta)^{-1}).$$ -As we know, $Sp(-\Delta)=\{4\pi^2k^2,k\in \mathbb{Z}\}$. We have -$$\mathrm{det}^{1/2}(1+(-\Delta)^{-1})=\Pi_{k=1}^{\infty}(1+\frac{1}{4k^2\pi^2})=2\sinh{1/2}$$ -It also follows the right answer. -So my question is "How to make it rigorous?" -First, it will need a gaussian measure on the periodic path. But I can not find a natural one. -Edit: Thanks to Alexander Chervov, he give a interesting measure by Fourier Analysis. It is a right one in some sense. But it is not even clear for me that its support is the contious path. And with this measure how can we get the final answer rigorousment. -Edit2 and Answer Thanks to A.J. Tolland and Glimm & Jaffe's book. I just complete the answer to my question. -Let $P$ is the measure of the brownian bridge. -Consider the operator forme $C^{-\infty}(S^1)$ to $C^{\infty}(S^1)$, -$$-\Delta+1, $$ -There is a unique gaussian mesure $Q$ on $C^{-\infty}(S^1)$(in fact its support is $C^0(S^1)$) whose matrix of covariance is -$$(-\Delta+1)^{-1}$$ -By the uniqueness, we have -$$\frac{dx}{\sqrt{2\pi}}e^{-\frac{1}{2}\int_{0}^1(x_s+x)^2ds}dP=\frac{\mathrm{det}^{1/2}(-\Delta)}{\mathrm{det}^{1/2}(-\Delta+1)}dQ$$ -After integral, we have get the resultat. -Remark, the existance and the uniqueness of the gaussian measure is the big theorem in the Appendix A.4 of Glimm & Jaffe's book. - -REPLY [8 votes]: Take a look at Appendix A of the 2nd edition of Glimm & Jaffe's book. They give a rigorous construction of the measure you're after aka, the Ornstein-Uhlenbeck measure, which is the cylinder measure you get by taking the continuum limit of the Euclidean signature harmonic oscillator). The key point is that the cylinder measure defined using the momentum basis (or the lattice approximation) 'vanishes at infinity', hence is an actual measure. -They also prove that this measure is supported on (distributions almost everywhere equal to) continuous functions.<|endoftext|> -TITLE: Which functions have all derivatives everywhere positive? -QUESTION [16 upvotes]: Consider the class of functions from $\mathbb R$ to $\mathbb R$, such that the function is positive everywhere and its $n$th derivative is positive everywhere for all $n$. -The only examples I can construct are the functions $ae^{bx}+c$ for $a,b,c>0$. -Are these functions the only examples? -If not, for which nonlinear functions $g$ does $e^{g(x)}$ have this property? - -REPLY [20 votes]: See completely monotonic in the literature. Function $f(x)$ is completely monotonic if and only if $f(-x)$ is the sort of function you're looking for. -S.N. Bernstein (1928). "Sur les fonctions absolument monotones". Acta Mathematica 52: 1–66. doi:10.1007/BF02592679. -http://mathworld.wolfram.com/CompletelyMonotonicFunction.html - -REPLY [14 votes]: Well, there are certainly more. If you look at the chain rule then you see that the $n$-th derivative is a linear combinations of products of derivatives of the two functions you compose with positive coefficients. Thus if you have two functions with your property, then their composition will again have only positive derivatives. So you can go on... - -REPLY [8 votes]: If $f(x)$ is a function with positive derivatives, then the function $f(-x)$ is completely monotonic. The completely monotonic functions are classified by Bochner's theorem; see Nimza's question On the generalisation of Bernstein's theorem on monotone functions.<|endoftext|> -TITLE: Bodies of constant width? -QUESTION [36 upvotes]: In two-dimensional case one can generalize figures of constant width as figures which can rotate in a covex polygon. -Here is one example which can be used to drill triangular holes: - -I would like to know what happens with this generalization in dimension $3$ and maybe higher. -Obviously body of constant width $1$ can rotate arbitrary in a unit cube. -More formally, given a body $B$ of constant width $1$ -and $A\in SO(3)$ there is $v\in \mathbb R^3$ such that -$$A(B)+v\subset\square,$$ -where $\square$ is unit cube. -On the other hand, except for the cube, I do not see any other examples of convex polyhedron which have nontrivial rotating bodies (i.e. distinct from the inscribed ball). -I hope that the answer is known. -(= I hope I should wait for the answer and I do not have to think.) -The question is inpired by this one: "Local minimum from directional derivatives in the space of convex bodies." - -REPLY [18 votes]: This adds nothing to Yoav Kallus' answer, but I was curious to see what these rotors look like -(Schneider's notes has no figures). -I found grainy photos of rotors for the cube, the regular octahedron, and the regular tetrahedron -in a 50-year old paper by -Michael Goldberg, -"Rotors in Polygons and Polyhedra," -Mathematics of Computation, Vol. 14, No. 71 (July, 1960), pp. 229-239: -      - - -(They remind me of stones found on a beach!) -Of course one can find much better examples of cube rotors, which as Anton points out, are just constant-width bodies. E.g., this is from the cover of Bryant and Sangwin's 2008 How Round Is Your Circle:<|endoftext|> -TITLE: Weyl's Equidistribution Theorem and Measure Theory -QUESTION [5 upvotes]: According to Rajendra Bhatia in his book Fourier Series, Weyl's Equidistribution Theorem states that if $x$ is an irrational number, then for every subinterval $[a, b]$ of $(0, 1)$ we have -$\lim_{n \rightarrow \infty} \frac{1}{N}\operatorname{card}\{k : 1 \leq k \leq N, \tilde{(kx)} \in [a, b]\} = b - a$ where $\tilde{(kx)}$ is the fractional part of the number $kx$. -My question is what happens if we generalise to measurable subsets of $(0, 1)$? -Does $\lim_{n \rightarrow \infty}\frac1N\operatorname{card}\{k : 1 \leq k \leq N, \tilde{(kx)} \in A\} = \mu(A)$ where $A$ is a measurable subset and $\mu$ the Lebesgue measure function? -Further, for non-measurable subsets $V$ is the sequence $\frac{1}{N}\operatorname{card}\{k : 1 \leq k \leq N$, $\tilde{(kx)} \in V\}$ bounded above and below and if so, does it have the same set of sublimits for all irrational $x$? -After my last question that revealed I had momentarily forgotten all my undergraduate real analysis, I hope this one is worthy of MathOverflow... thanks... - -REPLY [3 votes]: This is a very interesting question, which actually asks about the interplay between equidistribution (or harmonic analysis if you would like to call it that way) and ergodic theory. -As Vaughn mentioned, for any $L^{p}$ function ($p\geq 1$), the pointwise ergodic theorem would imply that for Lebesgue almost every point $x$, the value of the ergdoic averages -$$ \frac{1}{N}\sum_{n=0}^{N-1}f(x+n \alpha) \to \int fd\mu$$ -where $\mu$ is the Lebesgue measure. Hence one can take the characteristic function of your favourite measurable set, and get the "equidistribution" statement you would like, but only at the cost of convergence a.e. -Noam showed why a.e. is sharp here. -When one asks about continuous function (or more frequently in the field, takes a function from a suitable Sobolev space), one can get more information than the statement above (including for example estimation about the errors), see for example this inequality for your problem - https://en.wikipedia.org/wiki/Low-discrepancy_sequence , but this also comes at a cost, you usually have to limit the irrational number your dealing with to be Diophantine generic or so. -As you may know, Weyl's equidistribution was generalized to equidistribution of sequences of the form ${p(n)\alpha}$, where $p(n)$ is some polynomial in $Z[x]$. -Now one can ask about such "ergodic limits" for the sequence ${n^{2}\alpha}$, for general $L^{p}$ function (continuous functions will be dealt easily by the equidistribution theorem). -It was an open question for some time, but the main result here is by Bourgain, which showed that for any $p>1$, if the function is in $L^{p}$, then the averages will converge a.e. to the integral. -This is the analogue of the usual pointwise ergodic theorem for the quadratic case. -The interesting point here is that $p>1$ is sharp, there is today a counterexample for $p=1$ case.<|endoftext|> -TITLE: General Isoperimetric Inequality via Representation Theory of SO(n) -QUESTION [12 upvotes]: Is there a known proof of the $n$-dimensional isoperimetric inequality which generalizes Hurwitz's proof using Fourier analysis in the $2$-dimensional case? -Specifically, I imagine such a proof would replace $S^1 \cong \text{SO}(2)$ with the group $\text{SO}(n+1)$, and we would decompose a smooth function on $S^n$ according to the representation theory of $\text{SO}(n+1)$, and then express the volume and surface area in terms of the "Fourier coefficients". -If this doesn't work, why not? - -REPLY [13 votes]: For the second time today (see Bodies of constant width?), I give the answer: see "Geometric Applications of Fourier Series and Spherical Harmonics" by Helmut Groemer. That book deals exactly with your question, and the answer is yes.<|endoftext|> -TITLE: Flat sector in a proper cocompact CAT(0) space -QUESTION [5 upvotes]: Let $X$ be a complete CAT(0) space with a proper and cocompact group action by isometries, and suppose there are $\xi, \xi' \in \partial X$ with $\angle (\xi, \xi') < \pi$. Using proposition 9.5 (3) of Chapter II.9 from the book Metric spaces of non-positive curvature by Bridson and Haefliger, there exist two points $\\eta, \eta' \in \partial X$ and a point $y\in X$ such that $\angle_y (\eta,\eta') =\angle(\xi,\xi')$. Then by corollary 9.9 of the same chapter, the convex hull of two geodesic rays $c, c'$ issuing from $y$ with $c(\infty) = \eta$, $c'(\infty) = \eta'$ is isometric to a sector in $\mathbb{E}^2$ bounded by two rays meeting at the angle $\angle (\xi, \xi')$. -The above result gives geodesic rays with possibly different endpoints. So I would like to know whether we can have a stronger result that there are geodesic rays $c_1, c'_1$ issuing from some point $y_1\in X$ with endpoints $\xi,\xi'$ such that the convex hull of these rays is isometric to a flat sector. I think this should be true in general, but maybe there is a counter-example that I do not know. - -REPLY [2 votes]: Xie proved that this is true if $X$ is a 2 dimensional complex. I recommend reading his article: "The Tits Boundary of a CAT(0) 2-Complex", Trans. AMS., 357, no. 4, 1627-1661. I think that the answer to the more general question above is unknown. The techniques used in Xie's paper are further developed in an article by Bestvina, Kleiner, and Sageev: "Quasi-flats in CAT(0) complexes", arXiv:0804.2619v1 [math.GR].<|endoftext|> -TITLE: Metric Connections on a Lie Group -QUESTION [5 upvotes]: A Lie group has three standard Cartan connections; the (-)-connection, the (0)-connection, and the (+)-connection. The (0)-connection is Levi-Civita with the associated metric the bi-invariant metric. The other two connections aren't Levi-Civita due to the presence of torsion. However, there's nothing to stop them a priori from being metric connections. My question is; are the minus and plus connections compatible with the bi-invariant metric? This seems reasonable but I can't find a reference. - -REPLY [8 votes]: Yes. -Let $\nabla$ be an arbitrary connection on the tangent bundle of a Riemannian manifold $(M,g)$. -The standard trick for expressing the Levi-Civita connection in terms of $g$ gives you, -for any 3 vector fields $X$, $Y$, $Z$: -$$Xg(Y,Z)+ Yg(Z,X)- Zg(X,Y)= N(X,Y,Z) $$ -$$+ g(T(X,Z),Y)+ g(T(Y,Z),X)- g(T(X,Y),Z) $$ -$$ +2 g(\nabla_X Y,Z)- g([X,Y],Z) + g([X,Z],Y) + g([Y,Z],X),$$ -where -$$ T(X,Y)=\nabla_X Y- \nabla_Y X -[X,Y]$$ -is the torsion of $\nabla$ and -$$ -N(X,Y,Z)= \nabla_Xg(Y,Z)+ \nabla_Yg(Z,X)-\nabla_Zg(X,Y). -$$ -This is the "non-metricity": $N=0\Leftrightarrow \nabla g=0$. -Now, turning to the case at hand: we define the $\pm$ and $0$ connections by -$$ (\nabla_X Y)_e=\epsilon [X,Y],$$ -$ \epsilon = 1, 0, \frac{1}{2}$ respectively, so the torsion is -$$T(X,Y) = (2\epsilon -1)[X,Y]= \pm[X,Y]\textrm{ or } 0, $$ -hence the names of the connections. But then you get -$$ 0 = N(X,Y,Z) -2\epsilon\left[ g([Z,Y],X) + g(Y,[Z,X]) \right],$$ -and the second summand is zero due to bi-invariance, so $N=0$. - -REPLY [2 votes]: If I understand correctly, the answer is Yes. -the +/-/0 connections can be defined by, if $X,Y$ is the left invariant vector -$$\nabla_{X}Y=a[X,Y]$$ -where $a=1,-1,0$. -The connection is metric for left invariant metric iff -$$0=\langle\nabla_{X}Y,Z\rangle+\langle Y,\nabla_{X}Z\rangle.$$ -This is trival for the bi-invariant metric.<|endoftext|> -TITLE: Why hasn't anyone proved that the two standard approaches to quantizing Chern-Simons theory are equivalent? -QUESTION [64 upvotes]: The two standard approaches to the quantization of Chern-Simons theory are geometric quantization of character varieties, and quantum groups plus skein theory. These two approaches were both first published in 1991 (the geometric quantization picture here and the skein theoretic approach here), and despite a tremendous amount of development since then, it is still not known whether they are equivalent! I guess that it is reasonable to say that the problem of their equivalence has been around for 20 years. -There is (at least) one important theorem, namely the asymptotic faithfulness of the mapping class group representations produced by these two quantizations, which has proofs in both settings. The two proofs are of completely different character, and are of course logically independent, since the two representations are not known to be the same (this was proved for the quantum group skein representation by Freedman, Walker, and Wang, and for the geometric quantization representation by Andersen). -Is there a good reason why the equivalence of these two viewpoints is not yet a theorem? Is there an idea for a proof, which hasn't been completed because "it's just a long calculation" or "everyone knows it's true" or "it's nice to know, but it wouldn't actually help us prove theorems"? Or is it that it's actually a hard problem that no one knows how to approach? Is it an "important" problem whose solution would have lots of consequences and applications, or at least advance our understanding of "quantization"? - -REPLY [4 votes]: For future reference. The elegant argument sketched by Kevin Walker above involving pants decompositions and Bohr-Sommerfeld fibers was published by Andersen on the arXiv later that year, as part of the following paper: -Jørgen Ellegaard Andersen, Mapping class group invariant unitarity of the Hitchin connection over Teichmüller space, June 2012.<|endoftext|> -TITLE: Heegaard splitting of covering hyperbolic manifold. -QUESTION [13 upvotes]: I am curious about how the Heegaard genus changes after a finite covering. -Is there anyone constructing an closed hyperbolic 3-manifold $N$ such that -the Heegaard genus of a finite covering of $N$ is less than the Heegaard genus of $N$? -Thank you! -Note: Heegaard genus of a 3-manifold means the minimal genus of all Heegaard splittings. - -REPLY [6 votes]: Hyam Rubinstein and me have results about the behavior of the Heegaard genus under double covers for non-Haken manifolds, see http://arxiv.org/abs/math/0607145. Essentially, we show that the Heegaard genera of the two manifolds bound each other linearly. (The statement is a little more complicated for branched covers.)<|endoftext|> -TITLE: Smoothability of compact Alexandrov surfaces with curvature bounded from below -QUESTION [9 upvotes]: Let $(X,d)$ be compact metric space of curvature greater than $-1$ (in the sense of comparison triangles), assume that its Hausdorff dimension is $2$. Then a result of Perelman says that $X$ is a 2-dimensional manifold. -Claim -$(X,d)$ can be Gromov-Hausdorff approximated by a sequence of Riemannian surfaces $(M_i,g_i)$ such that $\int_{M_i}|K_{g_i}|dv_{g_i}$ is bounded. -To see this : - -A version of the Gauss-Bonnet Theorem holds (Machigashira, The Gaussian curvature of Alexandrov surfaces), which implies that $(X,d)$ is an Alexandrov surface with bounded integral curvature. - -Any such surface can be approximated by smooth Riemannian surfaces with bounded integral curvature. See Reshetnyak, Geometry IV, Encyclopaedia of Mathematical Sciences. - - -Question -Is any compact Alexandrov surface of curvature greater than $-1$ approximated by a sequence of smooth compact Riemannian surfaces with curvature bounded from below (by -1, or something else if this helps) ? -Maybe this is classic but I didn't find explicit results of this kind. -Thanks. - -REPLY [11 votes]: Edit: Addressing Igor's comment I'd like to correct the references I gave. The correct reference for the exact argument I sketch should be the original book by Alexandrov "Intrinsic Geometry of Convex Surfaces"(Chapter 7, section 6, Lemmas 1-3). In the later book by Alexandrov and Zalgaller a similar (but necessarily more complicated) argument is given for the case of surfaces with bounded integral curvature (Theorem 10, page 84). One can reconstruct the original proof in the easier case of curvature bounded below from that one but it requires some work. -This is indeed classical and is due to Alexandrov ( see the book by Alexandrov and Zalgaller "Intrinsic geometry of surfaces"). -The general structure the proof is as follows. You take a very fine triangulation of $X$ and substitute the curved triangles by triangles with the same sides in the space form of constant curvature $-1$. This will give you a polyhedral surface of curvature $\ge -1$ (you only need to check that the cone angles at vertices are $\le 2\pi$ which is immediate from the definition of an Alexandrov space). Away from the cone points the metric will be smooth of constant curvature $-1$. The cone points can then be easily smoothed to get a smooth metric of $\sec\ge -1$. The hard part is to show that the resulting polyhedral surface is close to the original space $X$. -A more interesting result is another theorem of Alexandrov ( "Intrinsic Geometry of Convex Surfaces") that locally any 2-dimensional Alexandrov space of $curv\ge -1$ is isometric to a level set of a convex function in $\mathbb H^3$. When the lower bound is 0 he proved an even sharper result that any 2-sphere of curvature $\ge 0$ is isometric to the boundary of a convex body in $\mathbb R^3$. -BTW, the result you attribute to Perelman that a 2-dimensional Alexandrov space is a topological manifold is actually due to Alexandrov too.<|endoftext|> -TITLE: Counting the number of subgraphs in a given labeled tree -QUESTION [5 upvotes]: Are there any results on the number of subgraphs in a labeled tree (or a general labeled graph)? I would also be happy to know any results on the number of subgraphs in an unlabeled tree. Cayley's formula says how many different trees I can form given n vertices, but it doesn't seem to relate to the problem of counting subgraphs in a given tree. Any help is appreciated. - -REPLY [5 votes]: Paul's idea can be improved into a linear algorithm to count the subtrees of a tree. Arbitrarily root the tree at some vertex $r$ and let $T_x$ denote the subtree rooted at vertex $x$. Define $A(x)$ to be the number of subtrees of $T_x$ that include $x$, and define $B(x)$ to be the number of subtrees of $T_x$ that don't include $x$. Do not include the null subtree in $B(x)$ (add 1 to the final answer if you want). -If $x$ is a leaf then $A(x)=1$ and $B(x)=0$. -If $y_1,\ldots,y_k$ are the children of a non-leaf $x$, then -\begin{align*} - A(x) &= \prod_{i=1}^k (1+A(y_i)) \\\\ - B(x) &= \sum_{i=1}^k (A(y_i) + B(y_i)), -\end{align*} -and the answer is $A(r)+B(r)$. -The two recurrences need to be applied once for each vertex. The time required for vertex $x$ is $O(1+\text{degree of }x)$, which adds up to $O(n)$ when summed over all $x$. So the total time is $O(n)$. Here we are cheating slightly by counting arithmetic operations as $O(1)$ even though numbers with a linear number of bits might be involved. -This is most unlikely to be original.<|endoftext|> -TITLE: A flag complex is contractible iff the underlying graph is....? -QUESTION [8 upvotes]: Let $G$ be a finite simple graph and let $C(G)$ be the flag complex associated to $G$ (the set of vertices of $C(G)$ is the vertex set of $G$ and the set of all cliques of $G$ are its simplexes). -Are there characterizations of contractibility of $C(G)$ ONLY in terms of the graph theoretical properties of $G$? - -REPLY [4 votes]: A nice graph theory lemma for showing homotopy equivalence that builds on the "clique starring" already discussed is stated as Lemma 3.2 of Alexander Engström's paper arXiv:math/0508148. I'll rephrase in terms of the language the question was asked in (clique complexes rather than independence complexes). -Lemma: If $v$ and $w$ are vertices of a graph $G$ with $N[v] \subseteq N[w]$, then $C(G)$ is homotopy equivalent to $C(G \setminus v)$. -(Here $N[v]$ is the closed neighborhood of $v$, i.e., $v$ and all its neighbors.) -The proof technique is that of elementary collapses. See -http://en.wikipedia.org/wiki/Collapse_(topology). -Even if the lemma of Engström doesn't give you what you need, the broader technique of collapsing can be quite useful. Collapses are the "engine" of discrete Morse theory, for example. -It's not too hard to write a computer program (or some are available) that does automatic collapsing, and you could generate a large number of examples this way. If you can collapse a complex to a point, then the complex is contractible. The converse is not true, but the program might at least give you a smaller complex to examine by hand. -On the other hand, as Benjamin Steinberg points out, the barycentric subdivision of any simplicial complex is flag, so classifying contractible flag complexes should be as hard as classifying contractible simplicial (and more generally CW) complexes.<|endoftext|> -TITLE: Homotopic morphisms between curved A-infinity algebras -QUESTION [5 upvotes]: I know how to think about (curved) $A_\infty$-algebras 'geometrically', i.e. via formal non-commutative geometry in the sense of Kontsevich etc. I also know how to think about $A_\infty$-morphisms in this way. But what if two $A_\infty$-morphisms are homotopic? Does anyone know how to interpret this fact geometrically? -This is particularly important in the curved situation, because then there's no such thing as a quasi-isomorphism, so we only have homotopy-equivalence. Also I have a vague memory of reading something about this (probably written by Kontsevich), but I've searched all the papers I can think of and not found it. - -REPLY [12 votes]: I don't know specifically about homotopies, but the notion of a curved $A_\infty$-algebra is generally problematic. In the conventional setting of algebras over a field, it is just trivial in the following strong sense. -Let $A$ and $B$ be two curved $A_\infty$-algebras over a field $k$ with nonzero curvature elements $m_{0,A}\ne0\ne m_{0,B}$. This is a sufficient condition to trivialize nonunital curved $A_\infty$-algebras; in the (strictly) unital case, assume that $m_{0,A}$ and $m_{0,B}$ do not belong to the one-dimensional vector subspaces generated by the units of $A$ and $B$ (which could happen in the $\mathbb Z/2$-graded case). -Then any isomorphism of graded vector spaces $f\colon A\to B$ taking $m_{0,A}$ to $m_{0,B}$ (and preserving also the units, in the unital case) can be extended to an $A_\infty$-isomorphism $(f_0,f_1,f_2,\dotsc)\colon A\to B$ with $f_0=0$ and $f_1=f$. So there are precisely as many curved $A_\infty$-algebras with nonzero curvature, up to $A_\infty$-isomorphism, as there are graded vector spaces; and any curved $A_\infty$-algebra with a nonzero curvature is $A_\infty$-isomorphic to a curved $A_\infty$-algebra with $m_1=m_2=m_3=\dotsb=0$. -Similarly, any curved $A_\infty$-module over a (nonunital or strictly unital) curved $A_\infty$-algebra with a nonzero curvature element is contractible. -These results are mostly due to Kontsevich; I learned them from conversations with him while visiting IHES and subsequently recorded them in what is now AMS Memoir vol.212 #996, 2011, http://arxiv.org/abs/0905.2621, Remark 7.3. -It appears that if you want to have a nontrivial theory of curved $A_\infty$-algebras, you have to do it over, say, a local ring and require the curvature elements in your algebras to be divisible by the maximal ideal of the local ring. I am presently working on this; the writeup is available from my homepage.<|endoftext|> -TITLE: What is the relation between L. Lafforgue and Frenkel-Gaitsgory-Vilonen results on Langlands correspondence ? -QUESTION [16 upvotes]: What is the relation between Lafforgue's result on Langlands -and Frenkel-Gaitsgory-Vilonen ? ( http://arxiv.org/abs/math/0012255 , http://arxiv.org/abs/math/0204081 ) -Does one imply other ? If not why ? -More technical: -Do FGV work only with unramified Galois irreps (Seems Yes) ? If Yes, is it difficult to cover ramified case ? If yes, what is the problem ? -Is there clear relation between irreps of GL(Adels) and Hecke-eigen-sheaves on BunGL ? -How to see in FGV setup that Hecke eigenvalues should correspond to Frobenius eigs ? - -Background -$GL_n$ Langlands correspondence is bijective correspondence between (1) and (2), where -(1) n-dimensional Irreps of (almost) Galois group -(2) Certain Irreps of GL_n(Adels). -Main requirement that Frobenius eigenvalues should be equal to Hecke eigenvalues -for each point "p". -Consider the case of "function fields" i.e. Galois group is taken for some curve over -finite field and adels over this curve. - -Lafforgue's proved the correspondence above for the curves over finite-fields. -His proof follows strategy proposed and worked out by Drinfeld in GL_2 case. -He considers moduli spaces of "schtukas" where both groups acts. -And proves that "functions" on it can be decomposed as $\sum V\otimes V^t \otimes W$ -V - irrep of GL(Adels), W - of Galois. -As far as I understand main difficulties are of "technical" nature - one should -find correct compactifications and introduce "negligible" pieces which are not interesting... -It is completely different from the strategy of FGV, proposed by Drinfeld(?) and Laumon. -In this setup starts from the Galois irrep (=local system on curve) and constructs -certain sheaf on BunGL which is Hecke-eigensheaf (with "eigenvalue" given exactly -by the local system from which one starts). - -REPLY [13 votes]: Let me try to answer. [FGV] is only about unramified representations of the Galois group -but they prove a stronger fact in this case (existence of certain "automorphic sheaf"). Lafforgue's result doesn't follow from there for several reasons: -a) Formally [FGV] use Lafforgue, but this was actually taken care of by a later paper of Gaitsgory ("On the vanishing conjecture..."). So that is really not a problem now. -b) Extending [FGV] to the ramified case is not trivial. I actually suspect that it can be done using the thesis of Jochen Heinloth but this has never been done (even the formulation is not completely clear in the ramified case) -c) In the unramified case what follows immediately from [FGV] is that you can attach a cuspidal automorphic form to a Galois representation. It is not obvious to me that the converse statement follows (Lafforgue's argument actually goes in the opppsite direction: -he proves that a cuspidal automorphic form corresponds to a Galois representation and then the converse statement follows immediately from the converse theorem of Piatetski-Shapiro et. al. -and from the fact that you know everything about Galois L-functions in the functional field case).<|endoftext|> -TITLE: Multi-dimensional moment problem -QUESTION [11 upvotes]: Let $\mu$ be a measure on $\def\r{\mathbb{R}}\r^n$, $1\le n \le \infty$. Given a (finite) multi-index $\bar{i} = (i_1, i_2, \ldots)$, one can define the moment -$$ m_{\bar i} = \int x_i^{i_1} x_2^{i_2} \cdots x_n^{i_n} d\mu .$$ -I'm interested in the inverse problem of reconstructing $\mu$ from the collection of moments $\{m_{\bar i}\}$. I'm mainly interested in the case $n=\infty$, but I would be happy to learn of good references for any $n$, even $n=1$. I'm vaguely aware that one way to tackle the $n=1$ moment problem is via the Stieljes transform. Is there something similar for $n>1$? -I'm mainly interested in this problem from a practical, computational point of view. I can generate various instances of $\{m_{\bar i}\}$, and I would like to know in each case what the corresponding measure $\mu$ is. -Summary: How, as a practical matter, does one go about solving moment problems, especially multidimensional moment problems? - -REPLY [6 votes]: The infinite-dimensional power moment problem was introduced by A.G. Kostyuchenko and B.S. Mityagin in 1960. -A description of this and other interesting generalizations of the power moment problem can be found in the following book: -[BK] Yu. M. Berezansky and Y. G. Kondratiev, "Spectral Methods in Infinite-Dimensional Analysis", 1988. -See Chapter 5, Section 2 in this book. -I have only the Russian paper original of this book. Probably it is translated by Kluwer Academic Publishers. -Some conditions which ensure existence of a representing measure are given in the book. I do not remember any -algorithms in [BK]. However, there are historical notes in [BK] and you can look at them. -On the other hand there are classical books: -[ST] J. A. Shohat and J. D. Tamarkin "The problem of moments", -[A] N. I. Akhiezer "The classical moment problem". -In the case of a finite interval there are several methods to recover the measure, see [ST, p. 90]. -For the two-dimensional moment problem I proposed an algorithm in -S.M. Zagorodnyuk, On the two-dimensional moment problem.- Ann. Funct. Anal., 1, no. 1 (2010), 80-104. -Essentially it consists of solving of infinite linear systems of equations with parameters. -However, I think this algorithm requires powerful computer and much additional technical work to implement it. -Some time ago, one scientist asked me a similar question: how practically reconstruct the measure? -I answered, that one can replace integrals by integral sums and then to solve linear equations. It is not -a rigorous way, but for some cases this will probably have sense.<|endoftext|> -TITLE: Is every finite group a quotient of the Grothendieck-Teichmuller group? -QUESTION [12 upvotes]: The Grothendieck-Teichmuller conjecture asserts that the absolute Galois group $Gal(\mathbb{Q})$ is isomorphic to the Grothendieck-Teichmuller group. I was wondering, would this conjecture imply the Inverse Galois Problem? I.e. is every finite group a quotient of the Grothendieck-Teichmuller group? - -REPLY [4 votes]: It's only a partial answer since this survey is already 5 years old, but it suggest that almost nothing is known about (non-abelian) finite quotients of $\widehat{GT}$ (question 1.7). -Edit: I should maybe recall what happen in the abelian case, and why it's encouraging: elements of $\widehat{GT}$ are pairs $(f,\lambda)$ where $f$ is in the derived subgroup of $\hat{F}_2$, and $\lambda \in \hat{\mathbb{Z}}^{\times}$, satisfying some complicated equations. It turns out that the set theoretic map $(f,\lambda) \mapsto \lambda$ induces a surjective group morphism $\widehat{GT}\rightarrow \hat{\mathbb{Z}}^{\times}$. And the good news is that the composite -$$G_{\mathbb{Q}} \hookrightarrow \widehat{GT} \rightarrow \hat{\mathbb{Z}}^{\times}$$ -is nothing but the cyclotomic character.<|endoftext|> -TITLE: Ratio of circumscribed/inscribed $(n{-}1)$-gons -QUESTION [14 upvotes]: As a discrete analog of the MO question, -"Löwner-John Ellipsoid: incribed and circumscribed," -I've been wondering what might be the maximum ratio -of this quantity? -Let $P$ be a convex polygon of $n$ vertices, -$P^+$ be a minimum area polygon of $n{-}1$ vertices circumscribing -$P$, and $P^-$ a maximum area polygon of $n{-}1$ vertices -inscribed in $P$. - -Over all polygons $P$ of $n$ vertices, what is the maximum - of the ratio area$(P^+)/$area$(P^-)$, as a function of $n \ge 4$? - -Here are possible optimal (en/in)closures for a square and a regular pentagon: -      - - -[Some updates below.] -I believe that it is known that the circumscribing $(n{-}1)$-gon must have one -edge "flush" with $P$ (i.e., including an edge of $P$ as a subset), and -that this may be proved in a 1984 paper by -Chang and Yap -"A polynomial solution for potato-peeling problem" -(Discrete & Computational Geometry -Volume 1, Number 1, 155-182), which I cannot access at the moment. -This is certainly true for enclosing triangles. -I am not sure if there is any analogous characterization of inscribed $(n{-}1)$-gons. -A key result for minimum area circumscribing is by Victor Klee in 1986: -"Facet-centroids and volume minimization," -Studia Scientiarum Mathematicarum Hungarica, Vol. 21, 143-147, 1986. -As the title indicates, he proved that each facet's centroid must touch $P$ (in any dimension). -The circumscribing polygons I drew above have their edge midpoints touching $P$. -For $n{=}4$, $a \times b$ rectangles have an area ratio of $(2 a b) / (a b / 2) = 4$, -illustrated with the square above; -perhaps this is the worst ratio over all $n$? -Any ideas would be appreciated, from a clean proof (or counterexample!) that one circumcribing -edge must be flush, to any constraints on the inscribed polygons, to an answer to the ratio -question, even for specific $n$, even for regular $n$-gons. Thanks! - -REPLY [2 votes]: Just a note about inscribed polygons. It is easy to show that area$(P^-)\ge$ area$(P)(1-1/(n-2))$. This follows from the fact that among the triangles formed by three vertices of $P$, (at least) one with minimum area must ``lie'' on a side of $P$. (Proof: Take minimum area triangle, use area$\cdot 2=$ height$\cdot$base and move vertices.) Now triangulate $P$ such that this minimum area triangle appears. Omitting it we get a polygon with $n-1$ vertices, done. This bound is also sharp for regular $n$-gons. (Proof is again by moving the vertices of maximum area $(n-1)$-gon. -I guess the next task is to determine the minimum of area$(P^+)/$area$(P)$.<|endoftext|> -TITLE: Homology generated by lifts of simple curves -QUESTION [17 upvotes]: Let $\Sigma$ be a compact connected oriented surface and $p:\tilde{\Sigma}\to\Sigma$ a finite regular cover. -Consider the set $\Gamma$ of simple closed curves on $\tilde{\Sigma}$ obtained as a connected component of $p^{-1}(\gamma)$ where $\gamma$ is a simple curve in $\Sigma$. -My question is: is it true that $\Gamma$ generates $H_1(\tilde{\Sigma},\mathbb{Z})$ and if not, can we identify the subspace it generates? - -REPLY [7 votes]: I just posted a paper with Justin Malestein entitled "Simple closed curves, finite covers of surfaces, and power subgroups of $\text{Out}(F_n)$" that gives a nearly complete answer to this question. It can be downloaded here. -In this paper, we construct examples of finite covers of punctured surfaces where the first rational homology is not spanned by lifts of simple closed curves. We can deal with any surface whose fundamental group is a nonabelian free group, though we do not know how to deal with closed surfaces. -In fact, we prove a much more general result. Let $\mathcal{O}$ be a subset of a nonabelian free group $F_n$ that is contained in the union of finitely many $\text{Aut}(F_n)$-orbits. For instance, if $F_n$ is the fundamental group of a punctured surface then $\mathcal{O}$ could be the set of all simple closed curves (or even for some fixed $m \geq 0$ the set of all closed curves with at most $m$ self-intersections). Then there exists a finite-index normal subgroup $R$ of $F_n$ such that $H_1(R;\mathbb{Q})$ is not spanned by the homology classes of powers of elements of $\mathcal{O}$. -I also should point out two related papers (see the introduction to my paper above for a more complete summary of the literature). The paper -T. Koberda, R. Santharoubane, -Quotients of surface groups and homology of finite covers via quantum representations, -Invent. Math. 206 (2016), no. 2, 269–292. -constructs examples of finite covers $\widetilde{S}$ of surfaces $S$ such that $H_1(\widetilde{S};\mathbb{Z})$ is not spanned by lifts of simple closed curves on $S$. They can deal with any orientable surface whose fundamental group is nonabelian (including closed surfaces). However, unlike us they cannot ensure that lifts of simple closed curves do not span a finite-index subgroup of $H_1(\widetilde{S};\mathbb{Z})$. This is why they only deal with integral homology as opposed to rational homology. -Another interesting paper along these lines is -B. Farb, S. Hensel, -Finite covers of graphs, their primitive homology, and representation theory, -New York J. Math. 22 (2016), 1365–1391. -Among other things, this paper proves that the "rational simple closed curve homology" is everything for abelian covers. It also constructs counterexamples to this for genus $1$ surfaces with $1$ or $2$ punctures.<|endoftext|> -TITLE: Complexity of matching red and blue points in the plane. -QUESTION [9 upvotes]: I'm just asking because I'm curious. -I was seeking references on the following problem, that a friend exposed to me last holidays : -Problem -Given $n$ red points and $n$ blue points in the plane in general position (no 3 of them are aligned), find a pairing of the red points with the blue points such that the segments it draws are all disjoint. -This problem is always solvable, and admits several proof. A proof I know goes like this : -Start with an arbitrary pairing, and look for intersections of the segments it defines, if there are none you're done. If you found one, do the following operation : -r r r r - \ / | | - X => | | - / \ | | -b b b b - -(uncross the crossing you have found), you may create new crossings with this operation. If you repeat this operation, you cannot cycle, because the triangle inequality shows that the sum of the length of the segments is strictly decreasing. So you will eventually get stuck at a configuration with no crossings. -Questions - -What is the complexity of the algorithm described in the proof ? -What is the best known algorithm to solve this problem ? - -I wouldn't be surprised to learn that this problem is a classic in computational geometry, however googling didn't give me references. Since some computational geometers are active on MO, I thought I could get interesting answers here. - -REPLY [10 votes]: The Ghosts and Ghostbusters problem can be solved in $O(n\log n)$ time, which is considerably faster than the $O(n^2\log n)$-time algorithm suggested by CLRS. -The ham sandwich theorem implies that there is a line $L$ that splits both the ghosts and the ghostbusters exactly in half. (If the number of ghosts and ghostbusters is odd, the line passes through one of each; if the number is even, the line passes through neither.) Lo, Matoušek, and Steiger [Discrete Comput. Geom. 1994] describe an algorithm to compute a ham-sandwich line in $O(n)$ time; their algorithm is also sketched here. Now recursively solve the problem on both sides of $L$; the recursion stops when all subsets are empty. The total running time obeys the mergesort recurrence $T(n) = O(n) + 2T(n/2)$ and thus is $O(n\log n)$. -This algorithm is optimal in the algebraic decision tree and algebraic computation tree models of computation, because you need $\Omega(n\log n)$ time in those models just to decide whether two sets of $n$ points are equal.<|endoftext|> -TITLE: Elements in a localization - category theoretic approach -QUESTION [14 upvotes]: This question is about the elements in a localization $S^{-1} A$ of a commutative ring $A$. Is it possible to derive $\frac{a}{1} = 0 \in S^{-1} A \Rightarrow \exists s \in S : sa = 0$ only using the universal property of $S^{-1} A$? In order to make this question clear enough I will have to digress a little bit. -Examples. a) Let's start with a related example. If $C$ is a category of algebras of some type, then every continuous functor $C \to \mathrm{Set}$ is representable (special case of SAFT). For example for $C=\mathbf{Ab}$ the functor $\mathrm{Hom}(A,-) \times \mathrm{Hom}(B,-)$ is representable for all abelian groups $A,B$, showing the existence of the coproduct $A + B$. I claim that we can find a description of its elements via the universal property. Namely, choose the coproduct injections $i : A \to A + B$ and $j : B \to A + B$. Then $C:=\mathrm{im}(i) + \mathrm{im}(j)$ is an abelian group such that $i,j$ factor through $C$ and still satisfy the universal property - this shows $A = \mathrm{im}(i) + \mathrm{im}(j)$, i.e. every element has the form $i(a) + j(b)$ for $a \in A, b \in B$. I claim that $a,b$ are unique: By the universal property there is some $f : A + B \to A$ with $fi = \mathrm{id}$ and $fj=0$. It follows $f(i(a) + j(b))=a$ and therefore $a$ is unique, similarly $b$. -Conclusion: We defined $A + B$ via some universal property (whose existence follows from general category theory) and found the structure of its elements - only by applying the universal property. We didn't need to know any specific constructed model of $A + B$ in advance for this! -Of course for abelian groups and coproducts this isn't really exciting. But what about more complicated (algebraic) structures (e.g. groups, rings, affine spaces, lie algebras, etc.) ? There is a general construction of $A + B$ with generators and relations of $A$ and $B$ (see e.g. Durov, 4.16.14/15), but this does not give us a criterion when two given elements of $A + B$ are equal. It gets even more complicated for other universal properties aka representable functors: For example you can write down a group defined by $10$ generators and $27$ explicitly (Wikipedia) with an unsolvable word problem. Let me mention two (of the many) positive results in this direction: -b) Let $M,N$ be modules over a commutative ring $k$ and define the tensor product $M \otimes_k N$ as the classifying object of $k$-bilinear maps on $M \times N$. If $E$ is a generating set of $M$, then the universal property implies that every element has the form $\sum_{e \in E} e \otimes n_e$ for some $n_e \in N$ (which vanish for almost all $e$). There is a criterion when this element is zero and it can be proved with the universal property of the tensor product, without using its explicit construction (Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs, online). -c) If $H$ is a subgroup of a group $G_1$ as well as of a group $G_2$, then it is clear how to represent elements in the amalgamated sum $G_1 *_H G_2$, which is defined as a pushout. The uniqueness of the representation is shown in Serre's Trees, Section 1.2, by application of the universal property to the symmetric group over all reduced words. There is no need to impose a group operation on the set of reduced words (which would be very tedious) - after this proof you get it for free! -Localization. Let's consider a commutative ring $A$ and a map $i : S \to |A|$ into the underlying set of $A$. Then we can consider the subfunctor of $\mathrm{Hom}(A,-)$ which is given by homomorphisms $g : A \to B$ such that $gi$ factors through $B^*$. By general theorems a representing object exists and is usually denoted the localization $S^{-1} A$ when $i$ is understood. If $S'$ denotes the multiplicative closure of the image of $i$, then $S'^{-1} A = S^{-1} A$ (they satisfy the same universal property), thus we always may assume that $S$ is just a multiplicative closed subset of $A$. -Let $\tau : A \to S^{-1} A$ be the universal homomorphism which maps $S$ to units. Then $S^{-1} A = \{\tau(a) \tau(s)^{-1} : a \in A, s \in S\}$ (they satisfy the same universal property). Let es write $\frac{a}{s} = \tau(a) \tau(s)^{-1}$. Thus every element in $S^{-1} A$ is some fraction $\frac{a}{s}$. Now when are two such fractions equal? Of course we know this from the usual construction using equivalence classes of pairs $(a,s)$, but I want to avoid this construction and derive it only with the universal property. -It is clear that $s'a = sa'$ implies $\frac{a}{s} = \frac{a'}{s'}$. From this one easily derives more generally that $ts'a = tsa'$ also implies $\frac{a}{s} = \frac{a'}{s'}$; here $a,a' \in A$ and $s,s',t \in S$. -Question. Is there a proof of the converse, i.e. $\frac{a}{s} = \frac{a'}{s'} \Rightarrow \exists t \in S : ts'a = tsa'$, which only uses the universal property of the localization? -First observe that $\frac{a}{s} = \frac{a'}{s'}$ iff $\frac{sa'-sa'}{1}=0$. Thus it is enough to show $\frac{a}{1} = 0 \Rightarrow sa=0$ for some $s \in S$, i.e. that the kernel of $\tau : A \to S^{-1} A$ equals $I=\cup_{s \in S} \mathrm{Ann}(s)$. Since $S$ is multiplicative, $I$ is an ideal, and we may replace $A$ by $A/I$, where localization commutes with quotients because of universal properties. Thus we may assume that $I=0$, i.e. that $S$ consists of regular elements. Our task is then to show that $A \to S^{-1} A$ is injective. But a priori we only know by the universal property that the kernel of $A \to S^{-1} A$ equals the intersection of all kernels of homomorphisms $A \to B$ which map $S$ to units. Of course we have to use this and construct some specific $A \to B$, but I hope that we can either avoid some nasty element construction of $B$ or that we can just use a ring which is built up out of $S^{-1} A$, but can be used to show that the kernel is zero - therefore to find a kind of self-referential proof. The examples above suggest that this might be possible after all. -Another approach is to use $S^{-1} A = A[\{X_s\}_{s \in S}]/(s X_s - 1)$ (which is clear since both sides satisfy the same universal property). Here the polynomial algebra should be defined by its universal property, as well as the quotient ring. Assuming that we know the structure of its elements, then one can also show that the kernel of $\tau$ is as desired; see the note "Rings of fractions the hard way" by José Felipe Voloch. I think this is quite interesting, but it basically just gives another construction of the localization in terms of polynomial algebras and quotients, and then computes the kernel of $\tau$ from this specific construction. So this is not really what I'm after. -Moosbrugger suggest to talk about the category of $A$-modules. Here is a basic observation which follows from the universal property: If $R$ is a possibly noncommutative $A$-algebra, then there is at most one algebra homomorphism $S^{-1} A \to R$, and it exists iff all elements of $S$ become invertible in $R$. This follows easily from the universal property of $S^{-1} A$ applied to the center of $R$. Now let $M$ be an $A$-module and apply the above to $R=\mathrm{End}_A(M)$. It follows that the category of $S^{-1} A$-modules is equivalent to the category of $A$-modules, on which the elements of $S$ act as isomorphisms. -Therefore a possible formalization of my question might be the following: Let $S \subseteq A$ as above and let $B$ be a commutative $A$-algebra such that scalar restriction $\mathrm{Mod}(B) \to \mathrm{Mod}(A)$ is fully faithful and whose image consists of those $A$-modules on which the elements of $S$ act as isomorphisms. Can we then compute the kernel of $A \to B$ (without refering to the explicit construction of the localization $S^{-1} A$, but only using this statement about module categories, which of course yields $B \cong S^{-1} A$ by Morita). -Motivation. I hope that a categorical proof makes the usual construction of the localization (via equivalence classes of pairs) redundant. Note also that it is rather nasty to prove all the details (equivalence relation, well-defined addition, well-defined multiplication, universal property) with the usual construction. Also the definition $(a,s) \sim (a',s') \Leftrightarrow \exists t : ts'a=tsa'$ is not motivated at all there. A categorical proof should show in particular that this is the right choice, and not some random definition which turns out to be correct only after some computation. But of course I don't claim that a categorical construction of the localization is the easiest or best one. -Although the universal property is probably the most important aspect of localization (which we get, as already mentioned, by abstract nonsense), the criterion for the equality of elements is essential even for basic properties in commutative algebra, for example in order to prove that every integral domain embeds into its field of quotients. -On the other hand, localization could be seen just as a toy example for other, more involved examples, where it is not clear at all how to understand some representing object of some functor, which exists my general nonsense. - -REPLY [4 votes]: I've had this question on my mind for some time now. While I appreciate its spirit, I confess that I've had a hard time imagining what a satisfactory answer would be. But I'm going to be brave and propose a solution anyways. Let me know if it misses the mark. -I'm interpreting the question as Andreas Blass did in the meta-discussion: given $a \in A$ not annihilated by anything in $S$, exhibit a ring $B$ and a homomorphism $A \to B$ that inverts $S$ but doesn't annihilate $a$. In fact, this homomorphism $A \to B$ will be outright injective. -This is a module-theoretic approach that you can find in greater detail in Theorem 6.2 of An introduction to noncommutative Noetherian rings (2nd ed.) by Goodearl and Warfield. (That theorem constructs the ring of fractions at a right Ore set of regular elements of a noncommutative ring. Things are a little easier here because $A$ is assumed to be commutative.) -As Martin has already pointed out, we may assume that $S$ consists of regular elements. Let $E(A)$ be the injective hull of the module $A_A$ (being a noncommutative algebraist, I can't help but use right modules) and let $M = \{x \in E(A) : xs \in A \textrm{ for some } s \in S\}$, which is a submodule of $E(A)$. Define $B = \mathrm{End}_A(M)$. Scalar multiplication by an element of $A$ is an endomorphism of $M$, and this gives us a ring homomorphism $\phi \colon A \to B$. This morphism is injective because the element $1 \in A \subseteq M$ serves to "separate elements" of $\phi(A)$. It remains to show that $\phi$ inverts $S$. -First notice that $E(A)$, hence $M$, is $S$-torsionfree. For let $0 \neq m \in E(A)$. There exists $a \in A$ such that $0 \neq ma \in A$ because $A$ is essential in $E(A)$. Since $A$ is $S$-torsionfree, for any $s \in S$ we have $(ms)a = (ma)s \neq 0$. It follows that $ms \neq 0$. -Next notice that any element of $B = \mathrm{End}_A(M)$ is uniquely determined by its value at $1 \in A \subseteq M$. For suppose $f,g \in B$ with $f(1) = g(1)$. Let $m \in M$, and fix $s \in S$ such that $ms \in A$. Then $f(m)s = f(ms) = f(1)\cdot(ms) = g(1)\cdot(ms) = g(ms) = g(m)s$. Since $M$ is $S$-torsionfree, we conclude $f(m) = g(m)$. Hence $f = g$. -Finally, let $s \in S$, and let's prove that $\phi(s)$ is invertible in $B$. By the previous paragraph, it suffices to construct $h \in B$ such that $h(s) = 1$. For then $h \circ \phi(s) \colon M \to M$ sends $1 \mapsto s \mapsto 1$, so this map must be equal to $\mathrm{id}_M$. -Because $s$ is regular, the ideal $sA$ is a free module, so there is a module homomorphism $h_0 \colon sA \to M \subseteq E(A)$ sending $s \mapsto 1 \in A \subseteq M$. By injectivity of $E(A)$, this extends along the embedding $sA \subseteq A \subseteq E(A)$ to an endomorphism $h' \colon E(A) \to E(A)$ that satisfies $h'(s) = h_0(s) = 1$. To see that $h'$ restricts to an endomorphism of $M$, let $m \in M$ and fix $t \in S$ such that $mt \in A$. Then $ts \in S$ satisfies $$h'(m)(ts) = h'(mt)s = h'(1)(mt)s = h'(s)(mt) = 1 \cdot mt = mt \in A,$$ proving that $h'(m) \in M$. Thus $h = h'|_M \in B$ is the desired morphism. -(Goodearl and Warfield's analysis actually proves that $B = S^{-1}A$ and that $B_A \cong M_A$ as modules. This requires only a little more work.)<|endoftext|> -TITLE: Combinatorial proof for the number of lattice paths that return to the axis only at times that are a multiple of 4 -QUESTION [13 upvotes]: Consider lattice paths consisting of $2n$ steps, each of which is either $(1,1)$ or $(1,-1)$. The number of such lattice paths that return to the horizontal axis only at times that are a multiple of $4$ is given by $2^n \binom{n}{n/2}$. Can someone provide a combinatorial proof of this fact? - -Background: A few months ago on math.SE I asked for a combinatorial proof of the identity $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2},$$ when $n$ is even. -The non-alternating version is $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} = 4^n.$$ There are several combinatorial proofs of the non-alternating version, and I hoped to adapt one of them. One such proof is that $\binom{2k}{k} \binom{2n-2k}{n-k}$ counts the number of paths of length $2n$ with $2k$ steps above the horizontal axis and $2n-2k$ steps below it. Summing up over all values of $k$ gives the total number of paths of length $2n$, which is $2^{2n} = 4^n$. (I believe I saw this argument in Feller's An Introduction to Probability Theory and Its Applications. It's also in this note by David Callan.) -If we take the alternating version, the paths with positive parity are those with $2k$ steps above the axis for $k$ even, and the paths with negative parity are those with $2k$ steps above the axis for $k$ odd. For each path, break it every time it returns to the horizontal axis. This partitions each path into a number of segments equal to the number of times it returns to the axis. For every path that has a last segment consisting of $2j$ steps for $j$ odd, flip this segment over the horizontal axis. This mapping is an involution and changes the path's parity. Since every odd-parity path must have at least one such odd segment, $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k$$ must count the number of paths that have no odd segment; i.e., the number of paths whose returns to the horizontal axis occur only at multiples of $4$. If $n$ is odd, there are no such paths without an odd segment, but if $n$ is even, there are apparently $2^n \binom{n}{n/2}$ of them. -However, I was unable to find an independent combinatorial proof that $2^n \binom{n}{n/2}$ counts the number of lattice paths of length $2n$ with no odd segments. (Again, an "odd" segment here is one of length $2j$, where $j$ is odd.) The math.SE question remained unanswered for over two months until I found a different way to prove the identity I was after combinatorially, but this other way doesn't involve lattice paths. After all the time I spent trying the lattice path approach I would like to see an independent combinatorial proof that $2^n \binom{n}{n/2}$ counts the number of lattice paths whose return times to the axis are multiples of $4$. - -REPLY [7 votes]: For the record, this question is answered in a pair of papers: -Gábor V. Nagy, A combinatorial proof of Shapiroʼs Catalan convolution, In Advances in Applied Mathematics, Volume 49, Issues 3–5, 2012, Pages 391-396 -Péter Hajnal and Gábor V. Nagy. A Bijective Proof of Shapiro’s Catalan Convolution. Electronic Journal of Combinatorics, 21(2), 2014 -In particular Lemma 7 of the first paper is, I think, a satisfactory answer to this question. The second paper gives a fully bijective proof. -Thanks to Richard Stanley. - -These paths also have an interesting interpretation as NSEW lattice paths (where the set of steps is the four cardinal directions N, S, E, W). They correspond, under the usual bijection between one-dimensional paths of length $2k$ and two-dimensional paths of length $k$, to NSEW lattice paths from $(0,0)$ of length $n$ that intersect the $x$-axis only at positions with an even $x$-coordinate. In other words, to paths on the perforated plane obtained by removing (punching out) all the odd-numbered lattice points on the $x$-axis.<|endoftext|> -TITLE: A 2-category of chain complexes, chain maps, and chain homotopies? -QUESTION [14 upvotes]: First-time here... I hope my question isn't silly or anything... anyway... -Consider the category of chain complexes and chain maps. We can also define chain homotopies between chain maps. Does this form a 2-category? I am able to construct vertical and horizontal composition of chain homotopies but am unable to prove that the horizontal composition is associative and that the interchange law holds. -I like to include a lot of details in case the reader isn't familiar with everything, so the following is the problem in detail. -To be a bit more concrete, let's let $(C_n, \partial_n), (C'_n, \partial'_n),$ and $(C''_n, \partial''_n)$ be chain complexes (sorry, my differential is going up in degree) and let $f_n, g_n, h_n: C_n \to C'_n$ and $f'_n, g'_n, h'_n : C'_n \to C''_n$ be chain maps. Suppose that $\sigma : f \Rightarrow g,$ $\tau : g \Rightarrow h,$ $\sigma' : f' \Rightarrow g',$ and $\tau' : g' \Rightarrow h'$ are chain homotopies, where $\sigma_n, \tau_n : C_n \to C'_{n-1}$ and similarly for $\sigma'$ and $\tau'.$ Recall that these definitions say -$f_n \circ \partial_{n-1} = \partial'_{n-1} \circ f_{n-1}$ -and similarly for primes, $g$'s, $h$'s, and -$\sigma_{n+1} \circ \partial_{n} + \partial'_{n-1} \circ \sigma_{n} = f_{n} - g_{n},$ -$\tau_{n+1} \circ \partial_{n} + \partial'_{n-1} \circ \tau_{n} = g_{n} - h_{n},$ -and similarly for primes. -We can define the vertical composition of $\sigma : f \Rightarrow g$ with $\tau : g \Rightarrow h$ by -$(\tau \diamond \sigma)_{n} := \tau_n + \sigma_n$ -(I denote vertical composition with a diamond, $\diamond$). -We can also define the horizontal composition of $\sigma : f \Rightarrow g$ with $\sigma' : f' \Rightarrow g'$ by -$(\sigma' \circ \sigma)_n := \partial''_{n-2} \circ \sigma'_{n-1} \circ \sigma_{n} - \sigma'_{n} \circ \sigma_{n+1} \circ \partial_{n} + \sigma'_{n} \circ g_{n} + f'_{n-1} \circ \sigma_{n}.$ -Now we just verify that these are indeed chain homotopies. The vertical composition is easy: -$(\tau \diamond \sigma)_{n+1} \circ \partial_{n} + \partial'_{n-1} \circ (\tau \diamond \sigma)_{n} = f_n - g_n + g_n - h_n = f_n - h_n$ -and the horizontal composition is a bit more challenging but doable (I won't include the derivation here). -The identity for vertical composition is the zero map and similarly for the horizontal composition (note that for the vertical composition, the zero map is a chain homotopy between any two chain maps provided they are the same while for the horizontal composition, the chain maps are both the identities). The vertical composition is easily seen to be associative. However, the horizontal composition satisfies -$( \sigma'' \circ ( \sigma' \circ \sigma ) )_{n} - ( ( \sigma'' \circ \sigma' ) \circ \sigma )_n$ -$= ( f''_{n-1} - g''_{n-1} ) \circ ( f'_{n-1} - g'_{n-1} ) \circ \sigma_{n} - \sigma''_{n} \circ ( f'_{n} - g'_{n} ) \circ ( f_n - g_n ).$ -The interchange law doesn't hold either and the difference is given by -$( ( \tau' \diamond \sigma' ) \circ ( \tau \diamond \sigma ) )_{n} - ( ( \tau' \circ \tau ) \diamond ( \sigma' \circ \sigma ) )_{n}$ -$= ( f'_{n-1} - g'_{n-1} ) \circ \tau_{n} + (g'_{n-1} - h'_{n-1} ) \circ \sigma_{n} - \tau'_{n} \circ ( f_{n} - g_{n} ) - \sigma'_{n} \circ (g_{n} - h_{n} )$ -So I'm not sure if chain complexes, chain maps, and chain homotopies are supposed to form a 2-category, but I would've liked this result to be true. Does anyone know what the correct categorical structure of this category is? I have not yet considered chain homotopies of chain homotopies, so if the answer requires all such higher morphisms, then an answer in that direction would also be acceptable. -Any thoughts? Thanks in advance. - -REPLY [5 votes]: Viewing homotopy via interval objects analogously to topological homotopy as done in -Qiaochu Yuan's answer is certainly the best way to consider the problem. Nevertheless it's also possible to get along directly with your definition of homotopy. -In this point of view the problem is that your horizontal composition isn't appropriate: First note that we have $\sigma: f \Rightarrow g,\; \sigma': f' \Rightarrow g'$. Thus (I just write $f'f$ for $f'\circ f$) -$$f'f - g'g = f'f - f'g + f'g - g'g = ...=\partial^{''}(f'\sigma+\sigma'g) + (f'\sigma+\sigma'g)\partial. $$ -Therefore $\sigma' \circ \sigma:= f'\sigma+\sigma'g: f'f \Rightarrow g'g$ is a homotopy. -Let $\sigma'': f'' \Rightarrow g''$ be another homotopy. By setting in the definition of the former homotopy, one finds that -$$(\sigma'' \circ \sigma') \circ \sigma = f''f'\sigma + f'' \sigma' g + \sigma'' g'g = \sigma'' \circ (\sigma' \circ \sigma)$$ -is a homotopy $(f'' f')f = f''(f'f) \Rightarrow g''(g'g) = (g''g')g$. -Hence the horizontal composition is associative.<|endoftext|> -TITLE: Is the category of metric spaces and continuous maps Quillen equivalent to Top? -QUESTION [5 upvotes]: I am looking for models of ${\mathsf{Top}}$ distinct from modifications of simplicial sets. The above question should be understandable to the reader. I'll add more details when I get access to a proper computer. Let ${\mathsf{Met}}$ be the category of metric spaces and continuous maps. Then there is an embedding ${\mathsf{Met}}\hookrightarrow {\mathsf{Top}}$. Is this embedding a Quillen equivalence? -Edit: Professor May explains below that my question is not precise, per se, as there is more than one Quillen inequivalent model categorial structures on ${\mathsf{Top}}$. -Edit: The answer is no. Tom commented that the category ${\mathsf{Met}}$ does not have all small colimits. Thus the embedding ${\mathsf{Met}}\hookrightarrow {\mathsf{Top}}$ cannot be a Quillen equivalence. - -REPLY [8 votes]: The phrasing of your question prompts me to emphasize a model categorical difference between spaces and simplicial sets. With the standard weak equivalences, there is just one standard model -structure on simplicial sets. But with spaces, there is a natural trichotomy of interrelated model structures, two of which are Quillen equivalent to each other and to simplicial sets. -There is an h-model structure with actual homotopy equivalences as weak equivalences -and with Hurewicz (or h) cofibrations and fibrations. There is a q-model structure with weak homotopy equivalences as weak equivalences, Serre fibrations as q-fibrations, and retracts of relative cell complexes as q-cofibrations. And there is a mixed (or m) model structure with the q-equivalences and h-fibrations as the m-equivalences and m-fibrations. The m-cofibrant objects are the spaces of the homotopy types of CW-complexes, and algebraic topology over most of its history has implicitly worked in the m-model structure. The trichotomy carries over to chain complexes. A recent exposition is in the book ``More concise algebraic topology: localization, completion, and model categories'' by Kate Ponto and myself.<|endoftext|> -TITLE: On two spectral sequences for the cohomology of a double complex -QUESTION [6 upvotes]: For a (bounded) double complex (of abelian groups or vector spaces) one can consider two spectral sequences that converge to the cohomology of the totalization: one can first compute either the cohomology of rows, or the cohomology of columns. Suppose that one of these spectral sequences degenerates at $E_1$ (i.e. the cohomology of rows yields the factors of the induced filtration of the limit). Do any 'nice' properties of the second spectral sequence follow? - -REPLY [28 votes]: There is a basic way to see whether things like this should be true. Any bounded double complex of vector spaces over a field $k$ is (noncanonically) the direct sum of complexes of the following two sorts: -Squares: $$\begin{matrix} k & \rightarrow & k \\ \uparrow & & \uparrow \\ k & \rightarrow & k \end{matrix}$$ -Staircases:$$\begin{matrix} -k & \rightarrow & k & & & & \\ - & & \uparrow & & & & \\ -& & k & \rightarrow & k & & \\ - & & & & \uparrow & & \\ -& & & & k & \rightarrow & k\\ -\end{matrix}$$ -We'll say that the "length" of a staircase is the number of nonzero entries in it, so the above stair case has length $6$. Staircases may have even odd or even length, and may start and end either with vertical or horizontal maps. -The operation of "forming the spectral sequence" commutes with direct sum, so it is enough to know what the spectral sequences of each of these look like. -The spectral sequence of a square is zero on every page except for the square itself; in particular, it converges in one step. -The spectral sequence of an odd length staircase is one dimensional on every page except for the stair case itself. The one nonzero term is at one end of the staircase for the horizontal spectral sequence and the other for the vertical spectral sequence. So it also converges in one page. -The spectral sequence of an even staircase is the one which violates your claim. In one direction, it is zero on every page after the staircase itself. In the other direction, it there are $m$ pages with two nonzero terms, where the length of the staircase was $2m$. These terms are at the two ends of the stair case, and they annihilate each other on the $(m+1)$st page. -In particular, a double complex which consists simply of a length $2m$ staircase will die on the first page in one direction, but will survive for $m$ pages in the other.<|endoftext|> -TITLE: measure theory and continuum hypothesis -QUESTION [5 upvotes]: let's assume $\neg CH$, then there's a set $X$ such that $|\mathbb N|<|X|<|\mathbb R|$. -i'm wondering about the lebesgue measure of such set... is it even possible to measure it? would it be possible that its measure is more than $0$? i think not, because all the subset of cantor set are measure $0$ sets, and there would be a set $K$ such that $|X|=|K|$ and $K\subset Cantor$. -is it correct? - -REPLY [5 votes]: That set only have two options: it would be measurable (of $0$ measure) or it will be non-measurable. You can prove that all the measurable sets of positive measure have the same cardinality as the Continuum. In case that the set is non-measurable it's outer measure would be positive.<|endoftext|> -TITLE: Persistent homology of Markovian dynamical systems -QUESTION [7 upvotes]: Consider a dynamical system $(T,X)$ that admits a Markov partition $\mathcal{M}$ (e.g., an Anosov map), and consider the corresponding 0-1 transition matrix $A$. It is commonplace to study information about the growth of the number of closed orbits as a function of iterates through a zeta function. -However, it seems to me that there may be another approach based on persistent homology (which would be interesting). Assume for convenience that $X$ is connected, so that $A$ corresponds to a connected graph. Consider the usual graph distance and the filtration of Vietoris-Rips complexes on $[|\mathcal{M}|]$. Intuitively I would guess that the dimension 1 persistent homology would encode the same sort of information as the zeta function. But this approach could give higher dimensional information. - -So my question is: using the graph distance for the transition matrix of a dynamical system, can we use the persistent homology of the corresponding Vietoris-Rips filtration to obtain useful information about the dynamical system? - -Addendum: I realize that this might involve generalized persistence, as the partition diameter could come into play. - -REPLY [5 votes]: The answer to the question as stated seems to be "no". -Consider a three-element Markov partition $\mathcal{M} = \{A, B, C\}$ with directed edges $(A,B)$, $(B,C)$ and $(C,A)$. There is an obvious periodic orbit $A \to B \to C \to A$ but the Vietoris-Rips complex never has non-trivial persistent homology in dimension $1$ and completely fails to detect this orbit. -In particular, the Vietoris-Rips complex $\text{VR}_t$ has three vertices for $t \in [0,1)$ and becomes a $2$-simplex for $t \geq 1$. One can generalize this construction to length $k$ periodic orbits to get a huge family of dynamical systems with non-trivial behavior but with contractible Vietoris-Rips complexes. The problem is that the Vietoris-Rips complex instantly "fills-up" all simplices whose $1$-skeleta are present. - -But if one uses something different from the Vietoris-Rips complex, then there is hope to extract topological invariants of your dynamics. Consider for example the following filtration: each vertex of the Markov graph is born at scale $0$, each edge is born according to the graph distance as usual, so up to the $1$-skeleton we agree with the Vietoris-Rips complex. But now, introduce each $d$-simplex (whose edges are all present) at the sum of all edge birth involved in its $1$-skeleton. -So for instance in the example above, we would have a length $2$ barcode in the $1$-dimensional persistent homology which captures the periodic orbit involving $A, B$ and $C$: the cycle forms at scale $1$ but the $2$-simplex does not make it a boundary until scale $3$!<|endoftext|> -TITLE: If $A\in \mbox{Rings}\subset E_\infty\mbox{-rings}$, what is the equivalence between objects of $\mathcal{D}(\mbox{Mod}_A)$ and $A$-module spectra? -QUESTION [7 upvotes]: In Lurie's "A Survey of Elliptic Cohomology", he writes on page 14 that if $A$ is an ordinary commutative ring considered as an $E_\infty$-ring, then $A$-module spectra are the same thing as objects of the derived category of $A$-modules. This is mysterious to me. On the one hand, to an $A$-module spectrum $M$ we might associate the $A$-modules $\pi_n(M)$, but I don't know of any interesting maps between these; perhaps this will just end up being the homology of any representative chain complex of $A$-modules. But then, I certainly don't see a natural way of getting from an object of $\mathcal{D}(\mbox{Mod}_A)$ to an $A$-module spectrum. -Incidentally, what does this induce on the level of categories? The obvious first guess is that $A$-module spectra actually form a topological category and that passing to $\mathcal{D}(\mbox{Mod}_A)$ applies $\pi_0$. - -REPLY [8 votes]: This question already has been answered in the comments. -(Tilson) We regard a commutative ring as an $E_\infty$ spectrum via the EM functor $H$. This is definitely what Jacob is doing. One could also use associative rings and $A_\infty$ spectra for what follows. -(Wilson) Many of the correspondences between algebra and stable homotopy theory are described in Chapter 7 in Lurie's higher algebra book. -(Muro) The correspondence between algebras/modules and the associated EM-spectra is laid out in math.uic.edu/~bshipley/zdga17.pdf (Cor 2.15) which depends on her paper with Stefan Schwede "Equivalences of monoidal model categories." -It is a bit technical, but is easier to work out the correspondence if you restrict to non-negatively graded $\mathbb{Z}$-chain complexes and connective $H\mathbb{Z}$-modules. The correspondence can be spread into two stages: -1) Use the Dold-Kan correspondence to move between chain complexes and simplicial abelian groups. -2) Take the geometric realization of your simplicial abelian group which is a topological abelian group and hence an infinite loop space, so we can take its associated connective spectrum (by repeatedly applying the bar construction). The fact that geometric realization preserves products can then be used to see that this spectrum is an $H\mathbb{Z}$-module. -Now given an $H\mathbb{Z}$-module $M$ we can form the associated simplicial abelian group $H\mathbb{Z}-mod(H\mathbb{Z}\wedge\Sigma^\infty_+ \Delta^i, M)$ to go back. -This equivalence induces an equivalence their associated stable infinity categories.<|endoftext|> -TITLE: What is the metamathematical interpretation of knot diagrams? -QUESTION [12 upvotes]: I am not a geometric topologist, but from looking over papers in the field, it's clear that knot diagrams are a major tool and we know how to use them in a way that is rigorous and trustworthy. My background is in model theory and I am having trouble fitting them into that framework. I'm hoping for pointers to references or a quick sketch of the logical status of these things. -Specific points that I'm hung up on: knot diagrams have nice properties analogous to terms in formulas, like substitutability a la planar algebras. On the other hand, they are relation-like, relating segments of a link to each other. On the third hand, the Reidemeister moves seem like a set of formulas in a model-theoretic interpretation (of a "theory" of knot isotopy into a "theory" of graphs.) Finally, there's the standard trick of calculating invariants by recursively applying certain skein relations to get to the unknot. -In other words, I can't see that a knot diagram is always being used as a relation, term, formula, or substructure - it seems like none of these is adequate to fully describe their use. -I can see this playing out in a number of ways: - -Knot diagrams are a tool that can be completely subsumed by algebra in an algorithmic way, and are just a convenience; -There is a theorem that says proofs using diagrams can be "un-diagrammed", but it's an existence proof -People keep finding new ways to use knot diagrams in proofs, often by decorating the diagrams with new features like orientations - reproving via alternate techniques is then a useful contribution, but always tends to work out; e.g., we don't completely understand the metamathematics of knot diagrams, but the general shape of things is clear; -There are important theorems with no known proof except via diagrams, and nobody knows why; -It will somehow all become obvious if I take the right course on planar algebras, or o-minimal structures, or category theory; -It's subtle, but was all cleared up by Haken in the 70's; -Dude, it's just Reidemeister's Theorem, and you need to go away and think about it some more. - -Community wiki, in case the right answer is a matter of opinion. -Update - Just to be clear, this is not in any way a brief for eliminating knot diagrams - quite the opposite. Knot diagrams are honest mathematical objects, while also serving as syntax for other objects. That seems like a ripe area for mathematical logic. -Also, I'm including partial diagrams, as in skein relations, when I use the phrase "knot diagram". - -REPLY [4 votes]: I see this as a general problem in higher dimensional algebra, that there will need to be "higher dimensional rewriting". John Baez has illustrated the higher dimensional thinking by displaying the picture -$$ ||| \;\; ||| || $$ -$$||| \;\; ||||| $$ -which is easily seen to illustrate $2 \times (3+5)= 2 \times 3 + 2 \times 5$ but the 1-dimensional formula involves various conventions, and is less transparent. We have found diagrammatic rewriting useful in dealing with rotations in double groupoids (with connections), and there is a 3-dimensional rewriting argument in Section 5 of -F.-A. Al-Agl, R. Brown, R. Steiner, `Multiple categories: the -equivalence between a globular and cubical approach', Advances in -Mathematics, 170 (2002) 71-118. -which proves a key braid relation (Theorem 5.2). -So there is the interesting question of how to cope say with a 5-dimensional rewrite? Maybe computers could handle it? -These situations could well occur in algebras with partial operations whose domains are defined by geometric conditions, and with strict axioms.<|endoftext|> -TITLE: Does the quaternion group Q_8 have a presentation of this form? -QUESTION [6 upvotes]: In The number of conjugacy classes and the order of the group, a key step in the proof of the congruence saying that $G$ being a $p$-group means that $|G| \equiv c(G) \mod{(p^{2}-1)(p-1)}$ is the construction of a finite group (denoted $P$ in that post) whose relations, as given, are satisfied by \emph{every} word in the generators. -Let $Q$ denote a finite $p$-group which has a presentation of the kind possessed by $P$. More precisely, suppose $Q$ is generated by $x_{1}, \ldots, x_{r}$ and that the relations defining $Q$ are given in infinite families, where each family is of the form $w( e_{1}, \ldots, e_{L} ) = 1$, where $w$ is some word in the free group on $L$ generators and $e_{1}, \ldots, e_{L}$ vary independently over all words in $x_{1}, \ldots, x_{r}$. -The reasoning from before should apply (though I cannot fill in all gaps) to describe $\mathrm{Aut}(Q)$ (and obtain its order): -Since $Q$ is finite, an endomorphism $\phi: Q \to Q$ is injective iff it is surjective. So an endomorphism of $Q$ is an Automorphism iff it is surjective. If $\phi(x_{1}), \ldots, \phi(x_{r})$ do not generate all of $Q$, then the subgroup they generate is contained (since $Q$ is finite) in some maximal subgroup $M$ of $Q$. $M$ contains the Frattini subgroup $\Phi(Q)$, so this endomorphism fails to be surjective when it is converted to an endomorphism of the maximal elementary abelian quotient $Q/ \Phi(Q)$. -Gap 1: I do not immediately see how to prove that $|Q/ \Phi(Q)| = p^{r}$. It is clear that, since $Q$ is generated by an $r$-element set, $|Q/ \Phi(Q)| \leq p^{r}$. The nature of the relation set should imply that $Q$ is trivial if $|Q/ \Phi(Q)| < p^{r}$. -Assuming Gap 1 is fillable, when the words expressing $\phi(x_{1}), \ldots, \phi(x_{r})$ are written in terms of $x_{1}, \ldots, x_{r}$, one can form the matrix whose $(i,j)$ entry is the sum of the exponents, regarded as an element of $\mathbb{Z}/(p)$, to which $x_{i}$ appears in $\phi(x_{j})$. -Gap 2: I do not immediately see how to prove that element of $\mathbb{Z}/(p)$ is well-defined. -Assuming Gap 2 is also fillable, the matrix thus obtained will be invertible iff $\phi$ is surjective, and thus an \mathrm{Aut}omorphism of $Q$. Since all $r \times r$ matrices over $\mathbb{Z}/(p)$ arise this way from endomorphisms of $Q$ (assuming Gap 1 is filled), construction of this matrix yields a surjective homomorphism from $\mathrm{Aut}(Q)$ to $\mathrm{GL}(r,\mathbb{Z}/(p))$. What is the kernel of this homomorphism? -The kernel consists of all preimages of the identity matrix under that abelianization modulo $p$ map. If $|Q| = p^{E}$, assuming Gap 1 is filled, $|\Phi(Q)| = p^{E-r}$. Then any kernel element can be obtained by sending $x_{i}$ to $x_{i}f_{i}$ for all $i$, where $f_{i} \in \Phi(Q)$. This means that the order of the kernel is $p^{r(E-r)}$, and that $|\mathrm{Aut}(Q)| = p^{r(E-r)} \Pi_{k=0}^{r-1} (p^{r}-p^{k})$. -Going back to this question for the quaternion group Q_{8}, $\mathrm{Aut}(Q_{8}) \cong S_{4}$ so $\mathrm{Aut}(Q_{8})$ has $\mathrm{GL}(2, \mathbb{Z}/(2) ) \cong S_{3}$ as a quotient. The kernel of this quotient map consists of the automorphisms of $Q_{8}$ which send each element of $Q_{8} \setminus \Phi(Q_{8})$ to itself, except possibly for sign. Since { $\pm{1}$ } $ = \Phi(Q_{8})$, this sounds exactly like what was covered in the discussion of $\mathrm{Aut}(Q)$ above. -Yet I still have trouble obtaining a presentation of the kind $Q$ satisfies for the quaternion group of order 8: -One may say that, in $Q_{8}$, all elements have order dividing 4: $e^{4} = 1$ for all words $e \in \langle x,y \rangle$. -One may also say that, in $Q_{8}$, all elements square to an element of the center: $e_{1}^{2}e_{2} = e_{2}e_{1}^{2}$ for all words $e_{1}, e_{2} \in \langle x,y \rangle$. -These two statements do not capture $Q_{8}$ completely, since the group with those relations turns out to have order $32$: -$T = \langle x,y| e^{4}=1, e_{1}^{2}e_{2} = e_{2}e_{1}^{2} \rangle$ is nonabelian, since the nonabelian group $Q_{8}$ is one of its quotients. -Now consider the element $x^{2}y^{2}(xy)^{2} \in T$. This element is clearly a product of squares, so it is a product of central elements of order dividing 2. It therefore has order dividing 2 itself. -Yet $T/\langle x^{2}y^{2}(xy)^{2} \rangle$ is abelian, since $x^{2}y^{2}(xy)^{2} = 1$ implies $x^{2}y^{2} = (xy)^{-2})$, which implies (since every element has order dividing 4) $x^{2}y^{2} = (xy)^{2} $, which implies (left-multiplying by $x^{-1}$ and right-multiplying by $y^{-1}$) $xy = yx$. -Therefore $x^{2}y^{2}(xy)^{2}$ is a nontrivial element of $T$, so it has order 2. $T/\langle x^{2}y^{2}(xy)^{2} \rangle$ is, therefore, the largest 2-generated abelian group of exponent 4 (and being abelian subsumes the centrality of squares). So $T/\langle x^{2}y^{2}(xy)^{2} \rangle \cong C_{4} \times C_{4}$ and it is now clear that $|T| = 32$. -The question indicated by the title asks whether more relations can be added, in this symmetric fashion, to $T$ to obtain $Q_{8}$. The natural extension of this question is whether or not there exists a 'nice' criterion for looking at an arbitrary $p$-group and determining whether or not it has a presentation of the kind discussed here. Clearly the discussion of $\mathrm{Aut}(Q)$ gives some preliminary necessary conditions for the existence of such a presentation. - -REPLY [11 votes]: A group given by your presentation for words $w_1,...,w_n$ is called relatively free in the variety of groups given by the identities (laws) $w_1=1,...,w_n=1$. For a characterization of relatively free groups see Hanna Neumann's book "Varieties of groups" . The characterization says that $G=\langle X\rangle$ is relatively free with free generating set $X$ iff every map $\phi\colon X\to H$ into a group $H$ satisfying the same identities as $G$ can be extended to a homomorphism $G\to H$ (the extension is then unique). The variety generated by the quaternion group $Q_8$ (i.e. the smallest variety containing that group) is also described in Hanna Neumann's book. It is defined by three identities $[x^2,y]=1, [[x,y],z]=1, x^4=1$ (the second identity actually follows from the other two and is redundant). Also $Q_8$ is 2-generated, so if it was relatively free, its rank would be 2. But the dihedral group of order 8 generates the same variety (see the book), and all homomorphisms from $Q_8$ into $D_8$ have Abelian images. Thus $Q_8$ is not relatively free.<|endoftext|> -TITLE: Weight lattice and the fundamental group -QUESTION [9 upvotes]: Let $G$ be a compact connected Lie group and let $T$ be a maximal torus of $G$, with Lie algebras $\frak{g}$ and $\frak{t}$ respectively. Then, $\frak{t}$ can be considered as a Cartan subalgebra of $\frak{g}$. It is known that the kernel of exponential map $exp : \frak{t} \to$ $T$ is the lattice of all integral weights of $\frak{g}$, i.e. weihts $\lambda \in (it)^*$ such that $\lambda(H)\in 2\pi i\mathbb{Z},$ whenever $exp H= I$ for $H\in\frak{t}$. -I have the following questions: -1) What is the relation between the fundamental group $\pi_{1}(G)$ of $G$ with the integral lattice described above? I am trying to find any good references about this fact, but it seems difficult. -2) How we can use the fibration $T\to G$$\to G/T$ to compute $\pi_{1}(G/T)$? (answered) -3) What we can say about the second homotopy group $\pi_{2}(G)$? (answered) -4) Is it true, that if $G$ is semisimple, then $\pi_{1}(G)$ is finite? (answered) -Thank you! - -REPLY [2 votes]: Regarding Q3, one more explanation of why $\pi_2(G)=0$ (which is similar to what Claudio Gorodski points out above) can be found here. A nice observation made there is that one can use the same arguments to show that $\pi_3(G)$ is torsion-free! (which I guess was not in your list of facts about homotopy of compact Lie groups).<|endoftext|> -TITLE: Kernel of the representation of the mapping class group to $Aut(F_n)$ -QUESTION [5 upvotes]: Let $S_{g,1}$ be a orientable compact surface of genus $g$ with one boundary component and $\Gamma_{g,1}$ the mapping class group. -By $F_n$ I denote the free group on $n$ generators. -One obtains a representation $\rho: \Gamma_{g,1} \rightarrow Aut(F_{2g})$. -What is the kernel of $\rho$? - -REPLY [7 votes]: The representation is faithful, since a mapping class is determined by its action on the fundamental group of the surface. A surface is a $K(\pi,1)$, so given any element $Aut(S_{g,1})$, one obtains a (pointed) map $\varphi:S_{g,1}\to S_{g,1}$ which is unique up to homotopy. Now one needs to know that two homotopic homeomorphisms of a surface are isotopic, which is classic (at least one may find this in a paper of Waldhausen). In fact, one may identify the image in $Aut(F_{2g})$ as the subgroup preserving the peripheral element. -Also, note that everything should be fixing a basepoint in the boundary, as in HW's comment.<|endoftext|> -TITLE: $p$-adic uniformization not from the Drinfel'd spaces? -QUESTION [7 upvotes]: It seems that when we talk about the $p$-adic uniformization, we typically mean those uniformized by either the Drinfel'd upper spaces (for which we think of the examples of Mumford curves and some local studies of Shimura varieties), or the Tate uniformizaation of abelian varieties with multiplicative reduction (which we roughly think of as uniformized by tori). I would like to know what other kinds of spaces have been used in $p$-adic uniformization other than these two cases, as I haven't yet found much in the literature. -And in general, what kind of spaces could one expect to appear in the $p$-adic uniformization as the "universal covering" spaces? what is the $p$-adic counterpart of "simply connected" spaces? Since smooth Berkovich analytic spaces are locally contractible, can we expect such spaces exist in the sense of Berkovich? - -REPLY [3 votes]: There is another type of uniformization introduced in Mochizuki's book Foundations of $p$-adic Teichmüller theory. It uses curves equipped with nilpotent indigenous bundles. -I don't see what local contractibility has to do with non-existence of simply connected spaces. The finite étale covers of the affine line and the punctured affine line are very different, even though the underlying Berkovich spaces look almost the same, and are both contractible in the topological sense.<|endoftext|> -TITLE: Normality for non-noetherian schemes -QUESTION [7 upvotes]: I am interested to know to what extent the notion of normality makes sense on a non-noetherian scheme. -Specifically, I can ask the following question: let $\pi:X\to Y$ be a formally smooth morphism of schemes. -Assume that $Y$ is noetherian and normal. Let $U\subset Y$ be an open subset such that the complement has codimension $\geq 2$. Let $f$ be a regular function on $\pi^{-1}(U)$. -$\mathbf{Question:}$ Is it true that $f$ extends to all of $X$? - -REPLY [2 votes]: No, it is not true. I'm going to describe everything in terms of commutative algebra, so take $X=Y=\operatorname{Spec} R$. -Let $G$ be the group $\mathbb Z \times \mathbb Z$ with the lexicographic ordering. Let $R$ be any valuation ring with value group $G$ and then $R$ has dimension $2$ since it has exactly three prime ideals, forming the chain: -$$\mathfrak m = \{ r \in R \mid v(r) \geq (0, 1) \} \supset \mathfrak p = \{ r \in R \mid v(r) = (a, b) \mbox{ where } a \geq 1\} \supset 0,$$ -where $v$ denotes the valuation. Let $t$ be any element of valuation $(0,1)$ so that the vanishing set of $t$ is the single closed point $\mathfrak m$. Thus, $1/t$ is a regular function on $U = X \setminus \mathfrak m$, but $1/t$ is not in $R$ even though $\mathfrak m$ has codimension 2. -I think about the result for Noetherian normal rings in terms of two separate results: First, if $R$ is a Noetherian domain, then $R$ is the intersection of $R_{\mathfrak p}$ as $\mathfrak p$ ranges over all associated primes of principal ideals. Second, in a normal Noetherian domain, all associated primes of principal ideals have codimension 1 (this is condition S2 of Serre). I suspect that the first might generalize to non-Noetherian rings, provided that you have the right definition of associated prime. The second fails pretty much as badly as possible, since a single element can generate a prime ideal of arbitrary codimension, or even infinite codimension.<|endoftext|> -TITLE: Can every Lie group be realized as the full isometry group of a Riemannian manifold? -QUESTION [35 upvotes]: Suppose a finite-dimensional Lie group $G$ is given. Does there exist a connected manifold $M$ and a Riemannian metric $g$, such that $G$ is the full isometry group of $(M,g)$? -For example if I try to do this for a connected $G$, then I often get a bigger group as the full isometry group, which includes e.g. the orientation reversing isometries. (Maybe one has to take a non--orientable space for that?) -Even if I try to realize $\mathbb R$ as a full isometry group, I fail. (One could take the full isometry group of $\mathbb R$ with the standard metric, which is given by $\mathbb R \rtimes \mathbb Z_2$ and divide out the $\mathbb Z_2$ action. But this leads to a fixpoint and the quotient is therefore not a manifold any more.) -There is an article of J. de Groot1 which proves that every abstract group can be realized as an isometry group of some metric space, but it is not clear to me, if this is true in the category of Lie groups and Riemannian manifolds. -1de Groot, J. "Groups represented by homeomorphism groups." -Math. Ann. 138 (1959) 80–102. -MR119193 -doi:10.1007/BF01369667 -TITLE: Eliminating 1st order terms in elliptic partial differential equation -QUESTION [14 upvotes]: Under what conditions is it possible, using a suitable change of variables, to eliminate 1st order terms in an elliptic partial differential equation, so that the equation involves the 2nd derivatives, the dependent variable, and independent terms only? -To be concrete, consider the elliptic equation $-\Delta u + \sum_i \frac{d u}{dx^i} a^i + f(x)=0$. -If the $a^i$ are constant, define $u(x) = v(x) e^{\frac{1}{2}\sum_j a^j x^j}$ and obtain -$-\Delta v - \frac{1}{4} v \sum_i a^i a^i + f(x)e^{-\frac{1}{2}\sum_j a^j x^j}=0$, an elliptic equation without 1st order terms. -If the $a^i$ are not constant or if the equation is quasilinear, the problem is harder. It can be approached using contact transformations and Cartan's method of equivalence, but I am not aware of results. - -REPLY [5 votes]: The necessary and sufficient conditions for transforming one second-order differential operator of the type you are interested in into another are given by the so-called Cotton theorem, see Theorem 1 in this paper by Finkel and Kamran. In your case, you want the linear part of the transformed operator to vanish ($\tilde{\mathcal{A}}=0$ in the notation of the above paper), whence you can readily extract the conditions you ask for. Roughly speaking, in order to have no first-order terms in your transformed operator, the linear part of your original operator should be "pure gauge up to a change of independent variables".<|endoftext|> -TITLE: When finitely generated free algebras are finite -QUESTION [8 upvotes]: The variety (in the sense of universal algebra) of Boolean algebras, for example, -has the property that finitely generated free algebras have finite cardinality; -in that case specifically $|F_n|=2^{2^n}$, in the obvious notation. -Can one usefully characterize varieties whose finitely generated free algebras have finite cardinality? -Can one characterize natural number sequences arising as $|F_n|$ in association with such varieties? - -REPLY [6 votes]: I asked George McNulty, and here is his answer. It partially coincides with my answer here, but is much more complete. -======== -I think the first result of -this kind is an immediate if unstated consequence of a result in Peter Perkins dissertation -P. Perkins, ``Decision Problems of Equational Theories of Semigroups and General Algebras'', University of California, Berkeley 1966. -In a signature with two binary operation symbols and two constant symbols, Perkins -proves (his Theorem 36) that the collection of finite sets of equations that are bases of finite algebras is undecidable. He does this by reducing the word problem on a particular finitely presented semigroup to this question. Loosely, if the word w is a consequence of the semigroup presentation, then the associated finite set of equations will be a base of a finite algebra, whereas if w is not a consequence then the free algebra on one generator in the variety based on the set of equations will be infinite. Of course, this also shows that the locally finiteness problem -is also undecidable. This part of Perkins dissertation was published as -P. Perkins, ``Unsolvable problems for equational theories'', Notre Dame Journal of Formal Logic, vol. 8 (1967) 175--185. -One of the things I did in my 1972 dissertation was to establish various extensions of Perkins work on this topic. In particular, I showed that the above result holds for any finite signature that has an operation symbol of rank at least two. I had a rather long list of properties of finite sets of equations or of the varieties based on finite sets of equations that I could prove to be undecidable, but I didn't put all the proofs even in my dissertation. By the time I came to write it up for publication, I had figured out a handful of results from which most of the undecidability results I knew would follow. I published these in -G. McNulty, ``Undecidable properties of finite sets of equations''. Journal of Symbolic Logic, vol. 41 (1976) 589-604. -You can find in that paper a long list of such properties, but local finiteness is not -on the list, while being the base of a finite algebra is. The local finiteness business -follows in the same way as it did from Perkins result. -There are only a handful of other papers that address undecidable properties of -finite sets of equations (mostly, I think, because undecidability seems to prevail---although -some result like the Adjan-Rabin Theorem is unknown). Here they are: -V.L. Murskii, ``Nondiscernible properties of finite system of identity relations'', Doklady Akademii Nauk SSSR vol. 196 (1971) 520--522. -This paper is independent of my work or Perkins work. There is a large overlap between -Murskii's findings and what is in my dissertation, although this 3 page account of Murskii's work is, of course, very terse. I don't think Muskii's work covers either being the base of a finite algebra or being the base of a locally finite variety, but it is very interesting. Murskii was the first to frame a general condition on collections of finite sets of equations that would ensure undecidability. It was Murskii's paper that spurred me -to frame other general conditions that you can find in the paper of mine above. (I also -include there a second proof of Murskii's general condition. -Douglas Smith in his 1972 Penn State dissertation found another undecidability -result. It is in -D. Smith, ``the non-recursiveness of the set of finite sets of equations whose theories are one-based.'' Notre Dame Journal of Formal Logic, vol. 13 (1972) 135--138. -Ralph McKenzie wrote -R. McKenzie, ``On spectra and the negative solution of the decision problem for identities having a nontrivial finite model.'' Journal of Symbolic -Logic, vol. 40 (1975) 186--196. -Don Pigozzi wrote -D. Pigozzi, ``Base-undecidable properties of universal varieties.'' -Algebra Universalis, vol. 6 (1976), no. 2, 193–223. -Among other things, Pigozzi shows that it is undecidable whether the -variety based on a finite set of equations has the amalgamation property -or the Schreier property (subalgebras of free algebras are free). -C. Kalfa, ``Decision problems concerning properties of finite sets of equations. -Journal of Symbolic Logic, vol. 51 (1986) 79--87. -Here Cornelia Kalfa shows that the joint embedding property is undecidable, as is -whether the elementary theory of the infinite models is model complete. -The latest paper I know about is -C. O'Dunlaing, ``Undecidable questions related to Church-Rosser Thue systems.'' -Theoretical Computer Science, vol. 23 (1983) 339--345. -Here it is shown that it is undecidable whether of finite set of -equations is logically equivalent of a finite confluent set of equations. -Recently Ralph Freese and some collaborators have found fairly quick algorithms -for a lot of the kind of properties above when the equations examined have certain -restricted forms. -That's about what I know about this line of research. Hope some of it is useful. -================= -George also wrote: -================= -Also I noticed the interest expressed about free spectra in the original posting. There are a lot of papers on this (and related topics). Perhaps Joel Berman is the person who knows all about it.<|endoftext|> -TITLE: nef Cone of a Toric Variety -QUESTION [7 upvotes]: Is there a systematic/standard way of extracting the nef cone of a toric variety $X$ from its fan (or polytope)? Can I tell from the basis of $A^1(X)$ induced by the one-skeleton, based on coefficients, when a given divisor class is nef? -In particular I am working working with blowups of $\mathbb{P}^n$. I am uncertain if that extra piece of information helps. -I imagine that the answer to my question is yes, but I havent yet found a place where this is cleanly articulated. -Thanks in advance. - -REPLY [7 votes]: What you are interested in is toric Mori theory. This was first written down by Miles Reid back in the 80s (Decomposition of toric morphisms, incidentally to the best of my knowledge the first paper which wrote out the main steps of Mori's programme.) If you google "toric Mori theory", there are plenty of other hits; I checked Wisniewski's nice Toric Mori theory and Fano manifolds which may well do for you.<|endoftext|> -TITLE: Is there a Kan-Thurston theorem for fibrations ? -QUESTION [28 upvotes]: Given a fibration $F \to X \to B$ with all spaces path-connected. Is there a (discrete) group $G$ with normal subgroup $H$ such that -$$H^\ast(BG;\mathcal{A}) = H^\ast(X;\mathcal{A})$$ -$$H^\ast(BH;\mathcal{A}) = H^\ast(F;\mathcal{A})$$ -$$H^\ast(B(G/H);\mathcal{A}) = H^\ast(B;\mathcal{A})$$ -for every local coefficient system $\mathcal{A}$ on $X$ ? - -REPLY [5 votes]: I think the positive answer to this question is given in the proof of Proposition 1.5 of - -A.J. Berrick and B. Hartley: Perfect radicals and homology of group extensions. Topology and its Applications 25 (1987), 165-173. - -The idea is to first apply the Kan-Thurston construction to replace $B$ by a classifying space $BQ$. Then pull back the fibration $X\to B$ along $BQ\to B$, and apply the Kan-Thurston construction again to replace the total space $X\times_BBQ$ by a classifying space $BG$. Then consider the classifying space $BN$ for $N$ the kernel of $G\to Q$. The result is a map from the fiber sequence $BN\to BG\to BQ$ to the fiber sequence $F\to X\to B$ such that all the maps (on base, total space, fiber) are acyclic. Then the Hochschild-Serre spectral sequence for the extension $BN\to BG\to BQ$ should be the Serre spectral sequence associated to the original fiber sequence $F\to X\to B$.<|endoftext|> -TITLE: Algebras with supremum-founded subalgebra lattice -QUESTION [5 upvotes]: I am interested in algebras whose subalgebra lattice is supremum-founded. Let us call those algebras small. -A complete lattice $(L, \leq)$ is called supremum-founded, if for any two elements $x < y$ from $L$ there is an element $s \in L$ which is minimal with respect to $s \leq y$, $s \not\leq x$. -Of course, every finite lattice is supremum-founded whence every finite algebra is small. Moreover, it is not very difficult to show that every $1$-locally finite algebra (that is, an algebra whose $1$-generated subalgebras are finite) is small. -Does anybody know some more/bigger classes of examples? Is there a suitable reference? - -REPLY [2 votes]: I just noticed this old question. -I have a sufficient condition for smallness, -which covers all cases mentioned so far. -When the subalgebra lattice is distributive, the -condition I give is both necessary and sufficient. - -Call a subalgebra $S$ of an algebra $A$ -completely join irreducible or CJI -if it is a completely join irreducible element -of the lattice $L=\textrm{Sub}(A)$. -This means that $S$ has a largest proper subalgebra, $S_*$. -Lemma. -A subalgebra $S\leq A$ is CJI iff -whenever a nonempty subset $X\subseteq S$ generates $S$, then -some single $x\in X$ already generates $S$. (in particular, any CJI -subalgebra is $1$-generated.) -Reasoning. -If $S$ is CJI with lower cover $S^*$, and $X\subseteq S$ -generates $S$, then $X\not\subseteq S_*$. Any $x\in X-S_*$ -will generate $S$. -Conversely, if $S$ has the generation property -of the lemma statement, let $S_*$ -be the subalgebra of $S$ generated by the set of -all singleton nongenerators of $S$. $S_*$ will be -the largest proper subalgebra of $S$. -\\\\\ -Now consider: -Condition I. Every element of $L$ is a join -of CJI elements. (I.e., every subalgebra of $A$ -is generated by the union of its CJI subalgebras.) -Condition II. $L$ satisfies DCC on its CJI elements. -($A$ satisfies DCC on its CJI subalgebras.) -Lemmas. -(1) If Conditions I and II hold, then $L=\textrm{Sub}(A)$ -is supremum founded. -(2) If $L$ -is supremum founded, then Condition I holds. -(3) If $L$ is distributive and -is supremum founded, then Conditions I and II both hold. -Hence Conditions I and II characterize the property -of supremum-foundedness when $L$ is distributive. -(4) If $A$ is 1-locally finite or if all elements of $\textrm{Sub}(A)$ -are dually compact, then Conditons I and II hold. -Reasoning. -For (1); If $XS_2>\cdots$ be a -strictly descending chain in $P$. Let $F$ be the order filter -generated by $\{S_i\}$, and let $I$ be the complementary -order ideal. -(It is the largest order ideal containing none of the -$S_i$.) Let $X=\bigvee I$ be the element of $L$ that -corresponds to the order ideal $I$. $X$ -is the largest element of $L$ -containing none of the $S_i$'s. Let $Y=A$ = the top of $L$. -We have $X -TITLE: History question - why h in the definition of derivative? -QUESTION [11 upvotes]: Does anyone have a clue where the "h" came from? - -REPLY [16 votes]: I think that use of $h$ in the definition of derivative is linked to the relationship between Calculus of Finite Difference and Differential Calculus. -In the book Leçons sur le Calcul des Fonctions, Councier, 1806, Lagrange: - -Assigns to Maclaurin and d'Alembert the origin of differentiation as the limit of finite differences, (pp. 1); -Writes "Considerons un fonction $fx$ d'une variable quelconque $x$. Si la place de $x$ on substitue $x + i$, $i$ étant une quantité quelconque indetermineé, elle divendra $f(x + i)$ ...", (pp. 8); -Develops $f(x + i)$ in series: -$$f(x+ i) = fx + i f'x + \frac{i^2}{2} f''x + \frac{i^3}{2 . 3}f'''x + \frac{i^4}{2.3.4}f^{iv}+ \mbox{etc}$$ as we can see at (pp. 15), and -Writes "Nous appelerons la fonction $fx$ fonction primitive... Nous nommerous de plus la fontion dérivée $f'x$, primière fonction dérivée ou fonction derivée du primier ordre...", (pp. 15). - -Notes B of Lacroix's book An Elementary Treatise on the Differential and Integral Calculus, Cambridge, 1816, pp. 599, , using $h$ instead $i$, is based on Lagrange's work. -In 1829, Dr. Martin Ohm, in Versuch eines vollkommen consequenten Systems der Mathematik, Vol. III, pp.53, Berlin, available here, uses $h$. He writes: -$$f(x+h)=f(x) + \partial f(x).h+\partial^2 f(x) .\frac{h^2}{2!} + \partial^3 f(x) .\frac{h^3}{3!} +.\ldots$$ -Also, as we can see, Dr. Martin Ohm uses factorials!! (Martin and Georg Ohm were brothers. Georg discovered the Ohm's Law). -In G. Boole, Treatise on the Calculus of Finite Differences, MacMillan, London, 1880, pp.1, we can read: " The Calculus of Finite Differences may be strictly defined as the science which is occupied about the ratios of the simultaneous increments of quantities mutually dependent. The Differential Calculus is occupied about the limits to which such ratios approach as the increments are definitely diminished" Boole1. -At pages 2 and 3 , we can see the definition of derivative using $h$. All arguments are based on Finite Differences Boole2. -The differentiation was developed based on trigonometric assumptions (method of tangents). A very good history can be found in H.Sloman, The Claim of Leibnitz to the Invention of the Differential Calculus, MacMillan, 1860 Sloman1. -A possible explanation for the use of h in the definition of derivative (and the link between Differential Calculus and Calculus of Finite Differences) can be found in this book at page 127, second paragraph Sloman2. -PS: Although $h$ has been used in the books mentioned above (1816, 1829, 1860 and 1880), Milne-Thomson, in his recent book (1933), uses $\omega$ instead Milne-Thomson. -Milne-Thomson's book can be considered an example of Euler's notation use. -In Institutiones calculi differentialis cum eius usu in analysi finitorum ac doctrina serierum, Chapter 1, De differentiis finitis, pp.1, 1787, Euler writes "variabilis x capiat incrementum $\omega$" !<|endoftext|> -TITLE: Absolute continuity on $R^{n}$ -QUESTION [8 upvotes]: I know the definition of absolute continuity if there is a function $f:(a,b)\rightarrow R$. -I wonder what is an analogy of this concept if we have a function $f:A\rightarrow R$, where $A\subset R^{n}$ is open set. - -REPLY [9 votes]: I guess it may depend on exactly which property of absolutely continuous functions you think is most important to keep, or to put it another way, exactly which definition you prefer in one dimension. For me the most commonly useful property of absolutely continuous functions is that they map sets of Lebesgue measure zero to sets of Lebesgue measure zero. -Pulling in roughly equal parts from my memory of real analysis and what Wikipedia and EoM tell me, the story seems to be that a function $f\colon [a,b] \to \mathbb{R}$ is absolutely continuous if and only if all three of the following hold (Banach–Zaretskii theorem): - -$f$ is continuous; -$f$ is of bounded variation; -the Luzin N property holds: if $E$ has Lebesgue measure $0$, then so does $f(E)$. - -Each of these 3 properties generalises to higher dimensions. The first and third are immediate; the second requires a slightly different definition of variation than in one dimension, but is a completely standard thing. -Thus one could say that a function $f\colon \mathbb{R}^m \to \mathbb{R}^n$ is "absolutely continuous" if those three properties hold, and to me this seems a very reasonable generalisation of the usual definition. (Of course one could extend this to maps between smooth manifolds, where you also have a notion of zero volume.) -This seems to be different from the definition that Malý uses in the paper Tapio Rajala referred to in his answer. From a quick glance at that paper, there seem to be a number of different generalisations out there, and this seems to be another example of the phenomenon wherein various notions that are distinct in higher dimensions happen to all coincide in the lowest-dimensional case, so that you can generalise some aspects of the familiar setting, but not all. Which generalisation is useful depends on what your purpose is.<|endoftext|> -TITLE: Geometric Interpretation of the Lower Central Series for the Fundamental Group? -QUESTION [26 upvotes]: For any group G we can form the lower central series of normal subgroups by taking $G_0 = G$, $G_1 = [G,G]$, $G_{i+1} = [G,G_i]$. We can check this gives a normal chain -$G_0 > G_1 > ... > G_i >...$ -In the case where $G = \pi_1(X)$ The first quotient $H^1 = G_0/G_1$ is well known to be the first homology group (which has well known geometric content). -Question 1: - -Are there geometric interpretations of further quotients $G_i/G_{i+1}$? -What does the length (finite or infinite) of the chain tell us geometrically about X? - -We can also form the mod-p central series by taking $G^p_0 = G$, $G_{i+1}^p = (G_i^p)^p[G,G^p_{i}]$ and then again form the quotients $V_i^p = G^p_i/G^p_{i+1}$. In this case these are modules (vector spaces) over $Z_p$. -Question 2: - -What are interpretations of these $V^p_i$? What can we say if we know their dimension (as a vector space) or if they're non-zero? I'm particularly interested in small i (= 1,2,3,4), and small p (= 2 say). - -Question 3: - -Are there good methods for calculating the $V_i^p$ (both direct and indirect)? For instance, a direct way would be to calculate them from a presentation of the fundamental group. Is this tractable (with software such as GAP) if the presentation is "small" in some sense? Can I bound their dimension (above or bellow)? -Are there indirect ways of calculating these vector spaces? As homology/cohomology of some other object on X? As something else? Group homology? - -Question 4: - -Is there a good source for these types of questions? Has somebody worked out the V's for compact surfaces (orientable or not)? - -Answers or (even better) references to work on these types of questions would be great. I'm especially interested in examples worked out for surfaces and small i. - -REPLY [2 votes]: As for question 4, the answer is well understood for the fundamental groups of surfaces, or more generally for "formal spaces" $X$. In summary: renumber your series so that $G_1=G$. Then $V_i:=G_i/G_{i+1}$ is a free $\mathbb Z$-module for all $i$, and the direct sum $L:=\bigoplus_{i\ge1}V_i$ is a graded Lie algebra over $\mathbb Z$. E.g. for surface groups, it admits as presentation the natural linearization of the group presentation: $$L=\langle A_1,B_1,\dots,A_g,B_g\mid [A_1,B_1]+\dots+[A_g,B_g]=0\rangle.$$ -The enveloping algebra of $L$ is a quadratic Koszul algebra, and its Koszul dual $U(L)^!$ is isomorphic to the cohomology algebra of $X$. The ranks of the sections $V_i$ may be recovered from the Betti numbers of $X$ using Koszul duality and Möbius invertion. -More precisely, if $b_i$ are the Betti numbers of $X$, and $b(t)=\sum b_i t^i$ is its Poincaré series, and $c(t)=\sum c_i t^i$ is the Poincaré series of $U(L)$, then Koszul duality gives $c(-t)b(t)=1$. Then $c(t)=\prod_{i\ge1}(1-t^i)^{-\dim V_i}$ lets you compute $\dim V_i$.<|endoftext|> -TITLE: Absence of Maps Between p-local and q-local spectra -QUESTION [8 upvotes]: Suppose $X$ and $Y$ are spectra (or homotopy classes thereof) such that $X$ is p-local and $Y$ is q-local, for primes $p\neq q$. Is it indeed true then, and if so how would one show that $[X,Y]_\ast=0$? -Thanks! - -REPLY [14 votes]: Responding to your comment on Charles's answer: -[I wish I could delete my comment to Charles's answer. My comment seems to have exploded and taken your comment down with it.] -Both localization of a spectrum at a prime and completion of spectra at a prime are examples of Bousfield localization. The one is (a little circularly) localization with respect to $p$-localized homotopy theory, while the other is localization with respect to mod $p$ homotopy theory. -Localizing a spectrum at a prime $p$ (or at a set of primes) is the kind of Bousfield localization that is most like localization in algebra. In this case $\pi_n$ of the localization of $X$ is the algebraic localization of $\pi_nX$, i.e. the result of inverting all the primes other than $p$. The local spectra are the spectra $X$ such that for every $n$ the abelian group $\pi_nX$ is one on which every prime other than $p$ acts invertibly, and the acyclic objects are the ones which become trivial upon localizing, i.e. the spectra $X$ such that for every $n$ the group $\pi_nX$ is a torsion group without $p$-torsion. Localization of $X$ is the same as smash product of $X$ with the localization of the sphere spectrum. If you want to describe the acyclic objects as the $E_*$-acyclic objects for some $E$, you can of course let $E$ be the localization of the sphere. Note that we are not starting with an $E$ and using the general machine of Bousfield to make the localization functor, although of course we could. -Completing at a prime is again not too far from being a purely algebraic matter. Here we can choose $E$ to be the mod $p$ Moore spectrum, i.e. the homotopy cofiber of the map $p:S\to S$ from the sphere spectrum to itself. The acyclic objects are the spectra for which each homotopy group has $p$ acting invertibly. The localization $LX$ (which is called the $p$-completion) can be described as the holim, over natural numbers $k$, of the smash product of $X$ with the mod $p^k$ Moore spectrum. If each homotopy group of $X$ is finitely generated, then $\pi_nLX$ can be described as the tensor product of $\pi_nX$ with the group $\mathbb Z_p$ of $p$-adic integers, and also in this case $LX$ is the smash product $X\wedge LS$. But neither of these statements is true for general $X$: for example, the $p$-completion of $H\mathbb Q$ is trivial while $\mathbb Q\otimes \mathbb Z_p$ is the field of $p$-adic numbers and $X\wedge LS=H(\mathbb Q\otimes \mathbb Z_p)$. And (therefore) $LH(\mathbb Q/\mathbb Z)=\Omega H\mathbb Z_p$. -Note that a spectrum which is local at a prime $q$ different from $p$ is trivial mod $p$. -In general maps from $E$-acyclic objects to $E$-local objects are trivial, so we see that maps from a $q$-local spectrum to a $p$-complete spectrum are trivial. In particular, as Charles says, maps from a $q$-complete spectrum to a $p$-complete spectrum are trivial.<|endoftext|> -TITLE: A question on infinite dimensional Gaussian measure and affine tranformations. -QUESTION [9 upvotes]: Let $\gamma_\infty$ denote the product Gaussian measure on $\mathbb{R}^\mathbb{N}$. Which $a,b \geq 0$ satisfy that for every Borel set $K\subseteq \mathbb{R}^\mathbb{N}$ of positive measure, $a K + b y$ has positive measure for $\gamma_\infty$-a.e. $y$? -The knee-jerk answer is ``only $a = 1$, $b=0$.'' The Cameron-Martin Theorem tells us that $x \mapsto x + y$ is absolutely continuous exactly when $y$ is in $\ell^2$ (a set of $\gamma_\infty$-measure $0$). Similar arguments apply to linear transformations like those above. -This does not answer the question, however. In fact Solecki and I have recently proved that if $a \in \mathbb{R}$ and $K \subseteq \mathbb{R}^\mathbb{N}$ is Borel and of positive measure, then $\gamma_\infty(\sqrt{1+b^2} K + b K) > 0$ for $\gamma_\infty$-a.e. $y$ (i.e. a sufficient conditions is that $a^2 = b^2 +1$). See http://arxiv.org/abs/1201.3947 for the preprint. It is not difficult to prove that a necessary requirement for a positive answer is that $a^2 + b^2 -2ab \leq 1 \leq a^2 + b^2 + 2ab$. This leaves some discrepancy, however, and that is -the question at hand. - -REPLY [3 votes]: The answer is that $(a,b)$ must satisfy $a^2 = b^2 + 1$. It is possible to verify (see the preprint above) that for any $b$ and Borel $K \subseteq \mathbb{R}^{\mathbb{N}}$, that -$$\gamma_\infty (K) = \int \gamma_\infty (\sqrt{1+b^2} K + b y) d \gamma_\infty (y).$$ -By replacing $K$ by $(a/\sqrt{1+b^2})K$ in this equation, we obtain -$$\gamma_\infty (\frac{a}{\sqrt{1+ b^2}}K) = \int \gamma_\infty (a K + b y) d \gamma_\infty (y).$$ -Let $K$ be all $x$ in $\mathbb{R}^{\mathbb{N}}$ such that -$$ -\lim_{n \to \infty} \left(\frac{1}{n} \sum_{i< n} x_n^2 \right)^{\frac{1}{2}} = 1. -$$ -Then $\gamma_\infty(K) = 1$ and therefore $(a/\sqrt{1+b^2}) K$ has measure $0$ unless $a^2 = b^2 + 1$. -Thus if $a^2 \ne b^2 + 1$, then -$\int \gamma_\infty (a K + b y) d \gamma_\infty (y) = 0$ and hence the integrand vanishes for almost -every $y$.<|endoftext|> -TITLE: Is the property of not containing $\mathbb{F}_2$ invariant under quasi-isometry? -QUESTION [25 upvotes]: Is the property of not containing the free group on two generators invariant under quasi-isometry? Amenability is, so if there is a counterexample it is also a solution to the von Neumann-Day problem (which of course already has a solution). - -REPLY [23 votes]: It is a famous open problem. Akhmedov in MR2424177 claimed he could prove that the answer is "no". No proof exists, so I guess he discovered a gap in his argument.<|endoftext|> -TITLE: Branching process survival probability -QUESTION [5 upvotes]: I have a time-inhomogeneous Galton-Watson binary branching process over a finite number of generations $n$. In each generation $i$, there is a probability $p_i$ of a child surviving; so each node has 2 children with probability $p_i^2$, 1 child with probability $2 p_i (1-p_i)$, and zero children with probability $(1-p_i)^2$. Furthermore $p_i$ is a decreasing function of $i$. So the process starts out as a super-critical branching and ends as a sub-critical branching. -I want to show that the process survives to time $n$ with probability say $\Omega(1)$ or $\Omega(1/\text{poly}(n))$. What is the easiest criterion to show this? -The process is inhomogeneous. The number of expected survivors at level $i$ is $\mu_i = 2^i p_1 \dots p_i$, which is a unimodular function of $i$. It seems that a sufficient criterion should be $\mu_n \geq 1$ or maybe $\mu_n = \Omega(\text{poly}(n))$. -Thanks! - -REPLY [2 votes]: A new answer for the new version of the question. -Under the constraint that $p_i$ are decreasing and $\mu_n\ge 1$, the minimal survival probability is obtained when all $p_i=1/2$. You can see this by showing that for any level $j$, if you fix all the $p_i$ except for $p_j$ and $p_{j+1}$ then the minimum is obtained when $p_j=p_{j+1}$.<|endoftext|> -TITLE: $(q,x)$-analog of $n!$ -QUESTION [12 upvotes]: While doing some work in geometric representation theory I have come across the following -sequence of polynomials in two variables $(q,x)$ which I would like to denote -by $n!_{q,x}$. For small $n$ these polynomials look as follows: -$2!_{x,q}=x+q$ -$3!_{x,q}=x^3+x^2(q^2+q)+x(q^2+q)+q^3$ -$$ -4!_{x,q}=x^6+x^5(q^3+q^2+q)+x^4(q^4+q^3+2q^2+q)+x^3(q^5+q^4+2q^3+q^2+q)+ -$$ -$$ -x^2(q^5+2q^4+q^3+q^2)+x(q^5+q^4+q^3)+q^6 -$$ -The polynomials are actually symmetric in $q$ and $x$ and when one puts $x=1$ one recovers -the usual $q$-analog of $n!$ (in particular, when both $q$ and $x$ are 1, we get $n!$). -My question is this: has anybody seen such polynomials before? What is the correct definition of those polynomials for general $n$? Any information will be greatly appreciated. - -REPLY [17 votes]: I was hesitating to write an answer since I don't have references at hand but let me mention that if you denote your polynomials $P_n(x,q)$ and look at $Q(x,q)=x^{\binom{n}{2}}P_n(x^{-1},q)$ then (my guess is that) you are looking at: -$$Q(x,q)=\sum_{\pi\in S_n}x^{maj(\pi)}q^{inv(\pi)}=\sum_{\pi\in S_n}x^{maj(\pi)}q^{maj(\pi^{-1})}$$ -where $maj(\pi)$ is the Major index of a permutation -$$maj(\pi)=\sum_{\pi(i) > \pi(i+1)}i.$$ -These polynomials satisfy the following -$$ -\sum _{n=0} ^{\infty}\frac{Q _n(x,q)}{(x) _n(q) _n}u^n = \prod _{i,j=0} ^{\infty}\frac{1}{1-x^iq^ju} $$ -where $(q)_n$ denotes $(1-q)(1-q^2)\cdots(1-q^n)$. From this relation you can get a useful recurrence relation or other properties which may help you compute these polynomials.<|endoftext|> -TITLE: Can one do without a classifying space when showing vanishing of cohomology -QUESTION [9 upvotes]: Let $G$ be a discrete group and $A$ an abelian group, then $H^n (G,A)$ can be defined as $$ H^n (G,A) = H^n (B_G, A)$$ -Where $B_G$ is the classifying space of $G$, i.e. $B_G = E_G / G$ where $E_G$ is a contractible space on which $G$ acts s.t. $\pi_1 (B_G) = G$. -Now say you have an action of $G$ on a simplicial complex $Y$ and that the action is simplicial, cocompact and free, but $Y$ is not necessarily contractible. Say you know that (for some fixed $n$) $H^n (Y / G,A) = 0$ can you deduce that $H^n (G,A) = 0$? -(I've tried to ask this question a couple of days ago, but my formulation was bad so I had to delete and re-ask) - -REPLY [3 votes]: To highlight the assumption of high connectivity in Johannes' answer, I should point out that for any $G$, you can always take $Y = G$ (a simplicial complex with only vertices, seeing as $G$ is discrete). Then you have a simplicial, cocompact, free action, and it is true that $H^n(Y/G, A) = 0$ for $n>0$. But that tells us nothing about $H^n(G, A)$.<|endoftext|> -TITLE: What are the known implications of "There exists a Reinhardt cardinal" in the theory "ZF + j"? -QUESTION [27 upvotes]: This is, alas, in large part a series of questions on unpublished work of Hugh Woodin; it's also quite frivolous if Reinhardt cardinals turn out inconsistent. -Definitions: - -Call $\kappa$ an $I-1(\kappa,\delta)$ cardinal if it is the least ordinal lifted by a nontrivial elementary embedding $j:V_{\delta+2} \to V_{\delta+2}$ where $\delta$ is an uncountable strong limit cardinal of countable cofinality and the supremum of the critical sequence $\kappa_n$. -Call $\kappa$ a weak Reinhardt cardinal if it is $I-1(\kappa,\delta)$, there is an ordinal $\gamma \gt \delta$, and elementary substructures $V_{\kappa} \prec V_{\delta} \prec V_{\gamma}$. -Call $\kappa$ a Reinhardt cardinal if it is the critical point of $j:V \to V$. -Call $ZFR$ the base theory $ZF$ in the language $(\in,j)$ augmented with a scheme asserting $\{\exists\kappa\mid \kappa\text{ is Reinhardt}\}$ - -Question 1: What does $ZFR$ imply? - -Woodin folklore mentions a proof that $ZFR$ implies the existence of models of all large cardinal axioms not known to refute Choice. What is the proof $ZFR \implies$ "There exists a transitive model of $ZFC + I1(\kappa,\delta)$", and why does it not immediately extend to a proof of "There exists a transitive model of $ZF + DC_{\delta} + I-1(\kappa,\delta)$"? - -Reinhardt cardinals (trivially) satisfy any large cardinal property derived from $j:V \to M$ and so are themselves at least $I2$ in some sense, but I don't know how to proceed further, and Woodin frequently mentions that there is no known proof of the expected implication Reinhardt $\implies$ weak Reinhardt. And while David Asperò has recently published in A short note on very large large cardinals a nice proof that Reinhardt cardinals imply many $j:V_{\mu} \to V_{\mu}$ with the target of the embedding moved arbitrarily high, these $\mu$ are poorly constrained and $V_{\mu}$ in general satisfies neither Replacement nor Choice. They may, therefore, not even be rank-into-rank in the usual sense. -Question 2: What's "next" after $ZFR$? - -The rare tome On Woodin's Investigations into "Here there be Dragons" contains a proof that $j:HOD \to HOD$ follows from a strong extension of $ZFR$. What is this extension? - -The strongest such axiom I recall encountering in the literature is a scheme asserting that there is an elementary substructure $V_{\kappa} \prec V$. However perilously strong, this axiom is not super $n$-helpful for determining what sort of cardinal structure the Reflecting Reinhardt cardinal is duplicating. Thus I'm hoping the answer might be something else, more useful for establishing answers to Question 1. -Question 3: What is the point? - -Do these very large large cardinals hint at whether the Axiom of Choice constrains the consistency strength of $ZFC$, or not? - -Here we arrive at the soft and murky motivation behind these questions. In light of Kunen's Inconsistency Theorem, it's possible to ask if the Axiom of Choice is close kin to the Axiom of Constructibility, something that places a sharp bound on the amount of consistency strength $ZF$ can support. Under this interpretation, if $I-1(\kappa,\delta)$ is consistent with $ZF$ then there is no equiconsistent large cardinal axiom consistent with $ZFC$. And although there are various shenanigans we can perform to sneak the strength back in (e.g. work in $ZFC +$ Con($ZFR$)), these don't tell us much about what $V$ is like. Further, they raise the question of why we oughtn't just work in $ZFC + V=L$ and get our measurables from some $j:M \to N$, since the Axiom of Constructibility is doing the same job as Choice in this context, only better. -Contrariwise, it's also possible to ask if it's the definability restriction of $V=L$ that renders it so inhospitable to large cardinals, and to assume that if a prospective axiom is too strong for the notoriously non-constructive Axiom of Choice, then it was outright inconsistent to begin with and Choice has merely facilitated the proof. Woodin's $HOD$ Conjecture is work in this vein (it refutes $I-1(\kappa,\delta)$), and even if the $HOD$ Conjecture is false, $I-1(\kappa,\delta)$ might be inconsistent for some currently unknown reason. -The sort of relevant evidence I'm hunting for is a large cardinal property or forcing axiom that: -1. is very strong in ZFC -2. is weaker but not trivial in ZF -3. indicates that some property a Reinhardt cardinal should have has leaked away through the absent wellordering of some $P(\aleph_{\alpha})$ -The first two criteria are satisfied by a scant few things I can think of, including for example some generalizations of Chang's Conjecture beyond $(\omega_{3},\omega_{2}) \to (\omega_{2},\omega_{1})$; these generalizations, however, either have a cardinal structure that doesn't illuminate the third requirement or they have a cardinal structure that I find baffling and unwieldy. (And yes, that third requirement is so vague that the question may not admit an answer even if an example exists. If I knew how to make it precisely state some such property I would, but in general we don't know what those properties are!) - -First post to this site; no snub intended if I don't respond in comments =) - -Edit: -Hello Joel, and thank you for the warm welcome. For $ZFR$ I wanted -1. All the Axioms of $ZF$ in the language $\{\in\}$ -2. Replacement for formulae in the language $\{j,\in\}$ -3. An Axiom scheme asserting $j:\langle V,\in \rangle \to \langle V,\in \rangle$ -...and not that there are many critical points. This is, however, probably a mistake (or at least overly specific) and I'll gladly substitute any formalization of "$ZF$ + There is a Reinhardt cardinal" that doesn't force the embedding, by elementarity, to be the identity, even if that formalization is implicit and be understood in a fully two-sorted metatheory. Particularly if that's what Woodin had in mind for his proofs. -And yes, the first two requirements should easy enough to formalize (though I also would like $ZF + \psi \vDash$ Con($ZF$) $\wedge$ $ZFC + \varphi \vDash$ Con($ZF + \psi$) at the very least). It's the third requirement that I don't know how to formalize without an applicable, rigorous notion of "large cardinal property". Woodin's won't suffice in full generality when I don't know how to express a given property as some $j:V \to M$ with appropriate constraints on $M$. And in the case of Reinhardt cardinals, I'm not even sure which properties to expect. -Corazza's Wholeness Axiom is almost exactly what I'm looking for, thank you for the reminder. The failure of Replacement for $j$-classes acts as a pressure valve, disallowing the critical sequence from forming a set, and thereby preventing the absurdity of cof$(Ord) = \omega$ and reducing the consistency strength of Reinhardt cardinals to something weaker than $I3(\kappa,\delta)$ and stronger than a cardinal super $n$-huge for all $n \in \omega$, which is quite manageable. Unfortunately, it rules out the same hugeness that Kunen's theorem does, and doesn't supply new information about how a fully Reinhardt cardinal must appear. -I think? It would be nice to be wrong about that! And you've studied it more than most. - -REPLY [3 votes]: I prove several new consequences of Reinhardt cardinals in my paper "Measurable cardinals and choiceless axioms," some of which are not known to be consistent relative to any weaker theory. TLDR: -Theorem (ZFR). There is a proper class of regular cardinals. -Proposition (ZFR). For all cardinals $\nu$, for all sufficiently large regular cardinals $\delta$, the club filter on $\delta$ is $\nu$-complete. -If $F$ is a filter, a set $S$ is an atom of $F$ if $\{A\cap S : A\in F\}$ is an ultrafilter on $S$. A filter $F$ is atomic if every $F$-positive set includes an atom of $F$. -Theorem (ZFR). The club filter on any sufficiently large regular cardinal is atomic. -Corollary (ZFR). For a club class of cardinals $\lambda$, either $\lambda$ or $\lambda^+$ is measurable. -In particular, there is a proper class of measurable successors of singular cardinals. Actually, much more is true: -Theorem (ZFR). For a club class of cardinals $\lambda$, every $\lambda$-complete filter on an ordinal extends to a $\lambda$-complete ultrafilter.<|endoftext|> -TITLE: Matrices with entries in a $C^*$-algebra -QUESTION [14 upvotes]: Let $\mathcal{A}$ be a $C^\ast$-algebra. Consider vector space of matrices of size $n\times n$ whose entries in $\mathcal{A}$. Denote this vector space $M_{n,n}(\mathcal{A})$. We can define involution on $M_{n,n}(\mathcal{A})$ by equality -$$ -[a_{ij}]^*=[a_{ji}^*],\qquad\text{where}\quad [a_{ij}]\in M_{n,n}(\mathcal{A}). -$$ -Thus we have an involutive algebra $M_{n,n}(\mathcal{A})$. It is well known that there exist at most one norm on $M_{n,n}(\mathcal{A})$ making it a $C^\ast$-algebra. This norm does exist. Indeed take universal representation $\pi:\mathcal{A}\to\mathcal{B}(H)$ and define linear injective $*$-homomorphism -$$ -\Pi:M_{n,n}(\mathcal{A})\to\mathcal{B}\left(\bigoplus\limits_{k=1}^n H\right):[a_{ij}]\mapsto\left((x_1,\ldots,x_n)\mapsto\left(\sum\limits_{j=1}^n\pi(a_{1j})x_j,\ldots,\sum\limits_{j=1}^n\pi(a_{nj})x_j\right)\right) -$$ -Hence we can define norm on $M_{n,n}(\mathcal{A})$ as $\left\Vert[a_{ij}]\right\Vert_{M_{n,n}(\mathcal{A})}=\Vert\Pi([a_{ij}])\Vert$. At first sight this definition depends on the choice of representation, but in fact it does not. -My question. This norm on $M_{n,n}(\mathcal{A})$ can be defined internally. Namely -$$ -\Vert[a_{ij}]\Vert_{M_{n,n}(\mathcal{A})}=\sup\left\Vert\sum\limits_{i=1}^n\sum\limits_{j=1}^n x_i a_{ij}y_j^*\right\Vert -$$ -where supremum is taken over all tuples $\{x_i\}_{i=1}^n\subset\mathcal{A}$, $\{y_i\}_{i=1}^n\subset\mathcal{A}$ such that $\left\Vert\sum\limits_{i=1}^n x_i x_i^*\right\Vert\leq 1$, $\left\Vert\sum\limits_{i=1}^n y_i y_i^*\right\Vert\leq 1$. -Is there a proof of this fact without usage of structural theorem for $C^*$-algebras, -a straightforward proof which can be made by simple checking axioms of $C^*$-algebras? -The same question on math.stackexchange.com - -REPLY [15 votes]: For $x=(x_i)_{i=1}^n, y=(y_i)_{i=1}^n \subseteq A$ define $(x,y) = \sum_i x_i y_i^* \in A$, and set $\|x\| = \|(x,x)\|^{1/2}$. - -Lemma: We have that $(x,y)^* (x,y) \leq \|x\|^2 (y,y)$ the order in the C$^*$-algebra sense. -Proof: (Copied from Lance's Hilbert C$^*$-module book). Wlog $\|x\|=1$. For $a\in A$ let $a\cdot x = (ax_i)$. Then - \begin{align*} 0 &\leq (a\cdot x-y, a\cdot x-y) \\ -&= a (x,x) a^* - (y,x)a^* - a(x,y) + (y,y) \\ -&\leq aa^*- (y,x)a^* - a(x,y) + (y,y) \end{align*} - The claim that $a(x,x)a^* \leq aa^*$ follows as if $c\in A^+$ then always $aca^* \leq \|c\|aa^*$. Now set $a=(x,y)^*=(y,x)$ and the claim follows. - -In particular, $\|(x,y)\|^2 \leq \|x\|^2 \|y\|^2$ and so $\sup\{ \|(x,y)\| : \|x\| \leq 1 \} = \|y\|$. -So you define -$$ \| a \| = \sup \{ (x, ay) : \|x\|\leq 1, \|y\|\leq 1 \} $$ -where $(ay)_i = \sum_j y_j a_{ij}^*$. Then from the observation above, -\begin{align*} \|a\|^2 &= \sup \{ \|ay\|^2 : \|y\|\leq 1 \} -= \sup\{ (ay,ay) : \|y\|\leq 1 \} \\ -&= \sup\Big\{ \Big\| \sum y_j a_{ij}^* a_{ik} y_k^* \Big\| : \|y\|\leq 1 \Big\} \\ -&= \sup\{ \|(y, (a^*a)y)\| : \|y\|\leq 1 \} \\ -&\leq \sup\{ \|(a^*a)y\| : \|y\|\leq 1 \} \\ -&= \sup\{ \|(z,(a^*a)y)\| : \|y\|\leq 1, \|z\|\leq 1 \} = \|a^*a\|. -\end{align*} -However, for any $z$, -$$ \|az\| = \sup\{ \|(x,az)\| : \|x\|\leq 1 \} -\leq \sup\{ \|(x,ay)\| : \|x\|\leq 1, \|y\|\leq \|z\| \} -= \|a\|\|z\|. $$ -It follows from this that the norm on $M_n(A)$ is an algebra norm (i.e. submultiplicative). From the very definition, the involution is an isometry on $M_n(A)$. So we have the usual trick: $$\|a\|^2 \leq \|a^*a\| \leq \|a^*\| \|a\| = \|a\|^2$$ -and so we have equality throughout, establishing the C$^*$-identity for $M_n(A)$. -The idea is to define a generalised Hilbert space of rows of $A$ with an $A$-valued inner-product, and then copy the usual proof that operators on a Hilbert space are a C$^*$-algebra.<|endoftext|> -TITLE: Morse-Kelley set theory consistency strength -QUESTION [20 upvotes]: I've come across several references to MK (Morse-Kelley set theory), which includes the idea of a proper class, a limitation of size, includes the axiom schema of comprehension across class variables (so for any $\phi(x,\overline y)$ with $x$ restricted to sets, there a class $X=(x : \phi(x,\overline y))$). -I have seen various statements about MK and how it proves the consistency of various things, including $Con(ZF)$, $Con(ZFC)$, $Con(NBG)$, and in fact, for any $T\subset MK$ finitely axiomatized, it proves $Con(T)$. -However, and quite frustratingly, I don't see any references to back up these claims, except occasionally links to other places where the claim was made, but not proven (or even proof-sketched). I would really appreciate a reference where I can see a proof of these claims, or (if it's easier) a quick sketch of why it should be true. -It's not obvious to me at all why quantifying across proper classes should allow this sort of thing, since all relevant sets (sets of proofs, or sets of statements, or whatever) should be contained in some subset of $\omega$, so should be able to be constructed in $ZF$. - -REPLY [13 votes]: Concerning the fact that Kelley-Morse set theory proves Con(ZFC) and much more, I wrote a post on my blog explaining one way to do it. - -Kelley-Morse implies Con(ZFC) and much more - -The basic outline of the proof is to show that KM proves the existence of a class truth predicate Tr for first-order truth (and this is what Andreas also suggests in his answer), and then to prove that this truth predicate decrees that each axiom of ZFC is true. It follows by reflection that there will be rank initial segments $V_\theta$ that has the same satisfaction relation, and so these are transitive models of ZFC, and indeed KM proves that the universe is the union of a closed unbounded elementary tower $$V_{\theta_0}\prec V_{\theta_1}\prec\cdots\prec V_\lambda\prec\cdots\prec V.$$ -Alternatively, a more syntactic argument proceeds by observing that the truth predicate is closed under deduction, complete and contains no explicit contradictions, and so it provides a complete consistent extension of ZFC. -Follow the link for the details.<|endoftext|> -TITLE: Are there one-dimensional ideals in any local ring -QUESTION [6 upvotes]: I would like to know if it is always possible to find a one-dimensional ideal in a local commutative ring... actually I am interested in finding a curve through a point on a scheme (locally). If the ring is of finite dimension it should be obvious, but does anybody know about the situation in more general rings? - -REPLY [2 votes]: The above answer says there is a ring such that for any nonmaximal prime ideal there is a nonmaximal prime ideal containing it. In fact valuation ring with infinite products of integers group as valuation group is an example. A paper about spectral space of commutative rings written by Hochster tells us there is also ring with spectral space which order just inverse the spectral space of a given ring. So there also exists a ring such that for any nonminimal prime ideal there exists another nonminimal prime ideal which is contained in it. That is there is a ring which has no prime with height 1.<|endoftext|> -TITLE: Reference for representation theory of SL_2(Z/n) -QUESTION [16 upvotes]: There are many references for the representation theory (say over $\mathbf C$) of $SL_2(\mathbf{F}_q)$ and $GL_2(\mathbf{F}_q)$, for instance lecture 5 in Fulton--Harris "Representation theory" and section 4.1 in Bump's "Automorphic forms and representations". Is there anywhere I can find a similarly explicit description of the representation theory of $SL_2$ and $GL_2$ over the ring $\mathbf Z/n$, where $n$ is not a prime? (By the Chinese remainder theorem, it suffices to consider the case when $n$ is a prime power.) - -REPLY [4 votes]: Another reference that goes along the lines of what Professor Humphreys mentioned in his answer is a paper by Bill Casselman "the restriction of a representation of $GL_2(k)$ to $GL_2(O)$"Math. Ann. 206, 311- 318 (1973). Also with respect Kutzko's results, mentioned in previous answers, even though they were not published one can find them in his Ph.D thesis at Madison. On convenient feature about Kutzko's thesis is that it contains at the end very explicit character tables for the groups in question. If you are interested I have a scan copy of such tables.<|endoftext|> -TITLE: Motivic cohomology vs. K-theory for singular varieties -QUESTION [13 upvotes]: As far as I understand, for a smooth variety $X$ its motivic cohomology could be described as the corresponding piece of the $\gamma$-filtration of (Quillen's) $K^*(X)$; this is completely true for $\mathbb{Q}$-coefficients, and true up to bounded denominators for $\mathbb{Z}$-coefficients. -My question is: is there a similar result for singular varieties? Here for motivic cohomology I would like to take $Hom_{DM}(M(X),\mathbb{Z}(p)[q]$; $DM$ is the category of Voevodsky's motives, and $M(X)$ is the motif of $X$ (I don't want to take the motif with compact support instead). These cohomology theories satisfy cdh-descent, but have no easy descriptions in terms of complexes of algebraic cycles. Unfortunately, I don't know much about the $\gamma$-filtration of $K$-theory. -Actually, I would like to prove the following fact: if a morphism $X\to Y$ of varieties induces an isomorphism for $K^*$, then the exponents of the kernels and cokernels of the corresponding morphisms for motivic cohomology are bounded (by a constant that depends only only on the dimensions of $X$ and $Y$). Any hints and/or references would be very welcome! - -REPLY [9 votes]: The precise relationship between K-theory and motivic cohomology for smooth schemes is the (analog of the) Atiya-Hirzebruch spectral sequence. This generalizes to non-smooth schemes if one uses K'-theory and higher Chow groups. -It is easy to see that motivic cohomology cannot be used to recover K-theory, -because it does not detect nilpotents. For example, $k[t]/t^2$ has K-theory different -from $k$, but motivic cohomology is the same. cdh-descent than gives examples for -reduced schemes (look at a cusp). -There are some approaches (Bloch-Esnault) to beef up cycles to get results in -the above examples. -As for your original question, recover motivic cohomology from K-theory, this might be -possible, but I have no result to offer.<|endoftext|> -TITLE: When do functors induce monadic adjunctions to presheaf categories -QUESTION [9 upvotes]: For a category $C$, let $C-Set$ denote the category of set-valued functors $\delta\colon C\to Set$. Given categories $C$ and $D$, and a functor $F\colon C\to D$, composition with $F$ yields a functor that I'll denote by $$\Delta_F\colon D-Set\longrightarrow C-Set.$$ The functor $\Delta_F$ has both a left adjoint, which I'll denote by $\Sigma_F\colon C-Set\longrightarrow D-Set$, and a right adjoint $\Pi_F\colon C-Set\longrightarrow D-Set$. One then has a monad $M = \Delta_F\Sigma_F$ on $C-Set$ and a monad $N=\Pi_F\Delta_F$ on $D-Set$. Let $M-alg=(C-Set)^M$ denote the category of $M$-algebras on $C-Set$, and similarly, let $N-alg=(D-Set)^N$ denote the category of $N$-algebras on $D-Set$. -I'll say that $\Delta_F$ is monadic if the obvious functor $D-Set\longrightarrow M-alg$ is an equivalence of categories, and I'll say that $\Pi_F$ is monadic if the obvious functor $C-Set\longrightarrow N-alg$ is an equivalence of categories. -Questions: - -Under what conditions on $F$ is $\Delta_F$ monadic? -Under what conditions on $F$ is $\Pi_F$ monadic? - -Thanks! - -REPLY [9 votes]: I will try to give an answer to your first question. -The functor $\Delta_F$ verifies automatically nearly all the conditions of the monadicity theorem: - -it is a right adjoint; -it is a left adjoint with cocomplete domain, and thus coequalizers exist in the source and are preserved by $\Delta_F$. - -It remains to see when the functor $\Delta_F$ is conservative, i.e. reflects isomorphisms. This is true, for example, in each of the following two cases: - -$F$ is essentially surjective; -$\Delta_F$ is full and faithful. This condition holds if $F$ induces an equivalence of Cauchy completions, i.e. $F$ is full and faithful and every object in the codomain is a retract of an object in the image. $\Delta_F$ is also full and faithful when $F$ is a reflection onto a reflective subcategory, or more generally when $F$ formally inverts some arrows (such as the map from a model category to its homotopy category).<|endoftext|> -TITLE: Higher homotopy groups of slice disk complement -QUESTION [9 upvotes]: Let $K \subseteq{\mathbb{S}}^3=∂\mathbb{D}^4$ be a non-trivial slice knot, i.e. $K$ bounds a slice disk $\Delta$ in $\mathbb{D}^4$. Let $N(\Delta)$ be a regular neighborhood of $\Delta$ in $\mathbb{D}^4$. - -What is known about $\pi_i(\mathbb{D}^4-N(\Delta))$, for $i\geq2$? - -For what it's worth, $\pi_1(\mathbb{D}^4-N(\Delta))$ is known to be normally generated by the meridian of $K$. The knot complement $\mathbb{S}^3 - K$ and $∂(\mathbb{D}^4-N(\Delta))=M_K$ (the zero-framed surgery on the knot $K$) are both known to be aspherical. - -REPLY [10 votes]: To add to Ryan's answer, -$2$--knots usually don't have aspherical complements, see Dyer & Vasquez, The sphericity of higher dimensional knots, Canad. J. Math. 25(1973), 1132-1136. This suggests a complicated answer for slice disks in general. -On the other hand, if the disk is ribbon, then the complement is aspherical, see Asano, Marumoto, and Yanagawa, T, Ribbon knots and ribbon disks, Osaka J. Math. -18 (1981), 161-174 - -REPLY [9 votes]: The homotopy groups can be pretty big things. For example, your $D^4 - N(\Delta)$ class of spaces contains the class of all $2$-knot complements -- simply remove a 4-ball neighbourhood of $S^4$ that intersects the $2$-knot in an unknotted disc. -$2$-knot complements have fairly complicated homotopy groups. For example, Cappell-Shaneson knot complements fiber over $S^1$ with fiber a once-punctured $(S^1)^3$. The universal cover of this space is $\mathbb R \times (\mathbb R^3 - \mathbb Z^3)$, so by the Hilton-Milnor theorem, rationally the homotopy groups are a free lie algebra with countably-infinite many generators (up to the action of $\pi_1$ there's just one generator, though). -Off the top of my head I don't know if slice disc groups are any more general than knot groups, they're probably not very far from each other. I think Kawauchi may not cover this but the references in his survey book should mention something.<|endoftext|> -TITLE: Reference Request: Steinberg's 1975 paper "On a paper of Pittie"(retrieved) -QUESTION [6 upvotes]: I am currently work on a senior project trying to prove for semisimple Lie groups, $R(T)$ is a free module over $R(G)$ by computing an explicit basis for all the A,B,C,D cases. The canoical reference is a paper by Pittie( H.V. Pittie: Homogeneous vector bundles on homogeneous spaces, Topology II (1972) 199-203), but I could not find it online or in any books available in the library. Steinberg generalized Pittie's statement in his paper (Robert Steinberg, On a theorem by Pittie, Topology Vol. 14. pp. 173-177. Pergamon Press, 1975, Printed in Great Britain. Received 1 October 1974). -Since they already proved this in the past, I would like to see their papers before I finish my project, even if at some monetary cost. But I could not access either of them. Not knowing their work would not hinder my research, for I work in a much more elementary level than they did, but I think their work might be related to my eventual results and I should acknowledge in case they proved some formula I proved again on my own. So I want to ask where I can find them in paper or electronically. I can read parts of Steinberg's paper via google books, but I would like a pdf file or something (so I may check). -ADDED: -With advisor's help and the links provided by all the people below, I retrieved the two papers. -ADDED: -Received Steinberg's replying email. He notes "A correction should be made on p.175, line 6 ( which starts with "Consider now ") by putting the exponent "n sub a" on the item over which the product is being taken.The paper by Pittie appears in Topology, vol. 11, 1972, pp. 199-203, and, if I remember correctly, does not contain an explicit basis for the quotient. " This is important so I put in here. - -REPLY [5 votes]: To supplement Barry's citations, I'd point out that the journal Topology was at that time managed by a company which eventually gave up on it after editors resigned partly in protest against the high prices charged. While the online rights now belong to the ScienceDirect conglomerate, it's expensive to access. This can be frustrating because each paper discussed here is only 4+ pages long. -On the other hand, Steinberg's paper is reprinted in the moderately priced one volume Collected Papers (AMS 1997). Though Steinberg is long retired from UCLA, he maintains an email link there, and might be able to supply a reprint of his article. Pittie is an Indian mathematician who has taught at one of the colleges of City University of New York but has not published for many years; his entry in the combined membership list CML (www.ams.org) does give a current mailing address in New York City. -Some users of MO including myself do have access to both papers and might be able to answer precisely stated questions about them. -ADDED: I hadn't heard previously about the recent death of Harsh Pittie. I was somewhat acquainted with him when we were both at NYU-Courant decades ago and recall hearing some of his lectures on topology of Lie groups. His paper from that period was grounded in topology and K-theory, but Steinberg's follow-up (in his typical concise style) rounded out the discussion of representation rings in a more algebraic framework. Moreover, Steinberg exhibits an explicit basis for $R(T)$ as a free $R(G)$-module in the crucial case where $G$ is a semisimple simply connected compact Lie group and $T$ any maximal torus. In particular, the rank here is the order of the Weyl group $W$. (He also observes that the same ideas work for algebraic groups over any algebraically closed field.) -Though I've never worked through the details of Steinberg's paper carefully, the underlying idea can be observed (in an oversimplified way) in the rank 1 case. Denoting the weight lattice (character group of $T$ in additive notation) by $X$, the respective representation rings look like $\mathbb{Z}[X]$ and $\mathbb{Z}[X]^W$. Then Steinberg's basis elements, one for each element $w \in W$, are defined by applying $w^{-1}$ to a product of symbols (in my notation $e^\lambda$) with $\lambda$ running over suitable fundamental weights. In rank 1, the basis just consists of $e^0, e^{-\rho}$. - -REPLY [2 votes]: If you're willing to pay, you can go to the Topology website and track the articles down. Here's a link that'll take you straight to the issue with the Pittie piece: http://www.sciencedirect.com/science/journal/00409383/11/2 -- you can find a link to the Steinberg issue there too. (Caveat: I don't know for a fact the articles are actually available; it's possible the site will say the order can't be filled. I didn't want to plunk down the coin to find out.)<|endoftext|> -TITLE: Inverse limit of spectral sequences -QUESTION [5 upvotes]: I find myself in the following situation: -I have a sequence of first quadrant spectral sequences, let's call them $ E(n)_{p,q}^* $, each convergent to $E(n)_{p,q}^\infty$, with spectral sequence morphisms $E(n)_{*,*}^* \to E(n-1)_{*,*}^*$, so we have an inverse directed system of spectral sequences. -Each module of each page is a locally finite graded vector space, so if you define $E_{p,q}^* = \varprojlim_n E(n)_{p,q}^* $, $E_{p,q}^* $ turns out to be a spectral sequence (differentials are the limit of differentials in the original sequences, etc.) by an argument found in a paper by John Carter. However, in this same paper he states that you can't say that $E_{p,q}^*$ converges to $\varprojlim E(n)_{p,q}^\infty$, and his counterexample rests on the fact that his spectral sequences can be non-bounded... Does anyone know if this "convergence" result is true for bounded (or, say, first quadrant) sequences? - -REPLY [11 votes]: The inverse limit of directed systems of locally finite dimensional graded vector spaces is an exact functor. That is why it takes directed systems of spectral sequences to spectral sequences of the inverse limits. -In the case of the first quadrant spectral sequences, the notion of convergence does not involve taking any limits (or more precisely, all the limits involved stabilize at the terms known in advance from the gradings $p$ and $q$; see Dylan Wilson' comment). So the inverse limit of the limit terms $E^\infty(n)$ will be the limit term $E^\infty$ of the inverse limit of the spectral sequences. -The reason spectral sequence limits may not commute with the inverse limits in general must be because defining the term $E^\infty$ may involve passing to the direct limits in some cases. This does not happen for first quadrant spectral sequences.<|endoftext|> -TITLE: On a remark of Tait on FLT for the exponent 3 -QUESTION [37 upvotes]: This is one of those recreational questions that aren't really about research. I found a curious remark in an old volume of American Mathematical Monthly (1922) which I'll quote below: - -In the Proceedings of the Royal Society of Edinburgh, vol.7, p.144, in some mathematical notes by professor P.G. Tait, it is stated: -"If $x^3+y^3=z^3$, then $(x^3+z^3)^3y^3+(x^3-y^3)^3z^3=(z^3+y^3)^3x^3$. -This furnishes an easy proof of the impossibility of finding two integers the sum of whose cubes is a cube." -How does this "easy proof" follow? Students are notoriously suspicious of those steps which an author announces as "easy", and are sometimes inclined to believe that the word is used in a humorous sense. [...] There are of course proofs in existence that the sum of two cubes cannot be a cube. - -Did anyone manage to find a proof of FLT for the exponent 3 using this identity or is the alluded proof another illusion that did not fit in the margin? - -REPLY [20 votes]: If $x^3 + y^3 = z^3$ then it is easy to prove that without loss of generality a co-prime version must exist such that $x, y, z$ have no common factor other than 1. Further it is easy to prove that there must be two co-prime positive numbers $p$ and $q$ such that one of the three, say $z$, has the form $z^3 = 2p(p^2 + 3q ^2)$ with $p$ and $q$ of opposite parity (one odd, the other even), therefore $p^2 + 3q^2$ being an odd cube and $2p$ being an even cube, and gcd $(2p, p^2 + 3q^2)$ being 1 or 3. As Euler has shown, it is possible then to find a smaller solution to Fermat's problem for $n$ = 3. The proof of the last step is somewhat tedious. In both cases it includes the lemma that there exist two co-prime numbers $a, b$ of opposite parity such that $p = a^3 - 9ab^2$. -Perhaps Tait thought to use the result of his equation (a), "every cube is the difference of two squares, one at least of which is divisible by 9" in order to easen this last step. -Concerning part (b), it is obvious that Tait's second equation has the same structure as the original one, but now containing three cubes that are definitely greater than the original cubes. -$((x^3 + z^3)y)^3 + ((x^3 - y^3)z)^3 = ((z^3+y^3)x)^3$ -This procedure of increasing the cubes can be repeated in infinity. Although it is not justified to reverse the derivation and to conclude that by this method every triple of cubes could be reduced to smaller cubes, it cannot be excluded that Tait was misled by this slip. -I don't believe that there is an "easy" proof, and I have to offer three reasons: -The first one (and admittedly the weakest) is that I have not suceeded in finding an easy solution. Any simplification like -$(2x^3 + y^3)^3y^3 + (x^3 - y^3)^3(x^3+y^3) = (x^3 + 2y^3)^3x^3$ -$((2x^3 + y^3)^3 + (x^3 - y^3)^3)y^3 = ((2y^3 + x^3)^3 + (y^3 - x^3)^3)x^3$ -leads easily to the result that the derived equation is correct, finally leaving 0 = 0, but no information about the impossibility of an integer character of the roots appeares obvious to me. -A certainly stronger reason is that this, if correct very interesting, note has, as far as I can see, never been mentioned again by Tait and has not been included in his collected papers, which were edited and furnished with a preface by Tait himself. (The note has however been mentioned in a great many of other sources including Ribenboim's popular book and Wikipedia.) -The third and most important reason however, outweighing all others, is that this question has been asked in the American Mathematical Monthly in 1914, 1916, 1919, 1920, 1921, and 1922, which at that time was edited by eminent mathematicians like Hurwitz and certainly read by many others. Last but not least, the problem has been open here in MathOverflow for more than one year. No easy way could be shown. -So it is very probable that this note belongs to the same category as Fermat's original one.<|endoftext|> -TITLE: An n!-dimensional representation of the symmetric group S_{n+2} -QUESTION [22 upvotes]: I have come across a sequence of representations $V_n$ of the symmetric group $S_{n+2}$ which has the property that restricting the action $S_n \subset S_{n+2}$ gives the regular representation: -$$ Res^{S_{n+2}}_{S_n} V_n = \mathbb{Q}S_n. $$ -In other words, there is some natural way to give the regular rep of $S_n$ an action of $S_{n+2}$. This (to me) is surprising, but I imagine this has already been observed. -For concreteness, here are the first few terms in the sequence, written as a sum of partitions (using the usual indexing of representations of $S_{n+2}$ by partitions of $n+2$): -[2] -[3] -[2,2] -[3,1,1] -[3,3]+[2,2,1,1]+[4,1,1] -... - -I have two questions: - -Is there already work on unrestrictions of the regular representation of a symmetric group? Is my particular sequence of representations $V_n$ well-known? -In general, are there circumstances under which a representation of $S_n$ has a canonical way to extend the action to $S_{n+1}$? - -REPLY [8 votes]: When $n+2$ is a prime $p,$ it is relatively easy to construct such a representation. For then the symmetric group $S_{p}$ contains a Frobenius group $F$ of order $p(p-1).$ Every element of $F$ either has order $p$ or has order dividing $p-1.$ Furthermore, each non-identity element of $F$ of order dividing $p-1$ has exactly one fixed point in the natural permutation action of $S_{p}.$ Consequently, if we induce the trivial module for $F$ to $S_{p},$ and restrict that module back to the natural copy of $S_{p-2},$ then we obtain the regular module for $S_{p-2}$ (using for example, Mackey's formula and the fact that $[S_{p}:F] = (p-2)!,$ and noting that $F^{x} \cap S_{p-2} = 1$ for all $x \in S_{p}).$ However, I do not see offhand an easy way to generalize this argument when $n+2$ is not prime -Note also that a rather simpler argument using the cyclic subgroup $X$ generated by an $n+1$ cycle shows that (for general $n$), the permutation module afforded by the action of $S_{n+1}$ on the (say right) cosets of $X$ restricts to the regular module of $S_{n}.$<|endoftext|> -TITLE: Mapping class group and property (T) -QUESTION [15 upvotes]: Does anyone know what the current expert consensus is concerning the status of the question as to whether the mapping class group of a surface has property (T)? -There is a short (21 page) paper by J. Andersen which purports to use quantum representations to prove that it does not. See here. It was released in 2007, but it does not seem to have yet been accepted by a journal. I have asked several experts (on the mapping class group and on property (T)), and none of them seem to understand the details of this paper. One or two of them alluded to issues they had heard might exist, but they were pretty vague as to what these issues might be. - -REPLY [30 votes]: Dear Michael, -There certainly is a complete proof of this result including a proof of theorem 5. Theorem 5 -follows essentially from my joint work with Kenji Ueno presented in a series of 4 joint papers. These papers are all on the archive and three of them has been published/accepted for publication and the four is submitted for publication. I am currently writing up the paper [A6] "Mapping class group invariant unitarity of the Hitchin connection over Teichmuller space", where a detailed argument for Theorem 5 will be given. It is Corollary 1 of that paper. As soon as that paper is finished I will put it on the archive and I will submit that paper together with my paper -"The mapping class group does not have Property T" for publication. -I am not aware of any problems with my proof.- I have ever only received one email suggesting there was a problem (from Mark Sapir), which claimed that Vaughan Jones knew of a problem with the paper. I immediately wrote to Vaughan to ask him what this problem was and he applied right away he didn't know of any mistakes in my argument. Mark right after acknowledged via email that he had misunderstood Vaughan. -I am more than happy to answer all questions via email regarding this result and its proof, so Michael, if you would please let me know who you are and your email address, I will get in touch right away. -Yours, -Jørgen Ellegaard Andersen -Note added by Ian Agol: Andersen's paper proving Theorem 5 appeared on the Arxiv recently.<|endoftext|> -TITLE: Normalizer of a von Neumann algebra -QUESTION [6 upvotes]: Let $(A,H)$ be a von Neumann algebra in standard form (which means that $H=L^2A$), and -recall that the automorphism group $Aut(A)$ acts on both $A$ and $H$. -Let -$$N:=\{u\in U(H): uAu^*=A\}$$ -be the normalizer of $A$, equipped with the strong topoloy (the subspace topology from $U(H)$). -The group $N$ is canonically isomorphic to the semidirect product $$N\cong Aut(A)\ltimes U(A').$$ -Equip $Aut(A)$ with Haagerup's u-topology, i.e. the topology of pointwise convergence on $A_*$, equivlalently, it is the topology of pointwise convergence on $H$. -Question: is the projection map $$N\to Aut(A)$$ continuous? - -REPLY [4 votes]: Julien's comment to my previous answer leads to an even easier solution. -Consider the isomorphism map $\psi:Aut(A)\ltimes U(A^\prime)\rightarrow N$. The $u$-topology on $Aut(A)$ is the topology of pointwise convergence on $H$, which is the same as strong convergence when we consider the elements of $Aut(A)$ as unitaries on $H$. Since $N$ is a topological group, it follows that $\psi$ is continuous. Julien's comment shows that it is in fact a homeomorphism. So certainly the quotient $N\rightarrow Aut(A)$ is continuous.<|endoftext|> -TITLE: Are acyclic subcomplexes of finite contractible 2-complexes contractible? -QUESTION [14 upvotes]: Let $Y$ be a contractible finite simplicial 2-complex. -Let $X$ be an acyclic subcomplex of $Y$ (i.e. $X$ connected, $H_1(X)=0$, $H_2(X)=0$). -Is $X$ contractible? (Equivalently, is $\pi_1(X)$ trivial?) -I asked this question among others here before, but it remained unanswered. -Apparently this question is equivalent to the following one. -If $m < n$, $G = \langle a_1,\dotsc,a_m \mid r_1 = \dotsb = r_m =1 \rangle$ and $J=\langle a_1,\dotsc,a_n \mid r_1 = \dotsb = r_n =1 \rangle$ are two groups given by (balanced) presentations, the presentation of $G$ is a sub-presentation of the presentation of $J$, and $J$ is trivial, is $G$ trivial? (It can be shown that the abelianization of $G$ is trivial.) - -REPLY [10 votes]: According to Sergei Ivanov, On balanced presentations of the trivial group, Invent. Math. 165, 525--549 (2006), for $n=m+1$, this is a particular case of Kervaire–Laudenbach conjecture which is also open, and according to one of Klyachko's results, the negative answer in this special case would imply the negative answer to Whitehead's asphericity conjecture. -It seems however that nothing is known in general, not even equivalence with the Kervaire–Laudenbach conjecture. - -REPLY [8 votes]: One can show the following: - -Theorem: If the group given by the subpresentation is hyperlinear (in particular if it sofic), then it must be trivial. - -The argument is based on an old Theorem of Gerstenhaber-Rothaus. I can only give a sketch here: Denote the group given by the subpresentation by $G$. We want to show that $G$ embeds into the group given by the full presentation to get a contradiction. Now, as $G$ is hyperlinear, one can satisfy the group multiplication on a finite subset of $G$ by unitaries up to any desired precision in the normalized Hilbert-Schmidt norm. Equivalently, $G$ embeds into a metric ultraproduct of unitary groups. Now, if unitaries $g_{m+1},\dots,g_n$ can be found which satisfy the relations $r_{m+1},\dots,r_n$, then the embedding of $G$ can be extended to homomorphism into the ultraproduct defined on $g_1,\dots,g_n$. In particular, $G$ embeds into the group given by the full presentation. Now, the fact that $g_{m+1},\dots,g_n$ can be found relies on the solvability of the equations $r_{m+1},\dots,r_n$ and can be proved along the lines of Gerstemhaber-Rothaus using degree theory and Hopf's computation of the cohomology of unitary groups. -Gerstenhaber-Rothaus proved some generalized form of the Kervaire-Laudenbach Conjecture for residually finite groups in -M. Gerstenhaber and O.S. Rothaus, The solution of sets of equations in groups, Proc. Nat. Acad. Sci. U.S.A. 48 (1962), 1531–1533. -The above argument is essentially only the observation that their argument extends without any problems to hyperlinear groups.<|endoftext|> -TITLE: Is the Lipschitz unique for close Alexandrov space other than dimension 4? -QUESTION [6 upvotes]: For a closed smooth $n$-manifold ($n\ne 4$), the Lipschitz structure is unique by the result of Sullivan. How about the Alexandrov spaces? -At first I was thinking it is trivially true by induction, since locally distance ball in an Alexandrov space is homeomorphic to cone over it's space of directions, which is also an Alexandrove space. But for 5-dimension and higher, the space of directions could be 4-dimensional Alexandrov spaces which admits a metric with curvature bounded from below by 1. -Now one can ask another question for manifold: "whether there exists a positively curved 4-manifold that admits two different Lipschitz structure?" - -REPLY [8 votes]: First, I want to point out that unlike for manifolds there is no notion of a geometric Lipschitz structure for general Alexandrov spaces as they are not locally modeled on a fixed space topologically. One can still ask whether you can have two Alexandrov spaces which are homeomorphic but not bi-Lipschitz homeomorphic but that is a much more metrically rigid question. This is an open problem for $n>4$. I believe your second question is open too mostly because there are so few examples of positively curved smooth manifolds. $\mathbb{CP}^2$ and $\mathbb S^4$ are all the known orientable ones in dimension 4 and as far as I know neither is known to admit an exotic lipschitz structure. -Edit: Actually, $\mathbb {RP}^4$ may be a candidate for this. If I understand things correctly there exists an exotic smooth $\mathbb{RP}^4$ by a result of Fintushel and Stern. I'm actually somewhat confused by this because Fintushel and Stern only say that their example is homotopy equivalent to $\mathbb{RP}^4$ but don't mention homeomorphism. But I've seen people claim that Fintushel and Stern example produces a topological $\mathbb{RP}^4$. -Perhaps someone in the area can clarify this. But assuming this is the case one can hope to show that their example is not bilipschitz to the standard $\mathbb{RP}^4$. I don't think this is known however. As far as I know the only tools for distinguishing lipschitz structures on 4-manifolds are due to Sullivan and Donaldson in "Quasiconformal 4-manifolds". They are Lipschitz versions of Donaldson invariants and I think that theory only works for manifolds with $b_2>1$ which is not the case here. -Note that if you could construct two homeomorphic but not bilipschitz positively curved Alexandrov manifolds $X$ and $Y$ of dimension 4 (or more generally positively curved Alexandrov spaces of dimension 4) then you'd also get a counterexample to your first question given by spherical joins $X* X$ and $Y*Y$. This is because any bilipschizt homeomorphism between $X* X$ and $Y* Y$ would need to preserve the topological strata. So it would send $X$ to $Y$ and since they are convex in $X*X$ and $Y*Y$ respectively the resulting homeomorphism $X\to Y$ would be bilipschitz with respect to the original metrics on $X$ and $Y$.<|endoftext|> -TITLE: Brauer Groups and K-Theory -QUESTION [22 upvotes]: Is there some a priori reason why we should expect the Brauer group of real [complex] super vector spaces to be closely related to periodicity in real [complex] K-theory? By "a priori" I mean a proof that does not involve computing that both are Z/8 [Z/2] and does not involve noticing that both are related to Clifford algebras. -Theo Johnson-Freyd posed a similar question in the comments of this MO question. -"Brauer group" above means the group of Morita equivalence classes of Morita-invertible super algebras over $\mathbb R$ [$\mathbb C$]. -I suspect the answer is some form of "No", since several bouts of googling have failed to turn anything up. But I would be happy to be proved wrong. - -REPLY [16 votes]: I'm a bit late to the party, but here's what I suspect the answer should look like. Forgive me for working somewhat to very heuristically throughout. -To star things off, here's a silly question: why is cohomology $\mathbb{Z}$-graded? Starting from a spectrum $E$ and a space $X$ we can consider the set $[X, E]$ of homotopy classes of maps from the suspension spectrum of $X$ to $E$. This only gives the zeroth part $E^0(X)$ of the $E$-cohomology of $X$. To get more interesting information one thing we can do is smash $E$ with invertible spectra. These are precisely the shifts $S^n, n \in \mathbb{Z}$ of the sphere spectrum, and -$$X \mapsto [X, S^n \wedge E]$$ -recovers $E$-cohomology as ordinarily understood. So we should think of $\mathbb{Z}$ here as being the Picard group $\text{Pic}(\text{Sp})$ of the symmetric monoidal $\infty$-category of spectra. That is: - -Cohomology in general is graded by the Picard group of invertible spectra. - -In more complicated situations we get more interesting Picard groups that we can grade by. For example, $\infty$-local systems of spectra over a space $X$ describe cohomology theories for spaces over $X$, and these cohomology theories are therefore graded by the Picard group $\text{Pic}(\text{Loc}_{\text{Sp}}(X))$. The fiberwise suspension spectrum of a spherical fibration over $X$ is in particular an example of an element of this Picard group. This should be relevant to André Henriques' observation in the comments. -One of the more complicated situations we could be in is the following. Let $R$ be an $E_{\infty}$ ring spectrum and let $E$ be an $R$-module spectrum. Rather than settling for smashing $E$ with invertible spectra, we can now smash $E$ with invertible $R$-module spectra over $R$ to get more interesting information of the form -$$X \mapsto [X, A \wedge_R E]$$ -where $A$ is invertible with respect to $\wedge_R$; that is, $E$-cohomology acquires a grading by $\text{Pic}(R\text{-ModSp})$, which I will abbreviate to $\text{Pic}(R)$. In particular, letting $R = E = KO$: - -Real K-theory is graded by the Picard group of invertible $KO$-module spectra. - -The punchline should now be that there is a natural group homomorphism from the super Brauer group of $\mathbb{R}$ to the Picard group of $KO$; possibly this is even an isomorphism. The construction should go something like this: - -Start with a finite-dimensional superalgebra $A$ over $\mathbb{R}$. -Construct the category of finite-dimensional (projective?) right $A$-supermodules, which is symmetric monoidal with respect to direct sum and enriched over supervector spaces. Topologize its homs as finite-dimensional real vector spaces. -Take the core of the above, which is a symmetric monoidal topological groupoid, and hence which presents a symmetric monoidal $\infty$-groupoid. -Take the group completion of the above, which is a grouplike $E_{\infty}$ $\infty$-groupoid, or equivalently an infinite loop space. - -But this isn't quite right, because we only get a connective spectrum this way. I don't know how to fix this. Maybe we should take chain complexes or something. -Anyway, let's pretend I did the right thing, and let's call the resulting spectrum $K(A)$. Let's also pretend that $K(\mathbb{R})$ ended up being $KO$. The categories I constructed above are all tensored over supervector spaces, or equivalently are all module categories over supervector spaces, so the resulting spectra should all be $KO$-module spectra. Moreover, taking tensor products should give natural equivalences -$$K(A) \wedge_{K(\mathbb{R})} K(B) \cong K(A \otimes_{\mathbb{R}} B)$$ -and since $K(A)$ only depends on the category of $A$-modules, any Morita-invertible $A$ should give an invertible $K(\mathbb{R})$-module spectrum $K(A)$ in a way that respects products. -So real K-theory has two gradings, one by $\text{Pic}(S)$ ($S$ the sphere spectrum) and one by $\text{Pic}(KO)$. The content of the Clifford-algebraic proofs of Bott periodicity should be that these gradings collapse into a single grading, and in particular that all of the $K(A)$ end up being shifts of $K(\mathbb{R})$ as $K(\mathbb{R})$-module spectra. -Of course we can replace $\mathbb{R}$ by $\mathbb{C}$ throughout. Actually it seems like we really ought to replace $\mathbb{R}$ by an arbitrary ring spectrum $R$ throughout, and try to relate the Brauer group of $R$ and the Picard group of $K(R)$; this might make it easier to figure out the right statements.<|endoftext|> -TITLE: The Higman group -QUESTION [27 upvotes]: The group of Higman: -$ -\langle -\ -a_0, a_1, a_2, a_3 \ | \ -a_0 a_1 a_0^{-1}=a_1^2, -\ a_1 a_2 a_1^{-1}=a_2^2, -\ a_2 a_3 a_2^{-1}=a_3^2, -\ a_3 a_0 a_3^{-1}=a_0^2 -\ \rangle . -$ -Is it simple? What is actually known about it except the fact that it does not have non-trivial homomorphims into a finite group? - -REPLY [23 votes]: Higman's group is not simple. Indeed, if you look at Higman's paper, you will see that his group is an amalgamated product of two groups $K_{1,2}=\langle a_1, a_2, b_2\mid a_1^{-1}a_2a_1=a_2^2, a_2^{-1}b_2a_2=b_2^2\rangle$ and $K_{3,4}=\langle a_3, a_4, b_4\mid a_3^{-1}a_4a_3=a_4^2, a_4^{-1}b_4a_4=b_4^2\rangle$ with free amalgamated subgroups $\langle a_1, b_2\rangle$ and $\langle a_3, b_4\rangle$ with $a_1=b_4, b_2=a_3$. Now take $K_{1,2}$ and add the relations $w(a_1,b_2)=1, w(b_2,a_1)=1$ for some "complicated" word $w$. You get a non-trivial group $K$. Similarly imposing $w(a_3,b_4)=1=w(b_4,a_3)$ on $K_{3,4}$ you get a non-trivial group $K'$. Then there is a homomorphism from the Higman group to the amalgamated product of $K$ and $K'$ with a non-trivial kernel. The fact that $K$ (and, equally, $K'$) is not trivial is not obvious but it is not very difficult to prove. - A simplification. Instead of adding two relations, in fact it is enough to add a relation $w(a_1,b_2)=w(b_2,a_1)$ and similarly for $a_3,b_4$. For example, $(a_1b_2)^3=(b_2a_1)^3$. - Update 1. Probably the easiest way to be convinced that $K$ ($=K'$) is non-trivial is to input its presentation into GAP or Magma (you do not need 3 in the simplified relation above, 2 is enough). - Update 2. An even easier way is to impose further relations $a_2=b_2=1$ on $K$ (as in the simplification above). The factor-group is the infinite cyclic group. Hence $K$ is infinite (and $K'$ is infinite too). - Update 3. I forgot to mention that one also needs that after we impose the relation as in the simplification, we should make sure that the subgroup of $K$ generated by $a_1,b_2$ has an automorphism switching $a_1$ and $b_2$ (otherwise we cannot form the proper amalgamated product of $K$ and $K'$). That is why we cannot replace 3 in the Simplification above by $1$. - Update 4. Gilbert Baumslag told me that Higman did not know whether his group is simple, but Paul Schupp proved it. And indeed, Schupp, Paul E. Small cancellation theory over free products with amalgamation. Math. Ann. 193 (1971), 255–264 proved much more: Higman's group is SQ-universal, so every countable group embeds into one of its quotients. - -REPLY [4 votes]: Reference: -Higman, Graham, A finitely generated infinite simple group, J. Lond. Math. Soc. 26, 61-64 (1951). ZBL0042.02201. -It is shown that $G$ is infinite and has no proper normal subgroups of finite index, except $G$. -It is easy to see that this group is perfect: it has trivial abelianization. -I have heard through the grapevine that the space $X$ with four one-cells and four two-cells (corresponding, respectively, to generators and relations) is the classifying space of the group (I don't have a reference).<|endoftext|> -TITLE: A Relative Algebraic Hartogs Lemma -QUESTION [16 upvotes]: The Algebraic Hartogs Lemma states that in a Noetherian normal scheme, a rational function that is regular outside a closed subset of codimension at least two, is in fact regular everywhere. -In a research problem I was working on recently, I was (following suggestions by my advisor) using this to prove that a particular section of a line bundle existed on a space $\mathbb{A}^n_H$. The argument worked well when $H$ was a point. For more general $H$, I could show that the section was defined except on a codimension-two subset of every fiber; but it was not immediately obvious to me how to go from there, to showing that the section was defined "on all fibers simultaneously," i.e., over $\mathbb{A}^n_H$. This was especially problematic in that the base scheme $H$ in question was a component of a Hilbert scheme, and thus (as shown by Ravi Vakil) capable of exhibiting arbitrarily bad behavior. In particular, the fact that a composition of normal morphisms is normal (see EGA IV.2, section 6.8) would not come close to covering my situation. -In order to deal with this, I have obtained what seems to be a proof of the following statement: - -Lemma (Relative Algebraic Hartogs' Lemma) Let $X \to S$ be a flat, finite-type morphism of Noetherian schemes such that every associated fiber is normal. Let $\mathscr{L}$ be a line bundle on $X$. Suppose that $U \subset X$ is an open subscheme such that - (i) $U$ contains all the associated points of $X$, - (ii) for every $s \in S$, $U \cap X_s$ contains all the associated points of $X_s$, and - (iii) for every associated point $\eta$ of $S$, $U$ contains all but a codimension-two closed subset of $X_{\eta}$. - Then the restriction map - $$\Gamma(X,\mathscr{L}) \to \Gamma(U,\mathscr{L})$$ - is an isomorphism. - -(I'm using "associated fiber" to mean "fiber over an associated point," by analogy with the standard term "generic fiber.") -I should note that suggestions of Will Sawin in this answer were invaluable in coming up with the proof. -This lemma is surprisingly strong, in that two key hypotheses--normality, and codimension-two-ness--only need to hold "generically" (i.e., in the fibers over the associated points). Another interesting feature is that, if you look at the proof, the fibers do not actually have to be normal, as long as they "satisfy the Hartogs property"--which I understand is equivalent to satisfying the S2 condition. I'm also not convinced that the finite-type hypothesis is necessary, but I have not yet verified the argument that I suspect would remove it. -The proof is not incredibly long (4 pages from me; probably less from a more experienced mathematician who knows what details to leave out), uses only techniques that are extremely well known, and so far as I can tell, does not use them in an especially clever way. Thus, it seems likely that someone, at some point, has already written up a similar result. However, neither I, nor my advisor, nor anyone else I have spoken to was familiar with it (which suggests to me that, if nothing else, the result is less well-known than it should be). Hence, my question: - -Question: Does anyone know of a similar statement in the literature? (To be safe, I'll also ask if anyone knows of any counterexamples, although I think the proof is solid.) - -One final note: From what I understand, the popularity of the name "algebraic hartogs lemma" is quite recent, possibly a result of its use in Ravi Vakil's notes, so a similar result in the older literature would probably not be called by the word "Hartogs." - -Update: I have posted a proof of the Lemma above. - -REPLY [13 votes]: Actually, a very similar statement can be found in the paper Reflexive pull-backs and base extension by Brendan Hassett and myself. See Proposition 3.5. Indeed you do not need normality, only that the fibers are $S_2$ and the sheaf does not need to be a line bundle only coherent and flat over the base. We develop a little bit of a relative theory in section 3, so you might find the rest useful as well, especially the statement regarding a characterization via local cohomology. We did assume the codimension two condition for every fiber, but it might not be necessary actually. I would have to check the proof (I will try to do that sometime). I would add that (as you discovered) for a Hartogs type statement you do not need normality and many things can be done for $S_2$ schemes that are usually done for normal ones. The main reason normality is more widely used (besides that it is easier to define and think that one understands it better) is that working with divisors on $S_2$ but not normal schemes has to be done very carefully. See this MO answer for more on this and perhaps this and this for more on Hartogs type questions.<|endoftext|> -TITLE: Flips of triangulations on non-orientable surfaces -QUESTION [11 upvotes]: Let $N_{k,r}$ be a non-orientable surface of genus $k$ (i.e the connected sum of $k$ projective planes) and with $p\geq 1$ punctures. -I'm looking at ideal triangulations of the surface, namely maximal collection of pairwise disjoint ideal arcs (joining punctures) on the surface. -Given a triangulation and an arc $\alpha$ in that triangulation, we can perform a flip along $\alpha$ : the arc belongs to two triangles forming a quadrilateral, the flip along $\alpha$ gives the new triangulation where the arc has been replaced by the other diagonal of the quadrilateral, and all the other arcs remains unchanged. (When the arc belongs to only one triangle, we cannot do the flip.) -A classical result in the orientable case states that given any two triangulations of an orientable surface, there is a sequence of flips that transform one into the other (Hatcher, On triangulations of surfaces). So the so-called triangulation graph, whose vertices are triangulations and edges are flips, is connected. -I am looking for the similar result for non-orientable surfaces. Does anyone have a reference ? My guess is that the result should also hold but I did not find anything on the litterature. - -REPLY [10 votes]: My article "Tiling the measured foliation space of a punctured surface", Trans. Math. 306 no. 1 (1988) contains a proof of this fact in the case of oriented surfaces. It is essentially the same as Hatcher's proof of contractibility, but focussing solely on the issue of connectivity, which introduces some simplifications. The proof works, with a little bit more effort, in the nonorientable case as well. -The crucial step in the proof for purposes of this discussion (carried out in the lemmas on pages 38 and 39) is to consider the situation that $\delta$ is an ideal triangulation, $h$ is an oriented ideal arc whose interior intersects $\delta$ transversely and efficiently ($h$ does not double back across the same arc in a triangle of $\delta$), $x_0$ is the first point where $h$ crosses $\delta$ transversely, and $\alpha$ is the edge of $\delta$ containing $x_0$. In this situation one wants to prove that $\alpha$ belongs to two distinct triangles of $\delta$ and so can be flipped; this is basically the inductive step for proving connectivity. -The proof of this step is to consider the possibility that $\alpha$ belongs to only a single triangle $T$ of $\delta$ --- meaning that $\alpha$ is the ideal arc obtained by identifying 2 sides of $T$ --- and to derive a contradiction. In the article this step is carried out only in the orientable category, where the result of gluing the 2 sides of $T$ must be a disc (depicted in the diagrams of that paper as a ``puncture piece''). -However, the conclusion of this argument remains true in the nonorientable category, and the proof requires just one more case to be considered, namely, when the result of gluing the 2 sides of $T$ is a Mobius band. In this situation, the efficient intersection condition would imply that $h$ is trapped in the interior of the Mobius band, winding more and more closely around its core and crossing $\alpha$ infinitely often, contradicting that the number of intersections must be finite. -With this consideration, the whole proof should go through unscathed; orientability is not otherwise used.<|endoftext|> -TITLE: A limit involving binomial coefficients: $\lim_{n\to\infty} (-1)^n\sum_{k=1}^n(-1)^k{n\choose k}^{-1/k}=\frac12$? -QUESTION [39 upvotes]: Experimentation suggests the limit -$$\lim_{n\rightarrow\infty} (-1)^n\sum_{k=1}^n(-1)^k{n\choose k}^{-1/k}=\frac{1}{2}\ .$$ -Does somebody have an idea for (a start of) a proof? -Added: There seem to be variations: -$$\lim_{n\rightarrow\infty} (-1)^n\sum_{k=1}^n(-1)^k{2n\choose k}^{-1/k}=\frac{1}{8}\ ,$$ -$$\lim_{n\rightarrow\infty} (-1)^n\sum_{k=1}^n(-1)^k{3n\choose k}^{-1/k}=\frac{2}{27}\ ,$$ -etc. -Moreover, the exponent $-1/k$ in the original identity can seemingly be replaced by $-2/k$, $-3/k$ (by $-\alpha/k$ -for strictly positive $\alpha$?) without changing the limit value. -Update: Given a strictly positive rational number $\frac{p}{q}\leq 1$ (with coprime natural integers $p,q$), there is perhaps -a number $\lambda(p/q)$ such that -$$\lim_{n\rightarrow\infty}(-1)^n\sum_{k=1}^{pn}(-1)^k{qn\choose k}^{-\alpha/k}= -\frac{(\lambda(p/q))^\alpha}{2}$$ -for $\alpha$ real and strictly positive. -A few values for $\lambda$ are seemingly -$$\lambda\left(\frac{1}{3}\right)=\frac{4}{27},\ \lambda\left(\frac{1}{2}\right)=\frac{1}{4},\ \lambda\left(\frac{2}{3}\right)=\frac{2}{3\sqrt{3}},\ \lambda\left(1\right)=1\ .$$ -Two more conjectures: $\lambda\left(\frac{1}{q}\right)=\frac{(q-1)^{q-1}}{q^q}$ -and $\lambda(x)^x=\lambda(1-x)^{1-x}$ for rational $x$ in $(0,1)$. -The correct formula for $\lambda$ is perhaps $\lambda(x)=x(1-x)^{(1-x)/x}$. - -REPLY [8 votes]: Here is a different approach. Write $$a_{n,k} = \binom{n}{k}^{-1/(n-k)}$$ (or whatever the term is) for $0 \le k < n$ and $a_{n,k} = 0$ for $k \ge n$. Then $$\sum_{k=0}^{n-1} (-1)^k \binom{n}{k}^{-1/(n-k)} = \sum_{k=0}^\infty (-1)^k a_{n,k} = \frac{a_{n,0}}{2} + \frac{1}{2} \sum_{k=0}^\infty (a_{n,2k} - 2 a_{n,2k+1} + a_{n,2k+2})\,.$$ -So if you can show that $$\lim_{n\to\infty} \sum_{k=0}^\infty |a_{n,2k} - 2 a_{n,2k+1} + a_{n,2k+2}| = 0\,,$$ the claim will follow. This should hold for any sufficiently "smooth" double sequence $(a_{n,k})$, since it just means that locally the sequence (for fixed $n$) varies nearly linearly. -If $a_{n,k} = f_n(k)$ for some nice function $f_n$, then the second order difference is bounded by $\max \{|f''_n(x)| : 2k \le x \le 2k+2\}$, which may help in the estimate. For example, taking $a_{n,k} = n^{-1/(n-k)}$, so $f_n(x) = \exp\left(-\frac{1}{n-x} \log n\right)$, we have $$f''_n(x) = \left(\left(\frac{\log n}{(n-x)^2}\right)^2 + \frac{2\log n}{(n-x)^3}\right) f_n(x)\,.$$ For $0 < \alpha < 1$, we find that $$\sum_{k=0}^{\lfloor n - \alpha \log n\rfloor} |a_{n,2k} - 2 a_{n,2k+1} + a_{n,2k+2}| \le C \frac{1}{\alpha^3 \log n}\,,$$and the tail can be estimated by a constant times $\exp(-1/\alpha)$. Taking $\alpha$ arbitrarily close to zero shows that $$\lim_{n\to\infty} \sum_{k=0}^{n-1} (-1)^k n^{-1/(n-k)} = \frac{1}{2}\,.$$ For the original problem, the estimates are likely to be more involved, but should be possible. -Addition: -The formula for $\sum_{k=0}^n (-1)^k a_{n,k}$ given near the beginning of this answer can (in the case $a_{n,k} = f_n(k)$) be interpreted as $a_{n,0}/2$ plus half the difference between the trapezoid-rule and midpoint-rule approximations to $\int_0^\infty f_n(x)\,dx$ using the even integers as subdivision points. For a well-behaved sequence of functions, this difference should tend to zero. -So the problem has only little to do with binomial coefficients as such.<|endoftext|> -TITLE: Detecting equivalences of (infinity) categories by nerves -QUESTION [6 upvotes]: I have two questions: - -Is there a way to tell if a functor $F:C \to D$ between two small categories is an equivalence in terms of the map $$N(F):N(C) \to N(D)$$ between simplicial sets? More generally, can we test separately when a functor $F$ is: essentially surjective, full, faithful etc., by an easy to verify property of $N(F)$? I am not interested in any answer involving applying the left-adjoint to $N;$ I am really looking for a description using simplicial sets. -If $\varphi:X \to Y$ are quasi-categories, is there way of saying when $\varphi$ is a weak equivalence in the Joyal model structure (categorical equivalence in the language of Lurie) akin to "full and faithful and essentially surjective?". I am most interested in definitions which are easily checkable, so not definitions like "the induced map on mapping complexes..." since these are hard to compute in practice. The definition I have seen just says that $\varphi$ becomes a weak equivalence of simplicial categories after applying the left-adjoint to the homotopy coherent nerve, which is somehow not a very satisfying definition. Is there a more explicit description, not involving simplicial categories? - -I hope this question is not too vague. Any comments to improve the wording etc. are welcome. Thanks! - -REPLY [4 votes]: Just a few ideas/observations... -For 1: -a. A functor is an isomorphism if and only if the induced map is an isomorphism of simplicial sets. -b. So a functor is an equivalence if and only if the induced map on the nerves of skeleta is an isomorphism. (This isn't very helpful though...) -c. Any map of simplicial sets $NC \rightarrow ND$ gives us a functor $C \rightarrow D$ (just look at the map on 0-simplices, 1-simplices, and 2-simplices to see what to do). So a map $NC \times \Delta^1 \rightarrow ND$ corresponds to functor $F: C \times [1] \rightarrow D$ (since $N[1] = \Delta^1$), but this is precisely a natural transformation of functors $C \rightarrow D$. Thus, if we have $NC \rightarrow ND$ and $ND \rightarrow NC$ such that the composites are "homotopic" to the identity (where I mean, use $\Delta^1$), then the original functor is an equivalence. -d. None of what we have said so far is any easier than just proving your original map is an equivalence. One thing you could do, that allows for homotopy theory, is associate to $C$ a stronger invariant. For example, take the bisimplicial set call it $\mathfrak{N}C$ given by: $\mathfrak{N}C_k = N(\text{iso }C^{[k]})$ (where "iso C" means "maximal groupoid" or "just take isomorphisms as your morphisms." This is a special case of a construction of Rezk's). It turns out that a functor is an equivalence if and only if the induced map of these special nerves is a weak equivalence. -e. I don't think that checking it's a weak equivalence works for the usual nerve, since (for example) any category with a final object has a nerve that's weakly equivalent to a point. However it is probably the case that if the induced map on nerves is a categorical equivalence then the original categories are equivalent. This brings us to point 2... -For 2 -a. I'm not sure if there is any way of showing that a functor between $\infty$-categories is an equivalence (in general) that does not amount to showing the induced map on mapping spaces is an equivalence. However, you can do all of this without mentioning simplicial categories. Given an $\infty$-category, $\mathcal{C}$, we can define something (or several things) equivalent to the mapping space between two objects in $\mathfrak{C}[\mathcal{C}]$; let Hom_C^R(x,y) be the simplicial set defined by requiring that a map $\Delta^n \rightarrow \text{Hom}_{\mathcal{C}}^R(x,y)$ be a map $z: \Delta^{n+1} \rightarrow \mathcal{C}$ such that $d_0z$ is the constant diagram on $y$ and $z$ evaluated on the initial vertex is $x$. (Actually this might be $\text{Hom}^L$, I can't remember off the top of my head). Anyway, this turns out to be a Kan complex (i.e. a space), and Lurie/Joyal show that it's the same as what you get after doing $\mathfrak{C}$. See Lurie for more. -b. Given the description above, we can say a functor is an equivalence if it induces a weak equivalence of Kan complexes for all mapping spaces as we defined above (i.e. fully faithful), and essentially surjective (i.e. every object is equivalent to something in the image). This isn't easy to do but... -c. In general, an equivalence of $\infty$-categories is kind of a big deal. In particular, it should be at least as hard as writing down various Quillen equivalences of model categories, right? Maybe slightly easier, but still quite hard. -Anyway, I realize this probably isn't what you want, but it's the best I can do at the moment! I'll let you know if I think of anything more helpful (but probably someone brighter will end up posting something more helpful soon...)<|endoftext|> -TITLE: Friedlander Conjecture -QUESTION [10 upvotes]: I just read on the ALGTOP discussion list that Morel has announced a proof of the Friedlander conjecture. Question: Are there other applications besides the Milnor conjecture $H_*(G,F_p)=H_*(BG,F_p)$ for complex algebraic groups $G$? And, for an outsider to algebraic geometry, what is the motivation to consider etale cohomology and how does it relate to ordinary cohomology, i.e. how does Friedlander imply Milnor? - -REPLY [5 votes]: As indicated by D. Roberts, see the video of Morel's talk at the Abel Conference in honour of Milnor held very recently! -Also, see Friedlander's own historical survey!<|endoftext|> -TITLE: Well-pointed space which is not locally contractible -QUESTION [13 upvotes]: I am looking for an example of a well-pointed space in which no (sufficiently small) neighbourhood of the base-point is contractible. As usual, a well-pointed space is a pointed space in which the inclusion of the base-point is a Hurewicz cofibration. -In the reverse direction, I am looking for conditions under which a well-pointed space has a basis of contractible neighbourhoods around the base-point. - -REPLY [12 votes]: My initial answer was fully ignorant of the literature on the subject; I'm now recalling that it does exist. In 1934, Borsuk and Mazurkiewicz constructed a $2$-dimensional AR (=contractible, locally contractible $2$-dimensional compactum) $X$ that is not a countable union of smaller ARs (Sur les rétractes absolus indécomposables, C.R.. Acad. Sci. Paris 199 (1934), 110-112; see Borsuk's "Theory of retracts", Section VI.4). The proof actually shows that $X$ is not a countable union of contractible proper subsets (open or not). Thus $X$ contains points that do not have a basis of contractible neighborhoods. -A simplified version $Y$ of $X$ (constructed below), which is an AR failing to have a basis of contractible neighborhoods at a certain point $\infty$, is also forward tame at $\infty$ and so answers Ricardo's revised question at the Algebraic Topology Discussion List. $Y$ looks somewhat like a $2$-skeleton of $W^+$, the example in my previous answer, except that $W^+$ is not forward tame. A google search for "Mazurkiewicz singularity" (this is how Borsuk called the kind of ANRs that do not have arbitrarily fine countable covers by ARs) returns a paper by Armentrout which has an example somewhat similar to $W^+$ (also a decomposition space and a manifold factor) though more laborious. -Note also a positive result of Begle proved using Zorn's lemma. - -Here's a construction of $X$ (which I hope is easier to read than Borsuk's account) and also of $Y$. -Let $Q=D^2/\{a,b\}$, the quotient of the $2$-disk where two distinct interior points $a,b$ are identified with each other. Let $\partial Q$ be the image of $\partial D^2$ in $Q$, and let $C$ be the image in $Q$ of a path joining $a$ and $b$ in the interior of $D^2$. Let $(Q_1,\partial Q_1)=(D^2,\partial D^2)$, and form $(Q_k,\partial Q_k)$ by attaching $Q_{k-1}$ to $(Q,\partial Q)$ along a homeomorphism $\partial Q_{k-1}\to C$. Then $Q_k$ is contractible, but no proper subset of $Q_k$ containing $\partial Q_k$ is contractible (or even acyclic). -Now $Y$ is just the one point compactification $(Q_\infty)^+$ of the direct limit $Q_\infty$ of the chain of inclusions $Q_1\subset Q_2\subset\dots$. -To see that $Y$ is locally contractible, observe that $D^2\times [0,1]$ collapses onto a copy of $Q_2$ such that $\partial Q_2=\partial D^2\times\{1\}$ and $Q_1=D^2\times\{0\}$. It follows that $D^2\times [0,\infty)$ admits a proper deformation retraction onto a copy of $Q_\infty$. Thus $Q_\infty$ is proper homotopy equivalent to $[0,\infty)$. Then $Q_\infty$ is forward tame in the sense of Quinn (see Definition 7.1, p.75 and Proposition 9.6, p.99 in Hughes-Ranicki), and therefore its one-point compactification is locally contractible. -For the construction of $X$ we will need $k$ copies $C_1^{Q_k},\dots,C_k^{Q_k}$ of $S^1$ embedded in $Q_k$, namely $C_1^{Q_k}=\partial Q_k$, and $C_{i+1}^{Q_k}=C_i^{Q_{k-1}}$. -Now $X$ is formed by attaching increasingly smaller copies of $Q_1,Q_8,Q_{64},...$ to the Sierpinsky carpet so that for each $k$ of the form $8^m$, the $k$ circles $C_1^{Q_k},\dots,C_k^{Q_k}$ in $Q_k$ are attached to the boundaries of the disks removed at the $m$th stage of the construction of the Sierpinski carpet, and additionally $(k-1)$ arcs connecting $C_i^{Q_k}$ with $C_{i+1}^{Q_{k+1}}$ are identified with analogous arcs in the Sierpinsky carpet. This can, and must be done so that every neighborhood of every point of the Sierpinsky carpet contains $C_1^{Q_k}=\partial Q_k$ for some $k$ (of the form $8^m$).<|endoftext|> -TITLE: An interesting double coset in the theory of automorphic forms -QUESTION [7 upvotes]: Does anyone have some idea to describe the double coset $P(F)\backslash G(F)/H(F)$ , say using Weyl group elements ? Here $G=GL_n\times GL_{n-1}$ is defined over a number field $F$ , $H=GL_{n-1}$ diagnoal embedded into $G$ as a subgroup and $P$ is some standard parabolic of $G$ . -The interesting point is that $H$ is not the fixed point set of some involution on $G$ so the quotient is not a symmetric space. Such example appers e.g. in the theory of Rankin-Selberg convolutions. -Let's start from a special case: say P is maximal parabolic. -Any comments and references will be welcome. Thank you ! - -REPLY [4 votes]: First of all some remarks: - -The pair that you discussed is spherical, so it is known that there is a finite number of such orbits (formally speaking it is implied only in char $0$ case, but it does not matter here) -A convenient way to think of the spherical space $G/H$ is as $GL_n$ where the action of $G$ is given by the left action of $GL_n$ and the right action of $GL_{n-1}$. - -Now to your question: -I'll try to give a set of representatives for the case when $P$ is the Borel. For arbitrary parabolic, the set of representatives should be a subset of the set I'll describe. It should be not so hard to find it, although it is not completely trivial even in the classical case of Bruhat cells. -After a suitable chose of the Borel the problem becomes equivalent to classifying the orbits in $GL_n$ under the left action of lower triangular matrices in $GL_n$ and right action of upper triangular matrices in $GL_{n-1}$. Here I consider $GL_{n-1}$ to be embedded into $GL_{n}$ as the upper left corner. -I think that the following set will do: -the set of matrices of the type $w+b$ where $w$ is a permutation matrix and $b$ is a matrix with first $n-1$ columns equal to $0$, in the last column all the entries below (and including) the $j$-th entry also $0$, and the others allowed to be either $0$ or $1$. Here $j$ is the index of the non zero entry in the last column of $w$. -I'm ~95% sure that this set covers all the orbits and ~75% sure that it covers each orbit once. Basically it is an easy exercise, but one has to be careful when doing it. I was not, so please double check me. -Additional remarks: - -This is only one possible choice, there are many others. I do not claim that this one is a good choice from some high level point of view, this is just the first one that came into my mind. -A similar question is the description of $K \backslash G(F)/H(F)$ where $K$ is a maximal compact subgroup of $G(F)$. We discussed this question in your case in http://arxiv.org/abs/0910.3199. -For general spherical case your question might be discussed in: - -P. Delorme, Constant term of smooth H -spherical functions on a reductive p-adic group. Trans. Amer. Math. Soc. 362 (2010), 933-955. - http://iml.univ-mrs.fr/editions/publi2009/files/delorme_fTAMS.pdf. -or -Y. Sakellaridis and A. Venkatesh, Periods and harmonic analysis on spherical varieties. In preparation. -Good luck.<|endoftext|> -TITLE: Nice proofs of the Poincaré–Birkhoff–Witt theorem -QUESTION [48 upvotes]: Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over a field $k$, with an ordered basis $x_1 < x_2 < ... < x_n$. -We define the universal enveloping algebra $U(\mathfrak{g})$ of $\mathfrak{g}$ to be the free noncommutative algebra $k\langle x_1,...,x_n\rangle$ modulo the relations $(x_ix_j - x_jx_i = [x_i,x_j])$. -The Poincaré–Birkhoff–Witt (PBW) theorem states that $U(\mathfrak{g})$ has a basis consisting of lexicographically ordered monomials i.e. monomials of the form $x_1^{e_1}x_2^{e_2}...x_n^{e_n}$. Checking that this basis spans $U(\mathfrak{g})$ is trivial, so the work lies in showing that these monomials are linearly independent. -One standard proof of PBW is to construct a $\mathfrak{g}$-action on the commutative polynomial ring $k[y_1,...,y_n]$ by setting $x_1^{e_1}x_2^{e_2}...x_n^{e_n}\cdot 1 = y_1^{e_1}y_2^{e_2}...y_n^{e_n}$ and verify algebraically that this gives rise to a well-defined representation of $\mathfrak{g}$. Details can be found in Dixmier's book on enveloping algebras. -What other proofs of PBW are there out there? -Are there nice reformulations of the above proof from a different perspective, such as one that emphasizes the universal property of $U(\mathfrak{g})$? -However, I would be especially interested in learning about proofs which are not just repackaged versions of the same algebraic manipulations used in the above proof (for example, geometric proofs which appeal to some property of $U(\mathfrak{g})$ as differential operators, etc.). If we allow ourselves more tools than just plain algebra, what other proofs of PBW can we get? - -REPLY [5 votes]: Recently, me and Vladimir Dotsenko formulated PBW in terms of maps of monads in https://arxiv.org/abs/1804.06485. As a by product, one gets the classical PBW theorem is the manifestation that, in characteristic zero, the associative operad is free as a right Lie module through the map $\phi:\mathsf{Lie} \to\mathsf{Ass}$ that sends the bracket to the commutator. -The left adjoint to the forgetful map from associative algebras to Lie algebras is the enveloping functor, and since $\mathsf{Ass} = \mathsf{Com}\circ\mathsf{Lie}$, one gets the desired natural isomorphism between $\phi_!(\mathfrak g) = U(\mathfrak g)$ and $\mathsf{Com}(\mathfrak g) = S(g)$. -To prove that $\phi$ is free, one observes that we're dealing with weight graded connected operads $P$, so we can prove a weight graded module $M$ is free by showing that the bar complex $B(M,P,1)$ is acyclic (i.e. concentrated in degree 0). -Since freeness is an homological statement, one can use homological methods to prove this: we filter $\mathsf{Ass}$ by the powers of the ideal generated by the Lie bracket, and note that the associated graded object that results is the Poisson operad, which is $\mathsf{Com}\circ\mathsf{Lie}$, giving the result.<|endoftext|> -TITLE: Specific Elliptic Curves: Rank -QUESTION [10 upvotes]: Here's a challenge for elliptic curve descenders/programmers. It seems no public software or public tables can determine if the rank is zero for the following curves (over rational x,y): - y^2 = x^3 - 9122*x + 106889 - y^2 = x^3 - x^2 - 42144*x + 66420 - y^2 = x^3 - x^2 - 168615*x + 21827700 - y^2 = x^3 - 210386*x + 32627329 - -Can anybody definitely say if any of their ranks are zero? -By the way, these arose from Heronian triangles for a given base and height, so there are equivalent quartic forms which might be easier to analyze...for example, the rank of the first curve above is zero iff there are no rational s,t solutions to this equation: - ( s^2 - t^2 )^2 = 25*( 2*( s^2 + t^2 ) - 509 ) - -Also, I have tried both Magma and Sage. Sage seems to be better at determining the rank in about 20% of similar cases. For example, "y^2 = x^3 + x^2 - 58055*x + 4135350" has rank 0 according to Sage, but Magma only bounds the rank from 0 to 2 (limited to one minute). Anyway, these 4 cases are unsolved. - -REPLY [21 votes]: Bob: although you say that Magma disagrees with my tables, none of your examples actually shows this. You only show that using the most obvious Magma command cannot prove that the rank is zero. As has been pointed out, one can prove that the ranks are zero by computing the analytic ranks and applying certain Theorems. Or, you just have to go beyond 2-descent. The reason why 2-descent is insufficient for these curves is that both the curve and the 2-isogenous curve have Sha of order4. But Magma can also do 3- and 4-descent! It just does not use them automatically. A 3-descent should be able to show that the rank is 0, by computing that the 3-Selmer group is trivial. I say "should" because the 3-descent algorithm (developed and implemented by me with Tom Fisher, Michael Stoll, Cathy O'Neil and Denis Simon) involves working in the 3-division field and so can be slow. (I am running ThreeDescent(E) on the first of your curves as I type). Alternatively you can get the everywhere locally soluble 2-coverings in Magma and then apply FourDescent() to those. -So it is not true that existing descent techniques are insufficient for these curves, but the tools available need to be used with some understanding of how they work<|endoftext|> -TITLE: Real integral which cannot be evaluated without complex analysis? -QUESTION [6 upvotes]: One motivation for studying contour integrals in complex analysis is that they can be used to give elegant evaluations of certain real integrals, which tend to have more direct physical and geometrical interpretations (and are therefore easier to motivate). The methods of complex analysis often seem to give a more elegant way to compute real integrals than if real analysis alone was used. However, in many cases where an integral is computable using complex analysis there also seems to exist a way to evaluate it using only real analysis (possibly in several variables). I therefore ask the following vague question (see below for an attempt to make it more precise): - -Is there a real integral for which it can be proved that it cannot be evaluated without using methods of complex analysis? - -To make the question a bit more precise, first of all assume that all the functions involved are elementary functions (over $\mathbb{Q}$, say). A real integral is meant to be an integral of a real valued function along (some, possibly unbounded, part of) the real line. For the sake of this question, let's say that 'methods of complex analysis' means anything involving Cauchy's theorem or any of its consequences such as Cauchy's Residue Theorem. It's a bit tricky to define precisely what 'evaluate' should mean, but let's say it means that the integral is assigned the value of an explicit elementary function (over $\mathbb{Q}$) at a rational number. -Unless the answer to the question is yes, perhaps there is some general way to show that the above question has a negative answer. For example, I can imagine the possibility of results in logic saying that any sentence expressing the evaluation of a real integral in some language incorporating Cauchy's theorem is equivalent to a sentence in a language including only real analysis. Is anything in this direction known? - -REPLY [5 votes]: I'm quite sure one could camuflage any complex analysis technique into a real calculus computation, leaving no appeal even to the notion of complex numbers. For instance, the Cauchy's formula on a circle may be translated into a formula for Fourier coefficients; the homotopy invariance of a path integral may be recovered by means of the Stokes theorem and so on. It is even possible that in some case this way one gets a simpler or more elementary computation, and sometimes, why not, it could be a reasonable and welcome operation to do (e.g. I'm thinking to the needs of an elementary course where a certain computation has to be done but the audience is supposed to have no or little Complex Analysis). -But, I'd say that in any case this would be by no means a piece of evidence against the utility of Complex Analysis techniques, nor would suggest that one can renounce to them. I think this is an a general issue about big theories. Big theories are important not only beacuse they provide useful tools to solve problems, but, even before, because they show us the way, like the Polar star. To make an example, you do not need the Open Mapping theorem of Functional Analysis to know that a certain concrete operator is invertible, if you are able to prove the convenient particular bounds, that is, just an inequality. But it's Functional Analysis who told you in advance that an inequality as the wanted one could be proved, and addressed you there.<|endoftext|> -TITLE: The ground state is signed and symmetric -QUESTION [12 upvotes]: Background -In Berestycki and Lions it is asserted that (on page 316), if I am not misreading, that the "ground state", i.e. action minimizer among nontrivial solutions, corresponding to the action -$$ S[u] = \int_{\mathbb{R}^d} |\nabla u|^2 + G(u) dx $$ -where $G$ satisfies certain conditions must be signed and spherically symmetric. In the paper they showed that under certain conditions on $G$ that such a ground state (that is, signed and spherically symmetric) exists; but they do not show that all grounds states must have these symmetries (by which I mean that they don't show the "action minimizing solutions" must be both single-signed and spherically symmetric). -The only reference I can find on this claim is the paper of Coleman, Glaser, Martin; but I am not entirely convinced that they have established the necessity. Their proof uses the fact that a ground state must be a minimizer of $\int |\nabla u|^2$ under the constraint that $\int G(u)$ is fixed. They then use the Polya-Szego principle: under spherical rearrangements the latter integral is unchanged, while the former can only decrease. -But the decrease is not necessarily strict. Brothers and Ziemer gave a counterexample in the case that the distribution function of $u$ is not absolutely continuous. -My question -I know how to complete the proof and get that ground states must be spherically symmetric, and monotonic radially, provided one assume that $G$ is $C^{1,1}$. This one can do by unique continuation principles for elliptic PDEs or equivalently by uniqueness theorems for ODEs. But in Berestycki-Lions or in Coleman-Glaser-Martin, $G$ is only assumed to be $C^1$, for which non-uniqueness, at least in the case of ODEs, is well known as a possibility. -So, is the "uniqueness" statement true for $G$ merely $C^1$? Are there known counterexamples? - -REPLY [3 votes]: A little bit more digging turned up this paper of Mihai Mariş. -He shows that under two technical conditions: - -Minimizers are $C^1$ (which we have from elliptic regularity) -If $u$ is an admissible function ($H^1$ in our case) and $v$ is a unit vector in $\mathbb{R}^d$, the function $\tilde{u}(x) = u(x)$ when $x\cdot v \geq 0$ and $\tilde{u}(x) = u(x - 2 x\cdot v v)$ otherwise is also admissible. - -The solution to a constrained minimisation problem with $k$ constraints must be radially symmetric about an axis of dimension $k-1$ if the functionals are translation and rotationally invariant. -A very bad description of the process is follows: -Roughly speaking, the main idea is the following: by the Borsuk-Ulam Lemma we can find inductively $\sim d-k$ mutually orthogonal hyperplanes each of which split all the constraints exactly in half. The reflection symmetry inherent in the problem implies that from each orthant we can extend to a solution that is symmetric across each of the planes, which implies that it is symmetric under $x\to -x$ inside a subspace. This however implies that any linearly dependent hyperplane of the given ones will also split all the constraints in half. -The last step is to show that using multiple reflections and a bit of geometry, one can show that the original solution much also be reflectively symmetric across each of the planes. This shows radial symmetry.<|endoftext|> -TITLE: Necessity of hypercovers for sheaf condition for simplicial sheaves -QUESTION [20 upvotes]: I'm trying to understand where the definition of simplicial sheaf on a space/site comes from. -For a presheaf $F$ of sets on a topological space $X$, the sheaf condition can be viewed as saying that $F$ is a 'local object' with respect to the maps of presheaves $colim(\coprod U_{ij} \underrightarrow{\rightarrow} \coprod U_{i}) \rightarrow X$'for every open cover $\{ U_{i} \rightarrow X \}$. That is to say, HOMing $colim(\coprod U_{ij} \underrightarrow{\rightarrow} \coprod U_{i}) \rightarrow X$' into $F$ gives an isomorphism, and that is easily seen to be the usual sheaf condition. (This seems to be a fairly well-known way to say the sheaf condition, and I learned it from the nlab.) -Now for $F$ a presheaf of simplicial sets, people require more for the sheaf condition, namely that $F$ is local with respect to $hocolim(U_{*}) \rightarrow X$', where $U_{*}$ is a so-called hypercover, and these are some sort of refinement of simplicial diagrams that can be built from multiple intersections of elements of an open cover $U_{i} \rightarrow X$. (I don't really know how to draw a simplicial diagram here. Sorry.) -My question is why in the definition of simplicial sheaf it is necessary to consider not just double intersections of the $U_{i}$, but all intersections (organized into a simplicial diagram). I've read people say, 'The condition is satisfied for spaces, so should be satisfied for general sheaves', but I don't find this a completely convincing reason, since we don't need (?) the whole simplicial diagram to reconstruct a space (at least if we take naive colimits instead of homotopy colimits). -I imagine good answers would be something like 'it is necessary for homotopy invariance of constructions' or 'here is a presheaf F that we'd like to be a sheaf, but HOMing only double intersections into it doesn't see enough information'. - -REPLY [6 votes]: This answer is not really different from David's or Marc's; I just want to boil things down. -In brief: -If you want "weak equivalence of simplicial sheaves" to mean the same thing as "induce weak equivalence on stalks", you must localize simplicial presheaves with respect to hypercovers. -You could only localize with respect to covers, but then it may be possible for there to be an object whose stalks are all weakly contractible, but which is not itself weakly equivalent to the terminal object. -Note: a "weak equivalence on stalks" amounts to the same thing as "isomorphism of homotopy-group sheaves", so I could have instead said: if you only localize with respect to covers, then it may be possible for there to be an object with trivial homotopy-group sheaves, but which is itself not weakly equivalent to the terminal object. -I highly recommend the paper of Dugger, Hollander, and Isaksen, Hypercovers and simplicial presheaves, which explains this, and gives an example exhibiting the phenomenon I just mentioned. -Added. -For "localize with respect to covers", I mean: from an open cover of an open set $X$ build a simplicial object $C_\bullet$ of presheaves, with $C_0=\coprod U_i$, $C_1=\coprod U_i\cap U_j$, etc. Then a simplicial presheaf $F$ has descent for covers if -$$F(X) \to \mathrm{holim} F(C_\bullet)$$ -is a weak equivalence of simplicial sets (for all open covers $\{U_i\}$ of open sets $X$). Note that I'm taking a homotopy limit of a cosimplicial diagram here. -If you just use the "first two steps of the diagram", you would be saying that the map $F(X)$ to the "homotopy equalizer" of $F(C_0) \rightrightarrows F(C_1)$ is a weak equivalence. This is a different condition, and is the wrong way. -Here is one way to see why it is wrong (I'll be a bit imprecise here): start with the constant presheaf whose value is an Eilenberg-MacLane space $K(G,n)$, and "localize" it to get a simplicial presheaf $F$ which has descent for covers. I want to be able to say that $F(X)$ has something to do with the cohomology of $X$. With the correct definition (full simplicial diagram), you can see that you are seeing something that looks similar to the classical definition of Cech cohomology. In fact, if $\{U_i\}$ is a "good cover", so that all finite intersections are either empty or contractible, this is what you get. If you just use the "two-step" definition, it's not clear that $F(X)$ will know anything about 3-fold intersections of open sets in the cover, so presumably can't compute Cech cohomology.<|endoftext|> -TITLE: What sort of large cardinal can $\aleph_1$ be without the axiom of choice? -QUESTION [20 upvotes]: Assuming the axiom of choice it is very easy to see that $\aleph_1$ is a regular Joe of a successor cardinal. It is not very large in any way except the fact that it is the first uncountable cardinal. -If however we begin with a model of ZFC+Inaccessible, we can construct models of ZF in which $\aleph_1$ is somewhat inaccessible in the sense that $\aleph_1\nleq 2^{\aleph_0}$ If, on the other hand, we start with a model of ZF whic has this property then there exists an inner model with an inaccessible cardinal. -It can be that $\aleph_1$ is a measurable cardinal, you can even have that every subset of $\omega_1$ contains a club, or is non-stationary; and it is possible for $\aleph_1$ to have the tree property (I only know of models by Apter in which all successor cardinals have the tree property; but that would require a proper class of very large cardinals). -In general we say that $\aleph_1$ is P-large for a large cardinal property P, if it is consistent with ZF that $\aleph_1$ has property P, and from such model we can produce a model of ZFC+$\kappa>\aleph_0$ has property P. -Question: Is there a limit on how P-large can $\aleph_1$ be? (e.g. P can be tree property/$\kappa$-complete ultrafilter/supercompact measures/etc.) and are there properties P such that for $\aleph_1$ to have them we require more than ZFC+P? - -REPLY [8 votes]: In the paper -" -The relative consistency of a "large cardinal'' property for $\omega_1$, -Rocky Mountain J. Math. 20 (1990), no. 1, 209–213." -it is shown than a model of ZFC with a huge cardinal can extend to a model of ZF in which $\omega_1$ is huge. -The link to the paper: http://projecteuclid.org/euclid.rmjm/1181073173<|endoftext|> -TITLE: Die-rolling Hamiltonian cycles -QUESTION [12 upvotes]: Let $R$ be a rectangular region of the integer lattice $\mathbb{Z}^2$, -each of whose unit squares is labeled with a number -in $\lbrace 1, 2, 3, 4, 5, 6 \rbrace$. -Say that such a labeled $R$ is die-rolling Hamiltonian, -or simply rollable, -if there is a Hamiltonian cycle obtained by rolling a unit die -cube over its edges so that, for each square $s \in R$, -the cube lands on $s$ precisely once, and when it does so, -the top face of the cube matches the number in $s$. -For example, the $4 \times 4$ "board" shown below -is rollable. -      - - - -Q. Is it true that, if $R$ is die-rolling Hamiltonian, then the -Hamiltonian cycle is unique, i.e., there are never two distinct -die-rolling Hamiltonian cycles on $R$? - -This "unique-rollability" -question arose out of a problem I posed in 2005, and was largely -solved two years later, in a paper entitled, -"On rolling cube puzzles" (complete citation below; -the $4 \times 4$ example above is from Fig. 17 of that paper). -Although the original question involved computational complexity, -the possible uniqueness of Hamiltonian cycles is independent -of those computational issues, so I thought it might be useful -to expose it to a different community, who might bring -different tools to bear. -It is known to hold for $R$ with side lengths at most 8. -If not every cell of $R$ is labeled, and unlabeled cells are forbidden -to the die, then there are examples with more than one Hamiltonian cycle. -Edit1. Rolling a regular tetrahedron on the equilateral triangular (hexagonal) lattice -is not as interesting. See the Trigg article cited below. -Edit2. -Serendipitously, gordon-royle posted a perhaps(?) relevantly related -question: -"Uniquely Hamiltonian graphs with minimum degree 4." - - - The computational version is - -Open Problem 68 at -The Open Problems Project. - - -"On rolling cube puzzles." -Buchin, Buchin, Demaine, Demaine, El-Khechen, Fekete, Knauer, Schulz, Taslakian. -Proceedings of the 19th Canadian Conference on Computational Geometry, Pages 141–144, 2007. -PDF download. - - -Charles W. Trigg. "Tetrahedron rolled onto a plane." J. Recreational Mathematics, 3(2):82–87, 1970. - -REPLY [3 votes]: While browsing the recreational-mathematics tag I noticed this old question again. -In 2012 I published on my blog a draft paper in which I give another example of a (bigger) board with two distinct Hamiltonian cycles (but domotorp was quicker to answer). -In the paper I also prove that the Rolling Cube Puzzle in labeled boards without free cells and with blocked cells is NP-complete (an open problem). -http://www.nearly42.org/cstheory/rolling-a-cube-can-be-tricky/ -... I never had the time to clean it up and submit it to a journal for a serious review (though I tested the gadgets with Mathematica). -If someone wants to give it a look ...<|endoftext|> -TITLE: "Archimedeanising" an ordered field -QUESTION [6 upvotes]: If $K$ is an ordered field, let $B$ be the subring comprising the $x \in K$ such that $|x| \le n$ for some $n \in N$, and let $I$ be the ideal of $B$ comprising the infinitesimal elements (i.e. the $x \in K$ such that $|x| \le 1/n$ for every non-zero $n \in N$). Then $I$ is a maximal ideal and the order on $K$ induces an order on $A = B/I$ making it into an archimedean ordered field. -Has this construction been studied? More specifically, when does the natural projection of $B$ onto $A$ split (in the category of rings or, better still, ordered rings)? I believe it always has an order-preserving splitting if $K$ is real closed. - -REPLY [5 votes]: Yes, the map is sometimes called the standard part map, and it turns out that one map define it in general for a group in an o-minimal structure. See here, for example. Regarding the splitting, if it should only respect the additive group structure, I think it exists since everything is a vector space over $\mathbb{Q}$ (as explained in the comment)<|endoftext|> -TITLE: explanation on a scheme which is not affine scheme -QUESTION [13 upvotes]: Hartshorne at the end of page 76 of his Algebraic Geometry book gives an example of a scheme which is not an affine scheme. The scheme is constructed by gluing two affine lines together along their maximal ideals obtained by removing a point P. There's also a figure accompanying the example: -___________:_________ -Can someone please explain how to show that this is not an affine scheme? - -REPLY [30 votes]: Call $X $ your scheme over the field $k$, $P_1$ and $P_2$ the two special closed points, $A_1$and $A_2$ their respective open complements and $A_{12}=A_1\cap A_2$, so that $A_i\simeq \mathbb A^1_k$ and $A_{12}\simeq\mathbb G_m$, all affine schemes. -Here are some (not independent) proofs that $X$ is not affine. -Proof 1 -The point $(P_1,P_2)\in X \times X $ is in the closure of the diagonal $\Delta_X\subset X \times X $, but $(P_1,P_2)\notin \Delta_X$ . So $\Delta_X$ is not closed, hence $X$ is not separated and a fortiori not affine -Proof 2 -The images of the restriction map $\Gamma(A_i,\mathcal O_X)=k[T] \to \Gamma(A_{12},\mathcal O_X)=k[T,T^{-1}]$ are both -$k[T]$, and together do not generate $ k[T,T^{-1}]$. However, in an affine scheme (or more generally in a separated scheme) the ring of regular sections on the intersection of two open affines is generated by the images of the regular sections on the two opens. -Proof 3 -The two open immersions $\iota_j:\mathbb A^1_k \to X$ with respective image $A_j\subset X$ coincide on the open subscheme $\mathbb G_m\subset \mathbb A^1_k$ but are nevertheless distinct. This couldn't happen if $X$ were affine (or just separated). -Proof 4 -The cohomology vector space $H^1(X,\mathcal O_X)$ is infinite dimensional, whereas the cohomology of a coherent sheaf on an affine scheme vanishes in positive degree. -In detail, consider the covering $\mathcal U=\lbrace A_1,A_2\rbrace$ of $X$. It is a Leray covering because $A_1,A_2,A_{12}$ are affine hence acyclic, for the coherent sheaf $\mathcal O_X$ (cf. Proof 2) . Thus Čech cohomology computes genuine cohomology. -The Čech complex is the linear map $$C^0=\Gamma(A_1,\mathcal O_X)\times \Gamma(A_2,\mathcal O_X)=k[T]\times k[T]\stackrel {d^0}{\to} C^1=\Gamma(A_{12},\mathcal O^*_X)=k[T,T^{-1}]\to 0$$ -given by $$d^0(P(T),Q(T)) =Q(T)-P(T) $$. -Hence we get $H^1(X,\mathcal O_X)=k[T,T^{-1}]/k[T]$ -Proof 5 -The Čech complex above proves that $\Gamma(X,\mathcal O_X)=k[T]$ so that the restriction to the strictly smaller open affine subscheme $A_1\subsetneq X$ is bijective: $res: \Gamma(X,\mathcal O_X)\stackrel {\simeq}{\to} \Gamma(A_1,\mathcal O_X)$. -This cannot happen for an affine scheme $X$. -[In categorical language: $\Gamma$ is an anti-equivalence from the category of affine schemes to that of rings] -Proof 6 -Every global function $P(T)\in \Gamma(X,\mathcal O_X)=k[T]$ (see Proof 5) takes the exact same value at $P_1$ and $P_2$, namely $P(0)\in \kappa(P_1)=\kappa (P_2)=k$. -In contrast given two closed points in an affine scheme , there exists a global regular function vanishing at the first one but not at the second.<|endoftext|> -TITLE: What is the correct statement of this Erdös-Gallai-Tuza problem generalizing Turan's triangle theorem? -QUESTION [8 upvotes]: In Zsolt Tuza's Unsolved combinatorial problems I, Problem 46 is the following conjecture: -Let $G$ be a graph on $n$ vertices. Let $\alpha_1$ be the maximum number of edges of $G$ such that every triangle in $G$ contains at most one of these edges. Let $\tau_1$ be the minimum number of edges of $G$ such that every triangle in $G$ contains at least one of these edges. Then, it is conjectured that $\alpha_1+\tau_1 \leq \dfrac{n^2}{4}$. -The reference given is Paul Erdös, Tibor Gallai, Zsolt Tuza, Covering and independence in triangle structures, Discrete Mathematics 150 (1996) pp. 89-101. However, in this reference, the conjecture is only formulated (Conjecture 11) for the case when $G$ is a triangular graph (i. e., every edge of $G$ is contained in a triangle). Maybe I am missing something completely obvious, but is this really equivalent, or did Tuza extend his conjecture after this publication? Or is one of the statements wrong resp. mistakenly specialized? (Note that the conjecture for generic $G$ clearly generalizes Turan's theorem for $K_3$'s, but in the triangular case I don't see how it yields Turan.) -(Also, are there any news on this question?) - -REPLY [9 votes]: I hope I am not violating any kind of MathOverflow etiquette by responding to a question 2 years after it has been asked, but your question is answered in the following preprint of mine that just hit the arXiv: http://arxiv.org/abs/1408.5176 -The preprint shows that any vertex-minimal counterexample to the Erdős--Gallai--Tuza Conjecture has minimum degree greater than n/2; in particular, such graphs are triangular. Thus, the "triangular" version and the unrestricted version are equivalent. -The proof is not difficult, but (at the risk of flattering myself) I don't think it is entirely trivial either.<|endoftext|> -TITLE: Is there a fusion rule in positive characteristic? -QUESTION [5 upvotes]: Verlinde's fusion gives a certain "tensor product" of representations of loop groups. The category of representations of loop groups has (essentially equivalent) two incarnations. One is analytic, based on $LG=\operatorname{Map}(S^1,G)$, and one algebraic, based on $L\mathfrak g=\mathfrak g[t,t^{-1}]$. The latter of these makes sense in positive characteristic. In both cases, one constructs the "fusion" of two positive energy representations of the loop group via holomorphic induction on a thrice punctured sphere (in the analytic model, this is a disc in $\mathbb C$ minus two interior discs, and in the algebraic model, this is $\mathbb P^1(\mathbb C)$ minus three points). - -Can one define a fusion product like this in positive characteristic? - -I have done some searches online, but haven't even managed to figure out whether there is a reasonable category (corresponding the category of representations of positive energy in characteristic zero) of representations of positive characteristic loop groups where we could expect such a fusion product to exist. - -REPLY [8 votes]: To answer the question in the header, there is certainly a relevant fusion rule in positive characteristic. This arises in a purely representation-theoretic context in the work of various people (Olivier Mathieu and Henning Andersen in particular). But along the way relationships have to be built among a number of representation categories in order to arrive at a transparent version of fusion rules. I'm not sure about online access, but two useful papers in Comm. Math. Physics from the early 1990s are: -H.H. Andersen, "Tensor products of quantized tilting modules" (1992) -H.H. Andersen and J. Paradowski, "Fusion categories arising from semisimple Lie -algebras" (1995) -These papers arise indirectly from the influential Verlinde paper in Nuclear Physics B. Loop algebras or affine Lie algebras have representation theory in negative levels shown by Kazhdan and Lusztig to share many features with the theory for quantum groups at a root of unity based on the same type of Lie algebra. (In turn, this quantum group theory transfers in a subtle way to modular Lie algebra settings in prime characteristic.) An essential shared ingredient is the organizing role of an affine Weyl group. -In order to get a rigorous mathematical framework for "truncated" tensor products appearing in the fusion rules of Verlinde, use is made of "tilting" modules which include for small highest weights the classical-looking "Weyl modules" whose tensor products are reasonably well understood. But the tilting modules are usually more complicated, having finite filtrations with quotients which are Weyl modules (and similar filtrations involving dual Weyl modules). A key fact is that the category of tilting modules is closed under tensoring. This is the refined setting in which it makes sense to tensor, break up into a direct sum of indecomposables, and then truncate, allowing only a finite number of indecomposable objects to survive: others have "quantum dimension zero" and disappear from the picture. Technically it gets fairly complicated, but the underlying ideas are transparent from the viewpoint of representation theory. -The "tilting" modules themselves continue to be studied, but roughly speaking their formal characters are predicted by Kazhdan-Lusztig theory in the settings I indicated. (The unsolved problems are mostly in prime characteristic, where a lot is known but not everything.) -After the progress made in the early 1990s there were only a few attempts at surveys of the work for mathematicians, including one by Andersen Tilting modules for algebraic groups (1995). In any case there is substantial literature out there, which I can comment further on if it's useful.<|endoftext|> -TITLE: Reference request: Simple facts about vector-valued Sobolev space -QUESTION [12 upvotes]: Let $V,H$ be separable Hilbert spaces such that there are dense injections $V \hookrightarrow H \hookrightarrow V^*$. (For example, $H = L^2(\mathbb{R}^n)$, $V = H^1(\mathbb{R}^n)$, $V^* = H^{-1}(\mathbb{R}^n)$.) We can then define the vector-valued Sobolev space $W^{1,2}([0,1]; V, V^*)$ of functions $u \in L^2([0,1]; V)$ which have one weak derivative $u' \in L^2([0,1], V^*)$. Such spaces arise often in the study of PDE involving time. -I would like a reference for some simple facts about $W^{1,2}$. For example: - -Basic calculus, like integration by parts, etc. -The "Sobolev embedding" result $W^{1,2} \subset C([0,1]; H)$; -The "product rule" $\frac{d}{dt} \|u(t)\|_{H^2} = (u'(t), u(t))_{V^*, V}$ -$C^\infty([0,1]; V)$ is dense in $W^{1,2}$. - -These are pretty easy to prove, but they should be standard and I don't want to waste space in a paper with proofs. -Some of these results, in the special case where $V$ is Sobolev space, are in L. C. Evans, Partial Differential Equations, section 5.9, but I'd rather not cite special cases. Also, in current editions of this book, there's a small but significant gap in one of the proofs (it is addressed briefly in the latest errata). So I'd prefer something else. -Thanks! - -REPLY [6 votes]: Herbert Ammann's book on parabolic problems contains an excellent introduction.<|endoftext|> -TITLE: 4-dim compact positively curved manifolds with a nontrivial Killing vector field. -QUESTION [7 upvotes]: Kleiner-Hsiang (JDG 1989) proved such a manifold is homeomorphic to $S^4$ or $CP^{2}$. an interesting corrollary is that $S^2 \times S^2$ does not admit positively curved metric with countinuous symmetry. They asked in their paper if it is diffeomorphic. I don't know much on this area, only noticed one subsequent work Searle-Yang. Seems that there are preprints on arXiv attempting to do the problem in full generality. Not sure if it was fully settled. Anyone knows the precise status of this problem? - -REPLY [5 votes]: The status of the problem is that it is solved: the classification up to (equivariant) diffeomorphism was obtained very recently by K. Grove and B. Wilking, see Thm A here. The proof uses the solution of the Poincare' Conjecture and some fancy machinery of Alexandrov geometry.<|endoftext|> -TITLE: definition of Hessian with respect to connection -QUESTION [5 upvotes]: Hi, -I am reading the lecture notes on Morse Homology written by M.Hutchings, in that notes definition of Hessian is defined in coordinate free way: given any connection $ H(f,p)= \nabla_v(df)$ where $v$ is the tangent vector at critical point $p$, and $df$ is differential of $f$. I need to show this definition does not depend on choice of connection. Hutchings says that the difference of two connection is a tensor and $df$ vanishes at p, so the above fact holds. I can not understand the meaning of "the difference of two connection is a tensor" and how this observation solves my problem. -Thanks, - -REPLY [7 votes]: There is no real need for a connection in this situation. -The underlying fact is that if $s$ is a section of a vector bundle $E$ over $X$, and $s$ vanishes at a point $x_0$, then it has an intrinsic derivative $Ds(x_0):T_{x_0}X\to E_{x_0}$, defined by -$Ds(x_0)\cdot v=\lim_{t\to 0} s(x_t)/t$, where $t\mapsto x_t$, $t\in]-1,1[$ is any curve in $X$ with velocity $v$ at $t=0$. -In your case $E$ is $T^*X$ and $s=df$.<|endoftext|> -TITLE: Uniquely hamiltonian graphs with minimum degree 4 -QUESTION [11 upvotes]: A graph is uniquely hamiltonian if it has exactly one Hamilton cycle. -As every edge in a cubic graph lies in an even number of Hamilton cycles, a cubic graph cannot be uniquely hamiltonian, and a famous conjecture, called Sheehan's conjecture asserts that a 4-regular graph can also not be uniquely hamiltonian. -Apparently, however, there are uniquely hamiltonian graphs with minimum degree equal to four - the latest edition of Bondy & Murty's Graph Theory even gives a reference to a paper by H. Fleischner entitled "Uniquely hamiltonian graphs of minimum degree 4", To Appear, Journal of Graph Theory and dates it at 2007. -But I cannot find this paper on MathSciNet and nor can I find any paper that appears to contains these graphs. Googling reveals a few references to this work, including a 2003 conference/workshop where the abstract claims that there are some uniquely hamiltonian Eulerian graphs with minimum degree four. -Does anyone know what these graphs are? Or where this paper is? Or anything at all? -(I have emailed Fleischner but had no reply yet, though I may still get one as the email was recent.) -EDIT: Due to some bizarre synchronicity, I received a reply from Herbert Fleischner literally 5 seconds after I finished writing this question; the paper exists and has been accepted by JGT but final revisions have not been made. - -REPLY [5 votes]: Here is the abstract of a talk that Herbert Fleischner gave at CombinaTexas: Combinatorics in the South-Central U.S., available at this link, in 2003. -The title of his talk was, "Uniquely Hamiltonian Graphs": - -Abstract: In the 1970s, J.Sheehan asked whether there are 4-regular simple graphs having precisely one Hamiltonian cycle (= uniquely Hamiltonian graphs). In the early 1990s, 4-regular loopless uniquely Hamiltonian multigraphs were constructed whose underlying uniquely Hamiltonian (simple) graphs have 3- and 4-valent vertices only. C.Thomassen showed that regular graphs of sufficiently high degree cannot be uniquely Hamiltonian; and J.A.Bondy, in his article for the Handbook of Combinatorics, asked whether uniquely Hamiltonian graphs must have a vertex of degree 2 or 3. Starting from the examples quoted above, one can construct Eulerian uniquely Hamiltonian graphs of minimum degree 4. - -See also The Open Problems Garden, "Uniquely Hamiltonian graphs," which highlights the -conjecture: - -If $G$ is a finite $r$-regular graph, where $r > 2$, then $G$ is not uniquely Hamiltonian.<|endoftext|> -TITLE: Basic examples of induction on scales arguments -QUESTION [9 upvotes]: An important ingredient in recent progress on Euclidean harmonic analysis has been that of "inductions on scales". A few examples are the papers of Wolff, Tao, and Bourgain and Guth. -Here is a (vague) description of what I mean by an induction on scales argument. You have a monotonically increasing quantitative $Q(r)$ which you wish to prove is $O(1)$. One then cleverly decomposes the problem at "scale" $r$ into smaller pieces at scale $r/K$ to obtain an upper bound on $Q(r)$ by quantities of the form $Q(r/K)$ where $K$ is a large parameter. Assuming one can get good enough control in the upper bound, say something like $Q(r) \ll C + K^{-1} Q(r/K) $, then it follows that $Q(r) = O(1)$. (In practice these arguments have a more complicated structure involving various additional parameters and dependencies.) Typically $Q(r)$ is an estimate over a ball of radius $r$, and $Q(r/k)$ is a rescaled version of the estimate over a ball of radius $r/k$ (these arguments tend to exploit symmetries). -These arguments always seem somewhat magical to me, and I would like to build a better intuition about them. However, the examples (such as those referenced above) that I am aware of are intertwined with many other inputs and clever ideas. Therefor I am interested to find more basic examples where the induction on scales technique (or something similar) is useful. - -Give (or point to) an example of a more basic/classical result that can be proved using induction on scale arguments. - -REPLY [2 votes]: Looking at a related question on the side bar, it appears that Greg Kuperberg's answer to this MO question has a very simple version of induction on scales to show that the volume of the unit n-ball tends to 0 as n increases.<|endoftext|> -TITLE: A programming language that can only create algorithms with polynomial runtime? -QUESTION [29 upvotes]: Has someone constructed a programming language that can construct all the algorithms in P, and no others? -I'm interested in this restriction coming from the syntax naturally, as opposed to just being a normal Turing machine with a step-timer attached. - -REPLY [3 votes]: Yet another perspective (and IMHO a more natural one) is descriptive complexity theory (check also this Wikipedia article). -They study the question from a perspective different from the one mentioned by Neel. There are various languages that capture exactly the polynomial time computable functions. One of the most famous ones is FO+LFP.<|endoftext|> -TITLE: Origin of the notation s=\sigma+it in analytic number theory -QUESTION [7 upvotes]: I was wondering if the standard notation of denoting a complex variable by "$s$" had an interesting origin, or if it dates back to Riemann or Weierstrass. Almost every book in analytic number theory seems to uses that alphabet with -$ s = \sigma + i t$ -denoting its real and imaginary parts. -I shall be happy if anyone could enlighten me about it. I tried searching MO for relevant questions but couldn't find it. - -REPLY [5 votes]: In skimming through Narkiewicz "The Development of Prime Number Theory", one sees a reference on p. 155 (footnote 38) to a R. Lipschitz, who in Crelle volume 54 in 1857 "studied the series $\sum_{n=1}^\infty\exp(nui)n^{-\sigma}$ for real values of $\sigma$." I checked the reference, which is here (there are two papers by Lipschitz in this volume of Crelle, and it's the second one that's relevant); Lipschitz was indeed using $\sigma$. -Lipschitz is referred to several times in this section of Narkiewicz for later work on functional equations of various $L$-functions.<|endoftext|> -TITLE: Smart elliptic curve rational point search given Reg*#Sha -QUESTION [8 upvotes]: Hi folks, -Let E be a global minimal model of an elliptic curve over QQ, with a -2-torsion point which generates the torsion subgroup, and with -Mordell-Weil rank 1 (under BSD). Let RegSha be equal to -Regulator(E)*#Sha(E) which has been computed using the BSD formula. -(Assume computing R the non-torsion rational point generator of the -Mordell-Weil group E(QQ) using descent by 2-isogeny is time -consuming.) -Does the RegSha information make it "easier" to compute R? If so -kindly supply search strategy. -Regards, -Ifti - -REPLY [11 votes]: Since I cannot (yet) post comments, I'll post this as an answer, even though it is more a comment on Joe Silverman's answer. -Let $H$ be the bound for the multiplicative height (so $H = e^h$ if $h$ is a bound for the logarithmic height) of the point you want to find. Then (as is pointed out in Joe's paper) a naive search on the elliptic curve has a complexity of roughly $H^{3/2}$, whereas his algorithm gives you $H$. On the other hand, searching for the point on a 2-covering can be done in time essentially $H^{1/2}$ (the logarithmic height goes down by a factor of 4), and if I remember correctly, this is what mwrank does. Using $n$-coverings for arbitrary $n$, the complexity drops to $H^{1/((n-1)n)}$, but of course the implied constant grows (there are roughly $n^r$ covering curves to consider when the rank is $r$, and the complexity of the lattice computations also goes up, since one has to use rank-$n$ lattices). In addition, of course, you have to add the time you need to compute the $n$-coverings in the first place, which for $n = p$ a prime usually involves computing class and unit groups of fields of degree about $p^2$. For composite $n$, the complexity tends to be better, though. -The exponent $1/((n-1)n)$ comes from two ingredients: the first is that the logarithmic height of the preimage of your point on the covering curve goes down by a factor of $1/(2n)$. The second is the lattice-based point search method mentioned by Noam Elkies in his comment that gives you a complexity of $B^{2/(n-1)}$ for a search for points up to multiplicative height $B$ on a smooth curve in ${\mathbb P}^{n-1}$. Picking a good value of $n$ should lead to something subexponential; I assume this is what Noam was having in mind (his ANTS IV paper describes a version of the lattice-based point search). -The point I want to make is that descent and searching on coverings usually gives you a faster way of finding points than Joe's method, even without knowledge of the exact canonical height (the height is still useful since it gives you a bound for the search). Tom Fisher has demonstrated that this can be very effective already for relatively small values of $n$ like 6 or 12.<|endoftext|> -TITLE: Which vector bundle are the Christoffel symbols sections of? -QUESTION [5 upvotes]: The collection of Christoffel symbols $\Gamma_{ij}^k$ of a connection (or of a metric) on a smooth manifold $M$ is not the collection of components of a tensor field in some local chart, i.e. they don't correspond to a section of a tensor bundle over $M$. - -Is there a vector bundle, naturally associated to $M$, of which the collection of Christoffel symbols represents a section? - -REPLY [8 votes]: Connections on a vector bundle $E\to M$ are sections of an affine bundle associated to $E$. -Namely there is a vector bundle $J^1E$ of $1$-jets of sections of $E$, and an exact sequence of bundles -$$0\to T^*M\otimes E\to J^1E \overset{p}\to E\to 0$$ -where the map $p$ is the evaluation ("$0$-jet"). -Then a connection is a section of the affine bundle of sections (sic) of $p$, namely the $s\in Hom(E,J^1E)$ such that $p\circ s=id_E$. The associated vector bundle is $Hom(E,T^*M\otimes E)\simeq T^*M\otimes End(E)$, where one can view the Christoffel symbols (if $E=TM$) as living : once (local) a trivialisation is chosen there is an associated "trivial" connexion, and any other connection differs from it by a section of this vector bundle.<|endoftext|> -TITLE: Can elliptic integral singular values generate cubic polynomials with integer coefficients? -QUESTION [9 upvotes]: For the elliptic integral of first kind, $K(m)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-m^2sin^2\theta}} $, it is well-known that $K(m)$ can be expressed in what Chowla and Selberg call "finite terms" (i.e. algebraic numbers and a finite product of Gamma functions of rational values) whenever $i\frac{K(\sqrt{1-m^2})}{K(m)}$ belongs to an imaginary quadratic field $\mathbb Q(\sqrt{d})$ -(see Theorem 7 in S. Chowla and A. Selberg, On Epstein's zeta function, J. reine angew. Math. 227, 86-110, 196). -Examples for these so called elliptic integral singular values are given on this Wolfram page (with some small typos) and in the note of J.M. Borwein and I.J. Zucker, "Elliptic integral evaluation of the Gamma function at -rational values of small denominator," IMA Journal on Numerical Analysis, 12 (1992), 519- -526. -See also what Tito Piezas has to say about this in his pleasant-to-read Collection of Algebraic Identities. -The following question arises: - -For these singular values, is there (always, or, if not always: when?) a polynomial $P(t)$ of degree 3 with integer coefficients such that $K(m)=c\int\limits_{t_0}^\infty\dfrac{dt}{\sqrt{P(t)}} $ with $c\in\mathbb Q$? -(EDIT: After Noam Elkies' remark, introduced $t_0$, the biggest real zero of $P$, instead of $0$ as the lower limit. Only "complete" integrals make sense here.) - -In particular for $d=-7$, we have by the Carlson symmetric form $$\frac12\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t+1)(t+\frac{8+3\sqrt 7}{16})}} =K(k_7)=\dfrac1{7^{1/4}4\pi}\Gamma\left(\dfrac17\right)\Gamma\left(\dfrac27\right)\Gamma\left(\dfrac47\right),$$ on the other hand I have seen somewhere (I can't remember the reference) $$\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+21t+112)}} =\dfrac1{4\pi\sqrt{7}}\Gamma\left(\dfrac17\right)\Gamma\left(\dfrac27\right)\Gamma\left(\dfrac47\right)=\frac{K(k_7)}{7^{1/4}}.$$ -I would like to get it straight at least for this example: - -Can the polynomial in the first integral be transformed into one with integer coefficients? And is there any sort of relationship between both above polynomials? -Note that the ratio of the discriminants of the two above polynomials is $-2^{24}\cdot7^3$, and both of them do not yield affirmative answers, as the second one would have to be divided by $\sqrt7$ to obtain $K(k_7)$ directly! -EDIT: Follow-up question: If $P(t)$ is an integer cubic polynomial such that $\int\limits_{t_0}^\infty\dfrac{dt}{\sqrt{P(t)}} $ (with $t_0$ its biggest real zero) can be written in "finite terms", is this value always an algebraic multiple of an elliptic integral singular value $K(m)$? - -REPLY [11 votes]: The flurry of comments did not yet produce an answer to the question -concerning the complete elliptic integrals -$$ -I_1 := \frac12 -\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t+1)(t+\frac{8+3\sqrt{7}}{16})}} -$$ -and -$$ -I_2 := \int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+21t+112)}}. -$$ -It turns out that (i) Yes, $I_1$ can be transformed to a complete -elliptic integral associated to a cubic with integer coefficients, and -(ii) The identity $I_1 = 7^{1/4} I_2$ can then be recovered via -a form of -Landen's transformation. -This could be surmised by calculating the $j$-invariants of the -corresponding elliptic curves -$$ -y^2 = x(x+1)(x+\frac{8+3\sqrt{7}}{16}) -\phantom{\cong}\text{and}\phantom{\cong} -y^2 = x(x^2+21x+112). -$$ -The first $j$-invariant is $j_1 = 16581375 = 255^3$, which is rational, -so there's a linear change of variable that transforms -$x(x+1)(x+\frac{8+3\sqrt{7}}{16})$ to a polynomial with integer coefficients. -The second $j$-invariant is $j_2 = -3375 = -15^3 \neq j_1$, so we can't -get immediately from $j_1$ to $j_2$. But $j_1$ and $j_2$ are still -related by a $2$-isogeny [indeed $j_1 = j(\sqrt{-7})$ and -$j_2 = j((1+\sqrt{-7})/2)$], so $I_1$ and $I_2$ are related by -a Landen transformation. -For (i), first translate $t$ by $(8+3\sqrt{7})/16$ to get -$$ -I_1 := \frac12 -\int\limits_{\frac{8+3\sqrt{7}}{16}}^\infty - \dfrac{dt}{\sqrt{t - \bigl(t-\frac{8+3\sqrt{7}}{16}\bigr) - \bigl(t-\frac{-8+3\sqrt{7}}{16}\bigr) - }} \phantom{0}. -$$ -Then observe that -$$ - \bigl(t-\frac{8+3\sqrt{7}}{16}\bigr) - \bigl(t-\frac{-8+3\sqrt{7}}{16}\bigr) -= t^2 - \frac{3\sqrt{7}}{8} + \frac{3^2 7 - 8^2}{16^2} -= t^2 - \frac{6\sqrt{7}}{16} - \frac1{16^2}. -$$ -Thus the change of variable $x = 16 \sqrt{7} \cdot t$ yields -$$ -I_1 = 2 \cdot 7^{1/4} -\int\limits_{21+8\sqrt{7}}^\infty - \dfrac{dx}{\sqrt{x (x^2-42x-7)}} -$$ -with the integrand having integer coefficients. -For (ii), we apply Landen's change of variable for elliptic integrals -$\int dt/\sqrt{t(t^2+at+b)}$ with $a,b>0$: if $x = (t^2+at+b)/t$ then -$dt/\sqrt{t(t^2+at+b)} = dx/\sqrt{x(x^2+Ax+B)}$ where $(A,B) = (-2a, a^2-4b)$. -Moreover the map $t \mapsto (t^2+at+b)/t$ maps the interval $0 < t < \infty$ -to $a + 2\sqrt{b} \leq x < \infty$, with each $x$ value arising twice -except for the left endpoint $x = a+2\sqrt{b}$ which has a single preimage -$t = \sqrt{b}$ of multiplicity $2$. Therefore -$$ -\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+at+b)}} = -2 \int\limits_{a+2\sqrt{b}}^\infty\dfrac{dx}{\sqrt{x(x^2-AX-B)}}. -$$ -The $I_2$ coefficients $(a,b) = (21,112)$ yield $(A,B) = (-42,-7)$ and -$a + 2 \sqrt{b} = 21 + 8 \sqrt{7}$. Therefore $I_1 = 7^{1/4} I_2$ as desired.<|endoftext|> -TITLE: Parity of class number of pure cubic fields -QUESTION [6 upvotes]: A pure cubic field is an algebraic number field of the form $K = \mathbb{Q}(\theta)$ with $\theta^3 = m$, $m \neq \pm 1$. -What can be said about the parity (odd or even) of the class number of a pure cubic field? In particular, is there some theorem (or simple algorithm) that tells you when the class number will be odd? -I have found this paper http://www.rzuser.uni-heidelberg.de/~hb3/publ/pcub2.pdf of Lemmermeyer, but it doesn't seem to have what I'm looking for. -I'm also interested in finding some good references with information relating to my questions or on the theory of cubic fields. - -REPLY [8 votes]: There is an algorithm for computing the 2-class number of pure cubic fields using $2$-Selmer groups of elliptic curves due to G. Frey and his diploma students Eisenbeis and Ommerborn: Computation of the 2-rank of pure cubic fields, Math. Comput. 32 (1978), 559-569. -This was generalized by U. Schneiders in Estimating the 2-rank of cubic fields by Selmer groups of elliptic curves, J. Number Theory 62 (1997), No.2, 375-396; the proof of the upper bound on the 2-class number in this article is, however, incorrect. -This being said, the 2-class number of pure cubics is as mysterious as the 3-class number of quadratic number fields. In particular, there is no genus theory for the 2-part of the class number.<|endoftext|> -TITLE: A simplicial complex which is not collapsible, but whose barycentric subdivision is -QUESTION [19 upvotes]: Does anyone know of a simplicial complex which is not collapsible but whose barycentric subdivision is? -Every collapsible complex is necessarily contractible, and subdivision preserves the topological structure, so we are certainly looking for a complex which is contractible, but not collapsible. The only complexes I know of which are contractible but not collapsible are the dunce cap and Bing's house with two rooms. Neither of these have any free faces, and so no iterated subdivision will result in a collapsible complex. - -REPLY [10 votes]: @Andy: there is now a small explicit example available, with 15 vertices and 55 tetrahedra, in case you're still interested: it's the 3-ball called $B_{15,66}$ in here. (Sorry for the huge delay and the shameless self-promotion.)<|endoftext|> -TITLE: Do smooth ind schemes have Dualizing sheafs? -QUESTION [11 upvotes]: Say I have an ind scheme $X = \cup_i X_i$ over a field $k$. I have its tangent bundle $\hom_k(k[\epsilon], X)$ which I can think of as ind scheme via $\cup_i \hom_k(k[\epsilon],X_i)$. The problem is even if $X$ is smooth it might be the case that most of the $X_i$ are not smooth. I believe this happens at least for polynomial loop groups $G[z^\pm]$. In this case the sheaf of differentials is not locally free. This seems to be an obstruction to constructing the canonical sheaf inductively. -Additionally if each $X_i$ is infinite dimensional, which happens for the formal loop group, then it seems like top exterior power doesn't make much sense. And finally if you looked at say $\mathbb{P}^\infty := \cup_n \mathbb{P}^n$ then it also seems unclear what a canonical sheaf should be. If it were a line bundle it could be described as a line bundle $L_n$ on each $\mathbb{P}^n$ which are compatible under pull backs. But then each $L_n$ would have the same degree $d$. But the canonical line bundles $O(-n-1)$ have a different degree on each $\mathbb{P}^n$! -So is there any sense in asking for something like a canonical sheaf or dualizing sheaf for smooth ind schemes? -UPDATE: Brian Conrad shared the following with me: -If $f:X \to Y$ is a map between finite type schemes over a field (or one can be much more general...) then for a relative dualizing complex $\omega_Y$ on $Y$ we have that $f^!(\omega_Y)$ is a relative dualizing complex on $X$ (for suitable functor $f^!$ at derived category level). In other words, one does have "compatibility" for relative dualizing complexes, but with respect to the appropriate "derived pullback" operation $f^!$. (One has to think about duality and "canonical sheaf" in a much broader sense than Serre duality over a field in order to define "relative dualizing object" in a derived category.) -The upshot is that one has to work in derived categories (and so develop a suitable formalism of direct/inverse limits in derived categories) to make a good theory of duality on ind-schemes. - -REPLY [3 votes]: By judicious use of Kan extension, one can define a functorial !-pullback, and hence a dualizing complex, for a broad class of geometric objects, including arbitrary DG prestacks (Gaitsgory section 10.1), and in particular smooth ind-schemes as you ask. -This is used in Theorem 10.1.1 of Gaitsgory-Rozenblyum, where tensoring with the dualizing complex yields an equivalence between quasicoherent sheaves and ind-coherent sheaves in the case of formally smooth DG ind-schemes that are weakly $\aleph_0$ and locally almost of finite type. In this case, the dualizing complex is an ind-coherent sheaf that presumably looks something like a shifted line bundle, but I couldn't extract that information directly from the paper. Serre duality for DG ind-schemes locally almost of finite type also appears as Corollary 2.6.2 of the same paper, but its statement doesn't seem to use the dualizing complex directly.<|endoftext|> -TITLE: Mathematics and cancer research -QUESTION [27 upvotes]: What are applications of mathematics in cancer research? -Unfortunately, I heard quite little about applications of mathematics, but I heard something about applications of physics, and let me put this story here, as it might be useful to be aware of it. Radiation therapy is well-known, but less well-known is proton therapy, which is much more rare and based on proton accelerators used in particle physics. - -The chief advantage of proton therapy is the ability to more precisely localize the radiation dosage when compared with other types of external beam radiotherapy. - -As far as I know, it is only practiced in physics research centers that have proton accelerators. It is highly useful for cancer of sensitive tissues when it is dangerous to use other radiation therapy because it will destroy everything surrounding the malignancy. -A colleague of mine told me this really helped his father with eye cancer. -Moreover, proton therapy is so rare that the leading cancer experts were not aware of it and he become aware of it only through friends. - -REPLY [2 votes]: Ricci flow is an analytic surgery. Prediction and control of cancer invasion Can be done by Ricci flow theory see this paper -Ricci Flow and Entropy Model for Avascular Tumor Growth and Decay Control -Tijana T. Ivancevic -https://arxiv.org/abs/0806.0691 -We can apply discrete Ricci flow like Ollivier Ricci flow, to study cancer and tumors -https://www.nature.com/articles/srep12323 -There are a list of discrete Ricci curvatures which based on we can define Ricci flow. Just we need to know which discrete Ricci curvature on the graph network is suitable for the tumor.<|endoftext|> -TITLE: Does every compact Hausdorff ring admit a decomposition into primitive idempotents? -QUESTION [5 upvotes]: Let $\mathbf{R} = (R,\mathcal{T},+,\cdot,0,1)$ be a compact Hausdorff topological unitary ring, and consider the set $I(\mathbf{R}) := \{ e \in R \mid e \cdot e = e \}$ (of idempotents in $\mathbf{R}$). In the sequel, two idempotents $e,f \in I(\mathbf{R})$ are said to be orthogonal if $e \cdot f = f \cdot e = 0$, and an idempotent $e \in I(\mathbf{R})$ is called primitive if it is nonzero and cannot be written as a sum of two orthogonal nonzero idempotents. Now, my questions are the following: -1.) Does there have to exist a primitive idempotent in $\mathbf{R}$? -2.) Does $\mathbf{R}$ necessarily admit a set $E \subseteq I(\mathbf{R})$ of pairwise orthogonal, primitive idempotents such that $1 = \sum_{e \in E} e$ (that is, the family $(e)_{e \in E}$ is summable in $\mathbf{R}$ and the corresponding limit is $1$)? -If so, can you outline a proof or at least give a suitable reference? Otherwise, do you know any counterexamples? Are the statements true for the ring compactification of $(\mathbb{Z},\mathfrak{P}(\mathbb{Z}),+,\cdot,0,1)$? -Since all the rings I am dealing with are commutative, I am particularly interested in the commutative case. Personally, I expect a negative answer even in the commutative case, but that is just a feeling. Hence, I would be most enthuasiastic if somebody could provide me with a counterexample for the commutative case. -Remark: As $(e)_{e \in E}$ is a family of pairwise orthogonal idempotents, it is summable, anyway. [see "Topological Rings Satisfying Compactness Conditions", page 139, by Mihail Ursul] Hence, in the second question, we only need to require that the limit value is $1$. - -REPLY [2 votes]: I have just found an answer - at least for the commutative case - and it is YES. -Proof of 2.): Since $\mathbf{R}$ is a compact Hausdorff ring, it is profinite due to a surprising theorem in ["Profinite Groups" by Ribes and Zalesskii]. Another theorem in this book states that every profinite commutative ring is isomorphic to a product of profinite local rings. Thus, we can assume that $\mathbf{R} = \prod_{i \in I} \mathbf{R_i}$ for a family of profinite local rings $\mathbf{R_i}$ $(i \in I)$. For each $i \in I$, let $e_{i}$ denote the element of $\mathbf{R}$ given by $e_{i}(j) = \delta_{ij}$ for $j \in J$. As each of the $\mathbf{R_i}$'s is local, $E := \{ e_{i} \mid i \in I \}$ is a set of primitive idempotents in $\mathbf{R}$. Evidently, any two distinct elements of $E$ are orthogonal, and we have $1 = \sum_{i \in I} e_{i}$. This completes the proof.<|endoftext|> -TITLE: Oriented double normals -QUESTION [7 upvotes]: Given an embedded two-torus in three-dimensional Euclidean space, paint the inside of the torus red and the outside blue. Show that there is an oriented line in ${\mathbb R}^3$ that cuts the torus perpendicularly in (at least) two points at which it crosses from red to blue. -This is true (I'll say why in a minute), but I'd like to know if there is a simple proof using standard critical point theory. -Here is a proof that works even if the torus is immersed: -Consider the space of geodesics of ${\mathbb R}^3$ which is well-known to be symplectomorphic to the cotangent of the two-sphere. The congruence of oriented lines normal to the immersed torus is an immersed exact Lagrangian manifold in the space of geodesics. The oriented double normal we're looking for is just a double (multiple) point for this immersed Lagrangian. In other words, we would like to know that this immersion cannot be an embedding. The result now follows from a theorem of Claude Viterbo (JDG 47 (1997) 120-168). -Theorem (Viterbo). There is no exact Lagrangian embedding of the two-torus in the cotangent space of the two sphere. -This theorem and the arguments I gave before settle the problem not only in Euclidean space, but also in hyperbolic space, three-dimensional Hadamard manifolds, three-dimensional normed spaces with smooth, quadratically-convex spheres and, in general, it works for any three-dimensional Finsler manifold whose space of geodesics is symplectomorphic to the cotangent of the two-sphere. -Isn't there some simpler argument that works in ${\mathbb R}^n$ $(n > 3)$ and yields something like: if an compact oriented manifold is immersed as a hypersurface in ${\mathbb R}^n$, then it either admits an oriented double normal or it is homeomorphic/diffeomorphic to a sphere ? -Relation to the standard double-normal problem. In the standard (non-oriented) double-normal problem, everything reduces to considering the critical points of the distance-squared function defined on the symmetric product of the immersed manifold with itself. What -I can't see is whether there is some minimax procedure that constructs critical values (and points) that correspond to oriented double normals. - -REPLY [4 votes]: I suggest to look at my paper - P. E. Pushkar', “Generalization of the Chekanov Theorem. Diameters of Immersed Manifolds and Wave Fronts” Local and global problems of singularity theory, Collection of papers dedicated to the 60th anniversary of academician Vladimir -Igorevich Arnold, Tr. Mat. Inst. Steklova, 221, Nauka, Moscow, 1998, 289–304 -You can find it here - http://wenku.baidu.com/view/0514da4d852458fb770b56c6.html?from=related -Diameters are double normals! -In particular, there is an estimate in the paper - number of double normals of generic immersed submanifold $M^n$ of the Euclidean space is at least $(B^2-B)/2+nB/2$. Here $B$ is $\dim H_*(M,Z_2)$. This estimation is exact for product of spheres, oriented surfaces. -There is a proof (by Maxim Kazaryan) of my estimate for the case of embeddings, which use only square-function - http://www.mi.ras.ru/~kazarian/papers/homology05.pdf in the section 16. Unfortunately it is in Russian.<|endoftext|> -TITLE: Restriction of representation from GL(n) to O(n) -QUESTION [5 upvotes]: So my question is somewhat similar to Restriction from $\mathfrak{gl}_{2n}$ to $\mathfrak{sp}_{2n}$; but I was having difficulty understanding the formula given in reference (Harris & Fulton) mentioned there. -Equation (25.37) in Harris & Fulton's "Representation Theory: A first course", says that if $m=2n$ or $m=2n+1$, $\lambda = (\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_m)$ is a $GL_m$ highest weight, and $\bar{\lambda}=(\bar{\lambda_1} \geq \cdots \geq \bar{\lambda}_n)$ is an $O_m$ highest weight, then the multiplicity of the irreducible highest-weight $O_m$ representation $V_{\bar{\lambda}}$ in the restriction of the irreducible highest-weight $GL_m$ representation $V_{\lambda}$ is equal to $\sum_{\eta} N_{\eta, \bar{\lambda}, \lambda}$; where the sum is over all partitions $\eta$ with all parts even, and $N_{\lambda_1, \lambda_2, \lambda_3}$ is the Littlewood-Richardson coefficient. I had some trouble getting this formula to work for $m=2$: -Q1: Since $O_m$ is disconnected, an irreducible $O_m$ representation is given by (a) if $m=2n+1$, a highest weight ${\lambda}=(\bar{\lambda_1} \geq \cdots \geq \bar{\lambda}_n)$ and a choice of $+$ or $-$ (b) if $m=2n$, a highest weight ${\lambda}=(\bar{\lambda_1} \geq \cdots \geq \bar{\lambda}_n)$ and, if $\bar{\lambda}_n=0$, a choice of $+$ or $-$. [See pg 53 of http://dspace.mit.edu/handle/1721.1/8642?show=fullfor more details.] In the above formula, the book seems to be considering $\mathfrak{o}_m$ representations instead, so how should the choices of $+$ or $-$ that arise be incorporated into the above formula? -Q2: Now take $m=2$. First let $\lambda=(1,1)$. If $\bar{\lambda} \neq 0$, then clearly $\sum_{\eta} N_{\eta, \bar{\lambda}, \lambda}=0$; if $\bar{\lambda}=0$, then $\sum_{\eta} N_{\eta, \bar{\lambda}, \lambda}=0$ also (since $N_{2,0,(1,1)}=0$). So this seems to be saying that the restriction of $V_{(1,1)}$ to $O(2)$ has no constituents, which is clearly false. The same problem occurs for $\lambda=(2k+1, 2k+1)$. Another example that was troubling me is $\lambda=(6,4), \bar{\lambda}=(4)$: then $N_{6, 4, (6,4)}=N_{(4,2), 4, (6,4)}=1$, so this says that the multiplicity of $V_{4}$ in the restriction of $V_{(6,4)}$ is $2$, which is also false. - -REPLY [4 votes]: For the exterior powers (highest weight $(1,1,\dots,1)$), the restriction to the orthogonal group is still irreducible -- I think you are forgetting to take $\overline{\lambda} = 0$. -There are also issues with stable ranges in this formula (you have to assume that $\lambda$ doesn't have more than $n$ parts). If you want the formulas to work in all cases, then you have to use modification rules for transforming partitions of length more than $n$ into those with length at most $n$ (sometimes you have to introduce signs to cancel terms). Details of this modification rule can be found in several sources. One such that I like is Koike--Terada's paper: http://www.sciencedirect.com/science/article/pii/0021869387900998 (Section 2.4)<|endoftext|> -TITLE: Topological vector spaces that are isomorphic to their duals -QUESTION [6 upvotes]: After reviewing the (locally convex) -topological vector spaces that I know, -the only examples I could find where there is an isomorphism from the -space to its (anti)dual, are Hilbert spaces. -So my question is : -Are there topological vector spaces $V$ such that the topology does not come from a Hilbert structure, and such that there exists an isomorphism $\chi : V \to V'$, where $V'$ denotes the antidual of $V$ (continuous antilinear forms on $V$) ? - -REPLY [7 votes]: An interesting family of examples comes from number theory (or algebraic geometry, depending on who you ask): If you have a field $k$, the Laurent power series field $k((t))$ has an ultrametric topology where $\{ t^n k[[t]] \}_{n \in \mathbb{Z}}$ form a neighborhood basis of zero. This space is isomorphic to its topological dual: by adjoining $(dt)^{1/2}$, one obtains the perfect residue pairing $$\langle f(t) (dt)^{1/2}, g(t) (dt)^{1/2} \rangle = \operatorname{Res} f(t)g(t) dt.$$ You can do a similar trick with finite dimensional vector spaces over $k((t))$. -If you want nontrivial antilinearity, you may choose $k$ to be a separable quadratic extension of some underfield $F$, and change the residue pairing to be sesquilinear: $$\langle f(t) (dt)^{1/2}, g(t) (dt)^{1/2} \rangle = \operatorname{Res} f(t) \bar{g}(t) dt.$$<|endoftext|> -TITLE: Solving PDE via Cellular Automata -QUESTION [8 upvotes]: Is there a theory for solving PDE by using Cellular Automata ? Something which is on the line of, passing to the limit (scale) i.e., if you increase the number of grid points the solution to the cellular automata will converge to the PDE ? If so, how successful is this approach ? What are the limitations of this approach. Also, given a PDE how does one go about finding the rules for the corresponding cellular automata and vice versa ? - -REPLY [5 votes]: There are a number of PDEs that have been fruitfully attacked using cellular automata. First among these are various incarnations of the Navier-Stokes equations, which can be (and in geophysical or other complex flow applications, frequently are) simulated with lattice gases and lattice Boltzmann methods. Other PDEs that CAs can handle include diffusion and reaction-diffusion equations and wave equations. Another one that is more widely known is the random walk treated as a random CA, which can be used to tackle the heat equation. -Will Jagy's guess about hexagons anticipates (if we think about triangles instead) the improvement that so-called FHP models offer over HPP models; in higher dimensions the lattice issues get trickier. I have some unpublished and cryptic notes about a possible new approach in 3D using the root lattice $A_4$ and permutohedral boundary conditions. -One of the main advantages of the CA approach is the ability to work with complicated boundary shapes, though on the other hand the boundary conditions are a very delicate issue in general. -A very nice (though somewhat dated) reference for this and related problems is Chopard and Droz (a PDF of the opening parts is here) and IIRC you can find a paper by one of the authors online that covers some of the same topics with a similar approach.<|endoftext|> -TITLE: What does the t in t-category stand for? -QUESTION [7 upvotes]: To my knowledge the notion of a t-category was first introduced Beilinson, Bernstein and Deligne's Faiseaux Pervers. But while they explain the name "perverse sheaf", they don't give any indication how they came up with the name t-category. -Does anyone know whether the "t" in "t-category"/"t-structure" stands for something specific? - -REPLY [9 votes]: Though I don't have inside information, it's clear that the notes in Asterisque 100 (1982) by BBD gave the first formal definition of t-category and t-structure on a triangulated category (section 1.3). Since the nearest t-word involved is "triangulated" and no direct rationale is offered for the language introduced, that probably suggested the terminology here. Of course, someone else may have a more subtle theory. -P.S. Reviewing the original text further, I see in their Example 1.3.2(1) the -suggestive word tronque, which converts me to Donu's viewpoint. It would have simplified things for the reader of BBD just to start with a definition of categorie tronque (abbreviated t-category).<|endoftext|> -TITLE: Do real vectors attain matrix norms? -QUESTION [7 upvotes]: I apologize if the following question ends up being too elementary for this website; I asked it on math.SE a week ago and it remains unanswered. -Let $A$ be an $n \times n$ matrix with real entries and let $p \geq 1$. I'm wondering if -$$ \max_{ x \in \mathbb{C}^N, \|x\|_p = 1} \|Ax\|_p$$ is the same as -$$ \max_{ x \in \mathbb{R}^N, \|x\|_p = 1} \|Ax\|_p.$$ -The only difference is the replacement of $\mathbb{C}^n$ by $\mathbb{R}^n$. Certainly, the answer is yes if $p=1,2,\infty$; on the other, it is pointed out in this answer that the answer is no for mixed $(p,q)$-norms. -Edit, Will: I can't get this stupid thing to work. On MSE, Robert Israel introduced a $p,q$ mixed matrix norm with suitable notation, where the vectors $x$ are measured with the $p$-norm, but the $Ax$ are measured with the $q$-norm. - -REPLY [7 votes]: Okay, so basically the answer can be found in here: http://arxiv.org/pdf/math/0512608v1.pdf -Here's how the argument works (a simplified version of what is done in the paper with finite dimensions and $p=q$): -(Note: we define "$\Re$" of a vector by taking the real part componentswise) -Lemma 3.4 says (applied to the finite dimensional situation) -$$ - \int_0^{2\pi} \| \Re(e^{i\varphi} x) \|_p^p d\varphi = \int_0^{2\pi} |\cos(\varphi)|^p d\varphi -$$ -for any $x\in \mathbb C^n$ with $\|x\|_p=1$. This is fairly elementary to verify. -Therefore, whenever $x,y\in \mathbb C^n$ both have norm $1$, we will find a $\varphi\in[0,2\pi]$ such that $$\|\Re(e^{i\varphi}x)\|_p \leq \|\Re(e^{i\varphi}y)\|_p$$ -since the integral -$$ - \int_0^{2\pi}\|\Re(e^{i\varphi}y)\|_p^p - \|\Re(e^{i\varphi}x)\|_p^p d\varphi = 0 -$$ -is zero und thus the integrand has to be non-negative somewhere. -Then Lemma 3.2 of the paper yields the result. What the authors do here is to take a vector -$0\neq x\in \mathbb C^n$ such that $\|Ax\|_p/\|x\|_p$ is maximal (assume also that $A\neq 0$; then $Ax\neq 0$ follows automatically and we can divide by its norm below). Then they take a $\varphi$ such that -$$ - \left\|\Re(e^{i\varphi} \frac{x}{\|x\|_p})\right\|_p \leq \left\| \Re(e^{i\varphi} \frac{Ax}{\|Ax\|_p})\right\|_p -$$ -which is possible by the above. If $\Re(e^{i\varphi}x)\neq 0$, this can then be rewritten as -$$ - \frac{\|Ax\|_p}{\|x\|_p} \leq \frac{\|A \Re(e^{i\varphi}x)\|_p}{\|\Re(e^{i\varphi}x)\|_p} -$$ -which shows that the maximum is also attained at the real vector $\Re(e^{i\varphi}x)\in \mathbb R^n$. -If $\Re(e^{i\varphi}x)=0$, then $i\cdot e^{i\varphi}x$ is a real vector at which the maximum is attained.<|endoftext|> -TITLE: Fraktur symbols for Lie algebras -QUESTION [32 upvotes]: Does anyone know when and why the Fraktur script was introduced for Lie and other algebras—$\mathfrak{g}$, $\mathfrak{gl}_n$, $X/\mathfrak{g}$, -$\mathfrak{g}\oplus\mathfrak{g}$, $\mathfrak{su}$, $\mathfrak{M}_g$, etc.? -And introduced by whom? -Is its use pretty much restricted to algebra, or was it in the past employed in, say, geometry as well, but has only survived to the current time within algebra? -(Or maybe it is currently used outside of algebra and I am just ignorant of those areas.) -The typeface itself goes back to the 15th century. -The generally illuminating website "Earliest Uses of Various Mathematical Symbols" seems not to shed light on this issue. -I find the Fraktur font adds a certain elegance and mystery to the subjects that utilize it! -I'm a bit envious, not working in those fields... —$\mathfrak{Joseph}$ :-) - -REPLY [3 votes]: Let me point out a counter-example, in which Fraktur is not used for Lie-algebras. -It's Непрерывные группы (literally "Continuous Groups") by Lev Pontryagin, published in 1946 in the USSR. (Sorry, in fact, I have a Japanese translation from the second Russian edition in 1954. It is "連続群論" in 1957. See Worldcat entry for the bibliography.) It uses Fraktur for classical Lie groups(!), and Roman for their Lie algebras. -See the middle of p. 521, the left photo. The word "リー群" means "Lie group(s)", as you can see from the Japanese Wikipedia page. At the bottom of the same page, it says "$\mathfrak{H}_r$ のリー代数 $H_r$" ($\mathfrak{H}_r$'s Lie algebra $H_r$). See also p. 545, the right photo. You can see the Dynkin diagrams of the classical and exceptional Lie algebras. -     - -Click to enlarge. These photo citations must be ok under the copyright law. -(I originally posted this as a MSE question. I think these photo citations are allowed under the copyright )<|endoftext|> -TITLE: schur weyl duality for real orthogonal groups and relation to hyperoctahedral groups -QUESTION [6 upvotes]: I am wondering whether the Lie groups $SO(n)$ and the hyperoctahedral groups $H_n$ form some sort of duality. I am mainly interested in how to parametrize the conjugacy classes of $H_n$ in terms of the dominant weights of root systems of type $B_n$ or $C_n$. So far I haven't found an accessible reference that draws complete analogy with the case of $U(n)$ and $S_n$. I am also interested in the analogue of symmetric functions in the case of $H_n$. Any pointer to literature would be greatly appreciated. - -REPLY [6 votes]: The $O(n)$ version of Schur-Weyl duality involves Brauer algebras, the structure of which was not worked out completely until the 1980s by Hans Wenzl (Ann. of Math. (2) 128 (1988), no. 1, 173–193.) So perhaps you are looking for $H_n$ as a subgroup of the Brauer algebras? Googling "Brauer algebra" and "Hyperoctahedral group" gives: this recent preprint for example.<|endoftext|> -TITLE: construct the elliptic fibration of elliptic k3 surface -QUESTION [20 upvotes]: Hi all, -As we know, every elliptic k3 surface admits an elliptic fibration over $P^1$, but generally how do we construct this fibration? For example, how to get such a fibration for Fermat quartic? -Moreover, as we know all (elliptic) k3 surfaces are differential equivalent to each other, does this mean: topologically the elliptic fibration we get for each elliptic fibraion is the same, which is just the torus fibration over $S^2$ with 24 node singularities? Or, the totally space is the same, but different complex data(structure) provides different way or "direction" of projection onto $S^2$, thus induces different type of fibrations? -Thanks! - -REPLY [7 votes]: Here is an alternative way to construct elliptic fibrations of quartic surfaces containing a line. -Let $X\subset \mathbb P^3$ be a smooth quartic surface and assume that it contains a line $L\subset X$. Let $\mathfrak d$ denote the $1$-dimensional linear system of hyperplanes in $\mathbb P^3$ containing $L$ restricted to $X$. Further let $H\in \mathfrak d$ and $E=H-L$. -Observe that $E^2=H^2-2H\cdot L L^2= 4 - 2\times 1 -2=0$ and that by construction $\dim |E|\geq 1$. It follows that for any $E_1,E_2\in |E|$, $E_1\cap E_2=\emptyset$ and hence $|E|$ is basepoint-free and $\dim |E|=1$. -Therefore $|E|$ defines a morphism $f: X\to \mathbb P^1$ and it follows from the construction that the fibers of $f$ are cubic plane curves, so this is indeed an elliptic fibration. - -Remark 1: One can get more data out of this construction. - -Since each fiber is a plane cubic, this construction limits the possible singularities of the fibers quite a bit. -The line we started with is a triple section. -If $X$ contains another line, say $C$, such that $C\cap L=\emptyset$ then $C$ gives a section of $f$: Indeed if $C$ is a line, then $C\cdot H=1$ and if $C\cap L=\emptyset$, then $C\cdot E=1$. -If $X$ contains another line, say $C$, such that $C\cap L\neq\emptyset$, then $C$ is contained in a fiber intersecting $L$ exactly once. -You can keep playing with this to get more. - -Remark 2: -For an elliptic K3 the Picard number is at least $2$ and if it is exactly $2$, then it admits at most two different elliptic fibrations. If the Picard number is at least $3$, then an elliptic K3 may admit infinitely many elliptic fibrations, so obviously in many cases you don't get all the elliptic fibrations this way. -Remark 3: -Finally, note that the Fermat quartic contains lines: Using diverietti's notation the 16 lines $$\ell_{\xi,\xi'}=\big\{[a:\xi a:b:\xi' b] \ \vert\ [a:b]\in \mathbb P^1\big\}$$ -where $\xi,\xi'$ are 4$^\text{th}$ roots of unity all lie on that Fermat quartic, so one obtains 16 different elliptic fibrations this way. -Further note that if $\xi\neq \xi''$ and $\xi'\neq \xi'''$, then $\ell_{\xi,\xi'}\cap \ell_{\xi'',\xi'''}=\emptyset$, so this way we get a couple of sections for each of the elliptic fibrations as described above.<|endoftext|> -TITLE: Application for Morse-Smale systems -QUESTION [14 upvotes]: All my life I do topological classification of Morse-Smale systems (flows and cascades). Today, fundamental science is not in fashion, to receive grants require an application. Can you please tell where to look. - -REPLY [5 votes]: I know this is an older question, but since it may interest some, here are some papers about Morse-Smale theory applications: - -This paper describes applications of Morse-Smale theory to sensor networks. It has great pictures, and describes how Morse-Smale theory allows one to avoid slow points in computer computations, among other benefits. -This paper describes applications of Morse-Smale theory to video segmentation. In part, it says, - - -In computer vision, detail has been often identi- fied with scale, and so-called scale-space approaches have achieved simplification by blurring. However, blurring obliterates small detail together with the boundaries of large regions. A more discriminating treatment of scale would instead eliminate small detail while keeping large parts crisply delineated. - In this paper, we propose a representation of video that captures structure and affords flexible control of detail, without sacrificing crispness. Specifically, we represent structure through the so-called Morse- Smale complex of a scalar function f(x,y,t) associated with the video data, and we obtain a hierarchy of increasingly simple complexes through the process of topological simplification. - - -This thesis also describes how Morse-Smale theory can be used to simplify data storage. - -All of these papers have references to other papers doing similar things. It seems like Morse-Smale theory has very practical applications to real-world issues! -The main ideas in these papers seem to be that having transverse stable and unstable manifolds gives a nice cell structure or helps us find saddle points more easily.<|endoftext|> -TITLE: Blow-up for the quasilinear heat equation $u_t= u \ u_{x x}$ or the related $w_t= \left(w_x e^w\right)_x$ -QUESTION [5 upvotes]: What kind of approaches can be used to study the following quasilinear parabolic pde -for a scalar function $u=u(x,t)$ ? -$$ -u_t= u \ u_{x x} -$$ -The physical problem where this pde comes from dictates that the Cauchy problem -of interest corresponds to an initial condition which is a third-order polynomial with no-constant term -$$ -u(x,0) = x \ (x^2-s x + p), \ \ \ x \in \mathbb{R}^+, \ \ \ s p \neq 0 -$$ -Note that $u(x,0)$ is initially $\propto x$, and that $u(x,t)$ will remain so during the evolution as long as the pde makes sense. -Indeed, and this is the crux of the matter, numerical experiments and heuristic arguments strongly indicate that there exists a blow-up time $0< T=T(s,p) < \infty$, where the solution explodes to infinity, with -$$ -u(x,t) \sim \frac{f(x)}{T-t}, \ \ \ t \to T^- -$$ - -Can one prove this, and relate $T$ and the amplitude $f(x)$ to the parameters $s,p$ entering the initial condition ? -It is possible (and how !) to continue the solution after the blow-up time ? -A colleague of mine suggested that I could exchange the role of the dependent and independent variables, and/or to look at a paper by Clarkson, Fokas, \& Ablowitz, "Hodograph transformations of linearizable partial differential equations", where they study similar equations and relate them to the Harry-Dym equation and "solvable" (by inverse scattering transform or Painlevé reductions) variants. -Needless to say, the functional form of the terms in my pde do not fall under the conditions of applicability of Clarkson-fokas-Ablowitz' main theorem, while the hodograph transformation: -$$ -(t,x,u(x,t)) \longrightarrow (t,y=u,x=v(y,t)) -$$ -so that (using $u_x = 1/v_y$, $u_{x x}=-v_y^{-3}v_{y y}$, $u_t= -v_t/v_y$) -$$ -v_t + y (v_y^{-1})_y =0 -$$ -does not seem to give a pde much easier to solve (or to recognize, such as the porous medium equation) - -Edit: an alternative perhaps preferable form (closer to the porous medium equation!) -is for the function $w=\frac{1}{2}{\rm Log}(u^2)$, which obeys -$$ -w_t = \left(w_x e^w\right)_x -$$ -The Lie-point symmetries of the latter are recalled in this review (section 5), but none would match the initial condition I have to deal with. - -REPLY [4 votes]: By dividing with $u$, you may formally write the equation as -$$\tag 1 -\partial_t \log{u} = \Delta u. -$$ -For the log-diffusion equation -$$ \tag 2 -\partial_t u = \Delta \log u -$$ -the trick is to write -$$ -\log u = \lim_{m \to 0} \frac{u^m-1}{m} -$$ -and then study the corresponding porous medium equation. The issue is of course the convergence of the equation, but here is a recent preprint concerning such issues for equation (2): -http://www.vanderbilt.edu/math/people/liao/sm_files/m_limit_112311.pdf -Now, a short while ago I actually discussed with some of the authors of the above paper whether similar methods could be used for equations of type (1). I gathered that there is more or less no literature for this equation (although you might still want to search a bit). However, something similar might still work and it is an reasonable research question, I think. However, this probably won't help you too much. -In any case, substituting $v=u^m/m$ into the corresponding approximative equation -$$ \tag 3 -\partial_t \frac{u^m-1}{m}= \Delta u -$$ -yields -$$ -\partial_t v = m^{1/m} \Delta v^{1/m}, \quad m \in (0,1) -$$ -which is a porous medium equation for $v$. Here the problem is that you would like to take the limit $m \to 0$ and of course the convergence issues need to be studied. But at least a lot is known for the porous medium equation so maybe there is some hope to get something out of it. -Just out of curiosity, what might be the applications for such equations? -edit: Let me just add, that there is also some literature for equation (3) directly. In particular, see this paper and the references therein: -http://arxiv.org/pdf/1208.0621v2.pdf<|endoftext|> -TITLE: Computing homotopy (co)limits in a nice simplicial model category? -QUESTION [10 upvotes]: I'm trying to learn about and compute homotopy (co)limits. Specifically, if $\mathcal{C}$ is some Grothendieck site and $\mathcal{P}$ the simplicial model category of simplicial presheaves (say with the model structure in which weak equivalences and cofibrations are defined level-wise), then I would like to understand how to compute the homotopy colimit of the simplicial diagram determined by a covering $\{ U_{i} \rightarrow X \}$ in $\mathcal{C}$. -In the answer to my previous question, David Carchedi gave a conceptual/$\infty$-categorical explanation of why this homotopy colimit gets identified with the sieve determined by the covering. It would also be nice to see an explicit, model-categorical computation of this homotopy colimit. But I'm almost overwhelmed by the general theory of homotopy colimits in model categories and don't see how to apply the general theory to this example. If I understand correctly, I could use the projective model structure on diagram categories, a cofibrant replacement of the Cech diagram, and then compute the colimit of that. But I don't know what a cofibrant replacement of the Cech diagram looks like, and besides, there is maybe a better way to do this, perhaps using that we are working in a nice simplicial model category, not just any old model category. - -REPLY [13 votes]: In practice, one tends to "compute" arbitrary homotopy colimits as bar constructions, especially when you have a simplicial model category. -If $X:J\to P$ is a simplicially enriched functor, where $J$ is small, then you get a "bar construction" $B=B(*,J,X)$. This is a simplicial object in $P$, with -$$B_0 = \coprod_{j_0\in \mathrm{ob}J} X(j_0),$$ -$$B_1=\coprod_{j_0,j_1\in \mathrm{ob}J} X(j_0)\times J(j_0,j_1),$$ -$$B_2=\coprod_{j_0,j_1,j_2\in \mathrm{ob}J} X(j_0)\times J(j_0,j_1)\times J(j_1,j_2),$$ -etc. -(Here "$\times$" really means the simplicial "$\otimes$"; if $P$ is simplicial sets, then it really is $\times$.) If $X$ is suitably cofibrant, then the realization $|B|$ of $B$ will be the homotopy colimit of $X$. -This bar construction I described above is really a special case of "use the projective model structure"; you can use a bar construction to an explicit construction of a projective cofibrant resolution of $X$ (typically under some hypothesis on $X$, such as that each $X(j)$ is cofibrant in $P$). In fact, -$$|B(*,J,X)| = \mathrm{colim}_J |j\mapsto B(J(-,j),J,X)|,$$ -and there is a weak equivalence $|B(J,J,X)|\to X$, which is a true projective cofibrant approximation given some mild hypothesis on $X$. -The standard references are oldies but goodies: Segal's paper "Classifying spaces and spectral sequences," IHES 1968, and the "yellow monster": Bousfield & Kan, "Homotopy Limits, Completions, and Localizations," LNM 304. -Added: -When $J=\Delta^{\mathrm{op}}$, you can say something easier: the homotopy colimit of $X: J\to P$ is computed by the realization $|X| \in P$ (again, up to the cofibrancy of the objects $X(j)$). I don't know an explicit reference offhand, though everybody uses this fact; it may be in the two that I cited.<|endoftext|> -TITLE: Independent evidence for the classification of topological 4-manifolds? -QUESTION [101 upvotes]: Is there any evidence for the classification of topological 4-manifolds, aside from Freedman's 1982 paper "The topology of four-dimensional manifolds", Journal of Differential Geometry 17(3) 357–453? The argument there is extraordinarily complicated and a simpler proof would be desirable. -Is there evidence from any other source that would suggest that topological 4-manifolds are so much simpler than smooth 4-manifolds, or does it all hinge on Freedman's proof that Casson handles are homeomorphic to standard handles? -My question is motivated from a number of points of view: - -The classification of topological 4-manifolds is now 30 years old and an easier version of the proof has not emerged. In contrast, Donaldson's invariants have been superseded by more easily computed invariants. This is a very unsatisfactory state of affairs for such a far-reaching topological result, particularly as it is so regularly used in proof-by-contradiction arguments against results in smooth 4-manifold theory. -As the Bing topologists familiar with these arguments retire, the hopes of reproducing the details of the proof are fading, and with it, the insight that such a spectacular proof affords. I am delighted to see that the MPIM, Bonn is running a special semester on this topic next year. Hopefully this will introduce these techniques to a new generation of mathematicians (and save them from having to reinvent them!) -It may be possible to refine the proof to gain more control over the resulting infinite towers - and perhaps get Hoelder maps rather than homeomorphisms, for example. This would require either a better exposition of the fundamental result or some new independent insight, which was the basis of my question. - -REPLY [66 votes]: The answer to this question might have changed since it was first asked nine years ago: a book is now available whose goal it is to give a detailed elaboration on Freedman's work: -The Disc Embedding Theorem, ed. Behrens, Kalmar, Kim, Powell, Ray (Oxford University Press, 2021). -Some excerpts from the Preface: - -We choose to follow the proof from [FQ90], using gropes, which differs in many respects from Freedman's original proof using Casson towers [Fre82a]. The infinite construction using gropes, which we call a skyscraper, simplifies several key steps of the proof, and the known extensions of the theory to the non-simply connected case rely on this approach. … - -We briefly indicate, for the experts, the salient differences between the proof given in this book and that given in [FQ90]. First, there is a slight change in the definition of towers (and therefore of skyscrapers). … - -Additionally, the statement of the disc embedding theorem in [FQ90] asserts that immersed discs, under certain conditions including the existence of framed, algebraically transverse spheres, may be replaced by flat embedded discs with the same boundary and geometrically transverse spheres. The proofs given in [Fre82a, FQ90] produce the embedded discs but not the geometrically transverse spheres. We remedy this omission by modifying the start of the proof given in [FQ90], as in [PRT20]. … Besides these points, the proof of the disc embedding theorem given in this book only differs from that in [FQ90] in the increased amount of detail and number of illustrations. - -In Section 1.5 they mention some things that are not covered in the book. In addition to bypassing the part of Freedman's original proof that "consisted of embedding uncountably many compactified Casson handles within the original Casson handle and then applying techniques of decomposition space theory and Kirby calculus," they say: - -Note that the ambient manifold is required to be smooth in the statement of the disc embedding theorem. There exists a category preserving version of the theorem, where ‘immersed’ discs in a topological manifold are promoted to embedded ones. However, the proof requires the notion of topological transversality and smoothing away from a point (see Section 1.6). These facts, established by Quinn, in turn depend upon the disc embedding theorem in a smooth 4-manifold stated above. The fully topological version of the disc embedding theorem is beyond the scope of this book, since we will not discuss Quinn's proof of transversality.<|endoftext|> -TITLE: Algorithmic Borel finiteness for hyperbolic manifolds -QUESTION [11 upvotes]: It is a theorem of Borel that there is a finite number of arithmetic hyperbolic manifolds of volume bounded above by $V.$ Is there any algorithm (or hope of an algorithm) to actually construct all of them, given $V?$ Also, if the number of such manifolds is $f(V)$ are there any bounds on $f?$ - -REPLY [9 votes]: There should be an algorithm in principle. There's a couple of approaches. Given the estimates in the paper Counting arithmetic lattices and surfaces one can bound from above the degree of the invariant trace field of an arithmetic hyperbolic 3-manifold with volume $\leq V$. This in turn leads to a lower bound on the injectivity radius $\epsilon(V)$ ( it is conjectured that there is a universal lower bound to the injectivity radius of closed arithmetic 3-manifolds; this is true if one restricts to arithmetic manifolds defined over a number field of bounded degree by Lemma 4.9 of Arithmetic of Hyperbolic Manifolds and the fact that the Mahler measure of an integral polynomial of bounded degree is bounded). Now construct all manifolds of volume $ \leq V$ with injectivity radius $\geq \epsilon(V)$. All arithmetic manifolds of volume $\leq V$ will appear among this list. One may perform this construction by bounding the number of tetrahedra in a triangulation (see e.g. Breslin's Thick triangulations of hyperbolic n-manifolds), then gluing tetrahedra together in all possible ways, and computing whether they are arithmetic e.g. via Snap. -Another approach would be to construct all quaternion algebras over number fields of bounded degree with the appropriate ramification data coming from Borel's volume formula, and maximal orders in the quaternion algebras. Then compute presentations of the groups of units of the orders by applying Riley's algorithm to find a fundamental domain, then compute all finite index subgroups of bounded order by finding permutation representations.<|endoftext|> -TITLE: heller functor of finite order -QUESTION [5 upvotes]: Hi, -let A be a finite dimensional selfinjective algebra. -Assume mod A is periodic.That is for every finite dimensional module with no projective summand we have: $ \Omega^{n} (M) =M $ for some n ,where $ \Omega $ is the Hellerfunctor(giving the kernel of a projective cover of M ). -When is it true that $ \Omega $ is of finite order,that is $ \Omega^{n} $ is isomorphic as a functor to the identity functor of the stable category of modules ?(Assume there is a common biggest period for all module if necassary) -Thanks for help - -REPLY [5 votes]: Another example: If $A$ is a bialgebra over a field $k$ then $\Omega^n(k) = k$ implies $\Omega^n = id$. -The proof is rather simple: Choose a projective resolution $P \to k$ such that $\Omega^i(k)$ is the kernel of $P_{i-1} \to P_{i-2}$. Since $M$ is $k$-projective, -$$P \otimes_k M\to k \otimes_k M \cong M$$ is a projective resolution of $M$ where $A$ acts diagonally on the tensor product (that's the place were the bialgebra structure is needed). If $f: M \to N$ is $A$-linear, then we obtain a commutative diagramm with exact rows -$$\begin{array}{cccccccc} -0 & \to & M & \cong & k \otimes M & \to & P_{n-1} \otimes M & \to P_{n-2} \otimes M & \to \cdots \newline - & & \scriptstyle f\; \displaystyle \downarrow & & \scriptstyle id \otimes f\;\displaystyle\downarrow & & \scriptstyle id \otimes f\; \displaystyle\downarrow & \scriptstyle id \otimes f\; \displaystyle \downarrow\newline -0 & \to & N & \cong & k \otimes N & \to & P_{n-1} \otimes N & \to P_{n-2} \otimes N & \to \cdots \newline -\end{array}$$ -Since up to a projective summand $\Omega^n(M)$ is just the kernel in a projective resolution, we have $$M = \Omega^n(M) \oplus(\text{projective}) \text{ and }f = \Omega^n(f) \oplus p,$$ -i.e. $\Omega^n(M) = M$ and $\Omega^n(f) = f$ in the stable category. Thus $\Omega^n = id$. -Note: If $k$ is a commutative ring and $A$ projective as $k$-module then the same proof shows $\Omega^n = id$ on the stable category of finitely generated, $k$-projective $A$-modules.<|endoftext|> -TITLE: Is there any expression for the Feigenbaum constants ? -QUESTION [12 upvotes]: It has puzzled me for a long time that the Feigenbaum constant $\delta$ and reduction parameter $\alpha$ do not seem to be related to other constants (that is, numerically), even not to each other. In fact I have never seen them expressed as an integral, any kind of series or product, a nested expression, ... The only thing I have found on the internet is this algorithmic approach which further links to this, but it seems rather like an "a posteriori" method, being interested more in the algorithm than in the nature of $\delta$. Anyway, I would not expect the prominent occurrence of the number $163$ in the algorithm to have a deeper meaning. -On the other hand, I wouldn't be too surprised to see $\delta$ written as a continued fraction with "defineable" terms, given the fact that it can be defined by -$\delta=\lim\limits_{n\to\infty}\dfrac{\mu_n-\mu_{n-1}}{\mu_{n+1}-\mu_n}$, where the $\mu_n$ are the bifurcation points of an iterated map. But I have never seen one either. Any leads? - -REPLY [9 votes]: The Feigenbaum constant is the largest eigenvalue of the derivative of the renormalisation operator at its unique fixed point. There is a beautiful article of Lyubich in the October 2000 Notices of the AMS, entitled "The Quadratic Family as a Qualitatively Solvable Model of Chaos", in which he summarises the connection between universality and renormalisation (pages 1046, 1049-1051). -That doesn't directly answer your question, but hopefully that interpretation of the Feigenbaum constant, together with Lyubich's article and/or the references therein, may be of interest.<|endoftext|> -TITLE: Partitions of Unity -QUESTION [5 upvotes]: Fix a metric $g$ on a smooth, closed manifold $\mathcal{M}$. Take a finite subcover of the manifold from its atlas. Is it true that any smooth partition of unity subordinate to this cover has uniformly bounded derivatives in $L^p (\mathcal{M})$ for each $p \geq 1$ ? - -REPLY [4 votes]: A counterexample: Let $M$ be the unit circle. Let the two charts be the arcs $A = (0,2\pi)$ and $B = (\pi/2, 5\pi/2)$. For each $n$, consider the partition of unity subordinate to $\{A,B\}$ given by -$$ \psi_{A,n} = \sin^2 ( (2n+1) \theta ) $$ -and -$$ \psi_{B,n} = \cos^2 ( (2n+1) \theta ) $$ -Check that $\psi_{A,n} (0) = 0 = \psi_{B,n} (\pi/2)$, and clearly $\psi_{A,n} + \psi_{B,n} = 1$. -By the scaling property it is easy to check that -$$ \| \partial^k\psi_{A,n} \| = (2n+1)^k \| \partial^k\psi_{A,1} \| $$ -And using that -$$ \sin^2(\theta) = \frac12 (1-\cos 2\theta) $$ -you see immediately that your desired uniform bound is impossible.<|endoftext|> -TITLE: How the Jacobi metrics may be useful in mechanics with or without constraints? -QUESTION [6 upvotes]: A mechanical system $(Q,K,V)$ is specified by the configuration space $Q,$ the potential energy $V\in C^\infty(Q),$ and the kinetic energy $K=K_g$ given by a Riemannian metric $g$ on $Q.$ -If $V{<}e$ then $g_J=(e-V)g$ is the Jacobi metric on $Q,$ and Maupertuis' principle states that, up to reparametrization, the trajectories of motion of $(Q,K,V)$ with energy $e$ are exactly the geodesics of $g_J.$ (which by the way are the trajectory of motion of $(Q,K_{g_J},0).$ -My question is motivated by what I read in Jair Koiller's paper "Reduction of Some Classical Non-Holonomic Systems with Symmetry", Arch. Rational Mech. Anal. 118 (1992). - -The Jacobi metric is useful for constrained systems as well, an observation which seems not to have been sufficiently explored in the theory of nonholonomic systems. - -Koiller goes on saying that, if $D,$ a tangent distribution on $Q,$ represents a linear constraint on the velocities, then the trajectories of motion of the constrained system $(Q,K,V;D)$ with energy $e$ are the same as the ones of $(Q,K_{g_J},0;D),$ with energy $1,$ up to a reparametrization. -Edited Question: - -What results of riemannian geometry can be transferred into implications on the dynamics of mechancal systems by way of Jacobi metric? -Furthermore, in the light of the quotation, I would know: what are examples of the usefulness of Jacobi metrics in nonholonomic mechanics? - -Obviously any feedback is welcome. Thank you. - -REPLY [7 votes]: Dear Giuseppe, -I think that the idea is that much more is known about the dynamics of geodesic flows in Riemannian and Finsler manifolds than about the dynamics of more general Hamiltonian systems. Here are two examples where the Jacobi metric is or has been useful: -1. A generalized spherical pendulum: If $V$ is a smooth potential function on the two-sphere and $e > \max V$, then there are infinitely many periodic orbits at the energy level $e$. -This is because of the Franks-Bangert-Hingston result stating that any Riemannian metric on the two-sphere has infinitely many prime closed geodesics. -2. Every smooth, compact, convex hypersurface in $({\mathbb R}^{2n}, \omega_{can)}$ carries a closed characteristic. -This is a celebrated theorem of A. Weinstein that has been widely generalized, but the original proof reduced the problem to finding a closed geodesic for non-reversible Finsler metrics on spheres, which you can do using Birkhoff's minimax procedure. -I'm less familiar with the constrained case, but I guess the answer may change if you consider non-holonomic systems (which are not Hamiltonian systems) or vakonomic systems.<|endoftext|> -TITLE: Leopoldt's conjecture and cup-products -QUESTION [17 upvotes]: Among the many equivalent formulations of Leopoldt's conjecture, this one is probably the shortest: For any number field $K$, prime number $p$, finite set $S$ of primes of $K$ containing the primes above $p$, one has - -Leopoldt's conjecture: $H^2(G_{K,S},\mathbb{Q}_p)=0$. - -Here $G_{K,S}$ is as usual the Galois group of the maximal algebraic extension of $K$ un ramified outside $S$ and places at infinity, the $H^2$ is continuous cohomology. -Now, one of the most natural way to get a class in an $H^2$ is as a cup-product of two classes in an $H^1$. For example, if $\chi : G_{K,S} \rightarrow Q_p^\ast $ is a continuous character, then there is a cup-product map -$$H^1(G_{K,S},\chi) \times H^1(G_{K,S},\chi^{-1}) \rightarrow H^2(G_{K,S},\mathbb{Q}_p),$$ -which, according to Leopoldt's conjecture, should be zero. - -Is it any easier to prove that the above morphism is zero than to prove Leopoldt's conjecture itself ? - -I would also be interested to know the answer in special cases (of $K$, $\chi$, $p$) -where Leopoldt's conjecture is not known. - -REPLY [3 votes]: I think your cup-product is always zero independently of Leopoldt (at least if $\chi$ is of finite order). Consider $\mathbb{Z}_p$-coefficients instead (enough by Neukirch-Schmidt-Wingberg, 2.3.10). If $\chi=1=\chi^{-1}$, then both $H^1$ degenerate to $H^1(\tilde{K}/K,\mathbb{Z}_p)$ where $\tilde{K}$ is the compositum of all $\mathbb{Z}_p$ extensions, because cocyles are Hom's. In particular, your cup-product factors through -$$ -H^1(\tilde{K}/K,\mathbb{Z}_p)\times H^1(\tilde{K}/K,\mathbb{Z}_p)\xrightarrow{\cup}H^2(\tilde{K}/K,\mathbb{Z}_p) -$$ -but the $H^2$ is trivial because a free $p$-group has $p$-cohomological dimension $1$ (and NSW, 2.3.5 tells you that cohomology with $\mathbb{Z}_p$-coefficients is the projective limit of those with $\mathbb{Z}/p^n$-coefficients). So, your cup-product is zero if $\chi=1$. -Now suppose $\chi$ is of non-trivial and set $\Delta=G_K/\mathrm{Ker}(\chi)$. -Assume $H^i(\Delta,\mathbb{Z}_p(\chi^{\pm 1}))=0$ for $i=1,2$ -Then $H^1(G_{K,S},\mathbb{Z}_p(\chi^{\pm 1}))=H^1(G_{L,S},\mathbb{Z}_p)^{\Delta}$ by Hochschild-Serre, where $L$ is the field attached to $\chi$. Since cup-product is compatible with restriction by NSW 1.5.3, it is compatible with the above identification and coincides with -$$ -H^1(\tilde{L}/L,\mathbb{Z}_p)^\Delta\times H^1(\tilde{L}/L,\mathbb{Z}_p)^\Delta\xrightarrow{\cup}H^2(\tilde{L}/L,\mathbb{Z}_p)^\Delta=0 -$$ -by the $\chi=1$ case. -Now when is $\Delta$-cohomology trivial? If $\chi$ is of finite order, its order is prime to $p$ and we win. If $\chi$ is the cyclotomic character, then $H^2(\Delta,\mathbb{Z}_p(\pm 1))=0$ because $cd_p(\Delta)=1$. For $i=1$, by Kummer theory we see that $H^1(\Delta,\mathbb{Z}_p(1))=\varprojlim F_\infty^{\times}/(F_\infty^{\times})^{p^n}$, where $F_\infty$ is the cyclotomic extension of $\mathbb{Q}_p$...and I do not know what to do ;-)<|endoftext|> -TITLE: Derandomizing random matrices -QUESTION [5 upvotes]: My question is rather general - what is known about derandomization of results in random matrix theory, high-dimensional geometry, Banach spaces etc. using probabilistic constructions (like estimates of eigenvalues, Dvoretzky's theorem, metric embeddings)? Here I'm interested both in fully explicit counterparts of random constructions as well as "pseudorandom" (in some sense) examples, using "less" randomness than, say, filling every entry of a matrix with a random variable etc. For example - suppose we know that for a fixed norm an n x n matrix with IID standard gaussian entries has "large" norm with high probability. How to find an explicit infinite family of such matrices? -My question is rather vague, of course I have a specific application of this kind of results in mind, but at this moment I am more interested in general methodology of constructing "derandomized" examples, where to start looking for such objects etc. My only contact so far with pseudorandomness has been in the context of spectral graph theory, expander graphs, property (T) etc., I'm not sure if this perspective is relevant for high-dimensional geometry. -I'd be grateful for any hints, references or advice on who may know this kind of things. - -REPLY [9 votes]: There is active interest in such results in high-dimensional geometry, and expander graphs have even been used explicitly as a tool. Take a look for example at this paper and the references on the second page. -Added: After prodding from the OP, here are some more references to various results of this type. If this question manages to recapture Bill Johnson's attention, maybe he'll contribute some more that I didn't think of. -Here is a survey paper by Davidson and Szarek ending with a short section on derandomizing various constructions. As the Indyk–Szarek paper I linked to above shows, the discussion of Kashin-type results is definitely out of date, but it also has references to work on approximation of quasidiagonal operatrors and approximately free (in the sense of free probability) matrices; I don't know the state of the art on those things. -Other results on reducing randomness in Kashin-type theorems (not all cited by Indyk–Szarek) followed this paper by Schechtman; try a Google Scholar or MathSciNet search of the papers citing it. -This paper by Artstein-Avidan and Milman includes results on reducing randomness in a number of different theorems in geometric functional analysis. -As I said in comments below, randomness reduction is a hot topic in compressed sensing, and not being an expert in the area I don't dare try to guess at the quickly-moving state of the art. -All of the above results, although not necessarily explicitly stated that way, can of course be phrased in terms of properties of some matrices (e.g., identifying a subspace with a matrix whose columns are a basis for the subspace).<|endoftext|> -TITLE: Does a "composite field" always exist? -QUESTION [5 upvotes]: Suppose $F$ is a field, and $F_1, F_2$ are two extension fields of $F$. Is it always the case that there is a field $L$, containing three subfields $F, K_1, K_2$ and two ring isomorphisms $\varphi_{i}:F_i\rightarrow K_1$ fixing $F$? -Note 1: We lose no generality assuming $F$, rather than an isomorphic copy of $F$, is a subfield of $L$. -I ask this because I was wondering if there is a way to combine the reals and the $p$-adic numbers into a single extension of $\mathbb{Q}$. -Note 2: I seem to recall someone telling me this couldn't be done (perhaps with additional topological data preserved). But I cannot seem to remember the reason why. In any case, I want to know if there is something other than topology which prevents it. - -REPLY [7 votes]: In the language of Model Theory, your question can be rewritten as: "does the theory of fields have the amalgamation property? and the answer is yes. -Well known examples of theories with the amalgamation property include: fields, ordered fields, groups, abelian groups and boolean algebras.<|endoftext|> -TITLE: Splitting principle for holomorphic vector bundles -QUESTION [14 upvotes]: Let $E \to X$ be a vector bundle over a decent space $X$. Then there is a space $Z$ together with a map $p: Z \to X$ which induces a split injection on cohomology and such that $p^* E$ splits as a direct sum of line bundles (take e.g. the flag bundle of $E$). Is the analog true for holomorphic vector bundles (if we stay purely in the category of complex manifolds)? That is, if $X$ is a complex manifold and $E$ a holomorphic vector bundle, can we get a holomorphic map $p: Z \to X$ (with $Z$ a complex manifold) with the same properties: the map on cohomology is a split injection, and $p^*E$ splits in the holomorphic category as a sum of line bundles? -(As a side question, I'm curious what additional invariants one can construct for holomorphic vector bundles, which don't make sense for an ordinary complex vector bundle. I'm vaguely aware of the Atiyah class, but are there other examples?) - -REPLY [18 votes]: The answer is positive. Let $P$ be the principal $\mathrm{GL}_n$-bundle associated with $E$; then the space of flags is the quotient $P/B$, where $B$ is the Borel subgroup of $\mathrm{GL}_n$ consisting of upper triangular matrices. Set $Z = P/T$, where $T$ is the maximal torus consisting of diagonal matrices. A point of $Z$ is a point of $X$, plus $n$ independent 1-dimensional linear subspaces of the fiber of $E$. The projection $Z \to P/B$ is a fibration with contractible fibers, hence the pullback from the cohomology of $P/B$ to that of $Z$ is an isomorphism. Since the cohomology of $X$ injects into the cohomology of $P/B$, it also injects into the cohomology of $Z$.<|endoftext|> -TITLE: Coherent states vs quantization of Lagrangian submanifold -QUESTION [13 upvotes]: Coherent states http://en.wikipedia.org/wiki/Coherent_states -are vectors in the Hilbert space which in certain sense are strongly localized -and "corresponds" to points in classical phase space (see below). -On the other hand vectors in Hilbert space in classical limit corresponds to Lagrangian submanifolds (see below). -How these two ideas correspond to each other ? -Is the following reasonable/known: coherent states corresponds to Lagrangian submanifolds in the COMPLEXIFICATION of the phase space, which intersect with real points of the phase space by the just one point ? - -Motivating example(NOT quite correct thanks to "Squark"): consider R^2 and H=p^2+q^2. Consider p^2+q^2=0 - the real slice is just one point. On the other hand the in complexification we have 1-dimensional Lagrangian submanifold. The wave function $\psi$ corresponding to Lagrangian submanifold H=0, -is constructed in a simple way $\hat H \psi=0$. $\hat H$ is hamiltonian of the harmonic oscillator and its eigenfuction is well-known to be coherent state. -So in this example idea seems to work. -Vague attempt for correction -It seems we need to take into account Sommerfeld's 1/2 correction (it is related to Maslov's index). -I mean we should consider H=r, and choose such "r" which satisfies Bohr-Sommerfeld quantization condition i.e. symplectic form integrated over interior of H=r should be integer +1/2. But we need somehow subtract this 1/2 from quantum eigenvalue to fit -into the desired picture... - -Background. -Quantization of Lagrangian submanifolds -Down-to-earth idea of the construction of the vector in Hilbert space from the Lagrangian submanifold. -Consider sumanifold defined by the equations $H_i=0$. -Consider "corresponding" quantum hamiltonians $\hat H_i $, -consider vector $\psi$ in the Hilber space such that $\hat H_i \psi = 0$. -This $\psi$ we are talking about. -Why it is important "Lagrangian" ? It is easy. If $A \psi =0$ and $B\psi = 0$ -then it is true for commutator $[A,B]\psi = 0$. -In classical limit commutator correspond to Poisson bracket so we see -that even if we start from $H_i$ which is not close with respect to Poisson -bracket we must close it - so we get coisotropic submanifold. -Lagrangian - just restiction on the dimension - that it should be of minimal possible dimension - so after quantization we may expect finite dimensional subspace (in the best case 1-dimensional). -There are related by slightly different points of view. -Lagrangian Submanifolds in Deformation Quantization - -Coherent states -Sometimes one can realize Hilbert space corresponding to classical symplectic manifold $M$, -as holomorphic functions on $M$. -Each point $p\in M$ defines a functional $ev_p: Fun(M) \to C$ just evaluation of the function at point $p$. -On the other hand in Hilbert space any functional corresponds to a vector $v$, -such that $= ev_p(f)$. -So $v$ is coherent state corresponding to $p$. -This approach is probably due to J. Rawnsley. -See Mauro Spera "On Kahlerian coherent states" http://www.emis.de/proceedings/Varna/vol1/GEOM19.pdf - -The idea that complexification of the phase space is important -plays a role in "Brane Quantization" approach by Witten and Gukov. -See e.g. http://arxiv.org/abs/1011.2218 Quantization via Mirror Symmetry -Sergei Gukov - -There are actually different generalizations of the coherent states. -In particular group theoretical approach was developed by -Soviet physicists Askold Perelomov who worked in ITEP: -A. M. Perelomov, Coherent states for arbitrary Lie groups, Commun. Math. Phys. 26 (1972) 222-236; arXiv: http://arxiv.org/abs/math-ph/0203002 -A. Perelomov, Generalized coherent states and their applications, Springer, Berlin 1986. - -REPLY [3 votes]: This is a partial answer. I can understand the correspondence in the special case of coadjoint orbits $\mathcal{O}$ of compact Lie groups and Hamiltonians consisting of symbols of magnetic Laplacians on their cotangent bundlles. -On one hand a solution of the Hamilton-Jacobi equation corresponding to these Hamiltonians define to a Lagrangian submanifold in $T^{*} \mathcal{O}$, and on the other hand the energy eigenspaces of the magnetic Laplacians are finite dimensional, thus the symbols of their orthogonal projectors in $L_2(\Gamma(L))$ (where, $L$ is the line bundle over $\mathcal{O}$ corresponding to the magnetic part of the symplectic form on $T^{*} \mathcal{O}$) are reproducing kernels which can be viewed as coherent states parametrized by points of the coadjoint orbit. -Now, for real coadjoint orbits, their complexifications are equivariently homeomorphic to their cotangent bundles, thus the correspondence can be passed to the complexification.<|endoftext|> -TITLE: Is the Cheeger constant of an induced subgraph of a cube at most 1? -QUESTION [8 upvotes]: It is known that the -Cheeger constant -of a -hypercube graph $Q_n$ -is exactly $1$, regardless of its dimension $n$. Is $1$ also an upper bound -on the Cheeger constant of nontrivial induced connected subgraphs of $Q_n$? -The Cheeger constant $h(G)$ is also known as the -edge expansion -or the isoperimetic number. -To indirectly address a comment, -here are the Cheeger constants of more graphs: -$h($$Q_n$$) = 1$ -$h($$P_n$$) = 1 / \lfloor n / 2 \rfloor$ -$h($$C_n$$) = 2 / \lfloor n / 2 \rfloor$ -$h($$K_n$$) = \lceil n / 2 \rceil$ - -REPLY [6 votes]: Your conjecture is true, every subgraph of the cube has expansion constant at most $1$. -Proof: Suppose we are given a subgraph $G\subset Q_n$, $n>1$ and cut the cube into $2$ $(n-1)$-dimensional subcubes $A_1,A_2$. (So that $A_1\cup A_2=Q_n$) The key is to notice that each vertex in $A_1$ is connected to one and only one vertex in $A_2$. Then split $G$ into two parts, $G_1=G\cap A_1$ and $G_2=G\cap A_2$. One of these will have size $\leq \frac{|G|}{2}$, suppose it is $G_1$. Because $G_1$ is in $A_1$, it is only connected to vertices in $A_2$, and we have $\partial G_1 \leq |G_1|$. Since the expansion constant is the minimum, we conclude that $$h(G)\leq 1.$$<|endoftext|> -TITLE: What's the link between the Toeplitz operators on H^2 and those used to define Cuntz-Pimsner algebras? -QUESTION [5 upvotes]: An alternate way to phrase this question might be, "How did the Toeplitz operators used in the definition of the Cuntz-Pimsner algebra come by their name?" or, "What's the relationship between the Toeplitz algebra generated by multiplication operators on the Hardy space $H^2$ and the Toeplitz algebra associated to a Hilbert C*-module?" -References would be greatly appreciated. -Thanks! - -REPLY [5 votes]: After thinking about it some more, here's what I came up with -- but if anyone else has another explanation I'd love to hear it! -The classical Toeplitz operators $T_f$ are built from multiplication operators $M: C(\mathbb{T}) \to B(L^2(\mathbb{T}))$, where $M_f(\phi) = f \phi$ is given by pointwise multiplication. Note that $C(\mathbb{T})$ is dense in $L^2(\mathbb{T})$, so one can think of this as an action of $L^2(\mathbb{T})$ on itself by multiplication. -When building a Cuntz-Pimsner algebra (or generalized Toeplitz algebra), one starts with a Hilbert module $X$ instead of a Hilbert space; the algebra is generated by a set of Toeplitz operators $\{T_\xi\}_{\xi \in X}$, which act as multiplication operators on $X$ in the sense that $T_\xi(x) := \xi \otimes x \in X \otimes X$. However, without modifications, this definition would put $T_\xi \in \mathcal{L}(X, X \otimes X)$, which is not a $C^*$-algebra. In order to have $\{T_xi\}$ generate a $C^*$-algebra, we want to find a way to have $\{T_\xi\} \subseteq \mathcal{L}(\mathcal{E})$ for some (single) Hilbert module $\mathcal{E}$. Hence, we define $\mathcal{E} = \bigoplus_{n=0}^\infty X^{\otimes n}$, defining $T_\xi$ on elementary tensors by $T_\xi( x_1 \otimes \ldots \otimes x_n) = \xi \otimes x_1 \otimes \ldots \otimes x_n$. -In other words, the aspect of the classical Toeplitz algebras that I see the Cuntz-Pimsner algebras as generalizing is the idea of a Hilbert space acting on itself via multiplication operators.<|endoftext|> -TITLE: On the determination of a quadratic form from its isotropy group -QUESTION [6 upvotes]: Let $F:\mathbf{R}^n\rightarrow\mathbf{R}$ be a non-degenerate quadratic forms. Let -$$ -O(F):=\{g\in GL_n(\mathbf{R}):F(gv)=F(v),\forall v\in \mathbf{R}^n\} -$$ -be the isotropy group of $F$. -Q: So how does one prove in the simplest way possible that if $O(F)=O(G)$ then -there exists $\lambda\in\mathbf{R}^{\times}$ such that $F=\lambda G$? -P.S. I would like to have a proof that could be explained to undergraduate students who take an advanced linear algebra class. - -REPLY [6 votes]: Robert's answer involves picking any $v$ and $w$ in ${\mathbf R}^n$ and getting an identity of polynomials of degree at most 2 in $t$ away from where $F(v+tw) = 0$, which excludes at most two values of $t$. He equates the coefficients in the identity to get the desired formula. -The same technique works for nondegenerate quadratic forms over any field with at least five elements (since a quadratic polynomial function on a field is completely determined away from two points if at least three points remain). Of course we can ignore fields of size 2 and 4, since that would be in characteristic 2, but what happens in the case of quadratic forms over a field of size 3? Admittedly the question was being asked over the real numbers for the purpose of a linear algebra class, so the whole point of checking finite fields isn't essential, but it's worth noting that the result is still true. I'll indicate the steps, but leave out computational details. -We have two nondegenerate quadratic forms $Q_1$ and $Q_2$ on a finite-dimensional vector space $V$ over a field $K$ of characteristic not 2 (EDIT: I will dicuss characteristic 2 at the end), and we assume the isotropy groups of $Q_1$ and $Q_2$ are the same. We want to show if $Q_1(v) \not= 0$ that $Q_2(v) \not= 0$ and $Q_2(w) = cQ_1(w)$ for all $w \in V$, where $c = Q_2(v)/Q_1(v)$. -a) Let $v$ in $V$ be nonzero, $L \colon V \rightarrow K$ be linear, and $B \colon V \times V \rightarrow K$ be a nondegenerate bilinear form. If $B(w,v)L(w) = 0$ for all $w \in V$, then $L(w) = 0$ for all $w \in V$. (Hint: There is a basis of $V$ in the complement of any hyperplane.) -b) Let $B_1$ and $B_2$ be the symmetric bilinear forms associated to $Q_1$ and $Q_2$. Suppose $v \in V$ satisfies $Q_1(v) \not= 0$. Then by part a and a reflection as in Robert's answer, $$Q_1(v)B_2(v,w) = Q_2(v)B_1(v,w)$$ for all $w \in V$. (The reflection $s_v(w) = w - (2B_1(v,w)/Q_1(v))v$ is in the isotropy group of $Q_1$, and therefore $Q_2(s_v(w)) = Q_2(w)$ for all $w$, which implies the above equation from nondegeneracy of $B_1$.) -c) Assume from now on that $Q_1$ and $Q_2$ have the same isotropy group. Using part b, for all $v \in V$ we have $Q_1(v) \not= 0$ if and only if $Q_2(v) \not= 0$, and when $Q_1(v) \not= 0$ we have $\{w : B_1(v,w) = 0\} = \{w : B_2(v,w) = 0\}$.` -d) If $Q_1(v) \not= 0$ and $B_1(v,w) \not= 0$ then $Q_2(w) = (Q_2(v)/Q_1(v))Q_1(w)$. (To prove this, by part c we can assume $Q_1(w) \not= 0$.) -e) If $Q_1(v) \not= 0$ and $B_1(v,w) = 0$ then $Q_2(w) = (Q_2(v)/Q_1(v))Q_1(w)$. (By part c, $B_2(v,w) = 0$. We have $B_1(v,v+w) = B_1(v,v) = Q_1(v) \not= 0$, so we can apply part d with $v+w$ in place of $w$.) -Now what happens in characteristic 2? The bijection between quadratic forms and symmetric bilinear forms breaks down, so the notion of isotropy group or nondegeneracy has to carefully distinguish between quadratic forms and symmetric bilinear forms. Let me set up the terminology that I'll use, to minimize the chance of confusion. -Let $K$ be a field of characteristic 2. A quadratic form on a finite-dimensional $K$-vector space $V$ is a function $Q \colon V \rightarrow K$ that looks like a homogeneous quadratic polynomial in one (equivalently, any) choice of basis of $V$, and this is the same thing as saying the function $B(v,w) := Q(v+w) - Q(v) - Q(w)$ from $V \times V$ to $K$ is a bilinear form, and it is obviously symmetric. (Classically we'd want to divide this $B$ by 2, so $B$ here is analogous to $2B$ outside characteristic 2.) We have $B(v,v) = Q(2v) - 2Q(v) = 2Q(v) $, and in characteristic 2 this is 0, so we can't recover values of $Q$ from values of $B$. The points is that many $Q$'s can have the same $B$. -Example: On $K^2$, let $Q(x,y) = ax^2 + bxy + cy^2$. Then you can check from the characteristic being 2 that $B((x,y),(x',y')) = b(xy' + yx')$. In particular, $a$ and $c$ don't appear in $B$, so $x^2 + xy$ and $xy$ are two different quadratic forms on $K^2$ that have the same associated symmetric bilinear form. -An isometry of $(V,Q)$ is defined to be a linear map $A \colon V \rightarrow V$ such that $Q(Av) = Q(v)$ for all $v \in V$. An isometry of $(V,B)$ is defined to be a linear map $A \colon V \rightarrow V$ such that $B(Av,Aw) = B(v,w)$ for all $v$ and $w$ in $V$. Because $B(v,w)$ can be expressed in terms of values of $Q$, any isometry of $(V,Q)$ is an isometry of $(V,B)$, but in characteristic 2 the converse is false: the isometry group (or what you call isotropy group) of a quadratic form in characteristic 2 is smaller than the isometry group of its associated bilinear form. -Example: Let $Q(x,y) = ax^2 + bxy + cy^2$ with $b \not= 0$. Up to scaling by $b$, the bilinear form attached to $Q$ is $B((x,y),(x',y')) = xy' + yx'$, which is the standard alternating pairing on $K^2$, so the isometry group of $B$ is the symplectic group of $K^2$, and ${\rm Sp}(K^2) = {\rm SL}_2(K)$. By a direct calculus, a matrix $(\begin{smallmatrix}\alpha&\beta\\ \gamma&\delta\end{smallmatrix})$ with determinant 1 is an isometry of $(V,Q)$ if and only if $Q(\alpha,\gamma) = a$ and $Q(\beta,\gamma) = c$. For instance, $(\begin{smallmatrix}0&1\\1&0\end{smallmatrix})$ is in ${\rm SL}_2(K)$ and it is an isometry of $(V,Q)$ if and only if $a = c$, so when $a \not= c$ (ex: $x^2 + xy$) this particular $2 \times 2$ matrix that is an isometry of $B$ is not an isometry of $Q$. -We can define reflection in characteristic 2 as follows: if $Q(v) \not= 0$ then we set -$s_v \colon V \rightarrow V$ by $$s_v(w) = w - \frac{B(v,w)}{Q(v)}v.$$ The $2B$ appearing in the classical reflection formula is exactly the same thing as the $B$ here, so this definition for a reflection is truly independent of characteristic (but depending on whether or not we're in characteristic 2 there are different conventions of what the bilinear form associated to a quadratic form is). You can check by a calculation that $Q(s_v(w)) = Q(w)$ for all $w$, so $s_v$ is in the isometry group of $(V,Q)$, and thus also is in the isometry group of $(V,B)$, which is a larger group in characteristic 2. And you can check by a direct calculation that $s_v^2$ is the identity on $V$, which is a property reflections should have. The catch is that $s_v$ might be the identity function. This happens if $v \in V^\perp = \{u : B(u,V) = \{0\}\}$, and here you're going to say "well, we are only working in the nondegenerate case". Now we have to deal with the issue of how you actually define nondegeneracy in characteristic 2, which is a bit different from what you're used to outside characteristic 2 (at least in some references you find a difference, so let's address it). -For a symmetric bilinear form $B$ the notion of nondegeneracy is the same in all characteristics: it means if $w \mapsto B(v,w)$ is identically 0 then $v = 0$. For a quadratic form $Q$ with associated symmetric bilinear form $B$, outside characteristic 2 we call $Q$ nondegenerate when $B$ is nondegenerate (and there is no difference between $B$ or $2B$ being called the bilinear form of $Q$ when $2 \not= 0$). But in characteristic 2 we do not define $Q$ to be nondegenerate when $B$ is nondegenerate. -Definition: If $Q$ is a quadratic form (in any characteristic) with associated symmetric bilinear form $B$, we call $Q$ nondegenerate when the two conditions $Q(v) = 0$ and $B(v,V) = \{0\}$ imply $v = 0$. -Outside characteristic 2, saying $Q(v) = 0$ is the same as saying $B(v,v) = 0$, so the condition $Q(v) = 0$ would be implied by $B(v,V) = \{0\}$ and thus that condition on $Q$ can be dropped from the above definition: outside characteristic 2 the above definition is the same thing as the usual notion of nondegeneracy for a quadratic form. But in characteristic 2 we can have $Q(v) \not= 0$ while $B(v,V) = \{0\}$. -Example: On $K^3$, when $K$ has characteristic 2, let $Q(x,y,z) = x^2 + xy + y^2 + z^2$. The associated symmetric bilinear form is given by $B((x,y,z),(x',y',z')) = xy' + y'x$. Notice there is no $z$ or $z'$ in this formula. If $B(v,K^3) = \{0\}$ then $v = (0,0,z)$, so $Q(v) = z^2$. Thus if $Q(v) = 0$ and $B(v,K^3) = \{0\}$ then $v = (0,0,0)$. But we can have $B(v,K^3) = \{0\}$ with $v$ being nonzero, such as $(0,0,1)$. The bilinear form $B$ is degenerate but we say the quadratic form $Q$ is nondegenerate. -Of course if it happens to be the case that $B$ is nondegenerate then $Q$ is nondegenerate by the above definition, so a quadratic form in char. 2 whose associated symmetric bilinear form is nondegenerate is a nondegenerate quadratic form, but we can have nondegenerate quadratic forms whose associated symmetric bilinear form is degenerate, as in the above example. -Returning to the reflection formula, we can have $Q(v) \not= 0$ while $B(v,V) = \{0\}$, and the so-called reflection $s_v$ is the identity function: $s_v(w) = w$ for all $w$. That is dumb. So we only define reflections $s_v$ when $Q(v) \not= 0$ and $B(v,V) \not= \{0\}$, since we need both conditions for $s_v$ to be defined and to have order 2. -Now it's time to go back to the 5 steps a, b, c, d, e above and see how they fare in characteristic 2: if $Q_1$ and $Q_2$ are nondegenerate quadratic forms with the same isometry group (or what you call isotropy group), does $Q_2 = cQ_1$ for some $c \in K^\times$? -Part a goes through without a problem since it isn't a property of quadratic forms at all. However, the nondegeneracy condition on $B$ there is going to be felt when we apply part a to later parts in characteristic 2. -As for part b, to make this go through you need to assume $Q_1(v) \not= 0$ and $B_1(v,V) \not= \{0\}$, which are two genuinely different conditions: the first does not imply the second like it would outside char. 2, where $Q$-values are special cases of $B$-values. -If you assume $Q_1$ and $Q_2$ have the same isometry group, then for any $v$ with $Q_1(v) \not= 0$ and $B_1(v,V) \not= \{0\}$ then you can derive -$$ -Q_1(v)B_2(v,w) = Q_2(v)B_1(v,w) -$$ -for all $w \in V$ using the reflection $s_v$, just like part b is done outside characteristic 2. (When I wrote in the original part b that you use nondegeneracy of $B_1$, what you need is that $w \mapsto B_1(v,w)$ is not identically 0 with the specific $v$ where $Q_1(v) \not= 0$. It is, strictly speaking, weaker than nondegeneracy to assume that about $v$.) Notice, alas, that the two conditions $Q_1(v) \not= 0$ and $B_1(v,V) \not= \{0\}$ are not implied by nondegeneracy of $Q_1$l for a nondegenerate $Q_1$ we could have a $v$ where $Q_1(v) \not= 0$ while $B_1(v,V) = \{0\}$. The 3-dimensional nondegenerate quadratic form with degenerate bilinear form that I mentioned earlier is an example of this, with $v = (0,0,1)$. -To get the conclusion of part c to work in characteristic 2, it seems to me that you need to assume $B_1$ and $B_2$ are nondegenerate (which is stronger than saying $Q_1$ and $Q_2$ are nondegenerate). With this assumption, if $Q_1$ and $Q_2$ have the same isometry group then the conclusions of part c remain true in char. 2 by the same proofs as one uses outside char. 2. -Part d goes through in characteristic 2 if you assume $B_1$ and $B_2$ are nondegenerate (stronger than assuming $Q_1$ and $Q_2$ are nondegenerate). -Now we're at the last step, part e. Unfortunately, here there appears to be a serious problem related to $Q$-values not being $B$-values. We assume $B_1$ and $B_2$ are nondegenerate, $Q_1$ and $Q_2$ have the same isometry group, $Q_1(v) \not= 0$, and $B_1(v,w) = 0$ (so also $B_2(v,w) = 0$). We want to derive $Q_2(w) = (Q_2(v)/Q_1(v))Q_1(w)$, as in part d. But the hint I gave in part e breaks down: $B_1(v,v+w) = B_1(v,v)$, which is 0, not $Q_1(v)$, since the bilinear form associated to a quadratic form in char. 2 is alternating. In fact, for any $a$ and $b$ in $K$ we have $B_1(v,av+bw) = aB_1(v,v) + bB_1(v,w) = aB_1(v,v) = 0$. Thus there is no $u$ in the subspace spanned by $v$ and $w$ such that $B_1(v,u) = 0$. -If you try to use Robert's method, assuming $K$ has characteristic 2 and at least 5 elements (so we give up on $K = {\mathbf F}_2$ or $K = {\mathbf F}_4$), things break down because when $B_1(v,w) = 0$ we have $B_1(x,y) = 0$ for all $x$ and $y$ in the $K$-span of $v$ and $w$. -If we could get the conclusion of part e to work by some other method then we'd have the result you want, but at this point I don't know what that other method would be. I suggest posting a question on MO for the experts on quadratic forms: in characteristic 2 does the isotropy group of a nondegenerate quadratic form determine the quadratic form up to a nonzero scaling factor? I would not be surprised if there might be weird counterexamples when $K$ is small and $V$ has small dimension. -By the way, you don't gain much by switching to the viewpoint of the isometry group of the symmetric bilinear forms, instead of quadratic forms, in characteristic 2: such bilinear forms are automatically alternating, so if they are nondegenerate then their isometry group is the symplectic group for that dimension, hence the group is determined up to isomorphism by the dimension. I don't think it's that interesting to ask about two symplectic isometry groups being literally equal rather than isomorphic.<|endoftext|> -TITLE: Who invented the Morse-Bott-complex? -QUESTION [9 upvotes]: In the "Morse-Bott theory and equivariant cohomology" paper by D.M. Austin and P.J. Braam, the authors introduce the Morse-Bott-complex to calculate the de-Rham-cohomology of a compact manifold (using a Morse-Bott-function). -Where does this complex "come from", i.e. is it Austin and Braam's original work? -In the introduction, they state: - -Floer's homology groups for homology 3-spheres asked for many generalizations to describe more general gluing laws governing Donaldson's polynomial invariants, both through incorporating other auxiliary information (see Fukaya [1] and Braam and Donaldson [2]) as well as by considering general 3-manifolds [3]. In the course of this work, new techniques were developed in these infinite dimensional cases relating to equivariant cohomology, cup products and various alternative approaches to problems of classical Morse theory. The purpose of this paper is to give a self-contained finite dimensional description of these new aspects. - -Does this paragraph also refer to the Morse-Bott-complex or only to the section on "Equivariant cohomology and Morse-Bott theory"? -(The referenced works in the citation, as good as I could guess from the (preprint) references listed: - -Fukaya: Floer homology for oriented 3-manifolds -Braam, Donaldson: Fukaya-Floer homology and gluing formulae for polynomial invariants -Austin, Braam: Equivariant Floer theory and gluing Donaldson polynomials ) - -REPLY [7 votes]: I do not know a precise answer to your question (and perhaps someone else does), but I'm fairly certain the idea can be traced back to the paper -Bott, Raoul, -The stable homotopy of the classical groups. -Ann. of Math. (2) 70 1959 313–337. -This is the paper where Bott proves his famous periodicity theorem, and the proof uses Morse-Bott theory, as it's now known. The relevant Sections are 3 and 4. In particular, I'm fairly certain that the spectral sequence of Corollary 4.2 is a special case of the "Morse-Bott complex" in the paper of Austin and Braam. -Disclaimer: I'm no historian, and I haven't thought about these things in some time!<|endoftext|> -TITLE: The critical value of percolation on Cayley graphs. -QUESTION [14 upvotes]: Let $\Gamma$ be a discrete group with a generating set $S$. Let $p_c(\Gamma,S)$ be the critical probability for percolation of the Cayley graph of $\Gamma$. Is it known that if $\Gamma$ is non-amenable then there exists a generating set $S$ such that $p_c(\Gamma,S)<\frac{1}{2}$ (or some other constant)? -The details on the question: -Let $0 -TITLE: Why is the identity element of a group denoted by $e$? -QUESTION [25 upvotes]: The question was asked by a student, and I did not have a ready answer. I can think of the German word ``Einheit'', but since in German that is not how the identity element of a group is called, I doubt that is the origin. Any ideas? - -REPLY [18 votes]: The Encyklopädie article of Heinrich Burkhardt, Endliche discrete Gruppen (1899), p. 218 ascribes the origin thus: - -15. Allgemeiner Gruppenbegriff. (...) die Gruppe enthält ein - Element $e$, die Einheit $^{73)}$, das mit jedem andern $a$ $ae = a$ und $ea = a$ - ergiebt; (...) - -73) (...) G. Frobenius u. E. Stickelberger, J. f. Math. 86, 1879[78], p. 219. - -The paper in question, Ueber Gruppen von vertauschbaren Elementen (1879), reads: - -§.1. Definitionen. -Die Elemente unserer Untersuchung sind die $\varphi(\mathbf M)$ Klassen von (reellen) ganzen Zahlen, welche in Bezug auf einen Modul $\mathbf M$ incongruent und relativ prim zu demselben sind. (...) Das Element $\mathbf E$ (so bezeichnen wir im folgenden die Zahlenklasse, deren Repräsentant 1 ist) heisst das Hauptelement.<|endoftext|> -TITLE: A polynomial recurrence involving partial derivatives -QUESTION [12 upvotes]: Define recursively polynomials $f_n(a,b)$ by - $$ f_0(a,b)=1,\ \ f_n(0,b)=0\ \mathrm{for}\ n>0 $$ - $$ \frac{\partial}{\partial a}f_n(a,b) = f_{n-1}(b-a,1-a). $$ - For instance, - $$ f_1(a,b) = a,\ \ - f_2(a,b) = \frac 12(2ab-a^2) $$ - $$ f_3(a,b) = \frac 16(a^3-3a^2-3ab^2+6ab). $$ -Is there a ``nice'' solution to this recurrence, e.g., a formula for -the generating function $\sum_{n\geq 0}f_n(a,b)x^n/n!$? What I am -really interested in is $f_n(1,1)$. For the motivation, see the -solution to Exercise~4.56(d) (pg. 645) of Enumerative -Combinatorics, vol.1, 2nd ed. - -REPLY [3 votes]: There seems to be a PDE for $g(a,b,x)=\sum_{n\ge0}f_n(a,b)x^n$, -which can be thought of as a boundary value problem in the triangle $0\lt a\lt b\lt1$. -$$g_{aab}+g_{abb}+x^3g=0$$ -($x$ is a parameter and subscripts are derivatives) -with boundary values $g(0,b,x)=1$, $g_a(a,a,x) = x$, and -$g_{ab}(a,1,x) = x^2$. This comes from iterating the $f_n$ recurrence, -after Pietro's remarks that $(a,b)\to(b-a,1-a)$ has period 3 -suggested looking at third derivatives. - Does that determine $g$ uniquely, nicely? I don't know yet. -[Edit: I wrongly wrote $g$ at first using $\frac{x^n}{n!}$.]<|endoftext|> -TITLE: Categorical semantics of W-types -QUESTION [9 upvotes]: Jacob's book titled "Categorical Logic and Type Theory" gives a nice description of Π and Σ types as adjunctions to substitution functors induced by display maps. Is there a similar categorical description of W-types (and maybe M-types while we are at it)? - -REPLY [3 votes]: Also, this may be helpfull — -http://ncatlab.org/nlab/show/polynomial+functor -http://ncatlab.org/nlab/show/W-type<|endoftext|> -TITLE: Cech nerve as homotopy colimit? -QUESTION [5 upvotes]: Given a category $\mathcal{C}$ with a notion of covering $\{ U_{i} \rightarrow X \}$ for an object $X$ (say $\mathcal{C}$ is a Grothendieck site), we can form the Cech nerve -$$ \cdots \coprod_{i}{U_{ijk}} \substack{\rightarrow \\ \rightarrow \\ \rightarrow}\coprod_{i}{U_{ij}} \rightrightarrows \coprod_{i} U_{i}$$ -(In the notation, I've suppressed degeneracy maps going from right to left.) -This can be viewed in two ways. 1. As a simplicial object in simplicial presheaves, by considering each $U_{i}$ as a simplicial presheaf constant in the simplicial direction. I'll denote that by $U_{\bullet}$. 2. As a simplicial object in presheaves and hence a simplicial presheaf. I'll denote that by $\check{U}_{\bullet}$. -The latter $\check{U}_{\bullet}$ can be shown to be level-wise weakly equivalent to $\operatorname{colim}(\coprod_{i}{U_{ij}} \rightrightarrows \coprod_{i} U_{i})$, where we consider this as a simplicial presheaf constant in the simplicial direction. -On the other hand, we could compute $\operatorname{hocolim}(U_{\bullet})$, and from things I read, this is supposed to be identified with/weakly equivalent to $\check{U}_{\bullet}$. One reason I am having difficulty seeing this is that I don't really understand $\operatorname{hocolim}(U_{\bullet})$. Since each object in $U_{\bullet}$ is cofibrant, I would guess I could just take the usual colimit, but this seems to produce something constant in the simplicial direction, which is clearly wrong. -So the question is: -How to see if there is a weak equivalence $\operatorname{hocolim}(U_{\bullet}) \rightarrow \check{U}_{\bullet}$? -Probably if I read through the many pages of material suggested in the comments to this question, I'd be able to figure this out. But a more direct answer would make that reading more fruitful for me, I think. At least, pointing out what things I need to know to figure this out would help. - -REPLY [3 votes]: More generally, let $X_\cdot$ be a simplicial presheaf. As such, we can consider it as a simplicial object in presheaves, which in particular may be thought of as a simplicial object in simplicial presheaves $X'_\cdot.$ So we have: -$$X_\cdot:\Delta^{op} \to Set^{C^{op}}$$ and $$X'_\cdot =\left( \mspace{3mu} \cdot \mspace{3mu}\right)^{(id)} \circ X_\cdot:\Delta^{op} \to Set_{\Delta}^{C^{op}}$$ where $$\left( \mspace{3mu} \cdot \mspace{3mu}\right)^{(id)}:Set^{C^{op}} \to Set_{\Delta}^{C^{op}}$$ is the evident inclusion of presheaves into simplicial presheaves. -The homotopy colimit of $X'_\cdot$ in simplicial presheaves is computed "object-wise". Hence, for all $c \in C$ we have $$hocolim\left(X'\right)(C)=hocolim \left( X'\left(C\right)\right).$$ The right-hand side is the homotopy colimit of a simplicial object in simplicial sets, and can be computed by taking the diagonal of the corresponding bisimplicial set. But the diagonal of $X'\left(C\right)_\cdot$ is simply $X\left(C\right)_\cdot$ since $X'_\cdot$ ` in constant in one simplicial direction. Hence $$hocolim\left(X'\right)=X.$$<|endoftext|> -TITLE: Sampling uniformly from a sphere -QUESTION [7 upvotes]: Let $B^{n} _p= ${$ (x_1, \dots, x_n) : |x_1|^p + \dots |x_n|^p = 1 $} be the unit ball in $\mathbb{R}^n$ in the $\ell^p$ norm. -If $X_1,\dots,X_n$ are iid $\exp(1)$ -distributed random variables, then $(X_1/D,\dots,X_n/D)$, where $D =X_1+ \dots + X_n $ is uniformly distributed in $B^{n}_1$. -If $X_1,\dots,X_n$ are iid normally distributed with mean 0, then $(X_1/D,\dots,X_n/D)$, where $D = (X_1^2+\dots+X_n^2)^{1/2}$, is uniformly distributed in $B^{n}_2$. -Is there a choice of $X_1,\dots , X_n$ iid such that -$ ( X_1 / D, \dots, X_n/D)$, where $D = (|X_1|^p + \dots + |X_n|^p)^{1/p} $ is uniformly distributed in $B^{n} _p$ for arbitrary $p$? -I would be happy with any sensible common generalization of the two statements above. I have no particular reason to believe there is such a generalization - I'm just hoping that two so similar and neat examples have similarly nice generalizations. - -REPLY [9 votes]: If by uniform measure you mean $(n-1)$-dimensional Hausdorff measure on the sphere, the answer is no. As a consequence of the results of this paper by Barthe, Csörnyei, and Naor, under mild regularity assumptions the only measure on the boundary of any convex body which can be generated in this way is the "cone measure" on the $\ell_p$ sphere for $1 \le p < \infty$, which coincides with uniform measure only for $p=1,2$.<|endoftext|> -TITLE: The coproduct on the cohomology of a Hopf algebra -QUESTION [10 upvotes]: If $H$ is a Hopf algebra over a field $k$, the Hopf algebra cohomology $\mathrm{Ext}_H(k,k)$ is —as usual— an algebra, but the Hopfness of $H$ turns it into a Hopf algebra. - -Is there a reference on this Hopf structure on $\mathrm{Ext}_H(k,k)$? - -I am hoping someone wrote down all the basic details about this so as to avoid doing it myself... - -REPLY [11 votes]: Well, this might not be the answer you expected: - -In general, there is no coproduct such that $Ext_H^\ast(k,k)$ (cup product) is a graded Hopf algeba. - -For, let $k$ be a perfect field and suppose $A = Ext_H^\ast(k,k)$ is a graded Hopf algebra of finite type. By Borel's structure theorem on connected graded commutative Hopf algebras [A-M, VI.2.8], $A$ is (as $k$-algebra) isomorphic -to the tensor product of algebras of the types $k[x]$ and $k[y]/(y^r)$. In particular, $A/rad(A)$ is a domain. -Now it's easy to find counterexamples. For instance let $\text{char}(k)=2$. Then the cohomology ring of the dihedral group $H^\ast(D_8;k) = k[x,y,z]/(xy),\;|x|=|y|=1, |z|=2$ has zero-divisors, but its radical is zero. -More generally: If $\text{char}(k) = p$ and $G$ is a $p$-group with at least two conjugacy classes of maximal elementary abelian subgroups, then there are non-nilpotent classes $x,y \in H^\ast(G;k)$ such that $xy=0$. Thus $Ext_{k[G]}^\ast(k,k) = H^\ast(G;k)$ can't be a Hopf algebra. - -However, if $H$ is commutative, then the product induces a coproduct that makes $Ext_H^\ast(k,k)$ a commutative, cocommutative Hopf algebra. I guess searching for a reference will probably last much longer than the straightforward proof: Let $P \to k$ be a projective resolution over $H$. Then the operations on $H$ and the uniqueness of induced mappings (up to homotopy) induce a (kind of) DG-Hopf algebra $(P,\mu,\Delta)$ where the usual diagramms commute up to homotopy. Passage to cohomology now makes the diagramms commute and you have your Hopf algebra. -Note: $(P,\mu,\Delta)$ can be seen as an algebraic $H$-space analog. - -One may wonder why the coproduct $\Delta: H \to H \otimes H$ always induces a product on $Ext_H^\ast(k,k)$ while the product $\mu: H \otimes H \to H$ induces a coproduct only if $H$ is commutative. The reason is that $\Delta$ is an algebra homomorphism while $\mu$ is an algebra homomorphism if and only if $H$ is commutative. - -[A-M] Adem, Milgram: Cohomology of finite groups.<|endoftext|> -TITLE: Continuous bijections vs. Homeomorphisms -QUESTION [40 upvotes]: This is motivated by an old question of Henno Brandsma. -Two topological spaces $X$ and $Y$ are said to be bijectively related, if there exist continuous bijections $f:X \to Y$ and $g:Y \to X$. Let´s denote by $br(X)$ the number of homeomorphism types in the class of all those $Y$ bijectively related to $X$. -For example $br(\mathbb{R}^n)=1$ and also $br(X)=1$ for any compact $X$. Henno´s question was about nice examples where $br(X)>1$. The list wasn´t too long, but all the examples in there also satisfied $br(X) \geq \aleph_0$. So here is my question: - -Is there a topological space $X$ for which $br(X)$ is finite and bigger than $1$? - -REPLY [17 votes]: Sorry to be coming to the conversation so late (over 3 years late!), but I'm new to MO and just saw it. I've done some research on this very topic, so I couldn't resist sharing. -Let me start by saying that I don't know a complete answer to the question as stated. By I do have a partial answer, some other results that are relevant to the question, and an answer to the question of Doyle and Hocking mentioned in Michal's answer. It's all too long for a comment, so here we go . . . - -The following theorem answers the Doyle and Hocking question, and also tells us that any example answering Ramiro's question affirmatively will be at least a little bit exotic. -Theorem: Suppose $X$ is a sequential space. Then either $br(X) = 1$ or $br(X) \geq 2^\mathfrak{c}$. -Proof: Let $Y$ be a space bijectively related to $X$, and let $f: Y \to X$ be a continuous bijection. Without loss of generality, we may assume that $Y$ and $X$ are actually just two topologies on the same set $A$ (because $f$ is a bijection), and $X$ is a refinement of $Y$ (because $f$ is continuous). Let us assume that $Y$ is strictly finer than $X$ (this must be true, in particular, if $Y$ is not homeomorphic to $X$). -Since $X$ is sequential and $Y$ is strictly finer than $X$, there is some sequence $\langle x_n \rangle$ such that $x_n \to x$ in $X$ but $x_n \not\to x$ in $Y$. -In $Y$, since $x_n \not\to x$, there is an infinite $S \subseteq \{x_n : n \in \omega\}$ such that $x \notin \overline{S}$. Thus (passing to a subsequence if necessary) we may assume that $x_n \to x$ in $X$ but $x \notin \overline{\{x_n:n \in \omega\}}$ in $Y$. -Let $\mathcal F$ be any free filter on $\omega$. Define the topology $Z_{\mathcal F}$ on $A$ as follows (we will define the space by defining its closure operator). If $B$ is any subset of $A$, then $\overline{B}$ is the same in $Z_{\mathcal F}$ as in $X$, except possibly for the point $x$. Then we put $x \in \overline{B}$ if and only if either $x \in B$, $x \in \overline{B \setminus \{x_n:n \in \omega\}}$, or $\{n : x_n \in B\} \in \mathcal F^+$. -It's easy to check that this defines a topology on $A$, and that this new topology is finer than $X$ and coarser than $Y$. This means that the identity on $A$ gives continuous bijections $Y \to Z_{\mathcal F} \to X$. So $Z_{\mathcal F}$ is bijectively related to $X$ for any filter $\mathcal F$. -Now let's check that we get $2^\mathfrak{c}$ non-homeomorphic spaces this way. Consider the following topological invariant: -$$Filt(Z) = \{\mathcal F : \mathcal F \text{ is a filter and, for some countable set } \{x_n: n \in \omega\},\text{ and some } x, \ \mathcal F = \{\{n : x_n \in U\}: U \text{ open and } x \in U\}\}$$ -In other words, $Filt(Z)$ is just a list of all the filters that describe the convergence of some countable set to some point. Since $X$ is sequential, $Filt(X)$ is just the filter of cofinite sets. Given how we've defined $Z_{\mathcal F}$, $Filt(Z_{\mathcal F})$ is equal to either all isomorphs of $\mathcal F$ (if $X$ has one non-isolated point) or the filter of cofinite sets plus all isomorphs of $\mathcal F$ (if $X$ has more than one non-isolated point). -By an "isomorph" of $\mathcal F$ I mean every filter that can be obtained from $\mathcal F$ by a permutation of $\omega$. This class has size $\mathfrak{c}$ for any fixed $\mathcal F$. But the number of filters on $\omega$ is $2^\mathfrak{c}$, so $Filt(Z_{\mathcal F})$ is different for $2^\mathfrak{c}$ different choices of $\mathcal F$. Since it's a topological invariant, we have lots of different spaces. -QED. - -Next, let's observe that for every infinite $\kappa$ there is some $X$ with $br(X) = \kappa$. -Fix two ultrafilters on $\omega$, say $p$ and $q$. Let $X_p = \omega \cup \{*\}$ be the space obtained by making every point of $\omega$ isolated, and making the neighborhoods of $*$ take the form $A \cup \{*\}$ for $A \in p$. Define $X_q$ and $X_{p \cap q}$ similarly (note that $p \cap q$ is a filter, so this definition still makes sense). -Let $\mathcal X$ be the topological space obtained by taking the disjoint sum of $\aleph_\kappa$ copies of $X_p$, $\aleph_\kappa$ copies of $X_q$, $\aleph_\kappa$ copies of $X_{p \cap q}$, and $\aleph_\kappa$ singletons. -As in the proof of the previous theorem, if $\mathcal Y$ is bijectively related to $\mathcal X$, then we may assume that $\mathcal Y$ is just a refinement of $\mathcal X$. By refining $\mathcal X$, we can obtain any space of the following form (for any $\mu,\lambda,\theta \leq \kappa$): a disjoint sum of $\mu$ copies of $X_p$, $\lambda$ copies of $X_q$, $\theta$ copies of $X_{p \cap q}$, and $\aleph_\kappa$ isolated points. -Using the fact that the only extensions of the filter $p \cap q$ are $p$ and $q$, it is not too hard to check that these are the only spaces possible. -Now suppose we also have $\theta = \aleph_\kappa$. There are $\kappa$ possibilities for choosing $\mu$ and $\lambda$, yielding $\kappa$ spaces of this form. But now we may refine again to get $\mathcal X$ back (leave $\aleph_\kappa$ of the $X_{p \cap q}$ alone, refine $\aleph_\kappa$ of them to $X_p$, and refine another $\aleph_\kappa$ of them to $X_q$). - -Now I'll give a really nice example of a space $X$ where $br(X) \geq \mathfrak{c}$, namely the Baire space. -Unfortunately, I don't know a really slick proof of this assertion. I'll have to refer you to a paper of mine for details. Below I'll list a few related results that are also relevant to your question. For most of the results about non-separable spaces, see my other paper, joint with Arnie Miller. -I studied the "bijectively related" relation on (nonempty) perfect completely ultrametrizable spaces (henceforth, PCU spaces). -[Sidebar: At first this may seem like a weirdly specific class to look at. The following folklore result might clear things up a bit: $X$ is a PCU space if and only if there is some pruned, perfect tree whose end space is homeomorphic to $X$. So you can think of these as "tree spaces." The original intent of the two papers above was to explore how the structure of the trees can be leveraged to prove topological results.] -The separable PCU spaces are precisely the perfect, zero-dimensional Polish spaces (see Chapter 2 of Kechris's book for more on this). For this class, we have a pretty good idea of what your relation looks like: -Theorem: The separable PCU spaces are partitioned into exactly three equivalence classes by the relation described in your question: the class of the Cantor space, and the class of the Cantor space minus a point, and the class of the Baire space. -The latter two classes each have $\mathfrak{c}$ members. This result isn't in my paper, but it's a good exercise. -Once you move to non-separable PCU spaces, it is consistent that the situation stays "nice" like it is for Polish spaces. For example, -Theorem: If CH holds, then there are exactly four equivalence classes of PCU spaces of size $\mathfrak{c}$. -More generally, -Theorem: It is consistent with ZFC that $\mathfrak{c} = \aleph_n$ and that there are exactly $n+3$ equivalence classes of size-$\mathfrak{c}$ PCU spaces. -However, if $\mathfrak{c} = \aleph_2$ then the number of equivalence classes is independent of ZFC: it can be $5$ by the above theorem, but it is more if MA holds (I don't know exactly how many). -For cardinals below $\aleph_\omega$, there is a strong connection between these equivalence classes and the question of whether you can partition the Cantor space into $\kappa$ closed sets for $\kappa < \mathfrak{c}$. -I don't know much about what happens after $\aleph_\omega$, and I consider it a very interesting problem. - -Two other papers of mine you might want to look at: this one deals with what happens when you refine a topology by just a little bit, and this one (joint with Chris Good, Robin Knight, and Dave McIntyre) deals with finite intervals in the lattice of topologies. (Notice that, by our arguments above, $X$ and $Y$ are bijectively related if and only if there is an interval in the lattice of topologies, with the top and bottom spaces both homeomorphic to $X$, and some space in between homeomorphic to $Y$. But then everything in the interval is also bijectively related to $X$ and $Y$. So one strategy for answering Romiro's question could be to find a finite interval like this).<|endoftext|> -TITLE: Cohomology of $T^n/W$ for compact Lie groups -QUESTION [11 upvotes]: Let $G$ be a compact, connected and simply connected Lie group. -Let $T\subset G$ be a maximal torus and let $W$ be the corresponding -Weyl group. -Then we have the diagonal action of $W$ on $T^{n}$ for $n\ge 0$. -I would like to know if the cohomology groups $H^{*}(T^{n}/W;\mathbb{Z})$ -have been computed or if anything is known about them. -The case $n=1$ is particularly simple as $T/W$ is contractible -when $G$ simply connected. However, if $n>1$, one gets a complicated space. Any ideas? - -REPLY [4 votes]: The case n=2 is well studied and has beautiful anwser. This quotient space is a weighted projective space (and for $SU(n)$ it is $CP^{n-1}$). See E. Looijenga, Root systems and elliptic curves, Invent. Math. 38 (1976), 17–32. And I.N. Bernshtein and O.V. Shvartsman, Chevalley’s theorem for complex crystallographic Cox- -eter groups, Funct. Anal. Appl. 12 (1978), 308–310. As pointed out in Sean's answer these quotient spaces are moduli spaces of flat $G$ connections in the trivial $G$ bundle over the torus. It is interesting to extend this story to non-trivial $G$ bundles. A wonderful book about some of what goes on "Almost commuting elements in compact Lie groups" by Borel, Friedman and Morgan.<|endoftext|> -TITLE: Does the Hirsch conjecture hold for $n < 2d$? -QUESTION [7 upvotes]: The Hirsch conjecture asserts that the graph (i.e. $1$-skeleton) of a $d$-dimensional convex polytope with $n$ facets has diameter at most $n - d$. -After being open for decades, Francisco Santos has recently proved that this fails in general. - -Is it possible that the conjecture - holds for $n < 2d$? Santos's counterexample had $(n,d) = (86, 43)$. - -One observation which may be relevant: If $n < 2d$, then every pair of vertices has a common facet. One can use this to show that the general Hirsch conjecture reduces to the $n \ge 2d$ case, (see Ziegler's book Lectures on Polytopes, p. 84). But this doesn't seem to answer the question here. - -REPLY [13 votes]: The answer is no, as follows from the following Lemma of Klee and Walkup: -Lemma: If P is a d-polytope with n facets and we perform a "wedge" over any facet F we get a (d+1)-polytope P' with n+1 facets and with diameter(P') $\ge$ diameter(P). -Corollary: since there is a 43-polytope with 86 facets and diameter (at least) 44, for every positive integer k there is a (43+k)-polytope with 86+k facets and diameter at least 44. These polytopes have n<2d. -I take the occasion to announce an update. My recent paper http://arxiv.org/abs/1202.4701 (joint with Matschke and Weibel) contains smaller counter-examples to the Hirsch conjecture. The current record is a polytope with dimension 20, 40 facets, and diameter 21.<|endoftext|> -TITLE: Is every sigma-algebra the Borel algebra of a topology? -QUESTION [83 upvotes]: This question arises from the excellent question posed on math.SE -by Salvo Tringali, namely, Correspondence -between Borel algebras and topology. -Since the question was not answered there after some time, I am -bringing it up here on mathoverflow in the hopes that it may find an answer here. -For any topological space, one may consider the Borel sets of the -space, the $\sigma$-algebra generated by the open sets of that -topology. The question is whether every $\sigma$-algebra arises in -this way. -Question. Is every $\sigma$-algebra the Borel algebra of a -topology? -In other words, does every $\sigma$-algebra $\Sigma$ on a set $X$ -contain a topology $\tau$ on $X$ such that $\Sigma$ is the $\sigma$ -algebra generated by the sets in $\tau$? -Some candidate counterexamples were proposed on the math.SE -question, but ultimately shown not to be counterexamples. For -example, my -answer there shows that the collection of Lebesgue -measurable sets, which seemed at first as though it might be a -counterexample, is nevertheless the Borel algebra of the topology -consisting of sets of the form $O-N$, where $O$ is open in the -usual topology and $N$ is measure zero. A proposed counterexample -of Gerald Edgar's there, however, remains unresolved. And I'm not clear on the status of a related proposed counterexample of George Lowther's. -Meanwhile, allow me to propose here a few additional candidate -counterexamples: - -Consider the collection $\Sigma_0$ of eventually periodic subsets of -$\omega_1$. A set $S\subset\omega_1$ is eventually periodic if -above some countable ordinal $\gamma$, there is a countable length pattern which is simply repeated up to -$\omega_1$ to form $S$. This is a $\sigma$-algebra, since it is closed under -complements and countable intersections (one may find a common -period among countably many eventually periodic sets by intersecting the club sets consisting of starting points of the -repeated pattern). -Consider the collection $\Sigma_1$ of eventually-agreeing -subsets of the disjoint union $\omega_1\sqcup\omega_1$ of two copies of $\omega_1$. That is, -sets $S\subset \omega_1\sqcup\omega_1$, such that except for -countably many exceptions, $S$ looks the same on the first copy as it does on the -second. Another way to say it is that the symmetric difference of -$S$ on the first copy with $S$ on the second copy is bounded. This is a $\sigma$-algebra, since it is closed under complement and also under countable intersection, as the countably many exceptional sets will union up to a countable set. - -Please enlighten me by showing either that these are not actually -counterexamples or that they are, or by giving another -counterexample or a proof that there is no counterexample. -If the answer to the question should prove to be affirmative, but only via strange or unattractive topologies, then consider it to go without saying that we also want to know how good a topology can be found (Hausdorff, compact and so on) to generate the given $\sigma$-algebra. - -REPLY [52 votes]: Unfortunately, I can only provide a reference but no ideas since I don't have the paper. -In "On the problem of generating sigma-algebras by topologies", Statist. Decisions 2 (1984), 377-388, Albert Ascherl shows (at least according to the summary to be found on MathSciNet) -that there are $\sigma$-algebras which can't be generated by a topology. -Robert Lang (same journal 4 (1986), 97-98) claims to give a shorter proof. -As suggested by Joel, I add the ideas of Lang's example. The underlying space is -$\Omega= 2^{\mathbb R}$, that is the space of all indicator functions, and the $\sigma$-algebra is $\mathcal A = \bigotimes_{\mathbb R} \mathcal P$ where $\mathcal P$ -is the power set of the two element set. It is generated by the system $\mathcal E$ of the "basic open sets" of the product topology (prescribed values in a finite number of points). -This generator has the cardinality $c$ of the continuum and since the generated $\sigma$-algebra can be obtained in $\omega_1$ (transfinite) induction steps the cardinality -of $\mathcal A$ is also $c$. On the other hand, if $\mathcal T$ is a topology with $\mathcal A=\sigma(\mathcal T)$ then $\mathcal T$ separates points (this should follow from the "good sets principle"), in particular, for two distinct points of $\Omega$ the closures of the -corresponding singletons are distinct. Hence $\mathcal T$ has at least $|\Omega|=2^c$ -elements.<|endoftext|> -TITLE: Does Rolle's Theorem imply Dedekind completeness? -QUESTION [10 upvotes]: I think the answer to the title question is "yes", but Gerald Edgar, in his comment on Does antidifferentiability of continuous functions imply Dedekind completeness? , points out an article (actually a solution to a Monthly problem) in which M. J. Pelling claims to construct a non-Archimedean ordered field in which Rolle's Theorem holds; see http://www.jstor.org/pss/2321145 . -Since I'm working on an article for the Monthly ( http://jamespropp.org/reverse.pdf ) that claims to prove that the Mean Value Theorem (which follows from Rolle's Theorem) implies Dedekind completeness, I'd like to know who's right! -First, I'll show that the Mean Value Theorem implies the Constant Value Theorem. Then I'll show that the latter implies the Cut Property. Then I'll show that the Cut Property implies Dedekind completeness. And then one of you will tell me where my mistake is, or where Pelling's mistake is. -Step 1: Suppose $f$ is a continuous function on $[0,1]$ that is differentiable on $(0,1)$, with $f'(x)=0$ for all $x$ in $(0,1)$. We'd like to show that $f$ is constant. Suppose not; consider any $a > 0$ with $f(a) \neq f(0)$. Then (by the Mean Value Theorem) there must exist $c$ in $(0,a)$ with $f'(c) = (f(a)-f(0))/(a-0) \neq 0$, which is a contradiction. Hence the Constant Value Theorem holds. -Step 2: Suppose $A$ and $B$ are disjoint sets whose union is the whole ordered field, such that $a < b$ for all $a \in A$, $b \in B$. We'd like to show that there exists $c$ in the ordered field such that $a \leq c$ for all $a \in A$ and $b \geq c$ for all $b \in B$. Suppose not. Then for all $a \in A$ there exists $a' \in A$ with $a' > a$, and for all $b \in B$ there exists $b' \in B$ with $b' < b$, from which it follows that the indicator function of the set $A$ is continuous and differentiable everywhere, with derivative constantly 0. This contradicts the Constant Value Theorem. Hence the Cut Property holds. -Step 3: Let $S$ be a bounded non-empty subset of the ordered field. We'd like to show that $S$ has a least upper bound. Let $B$ be the set of upper bounds of $S$, and let $A$ be the complement of $B$. Then $A$ and $B$ satisfy the hypotheses of the Cut Property, so there exists $c$ such that $a \leq c$ for all $a \in A$ and $b \geq c$ for all $b \in B$. It is easy to check that $c$ is a least upper bound for $S$. -This argument seems fairly straightforward to me, so I'm inclined to think that the error must be in Pelling's more complicated argument (which I won't pretend to understand). Though perhaps it will turn out that Pelling and I mean different things when we say than an ordered field satisfies Rolle's theorem. -Can anyone shed light on this? - -REPLY [7 votes]: You may want to take a look at the article Ordered fields satisfying Rolle's theorem by Brown, Craven and Pelling. Among other things they characterize those fields for which Rolle's theorem holds for polynomials and for rational functions. In the second case the characterization is beautiful: those are exactly the real-closed fields. Anyway, I agree with you in that Pelling is talking about Rolle's theorem for polynomials in his Monthly's article.<|endoftext|> -TITLE: Characteristic classes of a fibered sum -QUESTION [5 upvotes]: I will phrase this question in terms of attaching smooth manifolds along a submanifold, though it is certainly more general. -Let $M_1$ and $M_2$ be smooth $n$-manifolds (maybe closed, for simplicity), $N$ a closed $k$-manifold, $D$ a closed $(n-k)$-disk bundle over $N$ (so that $D$ is an $n$-dimensional manifold whose boundary is the associated $(n-k-1)$-sphere bundle), and suppose we have embeddings $f_i:D\rightarrow M_i$. -Now say $\tilde{M_i}=M_i\setminus f(N)$, let $E_0$ be the punctured-open-disk bundle associated with $D$, and let $\alpha:E_0\rightarrow E_0$ be defined by sending $v_x$ in the fiber of $x\in N$ to the point $(1-|v_x|)\frac{v_x}{|v_x|}$ in the same fiber (intuitively, $\alpha$ turns $E_0$ inside-out so that we can attach the manifolds with a "collar"). -Then if we form the Topological pushout of $\tilde{M_1}$ and $\tilde{M_2}$ using the smooth embeddings $f_1|_{E_0}$ and $f_2|_{E_0}\circ \alpha$, this in fact produces a pushout in the smooth category. The resulting manifold $M$ then has a tangent bundle, and in fact this tangent bundle can be formed by attaching the tangent bundles of $M_1$ and $M_2$ using the same recipe. -So, finally, here is the question: is there a formula for characteristic classes of $M$ (maybe just restrict attention to Stiefel-Whitney, Chern, Pontryagin classes) in terms of the characteristic classes of $M_1$, $M_2$, and $N$ (and the embeddings $f_1$, $f_2$)? More generally, is there a similar formula for attaching arbitrary bundles over arbitrary topological spaces (i.e. we would form a topological pushout on the base spaces and indicate how the fibers would be identified over points that are glued together)? - -REPLY [5 votes]: It depends what you mean by "formula", since you are talking about cohomology classes in different manifolds, so at the very least you need a way to relate them, which depends on context. So the "answer" is the Mayer-Vietoris sequence. Naturality of characteristic classes, together with the fact that the characteristic classes of (the tangent bundle of) a disk bundle $D(E)\to M$ can be computed using $TD=TM\oplus E$ gives you some information which you then have to put together. -For characteristic numbers, there are formulas, eg Novikov additivity gives a formula for the signature of $M$ in terms of those of $M_1, M_2$ and $D$. Similarly for the Euler characteristic. If you glue $D\times I$ to $(M_1\cup M_2)\times I$ you get a bordism from $M_1\cup M_2$ to $M$ and so bordism invariance of characteristic numbers can be helpful.<|endoftext|> -TITLE: Example of a compact Kähler manifold with non-finitely generated canonical ring? -QUESTION [31 upvotes]: A celebrated recent theorem of Birkar-Cascini-Hacon-McKernan and Siu says that the canonical ring $R(X)=\oplus_{m\geq 0}H^0(X,mK_X)$ of any smooth algebraic variety $X$ over $\mathbb{C}$ is a finitely generated $\mathbb{C}$-algebra. -On the other hand P.M.H. Wilson (using a construction of Zariski) gave an example of a compact complex manifold $X$ with $R(X)$ not finitely generated. However his manifold $X$ is not Kähler. -Does anyone know an example of a compact Kähler manifold $X$ with $R(X)$ not finitely generated? Or is this an open problem? - -REPLY [7 votes]: As pointed out by Ruadhai Dervan in the comments, a paper by Fujino contains the answer to this question: the canonical ring of any compact Kähler manifold is finitely generated. By bimeromorphic invariance of this ring, the result even holds for compact complex manifolds in Fujiki's class $\mathcal{C}$. -The idea is to consider the Iitaka fibration of the manifold, which has the obvious property that its base is always a projective variety. Thanks to Fujino-Mori finite generation upstairs can be deduced from finite generation downstairs (with a boundary divisor term), and this latter statement follows from BCHM. The details are in the paper of Fujino cited above.<|endoftext|> -TITLE: What is the name for a finite-group representation that is the sum of all the irreducibles (once)? -QUESTION [5 upvotes]: I vaguely remember seeing a paper studying the concept of a totally multiplicity-one representation of a finite group, which concept, I recall, had a particular name, which I forget. What is this name, and is there a reference paper (for example, the one I might have been reading) where I can find out what is known about these representations? - -REPLY [9 votes]: As others have commented, the most likely answer to the question in the header is Gelfand model, though you are apparently looking further for related literature. Probably the notion developed in the 1970s and 1980s in a series of papers by I.M. Gelfand and his collaborators. These are in Russian, but mostly published also in English translation journals and latter reprinted in the multi-volume collected works. -It's clear that the ideal notion of "model" of representations for a given group is a representation containing each irreducible representation exactly once as a constituent. But in some situations this is weakened to require only that "most" representations occur and/or that "most" of them have multiplicity just one. The groups studied usually are of Lie type (finite, compact, etc.) or perhaps closely related (symmetric groups, for example). And the results are varied, some more computable or usable than others. Finding a model for a specific group is a natural goal but difficult to reach. -MathSciNet and other databases provide good references and sometimes reviews. As a sample, here is a concise author summary which illustrates some of the typical motivation: -Gel′fand, I. M.; Zelevinski ��ı, A. V. [Zelevinsky, Andrei], Models of representations of classical groups and their hidden symmetries. Akad. Nauk SSSR Inst. Prikl. Mat. Preprint 1984, no. 71, 26 pp. -Authors’ summary: “For all complex classical groups G we construct new realizations of the representation model of G, i.e., the direct sum of all irreducible algebraic finite-dimensional representations of G occurring with multiplicity one. These realizations have hidden symmetry: the action of the Lie algebra of G on them extends naturally to the action of a larger Lie -(super-) algebra. The construction of hidden symmetries is based on a geometrical construction, similar to a twistor construction of Penrose.” -Another paper is part of a series in Russian by Bernstein-Gelfand-Gelfand: -Models of representations of compact Lie groups. (Russian) Funkcional. Anal. i Prilozˇen. 9 (1975), no. 4, 61–62. -[ADDED] Maybe I should emphasize that constructing an abstract model of representations for a group isn't by itself the goal, since it may be too big to provide further insight. As Junkie indicates (with reference to a paper that is also on the arXiv), symmetric groups and other finite Coxeter groups have been studied in this framework with partial success. For finite groups of Lie type, the ideas of Gelfand-Graev led to a simple construction which is "almost" a model of the ordinary characters but doesn't capture all of them. This is developed by Carter in his 1985 book Finite Groups of Lie Type, Chapter 8, and less completely by Digne-Michel in their 1991 text Representations of Finite Groups of Lie Type. Starting with any "regular" character of a maximal unipotent subgroup (the choices here don't matter), induction to the whole finite group yields a multiplicity-free character of large degree which has most other characters as constituents. Taken in isolation this is not so helpful, but combined with Deligne-Lusztig theory it leads to interesting results. -An underlying theme for groups of Lie type is that's it's fairly easy to construct big representations by induction from nice subgroups, but then it's not so easy to extract information about the irreducibles or multiplicities.<|endoftext|> -TITLE: Is the radical of an irreducible ideal irreducible? -QUESTION [39 upvotes]: I originally posted this to math.stackexchange.com -here. I got a partial answer, but I now suspect that the complete answer is much harder than I thought, so I'm posting it here. -Fix a commutative ring $R$. Recall that an ideal $I$ of $R$ is irreducible if $I = J_1 \cap J_2$ for ideals $J_1$ and $J_2$ only when either $I = J_1$ or $I = J_2$. -Question : Assume that $I$ is an irreducible ideal. Must the radical of $I$ be an irreducible ideal? -On math.stackchange.com, I learned that the answer is "yes" if $R$ is Noetherian. My guess is that there is a counterexample if $R$ is not assumed to be Noetherian, but I have no idea how to construct it. - -REPLY [21 votes]: I construct a counterexample for your question in the non-noetherian case: -(1) Let $A = k[[X,Y]]/(XY) = k[[x,y]]$, where $k$ be a field. Notice that $(0) = (x) \cap (y)$ so $(0)$ is reducible in $A$. -(2) We consider the injective hull $E(k)$ of $k$, and set $m \in E(k)$ be the element such that $mA \cong k$. Notice that every non-zero submodule of $E(k)$ contains $mA$ and $(0)$ is irreducible in $E(k)$ -(3) Set $R = A \ltimes E(k)$ be the indealization. We have that $(0 \ltimes E(k))^2 = 0$ so $\sqrt{(0)R} = 0 \ltimes E(k)$ is reducible by (1). -(4) We can prove that for every non-zero element $(a,s)$ of $R$, we have $0 \ltimes mA \subseteq (a,s)R$. So the ideal $(0)$ is irreducible in $R$. -EDIT (13/02): It should be noted that this example is also a counterexample for a non-noetherian ring with an ideal is irreducible but not primary. Indeed, we have $(0)$ is irreducible as above. However -$$(x,0).(y,m) = (0,0) \in R,$$ -and $(x,0)$ and $(y,m)$ are not nilpotent so $(0)$ is not primary in $R$<|endoftext|> -TITLE: Dedekind Zeta function: behaviour at 1 -QUESTION [19 upvotes]: Let $\zeta_F$ denote the Dedekind zeta function of a number field $F$. -We have $\zeta_F(s) = \frac{\lambda_{-1}}{s-1} + \lambda_0 + \dots$ for $s-1$ small. -Class number formula: We have $\lambda_{-1} = vol( F^\times \backslash \mathbb{A}^1)$, where $\mathbb{A}^1$ denotes the group of ideles with norm $1$. - -What is known or conjectured about $\lambda_0$? - -Tate's thesis can be copied word by word for function fields with the same class number formula, so: - -What is known for the zeta function of a function field? - -REPLY [20 votes]: We can actually do a good bit in the function field case, because the zeta function is of the form -$$\zeta_F(s)={P(q^{-s})\over (1-q^{-s})(1-q^{1-s})}$$ -where $P$ is a polynomial of degree equal to twice the genus of the underlying curve. When the genus of the curve is zero (e.g., $F=\mathbb F_q(t)$), $P(x)=1$. In this case, we can calculate the Laurent expansion of $\zeta_F(s)$ to be (using WolframAlpha to avoid thinking) -$${q\over (s-1)(q-1)\log(q)}+{(q-3)q\over 2(q-1)^2}+O(s-1)$$ -For the general case, we can multiply the above by the Laurent expansion for $P(q^{-s})$. For a genus-$g$ curve, the corresponding polynomial is $P(q^{-s})=1+a_1q^{-s}+\ldots+a_{2g}q^{-2gs}$. The Laurent expansion of $P(q^{-s})$ is -$$\big(1+a_1 q^{-1}+\ldots+a_{2g}q^{-2g}\big)-(s-1)\log(q)\big(a_1 q^{-1}+2a_2q^{-2}\ldots+2g\cdot a_{2g}q^{-2g}\big)+O\big((s-1)^2\big)$$ -Multiplying through, we get that the zero-th term in the Laurent expansion of $\zeta_F(s)$, where $F$ is the function field of a genus-$g$ curve, is -$${(q-3)q\over 2(q-1)^2}\cdot P(q^{-1})+{q\over (q-1)\log(q)}\cdot {d\over ds}P(q^{-s})\bigg|_{s=1}$$<|endoftext|> -TITLE: Jacobi's equality between complementary minors of inverse matrices -QUESTION [24 upvotes]: What's a quick way to prove the following fact about minors of an invertible matrix $A$ and its inverse? -Let $A[I,J]$ denote the submatrix of an $n \times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then -$$ |\det A[I,J]| = | (\det A) \det A^{-1}[J^c,I^c]|,$$ -where $I^c$ stands for $[n] \setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\lambda$ and the complete symmetric functions $h_\lambda$. -I also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :( -Edit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias. - -REPLY [10 votes]: Here is a short proof inspired by darij grinberg's answer and not using Schur complements (in particular, no invertibility of sub-matrices or continuity assumptions are needed). -WLOG let $I=J=[k]$. Let $A_i$ denote the $i$th column of a matrix $A$. Consider the product -$$ -A -\left[ -\begin{array}{c|c|c|c|c|c} -e_1 & \cdots & e_k & A^{-1}_{k+1} & \cdots & A^{-1}_n -\end{array} -\right] -= -\left[ -\begin{array}{c|c|c|c|c|c} -A_1 & \cdots & A_k & e_{k+1} & \cdots & e_n -\end{array} -\right] -$$ -which can also be written -$$ -A -\begin{pmatrix} -I_k & A^{-1}[J,I^c] \\ 0 & A^{-1}[J^c,I^c] -\end{pmatrix} -= -\begin{pmatrix} -A[I,J] & 0 \\ A[I^c,J] & I_{n-k} -\end{pmatrix} -$$ -Taking the determinant of both sides yields -$$ (\det A) (\det A^{-1}[J^c,I^c]) = \det A[I,J] $$.<|endoftext|> -TITLE: How to evaluate binomial coefficients efficiently and as correctly as possible? -QUESTION [8 upvotes]: This question is more precisely about evaluation with a computer, of a binomial coefficient of the form $ \binom{x}{m}$ where $x$ is a real number and $m$ a rational integer. -The reason why I ask is that I found out recently that sage is using the naive definition with the $\Gamma$ function, which means that it gets as a result NaN (not-a-number) with quite small parameters, for which the real result is pretty reasonable and should have been given (see the bug report). -I have proposed to change the implementation by returning zero in more cases than it already does, to reduce to a situation $\binom{x}{m}$ with $x\geq m\geq 0$, so we can write $x=m+k+u$ with $k$ a natural integer and $u\in[0;1[$, then computing the quotient $\Gamma(x+1)/\Gamma(m+1)$ with a Pochhammer symbol times the quotient $\Gamma(m+1+u)/\Gamma(m+1)$. For that last quotient, I was proposing a direct computation for small $m$ and a polynomial expansion in $u$ for big $m$. -There are two problems with this approach: - -I don't really know how big the error is, which for a numerical computation is a pretty big issue ; -I used the naive code as a starting point, and added naive ideas to the mix : there may exist better approaches (it's also because of this point that I didn't try to evaluate the error more precisely). - -It would be surprising if there existed no algorithm for this kind of computations, given how important those coefficients are in various situations... - -REPLY [2 votes]: Since you're using sage, I'd write a quick C or Cython function. Use the Arb library (I think this is already included in the latest release of sage) by computing the function in the naive way using the gamma function. -The benefit to doing so, is that the library uses "Ball" arithmetic and automatically finds rigorous error bounds for the computation. -Basically, the way I'd do it is by given a required precision...compute naively the result. If the relative accuracy in terms of bits isn't high enough, increase the working precision and compute again. Continue the process until the result is as accurate as required. Then, convert the arb type result to an mpfr variable by making an appropriate arb function call.<|endoftext|> -TITLE: Is there a formula for the size of Symplectic group defined over a ring $Z/p^k Z$? -QUESTION [6 upvotes]: Is there a formula for the size of Symplectic group defined over a finite ring $Z/p^k Z$? - -REPLY [15 votes]: From Christopher Perez's answer, we have $|Sp_{2n}(\mathbb{Z}/p\mathbb{Z})| = p^{n^2} \prod_{i=1}^n (p^{2i}-1)$. Following Johannes Hahn, we wish to determine the size of the kernel of the homomorphism $Sp_{2n}(\mathbb{Z}/p^k\mathbb{Z}) \to Sp_{2n}(\mathbb{Z}/p\mathbb{Z})$. -To compute this, we may induct on $k$: For $k \geq 1$, elements in the kernel of the homomorphism $Sp_{2n}(\mathbb{Z}/p^{k+1}\mathbb{Z}) \to Sp_{2n}(\mathbb{Z}/p^k\mathbb{Z})$ have the form $I + p^k A$ for a matrix $A = \left( \begin{smallmatrix} E & F \\ G & H \end{smallmatrix} \right)$ with values in $\mathbb{Z}/p\mathbb{Z}$. The symplectic condition (given in Wikipedia) is equivalent to the conditions that $F^t = F$, $G^t = G$, and $E^t + H = 0$. $F$ and $G$ are therefore symmetric matrices, while $H$ and $E$ determine each other, with no further conditions. The kernel of one-step reduction therefore has size $p^{n(n+1)/2} \cdot p^{n(n+1)/2} \cdot p^{n^2}$, or $p^{2n^2 + n}$. -The final answer is therefore: $|Sp_{2n}(\mathbb{Z}/p^k\mathbb{Z})| = p^{(2k-1)n^2 + (k-1)n} \prod_{i=1}^n (p^{2i}-1)$ -Edit: To address kassabov's complaint, I'll explain the calculation in a bit more detail. The symplectic condition on $I + p^k A$ is that $(I + p^k A)^t \Omega (I + p^k A) \equiv \Omega \pmod {p^{k+1}}$, where $\Omega = \left( \begin{smallmatrix} 0 & I_n \\ -I_n & 0 \end{smallmatrix} \right)$. Because $k \geq 1$, we may eliminate the $p^{2k} A^t \Omega A$ term when expanding, to get -$$ \Omega + p^k A^t \Omega + p^k \Omega A \equiv \Omega \pmod {p^{k+1}}$$ -By subtracting $\Omega$ from both sides, we get the conditions I mentioned on the blocks in $A$. As you mentioned, this coincides with the symplectic Lie algebra condition. George McNinch gave an elegant explanation in the comments, but a possibly more pedestrian reason is that $(p^k)$ is a square zero ideal in $\mathbb{Z}/p^{k+1}\mathbb{Z}$, so one has a canonical isomorphism between the kernel of reduction and the Lie algebra tensored with the quotient ring.<|endoftext|> -TITLE: Limits of intrinsically ergodic systems -QUESTION [5 upvotes]: Let $(X_i)$ be a sequence of compact metric spaces and $(f_i)$ a sequence of transitive transformations $f_i:X_i \to X_i$ with $0 < h_{top}(f_i) < \infty$. -The sequence of dynamical systems satifies: - -$X_i \subset X_{i+1}$, $h_{top}(f_i) < h_{top}(f_{i+1}) $; -$X_i$ converges to a compact metric space $X$; -$f_{i+1}\mid_{X_i} = f_i$ for every $i$; -Besides, there is a transformation $f:X \to X$ such that f is transitive, $0 < h_{top}(f) < \infty$ and $f\mid_{X_i} = f_i$. -$h_{top}(f_i)$ converges to $h_{top}(f)$ - -Assume now that the system $(X_i, f_i)$ is intrinsically ergodic for all $i\ge0$, i.e., it has a unique measure of maximal entropy. -QUESTION. Is $(X,f)$ intrinsically ergodic? -(If it helps, each $(X_i,f_i)$ in my set-up is a transitive subshift of finite type (SFT), but $(X,f)$ is not an SFT.) -If the answer is yes, does there exist a natural way to project the (unique) measure of maximal entropy $\mu$ on $X$ onto $X_i$ so that the projection of $\mu$ is the measure of maximal entropy $\mu_i$ on $X_i$? - -REPLY [6 votes]: The answer is no. It's based on a (un?)published example of Crannell, Rudolph and Weiss. -The example is the following shift: $X$ is the subset of $\lbrace 0,\pm 1\rbrace ^{\mathbb Z}$ with the property that $x_k\cdot x_{k+2^n}$ is not allowed to be $-1$ for any values of $k$ and $n$. -What they prove is that there are 2 measures of maximal entropy for $X$: one the Bernoulli (1/2,1/2) measure living on sequences of 0's and 1's; the other the Bernoulli (1/2,1/2) measure living on sequences of 0's and $-1$'s. In fact I showed with Ayse Şahin that these are the unique measures of maximal entropy. -Now if you let $X_i$ be the subset of $X$ where you can't have $i$ consecutive $-1$'s, then $X_i$ is intrinsically ergodic, but $X$ is not.<|endoftext|> -TITLE: Simple proofs for the existence of elliptic curves having a given number of points -QUESTION [16 upvotes]: Yesterday, after he gave a nice talk, Dick Gross and I were chatting and he brought up the following annoying problem: suppose that for $p$ a prime that $H_p$ is the "Hasse interval" $[p+1- 2 \sqrt{p},p+1+2\sqrt{p}]$. Then, for every point $r \in H_p$ there is an elliptic curve $E_{a,b}: y^2 = x^3 + a x + b$ over $\mathbb{F}_p$ such that $N_p(E_{a,b}) = r$, where $N_p(C)$ denotes the number of projective points of the curve $C$. But the only proof that we knew of this fact involved the whole theory of complex multiplication and Deuring's theorems about reduction. So the question arose if there is a simpler proof of this fact, say by using $p$-adic methods. I even asked for the weaker case: let $H_p' = [p+\sqrt{p},p+2\sqrt{p}]$. Can you prove the existence of an $E_{a,b}$ with $N_p(E_{a,b}) \in H_p'$ with a fairly simple proof? -On the converse side, there's Hasse's proof of the Riemann Hypothesis for elliptic curves over finite fields, that $N_p(E_{a,b}) \in H_p$, which does involve a fair amount of machinery (even though it's been simplified). Suppose that we're after the weaker statement: -There are absolute constants $0 < c_1 < c_2$ such that if $y^2 = x^3 + a x + b$ is an elliptic curve over $\mathbb{Q}$ then, for sufficiently large primes $p$ -$c_1 p \le N_p(E_{a,b}) \le c_2 p$. -More generally, if $f(x,y) \in \mathbb{Q}[x,y]$ is an absolutely irreducible polynomial of total degree $d$ that there are $0 < c_1 < c_2$ only depending on $d$ such that -$ c_1 p \le N_p(f) \le c_2 p$ for all sufficiently large primes $p$. -Again, how simple a proof is there for this statement? -When $f(x,y) = a x^2 + b y^2 + c$ which is genus 0, the simplest proof I know of consists in showing -1) If there is a point $P$ in $\mathbb{F}_p^2$ on $f$, then one can explicitly construct a one-to-one correspondence between the projective points on $f$ and the projective line, by using the pencil of lines through $P$. -2) Use the pigeon hole principle to show the existence of a point on $f$: -If $a \ne 0$ there are exactly $(p+1)/2$ values of $a x^2$, so we can see that the intersection $\{ax^2\} \cap \{-(c + by^2)\}$ has at least one point (we just barely made it). -I know of no such simple proof for an elliptic curve $E$. - -REPLY [2 votes]: You can compute the average and standard deviation of the number of points on y^2=f, -choosing f randomly. That should give you (something close to) what you wanted about H_p': -there must be an elliptic curve with that number of points, otherwise standard deviation would be too small. (Also, you can always twist so you have more than p+1 points rather than less.)<|endoftext|> -TITLE: Geometric interpretation of $2A$ conjugacy class in Conway group $Co_1$ -QUESTION [5 upvotes]: I am struggling with following problem. Consider $2A$ class in $Co_1$ having $819*759*75$ elements. Each element $a$ from $2A$ have two representatives in $Co_0$. Element $a$ corresponds to $E_8$ sublattice in Leech lattice defined as $\{v: av=-v\}$ where I call by $a$ also proper preimage in $Co_0$. Now the opposite: having $E_8$ sublattice $L$ in Leech lattice I can find element $a_L$ in $2A$ class. -My goal is to find relation between $Order(ab)$ for $a,b$ in $2A$ and corresponding geometry of two $E_8$ sublattices. The $Order(ab)$ can be $2,3,4,5,6$. (BTW in 2017 I have found this relation already). -Take any other sporadic group $g$ and certain conjugacy class $C$ of involutions. The possible values of $Order(ab)$ for $a,b$ in $C$ can be obtained from character table. Can this help to connect some "lattice" to the group $g$ ? - -REPLY [4 votes]: Here is the answer for $Co_1$ which I obtained in GAP. -For given $2A$ involution $a$ in $Co_1$ the number $n$ of such $2A$ involutions $b$ that product $ab$ is in conjugacy class $C$ is presented in format $[C, n]$. I call hook the order of product. -$[ [ "1a", 1 ], [ "2a", 12870 ], [ "2c", 60480 ], [ "3b", 573440 ], [ "3d", 491520 ], [ "4a", 34560 ], [ "4c", 4838400 ], [ "4d", 2419200 ], [ "5b", 12386304 ], [ "6e", 25804800 ] ]$ -Next the relation between two $E_8$ sublattices $L_a$ and $L_b$ of Leech lattice is presented in following format. -[hook of a and b, number of vectors belonging to intersection of $L_a$ and $L_b$, dimension of vector space generated by both $L_a$ and $L_b$, $[k,l]$ where $k$ is number of vectors in $L_a$ having exactly $l$ perpendicular vectors in $L_b$. -First two rows correspond to $2A$ hook in $Co_1$. These are two cases in $Co_0$. For given $E_8$ sublattice there are $270$ sublattices perpendicular to it (first row) and $12600$ intersecting in $D_4$ sublattice (second row). -$[ 2, 0, 16, [ [ 240, 240 ] ] ]$ -$[ 2, 24, 12, [ [ 192, 84 ], [ 24, 126 ], [ 24, 240 ] ] ]$ -$[ 2, 4, 14, [ [ 48, 84 ], [ 132, 126 ], [ 60, 240 ] ] ]$ -$[ 3, 0, 16, [ [ 240, 126 ] ] ]$ -$[ 3, 6, 14, [ [ 162, 84 ], [ 78, 126 ] ] ]$ -$[ 4, 0, 16, [ [ 240, 84 ] ] ]$ -$[ 4, 0, 16, [ [ 24, 84 ], [ 192, 126 ], [ 24, 240 ] ] ]$ -$[ 4, 2, 15, [ [ 120, 84 ], [ 114, 126 ], [ 6, 240 ] ] ]$ -$[ 5, 0, 16, [ [ 120, 84 ], [ 120, 126 ] ] ]$ -$[ 6, 0, 16, [ [ 66, 84 ], [ 168, 126 ], [ 6, 240 ] ] ]$ -The numbers $84$ and $126$ which appear on second $l$ position correspond to $D_7$ and $E_7$ sublattices of $E_8$. The first case happens when projection of the vector on $E_8$ is collinear with "type 4" vector which is sum of two perpendicular vectors from the lattice. The second case happens when projection of the vector on $E_8$ is collinear with lattice vector. -As we can see one of the 4-hook is obtained by intersection in one line of the two 8-planes. This feature can be used to define the Leech lattice from $Co_0$ in 24-dim space. I don't know whether this is useful for something yet.<|endoftext|> -TITLE: Difficulties with the mod 2 Moore Spectrum -QUESTION [15 upvotes]: I have been informed that there is a reference out there which specifically details what goes wrong with the mod 2 Moore spectrum, i.e. why it is not $A_\infty$ or something? I do not know the details, but I am interested in this from the point of view of trying to understand how to come up with the correct notion of ideals of spectra (in the sense of Smith and others), especially the ideal generated by multiplication by 2 on the sphere spectrum. I think it is a paper by Neeman. Does anyone know of this paper, or of other papers which might detail this situation carefully? -Thanks! -PS Is this question appropriate, as it is only a reference request, in the strongest sense of the phrase? - -REPLY [18 votes]: An argument from the old days. A unital multiplication gives a non-trivial splitting of $R\wedge R$ whereas its mod 2 cohomology is indecomposable as a module over the steenrod algenbra.<|endoftext|> -TITLE: Sequences without long arithmetic progressions -QUESTION [7 upvotes]: First, a bit of notation. If we have an arithmetic progression $a, a+k, a+2k, \ldots, a+(n-1)k$ we will call $k$ the distance, and $n$ the length. -While trying to find an example for a paper I'm writing in ring theory, I was led to ask the question: Is there a sequence of 0's and 1's for which if there is an arithmetic progression in the sequence which is constantly 0 (or 1), is there a bound for the length in terms of the distance? -I found that the answer is yes, and the Thue-Morse sequence works. Modifying the ideas of Corollary 2 in "Thue-Morse at multiples of an integer" (available here), we see that the length of a constant arithmetic progression on the Thue-Morse sequence of distance $k$ is bounded by $32k^3$. -So, here are my questions for you experts. -(1) Is there an easier sequence where one can prove this is true (possibly with citation in the literature)? -(2) If not, is there a straightforward citation for this fact for the Thue-Morse sequence? (The reference I gave above works for arithmetic progressions which start near the front of the sequence. But to get an arbitrary arithmetic progression, you need to increase the bound given in the paper a little, and also give a supplementary argument.) - -REPLY [6 votes]: How about this? Fix your favourite irrational number $\phi$. I like the golden mean. Let $x_s=[ s\phi ] - [(s-1)\phi]$ ($[t]$ means the integer part of $t$). These sequences are called Sturmian sequences. -Of course $x_s$ is 1 if and only if $s\phi \bmod 1$ lies in $[0,\phi)$. -Now for any $a$ and $k$, you're asking whether $(a+jk)\phi\bmod 1$ lies in $[0,\phi)$ for all $0\le j < n$ or lies in $[\phi,1)$ for all $0\le j < n$. -Provided $S_{k,n}:=\lbrace jk\phi\bmod 1\colon 0\le j < n\rbrace$ is $\delta$-dense in the circle, where $\delta=\min(\phi,1-\phi)$ this cannot happen. -This means you can "compute" the maximum length as a function of $k$, namely $L_{max}(k)=\max\lbrace n\colon S_{k,n}$ is not $\delta$-dense$\rbrace$. In the case of golden mean, $L_{max}(k)$ grows linearly in $k$, but I can't write down the proof of that here (the margin is too small). - -Some more stuff as requested by OP -Claim: If $d(k\phi,\mathbb Z/6)=\epsilon$ then $L_{max}(k)<1/\epsilon$. -Proof: Let $k\phi=a/6+\epsilon$, where $|\epsilon|<1/12$. Notice that $\delta<1/3$. If $a=0$, then $1/\epsilon$ steps produces an $\epsilon$-dense subset of the circle. If $a=1$ or 5, then 12 steps produces a $\delta$-dense subset of the circle. If $a=3$, then $1/\epsilon$ steps produces a $2\epsilon$-dense subset of the circle [because $2k\phi$ is $2\epsilon$-close to $\mathbb Z$] and if $a=2$ or 4, then $1/\epsilon$ steps produces a $3\epsilon$-dense subset of the circle. -Next notice $d(k\phi,\mathbb Z/6)\le d(6k\phi,\mathbb Z)$. From Hardy and Wright (5th ed, Theorem 194) plus a simple argument you get $d(k\phi,\mathbb Z)\le 1/(Ak)$ for a suitable $A$. -Combining these ingredients gives the linear growth of $L_{max}$.<|endoftext|> -TITLE: Ellipsoids and lattices: an enclosure problem. -QUESTION [9 upvotes]: $E \subset {\mathbb R}^2$ is an ellipse of area $1$ centered at the origin that contains no other point with integer coordinates. Is there a matrix $A \in SL(2,{\mathbb Z})$ such that the ellipse $A(E)$ is contained in a disc of radius 10? -Hopefully, this is really easy and it is only my ignorance in "reduction theory" (or other similar techniques) that is at the heart of my difficulties. -More generally, given an ellipsoid $E$ of unit volume in ${\mathbb R}^n$ centered at the origin and containing no other point with integer coordinates, I'm interested in a good upper estimate of the size of the ball in which I can enclose $A(E)$ for some matrix $A \in SL(n,{\mathbb Z})$. - -REPLY [2 votes]: This is an addition to Noam's answer. In higher dimensions the same type of reasoning---using explicit conditions on the coefficients of reduced quadratic forms---is not feasible. Nevertheless, it is possible to prove the following result: -Theorem. Let $E \subset \mathbb{R^n}$ be an $n$-dimensional ellipsoid centered at the origin and containing no other integer point. There exists a transformation $T \in GL(n,\mathbb{Z})$ such that $T(E)$ is contained in the ball of radius -$$ -\left(\frac{3}{2}\right)^{(n-1)(n-2)/2} \frac{2^n}{\epsilon_n}\sqrt{n^{n-1}} -$$ -centered at the origin. -Here $\epsilon_n$ is the volume of the unit ball of dimension $n$. -This theorem follows easily from the following two results: -Theorem. Given a positive-definite $n \times n$ matrix $A = (a_{ij})$, there exists a unimodular integer matrix $U$ such that the matrix $A' = (a'_{ij})$ -defined by $A' := U^t A U$ satisfies the following inequality: -$$ -\left(\frac{2}{3}\right)^{(n-1)(n-2)} \left(\frac{\epsilon_n}{2^n}\right)^2 \leq \frac{\det(A')}{a'_{11} a'_{22} \cdots a'_{nn}} . -$$ -Theorem. If $A = (a_{ij})$ is a positive-definite $n \times n$ matrix, then for every $x \in \mathbb{R}^n$ we have that -$$ -\frac{\det(A)}{n^{n-1}a_{11} a_{22} \cdots a_{nn}} \sum a_{ii} x_i^2 \leq \sum a_{ij} x_i x_j \leq n \sum a_{ii} x_i^2 . -$$ -The first of these two results is an important theorem of Minkowski on the reduction of positive quadratic forms (see Theorem 3 in page 69 of Lekkerkerker), while the second is an improved version of Lemma 4 in page 43 of Siegel's Lectures on -Quadratic Forms.<|endoftext|> -TITLE: Hausdorff measure and the volume form -QUESTION [16 upvotes]: There are two tools, generalizing a concept of a volume to the case of submanifolds in $\mathbb{R}^n$, namely the Hausdorff measure $H^k$ and the volume form. The question is how to show that if $M$ is an orientable $k$-submanifold in $\mathbb{R}^n$ with a volume form $dV$ then -$$ - \int\limits_{M} f(x) dV = \int\limits_{M} f(x) H^k(dx) -$$ -if the integrals exist. -P.S. Maybe the question is too silly for MathOverflow and more suitable for Mathematics Stack Exchange, but I have 2 reasons to post it here: - -In books on geometric integration theory (Krantz, Parks; Federer) I failed to find an answer. -I've already posted the question on Mathematics Stack Exchange and on one more forum, but I didn't receive any response. - -REPLY [17 votes]: I guess you know that it is true in $\mathbb R^k$. -Without loss of generality we can assume that $f\ge 0$. -Fix $\varepsilon>0$ and cover your manifold by $(1\mp\varepsilon)$-Lipschitz charts. -Break your integrals into pieces using subordinate partition of unity and put these pieces back together. -Since in $\mathbb{R}^{k}$ you have equality, you will get -$$ -\int\limits_{M} f(x)\cdot dV\ \ \lessgtr\ \ (1\pm \varepsilon)^k\cdot\int\limits_{M} f(x)\cdot H^k(dx) -$$ -Since $\varepsilon>0$ is arbitrary your statement follows.<|endoftext|> -TITLE: Functions of Pseudodifferential Operators -QUESTION [9 upvotes]: Suppose I have a self-adjoint pseudo-differential operator $A$ on $\mathbb{R}^n$ and a continuous function $f$ (possibly bounded, or Schwartz, or compactly supported) on its spectrum. Then I can consider the operator $f(A)$ defined by functional calculous. -Is $f(A)$ again a pseudo-differential operator and if yes, how are the symbols related? -In what way does the type of operator or the type of function matter? - -REPLY [3 votes]: Since the question posed is about the "In what way does the type of operator or the type of function matter?", I thought the following observation will be apt: -As pointed out by Liviu Nicolaescu in the comment above, Taylor's approach seems to have much wider applicability when it comes to functional calculus. In fact in page 295 of Taylor's book it is mentioned that Seeley's results form a special case of the result. -Moreover, these methods have gone beyond elliptic operators. For instance, Uhlmann, Melrose and Guillemin have developed a framework of distributions whose wavefront sets are in several Lagrangian intersecting manifolds (pseudodifferential operators with singular symbols) for a functional calculus on real principal-type operators, operators of double characteristics, wave operators. Principal symbols also have been computed for these operators and all the computations are purely symbolic.<|endoftext|> -TITLE: Whitney embedding for manifolds with boundary -QUESTION [7 upvotes]: Suppose one has a smooth manifold with boundary M and compact on top of it. Is it true that it can always be embedded in an upper half plane such that the boundary is embedded in the hiperplane $x_n=0$? Or are there obstructions to that? If yes, what are those obstructions? Thank you! - -REPLY [5 votes]: See Theorem 1.4.3. of - -M. Hirsch: Differential Topology, Springer Verlag, 1976<|endoftext|> -TITLE: Markov Processes with Given Marginals -QUESTION [6 upvotes]: Let $\mu_t, t \geq 0,$ be a family of probability measures on the real line. One can assume whatever one wishes about them, although typically they will be continuous in some topology (usually at least the topology of weak convergence of measures), and they will be absolutely continuous with respect to Lebesgue measure. The basic question is as follows: -Is there a Markov process $X_t$ such that its marginal distribution at each time is $\mu_t$? -An obvious example is when $$d \mu_t = \frac{e^{-x^2/2t}}{\sqrt{2 \pi t}} dx$$ and $\mu_0 = \delta_0$, in which case we know that Brownian motion is such a Markov process. I am curious to know if there is any general theory along these lines. -Edit -As per Byron's comment below, I would like the Markov process to be continuous. Ideally I would like to have an SDE description of the process. -The SDE description actually suggests one possible answer: simply compute and play with the time and space derivatives of the density function to see if they satisfy some sort of parabolic equation (like the heat equation), use this to get the adjoint of the generator, and then compute the generator itself. This is a very plausible option, but I was hoping that there might be something more systematic. - -REPLY [4 votes]: The following result was proved in -Kellerer, H.G. (1972) Markov-Komposition und eine Anwendung auf Martingale. Math. Ann., 198, -99–122. -Let $p(y, t)$ be a family of marginal densities, with finite first moment, such that -for $s , t$ the density at time $t$ dominates the density at time $s$ in the convex order. Then there exists a Markov process $X(t)$ with these marginal densities under which $X(t)$ is a submartingale. Furthermore, if the means are independent of $t$ then $X(t)$ is a martingale. -Constructive versions have then been studied by Madan and Yor in -http://projecteuclid.org/download/pdf_1/euclid.bj/1078681382 -The SDE description you mention is actually discussed in Section 2.3 of the paper.<|endoftext|> -TITLE: Is the reals the smallest connected ordered topological ring? -QUESTION [10 upvotes]: The real numbers is a locally compact Tychonoff connected complete ordered topological field. I am looking at minimal collections of adjectives that can characterize the reals. The one often used to define the reals is that it is (Dedekind- or Cauchy-) complete ordered field. -Consider the real numbers among other ordered topological rings. I am wondering if the real numbers is the smallest connected ordered topological ring? Here "smallest" means that it embeds into every other connected ordered topological ring. If this is not true, do you have some minimal collection of additional adjectives (advoiding the word "complete") that can characterize the real numbers among ordered topological rings? "Minimal" means that if you drop an adjective then you there are other ordered topological rings that also fulfil your definition. - -REPLY [13 votes]: The answer to the original question is positive, and even more is true. - -Proposition: $(\mathbb R,+,\le)$ is up to isomorphism the unique nontrivial connected ordered group. - -Proof: -Let $(G,+,\le)$ be such a group (not necessarily abelian despite the notation). First, observe that $G$ is archimedean, i.e., for every $a,b>0$, there is a natural number $n$ such that $b\le na$. This follows from the fact that -$$\{b:\exists n\in\mathbb N\,(-na\le b\le na)\}$$ -is a nontrivial convex subgroup, hence clopen. -Second, by a well-known theorem of Hölder, every archimedean group is isomorphic to a subgroup of $\mathbb R$. The argument is as follows. We fix a positive element $a$. For any $b\in G$ and $n\in\mathbb N$, there is a unique $b_n\in\mathbb Z$ such that $b_na\le nb<(b_n+1)a$. Then it is easy to show that -$$f(b):=\lim_{n\to\infty}\frac{b_n}n$$ -exists, and that $f$ is an ordered group embedding of $G$ into $\mathbb R$. -Now, $\mathbb R$ has no nontrivial proper connected subgroup: if $a\in\mathbb R\smallsetminus G$, then $(-\infty,a)\cap G$ and $(a,+\infty)\cap G$ form an open partition of $G$. (By the way, in general an ordered space is connected iff it is densely ordered and Dedekind complete.) -Since there is only one way how to expand $(\mathbb R,+,\le,1)$ to an ordered ring, we obtain - -Corollary: $(\mathbb R,+,\cdot,\le)$ is up to isomorphism the unique nontrivial connected ordered ring.<|endoftext|> -TITLE: contactomorphism of $S^{2n+1}$ for n>1 -QUESTION [5 upvotes]: Is there some result about contactomorphism groups of $S^{2n+1}$ or $T^{2n-1}$ for n>1? -For example, do we know the rank of $\pi_{i}(Cont(S^{2n+1})) \otimes \mathbb{Q}?$ where "Cont" means the contactomorphism group, and $S^{2n+1}$ with standard contact structure. - -REPLY [2 votes]: This might be of some help: arXiv:1304.5785. In Section 4, the evaluation map is used to detect a non--trivial element in higher homotopy groups of the contactomorphism group of the standard sphere. -Regarding the torus case, I am only aware of Tim's reference to Bourgeois work, Geiges paper on the 3--torus and its preceding work for torus bundles with J. Gonzalo. Yet again, this is 3--dimensional. -Note that the study of the contactomorphism group is tantamount to that of the space of contact structures, supposing that the diffeomorphism group is (or thought as if it were) known. This is the approach in the previous articles. In particular the contactomorphism group of the standard 3--sphere is known to retract to U(2) since the appearance of Eliashberg '92 "Contact 3--manifolds twenty years since J. Martinet's work".<|endoftext|> -TITLE: Question about prompt names of ordinals -QUESTION [5 upvotes]: I asked this question first on math SE and was told that it would better fit here. So: -The following concept is due to Shelah and I have some issues with a claim using this notion: -Suppose that $\nu$ is a limit ordinal and that $P_\nu$ is an iteration of forcing notions. We say that a $P_\nu$ name $\dot{\alpha}$ of an ordinal is $prompt$ iff the following two things hold: - -$\Vdash_\nu \dot{\alpha} \le \nu$ -If $p \Vdash_\nu "\dot{\alpha} = \xi"$ then even $p \upharpoonright \xi ^\smallfrown 1_\nu \upharpoonright [\xi, \nu) \Vdash_\nu \dot{\alpha} = \xi$ ( $1_\nu$ should be the largest element of the iteration, and $\xi$ is the hacek name of an ordinal though I refused to write the hacek) - -Then the following two things should hold: - -If $\dot{\alpha}$ is prompt $\eta \le \nu$ and if $p \Vdash_\nu \eta \le \dot{\alpha}$ then $p \upharpoonright \eta ^\smallfrown 1_\nu \upharpoonright [\eta, \nu) \Vdash_\nu \eta \le \dot{\alpha}$ -If $\dot{\alpha_i}$ are prompt then so is the supremum $sup$ and the minimum $min$ - -I have problems proving those two assertions so any help would be highly appreciated. Thank you! - -REPLY [6 votes]: I'm giving a second answer rather than editing the first, partly because the first is already unpleasantly long, and partly because I think the following is a better way to view the situation. Let me first recall some general information about iterated forcing. I'll write $P_\xi$ for the forcing resulting from the first $\xi$ steps of the iteration, and I'll write $B_\xi$ for its regular open algebra, i.e., the complete Boolean algebra that has $P_\xi$ as a dense subset. If $\xi<\eta$, then the obvious embedding $P_\xi\to P_\eta$, appending $(\eta-\xi)$ 1's to any condition, extends to a complete embedding of Boolean algebras $B_\xi\to B_\eta$. The completeness of these embeddings seems to be what underlies the results in the question. -I'll simplify notation by identifying all the algebras $B_\xi$ with their images in $B_\nu$ (where $\nu$ is, as in the question, the length of the iteration). To say that $\dot\alpha$ is prompt is just to say that, for each $\xi<\nu$ the Boolean truth value $\Vert\dot\alpha=\xi\Vert$ is in $B_\xi$. This implies, thanks to completeness of the subalgebras, that the following is in $B_\eta$ (for any $\eta\leq\nu$): -$$\Vert\eta\leq\dot\alpha\Vert= --\bigvee_{\xi<\eta}\Vert\dot\alpha=\xi\Vert,$$ -and this, when translated back from Boolean-algebra language to forcing language, gives conclusion 1 of the question. Similarly, for conclusion 2, using the assumption that $\Vert\dot\alpha_i=\xi\Vert\in B_\xi$ for all $i$ and $\xi$, we find that $B_\eta$ contains -$$\Vert\sup_i\ \dot\alpha_i=\eta\Vert= -\left(\bigwedge_i\bigvee_{\xi\leq\eta}\Vert\dot\alpha_i=\xi\Vert\right)\land -\left(\bigwedge_{\gamma<\eta}\bigvee_i\Vert\gamma+1\leq\dot\alpha_i\Vert\right)$$ -and -$$\Vert\min_i\ \dot\alpha_i=\eta\Vert= -\left(\bigvee_i\Vert\dot\alpha_i=\eta\Vert\right)\land -\left(\bigwedge_i\Vert\eta\leq\dot\alpha_i\Vert\right). -$$ -As far as I can see, all the work in my previous answer --- extending conditions, truncating them, and observing compatibility --- was just repeating (too many times) the argument for why each $B_\eta$ is a complete subalgebra of $B_\nu$ (and why promptness is equivalent to its Boolean formulation). The moral of the story is that the Boolean-valued viewpoint sometimes makes things considerably easier and clearer than the forcing viewpoint.<|endoftext|> -TITLE: Resolution of singularities for flat families. -QUESTION [10 upvotes]: Is there a resolution of singularities for flat families? -More precisely, if $X \rightarrow \mathbb{A} ^n$ is a flat map, does there exist a map $Y \rightarrow X$ such that, for every $p \in \mathbb{A} ^n$, the fiber map $Y_p \rightarrow X_p$ is a resolution of singularities? Can one require, moreover, that the map $Y \rightarrow \mathbb{A} ^n$ is smooth? - -REPLY [11 votes]: I assume you want $Y \to X$ to be proper. The answer is a definite no, in general. For example, take a polynomial $f: \mathbb A^2 \to \mathbb A^1$; such a $Y$ would have to be finite over $\mathbb A^2$, and birational, so $Y = \mathbb A^2$. There are lots of counterexamples in higher dimension too: for example, it follows from the purity theorem that you usually can't have a simultaneous resolution when $X$ is smooth. Thus, for example, in the very simple example $f\colon \mathbb A^3 \to \mathbb A^1$, $f(x, y, z) = x^2 + yz$, in which the only singular fiber is over the origin, and it has the simplest kind of surface singularity, of type $A_1$, you don't have a simultaneous resolution. -There are some non-trivial results, but they require the base to be 1-dimensional, and they require a base change on the base to get the resolution. For example, in the example above if one makes a base change $t \mapsto t^2$ on the base and normalizes, one has a simultaneous resolution. This is particular case of a theorem of Brieskorn: see for example M. Artin, Algebraic construction of Brieskorn's resolutions, Journal of Algebra 29 (1974). This is only possible in very particular cases, though. - -REPLY [7 votes]: This is false in general. Take $f:\mathbb A^2\to \mathbb A^1$, $(x,y)\mapsto xy$. Any attempt to resolve the singular fiber will bring in a new component in the fiber, so it remains singular.<|endoftext|> -TITLE: The existence of proper classes -QUESTION [6 upvotes]: Andreas Blass states that proper classes do not exist and emphasizes that this is only his philosophical opinion, and not a mathematical fact.(link text). -Is it really not a mathematical fact? I think there are some mathematical results that justify his philosophical opinion. Levy and Vaught prove that Ackermann's set theory proves the existence of the classes {V}, P{V}, PP{V},....(Pacific Journal of Mathematics, 11:1045-1062, 1961). Furthermore, Reinhardt proves that Ackermann's set theory equals ZF (Annals of Mathematical Logic, 2:189-249). My understanding of these results is that anything we can prove in Ackermann's set theory, we can prove in ZF as well. There is no need to assume the existence of the classes P{V}, PP{V}, PPP{V},...because there is nothing new math fact to obtain. -Is my understanding correct? - -REPLY [3 votes]: In a formal theory like ZF (or ZFC), one can say that proper classes don't exist, since the formal language can only speak of sets. Inside the formal theory, one can prove sentences stating for instance that there is no universal set. But if we are looking "from the outside" at a model for ZF, we can speak of collections of objects (for instance the collection of all sets) that do not correspond to sets in the model. We call them classes to distinguish the informal from the formal language. -From a technical point of view, the situation is no more paradoxical than for instance a formal theory of the rational numbers in which it can be proved that there is no square root of 2, but that can be embedded in a larger structure where $\sqrt{2}$ does exist. -But it's a bit disturbing if one wants to think of set theory as a foundation of mathematics, describing the real mathematical universe. It seems weird to be able to give specific examples of things that don't exist. With everyday non-existing things like circles with integer radius and perimeter you can't, because there simply aren't any. It reminds me of the great Master Cerebron in Stanislaw Lem's "The cyberiad", who lectured for 47 years on the three types of dragon, each of which doesn't exist, but in completely different ways. -To understand the situation, I think one has to see the problem that ZF was meant to solve. In the late 1800's, Gottlob Frege devised a much simpler formal theory of sets that turned out to be inconsistent due to the Russell paradox of the set of all sets that aren't members of themselves. Restoring Frege's logic turned out to be quite a challenge. Russell suggested a hierarchy of types, while ZF goes in another direction, squeezing set theory into first-order logic. I'm not familiar with other theories, but it seems that whatever we do, Rusell's paradox will keep haunting us one way or another. Finally, since I'm currently reading it, I can't resist mentioning Logicomix.<|endoftext|> -TITLE: The maximal "nearly convex" function -QUESTION [26 upvotes]: The following problem is only tangentially related to my present work, and I do -not have any applications. However, I am curious to know the solution -- or -even to see a lack thereof, indicating that the problem may be worth a -serious research. -Let $\mathcal F$ denote the class of all functions $f\colon[0,1]\to{\mathbb -R}$ satisfying the inequality - $$ f(tx+(1-t)y) \le tf(x) + (1-t)f(y) + |y-x|, \ x,y,t \in [0,1] $$ -and the boundary condition $\max\{f(0),f(1)\}\le 0$. -Substituting $x=0$ and $y=1$, we see that all functions from $\mathcal F$ are -uniformly bounded from above by $1$, and we let - $$ F(x) := \sup \{ f(x)\colon f\in {\mathcal F} \}. $$ -An interesting observation is that the function $F$ itself belongs to -$\mathcal F$; hence, it is the pointwise maximal function of this class. What -is this function, explicitly? -It is not difficult to see that $F$ is continuous on $[0,1]$, symmetric -around $x=1/2$, positive on $(0,1)$, and vanishing at $x=0$ and $x=1$. -Substituting $t=x$ and $y=0$ and renaming the variables in the resulting -estimate gives $F(x)\le 2\sqrt x$; hence, indeed, - $$ F(x) \le \min\{2\sqrt x,1,2\sqrt{1-x} \}. $$ -On the other hand, the functions $\min\{4x,1,4(1-x)\}$ and $ex\ln(1/x)$ -belong to $\mathcal F$, implying - $$ F(x) \ge \min\{4x,1,4(1-x)\} $$ -and - $$ F(x) \ge \max \{ ex\ln(1/x), e(1-x)\ln(1/(1-x))\}, $$ -for all $x\in[0,1]$. -The graphs of the bounding functions in the right-hand sides: - (source) -(Thus, the graph of $F$ resides somewhere between the highest of the green -curves and the red curve.) -Comparing the estimates, we see that $F(x)=1$ for $x\in[1/4,3/4]$, and -$0 -TITLE: Sylow subgroups of projective general linear groups -QUESTION [11 upvotes]: What is known about the Sylow 2-subgroups of $\rm{PGL}_n(\mathbb{F}_q)$, where $q$ is any prime power? For example, according to Theorem 7.9 in Isaacs's Character Theory, these Sylow 2-subgroups cannot be generalised quaternion. Is there a classification of all these Sylows available? -Concretely, let me make the following semi-conjecture, which, if true, would make me very happy: - -The irreducible complex representations of Sylow 2-subgroups of $\rm{PGL}_n(\mathbb{F}_q)$ - have trivial Schur index. - -Is this known? Any references to literature that discusses such questions would be very welcome. -If anyone has information on Sylow $l$-subgroups for arbitrary $l$, I would also be very interested. - -REPLY [12 votes]: The irreducibles of the Sylow 2-subgroups of ${\rm GL}(n,q)$, $q$ odd, have indeed trivial Schur indices: Let $S$ be such a Sylow 2-subgroup. First, observe that the natural module $(\mathbb{F}_q)^n$ splits into a sum of simple modules $U_1\oplus \dotsb \oplus U_l$, and the dimension of each simple module is a power of $2$. This shows that $S \cong S_1 \times \dotsb \times S_l $, where the $S_i$'s are Sylow 2-subgroups of a ${\rm GL}(2^k, q)$. Thus, w.l.o.g. we may assume that $n=2^k$. Then I use induction on $k$. First note that if $S$ is a Sylow 2-subgroup of ${\rm GL}(2^k, q)$, then -$$ T=\lbrace \begin{pmatrix} s & \\ & t \end{pmatrix} \mid s, t\in S \rbrace - \cup \lbrace \begin{pmatrix} & s \\ t & \end{pmatrix} \mid s,t\in S\rbrace - \cong S\wr C_2 $$ -is a Sylow 2-subgroup of ${\rm GL}(2^{k+1}, q)$, except when $k=0$ and $q\equiv 3\mod 4$. This follows from the description in Derek Holt's answer, but it can also be seen directly by observing that $T$ has the right order. -Write $N=S\times S$, so that $T= C_2 N$, and let $\chi\in {\rm Irr} (T)$. Three cases have to be considered: -1. We have $\chi_N \in {\rm Irr} (N)$. By induction, $\chi_N$ has trivial Schur index. By Lemma~10.4 in Isaacs' character theory book, for example, it follows that $\chi$ has trivial Schur index. -2. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are not Galois conjugate. -Then $\chi = \theta^T$ and $\theta$ have the same field of values and the same Schur index (again, see Lemma~10.4 in Isaacs). -3. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are Galois conjugate. This means that $\chi=\theta^T$, but $|\mathbb{Q}(\theta):\mathbb{Q}(\chi)|=2$. Now a representation $R\colon N\to {\rm GL}(\chi(1), \mathbb{Q}(\chi) )$ affording the character $\theta+\theta^g$ can be extended to a representation over the same field, since $N$ has a complement (of order 2) in $T$. -To get the induction going when $q\equiv 3\mod 4$, one has to check that the Sylow 2-subgroup of ${\rm GL}(2,q)$ has trivial Schur indices, but that is clear since it's a semidihedral group.<|endoftext|> -TITLE: Pseudoisotopy in low dimensions -QUESTION [22 upvotes]: Recall that the space $P(M)$ of (smooth) pseudoisotopies of the compact manifold $M$ is defined as the space of all diffeomorphisms $M\times I\to M\times I$ that fix every point in $(M\times 0)\cup (\partial M \times I)$. I have worked on these things in high-dimensional cases, but I realize that there are big gaps in my knowledge in low-dimensions. I am wondering how much of what I don't know is known. -If the dimension of $M$ is less than $2$ then $P(M)$ is contractible. According to the Smale Conjecture $P(D^2)$ (or equivalently $P(S^2)$) is contractible. I have the impression that Hatcher also proved that $P(M)$ is contractible when $M$ is a two-dimensional annulus. Is it perhaps known, or believed, that this holds for all compact $2$-manifolds? Or maybe it's not true. -(I could have started by just asking Allen Hatcher or Ryan Budney, but I thought others might be interested in any answers.) -And what is known in the next dimension? - -REPLY [18 votes]: When $M$ is a compact $2$-manifold, with or without boundary, $P(M)$ is known. When $M$ is a 3-manifold there's bits and pieces known, especially once you get to more fine detail like pseudo-isotopy embedding spaces. But at the level of $P(M)$ I don't think there's a complete description for a single $3$-manifold. For $4$-manifolds the situation is worse, but again there are some things known for pseudo-isotopy embedding spaces. -As you mention, $P(S^2)$ and $P(D^2)$ are contractible by the Smale Conjecture. I think $P(M)$ is contractible for an arbitrary $2$-manifold. The argument goes like this: You look at locally trivial fiber bundle $P(M) \to Diff(M)$. The fiber over the identity map is $Diff(M \times I)$, the group of diffeomorphisms fixing the entire boundary $\partial M \times I \cup M \times \partial I$. The other fibers are empty because diffeomorphisms are prescribed up to isotopy by their action on $\pi_1 M$. Except in a few cases, $Diff(M)$ has contractible components, so showing $P(M)$ is contractible is equivalent to showing $Diff(M \times I)$ is contractible. But $M \times I$ has incompressible annuli in it, corresponding to $C \times I$ where $C \subset M$ is a closed curve in $M$ that does not bound a disc in $M$. Hatcher's work on spaces of incompressible surfaces kicks in and tells you the space of these incompressible annuli is contractible. So you can reduce studying $Diff(M \times I)$ to things like $Diff(A \times I)$ where $A$ is an annulus or a pair of pants, but then you have vertical incompressible discs you can use and reduce to the point that $A$ is itself a disc. -The above argument works for any surface other than $\mathbb RP^2$, a torus or a klein bottle. But similar arguments cover these cases. -Here's the reference for the results of Hatcher's I'm using, it also defines the terminology like "incompressible annulus, disc" and so on. http://www.math.cornell.edu/~hatcher/Papers/emb.pdf -Allen had a student (Kiralis) who wrote some papers on pseudo-isotopy diffeomorphisms of 3-manifolds. That might be a place to look. -But the kind of things that I remember are mostly along the lines of pseudo-isotopy embeddings of knots and links in $D^3$ and $S^3$. I'm about to hop on an airplane. I'll edit this response sometime in the next few days and add some of these observations if they haven't already been made by someone else. -edit 1: Here are two unrelated observations. -(a) Let $N$ be a co-dimension zero solid torus in $M=\mathbb R^3$ or $M=S^3$. There's the pseudo-isotopy embedding fibration $P(N,M) \to Emb(N,M)$. If $N$ is an unknotted solid torus, the question of what the map the image of the map $\pi_0 P(N,M) \to \pi_0 Emb(N,M)$ is, this is a long-standing hard problem in knot theory. Another way to say it is `which knots in $S^3$ bound a disc in $D^4$?'. These knots are called slice knots. Ralph Fox has the Slice Ribbon Conjecture, which might be described as a hopeful combinatorial answer to the question. There are many useful tools for determining whether or not a given knot is slice,starting with the Alexander module and more recently tools from Heegaard Floer theory. -Two examples: - -Look at the class of knots that are a connect-sum of torus knots. Using the Alexander module Litherland proved such a knot bounds a disc in $D^4$ if and only if in the prime decomposition, the number of times a prime summand appears (like say a right-handed trefoil) is equal to the number of times its mirror-image appears (the left handed trefoil, in my example). Related previous MO thread -Paolo Lisca has used Heegaard Floer theory to determine when a connect sum of 2-bridge knots in $S^3$ bounds a disc in $D^4$ REF, but in this case the answer is more elaborate. - -(b) Just like how diffeomorphism groups $Diff(D^n)$ and spaces of long knots have actions of cubes operads, pseudo-isotopy embedding spaces and diffeomorphism groups also have such actions. I'm pretty sure there's a pseudo-isotopy splicing operad as well, but I haven't written out the details. -edit 2: Rick Litherland generalized a result of Zeeman (Deforming twist-spun knots TAMS 250 (1979) 311--331) showing that "Deform-spun knots" have complements that frequently fibre over $S^1$. This process called "Deform spinning" is just the boundary map in the pseudo-isotopy fibre sequence for spaces of knots. One of the nice things about this is Litherland gives a prescription for what the fibre is. In the case where you're looking at the pseudo-isotopy sequence for long knots in $\mathbb R^3$, it generates long embeddings of $\mathbb R^2$ in $\mathbb R^4$. So the fibre is a 3-manifold with boundary a sphere. This process produced some embeddings of 3-manifolds in the 4-sphere that nobody had known about at the time, like the once-punctured Poincare Dodecahedral Space (which without a puncture does not embed in $\mathbb R^4$, at least, not smoothly, it does admit a tame topological embedding). I got interested in this case largely because it represents sort of an extreme end of the terrain of your dissertation. -edit 3: I forgot to mention, I kept on pushing trying to understand why your dissertation broke down in co-dimension two. In some sense my paper "An obstruction to a knot being deform-spun via Alexander polynomials" is an answer. In a way it's not, since co-dimension two deform-spinning is more of a free loop space construction than a based loop space construction. But I found the exercise informative.<|endoftext|> -TITLE: Existing proofs of Rokhlin's theorem for PL manifolds -QUESTION [11 upvotes]: I'm looking for a comprehensive reference to existing proofs of Rokhlin's theorem that a 4-dimensional closed spin PL manifold has signature divisible by 16. -I'm specifically interested in direct proofs (if any such exist) which do not rely on the fact that $\pi_i(PL/O)=0$ for small $i$. -The most commonly cited reference seems to be the book by Kirby "The Topology of 4-manifolds". But the proof there is for smooth manifolds and I'm not sure why it works for PL manifolds although I've seen it claimed in various places that it does. The same is said about Rokhlin's original proof but I don't know why that's true either. I would also like to know if other proofs for PL manifolds exist. I'm particularly interested to know if there is a PL proof based on the Atiyah-Singer index theorem. - -REPLY [11 votes]: Another approach to the theorem that could probably be rewritten to work in the PL category is the approach of Kirby and Melvin in Appendix C of the following paper: -MR1117149 (92e:57011) -Kirby, Robion(1-CA); Melvin, Paul(1-BRYN) -The 3-manifold invariants of Witten and Reshetikhin-Turaev for sl(2,C). -Invent. Math. 105 (1991), no. 3, 473–545. -See Corollary C6. -The idea of this approach is as follows. There is a famous $\mathbb{Z}/2$-invariant of homology $3$-spheres called the Rokhlin invariant. The usual definition of this invariant is as follows. Letting $M^3$ be a homology $3$-sphere, there exists a compact spin $4$-manifold $W^4$ with $\partial W^4 = M^3$. Let $\sigma$ be the signature of $W^4$. Rokhlin's theorem implies that modulo $16$, the value of $\sigma$ is independent of $W^4$. Since $\sigma$ is divisible by $8$ for number-theoretic reasons (namely, van der Blij's lemma about quadratic forms), the value of $\sigma/8$ is well-defined modulo $2$. -Using the Kirby calculus, Kirby and Melvin give a $3$-dimensional'' construction of the Rokhlin invariant, avoiding all mention of $4$-manifolds. They then go backwards and use this to prove Rokhlin's theorem about $4$-manifolds. -Looking at their proof, Kirby and Melvin use smoothness in two ways. The first is to prove that the Rokhlin invariant is well-defined. But this is harmless since (by work of Moise) all PL $3$-manifolds can be smoothed in a unique way. The second use of smoothness is to obtain a handlebody decomposition of the $4$-manifold. But this should be easier in the PL category!<|endoftext|> -TITLE: Reference request: an algebraic stack whose closed points have no automorphisms is an algebraic space -QUESTION [5 upvotes]: The question is stated in the title. I think BCnrd states in a comment here -Is every (Artin/DM) algebraic stack fibered in sets an algebraic space? -that while the answer is not found in Laumon & Moret-Bailly, but that it is nevertheless true. Does anyone know of a reference? (Also: I only care about stacks over $\mathbb{C}$, if that makes a difference.) - -REPLY [5 votes]: Here's a counterexample to your title question, that is not a counterexample to BCnrd's claim: -Let $Y =\operatorname{Spec} \mathbb{C}[[t]]$. This scheme has a closed point and an open generic point. Let $Z$ be the scheme formed by gluing two copies of $Y$ by the identity map on generic points. This is a non-separated scheme with an action of a group of order two that fixes the generic point and switches the two closed points. Let $X$ be the stack quotient of $Z$ by this action. -$X$ is not an algebraic space, since the image of the generic point of $Z$ has a nontrivial stabilizer. In particular any geometric point (meaning a map from the spectrum of an algebraically closed field) that factors through the generic point of $Z$ will have nontrivial stabilizer. The topological space $|X|$ (defined in Champs Algebriques 5.5) only has one closed point, and the automophism group of any element in its equivalence class is trivial.<|endoftext|> -TITLE: Families of subsets containing every singleton as an intersection -QUESTION [5 upvotes]: For $X=\lbrace 0,\ldots,n-1\rbrace$, let $F\subseteq 2^X$ be a family of subsets of $X$ such that, for every $x\in X$, the singleton $\lbrace x\rbrace$ is the intersection of some elements of $F$. I am interested in the minimal families that have this property, in particular whether it is possible to have $|F|< n$. Can anyone (a) give an example where $|F|< n$, (b) provide an argument for why $|F|\ge n$, or (c) point me in the direction of some existing results. -Bonus question: For any $k< n$, the family $F= \lbrace \lbrace x,x+1,\ldots,x+k-1 \rbrace:x\in X \rbrace$, where addition is carried out modulo $n$, satisfies the stated condition (minimally) and contains exactly $n$ elements, so $|F|= n$ is always achievable. How does the structure of a general minimal family relate to these highly regular families? Is a minimal family always a disjoint union of some regular families? - -REPLY [7 votes]: You can achieve $\lvert F\rvert = 2\lceil\log_2 n\rceil$ by using all subsets of the form $\{x\in X \vert i^{\text{th}}\text{ bit of }x\text{ is }j\}$ for $i \in \{0,1,\ldots,\lceil\log_2 n\rceil-1\}$ and $j\in\{0,1\}$. -This rate is within a factor of two of best possible because there are at most $2^{\lvert F\rvert}$ different intersections you can form from $\lvert F\rvert$ sets. You require that all $n$ singletons be among this list of intersections, so $\lvert F\rvert \geq \log_2 n$. -Are you interested in the actual minimal value of $\lvert F\rvert$, or just asymptotics?<|endoftext|> -TITLE: How to make the move to a research university from a liberal arts school? -QUESTION [16 upvotes]: Some mathematicians are in positions that emphasize teaching at predominantly undergraduate colleges but find themselves desiring positions at research universities. This may happen because their career preferences have changed or because they always wished for research oriented positions but were unable to find one earlier. Having said this, I'll ask: - -1) Precisely how does one move from a teaching oriented job to a research oriented job? I'm aware that one should build relationships at universities and should write good papers, but, given the teaching and service expectations at most predominantly undergraduate institutions, is it really reasonable to expect that enough solid work can be done to make one's CV competitive? - -The prevailing wisdom is that one should try to make such a move before getting tenure at a predominantly undergraduate institution. So, additionally, I'll ask: - -2) Is it possible to move from a teaching job to a research job after receiving tenure at a predominantly undergraduate institution? If so, what advice do you have for someone trying to do this? - -REPLY [14 votes]: I can't answer this question, but perhaps can offer some info: -Edit 1: I realized that what appears below would perhaps be more suitable as an answer to a question like: "How do you cope with the reality of being at an undergraduate-focused school when you had your heart set on a research career?" This is markedly different from the OP's question, but evidently my answer seems useful for this one. Perhaps I've implicitly answered the question...It would be nice if this question received an actual explicit answer! (e.g. I think Joseph O'Rourke's deleted answer was closer to the mark than mine is!) -1') Important Edit 2 Recent economic pressures on undergraduate colleges connected with projected enrollment decline, health insurance laws, public perception of higher education and resulting changes in "product delivery", has led me to revisit the advice I gave in (1) below. It is very unlikely that any but perhaps a few elite undergraduate liberal arts colleges will be able to afford to support research substantially in a sustainable way. Small class sizes and reduced teaching loads are regarded by most colleges as mutually exclusive and unaffordable to maintain simultaneously. Since the former is a big selling point of the liberal arts college, the latter probably will not easily be found at a liberal arts college without financial problems at the institution following closely behind, if the institution is one of the many that are primarily enrollment-driven. If you happen to enjoy research and don't want to be forced out of it by economic reasons like teaching load or salary that needs supplement, do not look for a job at a liberal arts college with less than a billion dollar endowment. (I believe this edit is important, since the old answer might mislead future job candidates for such institutions…which I really do not want to do.) -1) Not all liberal arts schools are the same. Not considering the elite schools, some departments have 3 hour loads distributed over two courses. I work at such a place. Office hours and service are still more of a commitment than at a research school, but there is still time left over to work. My college is clearly trying to raise its research profile a little. My feeling is that due to the sheer number of qualified PhDs looking for jobs this is an emerging market, and so colleges that want to move up in the rankings will consider reducing teaching loads in order to take advantage of the oversupply of strong research candidates. You won't be able to write two good papers per year...perhaps not even one good paper per year...but you will be able to do research at such a place. Chances are, if you have tenure at a small school then you have a knack for engaging undergraduates (whether you like it or not). I'll bet you will be an attractive candidate for a move to such a department. -2) This is presumptuous to write, because it's pretty obvious, but getting a good research job usually requires making very serious sacrifices. Many of my research friends still are working hard on work/life balance issues. Some have had to be separated from a spouse (often on a different continent) in order to try to get a decent research job. I can't help but think that people who land these jobs often quite deserve them. I can say that my own unwillingness to expose myself to such risk is the main reason why I originally accepted a job at a liberal arts college instead of trying to go to a research university. -3) Having just chaired a search committee, I've recently reviewed more than 750 applications for a tenure-track liberal arts job. The gut-wrenching truth is that there were more excellent research candidates than the market can support. If we restrict our attention only to those with at least one postdoc, it is probably possible to fill all openings on MathJobs with candidates having more than 5 publications (and this is conservative). Budgets allow for bringing a few candidates to campus for an interview, and almost nothing is worse for a potential employer than blowing a campus interview on someone who is not serious about a position. To some extent, your vulnerability as an applicant is attractive to the interviewer. If you have tenure already, it is possible that you will be harder to secure...or worse, will be seen as collecting offers to bargain for a raise at your current job. Unfortunately, I have strong evidence that this happens. In one case, this evidence stopped us from talking to an untenured candidate. I think that this is the main reason why it is nearly impossible to move after tenure. (Sorry to kill hope...I hope I am wrong about this!) -4) One final important point I forgot to include before. As you mourn the fact that you never will teach graduate courses or supervise PhD students, keep in mind that there is an important positive side to this fact: You will not have to help such PhD students find an academic job in the current market. I imagine that having taken a job at an undergraduate institution, if you were to relocate to a more research-based position it probably will not be Princeton. Instead, you will probably find yourself at a lower-tier research school. I speculate that PhD students graduating from such schools are at a disadvantage for landing excellent postdocs (If I'm wrong about this, someone please contradict me) and subsequently finding permanent academic employment. I imagine watching students go through this would be very stressful for a conscientious thesis advisor.<|endoftext|> -TITLE: On the class number -QUESTION [8 upvotes]: If $K = \mathbb{Q}(\alpha)$ is a number field, where $\alpha$ is algebraic, and $\mathcal{O}_K$ the ring of integers in $K$, then the set of fractional ideals over $\mathcal{O}_K$ forms a group and if we mod out by the set of principal ideals, the resulting group is finite and we call its size the class number of $K$, which we denote $h(K)$. -I have two questions regarding the class number of imaginary quadratic fields: -If we consider the function $f(d) = h(\mathbb{Q}(\sqrt{d}))$ which maps the negative integers to the positive integers, do we know that this function is surjective? That is, can every positive integer be realized as the class number of an imaginary quadratic field extension of $\mathbb{Q}$? -We know that $h(\mathbb{Q}(\sqrt{d}))$ tends to infinity as $d$ tends to negative infinity, since there are at most finitely many imaginary quadratic fields with a given class number. However, do we have a rough estimate at how large class numbers can be relative to $|d|$? That is, if we consider the function $f(D) = \max_{|d| \leq D} h(\mathbb{Q}(\sqrt{d}))$ where $d < 0$, do we have any idea how large $f$ can be relative to $D$? -Thanks for any insights. - -REPLY [5 votes]: This is from Buell, Binary Quadratic Forms. From page 84, the class number for a negative discriminant $\Delta$ is about $$\frac{\sqrt{|\Delta|}}{\pi},$$ which comes from an $L$-function calculation on page 83. -Let's see, on page 101, he points out that for negative field discriminants, class group and narrow class group are identical. Then on page 103, the group of classes of binary quadratic forms is isomorphic to the narrow class group. So that works out. -I don't know about surjectivity of class numbers. I imagine so. See OEIS -I wrote a little program up to 1000, here it is up to 111. The first number that achieves a given class number tends to be squarefree, an exception being h=104. EDIT: I've run this up to 4000 so far. To get a class number $h,$ there was always some $k$ with $k < 4 h^2.$ The largest $h$ where $h^2$ did not suffice was $h=677,$ with smallest $k = 601247.$ For $678 \leq h \leq 4000,$ there was always some $k \leq 0.751517... h^2,$ equality at $h=857, k=551951.$ - 1 3 = 3 - 2 15 = 3 * 5 - 3 23 = 23 - 4 39 = 3 * 13 - 5 47 = 47 - 6 87 = 3 * 29 - 7 71 = 71 - 8 95 = 5 * 19 - 9 199 = 199 - 10 119 = 7 * 17 - 11 167 = 167 - 12 231 = 3 * 7 * 11 - 13 191 = 191 - 14 215 = 5 * 43 - 15 239 = 239 - 16 399 = 3 * 7 * 19 - 17 383 = 383 - 18 335 = 5 * 67 - 19 311 = 311 - 20 455 = 5 * 7 * 13 - 21 431 = 431 - 22 591 = 3 * 197 - 23 647 = 647 - 24 695 = 5 * 139 - 25 479 = 479 - 26 551 = 19 * 29 - 27 983 = 983 - 28 831 = 3 * 277 - 29 887 = 887 - 30 671 = 11 * 61 - 31 719 = 719 - 32 791 = 7 * 113 - 33 839 = 839 - 34 1079 = 13 * 83 - 35 1031 = 1031 - 36 959 = 7 * 137 - 37 1487 = 1487 - 38 1199 = 11 * 109 - 39 1439 = 1439 - 40 1271 = 31 * 41 - 41 1151 = 1151 - 42 1959 = 3 * 653 - 43 1847 = 1847 - 44 1391 = 13 * 107 - 45 1319 = 1319 - 46 2615 = 5 * 523 - 47 3023 = 3023 - 48 1751 = 17 * 103 - 49 1511 = 1511 - 50 1799 = 7 * 257 - 51 1559 = 1559 - 52 1679 = 23 * 73 - 53 2711 = 2711 - 54 2759 = 31 * 89 - 55 4463 = 4463 - 56 1991 = 11 * 181 - 57 2591 = 2591 - 58 2231 = 23 * 97 - 59 2399 = 2399 - 60 2159 = 17 * 127 - 61 3863 = 3863 - 62 2471 = 7 * 353 - 63 2351 = 2351 - 64 2519 = 11 * 229 - 65 3527 = 3527 - 66 3431 = 47 * 73 - 67 3719 = 3719 - 68 2831 = 19 * 149 - 69 3119 = 3119 - 70 3239 = 41 * 79 - 71 5471 = 5471 - 72 3311 = 7 * 11 * 43 - 73 2999 = 2999 - 74 4151 = 7 * 593 - 75 4703 = 4703 - 76 3071 = 37 * 83 - 77 6263 = 6263 - 78 5111 = 19 * 269 - 79 4391 = 4391 - 80 5183 = 71 * 73 - 81 3671 = 3671 - 82 3839 = 11 * 349 - 83 3911 = 3911 - 84 4031 = 29 * 139 - 85 4079 = 4079 - 86 6767 = 67 * 101 - 87 5279 = 5279 - 88 4199 = 13 * 17 * 19 - 89 6311 = 6311 - 90 5951 = 11 * 541 - 91 4679 = 4679 - 92 4991 = 7 * 23 * 31 - 93 5351 = 5351 - 94 7367 = 53 * 139 - 95 6959 = 6959 - 96 6071 = 13 * 467 - 97 5519 = 5519 - 98 6191 = 41 * 151 - 99 5591 = 5591 - 100 7991 = 61 * 131 - 101 5879 = 5879 - 102 9383 = 11 * 853 - 103 13799 = 13799 - 104 9359 = 7^2 * 191 - 105 6719 = 6719 - 106 7631 = 13 * 587 - 107 8231 = 8231 - 108 5759 = 13 * 443 - 109 5711 = 5711 - 110 7751 = 23 * 337 - 111 15359 = 15359 - -REPLY [3 votes]: As Will Jagy explained, $h(-D)$ is roughly $\sqrt{D}$, given by Dirichlet's class number formula -$$h(d) = \frac{w \sqrt{d}}{2 \pi} L(1, \chi_d)$$ -where $L(1, \chi_d)$ is the L-function associated to the quadratic character of $\mathbb{Q}(\sqrt{-D})$. Upper bounds for $L(1, \chi_d)$ are easy to prove; you can get $\log(d)$ by partial summation, implying the bound $h(-d) \ll d^{1/2} \log(d)$. Effective lower bounds are notoriously more difficult. -I am fairly sure that the function $f(d)$ you describe is not known to be surjective, although it is widely expected to be; class numbers are fairly difficult to get a handle on, although a variety of divisibility results are known. However I am not entirely sure of this!<|endoftext|> -TITLE: Is there a sheaf theoretical characterization of a differentiable manifold? -QUESTION [106 upvotes]: I'm going through the crisis of being unhappy with the textbook definition of a differentiable manifold. I'm wondering whether there is a sheaf-theoretic approach which will make me happier. In a nutshell, is there a natural condition to impose on a structure sheaf $\mathcal{C}^k_M$ of a topological space $M$ that can stand-in for the requirement that $M$ be second-countable Hausdorff? -Background -Most textbooks introduce differentiable manifolds via atlases and charts. This has the advantage of being concrete, and the disadvantage of involving an arbitrary choice of atlas, which obscures the basic property that a differential manifold "looks the same at all points" (for $M$ connected, without boundary: diffeomorphism group acts transitively). And isn't introducing local coordinates "an act of violence"? -I saw a much nicer definition of differentiable manifolds on Wikipedia, which I don't know a good textbook reference for. This definition proceeds via sheaves of local rings. The Wikipedia definition stated: - -A differentiable manifold (of class $C_k$) consists of a pair $(M, \mathcal{O}_M)$ where $M$ is a topological space, and $\mathcal{O}_M$ is a sheaf of local $\mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,\mathcal{O}_M)$ is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$. [$\mathcal{O}(U)=C^k(U,\mathbb{R})$ is the structure sheaf on $\mathbb{R}^n$.] - -Beautiful, really! Entirely coordinate free. But isn't there a General Topology condition missing? -I confirmed on math.SE (to make sure that I wasn't hallucinating) that this definition is indeed missing the condition that $M$ be second-countable Hausdorff. That indeed turned out to be the case, so I edited the Wikipedia definition to require $M$ to be second-countable Hausdorff. -Why am I still not happy? -The deep reason that we require a differentiable manifold to be paracompact, as per Georges Elencwajg's extremely informative answer, is that paracompactness makes sheaves of $C_M^k$-modules (maybe $k=\infty$) acyclic. This is a purely sheaf-theoretic property (a condition on the structure sheaf of $M$ rather than on $M$ itself), which quickly implies good things like that every subbundle of a vector bundle on $M$ be a direct summand. Is this in fact enough? -If it were enough to require that $\mathcal{O}_M$ be acyclic, or maybe fine, then the nicest, most flexible (and, in a strange sense, most enlightening) definition of differentiable manifold, would be: - -Definition: A differentiable manifold (of class $C_k$) consists of a pair $(M, \mathcal{O}_M)$ where $M$ is a topological space, and $\mathcal{O}_M$ is an acyclic sheaf of local $\mathbb{R}$-algebras defined on $M$, such that the locally ringed space $(M,\mathcal{O}_M)$ is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$. - -Maybe the word acyclic should be fine. Maybe soft and acyclic. Maybe a bit more, but still something that can be stated in terms of the structure sheaf. - -Question: Can I put a natural sheaf-theoretic condition on $\mathcal{O}_M$ (acyclic? fine?) which ensures that $M$ (a topological space) must be a second-countable Hausdorff space? If not, would such a condition at least ensure that $M$ be a generalized differentiable manifold in some kind of useful sense? - - -Update: This question really bothers me, so I've started a bounty. I'd like to narrow it down a little in order to make it easier to answer: -Does acyclicity (or a slightly stronger condition) on $\mathcal{O}_M$ of a topological (Hausdorff?) space imply paracompactness (or a slightly weaker but still useful condition)? -Hausdorff bothers me as well, of course; but a sheafy characterization of paracompactness somehow seems like it has the potential to be lovely and really enlightening. - -REPLY [4 votes]: This is more a long comment than an answer, but the comments concerning the question Hausdorff or non-Hausdorff triggered this... -The second countable condition is certainly desirable for many reasons (embedding theorems etc) but there are prominent examples of non-Hausdorff manifolds in differential geometry. If you have a Lie algebroid, i.e. a vector bundle $E \longrightarrow M$ such that the sections of $E$ are equipped with a Lie bracket which satisfies a Leibniz rule along a bundle map $E \longrightarrow TM$ (the anchor) then there is a notion of a corresponding Lie groupoid integrating this. The obstructions for existence have been described in a beautiful paper by Crainic and Fernandes. Nevertheless, the resulting Lie groupoid is typically not Hausdorff but the non-Hausdorffness (funny word) is not so bad. The fibers of the source/target as well as the base of the Lie groupoid are all Hausdorff, it is only the way they are glued together which makes it non-Hausdorff. -So this does not answer your question at all, but I think that an answer should also take care of these kind of situation as this is really important in many areas of differential geometry (Lie groupoids are everywhere...).<|endoftext|> -TITLE: Polygons uniquely inducing arrangements -QUESTION [11 upvotes]: A beautiful, relatively recent result is that, - -Every simple arrangement $\cal{A}$ of $n$ lines in the plane is induced by a simple $n$-gon $P$. - -In a simple arrangement, every pair of lines intersect in a point, -and no three lines intersect in a common point. -A polygon $P$ induces $\cal{A}$ if $\cal{A}$ is obtained by extending its -$n$ edges to lines. Thus $P$ "visits" each line of $\cal{A}$ exactly once; -it is a Hamiltonian-like cycle: -             - - -This is proved in the paper, -"On Inducing Polygons and Related Problems." -Eyal Ackerman, Rom Pinchasi, Ludmila Scharf, Marc Scherfenberg. - -Algorithms-ESA 2009. -Lecture Notes in Computer Science, Volume 5757, 2009, pp, 47-58. -(PDF link -) -Two natural question occur to me, neither of which is addressed in the -paper: - -Q1. Which arrangements $\cal{A}$, $n>3$, have a unique inducing polygon? -Q2. Does the theorem extend to $\mathbb{R}^3$, or higher -dimensions? I.e., does every simple arrangement of $n$ planes have an -inducing simple polyhedron of $n$ faces? - -It could be the answers are relatively easy: none and no respectively...? -If anyone sees quick arguments, I'd appreciate hearing them. -Thanks! -Addendum. -Here is an attempt to illustrate Gjergji Zaimi's idea, as I interpret it. -The hexagon induces the arrangement of lines in the horizontal plane, -and the polyhedron "attached" to the hexagon would be the intersection of the two tetrahedra. - -REPLY [4 votes]: Q1: The only arrangement with a unique inducing polygon is the arrangement with three lines. In fact it follows from the first proof in the paper you cite that the number of inducing polygons is $\geq \lfloor\frac{n}{2}\rfloor$. This is because one can pick a line so that every intersection lies on the same half-plane defined by this line. Then one can pick an arbitrary intersection point $P$ on this line and produce a path which visits every line once. This path will also lie on the same half-plane so their algorithm produces an inducing polygon with $P$ as a vertex. But $P$ was arbitrary.<|endoftext|> -TITLE: how to compute the henselization of some simple rings? -QUESTION [5 upvotes]: Hi,everyone. -I want to know that how to compute the henselization of some simple rings, for example: $k[x]_{(x)}$ and $R[X]_{(X)}$ where $k$ is a field and $R$ is a excellent DVR. -thank you very much! - -REPLY [6 votes]: Perhaps a better source for this result is Nagata's 1962 book, "Local Rings". Chapter VII begins with a section on constructing the henselization of a local ring and its properties. For a normal domain R the above result (note though that the separable algebraic closure, not the algebraic closure must be used in char p) follows since his definition is to take the separable closure of R in the algebraic closure of the function field k(R), intersect it with the fixed field for the decomposition group of the maximal ideal of R, and then localize the resulting subring.<|endoftext|> -TITLE: The fundamental theorem of Galois theory -QUESTION [12 upvotes]: Who proved the modern form of the fundamental theorem of Galois theory?. Was it in the original Galois' manuscript? - -REPLY [9 votes]: Galois's Proposition I (as translated by Edwards) is: -Let the equation be given whose $m$ roots are $a,b,c,\ldots$. There will always be a group of permuations of the letters $a,b,c,\ldots$ which will have the following property: 1) that each function invariant under the substitutions of this group will be known rationally; 2) conversely, that every function of these roots which can be determined rationally will be invariant under these substitutions. -As Edwards observes, it takes a lot of work to decipher the exact meanings of "substitution" and "invariant" here, but once you've done that, this can be translated into modern language as: -If an element of the splitting field of $K(a,b,c,\ldots)$ is left fixed by all the automorphisms of the Galois group then it is in $K$. -The fundamental theorem of Galois theory (i.e. the Galois correspondence) follows easily, though Edwards doesn't say who first stated it.<|endoftext|> -TITLE: universality of Macdonald polynomials -QUESTION [17 upvotes]: I have been recently learning a lot about Macdonald polynomials, which have been shown to have probabilistic interpretations, more precisely the eigenfunctions of certain Markov chains on the symmetric group. -To make this post more educational, I will define these polynomials a bit. Consider the 2-parameter family of Macdonald operators (indexed by powers of the indeterminate $X$) for root system $A_n$, on a symmetric polynomial $f$ with $x = (x_1, \ldots, x_n)$: -$$D(X;t,q) = a_\delta(x)^{-1} \sum_{\sigma \in S_n} \epsilon(\sigma) x^{\sigma \delta}\prod_{i=1}^n (1 + X t^{(\sigma \delta)_i} T_i),$$ -(mathoverflow doesn't seem to parse $T_{q,x_i}$ in the formula above, so I had to use the shorter symbol $T_i$, which depends on q). -where $\delta$ is the partition $(n-1,n-2,\ldots, 1,0)$, $a_\delta(x) = \prod_{1 \le i < j \le n} (x_i - x_j)$ is the Vandermonde determinant (in general $a_\lambda(x)$ is the determinant of the matrix $(a_i^{\lambda_j})_{i,j \in [n]}$). -$x^{\sigma \delta}$ means $x_1^{(\sigma \delta)_1} x_2^{(\sigma \delta)_2} \ldots x_n^{(\sigma \delta)_n}$. -Also $(\sigma \delta)_i$ denotes the $\sigma(i)$-th component of $\delta$, namely $n-i$. -Finally the translation operator $T_i = T_{q,x_i}$ is defined as -$$ T_{q,x_i}f(x_1, \ldots, x_n) = f(x_1, \ldots, x_{i-1}, q x_i , x_{i+1} ,\ldots, x_n).$$ -I like to think of the translation operator as the quantized version of the differential operator $I + \partial_i$, where $q-1$ is analogous to the Planck constant(?). -If we write $D(X;q,t) = \sum_{r=0}^n D_{n-r}(q,t) X^r$, then -Macdonald polynomials $p_\lambda(q,t)$ are simply simultaneous eigenfunctions of these operators. When $q=t$ they become Schur polynomials, defined by $s_\lambda = a_{\delta +\lambda} / a_\delta$. When $q= t^\alpha$ and $t \to 1$, we get Jack symmetric polymomials, which are eigenfunctions of a Metropolis random walk on the set of all partitions that converge to the so-called Ewens sampling measure, which assigns probability proportional $\alpha^{\ell(\lambda)} z_\lambda^{-1}$. When $q = 0$, they become the Hall-Littlewood polynomials and when $t=1$ they become the monomial symmetric polynomials etc. -I was told repeatedly by experts that Macdonald polynomials exhaust all previous symmetric polynomial bases in some sense. Does anyone know a theorem that says that every family of symmetric polynomial under some conditions can be obtained from Macdonald polynomials by specializing the $q$ and $t$? - -REPLY [14 votes]: There are plenty of polynomials, and only a few are specializations of Macdonald polynomials. As a polynomial botanist, the following is a very incomplete family tree. -A more comprehensive list of symmetric functions and generalizations can be found here. - -Edit: I added a few more generalizations, but it is getting crowded. Click on the image for a larger version.<|endoftext|> -TITLE: The connected components of the free loop space -QUESTION [8 upvotes]: I am trying to understand the topology (in terms of homology groups) of the free loop space $\Lambda M$ of nice spaces (Complete Riemannian connected finite dimensional manifolds $M$). I see the free loop space (of H^1 loops) as a Hilbert manifold, cf. Klingenbergs book. If the manifold $M$ has a non-trivial fundamental group, the free loop space has as many connected components as there are conjugacy classes in $\pi_1(M)$. How much do these components of $\Lambda M$ differ? Are these components all homotopy equivalent? For the circle the answer is yes, because all components of the free loop space are homotopy equivalent to the circle itself. -The following question is related to my question -Are the path components of a loop space homotopy equivalent? -However, I cannot seem to use the answer to this question directly, because I cannot concatenate two free loops, but maybe I am missing something obvious. - -REPLY [2 votes]: You might like to look at my preprint -http://arxiv.org/abs/1003.5617 -on the homotopy 2-type of a free loop space $LX$. It assumes that $X$ is a 2-type, i.e. the classifying space of a crossed module, and then gives precise formulae for crossed modules representing the 2-types of the components. -I am aware that the main interest in free loop spaces seems to be their homology, and I can't see how these results help on that. -The paper is planned to be revised with Chris Wensley and to include specific computer calculations, hence the delay. -I might as well quote the theorem. -Let $\mathcal M$ be the crossed module of groups $\delta: M \to P$ -and let $X=B\mathcal M$ be the classifying space of $\mathcal M$. Then the -components of $LX$, the free loop space on $X$, are determined by -equivalence classes of elements $a \in P$ where $a,b$ are equivalent -if and only if there are elements $m \in M, p \in P $ such that -$$b= p + a + \delta m -p. $$ Further the homotopy $2$-type of a -component of $LX$ given by $a \in P$ is determined by the crossed -module of groups $L\mathcal M [a]=(\delta_a: M \to P(a))$ where -(i) $P(a)$ is the group of elements $(m,p)\in M \times P$ -such that $\delta m= [a,p]$, with composition $(n,q)+(m,p)= -(m+n^p,q+p)$; -(ii) $\delta_a(m)= ( -m^a + m,\delta m)$, for $m \in M$; -(iii) the action of $P(a)$ on $M$ is given by $n^{(m,p)}= n^p$ for $n \in M, (m,p) \in P(a)$. -In particular $\pi_1(LX,a)$ is isomorphic to Cok $ \delta_a$, and -$\pi_2(LX,a) \cong \pi_2(X,*)^{\bar{a}}$, the elements of -$\pi_2(X,*)$ fixed under the action of $\bar{a}$, the class of $a$ -in $G=\pi_1(X,*)$.<|endoftext|> -TITLE: Where is the Euler/Goldbach correspondence? -QUESTION [6 upvotes]: I know that there is a 1965 volume containing the Euler/Goldbach correspondence, but I'm interested in looking at the original manuscripts. I'm not finding anything at University of Basel or Berlin-Brandenburg Academy of Sciences. Any help would be appreciated! - -REPLY [3 votes]: The Euler-Goldbach correspondence has been included in the Euler Archive, part of the MAA digital library, at: -http://eulerarchive.maa.org/correspondence/correspondents/Goldbach.html<|endoftext|> -TITLE: Comparables to Journal of Algebra, Journal of Pure and Applied Algebra -QUESTION [26 upvotes]: It was recently suggested to me to seek comparable, alternative, journals to the above two (I am not interested in discussing why one would want to do so here). I am wondering if anyone has suggestions of comparable journals to these two, which I rate as middle-of-the-road specialized journals in algebra. -Optimal answers will be names of journals whose likelihood of accepting any given paper, and esteem amongst people in the field, correlates highly with these two; so, they will consequently be specialized in algebra, and of approximately equal caliber. Also, it would be good if the rigor of the referee process is comparable. - -REPLY [4 votes]: The Journal of Commutative Algebra is a more specialized journal that I believe tries to be of similar quality to the Journal of Algebra. -In particular, they have recently asked their referees whether they would recommend that this paper would also be accepted at the Journal of Algebra and state that they are shooting for comparable quality.<|endoftext|> -TITLE: A p-adic analogue for a formula of Riemann? -QUESTION [13 upvotes]: This might be naive question but I was wondering whether a p-adic analogue of the following (shockingly) beautiful formula $$\zeta(s)\Gamma(s) = \int_0^\infty \frac{t^{s-1}}{e^t-1} dt$$ (vaild for $\mathrm{Re}(s)> 1$ and with all the usual notations) of Riemann is (well) known. -So with a p-adic analogue I'm of course aiming at an integral expression for the product of the classical p-adic zeta function $\zeta_p$ that interpolates $\zeta(s)$ in its values at the negative even integers and $\Gamma_p$ might be taken to be Morita's p-adic Gamma function. (Note that up to a power of $t$ the right hand side of the above formula can be said (in a catchy way) to be the Mellin transform of the generating function of the Bernoulli numbers (which essentially give the values of $\zeta(s)$ at the negative even integers...)). -What is "of course" lurking/hidden behind the above formula is the functional equation of the (completed) Riemann zeta function (for the time being I haven't learned/understood yet how this works in the p-adic setting, but I heard that in the work of Perrin-Riou, e.g., questions concerning functional equations of zeta functions in the p-adic world are dealt with (at least conjecturally)). -So, in a broader sense my question aims at understanding how the functional equation works in the p-adic setting and whether there are "nice functions" behind, that implement the mechanism (in the complex, global setting this is related, e.g., to theta functions and Bernoulli numbers, and the above formula is one incarnation of the eternal beauty of this area of the (mathematical) universe). -Thank you very much in advance for any help! -EDIT: I'd like to make one aspect of my question more precise after having a glance at the nice Bourbaki talk of Colmez suggested by Olivier below. -For example, the p-adic zeta function $\zeta_p$ of Kubota and Leopoldt can be constructed by constructing a "complicated" measure $\mu_{\zeta_p}$ on $\mathbb Z _p^\times$ by (making use of the Coleman map) and showing then that $$\int _{\mathbb Z_p^\times} \chi(g)^k d\mu_{\zeta_p} = (1-p^{k-1}) \zeta(1-k),$$ for $k$ an even and positive integer. (For $k$ odd we get 0). In the same spirit it is also possible to approximate p-adically special values of the completed zeta function (suitably normalized) which then satisfy a corresponding functional equation. My (very vague) question here is (maybe it's not what one should ask) whether one can write a (completed) p-adic zeta function in terms of an integral over a nice space with a simple measure but more interesting functions. In the description above of $\zeta_p$ the measure is very complicated but the function one integrates is quite easy. For example, is there some sort of p-adic analogue of theta functions that would do the job (suitable interpreted perhaps)? -(I know that to $\mu_{\zeta_p}$ there corresponds a certain (formal) power series via the Mahler transformation but I don't see a nice interpretation of this "function" either at the moment...) -(My apologies if this is way too vague for you...) - -REPLY [11 votes]: Yes, there is. -And in surmising correctly that finding a $p$-adic analogue of this formula will provide an understanding of the functional equation of $p$-adic $L$-functions, you have just got a glimpse of what are now called the Coleman map, the Bloch-Kato exponential map, explicit reciprocity laws, and from then on the $p$-adic Langlands program. The article -Théorie d'Iwasawa des représentations de de Rham d'un corps local Annals of Math 148 or the survey Fonctions $L$ $p$-adiques Séminaire Bourbaki 851 will tell you (much much) more.<|endoftext|> -TITLE: Is there an algorithm known to decompose quiver representation? -QUESTION [9 upvotes]: We have a finite dimensional representation of a finite quiver over, say, the rationals. Is there an algorithm known to decompose this representation into its irreducible components? -A related question: we have two (finite-dimensional) representations of a finite quiver. Is there an algorithm to check if one is a summand of the other? -Thanks a lot for hints and pointers! - -REPLY [8 votes]: If you want to decompose a finite-dimensional representation over, say, $\mathbb{Q}$ into its indecomposables over, say, $\overline{\mathbb{Q}}$, there is not only an algorithm but an efficient one (at least theoretically efficient: polynomial-time). See -Chistov, A., Ivanyos, G. and Karpkinski, M. Polynomial Time Algorithms for Modules over Finite Dimensional Algebras, ISSAC 1997. -It works more generally over finite fields, and the real or algebraic closures of number fields. -Not only that, but they also show how to test efficiently if two such representations are equivalent. These two pieces together answer your second question (when is one rep, say $R_1$, contained as a summand in another $R_2$): decompose both into indecomposables, then see if every indecomposable in $R_1$ appears in $R_2$ with greater or equal multiplicity (test each indecomposable of $R_1$ for equivalence to each indecomposable in $R_2$). -Update: Brooksbank and Luks (J. Algebra 2008, or freely available author's copy) provide another algorithm for these problems that is apparently efficient in both theory (polynomial-time) and practice (based on experiments in MAGMA). (This is all over finite-dimensional algebras; the OQ didn't specify whether the quiver could have directed cycles or not...)<|endoftext|> -TITLE: Locally Indicable Hyperbolic Groups -QUESTION [9 upvotes]: It is unknown whether a hyperbolic group is residually finite. Is it known under the additional hypothesis of locally indicability? Namely: Is a locally indicable hyperbolic group, residually finite? - -REPLY [11 votes]: The answer is "it is unknown". There are many potential counterexamples to the conjecture that all hyperbolic groups are residually finite. For example, let $\phi, \psi$ be two injective (but not surjective) emdomorphisms of the free group $F_2$. Consider the corresponding multiple ascending HNN extension of $F_2$, $G=\langle F_2,t,s| x^t=\phi(x), x^s=\psi(x), x\in F_2\rangle$. That group has a homomorphism onto the free group $\langle t,s\rangle$ whose kernel is locally free. So $G$ is locally indicable. It is an open problem whether $G$ is residually finite provided its presentation satisfies a small cancelation condition (see Problem 5.2 here ).<|endoftext|> -TITLE: free loop space and invariant forms -QUESTION [13 upvotes]: Cartan proved that for a connected compact Lie group $G$ the left invariant differential forms yield the correct cohomology of $G$. The same argument works for a connected compact $G$-manifold: the idea is to "average" left invariant forms on $G$ using a Haar measure. -Can we extend this result for non compact infinite dimensional manifolds? In particular, consider the free loop space $LM$ of a manifold $M$; this is an infinite dimensional $S^1$-manifold. Is there a way to compute the cohomology of $LM$ using a model of "invariant forms" and the idea of averaging? -By a result of Chen, we know that iterated integrals of differential forms in $M$ yield the correct cohomology of $LM$. Is this model related to Cartan's story of invariant forms? -These questions are a bit vague, but I guess how to make them precise is part of my question. - -REPLY [5 votes]: Iterated integrals define a map -$$ -\sigma: C(\Omega(M)) \to \Omega(LM) -$$ -where $C(\Omega(M))$ is the cyclic bar complex of $\Omega(M)$. It has various nice properties; for instance, it induces an isomorphism in cohomology, when $M$ is simply-connected. -Unfortunately, the forms in the image of $\sigma$ are not invariant with respect to the $S^1$-action on $LM$, which probably means that there is no relation of the kind you expect. -However, the forms in the image of $\sigma$ are basic (in particular, invariant) with respect to the action of $Diff([0,1])$ by reparameterization, but the image of $\sigma$ is not even dense in the space of all basic forms. (The image is characterized by a stronger invariance that cannot be formulated in terms of a group action: a kind of thin homotopy invariance). -Just in order to add a layer of confusion, I remark that if $G$ is a compact group acting on $M$, there exists a $G$-equivariant extension of the iterated integral map $\sigma$, introduced by Getzler-Jones-Petrack. It uses the Cartan model for $G$-equivariant forms on $LM$, and an equivariant version of the cyclic bar complex of equivariant forms on $M$.<|endoftext|> -TITLE: Does anyone recognize this quiver-with-relations? -QUESTION [6 upvotes]: Below I describe an infinite (but locally finite) quiver with relations. My question is whether anyone recognizes it and can provide appropriate pointers to the literature. I'm mainly interested in computing its Hochschild homology. The path category of the quiver (with relations) is related to the Temperley-Lieb category at non-generic (non-semisimple) values of the parameter. (At least I think it is -- I haven't double-checked the calculations.) -(This is not my area of expertise, so I apologize in advance if I'm not using standard terminology.) - -The quiver - -A vertex for each natural number $i=0, 1, 2, \ldots$. -An arrow $u_i : i \to i+1$ for each $i\ge 0$. -An arrow $d_i : i \to i-1$ for each $i \ge 1$. - -The relations - -$u_i u_{i+1} = 0$ for all $i\ge 0$. -$d_i d_{i-1} = 0$ for all $i \ge 2$. -$u_i d_{i+1} = d_i u_{i-1}$ for all $i\ge 1$. -$u_0 d_1 = 0$. - -Here's a more geometric description of the quiver. Start with a 2-dimensional mesh: A vertex for each pair of integers $(j,i)$ such that $j+i$ is even; "up" arrows connecting $(j, i)$ to $(j+1, i+1)$; "down" arrows connecting $(j, i)$ to $(j+1, i-1)$. Impose a commutativity relation for the boundary of each square of the mesh. Impose another relation that the composition of two "up" arrows is zero, and similarly for two "down" arrows. Declare that objects $(j, i)$ with $j<0$ are zero (i.e. any path which factors through such an object (vertex) is zero. Finally, mod out by the horizontal translation $(j, i) \mapsto (j+2, i)$. - -REPLY [4 votes]: The Hochschild homology (for the finite truncations) is computed in: -MR2248284 (2007j:16014) de la Peña, José A. ; Xi, Changchang . -Hochschild cohomology of algebras with homological ideals. -Tsukuba J. Math. 30 (2006), no. 1, 61--79.<|endoftext|> -TITLE: Is the support of a flat sheaf flat? -QUESTION [12 upvotes]: Note: in the following, all scheme/algebra morphisms should be assumed essentially of finite type. - -Geometric version: Let $X$ be a scheme flat over $S$ (both noetherian), and let $\mathscr{F}$ be a coherent sheaf on $X$, also flat over $S$. The scheme-theoretic support $\mathfrak{X}$ for $\mathscr{F}$ is a closed subscheme of $X$. Is it necessarily true that $\mathfrak{X}$ is flat over $S$? -Algebraic version: Let $B$ be a flat $A$-algebra (both noetherian), and let $M$ be a finitely generated $B$-module, also flat over $A$. Is it necessarily true that $B/\operatorname{Ann}(M)$ is flat over $A$? - -Motivation: the only way I know how to visualize a coherent sheaf is to visualize its support, which is a closed subscheme. I justify this by the fact that many of the properties of a coherent sheaf are shared by (the structure sheaf of) its scheme-theoretic support. For instance, they have the same associated points. In case $A$ is a DVR, this even provides a proof for the algebraic version above, since a module is flat over a DVR iff all its associated points map to the generic point. (see Angelo's comment below) -This general "visual intuition" tells me that the two (equivalent) statements above should be true. However, I cannot think of a good argument for this. Although it is not really essential to anything I am doing, it is bothering the heck out of me not to know whether this actually works, and distracting me from my other, more "essential" work. Thus, I would appreciate some help here. A positive answer will help me sleep at night (figuratively speaking); a negative answer will, hopefully, give me a useful counterexample against which to test my intuition in the future. -Second motivation: If the statement is true, then it provides evidence for a morphism from the Quot scheme to the Hilbert scheme, that--loosely speaking--takes a coherent sheaf to its support. (Thinking about it in these terms may also suggest solutions to mathematicians who--unlike me--have a great deal of experience with Quot and Hilbert schemes.) - -REPLY [9 votes]: Here is an algebraic construction. The way I think about it is based on these two facts: 1) when $A$ is regular domain, a module-finite A-algebra is flat iff it is Cohen-Macaulay (CM) of same dimension and 2) there are non-CM domains which admit a CM module of same dimension. -Now the concrete construction. Let $B = k[x,y,u,v]$, and we map $B$ onto $C = k[a^4, a^3b, ab^3, b^4]$ which is not CM (so $x$ maps to $a^4$ and $v$ to $b^4$). Let $P \subset B$ be the kernel. -Let $A = k[x,v]$ and $M = \bar C$, the integral closure of $C$. $M$ is actually $k[a,b]$ which is flat over $A=k[a^4, b^4]$. -However, the annihilator of $M$ over $C$ is zero, so the annihilator of $M$ over $B$ is $P$. But if $B/P \cong C$ is flat over $A$, it would be Cohen-Macaulay, contradicting our choice.<|endoftext|> -TITLE: Varieties cannot be isomorphic to proper open subsets -QUESTION [5 upvotes]: Let $X$ be an irreducible variety and let $U \subset X$ be a proper open set. Question : can there be a morphism $f : X \rightarrow U$ such that the composition $U \hookrightarrow X \stackrel{f}{\rightarrow} U$ is the identity? -My guess is that the answer is "no", but I can't seem to prove it. -One observation is that if $f$ exists, then it would give a procedure for taking a regular function $\phi : U \rightarrow k$ and extending it to all of $X$. But this can definitely be done in some cases; for instance, if $X = \mathbb{A}^2$ and $U = \mathbb{A}^2 \setminus \{(0,0)\}$. - -REPLY [19 votes]: http://en.wikipedia.org/wiki/Ax-Grothendieck_theorem<|endoftext|> -TITLE: Uniform setting for computing orders of algebraic groups over finite quotients of the integers? -QUESTION [11 upvotes]: A couple of recent questions on MO have involved the characters or the orders of specific finite groups of the form $G(\mathbb{Z}/n\mathbb{Z})$ for a familiar algebraic group $G$ defined over $\mathbb{Z}$ (implicitly as a group scheme) especially when $n$ is a prime power: here and -here. -It's reasonable to ask in what generality questions of the second kind can be treated uniformly. Here for example the question doesn't seem to require any knowledge of the structure of symplectic matrices, once one knows the orders of the finite symplectic groups over fields. (These have been written down in numerous papers and books.) In particular: - -For group schemes $G$ over $\mathbb{Z}$, in what natural generality is there a uniform procedure for computation of the orders of finite groups $G(\mathbb{Z}/p^k\mathbb{Z})$ as $p$ ranges over the prime numbrs? - -An old computation having a similar flavor occurs for the multiplicative group scheme (which is reductive but not simple) in the determination of the group of units in the ring $\mathbb{Z}_p$ of $p$-adic integers, written as an inverse limit of finite groups with successive quotients of order $p-1$ or $p$. Here the additive group scheme (viewed as the Lie algebra of the multiplicative group scheme) enters the picture, providing an iterated extension of the group of units of the residue field $\mathbb{F}_p$. See for example the book by Serre A Course in Arithmetic, II, section 3. -As George McNinch points out in his comment on Scott Carnahan's direct computation of the order for finite symplectic groups, a uniform approach is suggested by a paper of Serre, "Exemples de plongements des groupes PSL$_2(\mathbb{F}_p)$ dans des groupes de Lie simples", Invent. Math. 124 (1996), 3.1. Here G is a connected simple algebraic group over an algebraically closed field, essentially treated as a group scheme. Serre refers in turn to Demazure-Gabriel, Groupes Algebriques (1970), II, section 4, no. 3, a treatise influenced by the earlier Demazure-Grothendieck seminar SGA3. -In a Bourbaki talk, Chevalley showed how to view the simple adjoint groups, which he had constructed in a uniform way in 1955, as group schemes over $\mathbb{Z}$, a theme refined further by Kostant, and more recently by Lusztig in J. Amer. Math. Soc. 22 (2009). That seems to be a good setting for the question I've raised, though perhaps one can go further in the direction of reductive groups? -ADDED: The answers and comments are very interesting, but while I think further about them and the literature I should clarify that I'm taking for granted the standard (though nontrivial) formulas for the group orders over finite fields. And while it's natural (at least for Chevalley groups) to start over the ring of integers, there is certainly a passage to local rings implied here. The specific prime stays in the background, since unlike many questions in Lie theory this kind of computation doesn't distinguish "good" and "bad" primes: whatever is done should apply uniformly to all primes. Meanwhile a result in Appendix A.5 of the recent book by Conrad-Gabber-Prasad on pseudo-reductive groups has been pointed out to me. This suggests that groups aren't so essential to my question, but only a well-behaved class of schemes (again assuming that one can already count their points over finite fields). - -REPLY [4 votes]: I think the assumptions needed on a group scheme $G$ over $\mathbf{Z}$ are that $G$ -should be smooth, affine, and of finite type over $\mathbf{Z}$. Actually, -I'm going to take the point of view that $p$ is fixed and so I'll suppose that $G$ is smooth, affine, and of finite type over the local ring $\mathbf{Z}_{(p)}$. -I'm going to write $\Lambda = \Lambda_r = \mathbf{Z}/p^r\mathbf{Z}$ and $F = \mathbf{F}_p = \Lambda/p\Lambda$. -To give a formula for $|G(\Lambda)|$, you should have a formula for the number of points of the special fiber -- i.e. for $|G(\mathbf{F}_p)|$ (which is OK e.g. if the special fiber is reductive by well-known order formulas as mentioned in Marty's answer). -Let $H$ be an affine group scheme over $\Lambda$ (say, arising by base-change from -the group scheme $G$ over $\mathbf{Z}_{(p)}$). -The result in Demazure-Gabriel to which Serre refers in the article "Exemples de plongements..." (as discussed in the question) implies that the kernel $I$ -of the natural mapping $H(\Lambda_k) \to H(\Lambda_{r-1})$ -identifies naturally with the $F$-vector space -$$Hom_{\Lambda}(\omega_{H/\Lambda},p^{r-1}\Lambda)$$ -where $\omega_{H/\Lambda}$ is the module of Kahler differentials of $H$ over $\Lambda$. -In particular, $|I| = p^d$ where $d = \dim_F (\omega_{H/\Lambda} \otimes_{\Lambda} F)$. -If $H$ is smooth and of finite type over $\Lambda$ (e.g. if $H = G_{/\Lambda}$ for -$G$ smooth over $\mathbf{Z}_{(p)}$) then the natural mapping -$H(\Lambda_r) \to H(\Lambda_{r-1})$ is surjective (say, by [SGA1, Corr. III.5.3]) -and $\omega_{H/\Lambda}$ is a free $\Lambda$-module of rank $\dim H_{/F}$. -And in this case, one gets the formula for $|H(\Lambda)|$ given in Stasinski's answer. -And if $H = G_{/\Lambda}$ then of course $H(\Lambda) = G(\Lambda)$. -Remarks: -(1) It seems to me (?) that the result just quoted from Demazure-Gabriel answers -the conjecture from the end of Greenberg's 1963 paper mentioned in Stasinski's answer. -(2) Let $k$ be (say) an algebraically closed field, and let $W = W_r$ be the -ring of Witt vectors of length $r$ over $k$. Then our smooth group scheme $G$ as before determines a group scheme -$H = G_{/W}$ over $W$. By "Greenberg's functor" we can view the group -of points $G(W) = H(W)$ as a linear algebraic group over $k$. -If say $r=2$, then there is a strictly exact sequence of linear algebraic groups -$$ (*) \quad 0 \to V \to H(W) \to G_{/k} \to 1$$ -where $V = $ Lie$(G_{/k})^{[1]}$ is the vector group on which $G_{/k}$ acts via the first Frobenius twist of the adjoint representation Lie$(G_{/k})$. -I mention this because one somehow doesn't "see" the Frobenius twist when the residue -field is $\mathbf{F}_p$, but the Frobenius twist -is important in general. -Indeed, if say $G_{/k}$ is a simple alegbraic group, the exact sequence $(*)$ is not split -- it corresonds -to a non-zero class in $H^2(G_{/k},$Lie$(G_{/k})^{[1]})$. -So the Frobenius -twist is crucial since most of the time $H^2(G_{/k},$Lie$(G_{/k})) = 0$.<|endoftext|> -TITLE: Reference request: Ehrhart's conjecture on the geometry of numbers -QUESTION [9 upvotes]: Conjecture (Ehrhart). If a convex body $K \subset {\mathbb R}^n$ has its barycenter at the origin and contains no other point with integer coordinates, the volume of $K$ is less than or equal to $(n + 1)^n/n!$. -Ehrhart proved this for $n = 2$ and for simplices in any dimension (see his paper in J. Reine Angew. Math. 305, (1979) 218-220 and the references therein). These results and the conjecture are also cited in the book "Unsolved Problems in Geometry" by Croft, Falconer, and Guy. -Does anybody know whether anything interesting has been done on this conjecture? -For example, is the following weaker version of the conjecture known to be true? -Weaker version of Ehrhart's conjecture. There exists a universal constant $C > 1$ such that the volume of every convex body $K \subset {\mathbb R}^n$ satisfying the hypotheses of the conjecture is less than or equal to $C^n (n + 1)^n/n!$. -Addendum 28/08/2013: The paper with Balacheff and Tzanev on which the proof of the weak version is based is now available as arXiv:1308.5522. - -REPLY [10 votes]: I guess it must be in bad taste to answer my own question, but I now know a bit more about this problem and maybe there are other people interested in references to it. -As I mentioned in the question, Ehrhart himself settled the $2$-dimensional case and the case where the convex body is a simplex. Apparently nothing interesting was done until late last year when Robert J. Berman and Bo Berndtsson http://arxiv.org/abs/1112.4445 used PDE techniques to settle the case where the convex body is dual to a Fano polytope. -Reminder. A Fano polytope is a lattice polytope such that the vertices of each facet define a basis of the lattice. -As for the weak version of the Ehrhart conjecture: it is true. The preprint will be soon on the ArXiv. -Addendum. Here is the proof of the weak (or asymptotic) version of the Ehrhart conjecture modulo the following result that will appear in a forthcoming paper by Balacheff, Tzanev, and myself. -Theorem (ABT). If the origin is the unique integer point in the interior of a convex body $K \subset \mathbb{R}^n$, then the volume of the dual body $K^*$ is at least $(\pi/8)^n (n+1)/n!$. -The conjecture is that the volume of $K^*$ is at least $(n+1)/n!$, but we were able to show this only for $n=2$. -Now for the weak version of the Ehrhart conjecture: let $K \subset \mathbb{R}^n$ be a convex body with its barycenter at the origin and containing no other integer point. By the Blaschke-Santalo inequality (actually a version of it that holds for asymmetric bodies and their duals with respect to the barycenter, but I don't remember to whom it is due!!), we have -that $\epsilon_n^2 \geq |K||K^*|$, where $\epsilon_n$ is the volume of the Euclidean unit ball. -This inequality, together with the ABT theorem yields -$$ -\epsilon_n^2 \geq |K||K^*| \geq |K| (\pi/8)^n (n+1)/n! -$$ -or $|K| \leq (8/\pi)^n\epsilon_n^2 n!/(n+1)$. -Now, the proof reduces to showing that the quantity -$$ -\left(\epsilon_n^2 (n!)^2/(n+1)^{(n+1)}\right)^{1/n} -$$ -is bounded above by a quantity $C$. This is just the $n$th root of the quotient of the volume product of the ball and the volume product of the simplex and, therefore, standard fare in asymptotic geometry. This quantity is asymptotically $2\pi / e$ and seems to be always less than $3$. I'll check and edit when I have a bit more time. -I worked out this proof at the early stages of my collaboration with Balacheff (before Tzanev joined us), but this result got dropped out of the (forthcoming) paper. I reproduce it here from my notes, but I'm a bit rusty on the details. Very possibly, one can bypass ABT and apply a judicious mixture of Minkowki's lattice point theorem, Rogers-Shephard inequality, the Bourgain-Milman theorem, and the asymmetric version of Blaschke_Santalo to get the proof. In any case, those are the basic ingredients that go in the proof above.<|endoftext|> -TITLE: Erdos Kac for imaginary class number -QUESTION [6 upvotes]: In answer to A coverage question Cam mentions an article by SOUND. I have been running a computer program for THIS and would like to know if there are a reasonable average and standard deviation for the class number, related to Dirichlet's formula for $d > 4$ -$$ h(-d) = \sqrt d \; L(1, \chi_{-d}) \; / \; \pi. $$ -Sound writes - -Typically $L(1, \chi_{-d})$ has - constant size; rarely does it fall - outside the range $(1/10,10).$ - -This suggests a possible calculation of standard deviation, as I am seeing articles about "second moments" of the zeta function and $L$-functions, although nothing I can interpret. -Note: The original Erdos-Kac may be of an entirely different nature; it says that, in the long run, the number of prime divisors of a number $n$ is normally distributed with mean $\log \log n$ and standard deviation $\sqrt{ \log \log n},$ this being the colloquial description of a precise statement. -So, that is the question, average and variance for the class number of imaginary quadratic fields. -P. S. The computer program I am running is restricted to the above with $d \equiv 3 \pmod 4,$ but does not rule out square factors of $d$ ahead of time. In the first occurrence of such $d$ with a target class number, $d$ is almost always squarefree. Indeed, with class numbers up to 4000, the only exception is class number 104, which first occurs at $ d= 9359 = 7^2 \cdot 191.$ If that issue matters, I would be delighted to hear about it... -EDIT: Based on Noam's comment, maybe it is $\log h(-d)$ that has a nice mean and variance. -EDIT ANOTHER: the most interesting case is $d \equiv 3 \pmod 4$ where $d$ is prime. Noam had pointed out in one of the threads that primality is required to achieve an odd class number. - -REPLY [11 votes]: There are many papers about strong probabilistic models of $L(1,\chi_d)$, in particular by Granville and Soundararajan. These are quite precise (basically, because the Euler product at 1 is "almost" absolutely convergent, one can model its value by a random Euler product, and even prove that this model is close to the truth when taking discriminants of bounded size.) -See for instance: http://www.dms.umontreal.ca/%7Eandrew/PDF/L1chi2002.pdf<|endoftext|> -TITLE: Efficiently computing a few localized eigenvectors -QUESTION [8 upvotes]: Let $H = \triangle + V(x) : \mathbb{R}^2 \rightarrow \mathbb{R}^2$. I am interested in domain decomposition for an eigenproblem involving $H$. -The lowest 1000 eigenfunctions of $H$, $ \psi_i $, can be partitioned using a region, $\Omega \subset \mathbb{R}^2$, such that each $\psi_i$ localizes either inside of $\Omega$ or outside of $\Omega$. $\Omega$ is not a subspace of $\mathbb{R}^2$ as it may be an oddly shaped region. -Label the inner eigenfunctions $\psi_i^{in}$ and the outer ones $\psi_i^{out}$. There's only about 10 $\psi_i^{in}$s. Given $\Omega$, my goal is to efficiently compute the $\psi_i^{in}$. -One way to find the $\psi_i^{in}$ would be to discretize, compute all 1000 $\psi_i$s, and then partition. This is what I do now (5-point stencil for $\triangle$ on a $10^3 \times 10^3$ grid). The problem is that this requires diagonalizing over a 1000 dimensional space in order to get 10 eigenvectors. It seems like there would be a cheaper way to compute the $\psi_i^{in}$. -Edit: I reposted to https://scicomp.stackexchange.com/questions/1396/efficiently-computing-a-few-localized-eigenvectors#comment2200_1396 and hopefully clarified the problem statement. -Edit I think I can solve this if I can at least figure a way to solve -\begin{equation} -\max \psi^T H \psi \text{ subject to } P\psi = \psi \text{ and } \psi^T \psi = 1 -\end{equation} -where $P$ is projection onto the space of functions localized over $\Omega$. My guess is that this will end up looking like power iterations with a projection step built in between matrix applies. If this is doable then something like inverse iteration should be doable which will give me what I want. - -REPLY [9 votes]: To build on Federico's answer, why not run a restarted Lanczos iteration but compute harmonic Ritz vectors to get approximations to the interior eigenpairs? For something so small, though, why not just diagonalize, as has already be stated.<|endoftext|> -TITLE: Proofs in the same vein as Ax-Grothendieck -QUESTION [10 upvotes]: I would like to see other examples of (ideas of) proofs and results in the same vein as the proof of the Ax-Grothendieck theorem. To explain what I mean by "in the same vein", I will quote from the corresponding wikipedia article: - -This method of proof is noteworthy in that it is an example of the idea that finitistic algebraic relations in fields of characteristic $0$ translate into algebraic relations over finite fields with large characteristic.Thus, one can use the arithmetic of finite fields to prove a statement about $\mathbb{C}$ even though there is no non-trivial homomorphism from any finite field to $\mathbb{C}$.The proof thus uses model theoretic principles to prove an elementary statement about polynomials.The proof for the general case uses a similar method. - -Edit(after the comment by Angelo to Martin's answer below): answers along the line of "using finite fields for problems concerning infinite fields" are also welcome. - -REPLY [2 votes]: I saw a talk two weeks ago by Michel Brion that uses reduction to finite fields to prove that every algebraic semigroup (not necessarily affine) has an idempotent. Over a finite field it is clear because finite semigroups have idempotents.<|endoftext|> -TITLE: What is (explicitly) known about the SL(n,C) character varieties of 3-manifolds? -QUESTION [9 upvotes]: The $SL(2,{\bf C})$ character variety of a 3-manifold with 1-cusp $M$ (like a knot complement in the 3-sphere) essentially coincide with the variety defined by the A-polynomial. Those polynomials are known explicitly for torus knots, and tabulated for many hyperbolic knots, e.g. on Culler's webpage. -I am wondering if the $SL(n,{\bf C})$ character varieties (denote it $X_n(M)$) are also known (by a set of polynomial equations for instances) in some examples, like the complement of torus knots, or of the figure-8 knot ? -Another question: it is known that $X_n$ at a smooth point has complex dimension $(n - 1)$. Does there exist "remarkable" subvarieties of complex dimension $1$ in $X_n(M)$ ? - -REPLY [4 votes]: The answers at this related question might be of interest. -As implied by the comments, there is a large body of work on this topic. -Here are some authors (definitely not exhaustive) who have worked out the exact structure of character varieties of 3-manifold groups: - -Michael Heusener -Emily Landes -Melissa Macasieb -Vicente Muñoz -Kate Petersen -Joan Porti - -In particular, the answer to your first question is yes. See: - -Geometry of the SL(3,C)-character variety of torus knots, by -Vicente Muñoz, Joan Porti for torus knots and $n=3$, and -The SL(3,C)-character variety of the figure eight knot, by -Michael Heusener, Vicente Munoz, Joan Porti for the figure eight knot and $n=3$. - -For your second question, I recommend reading generalities about tangent spaces to character varieties in: -Character Varieties, by Adam Sikora. -With respect to local deformations for (finite volume hyperbolic) 3-manifold groups, the following references answers your second question positively: - -Local coordinates for SL(n,C) character varieties of finite volume hyperbolic 3-manifolds, by Pere Menal-Ferrer, Joan Porti and, -Twisted cohomology for hyperbolic three manifolds, by Pere Menal-Ferrer Joan Porti - -As to the third question, I am not sure what "remarkable" means here, so I will just leave that one alone. -Another interesting part of the story of character varieties of 3-manifold groups concerns dynamics. See the very nice exposition by Dick Canary titled Dynamics on character varieties: a survey (and references therein).<|endoftext|> -TITLE: Error correcting codes - basic question -QUESTION [7 upvotes]: Hi, -I have a basic question regarding error correcting codes. Suppose I want to encode a finite information $F$ (say a finite string) into a string $x$ of $n$ bits ($n$ can be as large as you want), with the following requirement: knowing any $k$ bits of the string $x$ (together with their respective positions in $x$), one can fully retrieve the information $F$. For which $(n,k)$ can this be achieved? What I am hoping to get is $k=o(n)$, but I do not know whether this is possible. -Thanks in advance! - -REPLY [3 votes]: To elaborate on Anadim's answer, if $C$ is a code of length $n$, then any codeword in $C$ is uniquely determined by any $k$ of its positions if and only if $C$ has minimum distance at least $n-k+1$. The Singleton bound says that, over an alphabet of size $q$, a code of length $n$ and minimum distance $n-k+1$ has size at most $q^k$. So if $k = o(n)$ then there are no asymptotically good codes satisfying your requirements. -If $q$ is large then the Singleton bound is attained by any MDS (maximum distance separable) code, for example, by the Reed–Solomon codes mentioned by Anadim. So if you have $q^m$ possible strings then, using Reed–Solomon codes, the feasible pairs are $(n,k)$ for $n$ and $k$ such that $q \ge n \ge k \ge m$. The restriction $q \ge n$ means this is not an asymptotic result, but for practical purposes, if you can take $q$ large, you can then make $k$ small compared to $n$. -Over the binary alphabet, the only MDS-codes of length $n$ are the repetition code of size $2$, the parity check code of size $2^{n-1}$, and the complete code consisting of all $2^n$ words of length $n$. In this case Hamming's packing bound is stronger than the Singleton bound, and the Plotkin bounds are stronger still when the minimum distance is large compared to $n$. In particular, one of the Plotkin bounds states that if $k < n/2$ then a binary code of length $n$ and minimum distance $n-k+1$ has size at most $2(n-k+1)$. This shows that if $k=o(n)$ then any binary code satisfying your requirements is necessarily very small. -If you are willing to take $k\approx n/2$ then it might work well to use a shortened Hadamard code of length $2k-1$, size $2k$ and minimum distance $k$. These attain one of the Plotkin bounds, so are the largest possible binary codes of length $2k-1$ that allow any codeword to be reconstructed from any $k$ of its positions. (A special case when $k=4$ is the subcode of the binary Hamming code of length $7$ consisting of all words of even weight.)<|endoftext|> -TITLE: random hyperharmonic series -QUESTION [5 upvotes]: The Harmonic Series is defined as: -$\sum_{n} \frac{1}{n}$ where $n=1,2,3,4....$. -This series is known to be divergent. -A generalization of this series can be made by raising each term to $p$: -$\sum_{n} \frac{1}{n^p}$ which is also known as the hyperharmonic series and is known to be convergent when $p>1$. -On the other hand, for $p=1$, if the signs of the terms are alternating the sum: -$\sum_{n} \frac{(-1)^n}{n}$ is convergent and approaches $\ln{2}$. -A natural extension would be to introduce randomness in the sign of each terms. -$\sum_{n} \frac{\epsilon_{n}}{n}$ where $\epsilon_{n}$ is defined by the probability of its outcome: $P(\epsilon_{j} = 1)=P(\epsilon_{j} = -1)=1/2$. This is called the random harmonic series in Schmuland (http://www.stat.ualberta.ca/people/schmu/preprints/rhs.pdf). -My question is: What would happen if we generalize this to the case of the random HYPERharmonic series? What would be the distribution of the result of the summation? - -REPLY [5 votes]: The case $p=2$ is treated briefly in the final section of the cited -Schmuland paper, which gives a picture of the distribution. The observations -in the paper's first few sections adapt to arbitrary $p>1$: the distribution -is not expected to have a simple formula, but the moment-generating function -has a product formula (exhibited below), -which we can use to compute each power moment ${\mathbb E}(X_p^{2k})$ -of this random variable $X_p = \sum_n \epsilon_n/n^p$ -as a polynomial in $\zeta(2p), \zeta(4p), \zeta(6p), \ldots, \zeta(2kp)$. -[The odd-order moments vanish by symmetry.] -This is because the distribution of $X_p$ is the convolution of -an infinite series of distributions, the $n$-th of which is supported on -$\pm 1 /n^p$ each with probability $1/2$; therefore -$$ -{\mathbb E}(\exp(tX_p)) -= \prod_{n=1}^\infty \left(\frac12 e^{t/n^p} + \frac12 e^{-t/n^p} \right) -= \prod_{n=1}^\infty \cosh(t/n^p). -$$ -We can recover the power moments from the expansion of -${\mathbb E}(\exp(tX_p))$ in a Taylor series about $t=0$, -writing -$$ -\begin{eqnarray} -\log {\mathbb E}(\exp(tX_p)) -&=& \sum_{n=1}^\infty \log(\cosh(t/n^p)) \cr -&=& \sum_{n=1}^\infty -\frac12 \left(\frac{t}{n^p_{\phantom1}}\right)^2 -- \frac1{12} \left(\frac{t}{n^p_{\phantom1}}\right)^4 -+ \frac1{45} \left(\frac{t}{n^p_{\phantom1}}\right)^6 -- \frac{17}{2520} \left(\frac{t}{n^p_{\phantom1}}\right)^8 -+ - \cdots \cr -&=& -\frac{\zeta(2p)}{2} t^2 -- \frac{\zeta(4p)}{12} t^4 -+ \frac{\zeta(6p)}{45} t^6 -- \frac{17\zeta(6p)}{2520} t^8 -+ - \cdots -\end{eqnarray} -$$ -and exponentiating. -While the random variables $X_p$ may be no more than objects of curiosity, -similar constructions arise naturally in analytic number theory, such as -the value at a fixed $s>1$ of the Dirichlet $L$-series associated to a random -real character $\chi$. In this example, the terms $\chi(n)/n^s$ in the -Dirichlet series are correlated, but the Euler product -$L(s,\chi) = \prod_l (1 - \chi(l)/l^s)^{-1}$ has independent factors, so -$\log L(s,\chi)$ is an infinite convolution of the same kind. Likewise -if the $\epsilon_n$ were random complex numbers of unit length: -the analogue could be $\log L(s,\chi)$ for a random Dirichlet charater $\chi$ -that need not be real, or $\log \zeta(\sigma + it)$ for fixed $\sigma>1$ -and random real $t$. For these sums of complex-valued random variables, -the factors in the moment generating function get more complicated than -hyperbolic cosines, but are still tractable.<|endoftext|> -TITLE: Special arithmetic progressions involving perfect squares -QUESTION [13 upvotes]: Prove that there are infinitely many positive integers $a$, $b$, $c$ that are consecutive terms of an arithmetic progression and also satisfy the condition that $ab+1$, $bc+1$, $ca+1$ are all perfect squares. -I believe this can be done using Pell's equation. What is interesting however is that the following result for four numbers apparently holds: -Claim. There are no positive integers $a$, $b$, $c$, $d$ that are consecutive terms of an arithmetic progression and also satisfy the condition that $ab+1$, $ac+1$, $ad+1$, $bc+1$, $bd+1$, $cd+1$ are all perfect squares. -I am curious to see if there is any (decent) solution. -Thanks. - -REPLY [4 votes]: Michael Stoll has given a nice answer, but here is a 17th century argument. - Let $a$, $b$, $c$, $d$ be an arithmetic progression - with common difference $\Delta \ne 0$. - Suppose that - $$1 + a b = z^2_1, \quad 1 + a c = z^2_2, \qquad 1 + a d = z^2_3,$$ - $$1 + b c = z^2_4, \quad 1 + b d = z^2_5, \qquad 1 + c d = z^2_6.$$ - Consider the following quantities: - $$A = 2 z^2_2 z^2_5 - z^2_1 z^2_6,$$ - $$B = z_2 z_3 z_4 z_5 - \Delta \cdot z_1 z_6,$$ - $$C = z_1 z_3 z_4 z_6 - 2 \Delta \cdot z_2 z_5,$$ - $$D = z_1 z_2 z_5 z_6 - 3 \Delta \cdot z_3 z_4.$$ - Then one may easily compute that - $$A^2, B^2, C^2, D^2$$ - is an arithmetic progression. In particular, if the $z_i$ are rational, then we have - constructed an arithmetic progression of four squares, contradicting a theorem of Fermat. -To be careful, we also have to check that this arithmetic progression of squares has non-trivial difference. A little algebra shows this can only happen (assuming $\Delta \ne 0$) if one of the following equations holds: - $$ (1 + a d)(1 + b c) = 0, \quad (1 + a d)(1 + b c) = 2 \Delta^2.$$ -The first of these equations leads to no rational solutions (as in Michael's answer), the second says that a square ($z^2_3 z^2_4$) is equal to twice a (non-zero) square, which is also impossible.<|endoftext|> -TITLE: Is there an "Abelian envelope" 2-functor? -QUESTION [11 upvotes]: I'm looking for a notion of an Abelian category $\mathcal{A}$ "generated" by a given category $\mathcal{C}$ -More precisely I need something along the following lines. Denote $\mathcal{Ab}_2$ the 2-category of Abelian categories and $\mathcal{Cat}$ the 2-category of categories. We have the forgetful 2-functor $\mathcal{F}: \mathcal{Ab}_2 \rightarrow \mathcal{Cat}$. Is there an adjoint 2-functor $\mathcal{G}: \mathcal{Cat} \rightarrow \mathcal{Ab}_2$ ? -I suspect the answer is "yes" because it can be constructed along the following lines. Denote $\mathcal{Ab}$ the category of Abelian groups. For any category $\mathcal{C}$, the category $\mathcal{Hom(C, Ab)}$ is Abelian. Moreover, we have the natural functor $\mathcal{i:C \rightarrow Hom(Hom(C,Ab),Ab)}$. Thus $\mathcal{C}$ is embedded in the Abelian category $\mathcal{D:=Hom(Hom(C,Ab),Ab)}$ and we can take the Abelian category generated by $\mathcal{C}$ within $\mathcal{D}$. The result is supposed to be $\mathcal{G(C)}$ -However, the only construction I managed to search up is the Karoubi envelope which generates a pseudo-Abelian category. So either my purported construction is wrong or simply not popular. Which is it? -EDIT: I realized my construction amounts to $\mathcal{G(C):=Hom(C,Ab^{op})}$. At least for small $\mathcal{C}$ this is indeed adjoint to $\mathcal{F}$, provided we interpret $\mathcal{Ab_2}$ as having right exact functors for 1-morphisms. Here $\mathcal{C}$ embeds by applying opposite Yoneda and taking the freely generated Abelian group. - -REPLY [13 votes]: I'd be inclined to go in a slightly different direction in constructing the free abelian category generated by a category. I would do it in stages: - -First construct the free $Ab$-enriched category $F_1(C)$ generated by a category $C$. This would have the same objects as $C$, but the hom $F_1(C)(a, b)$ between two objects would be the free abelian group generated by the hom $C(a, b)$ in $C$. -Next, to any $Ab$-enriched category $C$, we may freely adjoin finite limits. We can do this by embedding $C$ into $(Ab^C)^{op}$ by the dual of the Yoneda embedding, and then cutting down to the full subcategory $F_2(C)$ of $Ab$-enriched functors which arise as limits of finite diagrams of opposites of representables $C(c, -)^{op}$ in $(Ab^C)^{op}$. -Next, to any finitely complete $Ab$-enriched category $C$, pass to the so-called ex/lex completion $F_3(C)$. The "lex" here refers to left exactness (i.e., closed under finite limits, and functors preserving such), whereas the "ex" refers to Barr exactness and functors preserving such. - -The point is that an abelian category is precisely a Barr-exact (finitely complete) $Ab$-enriched category (see Freyd-Scedrov's Categories, Allegories, page 90, where a Barr-exact category is what they call an effective regular category). So the free abelian category generated by a category $C$ is $(F_3 \circ F_2 \circ F_1)(C)$. I did not check carefully that the ex/lex completion, as described in the nLab article, lifts from the $Set$-enriched world to the $Ab$-enriched world, but I think it's okay. -Not yet sure how this relates to your proposed construction (there are some size considerations to, er, consider), but I think the answer to your first question is 'yes'. -Addendum: In response to Martin's comment, and as confirmation that the description above is correct, here is a paper by Carboni and Magno which in the course of things constructs the free abelian category generated by an additive category: see page 300. The free abelian category on the terminal category $\mathbf{1}$ is $C_{ex}$, where $C$ is the category opposite to that of finitely generated abelian groups. There may be simpler descriptions, but I'm not prepared to say more on that now.<|endoftext|> -TITLE: how many consecutive integers $x$ can make $ax^2+bx+c$ square ? -QUESTION [10 upvotes]: The following problem was raised in a Mathlinks thread: -If $a,b,c\in\mathbb Z$ such that $a\ne0$ and $b^2-4ac\ne 0$, for how many consecutive integers $x$ can $ax^2+bx+c$ ba a perfect square ? -The polynomial $-15x^2+64$ is obviously a square for the five numbers $x=-2,...,2$, but the method used for finding this in the above thread cannot be extended further. Should $5$ really be the best possible answer? -Has this problem been treated somewhere else? - -REPLY [13 votes]: To resonate with Noam Elkies' comments, it is conjectured that $8$ squares is the maximum for arbitrary $a$, and $4$ squares is the maximum for $a=1$. For $5$ symmetric squares the smallest known leading coefficients are $a=15$ and $a=-20$, while for $5$ increasing squares they are $a=60$ and $a=-56$. It is known that there are infinitely many examples with $5$ or $8$ symmetric squares, or with $6$ increasing squares. It is also known that there is no symmetric sequence of $7$ squares, and only finitely many of $10$ squares up to obvious equivalences (this one follows from Falting's theorem applied to a specific hyperelliptic curve). -Good starters are: -Browkin-Brzeziński: On sequences of squares with constant second differences, Canad. Math. Bull. 49 (2006), 481–491. -Bremner: On square values of quadratics, Acta Arith. 108 (2003), 95–111.<|endoftext|> -TITLE: What is "augmented algebra"? -QUESTION [8 upvotes]: Really sorry for this question, but googling for some time did not help me. I was trying to understand the meaning of the following phrase: -Let B be an augmented algebra over a semi-simple algebra T. -But I am stuck already with "augmented algebra"... -- can not find a definition on the web. - -REPLY [14 votes]: An augmented ring is simply a triple $(A,M,\epsilon)$ with $A$ ring, $M$ a left $A$-module and $\epsilon:A\to M$ a surjection of $A$ modules. One says then that $A$ is augmented over $M$. You can find this defined in Cartan-Eilenberg, for example. -Often, $M$ is itself a ring and the map $\epsilon$ is also a ring morphism. This is probably the situation you have.<|endoftext|> -TITLE: What can be proven in Peano arithmetic but not Heyting arithmetic? -QUESTION [25 upvotes]: Hi. I'll confess from the start to not being a logician. In fact this question came up not from research but during a discussion with a friend about whether the classical proof that $\sqrt{2}$ is irrational can be made acceptable to an intuitionist. (It can be.) -The question is: Are there any "natural" statements which can be proven in Peano Arithmetic, but not in Heyting Arithmetic (Peano Arithmetic but with a logic that does not admit the law of the excluded middle)? -In fact, any statements -- even pathological ones -- that can be proven in one but not the other would be interesting to me, since I wasn't able to come up with any. (Even after doing a few web searches!) But of course, the closer to the surface the better. - -REPLY [15 votes]: According to Harvey Friedman, the following theorem is provable in PA but not HA: - -Every polynomial $P:\mathbb{Z}^n \to \mathbb{Z}^m$ with integer coefficients assumes a value closest to the origin. - -That is, there is a value which is at least as close to the origin, in the -Euclidean distance, than any other value. This is unprovable in HA even when $m=1$.<|endoftext|> -TITLE: partition into the orbits of a dynamical system -QUESTION [5 upvotes]: Let $T$ be a measure-preserving invertible transformation of a Lebesgue space, and let $P$ be the partition of the Lebesgue space into the orbits of $T$. -1) Is it true that $P$ is nonmeasurable (in the sense of Rokhlin) when $T$ is ergodic ? Why ? -2) If true, is it a necessary and sufficient condition for ergodicity ? -3) Is it true that the (complete) $\sigma$-field generated by $P$ is the $\sigma$-field of $T$-invariant sets ? - -REPLY [6 votes]: Although it appears you've already settled matters with the information in Jon's answer, I'll offer a quick summary and elaboration. -Let $(X,\mathcal{B},\mu)$ be a Lebesgue space (set + $\sigma$-algebra + probability measure) and $P$ the partition into orbits $\mathcal{O}(x) = \{T^n x \mid n\in \mathbb{Z}\}$ for an invertible measure-preserving transformation $T$. -For (3), it's exactly as you say: $\mathcal{B}$ contains each orbit $\mathcal{O}(x)$, and since a set $A$ is $T$-invariant if and only if it is a union of complete $T$-orbits, we get that the $\sigma$-algebra generated by $P$ is precisely the collection of $T$-invariant sets. The completed $\sigma$-algebra generated by $P$ is the collection of sets that are $T$-invariant mod $0$. -Consequently the answer to (1) is yes unless a single orbit carries full measure: ergodicity implies that every element of the $\sigma$-algebra generated by $P$ has measure $0$ or $1$, and consequently this $\sigma$-algebra is equivalent mod $0$ to the trivial $\sigma$-algebra. Thus for an ergodic transformation, $P$ is measurable if and only if a single partition element has full measure, which happens exactly when $\mu$ is supported on a single periodic orbit. -Finally, for (2), you observe correctly that non-measurability of $P$ follows as soon as $\mu$ has an ergodic component that is not a periodic orbit. For a concrete example, one may consider the map $T\colon [0,1]\to[0,1]$ that takes $x$ to $2x \pmod 1$, and the measure $\mu = \frac 12(\delta_0 + \lambda)$, where $\delta_0$ is the point mass on the fixed point at $0$, and $\lambda$ is Lebesgue measure. In this case $\mu$ is non-ergodic but $P$ is non-measurable. Or, if you prefer a completely non-atomic example, you can let $\nu_p$ denote the $T$-invariant measure on $[0,1]$ that comes from the $(p,1-p)$-Bernoulli measure on the full two-shift (where $p\in (0,1)$) and let $\mu$ be any convex combination of $\nu_p$ and $\nu_q$ for $p\neq q$.<|endoftext|> -TITLE: Real valued function whose derivative is nowhere continuous? -QUESTION [6 upvotes]: Let $f:\mathbb{R} \rightarrow \mathbb{R}$. Is it possible that $f$ has a derivative that nowhere continuous on its domain? Please provide an example if possible. -Thanks - -REPLY [11 votes]: I would say "no". $f$ is continuous, and that derivative is therefore of first Baire class. Such a function is continuous except on a set of first category. -Now I'll investigate Pietro's reference and see where (or if) I went wrong. - (checked) In fact, Pompeiu's derivative is continuous on a dense $G_\delta$ -set (where it vanishes) after all.<|endoftext|> -TITLE: A^1-invariant for Vector Bundles? -QUESTION [5 upvotes]: We know that if $X$ is a smooth connected variety over a field $k$, then any line bundle on $X\times_k\mathbb{A}^1$ is from a line bundle on $X$. This is simply because they have the same Picard group. This has a generalization, in fact, $X\times_k\mathbb{A}^1\to X$ induces an isomorphism of $CH_r(X)\to CH_{r+1}(X\times_k\mathbb{A}^1)$. -But I am thinking about vector bundles on $X\times_k\mathbb{A}^1$, I would not believe that any vector bundle on $X\times_k\mathbb{A}^1$ is from $X$. Can anyone give me a counter example ? I think it might be easy to take $X=\mathbb{P^1}$ where the vector bundles are completely split (into line bundles), while the same should not be true for $X\times_k\mathbb{A}^1$. Can anyone give me an example when I take $X$ to be an Abelian variety? for example an elliptic curve. - -REPLY [9 votes]: One way to think about $\mathbb{A}^n$-families of bundles on $X$ is in terms of $Ext^1$ groups of bundles on $X$. That is, suppose that $Ext^1(E,F) \simeq k^n$. Then, as Chris Brav mentions in the comments, we can think of this as an $\mathbb{A}^n$-point of the moduli of vector bundles on $X$, which over each point of $\mathbb{A}^n$ restricts to the bundle which is the corresponding extension of $E$ by $F$. -For example, if $X = \mathbb{P}^1$, then $Ext^1(\mathcal{O}(1),\mathcal{O}(-1))$ is $1$-dimensional. We can then think of the corresponding vector bundle on $\mathbb{P}^1 \times \mathbb{A}^1$ as being given by the transition function $g = \begin{pmatrix} z & t \\\ 0 & z^{-1} \end{pmatrix}$, where $t$ is the coordinate on $\mathbb{A}^1$. This gives a family of rank $2$ bundles which are isomorphic to $\mathcal{O} \oplus \mathcal{O}$ when $t \neq 0$, but which is isomorphic to $\mathcal{O}(1) \oplus \mathcal{O}(-1)$ when $t=0$. -You could do something similar on the elliptic curve, where you have a family of rank $2$ bundles generically isomorphic to the unique non-split extension of $\mathcal{O}$ by $\mathcal{O}$, but which specializes to the split extension. -In both of these examples, you can see that these bundles are not pulled back from $X$ because their transition functions cannot be the pull-back of a transition function for a bundle on $X$.<|endoftext|> -TITLE: The sequence $a_{n+1}=$ the greatest prime factor of $(xa_n+y)$ -QUESTION [8 upvotes]: Let $\operatorname{ GPF}(n)$ be the greatest prime factor of $n$, eg. $\operatorname{ GPF}(17)=17$, $\operatorname{ GPF}(18)=3$. -Is there a way to prove that the sequence $a_{n+1}=\operatorname{ GPF}(xa_n+y)$, eventually enter a cycle for all positive integers $x,a_0,y>0$? -Is there any set of positive integers $x,a_0,y>0$ such that -$a_{n+1}=\operatorname{ GPF}(xa_n+y)\operatorname{ LPF}(xa_n+y)$ diverges? -Where $\operatorname{ LPF}(n)\geq2$ is the least prime factor of $n$. - -REPLY [4 votes]: There is a paper on this problem, Mihai Caragiu, Recurrences based on the greatest prime factor function, JP J. Algebra Number Theory Appl. 19 (2010), no. 2, 155–163, MR2796479 (2012a:11010). The summary begins, -We introduce and discuss a generalized ultimate periodicity conjecture for prime sequences $\lbrace q_n\rbrace_{n\ge0}$ in which every term $q_n$ is recursively defined as the maximum element of the finite set $\lbrace P(A_jq_{n-1}+B_j)\mid j=1,\dots,k\rbrace$, where $P(x)$ represents the greatest prime factor of $x$, while $A_j$ and $B_j$ are fixed positive integers for $1\le j\le k$. -I haven't seen the paper, just the summary in Math Reviews.<|endoftext|> -TITLE: class numbers of $\mathbf{Q}(2^{1/n})$ -QUESTION [9 upvotes]: Calculating the class numbers of $\mathbf{Q}(2^{1/n})$ for small $n$ always yields $1$. Is it true for an infinite number of $n$s? Does applying Iwasawa theory to the false Tate curve tower $\mathbf{Q}(2^{1/p^n}, \mu_{p^n})$ help? - -REPLY [3 votes]: As KConrad says, the answer to the first question is "not yet known". For Iwasawa theory, you could consider the extension $L_\infty=\mathbb{Q}(\sqrt[p^\infty]{2},\zeta_{p^\infty})$ which is Galois over $\mathbb{Q}$ with Galois group isomorphic to $G=\mathbb{Z}_p^\times\ltimes \mathbb{Z}_p$ and some Iwasawa theory can tell you the structure (may be assuming that $2$ is a multiplicative generator $\pmod{p^2}$) of the $p$-part of the class group of your field, but nothing for $\ell\neq p$. -Already in the most basic situation where you consider the cyclotomic $\mathbb{Z}_p$-extension of the rationals, Iwasawa theory tells you that $p$ never divides the class number, and also that for all $\ell\neq p$ the power of $\ell$ dividing the class number of the $n$-th layer is bounded independently of $n$ (but may depend on $\ell$): this is a theorem of Washington. But it is still unknown whether only finitely many such $\ell$ do actually occur.<|endoftext|> -TITLE: Non-ZFC set theory and nonuniqueness of the hyperreals: problem solved? -QUESTION [21 upvotes]: The reals are the unique complete ordered field. The hyperreals $\mathbb{R}^*$ are not unique in ZFC, and many people seemed to think this was a serious objection to them. Abraham Robinson responded that this was because ZFC was tuned up to guarantee the uniqueness of the reals. Ehrlich wrote a long paper in 2012 (ref and link below), which I've only skimmed so far. It's mainly about the surreals $\textbf{No}$, not the hyperreals, but it seems to suggest that Robinson's idea has been carried forward successfully by people like Keisler and Ehrlich. Apparently NBG set theory has some properties that are better suited to this sort of thing than those of ZFC. -Section 9 of the Ehrlich paper discusses the relationship between $\mathbb{R}^*$ and $\textbf{No}$ within NBG. He presents Keisler's axioms for the hyperreals, which basically say that they're a proper extension of the reals, the transfer principle holds, and they're saturated. At the end of the section, he states a theorem: "In NBG there is ... a unique structure $\langle\mathbb{R},\mathbb{R}^*,*\rangle$ such that [Keisler's axioms] are satisfied and for which $\mathbb{R}^*$ is a proper class; moreover, in such a structure $\mathbb{R}^*$ is isomorphic to $\textbf{No}$." -My question is: Does this result indicate that Robinson's program has been completed successfully and in a way that would satisfy mathematicians in general that the nonuniqueness of the hyperreals is no longer an argument against NSA? It seems to me that this would depend on the consensus about NBG: whether NBG is expected to be consistent; whether it is a natural way of doing set theory with proper classes; and whether a result such as Ehrlich's theorem is likely to be true for any set theory with proper classes, or whether such results are likely to be true only because of some specific properties of NBG (in which case the nonuniqueness has only been made into a new kind of nonuniqueness). Since I know almost nothing about NBG, I don't know the answers to these questions. -One thing that confuses me here is that I thought the surreals lacked the transfer principle, so, e.g., where the hyperreals automatically inherit $\mathbb{Z}^*$ from $\mathbb{Z}$ as an internal set, a specific effort has to be made to define the omnific integers $\textbf{Oz}$ as a subclass of $\textbf{No}$, and $\textbf{Oz}$ doesn't necessarily have the same properties as $\mathbb{Z}$ with respect to, e.g., induction and prime factorization (see Can we axiomatize Omnific Integers without the Surreal Number system? ). Would the idea be that according to Ehrlich's result, $\mathbb{Z}^*$ would be (isomorphic to) a subclass of $\textbf{Oz}$? -I'm a physicist, not a mathematician, and if this seems inappropriate for mathoverflow, please add a comment saying so, and I'll move it to math.SE. I posted here because it relates to current research, but I'm not a competent research-level mathematician. -Philip Ehrlich (2012). "The absolute arithmetic continuum and the unification of all numbers great and small". The Bulletin of Symbolic Logic 18 (1): 1–45, http://www.math.ucla.edu/~asl/bsl/1801-toc.htm - -REPLY [8 votes]: As part of his question, Bell Crowell correctly observes: -"Section 9 of the Ehrlich paper discusses the relationship between R∗ and No within NBG. He presents Keisler's axioms for the hyperreals, which basically say that they're a proper extension of the reals, the transfer principle holds, and they're saturated. At the end of the section, he states a theorem: "In NBG [with global choice] there is (up to isomorphism) a unique structure ⟨R,R∗,∗⟩ such that [Keisler's axioms] are satisfied and for which R∗ is a proper class; moreover, in such a structure R∗ is isomorphic to No."" -At that time I made it absolutely clear that the first part of the result is due to H.J. Keisler (1976) and that my modest contribution is to point out the relation (as ordered fields) between R* and No. The work of Keisler and the relation of my work to it seem to be lost in the remarks of Vladimir. -Of course, attributing the result to Keisler, as I remain entirely confident I correctly did, does not diminish the subsequent important contributions of others. -Edit: Readers interested in reading the paper including the discussion of Keisler's work may go to: http://www.ohio.edu/people/ehrlich/ -EDIT: Since Vladimir appears to insist in his comment below that Keisler DOES NOT discuss proper classes in 1976, I am taking the liberty to quote Keisler and some of the relevant discussion from my paper. I will leave it to others to decide if I am giving Keisler undue credit. -Following his statement of his Axioms A-D of 1976--the function axiom, the solution axiom, and the axioms the state that R* is proper ordered field extension of the complete ordered field R of real numbers--Keisler writes: -“The real numbers are the unique complete ordered field. By analogy, we would like to uniquely characterize the hyperreal structure ⟨R,R∗,∗⟩ by some sort of completeness property. However, we run into a set-theoretic difficulty; there are structures R* of arbitrary large cardinal number which satisfy Axioms A-D, so there cannot be a largest one. Two ways around this difficulty are to make R* a proper class rather than a set, or to put a restriction on the cardinal number of R*. We use the second method because it is simpler.” [Keisler 1976, p. 59] -With the above in mind, Keisler sets the stage to overcome the uniqueness problem by introducing the following axiom, and then proceeds to prove the subsequent theorem. -AXIOM E. (Saturation Axiom). Let S be a set of equations and inequalities involving real functions, hyperreal constants, and variables, such that S has a smaller cardinality than R*. If every finite subset of S has a hyperreal solution, then S has a hyperreal solution. -KEISLER 1 [1976]. There is up to isomorphism a unique structure ⟨R,R∗,∗⟩ such that Axioms A-E are satisfied and the cardinality of R* is the first uncountable inaccessible cardinal. -If ⟨R,R∗,∗⟩ satisfies Axioms A-D, then R* is of course real-closed. It is also evident that, if ⟨R,R∗,∗�� further satisfies Axiom E, then R* is an $\eta_{\alpha}$-ordering of power $\aleph_{\alpha}$, where $\aleph_{\alpha}$ is the power of R*. Accordingly, since (in NBG) No is (up to isomorphism) the unique real-closed field that is an $\eta_{On}$-ordering of power $\aleph_{On}$, R* would be isomorphic to No in any model of A-E that is a proper class (in NBG). -Motivated by the above, in September of 2002 we wrote to Keisler, reminded him of his idea of making “R* a proper class rather than a set”, observed that in such a model R* would be isomorphic to No, and inquired how he had intended to prove the result for proper classes since the proof he employs, which uses a superstructure, cannot be carried out for proper classes in NBG or in any of the most familiar alternative class theories. -In response, Keisler offered the following revealing remarks, which he has graciously granted me permission to reproduce. -"What I had in mind in getting around the uniqueness problem for the hyperreals in “Foundations of Infinitesimal Calculus” was to work in NBG with global choice (i.e. a class of ordered pairs that well orders the universe). This is a conservative extension of ZFC. I was not thinking of doing it within a superstructure, but just getting four objects R, R*, <*, * which satisfy Axioms A-E. R is a set, R* is a proper class, <* is a proper class of ordered pairs of elements of R*, and * is a proper class of ordered triples (f,x,y) of sets, where f is an n-ary real function for some n, x is an n-tuple of elements of R* and y is in R*. In this setup, f*(x)=y means that (f,x,y) is in the class *. There should be no problem with * being a legitimate entity in NBG with global choice. Since each ordered triple of sets is again a set, * is just a class of sets. I believe that this can be done in an explicit way so that R, R*, <*, and * are definable in NBG with an extra symbol for a well ordering of V." [Keisler to Ehrlich 10/20/02] -Moreover, in a subsequent letter, Keisler went on to add: -I did not do it that way because it would have required a longer discussion of the set theoretic background. [Keisler to Ehrlich 5/14/06]<|endoftext|> -TITLE: Associated vector bundles of infinite rank and induced connections -QUESTION [7 upvotes]: Let $\mathbb{V}$ be a representation of a Lie group $G$ and let $P \to M$ be a principal $G$-bundle with a principal connection. If $\mathbb{V}$ is finite-dimensional, then one can associate to this data an associated vector bundle $P\times_G \mathbb{V}$ with linear connection. I thought that basically the same construction should work also when $\mathbb{V}$ is an infinite-dimensional representation, but I haven't found any textbook that would not constrain itself to finite rank. All the textbooks concerning to infinite-dimensional differential geometry that I know of (Michor, Lang, Neeb) doesn't treat associated bundles and induced connections. - -Edit: -I now realize that it may not be as straightforward as it seems on a first glance. I want to, in fact, generalize a slightly more complicated construction -- the so called tractor connection induced by a Cartan connection. -Changing the notation a little bit, given a finite-dimensional Lie group $G$ with a closed subgroup $H$, I need to work with an infinite-dimensional vector space $\mathbb{V}$ which is a representation of $\mathfrak{g}$ and also a representation of $H$ (so I can form associated bundles to $H$-principal bundles) with these two representation being compatible. Practically, I am interested mainly in Harish-Chandra modules and their globalizations. I think I am also fine with just a "sort of connection" working on some dense subbundle of the associated bundle and so $L^2$-globalizations are also OK. -I can briefly describe the construction for $\mathbb{V}$ being finite-dimensional representation of $G$ if it is needed. - -REPLY [6 votes]: If you use the cocycle description of principal bundles (for finite dimensions) as in 18.7.4 of 1, you can describe the associated bundle using convenient calculus, since then you can flip coordinates freely. -Inducing connections is then described in 19.8 (for associated fiber bundles) and in 19.10 (for associated vector bundles) of 1; use also 19.9 ("recognizing induced connections") where the cocycle description is spelled out. -Idea: You can carry over to infinite dimensions all constructions of finite dimensional differential geometry for which you have direct chart descriptions. But be very careful whenever you have to solve equations (ODE's, implicit functions, etc.) -1 -Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008. -(pdf).<|endoftext|> -TITLE: Which functors between multicategories that come from monoidal categories are monoidal? -QUESTION [9 upvotes]: Given any monoidal category $(\mathcal{C}, \otimes, I)$ we have an associated multicategory -$M_{\mathcal{C}}$ with underlying category $\mathcal{C}$, and $k$-morphisms $M_{\mathcal{C}}(A_1,\ldots,A_k; B) = \mathcal{C}(A_1\otimes\cdots\otimes A_k, B)$. -Given any (lax) monoidal functor $F\colon \mathcal{C}\rightarrow \mathcal{D}$ we get a multifunctor $M_\mathcal{C}\rightarrow M_\mathcal{D}$. -Now suppose that we have two monoidal categories $(\mathcal{C}, \otimes, I)$ and $(\mathcal{D}, \boxtimes, I')$, and a multifunctor $F\colon M_\mathcal{C}\rightarrow M_\mathcal{D}$. On the underlying categories $\mathcal{C}$ and $\mathcal{D}$, is the functor $F$ monoidal? -In order for $F$ to be monoidal we need two types of structure maps. It's clear how to get the structure map $F(A)\boxtimes F(B) \rightarrow F(A\otimes B)$: just look at the image under $F$ of the identity in $M_\mathcal{C}(A,B;A\otimes B)$. But we also need a morphism $I'\rightarrow F(I)$ (which is nicely coherent), and I don't see where in the structure of the multicategory that could be encoded. -So does this mean that there are $F$'s that don't come from monoidal functors, or is there some other structure that guarantees that there is an underlying monoidal functor? - -REPLY [16 votes]: See Leinster's "Higher Operads, Higher Categories" Example 2.1.10. Maps of multicategories which come from monoidal categories are precisely the same as lax monoidal functors. - -REPLY [7 votes]: I guess that you can take the image under $F$ of the identity in $M_C(;I)$ (the arity 0 is allowed). - -REPLY [7 votes]: Lax monoidal functors $\mathcal{C} \to \mathcal{D}$ correspond one-to-one with multifunctors $M_\mathcal{C} \to M_\mathcal{D}$. Indeed, the notions of transformation are compatible too, so that you get an isomorphism of categories. As Guillaume says, there's nothing special about the arity $0$.<|endoftext|> -TITLE: Coproduct on coordinate ring of finite algebraic group -QUESTION [6 upvotes]: I'm reading Mukai's book "An introduction to invariants and moduli", and I am having trouble understanding one of his examples. It is example 3.49 on page 101. -The setup is as follows. Let $G$ be a finite group, considered as an algebraic group over a field $k$. The coordinate ring of $G$ is then just the set of functions $G \rightarrow k$ with the usual pointwise addition and multiplication. This can be identified with the group ring $k[G]$ in the obvious way (an element $[g] \in k[G]$ corresponds to the function $G \rightarrow k$ that takes $g$ to $1$ and $h$ to $0$ for $h \neq g$). Under this identification, it seems to me that the coproduct is the function -$$\phi : k[G] \rightarrow k[G] \otimes k[G]$$ -$$\phi([g]) = \sum_{h \in G} [h] \otimes [h^{-1} g]$$ -However, Mukai asserts that if $G$ is the finite cyclic group of order $n$, then the coordinate ring of $G$ is $k[t]/(t^n-1)$ with the coproduct $t \mapsto t \otimes t$. These do not seem like the same thing to me -- what am I doing wrong? - -REPLY [6 votes]: I think the author accidentally described the dual of the Hopf algebra you're thinking of. -Finite group rings are usually endowed with multiplication $(g,h)\mapsto gh$ and comultiplication $g \mapsto g\otimes g$ (see here). -The coordinate ring $k[G]$ is obtained by dualizing. Then $g \mapsto g\otimes g$ becomes -$e_g^2 = e_g$, where $e_g$ is the function on $G$ that maps $g$ to $1$ and all other group elements to $0$. Comultiplication will look exactly the way you described it (i.e. $e_g \mapsto \sum_h e_{gh^{-1}}\otimes e_h$). - -REPLY [3 votes]: The book presumably assumes the field $k$ contains $n$ distinct roots of unity (in particular, that the characteristic of $k$ is coprime to $n$). Then you get a $k$-algebra isomorphism between $k[x]/(x^n-1)$ (isomorphic to the group ring $k[G]$ by sending $x$ to a generator) and the coordinate ring $\bigoplus_{g \in G} k$ by a finite Fourier transform. This reflects the fact that finite abelian groups are Pontryagin self-dual. - -REPLY [2 votes]: The Hopf algebra structure here involves a coproduct taking the function $f$ on $G$ to $\sum_i g_i \otimes h_i$, where $f(xy) = \sum_i f_i(x) g_i(y)$ when $x,y \in G$. Whatever Mukai is doing for a cyclic group should be consistent with this formulation of the coproduct, but I'm unfamiliar with his book. -More generally, this kind of formalism occurs when you consider a finite group -scheme as in Jantzen Representations of Algebraic Groups (AMS, 2003), I.2.3.<|endoftext|> -TITLE: Blowing-up an ordinary double point, then contracting the exceptional locus to a curve -QUESTION [6 upvotes]: Let $X\subset\mathbb P^4$ a projective hypersurface with an ordinary double point at $o\in X$. -Blow-up $\mathbb P^4$ at $o$ and let $E\simeq\mathbb P^3$ the exceptional divisor of this blow-up. Consider the strict transform $\widetilde X$ of $X$. Then, it is easy to verify that $Y=\widetilde X\cap E$ is a quadric hypersurface in $E\simeq\mathbb P^3$, thus isomorphic to $\mathbb P^1\times\mathbb P^1$. -Now, take an hyperplane (if you want, generic) section $C$ of $Y$ in $E$, that is $C=Y\cap H(\simeq\mathbb P^1)$, where $H$ is a hyperplane in $E\simeq\mathbb P^3$. Obviously, the curve $C$ meets each line of the two rulings of $Y$ in exactly one point. So, choose one ruling and define the obvious morphism $\varphi\colon Y\to C$ obtained by collapsing any line of the choosen ruling to its intersection point with $C$. -Here is my question. -Does there exist a smooth projective variety $\overline X$, containing a curve isomorphic to $C$ (which I still call $C$), and a regular morphism $\Phi\colon\widetilde X\to\overline X$ such that: - -$\Phi|_{\widetilde X\setminus Y}\colon\widetilde X\setminus Y\quad\overset{\simeq}\to\quad\overline X\setminus C$ -$\Phi|_{Y}\colon Y\to C$ is the $\varphi$ above. - -I guess this is very classical, but I wasn't able to find it. -Thanks in advance! - -REPLY [8 votes]: I guess the answer in no when $\deg(X) \geq 3$. -In fact, in this case Griffiths showed that the two rulings $L_1$ and $L_2$ in $E$ are numerically equivalent in $\widetilde{X}$ (although not algebraically equivalent in general), so the corresponding $K_{\widetilde{X}}$-negative rays $R_1$ and $R_2$ in the Mori cone of $\widetilde{X}$ coincide. -It follows that the contraction of $R_1$ is actually the contraction of the whole quadric, that is it is not possible to contract the two rulings separately. -The situation is very subtle: in fact, the completed local ring $\widehat{\mathcal{O}}_{X,o}$ is not factorial, but the fact that $L_1$ is numerically equivalent to $L_2$ implies that $\mathcal{O}_{X,o}$ is factorial. -See [Debarre, Higher Dimensional Algebraic Geometry, p. 160] and the references given therein.<|endoftext|> -TITLE: Labeling a Square Array -QUESTION [7 upvotes]: Suppose that the $n^2$ cells of an $n\times n$ array are labeled with the integers $1, \dots, n^2$. Under the traditional left-to-right and top-to-bottom labeling, the labels of horizontally adjacent cells differ by $1$, and the labels of vertically adjacent cells differ by $n$. Is it possible to relabel the array so that the labels of adjacent cells (horizontally or vertically) differ by less than $n$? -I suspect the answer is "no" but do not have a proof. I made up this problem while contemplating a similar Putnam Competition problem (1981, A-2). In this problem, adjacent cells are horizontal, vertical, or diagonal neighbors. - -REPLY [10 votes]: Suppose such a labeling exists. Let $F_k$ the region formed by the cells labeled $\{1,2,\dots,k\}$. There is a minimum $k>1$ such that $F_k$ connects two opposite sides of the checkerboard (wlog suppose left and right side). Color in black the cells belonging to $F_{k-1}$, and in white the rest. -Then, $k$ is in a white cell touching a black cell, and in every other column there is at least one black and one white cell. So overall there are at least $n$ white cells touching black cells (one per column). One of them must contain a number $\geq k-1+n$, and (since it touches the black region) it touches a number $\leq k-1$. Contradiction!<|endoftext|> -TITLE: Zograf's bound on the index of a modular curve for Shimura curves -QUESTION [7 upvotes]: I've been reading Voight's paper on Shimura curves and it prompted the following question; see http://www.cems.uvm.edu/~voight/articles/shimura-clay-proceedings-071707.pdf for which notes I'm talking about -Let $F$ be a totally real number field, $B$ a quaternion algebra which splits at exactly one real place, and $\mathcal{O}\subset B$ a maximal order. Let $\Gamma^B(1)$ be the associated discrete arithmetic subgroup of $\mathrm{PSL}_2(\mathbf{R})$. This group is the analogue of $\Gamma(1)=\mathrm{SL}_2(\mathbf{Z})$. Let $X^B(1) = \Gamma^B(1)\backslash \mathbf{H}$. -Question 1. Let $\Gamma \subset \Gamma^B(1)$ be a finite index subgroup. Do I understand correctly that $X^B(\Gamma) = \Gamma \backslash \mathbf{H}$ is a compact Riemann surface, and that the inclusion $\Gamma \subset \Gamma^B(1)$ induces a finite morphism $X^B(\Gamma) \to X^B(1)$? What are the branch points of this morphism? -Zograf showed that, if $\Gamma \subset \mathrm{SL}_2(\mathbf{Z})$ is a congruence subgroup, the index of $\Gamma$ in $\mathrm{SL}_2(\mathbf{Z})$ is bounded by $128(g+1)$, where $g$ is the genus of the compactification of the Riemann surface $\Gamma \backslash \mathbf{H}$. (This is how he proved that there are only finitely many congruence subgroups of $\mathrm{SL}_2(\mathbf{Z})$ of given genus.) -Question 2. Is there a similar theorem for Shimura curves? That is, assume that $\Gamma \subset \Gamma^B(1)$ is a congruence subgroup (does this term make sense? what should it mean?). Is the index of $\Gamma$ in $\Gamma^B(1)$ bounded by the genus of $X^B(\Gamma)$? If yes, can we make this bound explicit? -I'm not sure what I mean by a congruence subgroup of $\Gamma^B(1)$ yet, but I want the curve $X_\Gamma(B)$ to be a Shimura curve. - -REPLY [6 votes]: Here are the answers to some of your questions. If $P$ is a (non-zero) prime ideal of the integers of $F$ then $B$ will either be split or ramified at $P$, depending on whether $B\otimes_F F_P$ is isomorphic to $M_2(F_P)$ or not. Now $B$ will only be ramified at a finite set of primes, and if $P$ is a prime which splits $B$ then $\mathcal{O}/P^n$ will be isomorphic to $M_2(R/P^n)$ with $R$ the integers of $F$. More generally if $N$ is an ideal of $R$ which is coprime to all the ramified primes, $\mathcal{O}/N$ will be isomorphic to $M_2(R/N)$. In particular you can define congruence subgroups as pre-images of upper-triangular matrices mod $N$ just as in the classical case. They all have finite index. -The quotient you talk about is a Riemann surface, and is compact except for the special case $F=\mathbb{Q}$ and $B=M_2(\mathbb{Q})$, when it is the usual non-compact modular curve. The inclusion from your finite index subgroup into $\Gamma^B(1)$ will induce a finite flat covering of your Riemann surface as you say. The ramification is controlled, just as in the classical case, by the amount of torsion you lose when passing to the finite cover. Any point in the upper half plane will have a finite stabiliser in $\Gamma/Z$ with $Z$ the scalars and this finite group is always cyclic. Vaguely, you can compute the ramification index in the covering by noting how the stabilizer changes -- classically this is like noting that elliptic points have stabilisers of size greater than 1 in $PSL(2,\mathbb{Z})$ but may have stabilizers of size 1 in a finite index subgroup of this. This is easy to understand -- it's just a local calculation -- you might want to think about how to put a Riemann surface structure on the quotient at a point where the stabilizer is non-trivial (it locally acts as rotations). -About the analogue of Zograf's theorem for Shimura curves: I didn't even know that theorem for modular curves so I can't possibly answer. But you might want to think about what Riemann-Hurwitz gives you. Often the base genus of these things is at least 2, even for full level, and Zograf's theorem looks very much like an application of Riemann-Hurwitz in some sense, but complicated by the fact that one has to control elliptic points.<|endoftext|> -TITLE: Monadicity theorem in homotopy theory. -QUESTION [6 upvotes]: Let $\mathbf{C}$ be a cofibrantly generated model category (assume for simplicity that all objects are fibrant) and $\mathbf{C}^{\mathrm{T}}$ the category of $\mathrm{T}$-algebras with the induced model structure (same weak equivalences and fibrations as in the underlying model category $\mathbf{C}$). -By definition, the adjuction $\mathrm{T}:\mathbf{C}\rightleftharpoons\mathbf{C}^{\mathrm{T}}: \mathrm{U}$ is monadic. How about the homotopical version, i.e, - $\mathbb{L}\mathrm{T}:Ho\mathbf{C}\rightleftharpoons Ho(\mathbf{C}^{\mathrm{T}}): \mathbb{R}\mathrm{U}$ -is there any result about the "homotopical" monadicity theorem, which compares $Ho(\mathbf{C}^{\mathrm{T}})$ and $Ho(\mathbf{C})^{\mathbb{L}\mathrm{T}}$. - -REPLY [5 votes]: Maybe an update on the literature on homotopical refinements of monadicity: -The article - -Kathryn Hess, A general framework for homotopic descent and codescent, (arXiv:1001.1556) - -discusses homotopical monadicity in terms of simplicial model categories. -The article - -Emily Riehl, Dominic Verity, Homotopy coherent adjunctions and the formal theory of monads (arXiv:1310.8279) - -discusses it in terms of quasi-categories. -Finally, as mentioned in the comments above - -Jacob Lurie, section 6.2 of Higher Algebra - -discusses it more abstractly in $\infty$-category theory. -Maybe as a caveat, in Hess's nice article the monads are ordinary (if maybe simplicially enriched) monads on the underlying categories, so that I suppose that there should be some extra discussion of "rectification", namely discussion of under which conditions this presents an $\infty$-monad with all its higher coherence data. See the comments on the nLab at infinity-Monad -- Properties -- Homotopy coherence.<|endoftext|> -TITLE: is tensor square of a reduced ring reduced? -QUESTION [5 upvotes]: Let $R$ be a reduced algebra of finite type over a field $k$ of characteristic 0. Let $S$ be a reduced finite $R$-algebra. Is $S \otimes_R S$ reduced? -(In positive characteristic one can get non-reduced tensor products of reduced algebras even over a field.) -I have failed to find a counterexample so I thought that the statement might be true after all. The question is motivated by the discussion in the comments to this question. -If $S=R[x_1, \ldots, x_n]/I$, what one has to show is that the ideal $I \otimes 1 + 1 \otimes I$ in $S \otimes_R S=R[x_1, \ldots, x_n, y_1, \ldots, y_n]$ is radical. However, I have no good ideas as to how to approach this. - -REPLY [13 votes]: No. Let $R = k[x,y]/(y^2-x^3)$. Let $S = k[t]$, with the map $R \to S$ given by $(x,y) \mapsto (t^2, t^3)$. So $S \otimes_R S = k[t,u]/(t^2-u^2, t^3-u^3)$. The element $t-u$ is not zero in the tensor product, because all the relations are in degree $>1$. But $(t-u)^3 = 3 (t-u)(t^2-u^2) - 2 (t^3-u^3)$ is in the ideal.<|endoftext|> -TITLE: A Conjecture on the Density of a subset of integers -QUESTION [10 upvotes]: Let $X$ denote the largest subset of odd integers with the property that -every exponent in the prime factorization of any $x \in X$ belongs to $X$. -The conjecture states that the density of $X$ among the integers is -$1-\frac{1}{\sqrt{3}}$. -Is this conjecture correct ? - -REPLY [13 votes]: This is an update of my original response which was incorrect. -The conjecture is false. I will show that any $X$ in the conjecture has density at most -$$ 0.4226496 < 1-\frac{1}{\sqrt{3}}= 0.4226497...$$ -Let $Y$ be the set of odd numbers whose prime exponents are odd and different from $9$. Clearly, any $X$ in the conjecture is a subset of $Y$. Yet, the density of $Y$ equals -$$ \frac{1}{2}\prod_{p>2}\left(1-\frac{1}{p}\right)\left(1+\frac{1}{p-p^{-1}}-\frac{1}{p^9}\right) $$ -which is less than -$$ 0.42264954363092750400907132916 $$ -by the SAGE command -(1/2)*prod([RealField(100)(1-1/p)*(1+1/(p-1/p)-1/p^9) for p in prime_range(3,1000000)]) - -REPLY [5 votes]: Let $A$ be any set of positive integers such that $1\in A$, and let $B$ be the set of integers such that every exponent in the prime factorization of any $x\in B$ belongs to $A$. Then the density of $B$ exists and equals -$$ -\prod_{\text{primes }p} \bigg( 1-\frac1p \bigg) \bigg( 1 + \sum_{a\in A} \frac1{p^a} \bigg). -$$ -In this case, $A=B=X$ (which is kind of cool).<|endoftext|> -TITLE: permutation of projective limits with inductive limits -QUESTION [6 upvotes]: Hi everybody, -I have a lack of references concerning projective limits and injective limits. Up to my faults in Bourbaki there are only proj and inj limits indexed by a partially ordered set (not a category), and this set is almost systematically directed (i.e. for all $i,j$ there exists $k$ such that $k\geq i$, and $k\geq j$). The problem of combining projective limits with inductive limits is not treated at all as far as I can see. -I need a reference for this. Can someone help me ? Thank you ! -My specific problem is the following: Assume given a system of modules $(M_{i,j})_{i,j}$ and arrows among them, over the same ring $A$. -We consider both -1) the projective limit with respect to the index $j$ of the injective limits with respect to the indexes $i$. Call it -$\projlim_j \;\;(\injlim_i \;\; (M_{i,j}))$ -2) the injective limit with respect to the index $i$ of the projective limits with respect to the indexes $j$. Call it -$\injlim_i \;\;(\projlim_j\;\; (M_{i,j}))$ --- (Up to errors) There exists a canonical map -$\displaystyle CAN : \injlim_i \;\;(\projlim_j\;\; (M_{i,j})) \;\to\; \projlim_j\;\; (\injlim_i\;\; (M_{i,j})) $ -Under which assumptions this map is INJECTIVE ? -As an example if there is no arrows at all between the $M_{i,j}$, then one has a direct sum instead of the injective limit, and a product instead of a projective limit. The arrow CAN becomes -$\displaystyle CAN : \oplus_i \;\;\prod_j \;M_{i,j} \;\to\; \prod_j \;\;\oplus_i \;M_{i,j}$ -where -$((a_{i,j})_i)_j \mapsto ((a_{i,j})_j)_i$ -In this case the arrow CAN is always injective independently on the nature of the $M_{i,j}$ (and up to errors it is an isomorphism if and only if one of the sets of index "$i$" or "$j$" is finite). I suspect that in the general case the injectivity only depends on the nature of the arrows, but not of the nature of the objects. Does anyone have a useful comment or a reference? -Many thanks ! - -REPLY [7 votes]: The canonical morphism $\alpha : \mathrm{colim}_{i \in I} ~ \mathrm{lim}_{j \in J} M_{ij} \to \mathrm{lim}_{j \in J} ~ \mathrm{colim}_{i \in I} M_{i,j}$ is injective in the following situation: -1) $I$ is a directed set. -2) For every $j \in J$ and every $i \to i'$ in $I$ the morphism $M_{i,j} \to M_{i',j}$ is injective. -More detailed: Let us still assume 1) and let $x \in \mathrm{ker}(\alpha)$. Choose $i$ such that $x$ comes from an element in $\lim_j M_{i,j}$, say $x = (x_{ij})_j$. Now $\alpha(x)=0$ says that for all $j$ there is some $i(j) \geq i$ such that the image of $x_{ij}$ in $M_{i(j),j}$ vanishes. If we have 2), this already gives us $x_{ij}=0$ for all $j$, thus $x=0$. -More generally, if $j \mapsto i(j)$ is a bounded function, say by $i_{\infty}$, we see that the image of $x$ in $\lim_j M_{i_{\infty},j}$ vanishes, which implies $x=0$ in the colim-lim. Now it's easy to find an example where this function is not bounded and, in fact, $\alpha$ is not injective: -Let's take $I=\mathbb{N}$ as a partial order and $J = \mathbb{N}$ as a discrete category, so that we consider the canonical map $\mathrm{colim}_n \prod_m M_{nm} \to \prod_m \mathrm{colim}_n M_{n,m}$. For all $n,m$ let $M_{n,m} = A[X]$ and let $M_{n,m} \to M_{n+1,m}$ be the formal derivative of polynomials. Then $(1,X,X^2,X^3,\dotsc)$ represents a nontrivial element in the kernel of $\alpha$, since every $X^m$ satisfies $\partial^{m+1} X^m = 0$, but there is no $n$ with $\partial^{n} X^m = 0$ for all $m$.<|endoftext|> -TITLE: Is this a situation where triple mutual information is always non-negative? -QUESTION [9 upvotes]: Suppose I have three identically-distributed homogeneous continuous-time discrete state space Markov chains $X_1(t), X_2(t), X_3(t)$, $t\geq 0$. They evolve independently but share a common random variable $X_0$ as an initial condition. - I let $$X_1=X_1(t_1), \ \ \ X_2=X_2(t_2), \ \ \ X_3=X_3(t_3)$$ for some times $t_1, t_2, t_3\geq 0$. -I want to show that -$$ -I(X_1; X_2; X_3) \geq 0 -$$ -where $I$ is the multivariate mutual information (or information interaction) -$$ -I(A,B,C)= H(A,B,C) -H(A,B) - H(B,C) - H(A,C) + H(A) + H(B) + H(C) -$$ -where $H$ is the usual Shannon entropy. -Background/Motivation -There are well-known situations where $I(A;B;C)<0$, a famous one being if $A$ and $B$ are independent random variables, each $\pm 1$ with probability $1/2$, and $C=AB$. But I conjecture that in the case I have described above $I(X_1;X_2;X_3)\geq 0$. I believe that the Markov chains being continuous-time and homogeneous is essential. -The more general motivation is that I want to find very general situations where multivariate mutual information is non-negative. (One well-known example is if $A,B,C$ form a Markov chain.) - -REPLY [5 votes]: You are touching a treacherous place that coincidentally intersects with some of my own research. I have several responses. -(1) If at all possible---stay out the synergy/redundancy waters. Instead, see if either of the two known non-negative generalizations of mutual information fit your needs. They are: -(a) "Total Correlation": http://en.wikipedia.org/wiki/Total_correlation -(b) "Dual Total Correlation": http://en.wikipedia.org/wiki/Dual_total_correlation -I personally think the Dual Total Correlation makes a lot more sense, but that's just my opinion. -If you really want to go into the synergy/redundancy waters, here's the deal--- -(2) The "triple mutual information" you refer to is actually the redundant information minus the synergistic information. Therefore the triple mutual information will be nonnegative anytime redundancy >= synergy. Here's a paper that describes this, http://arxiv.org/pdf/1004.2515.pdf . Note that the above paper doesn't properly define the "redundant information" among variables, but it does correctly show that the "triple mutual information" is redundancy minus synergy. -(3) There have been attempts since the above-cited paper for how to properly define the "redundant information". As of this writing it's an unsolved problem. Here's a place to start if you want to get into this. http://arxiv.org/abs/1310.1538<|endoftext|> -TITLE: Probabalistic questions about singularities and exotic spheres -QUESTION [5 upvotes]: I was looking a bit at the connection between singularities of complex algebraic varieties and exotic spheres. I find it quite remarkable that you can obtain all 28 differentiable structures on the 7 sphere by intersecting a small sphere with some very innocent looking algebraic variety. It makes me wonder how common that is: if you take a "random" algebraic variety with a singularity at the origin in $\mathbb{C}^5$, and intersect it with a small sphere, what are the chances you get an exotic structure on the 7 sphere? Are some smooth structures more likely to appear than others? Has anyone actually computed probabilities? - -REPLY [4 votes]: Part of the problem is lack of precise probabilistic model for this question (i.e., what is a "random" singularity?). You also have to restrict to isolated hypersurface singularities in ${\mathbb C}^n$. The question starts to make sense if you further restrict to Brieskorn singularities of the form -$$ -z_0^{a_0}+...+ z_n^{a_n}=0 -$$ -where all $a_k\ge 2$ are integers. This situation was analyzed in great detail, for instance, in Milnor's book "Singular points of complex hypersurfaces," sections 8 and 9. -(See also Hirzebruch's paper accessible here.) For isolated singularities, provided that $n\ge 3$, the link $K$ of the singularity is a topological sphere $\iff$ it is an (integer) homology sphere $\iff \Delta(1)=\pm 1$ (theorem 8.5 in Milnor's book). Furthermore, say, for odd $n$, one can decide if $K$ is the exotic or not by looking at $\Delta(-1)$ mod $8$. For Brieskorn singularities, $\Delta$ is quite computable, so you can get your hands dirty and start computing probabilities. It is also, probably, easier, to use for this computation the graph-theoretic criterion on page 18 of Hirzebruch's paper. -I looked at the case when $n=2$ and the relevant question is about the probability of $K$ to be an integer homology sphere, equivalently, the numbers $a_0, a_1, a_2$ being pairwise coprime. Such probabilities were computed here and the answer is approximately $0.286$.<|endoftext|> -TITLE: Can a group be a universal Turing machine? -QUESTION [80 upvotes]: This question was inspired by this blog post of Jordan Ellenberg. -Define a "computable group" to be an at most countable group $G$ whose elements can be represented by finite binary strings, with the properties that - -There is an algorithm (by which I mean "Turing machine") which, when given a binary string, can determine whether that string lives in the group $G$ in finite time (i.e. $G$ is a computable set); -There is an algorithm which, when given binary strings $x,y$ in $G$, can compute $x^{-1}$ and $xy$ in finite time (i.e. the group laws on $G$ are computable functions). - -In the blog post mentioned above, the group $SL_3({\bf Z})$ was discussed, which is certainly an example of a computable group, but one can of course devise many other examples of computable groups. But note that this condition is stronger than being recursively presented, since I am requiring the word problem to be decidable. -Observe that if $G$ is a computable group, then there is an obvious algorithm which, when given any two elements $x,y$ of $G$ as input, which will halt in finite time if and only if $x,y$ fail to generate a free group, simply by evaluating all the nontrivial words of $x,y$ in turn and checking if any of them are the identity. -My first question is if there is any computable group $G$ which is "Turing complete" in the sense that the halting problem for any Turing machine can be converted into a question of the above form. In other words, there would be an algorithm which, when given a Turing machine $T$, would return (in finite time) a pair $x_T, y_T$ of elements of $G$ with the property that $x_T, y_T$ generate a free group in $G$ if and only if $T$ does not halt in finite time. Or more informally: can a group be a universal Turing machine? -I suspect the answer to this question is "yes", by doing something like taking G to be something like the group of reversibly computable operations on an infinite string of bits, though I wasn't able to quite push through the details. One might also modify the question by taking G to be a computable semigroup rather than a computable group, though I think this should not make too substantial of a difference. -My second (and more vague) question is whether one can take $G$ to be a "non-artificial" group, which can be defined without reference to computability or Turing machines. (For instance, a group which is somehow constructed using Diophantine equations would qualify.) - -REPLY [3 votes]: This question pops up constantly when I visit the site, and I'm not sure the second question has been answered, it seems the existing answers do not give very natural groups since halting properties of Turing machines are mentioned when the group is constructed, so I'll give this somewhat belated answer. -If we take "natural" to just mean that the group arises from a natural action on some natural objects, and was not defined just for the purpose of a decidability problem, then there are plenty of examples. First off, I would not be surprised if this problem is undecidable even in something as natural as right-angled Artin groups or linear groups, e.g. for $F_2 \times F_2$ it reminds one of PCP (I'm not sure it's quite the same as what one calls PCP in groups though). It would be interesting if experts could comment, I don't know an official name for this problem (I suggest one below) so it's hard to search for it. (edit: I just noticed Is it decidable whether or not a collection of integer matrices generates a free group? in the comments. So I suppose this is open.) -In the meantime, I'll present a cellular automaton solution, mostly based on my own research since it's what I know best. -Pick $A$ to be an alphabet of any composite cardinality except $|A| \neq 4$, and pick a Cartesian product decomposition $A = B \times C$. On $A^{\mathbb{Z}}$, we define the following finite set of homeomorphisms: - -For all $\pi \in \mathrm{Sym}(A)$, let $f_{\pi} : A^{\mathbb{Z}} \to A^{\mathbb{Z}}$ apply $\pi$ cellwise, meaning $f_{\pi}(x)_i = \pi(x_i)$. Let $F \cong \mathrm{Sym}(A)$ be the group of such $f_{\pi}$. -$\sigma_1 : A^{\mathbb{Z}} \to A^{\mathbb{Z}}$ is defined by identifying $x \in A^{\mathbb{Z}}$ with $(y, z) \in B^{\mathbb{Z}} \times C^{\mathbb{Z}}$ in the obvious way, and defining $\sigma_1(y, z) = (\sigma(y), z)$ where $\sigma(y)_i = y_{i+1}$ is the left shift. - - -Definition. Let $G = \langle F, \sigma_1 \rangle$ be the group the above maps generate. - -In other words, $G$ is the group of homeomorphisms on infinite sequences of symbols from $B \times C$, which are generated by shifting the first track, and permuting the individual symbols (by any permutation of the set $B \times C$). The group $G$ is a subgroup of the reversible cellular automata, i.e. the shift-commuting self-homeomorphisms of $\mathrm{Aut}(A^{\mathbb{Z}})$, and thus is a computable group. The bit string can be taken to be the minimal local rule of the CA. -I think the terminology "is a universal Turing machine" is not very descriptive, I think this property can be stated in standard terms, if we take the "free group problem" for a group to be the problem of, given two elements, deciding whether they generate a nonabelian free group (equivalently, freely generate one). - -Theorem. The group $G$ has $\Pi^0_1$-complete free group problem under many-one reductions. Thus, it is a universal Turing machine in the sense of the question. - -Proof. The following things are known: - -Given a Turing machine, you can construct a reversible cellular automaton in $\mathrm{Aut}(D^{\mathbb{Z}})$ for some $D$, such that the latter is periodic if and only if the Turing machine halts. This is proved in [1]. -There is an effective embedding $\phi : \mathrm{Aut}(D^{\mathbb{Z}}) * \mathrm{Aut}(D^{\mathbb{Z}}) \to \mathrm{Aut}(D^{\mathbb{Z}})$. This is proved in [2]. -Let $f_1, ..., f_k \in \mathrm{Aut}(D^{\mathbb{Z}})$, for any finite alphabet $D$. Then you can (effectively) construct an embedding $\psi : \langle f_1,...,f_k \rangle \to G$. This is proved in the preprint [3]. - -Combining these, given Turing machine $T$, using the first point above construct a reversible cellular automaton $f_T$ which is periodic iff $T$ halts. Then, using the second point above construct $\phi(f_T, \mathrm{id})$ and $\phi(\mathrm{id}, f_T)$. Then the group $H = \langle \psi(\phi(f_T, \mathrm{id})), \psi(\phi(\mathrm{id}, f_T)) \rangle \leq G$ is free if $T$ never halts, while if $T$ halts, $H$ is not even torsion-free. End of proof. -In general if $G * G \to H$ by an effective construction and $G$ has undecidable torsion problem, then the free group problem is undecidable for $H$. -About some other groups: - -For any uncountable sofic shift $X \subset A^{\mathbb{Z}}$, the automorphism group has undecidable free group problem. This follows easily from the above and [2]. (For countable sofics, this problem is decidable.) -For the topological full group of the full shift $A^{\mathbb{Z}}$, I have thought about this problem quite a bit, and I don't know if it's decidable. It should not be hard to show (using the results of [4]) that at least for the topological full group of $A^{\mathbb{Z}^d}$ with $d \geq 3$, this problem is undecidable. I am not sure about dimension $2$. -For automata groups (and the bigger group of all reversible transducers [5]), this problem is undecidable. Namely, there is an automata group $G$ where the torsion problem is undecidable [6, 7], and $G * G \to H$ for some other automata group $H$ [8], so this problem will be undecidable for $H$. -For the group of reversible Turing machines in the sense of [4], I don't know if the group's free product with itself embeds in it, but I think it shouldn't be hard to show this is undecidable using undecidability of the torsion problem. I didn't try it though. - -This idea will not work in groups where the order problem is decidable, which limits how natural an example one can find this way. -References -[1] Kari, Jarkko; Ollinger, Nicolas, Periodicity and immortality in reversible computing, Ochmański, Edward (ed.) et al., Mathematical foundations of computer science 2008. 33rd international symposium, MFCS 2008, Toruń Poland, August 25–29, 2008. Proceedings. Berlin: Springer (ISBN 978-3-540-85237-7/pbk). Lecture Notes in Computer Science 5162, 419-430 (2008). ZBL1173.68521. -[2] Salo, Ville, A note on subgroups of automorphism groups of full shifts, Ergodic Theory Dyn. Syst. 38, No. 4, 1588-1600 (2018). ZBL1390.37021. -[3] Salo, Ville, Universal groups of cellular automata, https://arxiv.org/abs/1808.08697 -[4] Barbieri, Sebastián; Kari, Jarkko; Salo, Ville, The group of reversible Turing machines, Cook, Matthew (ed.) et al., Cellular automata and discrete complex systems. 22nd IFIP WG 1.5 international workshop, AUTOMATA 2016, Zurich, Switzerland, June 15–17, 2016. Proceedings. Cham: Springer (ISBN 978-3-319-39299-8/pbk; 978-3-319-39300-1/ebook). Lecture Notes in Computer Science 9664, 49-62 (2016). ZBL1350.68101. -[5] Grigorchuk, R. I.; Nekrashevich, V. V.; Sushchanskii, V. I., Automata, dynamical systems, and groups, Grigorchuk, R.I. (ed.), Dynamical systems, automata, and infinite groups. Transl. from the Russian. Moscow: MAIK Nauka/Interperiodica Publishing. Proc. Steklov Inst. Math. 231, 128-203 (2000); translation from Tr. Mat. Inst. Steklova 231, 134-214 (2000). ZBL1155.37311. -[6] Gillibert, Pierre, An automaton group with undecidable order and Engel problems, ZBL06824269. -[7] Bartholdi, Laurent; Mitrofanov, Ivan, https://arxiv.org/abs/1710.10109 -[8] Fedorova, Mariia; Oliynyk, Andriy, Finite automaton actions of free products of groups, Algebra Discrete Math. 23, No. 2, 230-236 (2017). ZBL1375.20028.<|endoftext|> -TITLE: Is the bounded derived category of coherent sheaves of a variety a small category? -QUESTION [7 upvotes]: The question is in the title. -I am trying to apply the Mitchell (Freyd-Mitchell?) embedding theorem, which states that for every small abelian category $A$, there exists a ring $R$ such that A embeds into the category $R$-mod. The derived category is not abelian, of course, but I have a particular subcategory that is abelian, and life would be easiest if the derived category was smal, so that the subcategory was small and abelian. - -REPLY [5 votes]: The answer to your question really depends on what you mean by the word "the". An unhelpful answer is that the coherent sheaves over any variety form a proper class (hence "no"). A more useful answer is (as mentioned in the comments) that there exists a small category that is equivalent to any category that can be reasonably called the bounded derived category of coherent sheaves (hence "yes"). -Furthermore, the construction of such a category can be accomplished without the use of replacement. In particular, the category lives in the same ZC universe (i.e., $V_\alpha$ for $\alpha$ a not-necessarily-inaccessible limit ordinal greater than $\omega$ - see e.g., Wikipedia or the set theory section of the Stacks project) as the defining field.<|endoftext|> -TITLE: Is the restriction map an epimorphism of commutative rings? -QUESTION [9 upvotes]: Let $i : U \to X$ be a quasicompact open immersion of schemes. I would like to know whether the canonical morphism $i_* \mathcal{O}_U \otimes_{\mathcal{O}_X} i_* \mathcal{O}_U \to i_* \mathcal{O}_U$ is an isomorphism. -Remark that this is trivial if $i$ is an affine morphism (base change formula), so we're good with semi-separated $X$. But I don't know what happens in general. Of course we may assume that $X$ is affine. Then $U$ is a finite union of basic open subsets of $X$ and the question is equivalent to: Is the canonical map $\mathcal{O}(U) \otimes_{\mathcal{O}(X)} \mathcal{O}(U) \to \mathcal{O}(U)$ an isomorphism? -In other words, is the restriction map $\mathcal{O}(X) \to \mathcal{O}(U)$ an epimorphism in the category of commutative rings? For example, if $U$ is basic open, then this is just a localization, thus an epimorphism. More generally, if $U$ is affine, it is an epimorphism (if you don't like the argument above) since the map on spectra is the open immersion $U \to X$, which is a monomorphism of schemes. For a simple positive example where $U$ is not affine, take $X=\mathbb{A}^2=\mathrm{Spec}(k[x,y])$ and $U = D(x) \cup D(y)$. Then the ring homomorphism is just the identity $k[x,y] \to k[x,y]$. -The best source for the theory of epimorphisms of commutative rings is: Séminaire Samuel. Algèbre commutative, 2, 1967-1968: Les épimorphismes d'anneaux, available on nundam. See also this mathoverflow discussion. -There are many characterizations of epimorphisms, but no one seems to answer the question directly. - -REPLY [3 votes]: Here is a different view on your original question whether the canonical morphism $$ \tag{$\star$} -i_* \mathscr{O}_U \otimes_{\mathscr{O}_X} i_* \mathscr{O}_U \to i_* \mathscr{O}_U$$ is an isomorphism. -It seems that the question is very different depending on the dimension of the complement of $U$. Let $Z=X\setminus U$ and observe that we may break up the embedding $U\to X$ into intermediate embeddings and so assume that $Z$ is irreducible. Then we will distinguish two cases, whether $Z$ is a divisor or has higher codimension. -1 -Assume that $\mathrm{codim}(Z,X)\geq 2$. -Claim: If $X$ is noetherian and $U$ is generically reduced (reduced at all the generic points), then $i_*\mathscr O_U$ is a coherent $\mathscr O_X$-module. -Proof: Consider the normalization $\nu:X'\to X$ and let $i':U'=\nu^{-1}U\to X'$. Then we have an injective natural map: -$$ -i_*\mathscr O_U \to i_*\nu_*\mathscr O_{U'}\simeq \nu_*i'_*\mathscr O_{U'}\simeq \nu_*\mathscr O_{X'}. -$$ -Since $\nu$ is proper (it is finite), $\nu_*\mathscr O_{X'}$ is a coherent $\mathscr O_X$-module and this exhibits $i_*\mathscr O_{U}$ as a submodule. $\square$ -Let $p\in X$ be a point with residue field $\kappa(p)$ and consider the reduction of $(\star)$ at $p$: -$$ -i_* \mathscr{O}_U \otimes_{\mathscr{O}_X} i_* \mathscr{O}_U\otimes\kappa(p) \to i_* \mathscr{O}_U\otimes\kappa(p)$$ -For simplicity let's introduce the notation $V_p=i_*\mathscr O_U\otimes\kappa(p)$ so the above morphism is essentially -$$ \tag{$\star\star$}V_p\otimes_{\kappa(p)}V_p\to V_p$$ -It is clear that if $(\star)$ is an isomorphism, then so is $(\star\star)$. -By the above Claim $V_p$ is finite dimensional and hence for $(\star\star)$ to be an isomorphism it is necessary that $\dim_{\kappa(p)}V_p=1$. -In the example of the two planes meeting in a single point this fails. -On the other hand if $X$ is reduced, then $(\star\star)$ being an isomorphism is actually equivalent to $(\star)$ being an isomorphism near $p$. Indeed, if $\dim_{\kappa(p)} V_p=1$ then this holds in a neighbourhood of $p$ and then $i_*\mathscr O_U$ is a line bundle (in the same neighbourhood) and therefore $(\star)$ is an isomorphism (in the same neighbourhood). -So (at least if $X$ is reduced and noetherian) then the remaining question is when $i_*\mathscr O_U$ is a line bundle. This happens for example if $X$ is $S_2$. In this case the natural map $\mathscr O_X\to i_*\mathscr O_U$ is an isomorphism, so the latter is trivially a line bundle. This happens for example if $X$ is normal. -In fact, being a line bundle means being locally isomorphic to $\mathscr O_X$ so this condition is actually equivalent to being $S_2$ (assuming $X$ is reduced). (To see why, consider the natural map $\mathscr O_X\to i_*\mathscr O_U$, and observe that any section of $\mathscr O_X$ would vanish on a codimension $1$ subset, and hence this map induces an embedding on the residue fields $\kappa(p)\to V_p$, but if it is an embedding, then it is an isomorphism and hence surjective. Then by Nakayama's lemma the original map is surjective on stalks and hence an isomorphism). -If we do not require $X$ to be reduced, then a little less is enough. We always have an exact sequence of sheaves: -$$ -\mathscr H^0_Z(X,\mathscr O_X) \to \mathscr O_X\to i_*\mathscr O_U \to \mathscr H^1_Z(X,\mathscr O_X). -$$ -This shows that as soon as $\mathscr H^1_Z(X,\mathscr O_X)=0$, it follows that $\mathscr O_X\to i_*\mathscr O_U$ is surjective. Of course, if $X$ is reduced, then $\mathscr H^0_Z(X,\mathscr O_X)=0$, so we're back at assuming that $X$ is $S_2$. I think this gives a pretty good description of what happens in case $\mathrm{codim}(Z,X)\geq 2$. -2 -Assume that $Z\subset X$ is a divisor. -2a) -If $Z$ is a $\mathbb Q$-Cartier divisor, then $U$ is locally equal to the basic open set corresponding to the defining equation of $mZ$ (where $mZ$ is a multiple which is Cartier) and hence the natural map $\mathscr O_X\to i_*\mathscr O_U$ is an epimorphism since it is a localization. -2b) -If $Z$ is a non-$\mathbb Q$-Cartier divisor, then it seems a little more complicated, but it seems to me that it still might work. -Perhaps one can do this: blow up $X$ along $Z$ and get $\sigma: Y\to X$. This makes the preimage of $Z$ a Cartier divisor whose complement is still $U$, say $j:U\to Y$ is the embedding. Then $\mathscr O_Y\to j_*\mathscr O_U$ is an epimorphism by part 2a). Now it seems to me that the push-forward of this to $X$ remains an epimorphism since this happens on the sections on open sets and not just on the stalks. -Then if we assume that $X$ is normal, then $\mathscr O_X\to \sigma_*\mathscr O_Y$ is an isomorphism, so the desired condition follows. I am not entirely sure about this last part, but I have already written too much, so I will just leave this as is for now.<|endoftext|> -TITLE: Why no morphisms from the contradictory proposition to the inconsistent context? -QUESTION [6 upvotes]: Consider Higher order predicate logic over dependent type theory (DPL) as defined in Chapter 11 of B. Jacobs's book "Categorical Logic and Type Theory" (though I think this question applies to first-order predicate logic too). -In the category $\mathbb{P}$ of propositions-in-dependent-type-contexts, an object is a well formed proposition $\Gamma \vdash \varphi:Prop$ and a morphism $(\Gamma \vdash \varphi:Prop) \rightarrow (\Delta \vdash \psi : Prop)$ consists of a context morphism $\vec{M} : \Gamma \rightarrow \Delta$ such that $\Gamma | \varphi \vdash \phi(\vec{M})$ is derivable. This category $\mathbb{P}$ is fibred over the category of dependent type contexts $\mathbb{C}$ via $(\Gamma \vdash \varphi:Prop) \mapsto \Gamma$. -Consider the "contradictory proposition" object, $A$, in $\mathbb{P}$ which I define as $\emptyset \vdash \perp : Prop$ and the "inconsistent context" object, $B$, which I define as $x : 0 \vdash \top:Prop$ (where $0$ is the empty type). -I would expect $A$ and $B$ to be isomorphic in $\mathbb{P}$ since both objects seems equally void to me in the sense that neither appear to have any models, i.e. there are no morphisms from the terminal object, $1 := \emptyset \vdash \top : Prop$, to either $A$ or $B$. But, of course, $A$ and $B$ cannot be isomorphic since a morphism from $A$ to $B$ entails the existence of a context morphism $M : \emptyset \rightarrow x : 0$ which would imply that the void type is inhabited. -Can someone assuage my concerns about $A$ and $B$ being equally void yet not isomorphic? - -REPLY [6 votes]: My understanding of the situation is that you have to correctly understand what the morphisms do. A geometric picture might be helpful. -Let us interpret a context $\Gamma = x_1 : A_1, \ldots, x_n : A_n$ as a cartesian product $|\Gamma|$ of topological spaces. A proposition in a context $\Gamma \vdash \phi$ is interpreted as a subspace of a cartesian product $|\phi| \subseteq |\Gamma|$. I think the confusion arises if you think of a morphism $M : (\Gamma \vdash \phi) \to (\Delta \vdash \psi)$ as a continuous map $|\phi| \to |\psi|$, which it is not. It is an ambient map, i.e, a continuous map $|\Gamma| \to |\Delta|$ which also happens to map $|\phi| \to |\psi|$. Now it is clear that even though $|\phi|$ and $|\psi|$ may be homeomorphic as spaces, the may be embedded in $|\Gamma|$ and $|\Delta|$ in wildly different way (the wild Cantor sets come to mind), so it won't be possible to extend a homeomorphism between them to an ambient map. -In your example, we are comparing $\emptyset$ as a subspace of $\emptyset$ versus $\emptyset$ as a subspace of $1$. Even though the two empty subspaces are homeomorphic, there is no ambient map taking the latter to the former.<|endoftext|> -TITLE: Maximal disarrangement of $n \times n$ numbers -QUESTION [7 upvotes]: This question is inspired by -Martin Erickson's -question, -"Labeling a Square Array." -I'll start by quoting Martin: - -the $n^2$ cells of an $n \times n$ array are labeled with the integers - $1, \dots, n^2$. - Under the traditional left-to-right and top-to-bottom labeling, - the labels of horizontally adjacent cells differ by $1$, - and the labels of vertically adjacent cells differ by $n$. - -Let $\delta$ be the minimum absolute label difference -between any cells adjacent horizontally -or vertically in a particular filling of the array -by those $n^2$ numbers. -So $\delta=1$ in the standard filling. -What is the maximum of $\delta$ as a function of $n$? -In a sense, this is asking for a maximal "disarrangement" -of the $n^2$ numbers. -Phrased that way, it sounds like it may have applications -and therefore be well-studied: perhaps in the discrepancy theory literature? -Or perhaps it is entirely elementary... -Examples. For $n=2$, $\delta=1$, e.g., -$$ -\left( -\begin{array}{cc} -3&2\\ 1&4 \\ -\end{array} -\right) \;. -$$ -For $n=3$, $\delta=3$, e.g., -$$ -\left( -\begin{array}{ccc} -1&4&7\\ 5&8&2 \\ 9&3&6 -\end{array} -\right) \;. -$$ -Addendum. Here is an illustration of Gjergji Zaimi's solution, achieving, -for $n=5$, $\delta=10=\binom{5}{2}$: - -REPLY [15 votes]: The problem of minimizing the maximum difference of adjacent values in a labelled graph is called "bandwidth minimization". Recently there was interest in the dual problem of maximizing the minimum such difference, and it was originally called the separation number of a graph. Some recent papers call this the antibandwidth problem. -You are interested in the antibandwidth number for $P_n\times P_n$, where $P_n$ denotes the path of length $n$. Recently the antibandwidth problem was solved for all meshes and similar toroidal graphs. See "Antibandwidth and Cyclic Antibandwidth of Meshes and Hypercubes" by A. Raspaud, H. Schroder, O. Sykora, L. Torok, and I. Vrto. -The answer is that the antibandwidth number of $P_n\times P_m$, $m\geq n$ is $$\lceil \frac{n(m-1)}{2}\rceil.$$ -In particular the answer to your question is $\binom{n}{2}$. To see one construction which achieves this bound, simply color the squares in a chessboard fashion and then do a traditional labelling on the white squares first, and the black squares second.<|endoftext|> -TITLE: Decay of Relative Growth in Conway's Game of Life -QUESTION [5 upvotes]: Intro -The question is about Game of Life. -Let us denote the set of points obtained from initial configuration $A$ after $m$ steps as $A(m)$ (we are only interested in finite initial configuration, i.e. those one which formed by finite number of marked cells). -Let us denote the number of marked cells at configuration $A$ as $N(A)$. Then increment at $m$-th step could be computed as $i(A, m) = \max(0, N(A(m+1)) - N(A(m))$. -Majority of configurations doesn't grow in size after some number of steps. There are -known examples which grows linearly or quadratically in time. For all of this -examples the relative increment decay as times goes by: $\frac{i(A,m)}{N(A(m))} \to 0$ as -$m \to \infty$. -Question -Is it possible to prove some uniform result of this type: -is it true that $\forall \epsilon > 0$ (may be some other restriction) $\exists M_\epsilon$ such that $\forall A$ (for any arbitrary chosen initial configuration) and $\forall m > M_\epsilon$: $\frac{i(A,m)}{N(A(m))} < \epsilon$? -May be it makes sense for some good family of initial configurations $A$? - -This is rather a probe question which I asked with a hope to find some references or ideas in answers which will direct me in more useful settings. - -REPLY [9 votes]: [expanded from the comment above] -There cannot be such a result. The simplest aperiodic counterexample -is a "lightweight spaceship gun" of even period $p$, whose -$m$-th generation has population $9m/p + O(1)$ or $12m/p + O(1)$ -according to the parity of $m$, whence -$i(A,m)/N(A(m)) \rightarrow 1/3$ in one congruence class mod $2$. -There are also more complicated counterexamples such as -"sawtooth" -patterns whose population at time $m$ oscillates between $O(1)$ and $cm+O(1)$, -and for which the same pair $\bigl(N(A(m)),N(A(m+1))\bigr)$ and the same $i(a,M)>0$ -occurs infinitely often but for different configurations. -In general it's a reliable heuristic that if you imagine any kind of -behavior of a pattern in Conway's Game of Life then either it's -obviously impossible (e.g. exponential growth, or an aperiodic pattern -that always stays within a fixed rectangle), or somebody has -wasted devoted enough time and/or ingenuity to -construct a pattern showing that behavior. -[For the benefit of those who know little more of Conway's Game of Life -than the basic rules: a "spaceship" or "ship" of period $n$ is a finite pattern -$S$ that after $n$ steps reappears shifted by some translation $T$, -and thus moves through space with speed $T/n$. A period-$p$ "gun" -$G$ for a spaceship $S$ is a finite pattern whose $p$-th generation is -the disjoint union of $G$ with a copy of $S$ that does not further -interact with $G$; thus $G$ produces a stream of ships of type $S$. -The first few Life ships have period $4$; the smallest of these, the -glider, -has constant population (namely 5) so cannot be used here, but the next-smallest, the -"lightweight -spaceship" (LWSS), alternates between populations of $9$ and $12$, -so works for any even $p$. If you know a glider gun of sufficiently large period $p$ -(and the classic $p=30$ of -Gosper's gun -is large enough) then you can position three of them to make an LWSS gun -by colliding their glider streams.]<|endoftext|> -TITLE: An $n!\times n!$ determinant -QUESTION [19 upvotes]: Let us consider the matrix $A$ with its rows and columns enumerated by the elements of $S_n$ with $A_{\sigma\tau}=x^{c(\sigma\tau^{-1})}$ where $c()$ is the number of cycles in a permutation's decomposition. I'm interested in $|A|$. More specifically I aim to prove that all of its roots as of a polynomial in $x$ are integers between $-n+1$ and $n-1$ but the roots' multiplicities would also be nice to know. - -REPLY [9 votes]: Here (Darij Grinberg, A representation-theoretical solution to MathOverflow question #88399) is the proof that I hinted at in the comments section in more details. Repeated mistakes absorbed most of the time I spent writing it, which is why it took four days; let me apologize for this.<|endoftext|> -TITLE: Meaning/Origin of Seiberg-Witten Equations/Invariants -QUESTION [39 upvotes]: Having now seen and "understood" (quotes necessary) the Seiberg-Witten equations on a closed oriented Riemannian 4-manifold $X$, I have no real understanding of where they came from. -We take an orthogonal frame bundle $P$ of $TX$, a $\textrm{spin}^\mathbb{C}$ structure $\tilde{P}$ with determinant line bundle $\mathcal{L}$, the complex $\pm$ spin bundles $S^\pm(\tilde{P})$ associated to $\tilde{P}$, a unitary connection $A$ on $\mathcal{L}$, and then BAM: -$F_A^+=\psi\otimes\psi^*-\frac{1}{2}|\psi|^2$ -$D_A\psi=0$ -for a spinor $\psi\in C^\infty(S^+(\tilde{P}))$. From here we can consider the space of solutions (monopoles) and do some Floer theory stuff and whatnot. -I only read that these equations come from Witten's famous paper Monopoles and 4-Manifolds (along with two others joint with Seiberg)... however, unless I am mistaken, he simply writes them down and starts arguing for their similarity/duality to Donaldson's theory (with instanton solutions). I then try and go to the standard references of Donaldson, which don't seem to suggest how the SW equations come about (nor do I even really see how the instantons come about). Although I have studied physics for a long time, I seem to just juggle around these papers, without ever finding a natural "blooming" of the SW equations. -Even if it's in the language of String Theory, I would like to know the general story / understanding of the "blooming" of the SW equations, and how exactly they are "dual" to the instanton-scenario of Donaldson, perhaps even for the "blooming" of these instantons. (For instance, I don't see a set of equations for instantons). This post may not be stated in its clearest form, but I will try my best to make appropriate edits. - -REPLY [11 votes]: After thinking, and reading other references and re-reading the papers I mentioned, I may have found a sufficient explanation (at least to my care): Both instantons/monopoles are solutions to corresponding equations of motions from associated actions, and they "bloom" from an overarching SUSY action. -Witten formulated "twisted N=2 Supersymmetric Yang-Mills", a TQFT with SUSY (supersymmetry), which leads to the Donaldson invariants. This used an $SU(2)$-bundle over $X$ along with a gauge field (connection $\omega$) and matter fields (bosonic $\phi,\lambda$ and fermionic $\eta,\psi,\zeta$), and gave the Donaldson-Witten action functional $S_{DW}=\int_Xtr(\mathcal{L})$, -$\mathcal{L}=\frac{1}{4}F_\omega\wedge(\ast F_\omega+F_\omega)-\frac{1}{2}\zeta\wedge[\zeta,\phi]+id^\omega\psi\wedge\zeta-2i[\psi,\ast\psi]\lambda+i\phi d^\omega{\ast d^\omega}\lambda-\psi\wedge\ast d^\omega\eta$. -This has associated partition function $Z_{DW}=\int e^{-S_{DW}/g^2}D\Phi$ (here $\Phi$ denotes the space of aforementioned fields), where $g$ is a coupling constant that is the key here for answering our question. The "blooming" of this action functional is beyond the scope of my intentions and probably of MathOverflow, so I won't question it. -In weak coupling ($g\rightarrow 0$, known to physicists as the ultraviolet region), the action localizes to the classical Yang-Mills $S_{YM}=\int_Xtr(F_\omega\wedge\ast F_\omega)$ and have the equations of motion $d^\omega\ast F_\omega=0$. The global-minima solutions are $F_\omega=\pm\ast F_\omega$ (as Oliver clarifies in a comment). These solutions are the Donaldson instantons. -Now apparently, when we instead look at strong-coupling ($g\rightarrow\infty$, known to physicists as the infrared-region), the Seiberg-Witten equations should arise (a "duality" in Witten's TQFT). Indeed, Seiberg and Witten showed that this infrared limit of the above theory is equivalent to a weakly-coupled $U(1)$-gauge theory (the $SU(2)$-gauge group is spontaneously broken down to the maximal torus). Perhaps here is where a better understanding would be desirable (buzzwords 'asymptotic freedom' and 'symmetry breaking' appear). -Anyway, some physics-technique stuff happens (the previous paragraph can be described as "condensation of monopoles"), and we must consider a spin-c structure (which all of our oriented 4-manifolds have, whereas a spin structure would not allow us to consider all 4-manifolds); note that $Spin^c(4)=(SU(2)\times SU(2))\times_{\mathbb{Z}_2} U(1)$. This gives the data: $U(1)$-gauge field $A$ and positive spinor field $\psi$ (as written in the original post). The pair $(A,\psi)$ is a monopole when it minimizes an action $S_{SW}$, i.e. are time-independent solutions to equations of motions (the Seiberg-Witten equations). The action here is $S_{SW}=\int_X(|d^A\psi|^2+|F_A^+|^2+\frac{R}{4}|\psi|^2+\frac{1}{8}|\psi|^4)$, with scalar curvature $R$. -I hope this post is not too confusing. -[[Edit/Update]]: I just came across a book chapter by Siye Wu, The Geometry and Physics of the Seiberg-Witten Equations. These lectures tell the physical origin completely! (i.e. completely details my sketch). The SW equations and action functional pops up on pg191. -http://www.springerlink.com/content/q37322037j466218/<|endoftext|> -TITLE: A criterion for the sum of two closed sets to be closed ? -QUESTION [11 upvotes]: Let $V$ and $I$ be two closed subsets of a Banach space $A$. -The set $V$ is a convex cone, and $I$ is a linear subspace of $A$. I also know that $V\cap I=\{0\}$. -I would like to know whether $I+V$ is closed. I've seen that there is a criterion of Dieudonné which I can't use here because I know that neither $V$ nor $I$ is locally compact. -So my question is : are there any other criteria that I could try to use ? - -REPLY [2 votes]: In finite dimensional space, your result is true. Take a linear functional with $\text{ker } T = I$. It is not hard to prove, using $I\cap V={\boldsymbol 0}$, that $T(V)$ is closed by seeing that $$\text{inf }_{\begin{array}{c}\boldsymbol x \in V \\ \|\boldsymbol x\| = 1\end{array}} \|T(\boldsymbol x)\| > 0,$$ with the minimum achieved. Thus, by Lemma H1 of the book Geometric Functional Analysis and its Applications, $I+V$ is closed.<|endoftext|> -TITLE: Example of a weak Hausdorff space that is not Hausdorff? -QUESTION [17 upvotes]: I've looked on the web and haven't found a simple example. - -REPLY [4 votes]: Steen & Seebach's counterexample #99: Maximal Compact Topology is another example. This is also a KC space (every compact set is closed) but not Hausdorff.<|endoftext|> -TITLE: distribution of Young diagrams -QUESTION [15 upvotes]: Consider $\Lambda^p(C^n\otimes C^n)=\oplus_{\pi}S_{\pi}C^n\otimes S_{\pi'}C^n$ as -a $GL_n\times GL_n$-module. This space has dimension $\binom {n^2}p$. I would -like any information on the shapes of pairs of Young diagrams $(\pi,\pi')$ that -give the largest contribution to the dimension asymptotically. I am most interested in the case where $p$ is near $n^2/2$. Is there a slowly growing -function $f(n)$ such that partitions with fewer than $f(n)$ steps contribute negligibly? -If so, can the fastest growing such $f$ be determined? - -REPLY [8 votes]: For more details on this answer, together with two proofs of the following theorem, see arXiv:1705.07604 preprint -``External powers of tensor products as representations of general linear groups'' by Greta Panova and myself. -The following result converts the original question into a problem about the representation theory of the symmetric groups for which more asymptotic tools are available. -Theorem. -The random irreducible component of the external power -\begin{equation} -\label{eq:decomposition} - \Lambda^p(\mathbb{C}^m\otimes \mathbb{C}^n)=\bigoplus_{\lambda}S^{\lambda}\mathbb{C}^m\otimes S^{\lambda'}\mathbb{C}^n -\end{equation} -regarded as a $\operatorname{GL}_m\times \operatorname{GL}_n$-module -corresponds to a pair of Young diagrams $(\lambda,\lambda')$, -where $\lambda$ has the same distribution as -the Young diagram which consists of the boxes with entries $\leq p$ -of a uniformly random Young tableau with rectangular shape $n^m$ -with $m$ rows and $n$ columns. -Alternatively: the random Young diagram $\lambda$ has the same distribution -as a Young diagram which corresponds to a random irreducible component of -the restriction $V^{n^m}\big\downarrow^{\mathfrak{S}_{mn}}_{\mathfrak{S}_{p}}$ -of the irreducible representation $V^{n^m}$ of the symmetric group -$\mathfrak{S}_{mn}$ which corresponds to the rectangular diagram $n^m$. -Above, when we speak about random irreducible component of a representation we refer to the following concept. -For a representation $V$ of a group $G$ we consider its decomposition into -irreducible components -$$ V = \bigoplus_{\zeta \in \widehat{G} } m_\zeta V^\zeta, $$ -where $m_\zeta\in\{0,1,\dots\}$ denotes the multiplicity of an irreducible representation $V^\zeta$ in $V$. -This defines a probability measure $\mathbb{P}_V$ -on the set $\widehat{G}$ of irreducible representations -given by -$$ \mathbb{P}_V(\zeta) -:= \frac{m_\zeta \operatorname{dim} V^{\zeta}}{\operatorname{dim} V}.$$ -As I mentioned, the above theorem converts the original question into a problem about the representations of the symmetric groups for which several results are already available. -In particular, the law of large numbers for the corresponding random -Young diagrams has been proved in much wider generality by Biane -["Representations of symmetric groups and free probability", -Adv. Math., 138(1):126--181, 1998, Theorem 1.5.1] -using the language of free cumulants of Young diagrams. -The asymptotic Gaussianity of their fluctuations around the limit shape has been proved in [Piotr Śniady, "Gaussian fluctuations of characters of symmetric groups and of Young diagrams". Probab. Theory Related Fields 136 (2006), no. 2, 263–297, Example 7 combined with Theorem 8] -using the same language. -In the specific case of the restriction $V^{n^m}\big\downarrow^{\mathfrak{S}_{mn}}_{\mathfrak{S}_{p}}$, -the above-mentioned generic tools -can be applied in the scaling when -$m,n,p\to\infty$ tend to infinity -in such a way that the rectangle ratio $\frac{m}{n}$ converges to a strictly positive limit -and the fraction $\frac{p}{mn}$ converges to some limit. -[Boris Pittel and Dan Romik, "Limit shapes for random square Young tableaux", -Adv. in Appl. Math. 38 (2007), no. 2, 164–209] have worked out this specific example -and, among other results, found explicit asymptotic limit shapes of typical Young diagrams -which contribute to such representations. -If you like this answer, you may enjoy a related problem on MathOverflow.<|endoftext|> -TITLE: Diagonal invariants of the symmetric group on $k[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n]$ -QUESTION [12 upvotes]: This sounds like something that must have been answered long ago, but for some reason I can find nothing on it in the internet. (There has been lots of recent activity in diagonal covariants, related to the $n!$ conjecture, but invariants seem to have become a stepchild in this process.) -Let $k$ be a field of characteristic $0$. Let $n\in\mathbb N$. The group $S_n\times S_n$ acts on the polynomial ring $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]$ by -$\left(\sigma,\tau\right)\left(P\right) = P\left(X_{\sigma\left(1\right)},X_{\sigma\left(2\right)},...,X_{\sigma\left(n\right)},Y_{\tau\left(1\right)},Y_{\tau\left(2\right)},...,Y_{\tau\left(n\right)}\right)$. -Thus, the symmetric group $S_n$ also acts on $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]$ due to the diagonal embedding $S_n\to S_n\times S_n$. -Since the action of $S_n$ is not generated by pseudoreflections, it follows from the converse of the Chevalley-Shephard-Todd theorem that $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]$ is not a free $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]^{S_n}$-module. But it is easy to see that $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]$ is a free $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]^{S_n\times S_n}$-module of rank $n!^2$. -Question: Is $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right]^{S_n}$ a free $k\left[X_1,X_2,...,X_n,Y_1,Y_2,...,Y_n\right] ^ {S_n\times S_n}$-module? - -REPLY [9 votes]: I finally got around to writing the promised details. I tried to make this a bit instructive, I hope you still find it useful. -First I will expand a bit on my comment above. A good reference is Stanley's article "Invariants of finite groups and their applications to combinatorics". There is a folklore theorem (which first appeared in print in M. Hochster and J. A. Eagon, Cohen-Macaulay rings, invariant theory, and the generic perfection of determinantal loci, Amer. J. Math. 93 (1971), 1020-1058) which says that for a finite subgroup $G$ of $Gl_n(\mathbb C)$ the algebra of invariants $\mathbb C[x_1,x_2,\dots,x_n]^G$ is Cohen-Macaulay. Therefore if $G$ is a subgroup of $G'$ and $G'$ is generated by pseudoreflections we get as a corollary of the Chevalley-Shephard-Todd theorem that $\mathbb C[x_1,\dots,x_n]^G$ is free over $\mathbb C[x_1,\dots,x_n]^{G'}$. In particular, this holds for $G'=S_n\times S_n$ and $G$ its diagonal subgroup. - -Now, from a combinatorics perspective, we aren't simply satisfied by calculating the dimension of a polynomial algebra over another, but we would also like to exhibit a nice basis. I think it's worth spending sometime understanding the case of $R=\mathbb C[x_1,\dots,x_n]$ over $\mathbb C[x_1,\dots,x_n]^{S_n}$, first. Because the multivariate cases are similar in nature. -It's not hard to prove that the dimension of $R$ over $R^{S_n}$ is $n!$ and moreover there are two standard bases one learns about: - -The Artin Basis, consisting of monomials $x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$, with $0\le a_i\le n-i$ for all $i$. -Schubert polynomials. - -Schubert polynomials are a nice basis, because they are indexed over combinatorial objects, and satisfy many combinatorial and geometric properties. The Artin basis, on the other hand, makes it easy to see that the Hilbert series is $(1+q)(1+q+q^2)\cdots (1+q+\cdots+q^{n-1})=[n]_q!$, yet it somewhat hides the presence of the symmetric group. -We know that $[n]_q!$ is the generating function of any Mahonian statistics on the symmetric group, so it would be nice to have a basis to reflect that. We can do this by the so called "descent" basis, and we will see that this is a construction that generalizes to the bivariate case (this is essentially the content of Bergeron and Lamontagne's paper). -The most famous Mahonian statistic is the Major index. Our basis is modeled after this statistic and is indexed over permutations. Since we have -$$\operatorname{maj} (\sigma) = \sum _{\sigma _{i+1} < \sigma _i} i ,$$ -the most natural thing to try is the collection of monomials -$$b _{\sigma} = \prod _{i = 1} ^{n-1} (x _{\sigma_1} \cdots x _{\sigma _i}) ^{\chi (\sigma _i > \sigma _{i+1})}.$$ -To prove that the $b_{\sigma}$'s form a basis it is enough to show that the polynomials $m_{\lambda}b_\sigma$ where $m_\lambda$ ranges over all symmetric monomials parametrized by partitions $\lambda$ are linearly independent. And the proof goes like this: We will construct a bijection $(\lambda,\sigma)\leftrightarrow \lambda$ where $\mu,\lambda$ are partitions with at most $n$ parts and $\sigma\in S_n$, and then use a Grobner type argument, i.e. show that the leading monomial in $m_{\lambda}b_{\sigma}$ is precisely $X^{\mu}=x_1^{\mu_1}\cdots x_n^{\mu_n}$. You will find this argument spelled out in detail in section 2 of Allen's "The Descent Monomials and a Basis for the Diagonally Symmetric Polynomials". - -To guess a basis for the bivariate case Bergeron and Lamontagne, play a similar game, where they first calculate the Hilbert series of $R^{S_n}$ over $R^{S_n\times S_n}$. They actually calculate the Frobenius series and obtain the expression $$(q;q)_n(t;t)_n h_n\left[\frac{1}{(1-q)(1-t)}\right]$$ -in plethystic notation. But this is a well known generating function over $S_n$. Namely it is -$$\sum _{\sigma\in S _n} q^{\operatorname{maj}(\sigma)}t^{\operatorname{maj}(\sigma^{-1})}.$$ -So they construct a basis $$B _{\sigma}=\rho b _{\sigma}(X)b _{\sigma^{-1}}(Y)$$ -similar to the construction above, where $\rho$ is the Reynolds operator. To be able to use a Grobner type argument, they construct a bijection $(\lambda _1,\lambda _2,\sigma)\leftrightarrow (\mu _1,\mu _2)$ (section 12), and they are able to show that the polynomials $m _{\lambda _1}(X)m _{\lambda _2}(Y)B _{\sigma}(X,Y)$ are linearly independent (section 13). -Finally, a word on the case of diagonal coinvariants. It is a big open problem to exhibit a basis for the space of diagonal coinvariants, $\mathbb C[X,Y]$ over $\mathbb C[X,Y]^{S_n}$. Even though, there is a conjectured form for the Hilbert series as a generating function of two statistics over parking functions (the dimension here is no longer $n!$, rather $(n+1)^{n-1}$, proved by Haiman in 2001), these statistics are not natural enough to let one guess what the corresponding basis will be.<|endoftext|> -TITLE: mapping spaces of diagrams -QUESTION [5 upvotes]: I'm studying a small category $A$ and diagrams of based spaces or spectra indexed by $A$ (so let's say diagrams in a category $C$ that's closed symmetric monoidal, has a compatible model structure, etc.). I'm told that in this setting there's a projective model structure on diagrams $A \rightarrow C$, where the fibrations and weak equivalences are defined objectwise; and an injective model structure, where the cofibrations and weak equivalences are defined objectwise. -I want to take two diagrams $X,Y$ and form a mapping space $\text{Map}(X,Y)$ in a homotopy-invariant way. (The mapping space should be a subspace of the product of the mapping spaces for each level. In a more formal setting, I can pass to the twisted arrow category on $A$ and take an equalizer.) I should be able to accomplish this by taking a cofibrant replacement of $X$ and a fibrant replacement of $Y$, but there are two model structures available for doing this. This gives me two different mapping spaces; I want to show that they are equivalent. Any suggestions for approaches/tools/references? -(An intermediate step might be to show that they are both equivalent to the space I get by taking a projective-cofibrant replacement of $X$ and an injective-fibrant replacement of $Y$.) - -REPLY [5 votes]: (I deleted my first attempt at an answer, as I had right and left reversed and anyway I wanted to try to say it better.) -I would advocate the following broad and comparatively low-tech view of the matter. -You are asking why two homotopy invariant versions of Hom of diagrams should be equivalent. -In general if you are dealing with a functor $F:C\to D$, where both $C$ and $D$ have some maps called weak equivalences, you can define the notion of "right derived functor" of $F$ as follows: Take the category of all functors $C\to D$. Make the slice category of all functors $(G,T:F\to G)$ under $F$. In it, consider the full subcategory whose objects are those $(G,T)$ such that $G$ preserves weak equivalences. In this category call a morphism $(G_1,T_1)\to (G_2,T_2)$ a weak equivalence if the map $G_1\to G_2$ is a weak equivalence (objectwise in $C$). By definition a right derived functor of $F$ is a homotopically initial object. That is, it is an object $(G,T)$ that becomes initial when you invert these equivalences. -From this definition it is clear that any two right derived functors of $F$ are equivalent, in the sense that there is zigzag of weak equivalences connecting them. -You care about the case when $C=Fun(A,Top)^{op}\times Fun(A, Top)$ and $F:C\to Top$ is the straightforward Hom functor, taking $(X,Y)$ to the set of natural maps between diagrams, topologized as a subspace of a product of function spaces, one for each $A$-object. Now you need to know why the various homotopy invariant replacements that you can think of for $Hom$ are in fact derived functors in the sense above. -The usual way of making a right derived functor for $F$ goes like this: think up a class $C_0$ of special $C$-objects such that $F$ takes equivalences between such objects to equivalences. Think up an endofunctor $r$ of $C$ such that - -$r$ takes every object to an object in $C_0$, -there is a natural weak equivalence $e:1\to r$ to $r$ from the identity. -$r$ takes weak equivalences to weak equivalences - -As derived functor of $F$ use the composition $F\circ r$, equipped with the map $F\to F\circ r$ induced by $e$. This is in fact homotopically initial in that category of homotopy-invariant functors under $F$, under extremely weak axioms about $C$ and $D$ and their weak equivalences. -In your case you ought to be able to make $r$ and $e:1\to r$ by using cofibrant replacement of $X$ and fibrant replacement of $Y$ in either the projective or the injective model structure. The point to check, either way, is that maps $X_2\to X_1$ and $Y_1\to Y_2$ of diagrams induce equivalences $Hom(X_1,Y_1)\to Hom(X_2,Y_2)$ if the $X_i$ are cofibrant and the $Y_i$ are fibrant.<|endoftext|> -TITLE: Continued fractions and projective resolutions -QUESTION [14 upvotes]: Hello, -This question might be vague and not thought-through enough. -If we have a real positive number $x$, we can start to write it as a continued fraction: -$x = a_0 + \frac{1}{x_1} , \ldots , x_n=a_n + \frac{1}{x_{n+1}}$ where $a_i$ are non-negative integers, $x_i$ non-nogative real numbers less than $1$. -So we can write $x=[a_0,a_1,\ldots]$. But we also may write $x=[a_0,\ldots,a_{n-1},x_n]$, i.e. we might decide that our continued fraction is finite, allowing the last term to be non-integer. -If we have a module, we can start taking a projective resolution of it. Again, we can take an infinite projective resolution, or decide to truncate it at some finite level, but then the last term will not be maybe projective. -The last term will be projective if our module was "good", i.e. of small enough (in particular, finite) cohomological dimension. In the continued fraction setting, the last term will be integer if our number was "good", i.e. rational... -Is there any (wild?) relation between rational numbers among real numbers, and modules of finite cohomological dimension among modules? -Sasha - -REPLY [6 votes]: Hi Sasha, -As I understand it your question have 2 parts: - -Is there some relation between continued fractions and projective resolutions? -Is there some relation between rationality of a number and finite cohomological dimension of an object in an (abelian) category? - -Now let me make those questions more concrete (and of course lose information): -Can one construct a natural example of an abilian category $C$ and a real invariant $f:C \to \mathbb R$, s.t. - -A projective resolutions of an object $X$ in $C$ will give a continued fraction representation of $f(X)$? -$f$ will map finite cohomological dimension objects to rational numbers? - -Here I think that the answer for 1 is no and for 2 is yes. Let me explain why: - -The recursive definition of projective resolution ${P_i}$ of an object $X$ is: -$X_0=X$ and an exact sequence $0 \to X_{i+1} \to P_i \to X_i \to 0$ -The recursive definition of continued fraction ${a_i}$ of a number $x$ is: -$x_0=x$ and $x_n=a_n+1/x_{n+1} $. -It is unlikely to find an invariant that maps an exact sequence $0 \to A \to B \to C \to 0$ to numbers satisfying $c=b+1/a$, -one reason is that if $B=A+C$ then $0 \to C \to B \to A \to 0$ is also exact but the relation $c=b+1/a$ is not symmetric. You can try to go to derived categories this might be relevant for squark remark but I doubt that it will help here. -Consider the Grotendic group $K(C)$ and it subgroup $K_0(C)$ generated by images of protective objects (it coincide with the one generated by images of finite cohomological dimension objects). In case that you category is monoidal, I believe that one can show that $K_0(C)$ is a subring. -Now you just need to find an homomorphism $f:K(C) \to \mathbb R$ s.t. $f(K_0(C)) \subset \mathbb Q$. This sounds as a reasonable task. You can ask some experts in K-theory about it, you can also as this question in MO and label it as K-theory question. You can also ask Inna Zakharevich, (http://math.mit.edu/~zakh/) she is doing K-theory and she had worked with invariant that have to do with real and rational numbers. -Good luck and keep us updated.<|endoftext|> -TITLE: Itô-like calculus for $\alpha$-stable processes $\alpha \neq 2$. -QUESTION [5 upvotes]: After searching the net, I couldn't find a suitable reference to some extension of the Itô calculus that involves wilder sources of randomness than mere brownian motion. -Motivation : Itô calculus is used extensively by option traders and quants, and the basic assumption is that the price of a stock follows roughly a geometric brownian motion. But - -financial markets are notoriously non brownian due to the fat tailedness of their increments as random processes. -Itô calculus isn't suitable to something wilder than brownian motion (which is well enough for the vast majority of applications) - -But in this particular case, does anyone knows of something equivalent but applicable to $\alpha$-stable Levi processes with $\alpha \neq 2$ (i.e. non- brownian increments). - -REPLY [3 votes]: David Applebaums' "Lévy processes and stochastic calculus" -It looks like integrating with respect to an $\alpha$-stable process is addressed directly on page 212 with the Ito formula to follow starting on page 218. Sadly the Ito formula section is not included in the Google preview. -I have not read the new version of this book but I did read the first edition in graduate school and I found it quite readable.<|endoftext|> -TITLE: Erdős-Szekeres empty pseudoconvex $k$-gons -QUESTION [7 upvotes]: I am wondering if the -Erdős-Szekeres -empty convex $k$-gon question has a different answer if -convexity is replaced by a pseudoline-version of convexity. -The empty convex $k$-gon question -is a variant on the -Happy Ending Problem. -The last remaining case, $k=6$, was settled about four years ago. -It is now known that, although there -arbitarily large point sets in the plane with no empty convex heptagon, -every sufficiently large set contains -an empty convex hexagon. -Here is my attempt to generalize this to pseudolines. -An arrangement of pseudolines -is a collection of curves each pair of which intersects in exactly one -point, at which they cross. -There are nonstretchable pseudoline arrangements, i.e., those not -combinatorially equivalent to any straight-line arrangement. Here's one: - -         - -         -(Image by David Eppstein) - -In fact, -there are many more pseudoline arrangements— $2^{\Omega(n^2)}$, -than straight-line arrangements— $2^{O(n \log n)}$, -for $n$ lines and simple arrangements. -All the above are facts. Caveat: my attempt at defining convexity in this -context might not make sense. -Given $n$ points in the plane, say that they contain -an empty pseudoconvex $k$-gon -if - -there is an arrangement $\cal{A}$ of -$\binom{n}{2}$ pseudolines through the $n$ points. - -there is an empty $k$-gon $K$, a region of the plane bounded -by $k$ pseudolines containing no points in its interior. - $K$ is convex in the sense that for any two -additional points $a,b$ inside $K$, one can find a pseudoline -through $a,b$ compatible with the arrangement $\cal{A}$, -such that the pseudosegment $ab \subset K$. -Assuming this definition is not inconsistent, -does the -Erdős-Szekeres -empty convex $k$-gon question have a different answer? -For example, perhaps every sufficiently large point set always -has a pseudoconvex heptagon? -Aside from this question, I would be interested in learning -of convexity definitions analogous to what I tried to define above. -Thanks for pointers/ideas! - -REPLY [3 votes]: (Long comment, not an answer to the question posed.) -One way to generalise "general position" configurations of lines in the plane is to assign an orientation "clockwise" or "anticlockwise" to each set of three points and require a small set of axioms to hold. These purely combinatorial objects were called CC systems by Don Knuth in his book "Axioms and Hulls". In the book he proves a correspondence with geometric configurations of pseudolines. There is also a combinatorial definition of "convex polygon" and "interior", so the problem can be stated in a purely combinatorial form. I suggest you check that out as it will probably match your intuition. -Note that the original happy ending problem for 17 points, proved by Peters and Szekeres by computer, also holds for CC systems using much the same computation.<|endoftext|> -TITLE: probability of IID sum being positive -QUESTION [7 upvotes]: Let $X_1,X_2,...$ be iid random variable with mean zero. If $X_1$ has second moment then by the CLT we have $P(X_1+X_2+...+X_n\geq 0)\rightarrow \frac{1}{2}$, as $n$ goes to infinity. I am curious about whether does this hold without the second moment assumption? I think this might be a well studied problem. Could anyone provide an answer or reference to me? -Many thank! - -REPLY [9 votes]: No, this does not hold without the finite second-moment assumption, in general. -Consider the Levy $\alpha$-stable distributions, which will yield a whole family of examples. Using the parametrization, e.g., on the Wikipedia page, if $X$ is $\alpha$-stable, for $1 < \alpha < 2$, then it has characteristic function -$$ -\varphi(t) = \mathbb E e^{i t X} = \exp(i t \mu - |ct|^{\alpha} (1 - i \beta \mathrm{sgn}(t) \tan(\pi \alpha/2))) ~. -$$ -For the given range of $\alpha$, we also have that $\mathbb E X = \mu$ and $\mathbb E X^2 = \infty$. -Take $\mu = 0$, $c = \beta = 1$ and $\alpha = 3/2$, for simplicity and concreteness. Then, we get -$$ -\varphi(t) = \exp(- |t|^{3/2} (1 + i \mathrm{sgn}(t))) ~. -$$ -Clearly, $\varphi(t) \neq \varphi(-t)$, so, in particular, $X$ is not symmetric. -However, observe that if $X_1,\ldots,X_n$ are iid with this distribution and $S_n = X_1 + \cdots + X_n$, then $n^{-2/3} S_n$ has the same characteristic function as $X$ since $(\varphi(n^{-2/3}t))^n = \varphi(t)$, and, hence the same distribution. -Therefore, $\mathbb P(S_n \geq 0) = \mathbb P(X \geq 0) \neq 1/2$.<|endoftext|> -TITLE: Stable triviality of fiber bundles -QUESTION [6 upvotes]: This is probably has an obvious proof or a straightforward counterexample, but I'm having trouble finding either. -Let $p:E \to B$ be a fibre bundle, with fibre $F$. Assume that there is a spectrum $X$ and a homotopy equivalence of spectra -$$f: X \wedge \Sigma^{\infty} B_+ \to \Sigma^{\infty} E_+$$ -(in particular, there is a stable section to $p$). -Can we conclude that $X \simeq \Sigma^{\infty} F_+$? If so, we may conclude that the bundle is stably trivial; that is, -$$\Sigma^{\infty} E_+ \simeq \Sigma^{\infty} (B\times F)_+.$$ -If this is not the case, what sort of conditions do we need to demand of the fibration to make it hold? I'm happy to localize everything in site at your favorite prime, or for that matter, cohomology theory... -Edit: There is an immediate type of counterexample, gotten by taking $E = B \times F$ to be trivial, but where $\Sigma^\infty F_+ \simeq X \vee Y$, where $Y$ is $\Sigma^\infty B_+$-acyclic; i.e., -$$\pi_*(Y \wedge \Sigma^\infty B_+) = 0$$ -One can define this problem away by assuming that $X$ and $\Sigma^\infty F_+$ are $\Sigma^\infty B_+$-local, which I'm happy to do for now. - -REPLY [3 votes]: This is often not true. Here's an example where all components are finite complexes. -Let $E = S^3 \times S^2$ and $B = S^2$, with the map given by $S^3 \times S^2 \to S^3 \to S^2$ where the latter is the Hopf fibration. The suspension spectrum is equivalent to $\Sigma^\infty_+ S^3 \wedge \Sigma^\infty_+ S^2$, but the fiber is $S^1 \times S^2$ and doesn't have $\Sigma^\infty_+ S^3$ as the suspension spectrum. -(This doesn't come with a stable section to $p$, unfortunately. If you take $(S^3 \times S^2) \coprod S^2$, you can get an example with a stable section.) -What went wrong here is that we didn't take the "over $B$" structure into account. The suspension spectrum of $B$ inherits a coalgebra structure from the diagonal, and the suspension spectrum of $E$ inherits a comodule structure using the structure map. If $B$ is 1-connected and the weak equivalence $X \wedge \Sigma^\infty_+ B \simeq \Sigma^\infty_+ E$ is a weak equivalence of comodules, then you can construct a map of Eilenberg-Moore complexes -$$ -C(\Sigma^\infty_+ E, \Sigma^\infty_+ B, \Sigma^\infty_+ *) \to C(X \wedge \Sigma^\infty_+ B, \Sigma^\infty_+ B, \Sigma^\infty_+ *) -$$ -which becomes a weak equivalence $\Sigma^\infty_+ F \to X$ on Tot. If the base is not 1-connected, convergence is obviously a lot more delicate. -Unfortunately this basically assumes stable triviality of the bundle, rather than going in the reverse direction. -(In order to avoid saying a lot about homotopy theory for coalgebras and comodules, I'm not going to characterize such weak equivalences as anything better than "something where you can ultimately compare the cobar complexes".)<|endoftext|> -TITLE: "Converse" of Taylor's theorem -QUESTION [25 upvotes]: Let $f:(a,b)\to\mathbb{R}$. We are given $(k+1)$ continuous functions $a_0,a_1,\ldots,a_k:(a,b)\to\mathbb{R}$ such that for every $c\in(a,b)$ we can write $f(c+t)=\sum_{i=0}^k a_i(c)t^i+o(t^k)$ (for any $t$ in a neighbourhood of $0$). Can we conclude that $f$ is of class $C^k$? - -REPLY [3 votes]: Another answer: You are prescribing a Whitney jet, and the Whitney extension theorem gives the answer: - -Whitney, Hassler: Analytic extensions of differentiable functions defined in closed sets, Trans. AMS 36 (1934), 63--89. - -Also, another reference building on the paper of Marcinkiewicz and Zygmund, is - -G. Glaeser: Etude de quelques Algebres Tayloriennes, J. Anal. Math. 11 (1958), 1-118.<|endoftext|> -TITLE: What is the relation of the absolute Galois group and classical profinite groups? -QUESTION [5 upvotes]: Consider the absolute Galois group $G = \mathrm{Gal}(\overline{\mathbb{Q}}: \mathbb{Q})$ and $G_p = \mathrm{Gal}(\overline{\mathbb{Q}_p}: \mathbb{Q}_p)$. -Abelian class field theory gives us for the abelinisation -$$ G^{ab} = G / [G, G] = \prod\limits_{p} GL_1( \mathbb{Z}_p).$$ - -How can we relate the groups $GL_n( \mathbb{Z}_p)$ to $G_p$ or $G$? - -The motivation of my question in a nutshell: Langlands program relates automorphic representations to Galois representations. Automorphic representations factor in representations of $GL_n( \mathbb{Q}_p)$. Some representations (the cuspidal ones) are induced representations from the maximal compact subgroup $GL_n(\mathbb{Z}_p)$. It seems easier to relate them directly. Since the local Langlands conjecture have been proven for $GL_n(\mathbb{Q}_p)$ and the dual of $GL(n, \mathbb{Q}_p)$ is described by the dual of $GL_n(\mathbb{Z}_p)$, what can we deduce about the relation between $G_p$ and $GL_n(\mathbb{Z}_p)$? - -REPLY [9 votes]: This question has been solved by Paskunas in his PhD thesis, Unicity of types for supercuspidals, arXiv:math/0306124. -Corollary 8.2 of this reference gives an "inertial Galois correspondence" between supercuspidal types of the form $({\rm GL}(n,{\mathbb Z}_p), \lambda )$ and certain representations of the local inertial group. -The case of non-supercuspidal types is a work in progress by Shaun Stevens and William Conley, certainly to appear somewhere soon.<|endoftext|> -TITLE: Fubini Study Metric and Einstein constant -QUESTION [5 upvotes]: Hi all, -it is well known that the complex projective space with the fubini study metric is Einstein, but what is the explicit value, i.e. for which $\mu$ does $Ric=\mu g$ hold? -Moreover, I would like to know how to calculate the sectional cuvature explicitly, because I would like to calculate the number $\sqrt{\sum K_{ij}}$ explicitly for a given orthonormal basis. ($K_{ij}$ is the sectional curvature of the plane spanned by $e_i$ and $e_j$) - -REPLY [6 votes]: As suggested by Anton, you can use the O'Neill formulas in the Riemannian submersion $\mathbb C^{n+1}\to \mathbb{C} P^n$ that defines the Fubini-Study metric on $\mathbb C P^n$. This gives the following: suppose $X,Y$ are orthonormal tangent vectors at some point in $\mathbb C P^n$, and denote by $\overline X,\overline Y$ their horizontal lifts to $\mathbb C^{n+1}$ (which are also orthonormal). Then $$sec(X,Y)=1+\tfrac34\|[\overline X,\overline Y]^v\|^2=1+3|\overline g(\overline Y,J\overline X)|^2,$$ where $\overline g$ is the canonical Euclidean metric on $\mathbb C^{n+1}$, $()^v$ denotes the vertical component wrt the submersion and $J$ is the complex structure, i.e., multiplication by $\sqrt{-1}$. Note that this immediately implies that $\mathbb CP^n$ is $\tfrac14$-pinched. - -With the above formula, you can easily compute the Einstein constant of $\mathbb C P^n$ to be equal to $\mu=2n+2$, see e.g. Petersen's book "Riemannian Geometry", chapter 3. - -Another possible way of doing it is using that this is a Kahler manifold. The Fubini-Study metric can be thought of as $\omega_{FS}=\sqrt{-1}\partial\overline\partial\log\|z\|^2$, where $\|z\|^2$ is the square norm of a local non vanishing holomorphic section (it is independent of the choice of section by the $\partial\overline\partial$-lemma). You can then compute in local normal (holomorphic) coordinates the coefficients $g_{i\bar j}$ and use that the Ricci form is given by $Ric(\omega)=-\sqrt{-1}\partial\overline\partial\log\det(g_{i\bar{j}})$. This will obviously give you the same result, but in the form $Ric(\omega_{FS})=(n+1)\omega_{FS}$. As pointed out in the comments below, the reason for the missing factor $2$ in this computation is that we have to change from real orthonormal frames to complex unitary frames.<|endoftext|> -TITLE: H-principle and PDE's -QUESTION [29 upvotes]: According to Wikipedia: "In mathematics, the homotopy principle (or h-principle) is a very general way to solve partial differential equations (PDEs), and more generally partial differential relations (PDRs)". -I'd like to know if h-principle and theory from M. Gromov's "Partial Differential Relations" is a useful tool in the field of nonlinear PDE's. -What type of problems can be attacked using h-principle? -What type of results can be obtained? - -REPLY [17 votes]: You asked about the $h$-principle, but I'll say something about convex integration instead. -Here is a survey by DeLellis and Szekelyhidi about instances of the "h-Principle" where convex integration is used to construct low regularity solutions to many equations of fluid mechanics: -http://arxiv.org/abs/1111.2700 -These analytic results, however, are different in flavor to what you usually call the $h$-principle in topology and geometry. In topology a nontrivial instance of the $h$-principle might say something like "you can invert the sphere $S^2 \subseteq {\mathbb R}^3$ through a regular family of immersions"; what makes it non-trivial is that there could have been a topological obstruction to doing so (for instance, you can't invert $S^1 \subseteq {\mathbb R}^2$ because the inclusion map $i$ and $-i$ have different degrees). In these analytic results, you're not exactly interested in homotopies. -You can use the method of convex integration (at least, basically the same kind of convexity argument -- Gromov himself prefers not to call this convex integration) to construct wild solutions to PDE. For example, there are bounded solutions to incompressible Euler which are in the energy space and can have any prescribed energy density $\frac{1}{2} |v|^2(x,t)$ (in particular, they can be compactly supported in space and time). This is a bit shocking because sufficiently regular solutions to Euler conserve energy. The fact that weak solutions need not conserve energy is tied to ideas regarding the theory of turbulence, which is the main motivation for all these studies. -If you read Springer or Gromov you may not immediately recognize the similiarities between the analysis convex integration and the topology/geometry version (for example, sometimes Baire category arguments are used in analysis to simplify the technical arguments, sometimes at the expense of some regularity in the solution). But the arguments closely parallel Nash's proof that short maps can be approximated by $C^1$ isometric embeddings, which is where the story of convex integration begins. More recent developments regarding isometric embeddings can be found in the references to the survey linked above. One main challenge regarding both Euler and the isometric embedding problem is to find the degree of regularity at which there is a transition from flexibility to rigidity. -Preceding the developments in fluid mechanics, convex integration was also used by Kirchheim, Muller and Sverak to exhibit elliptic systems coming from Euler Lagrange equations with solutions that are Lipschitz but nowhere $C^1$ -- this flexibility result contrasts the result of Evans that minimizers of the same kinds of functionals are smooth off a closed set of measure 0. There are also many related investigations in the calculus of variations tied to the stability of differential inclusions $\nabla u \in K$, especially regarding how they arise in the mathematical theory of materials. For example, James and Ball presented the idea that if $u : \Omega \subseteq {\mathbb R}^3 \to {\mathbb R}^3$ is the configuration of a crystal, its deformation gradient $\nabla u$ minimizes free energy $\int_\Omega W(\nabla u) dx$ by taking values pointwise in the set $K$ of critical points of $W$. Muller's book "Variational Models for Microstructures and Phase Transitions" has more on this topic (for example regarding how you can explain microstructures as patterns which are "trying to minimize" such a functional), but I think this is a bit more distant from the original question. The relevance is only that convex integration can be used to produce wild solutions to $\nabla u \in K$; but here $K$ might even be a finite set, and $u$ is only Lipschitz, so it's fairly different from the topological setting.<|endoftext|> -TITLE: Can we decide if an abelian variety is simple by knowing its Zeta function ? -QUESTION [5 upvotes]: Let $A$ be an Abelian variety defined over the finite field with $q$ elements. Let $P_i(T)$ be the characteristic polynomial of the action of the Frobenius on the $i^{th}$ étale cohomology group. -Is the following assertion true: -"The Abelian variety $A$ is simple over the finite field with $q$ elements if and only if $P_1(T)$ is irreducible over $\mathbb Q$" -? -One implication is obvious, what about the other one ? - -REPLY [14 votes]: The following result follows from Tate-Honda theory - -Let $A$ be an abelian variety over a finite field $k$, and let $f_A$ be the characteristic polynomial of $A$. Then $A$ is isogenous to a power of a simple abelian variety if and only if $f_A$ is a power of an irreducible polynomial. - -I can't find a set of online notes which contains this statement. Kirsten Eisenträger's notes -are generally very good, but they get this result wrong on the first page -- Theorem 1.1 claims that, if $f_A$ is a power of an irreducible polynomial, then $A$ is simple, ignoring the possibility that -$A$ is a power of a simple variety. -Let $A$ be isogenous to $\bigoplus A_i^{n_i}$, where the $A_i$ are simple and mutually non-isogenous. Every abelian variety has such a decomposition. Then $f_A = \prod f_{A_i}^{n_i}$. -Suppose that $f_A$ is a power of an irreducible polynomial. Then all of the $f_{A_i}$ must also be powers of that polynomial. In particular, for any $i$ and $j$, either $f_{A_i}$ divides $f_{A_j}$ or vice versa; without loss of generality, suppose $f_{A_i} | f_{A_j}$. By a result of Tate, this means that $A_i$ is isogenous to a subvariety of $A_j$. Since $A_i$ and $A_j$ are simple, this means that $A_i$ and $A_j$ are isogenous. Since we assumed that the $A_i$ were mutually nonisogenous, there must in fact be only one summand in our decomposition of $A$, and $A$ is isogenous to $A_1^{n_1}$ for some simple $A_1$ and some $n_1$. -Suppose now that $A$ is isogenous to $B^{n}$ for $B$ simple. Then $f_A = f_B^n$. So our goal is to show that $f_B$ is a power of an irreducible polynomial. If not, write $f_B = gh$ where $g$ and $h$ are relatively prime of positive degree. By a result of Honda, there exist abelian varieties $C$ and $D$ with characteristic polynomials $g$ and $h$. By the result of Tate cited above, $C$ and $D$ are isogenous to subvarieties of $B$, contradicting that $B$ is simple. $\square$ - -The answer to the question in your title is "yes", we can decide whether $A$ is irreducible by knowing its $\zeta$ function. Fix a prime power $q$. Let $k$ be the field with $q$ elements. Let $W(q)$ be the set of irreducible monic polynomials over $\mathbb{Q}$ all of whose roots have norm $q^{1/2}$. The main result of Honda-Tate theory (Theorem 4.1 in Kirsten's notes) is that there is a bijection between isogeny classes of $k$-simple abelian varieties over $k$ and $W(q)$. For each polynomial $g$ in $W(q)$, there is some positive integer $n(g,q)$ such that the characteristic polynomial of the corresponding simple abelian variety is $f^{n(g,q)}$. The tricky point is that $n(g,q)$ is not always $1$. For example, in Denis's answer, what is going on is that $n(x-p, p^2)=2$. So it is true that $A$ is $k$-simple if and only if $f_A$ is of the form $g^{n(g,q)}$; you just need to know how to compute that $n$ function. I think you should be able to extract this from sections 4 and 5 of Kirsten's notes, but I don't know the details. -UPDATE: Brian Conrad e-mails to spell out the recipe (hope I copied this correctly). Let $f$ be irreducible of the required form. Let $\pi$ be a root of $f$ and let $F$ be the field $\mathbb{Q}(\pi)$. For every $p$-adic place $v$ of $F$, let $d_v$ be the denominator of $v(\pi) [F_v:\mathbb{Q}_p]/v(q)$ when written in lowest terms. Let $d = LCM(d_v)$ where the LCM ranges over all possible $v$'s. Then $f^d$ is the characteristic polynomial of the simple abelian variety. -If I'm not mistaken, this condition can be stated in an elegant geometric way. For any polynomial $g$ over $\mathbb{Q}_p$, let $N(g)$ be the $p$-adic Newton polytope of $g$. We will subdivide the path $N$ as follows: Recall that, if $h$ is irreducible over $\mathbb{Q}_p$, then $N(h)$ is a line segment, and that, if $g$ factors as $\prod h_i^{r_i}$, then $N(g)$ is the concatenation of $r_i$ copies of each $N(h_i)$, ordered with increasing slope. We will decompose $N(g)$ into one piece for each distinct irreducible factor, with that piece being $r_i$ times $N(h_i)$. For example, $x^2-p^2$, $x^2+p^2$ and $x^2-2xp+p^2$ all have Newton polytope a line segment from $(2,0)$ to $(0,2)$. In the first case, we would subdivide this line segment into two line segments, touching at $(1,1)$, because the two factors $x+p$ and $x-p$ are distinct. In the second case, we would subdivide if $x^2+p^2$ factored in $\mathbb{Q}_p$ (i.e. if $p$ is $1 \mod 4$) but not if it remained irreducible (if $p$ is $3 \mod 4$). In the third case, we would not subdivide, because the factor $(x-p)$ is repeated. -Then I believe the condition is that $f$ is the characteristic polynomial of an abelian variety if and only if all the vertices of $N(f)$, subdivided as above, have heights that are integer multiples of $v_p(q)$. - -REPLY [9 votes]: No, an elliptic curve over a field with $p^2$ elements may have zeta function $(X-p)^2=X^2-2pX+p^2$. -(See, e.g., section 4 of Waterhouse's paper "Abelian varieties over finite fields", though this probably goes back to Deuring.)<|endoftext|> -TITLE: Mixing time of unitary Brownian motion -QUESTION [9 upvotes]: Let $B_t$ be the unitary Brownian motion, i.e. Brownian motion on the unitary group $U(N)$. What is known about the mixing time of $B_t$, that is, how fast does the measure $B_t(\delta_{\{Id\}})$ converge to the Haar measure? (the question is motivated by trying to say something about mixing times for some discrete approximations of unitary Brownian motion) - -REPLY [5 votes]: If I understand correctly after a cursory check of the literature, the mixing time for Brownian motion on a compact simple Lie group $G$ is conjectured to behave asymptotically (in the limit of large dimension) as $\log \dim G$. See section 3 (and 3.2 in particular) of this paper.<|endoftext|> -TITLE: sums of rational squares -QUESTION [20 upvotes]: It is a well-known fact that if an integer is a sum of two rational squares then it is a sum of two integer squares. For example, Cohen vol. 1 page 314 prop. 5.4.9. Cohen gives a short proof that relies on Hasse-Minkowski, but he attributes the theorem (without reference) to Fermat, who didn't have Hasse-Minkowski available. So my question is, how did Fermat prove this theorem? and part 2 of the question is, what is the simplest direct proof? I googled for this result and found a manuscript with a proof that doesn't use Hasse-Minkowski, but it's not very short. - -REPLY [2 votes]: This is an answer to the second question, "what is the simplest direct proof?" -Let m be an integer satisfyng $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}=m$. $a,b,c,d \in \mathbb{Z}$, gcd(a,b)=1, gcd(c,d)=1. -$\frac{a^{2}d^{2}+b^{2}c^{2}}{b^{2}d^{2}}=m$ -$a^{2}d^{2}+b^{2}c^{2}=mb^{2}d^{2}$ -$b^{2}(md^{2}-c^{2})=a^{2}d^{2}$. -As our assumtion $b^{2} \not | a^{2}$ so $b^{2} | d^{2}$, $d^{2}=kb^{2}$. Then $\frac{1}{b^{2}}(a^{2}+\frac{c^{2}}{k})=m$. -$\frac{c^{2}}{k}=mb^{2}-a^{2}$ which is an integer. Thus we have $k=1$ as gcd(c,d)=1. -We have $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{b^{2}}=m$, $a^{2}+c^{2}=mb^{2}={p_{1}}^{\alpha_{1}}...{p_{k}^{\alpha_{k}}}$ [Unique prime factorization]. -If $a^{2}+c^{2}=wp$, p is prime and gcd(w,p)=1, then we can find integers x, y such that $x^{2}+y^{2}=p$. -Thus for every $p_{1}$, $p_{2}$,...,$p_{k}$, we can find $a_{1}, a_{2},...,a_{k}$ and $c_{1}, c_{2},...,c_{k}$ such that -${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$ -${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$ -$\vdots$ -${a_{k}}^{2}+{c_{k}}^{2}=p_{k}$ -Now if $m=p_{1}p_{2}$, ${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$, ${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$. -$({a_{1}}^{2}+{c_{1}}^{2})({a_{2}}^{2}+{c_{2}}^{2})$ = ${(a_{1}a_{2}+c_{1}c_{2})}^{2}+ {(a_{1}c_{2}-c_{1}a_{2})}^{2}=m$. -We can continue in this way to show the claim if $m=p_{1}p_{2}...p_{k}$.<|endoftext|> -TITLE: How to motivate and present epsilon-delta proofs to undergraduates? -QUESTION [23 upvotes]: This would seem to be a common question, but I am surprised not to see it already asked and answered on MO! -I am teaching an undergraduate course, and I want to teach them to construct basic epsilon-delta proofs, say, of $\lim_{x \rightarrow 3} x^2 = 9$ and $\lim_{x \rightarrow 4} x^2 \neq 17$. (Elementary, continuous functions only.) This is a serious stumbling block for many students, with good reason, and I anticipate it will be for mine as well. -Do other MO'ers have suggestions (beyond what I can find in typical calculus books) for presenting epsilon-delta for the first time? Any success stories to share? -Thank you very much! --Frank -(Background: I am teaching a discrete math course to American undergraduates who have already had a year of calculus. Whether $\epsilon-\delta$ is on topic for discrete math is perhaps questionable, but we did material on making sense of statements with lots of quantifiers, and also an introduction to techniques of proof, and so the material seemed like a natural fit. I should also mention that I intend to test the students on this material and not just expose them to it.) - -REPLY [4 votes]: Hi! -One of my professors back in Iran used to say everybody by heart has an intuitive understanding of "epsilon-delta" definition of limit/continuity based on the following everyday life example: -if you want to increase the amount of water coming out of a water faucet by epsilon, you -know that there is delta that if you turn the faucet by delta you get epsilon change in the output. This I guess everybody has experienced specially adjusting water temperature in shower where usual relationship between amount of output water and how much you need to turn is -not linear/uniform but nevertheless always continuous.<|endoftext|> -TITLE: Coreflective Subcategories of the Stable Homotopy Category -QUESTION [6 upvotes]: Here by stable homotopy category I mean the homotopy category of spectra, or maybe just some monogenic, Brown, algebraic, etc. stable homotopy category (in the language of Hovey, Palmieri and Strickland). -It is my understanding that it is still open as to whether or not a given localizing subcategory of the stable homotopy category can in fact be realized as the set of acyclics of some homology theory. Is this indeed the correct framing of the question? It seems to me that a localizing subcategory necessarily induces a localization functor (the left adjoint of the inclusion), but it not known whether or not this functor corresponds to a homology theory. Perhaps this is not correct, please let me know. -Assuming the correct understanding of the above situation, how does this fit into the framework of (co)reflective subcategories? We have Casacuberta et al.'s proof that Vopenka's principle implies that every localizing subcategory is co-reflective. Does that answer the above question (which is discussed in various papers of Hovey, Palmieri, Strickland and others) modulo the necessary set-theoretic assumption, or is this notion of coreflectivity unrelated? -And lastly, does the above paper, in answering questions about colocalizing subcategories, make strong statements regarding cohomological Bousfield classes? - -REPLY [6 votes]: I'm not sure if I constitute an expert or this constitutes a real answer but let me try. -If I understand correctly your first question is whether it is open that every localizing subcategory of an algebraic stable homotopy category $\mathcal{S}$ (so it should be triangulated, symmetric monoidal in a way compatible with the triangulation, be compactly generated and the compact objects should be rigid, and monogenic say) is a homological Bousfield class. This is, as far as I know, open - it is not even known in general that there are a set of localizing subcategories rather than a proper class. On the other hand it is known, under more general hypotheses, that there is a set of homological Bousfield classes (see Theorem 3.1 here). -It is possible to produce counterexamples i.e., to build rather simpleminded tensor triangulated categories where one can write down both the localizing subcategories and homological Bousfield classes and check that they don't agree but I don't know of any monogenic counterexample. -It is not true, without some further hypotheses, that the inclusion of every localizing subcategory $\mathcal{L} \to \mathcal{S}$ has a right adjoint (i.e. that every localizing subcategory is coreflective). If $\mathcal{L}$ is generated by a set, and hence by a single object, Brown representability guarantees us a right adjoint, but there is no reason a priori that $\mathcal{L}$ should have a generating set if it is just handed to us. This is where the work of Casacuberta, Gutiérrez, and Rosický comes in. They prove that, assuming Vopěnka's principle, every localizing subcategory of the homotopy category of a combinatorial stable model category is generated by a set of objects and hence is coreflective. This doesn't settle the question of whether all such subcategories are Bousfield classes though (or if there is a set of them) and as Tom points out this question seems to be of a different nature. -For your final question the CGR paper does not make strong statements about cohomological Bousfield classes that I am aware of. Under Vopěnka (with the same combinatorial model hypothesis) they prove that there is a bijection between localizing and colocalizing subcategories given by taking orthogonals. Thus if every colocalizing subcategory were singly generated this would prove that every localizing subcategory is a cohomological Bousfield class. However, they say they were not able to prove this, even under Vopěnka, and in the final part of the paper discuss the analogy with torsion classes where the analogous generation result is known not to hold (already in abelian groups).<|endoftext|> -TITLE: Strong Kodaira vanishing -QUESTION [7 upvotes]: Let $X$ be a smooth projective variety (say, over a field of characteristic zero). -Let us say that strong Kodaira vanishing holds for $X$ if -$$ -H^q(X,\Omega^p\otimes L)=0 -$$ -for every $p\geq 0$, $q>0$ and an ample line bundle $L$ on $X$. -My questions are now these: -1) Does strong Kodaira vanishing hold for $X={\mathbb P}^N$? -2) Does it hold for partial flag varieties of a semi-simple group $G$? -3) What tools are there for proving that strong Kodaira vanishing holds for -a given variety $X$? - -REPLY [7 votes]: Perhaps I might add that the "strong Kodaira vanishing" holds more generally for smooth -projective toric varieties in any characteristic. This goes back to Danilov. This includes -your case 1 of course. I can't remember how he did this, but an argument observed by number of people (Fujino, myself,...) is to use a what I might call a mock Frobenius splitting argument. The idea is to exploit the map $\phi$ given by multiplication by $r$ on the fan. -For projective space, this is just $[x_0,\ldots, x_N]\mapsto [x_0^r,\ldots, x_N^r]$. -If $r>1$ is prime to the characteristic, then $\phi^*$ can -be shown to give an injection -$$H^q(X,\Omega_X^p\otimes L)\hookrightarrow H^q(X,\Omega_X^p\otimes L^r)$$ -So choosing $r\gg 0$, we get the desired result by Serre vanishing.<|endoftext|> -TITLE: Can we change the Lebesgue measure by forcing? -QUESTION [11 upvotes]: Suppose $M$ is a model of ZFC, and $\mu^M$ is the Lebesgue measure on $\mathbb R^M$ such that $\mu^M(\mathbb R^M)=1$. It is known that if $r$ is a Cohen real over $M$ and $N=M[r]$ then $\mu^N(\mathbb R^M)=0$. -This is a very strong transition from having sets with a full measure being annihilated into nullity. Is it possible that for some[every?] $x\in(0,1)$ there exists $N=M[G]$ a generic extension of $M$ such that $\mu^N(\mathbb R^M)=x$? -If the answer is negative in its full generality ($M$ is just any model of ZFC) can we add some assumptions for a positive answer? (e.g. $M\models CH$) -This is really just idle curiosity which could not be satisfied via Google, references to a possible answer would be just as welcomed as a complete answer - -REPLY [13 votes]: The answer is no. If $\mathbb{R}^V$ is measurable in a forcing extension $V[G]$ having new reals, then the measure must be $0$. The point is that every new real $x$ in $V[G]$ but not in $V$ is transcendental over $\mathbb{R}^V$, since one cannot add algebraic numbers by forcing. It follows that the translates of $\mathbb{R}^V$ by the powers of $x$ are disjoint. Thus, a Vitali-style argument with wrapped translations of the unit interval of $V$ shows that if $\mathbb{R}^V$ is measurable, it must have measure zero. -Gunter Fuchs and I had made this observation in connection with Sebastian's question Probabilities independent of ZFC?, which is very much related to your question.<|endoftext|> -TITLE: $U\left(\mathfrak a\right) \otimes_{U\left(\mathfrak a\cap\mathfrak b\right)} U\left(\mathfrak b\right) \cong U\left(\mathfrak a + \mathfrak b\right)$ over a ring containing $\mathbb{Q}$ -QUESTION [7 upvotes]: While the Poincaré-Birkhoff-Witt theorem is usually proven (and sometimes even formulated) for free modules only, it is known (see also here) that it holds for arbitrary modules if the ground ring is a $\mathbb Q$-algebra. I am wondering if a similar generalization holds for the following fact, which I learnt today from Pavel Etingof: -Let $\mathfrak c$ be a Lie algebra. Let $\mathfrak a$ and $\mathfrak b$ be two Lie subalgebras of $\mathfrak c$ such that $\mathfrak a + \mathfrak b = \mathfrak c$. Clearly, $\mathfrak a \cap \mathfrak b$ is also a Lie subalgebra of $\mathfrak c$. Now, the map -$U\left(\mathfrak a\right) \otimes_{U\left(\mathfrak a\cap \mathfrak b\right)} U\left(\mathfrak b\right) \to U\left(\mathfrak c\right),$ -$\alpha\otimes_{U\left(\mathfrak a\cap \mathfrak b\right)} \beta\mapsto\alpha\beta$ -is an isomorphism (not of algebras, but of $\left(U\left(\mathfrak a\right),U\left(\mathfrak b\right)\right)$-bimodules). -This is proven for free modules using PBW and appropriate bases. I had no time to do any research on this. - -REPLY [4 votes]: For some reason, I had forgotten about this question even as I found the answer long ago. -Yes, the fact holds whenever the ground ring is a $\mathbb{Q}$-algebra. For the proof, see the First proof of Proposition 2.4.1 in my notes for Pavel Etingof, 18.747 Infinite-dimensional Lie Algebras, Spring term 2012 at MIT. (Of course, read "commutative $\mathbb{Q}$-algebra" for "field". In the footnote that gives two proofs for $\sigma$ being an algebra isomorphism, read the second of these two proofs; the first requires the ground ring to be a field.) -More generally, this shows that the fact holds whenever the PBW theorem holds for both $\mathfrak{a}$ and $\mathfrak{b}$ (that is, all three canonical maps $\operatorname{PBW}_{\mathfrak{a}} : \operatorname{Sym}\left(\mathfrak{a}\right) \to \operatorname{gr}\left(U \left(\mathfrak{a}\right)\right)$ and $\operatorname{PBW}_{\mathfrak{b}} : \operatorname{Sym}\left(\mathfrak{b}\right) \to \operatorname{gr}\left(U \left(\mathfrak{b}\right)\right)$ and $\operatorname{PBW}_{\mathfrak{c}} : \operatorname{Sym}\left(\mathfrak{c}\right) \to \operatorname{gr}\left(U \left(\mathfrak{c}\right)\right)$ are module isomorphisms). (The proof pretends to also require $\operatorname{PBW}_{\mathfrak{a} \cap \mathfrak{b}}$ to be an isomorphism; but if you dig a little bit deeper, you'll see that $\operatorname{PBW}_{\mathfrak{a} \cap \mathfrak{b}}$ merely has to be surjective, which it always is.) -But this is not an "if and only if"! The question of whether the fact holds more generally is still open.<|endoftext|> -TITLE: Dehn Twist in the sense of Geometric Group Theory and a Graph of Groups -QUESTION [5 upvotes]: Hello! -Does anyone knows a good book or script about Dehn Twists in the sense of graphs. More precisely: I need to know how a Dehn Twist yields an automorphism of a group or a subgroup of a group. I want to know something about the realtionship between dehn twists and graph of groups. -I know that my request is not nice formulated, but i didn't found any good introduction about this theme in the internet yet. I always get defintions for dehn twists in the sense of surfaces. But i don't know, how to conclude or educe something about groups in the sense of geometric group theory. -Thanks for help. - -REPLY [7 votes]: In the Handbook of Geometric Topology, Bestvina's article on R-trees has a few pages on this under Applications. There is also the paper Cyclic splittings of finitely presented groups and the canonical JSJ decomposition by Rips and Sela but this is a research paper and I don't think that's what you want. -Informally, a Dehn twist of a group is an element of the automorphism group which functions analogously to a Dehn twist in the mapping class group of a surface. If you have such a surface, you can separate it into smaller surfaces by finding a collection of disjoint simple closed curves and cutting along them. Now, each of the remaining surfaces has a free fundamental group and the fundamental group of the whole space can be computed by van Kampen's theorem. In other words, the original surface is a collection of components and separating annuli which gives you a graph of groups decomposition of the surface group. This is a Z-splitting of the fundamental group. -A Dehn twist, topologically, is performed by taking any loop which intersects more than one of the components above and twisting that curve around the simple closed curve which separates those components. The effect of the Dehn twist in the topological setting is by inner automorphism on the component spaces (you should work out the specific details here). -In the group setting, you have an edge group (which is analogous to the s.c.c. above) by which you conjugate. This gives an automorphism of the group which generalizes the situation described for surfaces above (ie inner on the vertex groups).<|endoftext|> -TITLE: How to compute the étale cohomology of the quotient of a surface by a finite group of automorphisms ? -QUESTION [7 upvotes]: Let $S$ be a smooth surface defined over a finite field $K$ of char. $p$. Let $G$ be a finite group of automorphisms of $S$. Let $Z\to S/G$ be the minimal resolution of the quotient of $S$ by $G$. Suppose that the fixed point set of every elements of $G$ is defined over $K$. -Let $\ell$ be a prime, $\ell\not=p$. -Is it true that $$H^{i}(Z,\mathbb Q_\ell) \simeq H^{i}(S,\mathbb Q_\ell)^G$$ for $i=1,3$ and -$$H^{2}(Z,\mathbb Q_\ell)\simeq H^{2}(S,\mathbb Q_\ell)^G+\mathbb Q_\ell C_1+\dots +\mathbb Q_\ell C_k$$where the $C_i$ are the exceptional curves of the resolution $Z\to S/G$ ? -This question is an echo of the question "Are there any known formulas about the Hodge-Deligne structure of quotients by action of groups ?" formulated in this forum 2 or 3 weeks ago. - -REPLY [3 votes]: Xavier, -As per your request, I'll give a partial answer on mathoverflow assuming, for simplicity, that -$G$ has isolated fixed points on $S$. Let $H^i(S)$ etc. denote the $\ell$-adic cohomology of -$S\otimes\bar K$. I'll ignore Tate twists. Rather than taking the minimal resolution $Z$, -let me take the minimal "log" resolution, so that the exceptional divisor $C\subset Z$ has normal crossings -Let [M] = Milne, Etale Cohomology. I'm referring to the book, but the online notes should work just as well. -Let $\iota:\Sigma\subset S$ be the set of fixed points, $U=S-\Sigma$ and $V=U/G$. Then by -Hochschild-Serre [M, p105] $H^i(V)= H^i(U)^G$. By Poincare duality $H^i(V)= H_c^{4-i}(V)$. -From the sequence -$$\ldots H_c^i(U)\to H^i(S)\to H^i(\Sigma)\ldots$$ -coming from $0\to j_!\mathbb{Q}_\ell\to \mathbb{Q}_\ell\to \iota_*\mathbb{Q}_\ell\to 0$ -we deduce that - -$H^i(V)= H^i(U)^G=H^i(S)^G$ for $i=1,2$. - -If you combine this with the Gysin sequence -$$0\to H^1(Z)\to H^1(V)\to \bigoplus H^0(C_k)\to H^2(Z)\to H^2(V)$$ -you should be in good shape. -To get the sequence above, apply the version of the Gysin sequence on [M, p 244] to -the complement of the set of double points $Z'=Z-C_{sing}$. Use the isomorphism -$H^2(Z)\cong H^2(Z')$ and do a diagram chase.<|endoftext|> -TITLE: Is there a coordinate-free proof of the hamiltonian character of the geodesic flow? -QUESTION [17 upvotes]: I do not know if this question is appropriate for this site, but I posted here without having answers, so now I make this attempt. -Let be $(M,g)$ a pseudoriemannian manifold. Let us identify the tangent and the cotangent bundles through the musical isomorphism $g^\flat:u\in TM\to g(u,\cdot)\in T^\ast M.$ -It is well known that: - -The geodesics of $(M,g),$ i.e. the solutions of - $\frac{D}{dt}\gamma=0,$ are integral curves for the hamiltonian vector - field of $K:u\in TM\to \tfrac{1}{2}g(u,u)\in\mathbb{R}$ w.r.t. the - canonical symplectic form. - -Question Knowing how to show it using coordinates and Christoffell symbols, I am wondering how to prove it in an intrinsic way. - -REPLY [3 votes]: Suppose that $G$ and $G'$ be the vector fields on $\mathrm{T}M$ that describe the first and the second definition of the geodesic flow; we need to show that $G=G'$. -If the metric tensor $g$ is flat, then $G=G'$ (it should be obvious). -Now assume that two metric tenors $g_0$ and $g_1$ on $M$ coincide at a point $p$ up to first order. -Observe that $G_0=G_1$ and $G'_0=G'_1$ on $\mathrm{T}_p$. -Finally, observe that given a Riemannian metric $g_1$ there is a flat metric $g_0$ that coincides with $g_1$ up to first at a given point $p$.<|endoftext|> -TITLE: On compact Kähler manifold diffeomorphic to complex projective space -QUESTION [10 upvotes]: In the paper On the complex projective spaces, Hirzebruch and Kodaira prove the following: - -If $X$ is compact Kähler manifold diffeomorphic to $\mathbb{CP}^n$, then $X$ is biholomorphic to $\mathbb{CP}^n$ if $n$ is odd or $n$ is even but $c_1 \neq -(n+1)g$. - -Here, $c_1$ is the first Chern class of $X$, and $g$ is a generator of $H^2(X,\mathbb{Z})$ with the same sign as the Kähler class. -My question is: Has the case $X$ is diffeomorphic but not biholomorphic to $\mathbb{CP}^n$ with $n$ even and $c_1=-(n+1)g$ been ruled out in the following years? Or do we have some constructions of manifolds of this type? - -REPLY [24 votes]: The case when $n$ is even has been famously ruled out by Yau as a consequence of his proof of the Calabi Conjecture, see his original paper or these notes. -In fact, thanks to results of Novikov, you can replace diffeomorphism by homeomorphism and the statement remains true. -When $n=2$ in the same paper Yau proved that you can also drop the Kähler condition and replace homeomorphism by homotopy equivalence. -When $n>2$ the statement for Kähler manifolds but assuming only homotopy equivalence is true up to $n=6$ thanks to Libgober and Wood (and true in general if you assume furthermore that the Pontryagin classes are the same as $\mathbb{CP}^n$), see also this question. The case of higher $n$ is probably open. -Finally, it is still an open problem to decide whether a compact complex manifold diffeomorphic to $\mathbb{CP}^n$ is biholomorphic to it (when $n\geq 3)$. If this were true, it would imply that the differentiable manifold $S^6$ does not admit a complex structure (otherwise, blowing up a point you'd get a complex manifold diffeomorphic but not biholomorphic to $\mathbb{CP}^3$), see for example here.<|endoftext|> -TITLE: Do weak equivalences in a model category admit a calculus of fractions? -QUESTION [6 upvotes]: Suppose it is given a model category $M$, $W$ being the set of its weak equivalences. One can define the localization of $M$ along $W$, denoted by $W^{-1}M$. In order to work with $W^{-1}M$, it would be of great help if $W$ admitted a calculus of fractions. So, my question is: is that true in general, or perhaps under some appropriate assumptions? I suspect the answer in general is "no", but I really hope it is "yes under appropriate assumptions": I'm currently working on dg-categories and craving for a "simple" description of the category $\mathrm{Ho}(\mathbf{dg\text{-}cat})$ of small dg-categories localized along quasi-equivalences. -Thanks in advance for any answer! - -REPLY [3 votes]: At the end I'll give you a reference to a counterexample in the paper of Meier and Ozornova. But before that, I want to talk in the other direction: besides the 3-arrow calculus that Charles Rezk mentions, there's another sense in which $W$ "almost" admits a calulus of fractions. It's a result in the same paper. -Recall the functor $\mathrm{Ex}: \mathsf{sSet} \to \mathsf{sSet}$, which is right adjoint to barycentric subdivision. There is a canonical natural transformation $1 \Rightarrow \mathrm{Ex}$, allowing to define maps $\mathrm{Ex}^n \to \mathrm{Ex}^{n+1}$ for every $n$, and most famously the colimit $\mathrm{Ex}^\infty$ was shown by Kan to be a fibrant replacement functor for the Kan-Quillen model structure on $\mathsf{sSet}$; in particular $\mathrm{Ex}^\infty X$ is a Kan complex for every simplicial set $X$ (I don't know if this extends further -- is $\mathrm{Ex}^\infty f$ a Kan fibration for every simplicial map $f$? Is $\mathrm{Ex}F$ injective-fibrant for every diagram $F$?). But moreover, a functor $F: \mathcal{C} \to \mathcal{D}$ between categories is fibrant in the Thomason model structure on $\mathsf{Cat}$ iff $\mathrm{Ex}^2 F$ is a Kan fibration (where we identify a category with its nerve). -All of this is just to motivate looking at $\mathrm{Ex}^2 \mathcal{C}$ for a category $\mathcal{C}$. Meier and Ozornova show that if $W$ is the category of weak equivalences in a model category (actually, only a substantially weaker structure called a "partial model category" is necessary), then $\mathrm{Ex}^2 W$ is a Kan complex (i.e. $W$ is Thomason-fibrant). -Why am I telling you this? Well, Meier and Ozornova also show something surprisingly (to me) clean: - - -Theorem: If $\mathcal{C}$ is a category, then $\mathrm{Ex}^1 \mathcal{C}$ is a Kan complex if and only if $\mathcal{C}$ admits a left calculus of fractions. - - -It follows that $\mathcal{C}$ admits a two-sided calculus of fractions if and only if both $\mathrm{Ex}^1 \mathcal{C}$ and $\mathrm{Ex}^1 \mathcal{C}^\mathrm{op}$ are Kan complexes. Meier and Ozornova also suggest (Question 2 at the end) that there might be notions of ``$n+1$-arrow calculus" giving clean characterizations of $\mathrm{Ex}^n \mathcal{C}$ being a Kan complex -- which in particular would recover the result that the weak equivalences in a model category admit a 3-arrow calculus, since they are Kan after applying $\mathrm{Ex}^2$. -Finally, Meier and Ozornova provide examples answering your actual question in the negative. Finite sets and injections (Example 5.1) form a partial model category whose weak equivalences do not admit a left calculus of fractions. More pertinently (Example 5.3), $\mathsf{Cat}$ with the Thomason model structure is itself a model category which does not admit a calculus of left fractions. - Note, though, that the weak equivalences in a left (resp. right) proper model category admits a left resp. right) caculus of fractions.<|endoftext|> -TITLE: Asymptotic behaviour of $\int f(t)^a\cos(at)dt$ -QUESTION [5 upvotes]: Are there any known necessary or sufficient conditions such that -$$\lim_{a\rightarrow \infty}\int_{-1}^1f(t)^a\cos(at)dt=0$$ -where $f:[-1,1]\rightarrow[1,\infty)$ is an even smooth concave real function such that $f(-1)=f(1)=1$? - -REPLY [5 votes]: Sorry for the late reply (for the last month I hardly had any time for anything like MO). The answer (somewhat vague) is that the curve $t\mapsto (t,\log f(t))$ ($-1\le t\le 1$) has to be the image of the upper half circle under some analytic in $\mathbb C\setminus[(-\infty,-1]\cup[1,+\infty)]$ mapping $F$, which is symmetric ($F(\bar z)=\overline{F(z)}$), one to one in the unit disk, and whose derivative $F\,'$ has decent boundary behavior. The "decent boundary behavior" is the vague part here. Unfortunately, I cannot make it less vague unless someone tells me how exactly to recognize the distributions on $[-1,1]$ whose Fourier transform tends to $0$. It is clear that they are necessarily fairly tame (of not more than the first order, etc.) and that all $L^1$-functions are there but where exactly you are in between is a mystery to me. On the crudest level, it tells you that $f$ must be analytic. -Let me know if such description is of any interest to you. If it is, I'll post the details. -EDIT: OK, here go the details. It is a somewhat long story, so I may need more than one patch of free time to type it. I apologize in advance for bumping this thread. Also, since the integral against $\sin at$ is zero, we can just as well talk about the full Fourier transform, i.e., the integration against $e^{-iat}$ -1) It is actually quite surprising that such functions exist at all. After all, the jump discontinuity normally means that the best rate of decay of the Fourier transform is $1/a$ and that slow rate of decay is played against the exponential growth of the integrand. So, I'll start with constructing one such function. It'll be easier to work with $g(t)=\log f(t)$, which is a smooth non-negative function with endpoint values $0$. Put $g(t)=\delta(1-t^2)$ with small $\delta>0$. Then the integral can be written as the path integral $\int_\gamma \frac{dt}{dz}e^{-iaz}\,dz$ where $\gamma$ is the curve $t\mapsto z(t)=t+ig(t)$. Note that $z(t)=t+i\delta(1-t^2)$ is an analytic function of $t$ and for small $\delta>0$, it is invertible in a fairly large disk. Thus, we can talk about its analytic branch $t(z)$ that coincides with $\Re z$ on $\gamma$ and is analytic in (a neighborhood of) the region $D$ bounded by $[-1,1]$ and $\gamma$. So, $t'(z)$ is also analytic there and we can shift the contour of integration from $\gamma$ to $[-1,1]$, which results in the representation $\int_{-1}^1 t'(z)e^{-iaz}\,dz$, which is just the ordinary Fourier transform of the integrable (and even smooth) function $t'(z)$ restricted to $[-1,1]$, so the integral, indeed, tends to $0$ in this case. -2) What I'd like to show now is that this contour integral representation and the possibility to shift the contour is the only possible reason for this effect. The starting point is that if the integral is bounded on the entire real line (the boundedness on the negative semi-axis is trivial and the boundedness on the positive one is less than what has been requested), then there exists a distribution $T$ supported on $[-1,1]$ such that the integral equals $\langle T, e^{-iat}\rangle$ for all $a\in\mathbb C$ (that is just a version of Paley-Wiener). Thus, the difference $\langle T, e^{-iat}\rangle-\int_\gamma \frac{dt}{dz}e^{-iaz}\,dz$ vanishes for all $a\in \mathbb C$. Now, the linear span of functions $e^{-iaz}$ is dense in the space of functions analytic in any fixed neighborhood of $D$ meaning that $\langle T, \psi(t)\rangle-\int_\gamma \frac{dt}{dz}\psi(z)\,dz=0$ for every function analytic in some neighborhood of $D$. -We will take $\psi(z)=\frac{1}{z-\zeta}$ with $\zeta\notin D$ and get the Cauchy integral plus something analytic in $\mathbb C\setminus[-1,1]$ vanish outside $D$. Note that this Cauchy integral and the distribution part are also well-defined for $\zeta\in D$ and give an analytic function of $\zeta$ there. Moreover, by Plemelj's jump formulae, the boundary values of that function on $\gamma$ are just $\frac{dt}{dz}$ (up to $2\pi i$ and $\pm$, which we aren't concerned with here). The upshot is that $\frac{dt}{dz}$ has an analytic extension to $D$ continuous up to $\gamma$ (here we use that the curve is assumed to be of some decent smoothness; otherwise we'll have to sing a long song of non-tangential boundary values a.e., etc.) -The behavior on $[-1,1]$ may be more complicated in general and the boundary values there exist only in the sense of distributions. The possibility of analytic continuation to the open domain $D$ guarantees only the possibility to shift the contour to something hovering as low over $[-1,1]$ as we wish, i.e., to the subexponential growth of the integral (for which it is necessary and sufficient). However, if you settle for some more reasonable class than $C_0$, say, $L^2$, then $T$ will be just an $L^2$ function and you'll have the classical theory of boundary values that will allow you to show that our distribution is, indeed, the boundary value of the analytic extension of $\frac{dt}{dz}$ and the reason for smallness of the integral is the possibility of the ordinary contour shift. I have no idea what you are going to use all this for, so I prefer to avoid the discussion of all those technical issues. Instead, I'll discuss in detail what the possibility of this analytic extension of the derivative means for the curve $\gamma$ itself. -3) Let $Q$ be the lower unit half-disk $\{z:|z|<1,\Im z<0\}$. Let $\varphi$ be the conformal mapping from $Q$ to $D$ such that the interval $[-1,1]\subset\partial Q$ is mapped to $\gamma$ and the lower semicircle is mapped to $[-1,1]$. The derivative $\varphi'$ is a continuous up to the boundary (except for the points $-1,1$ where it has an easy-to-control power singularity) non-vanishing function in $D$ (here we use reasonable smoothness of $f$ again). Note that after the composition with $\varphi$, the function $\frac{dt}{dz}$ on $\gamma$ becomes $\frac{(\Re\varphi)'}{\varphi'}$ on $[-1,1]$. This should be extendable analytically to the lower half-disk with "decent boundary values". Since $\varphi'$ has such extension, we conclude that so does $(\Re\varphi)'$. But this function is real-valued, so the Schwarz reflection principle applies and we conclude that it extends analytically to the entire unit disk. Let $(\Re\varphi)'=F$ where $F$ is a symmetric analytic function in the unit disk. The function $\varphi'-F$ is purely imaginary on $[-1,1]$ and extends analytically to the lower half-circle. Thus, using the reflection principle again, we conclude that $\varphi'=F+iG$ where $F,G$ are symmetric analytic functions in the unit disk and $F$ has decent boundary values on the lower semicircle. Thus, $\varphi'$ and $\varphi$ are analytic in the unit disk. Moreover, since $F+iG=\varphi'$ is nice on the lower semicircle, $G$ also has decent boundary values there and therefore, after reflecting, we see that $\varphi'$ has decent boundary values on the upper semicircle. To get the proclaimed description, it suffices now to map the unit disk to the upper half-plane so that the lower semicircle is mapped to $[-1,1]$ and use the reflection principle again for the last time. -That's it (modulo minor technicalities that I swept under the rug, but, as I said, to get into those would make no sense without knowing what exactly you are after).<|endoftext|> -TITLE: Traversing the infinite square grid -QUESTION [17 upvotes]: Starting somewhere on an infinite square grid, is it possible to visit every square exactly once, if at move $n$, one must jump $a_n$ steps in one of the directions north,south,east or west, and mark the ending square as visited? -If $a_n=n$ or if $a_n=n^2$? -Allowing diagonal moves as well, is there a general algorithm, given $a_n$, to check if a path exists? -Note: -I am asking if given $a_n$, there exists an infinite sequence of directions, $d_n\in(N,S,W,E)$, such that for all $(x,y)\in Z^2$, there exists a finite integer $k(x,y)$, such that starting at the unit square with center $(0.5,0.5)$, marked as visited, we have after moving sequentially $a_i$ steps in direction $d_i$, for $i=1,2,3,...,k$, visited $k+1$ different unit squares, and are situated at $(x+0.5,y+0.5)$. - -REPLY [6 votes]: It's possible for $a_n=n$ and probably most stepsizes without modular or growth obstructions. -We have covered some subset of an mxm square, are situated at the boundary, and want to visit a cell (x,y) in our square. Choose one of the x,y axes and move far away along it, (but not upon it), until stepsize s>>m and distance is some d from the axis. Then take either 1,2, or 4 more steps along the axis. Then alternately move away, and towards the axis, 2*d steps, until we land on it. Then by moving away and towards (x,y), n times, we can reach every point of the form j-1-3n on the axis, by just moving one more step towards (x,y) where j is our current coordinate, which we could shift to anything modulo 3 when we chose one of the 1,2,4 steps. And if 3n-n>m, we dont use any other squares within the the mxm square, to visit (x,y), and emerge on the opposit side. And since s>>m, if we take one more step we are at a boundary of a new square. -WLOG suppose we are at $(0,0)$, with stepsize $s(0)$, and want to visit $(x,y)$, $0\leq x\leq m$, $0\leq y \leq m$, The full path we take consist of these moves. We move south for $k$ squares (or $j(k)$ steps), then alternate west,east, $x$ times, now we are at $(x,-k)$ with stepsize $s(j(k)+2x)$. Then we alternate south, east, $n$ times, now we are at $(x,n-k)$ with stepsize $s(j(k)+2x+2n)$, select $n$ and $k$ such that $y=s(j(k)+2x+2n)+n-k$ and $s(j(k)+2x+2n)>m$. Then take two steps north, we are now at $(x,y+s(j(k)+2x+2n)+1)$. Move 1 step east, you are now at a corner of a square bounding all visited squares, define the new m to be the side of this new square, let (0,0) your position, pick a new point (x,y) and repeat. -PS. I asked a question about the less trivial 1-D version here: -https://math.stackexchange.com/questions/111377/self-avoiding-walk-on-mathbbz<|endoftext|> -TITLE: Status of conjectures in Serre's 1969 expose on Galois representations on l-adic cohomology -QUESTION [19 upvotes]: In -[S]: Serre, Jean-Pierre. Facteurs locaux des fonctions zeta des varietes algebriques (definitions et conjectures), Seminaire Delange-Pisot-Poitou, 1969-70 -Serre presents nine conjectures *C*$_1$-*C*$_9$ concerning Galois representations on $l$-adic cohomology of nonsingular projective algebraic varieties defined over a local or global field. What is the status of those conjectures today? The first two of them are part of the Weil conjectures but how about the rest? What (partial) progress has been made towards their resolution? Let me recall the conjectures below (skipping the first two). -Let $K_v$ be a local field of residue characteristic $p$, let $Y$ be a nonsingular projective variety over $K_v$, and let $m\ge 0$. If $G_v$ is the absolute Galois group of $K_v$ the functoriality of $l$-adic cohomology ($l \neq p$) gives us a representation $\rho_l\colon G_v \rightarrow H^m(\overline{Y}, \mathbb{Q}_l) =: V$ (here $\overline{Y}$ is the base change of $Y$ to the separable closure of $K_v$). One can measure how ramified $\rho_l$ is by introducing $\epsilon = \dim V - \dim V^{I_v}$ ($I_v$ is the inertia in $G_v$) and $\delta$ which is a little bit more involved to define (it is the inner product of $\mathrm{Tr }\ \rho_l|_{I_v}$ with the "Swan character" of $\rho_l$, see 2.1 in [S]). The conductor exponent of $\rho_l$ is then $f = \epsilon + \delta$. -*C*$_3$: The integers $\epsilon$, $\delta$ and $f$ are independent of $l$. -Status: ??? -*C*$_4$: $\mathrm{Tr }\ \rho_l|_{I_v}$ takes values in $\mathbb{Z}$ and is independent of $l$. -Status: ??? -Consider the geometric Frobenius $\pi \in G_v/I_v$. Then $\rho_l(\pi)$ acts on $V^{I_v}$ and we get a polynomial $P_{\rho_l}(T) = \det(1 - \rho_l(\pi) T)$. -*C*$_5$: $P_{\rho_l}$ has coefficients in $\mathbb{Z}$ and is independent of $l$. -Status: ??? -Assuming the latter conjecture split $P_{\rho_l}(T) = \prod (1 - \lambda_\alpha T)$ and let $Nv$ denote the cardinality of the residue field of $K_v$. -*C*$_6$: For each $\alpha$ there is an integer $m(\alpha)$ between $0$ and $m$ such that $|\alpha| = (Nv)^{m(\alpha)/2}$. -Status: ??? -*C*$_7$: If $\epsilon = 0$ (i.e., if $\rho_l$ is unramified) then all $m(\alpha)$ are equal to $m$. -Status: ??? -*C*$_8$: Let $g$ be an element of $G_v$ whose image in $G_v/I_v$ is an integral power of Frobenius. Then the characteristic polynomial of $\rho_l(g)$ has coefficients in $\mathbb{Q}$ and is independent of $l$. -Status: ??? -The last conjecture *C*$_9$ concerns the zeta function $\zeta(s)$ of a nonsingular projective variety $X$ defined over a global field $K$ (and fixed $m \ge 0$ as above), as well as the completed version $\xi(s)$ of $\zeta(s)$. It's a bit of a trek to define $\zeta(s)$ and $\xi(s)$ so I'll refer to [S] $\S$3, $\S$4 for that. Both $\zeta(s)$ and $\xi(s)$ are holomorphic functions on some right half-plane. -*C*$_9$: $\zeta(s)$ and $\xi(s)$ admit meromorphic continuations to the complex plane. In addition, $\xi(s)$ satisfies the functional equation $\xi(s) = w\xi(m + 1 - s)$ with $w = \pm 1$. -Status: Open. Afterall, a special case of this is meromorphic continuation of $L$ functions of elliptic curves over number fields. -As the answers come in feel free to edit the status fields above adding more information (and I will try to do that myself). - -REPLY [5 votes]: Because I recently had to think about this, let me sum up the results I know about conjecture C5. -This conjecture is known to hold for any $m\in\mathbb N$ if the dimension of $Y$ is less than 2 by Takeshi Saito, Weight spectral sequences and independence of $\ell$, J. Inst. Math. Jussieu 2 (2003). It also holds for any $m$ if $Y$ is an abelian scheme by Alexander Grothendieck Exposé IX of SGA7 Groupes de monodromie en géométrie algébrique. Hence, it holds for general $Y$ if $m\leq 1$. -Beyond these results and the few very special cases of Shimura varieties and varieties which are known to be quotients thereof, I don't think much is known.<|endoftext|> -TITLE: Why should I care about Heegaard-Floer theory? -QUESTION [43 upvotes]: I would like to collect a list of applications of Heegaard-Floer theory. By applications, I don't mean things like "it can detect the unknot" or "it can detect knot genus". Algorithms for these kinds of things have been known since the '60's and '70's. Instead, I mean two kinds of things. - -Questions that make no reference to Heegaard-Floer theory that can be answered using Heegaard-Floer theory (and, preferably, cannot be answered in other ways). -Things about knots or 3-manifolds that can be computed with Heegaard-Floer theory for which algorithms did not previously exist. - -I'm asking this question here in response to a large number of talks about Heegaard-Floer theory I've attended over the years. It seems like a hot subject and lots of talented young people are working on it, but most of the talks I've attended about it addressed what seemed to me to be technical questions internal to the subject. And when I've asked the speakers this question, I never seem to get a good answer. But since it is such a hot subject, I assume there must be some killer applications. - -REPLY [9 votes]: Yi Ni used Heegaard Floer Homology to prove, among many other things, that a knot admitting a lens-space surgery is fibred. I believe that no 'conventional' proof of this is known. -Ni, Yi, Knot Floer homology detects fibred knots. Invent. Math. 170 (2007), no. 3, 577–608.<|endoftext|> -TITLE: Asymptotic Geodesic Flow on Planar Graphs -QUESTION [5 upvotes]: Assume you have a simple and infinite graph. Choose $x_{0}$ an arbitrary vertex and consider -$$ -G_{n}:=\{x\in G:d(x_0,x)\leq n\} -$$ -with the graph metric (hop metric). Now for each pair of nodes there is a unit flow that travels through the minimum path between nodes (if there is more than one minimum path it splits equally along all such paths). The total flow in $G_{n}$ is equal to $|G_{n}|^2$. -Given a node $v\in G_{n}$ we define $T_{n}(v)$ as the total flow generated in $G_{n}$ passing through $v$. In other words, $T_{n}(v)$ is the sum off all the geodesic paths in $G_{n}$ which are carrying flow and contain the node $v$. Let $M_{n}$ be the maximum flow -$$ -M_{n}(v):=\max_{v\in G_{n}}{T_{n}(v)}. -$$ -For any graph $|G_{n}|\leq M_{n}\leq |G_{n}|^2$. It is not very difficult to see that for $G=\mathbb{Z}^2$ -$$ -M_{n}=O(|G_{n}|^{3/2}). -$$ - -My question is: can we do better than - this in a planar graph? In other - words, does there exist a infinite - planar graph $G$ such that - $M_{n}=o(|G_{n}|^{3/2})$? If so, how - low we can go. - -Note that the same problem could be reformulated by instead of having an infinite graph $G$ having a sequence of graphs $\{G_{n}\}$ such that $|G_n|\to\infty$. - -REPLY [7 votes]: You can't do better. Lipton and Tarjan's planar separator theorem says that any $n$-node planar graph $G=(V,E)$ contains a set $S$ of $O(\sqrt{n})$ vertices whose removal separates the graph into components all of which have size at most $2n/3$. We can then partition $V \setminus S$ into sets $X,Y$ each containing at least $(1/3-o(1))n$ vertices; there are order $n^2$ pairs $(u,v) \in X \times Y$, and any path between such a pair $(u,v)$ contains a vertex of $S$. Since $|S|=O(\sqrt{n})$, by the pigeonhole principle it follows that some element of $S$ is in order $n^{3/2}$ paths. between $X$ and $Y$.<|endoftext|> -TITLE: Mirror to the dualizing sheaf -QUESTION [5 upvotes]: I wonder - is there a general characterization of the object in the Fukaya category that is mirror to the dualizing sheaf on the other side? -This question has two cases: -1. CY -2. Non-CY -In 1. what I know is that by Polischuk-Zaslow the mirror of the dualizing sheaf in case of the 2-torus is a linear Lagrangian torus of type (1,0). -In 2. there is a subquestion - what is the analogue of the dualizing sheaf for the category of matrix factorizations of a Landau-Ginzburg model? - -REPLY [13 votes]: I'll comment on the related question "what is the Serre functor for the Fukaya category?" -Calabi-Yau setting -The Serre functor $S$, by definition, satisfies $\mathsf{Hom}(X,SY) \cong \mathsf{Hom}(Y,X)^\vee$; since it's characterized categorically, it's preserved by the derived equivalences which arise in mirror symmetry. For the derived category $D^b\mathsf{Coh}(X)$ of a non-singular projective $n$-variety, $S= \cdot \otimes \omega_X [n]$, where $\omega_X$ is the dualizing sheaf. So, when $X$ is Calabi-Yau, it's simply a shift. -For a Fukaya category of compact Lagrangians (of dimension $n$) in a symplectic manifold, $S$ is just a shift by $n$, because of the Floer-theoretic Poincare duality $HF^\ast(X,Y) \cong HF^{n-\ast}(Y,X)^\vee$. At the fully precise $A_\infty$-level, the claim that the Serre functor is a shift is partly conjectural. -The mirror to $\mathcal{O}$ is a section $\sigma$ of the SYZ fibration, so the mirror to the canonical sheaf is $\sigma[n]$. -LG models -To get a more interesting answer, consider Fukaya categories of Landau-Ginzburg models, a.k.a. Fukaya-Seidel categories. These arise as mirrors to Fano manifolds. Out of caution, I'll assume that the L-G model is a symplectic Lefschetz fibration $E\to \mathbb{C}$. The objects of the category are Lagrangian submanifolds which map to eventually-horizontal paths in $\mathbb{C}$ (for instance, Lefschetz thimbles). Kontsevich proposed that the Serre functor should then be the "wrapping" or "monodromy" functor. This has been proved (at least at the level of objects, probably more), by Seidel (cf. his Symplectic homology as Hochschild homology and Vanishing cycles and mutation). -The wrapping functor is defined as follows. Take a circle of large radius $R$ in $\mathbb{C}$, and consider the Dehn twist $\delta$ along this circle. So $\delta(z)=e^{i\rho(|z|)}z$, where the angle $\rho(|z|)$ runs from $0$ when $|z| < R-1$ to $2\pi$ when $|z| > R+1$. There's a symplectomorphism $\Phi$ of $E$, covering $\delta$, given by symplectic parallel transport of the fibration over the arc from $z$ to $\delta(z)$. The wrapping functor takes a Lagrangian $L$ to $\Phi(L)$. It takes a standard Lefschetz thimble (fibering over a ray) to a "once-wrapped thimble", i.e. a thimble for a path that wraps once around the circle. -In the case of the LG mirror to $\mathbb{CP}^2$, the mirror to $\mathcal{O}$ (which is one of the Beilinson generators of $D^b \mathsf{Coh}(\mathbb{CP}^2)$) is a thimble, so the mirror to the canonical sheaf is a once-wrapped thimble. -The proof that the wrapping functor is the Serre functor invokes a general characterization of the Serre functor in triangulated categories with full exceptional collections in terms of the algebraic process of "mutation". The thimbles associated with a collection of vanishing paths form a full exceptional collection, and mutation corresponds to Hurwitz moves on vanishing paths.<|endoftext|> -TITLE: Classification of smooth atlases -QUESTION [5 upvotes]: Let $\mathcal{A}$ be a smooth maximal atlas on a manifold $M$. Let $f:M\to M$ be a smooth invertible function, whose inverse is not smooth (for example $f:\mathbb R\to \mathbb R$, $f(x)=x^3$). Then $f$ induces an atlas $\mathcal{A}'$ which is not compatible with $\mathcal{A}$. -Since the relation of compatibility between atlases is an equivalence, the problem of classifying them appears. To simplify, we can try to classify the smooth atlases which are compatible on $M-N$, where $N\subset M$, but are incompatible on $N$. The simplest case seems to be when $N$ contains only a point. -Another problem is to find all maximal atlases for which a given function on $M$, or other object, for example a tensor, is smooth. -Are there any studies of these kinds of classifications of atlases? - -Update: I don't ask about exotic smooth structures. The compatibility of atlases is "finer", but both are named "smooth structures", which leads to confusions (see http://en.wikipedia.org/wiki/Smooth_structure#Confusion_about_terminology). - -Update 2 (example): -Let's consider the manifold $\mathbb R$ with the maximal atlas $\mathcal A$ generated by $id:\mathbb R\to \mathbb R$. Let the atlas $\mathcal A'$ be generated by $f(x)=x^3$. Then not all functions on $\mathbb R$ which are smooth in one atlas are smooth in the other too. -Now let's consider the set $N$ containing only the origin. When we restrict the charts from $\mathcal A$ and $\mathcal A'$ to $M-N$, we obtain two compatible atlases, although $\mathcal A$ and $\mathcal A'$ are not compatible. This justifies the first problem: -Q1: Find all the maximal atlases on $\mathbb R$ whose restriction on $M-N$ is compatible with $\mathcal A|_{M-N}$. -Since not all functions on $\mathbb R$ which are smooth in one atlas are smooth in the other, the following problem arises: -Q2: Find all the maximal atlases in which a given function is smooth. - -REPLY [9 votes]: To me it looks like you're asking for the equivalence relation on the set of atlases on $M$, where one atlas is considered equivalent to the other if the identity map $Id : M \to M$ (where $Id(x)=x$ always) is a diffeomorphism between $M$ with one atlas and $M$ with the other. -In the case of manifolds that admit only one smooth structure, like $\mathbb R$, the group of homeomorphisms of $M$, $Homeo(M)$ acts transitively on the maximal atlases of $M$. This is basically just a restatement that there is only one smooth structure up to diffeomorphism. The point stabilizers of the action of $Homeo(M)$ on the maximal atlases on $M$ is $Diff(M)$ ($M$ having a particular smooth structure). So the object that classifies your maximal atlases up to the identity map being a diffeomorphism is just -$$Homeo(M) / Diff(M)$$ -the above is in the case that $M$ admits precisely one smooth structure up to diffeomorphism. In general, you'd have a disjoint union of these, one for every possible smooth structure on $M$. -$Homeo(M)/Diff(M)$ is always a big, infinite object provided $dim(M)\geq 1$. For example, the points of non-differentiability of a homeomorphism of $M$ is a well-defined invariant of $Homeo(M)/Diff(M)$ which is equipped with a smooth structure. I'm not aware of people that study this object specifically since it's not clear to me what one might want to do with it, but of course there's lots of related things. For example, germs of homeomorphisms are a related topic that has to do with smoothing theory. The classical reference of Kirby and Siebenmann covers some of this topic. The difference in the homotopy-types of $Homeo(M)$ and $Diff(M)$ is also a classical topic of study in manifold theory. For example, there's relatively little known about the homotopy-type of $Diff(S^n)$ when $n \geq 4$. When $n \geq 5$, $\pi_0 Diff(S^n)$ is isomorphic to the group of homotopy $(n+1)$-spheres. But there's not much known about higher homotopy.<|endoftext|> -TITLE: Good book on Riemann surfaces and Galois theory? -QUESTION [16 upvotes]: I'm supervising an undergraduate project on Galois theory and covering spaces. I want to have him read about the fact that from a branched cover of a Riemann surface you get an extension of its field of meromorphic functions, and the Galois groups are the same, but I'm having trouble finding a good book. Fulton's "Algebraic Topology" is OK but rushes through this point. Forster's "Lectures on Riemann Surfaces" looks good but I'd rather not make him learn sheaves. Any recommndations? - -REPLY [4 votes]: Try Askold Khovanskii: Galois Theory, Coverings and Riemann Surfaces.<|endoftext|> -TITLE: Methods for determining domains of influence -QUESTION [8 upvotes]: Given a hyperbolic PDE, the domain of influence of a spacetime point $x$, say $I_x$ though $x$ could be replaced by any set, can be defined in two ways. Lets call one of them geometric ($I_x^G$) and the other analytical ($I_x^A$). In Lorentzian geometry, the geometric domain of influence consists of the interior of the cone of null geodesics emanating from $x$ (let me not bother about whether to include the boundary in the definition of $I_x^G$ or not). In general, a similar definition can be given using characteristic cones instead of null cones. The analytical domain of influence can be defined as the set of all spacetime points $y$ such that for every neighborhood neighborhood $O$ of $x$ there exist two solutions $u_1$ and $u_2$ satisfying the condition $u_1(x')=u_2(x')$, for all $x'$ on a Cauchy surface passing through $x$ except for $x'\in O$, and also the condition $u_1(y)\ne u_2(y)$. The latter one is the definition used in Lax's book on Hyperbolic PDEs. -Similar defintions can be given for the geometric and analytical domain of dependence, say $D_S^G$ and $D_S^A$. Such a definition should capture the desired equality $D_K = I_{S\setminus K}$, for a Cauchy surface $S$ and $K\subset S$ (once again, being sloppy with boundaries). I know that energy methods can be used to establish that $D^G_K \subseteq D^A_K$ (the analytical domain of dependence is at least as large as the geometric one). Hence, by duality, the same methods establish $I_K^A \subseteq I_K^G$ (that the geometric domain of influence is at least as large as the analytical one). -My question is about the reverse inclusion, $I_K^G \subseteq I_K^A$ or by duality $D_K^A \subseteq D_K^G$. Maybe it's too much to ask for the analytical and geometric definitions to coincide. But when they do, what methods are used to establish that? When they don't what methods can identify the obstruction? Except briefly in Lax's book, I don't know what references discuss this problem explicitly, so those would also be appreciated! - -REPLY [3 votes]: I highly doubt the result you actually asked for is true. -Consider the linear wave equation on $(1+3)$ Minkowski space. The analytic domain of influence of a point $x$ as Lax defined it, which morally says that $y$ is in the analytic domain only if one can find perturbations in arbitrary small neighborhoods of $x$ that change $y$ (if I interpret your question statement correctly), actually consists of only the null cone emanating from $x$ and nothing more, since strong Huygen's principle holds. -The same is true for the linear wave equation on $(1+(2k+1))$ Minkowski spaces. -The opposite conclusion can be drawn on $(1+2k)$ dimensional Minkowski spaces, where the Green's function have support inside the cone. -For more general situations, you may want to consult the classical result of Atiyah-Bott-Garding on the existence of Petrowsky lacunae. For any linear hyperbolic equation that admits a lacuna, the analytic domain of influence cannot cover the entirety of the geometric one. -But for the result that you seem to actually want, where you should replace the analytic domain of dependence by a suitable "convex" envelope of it, I don't know if such a result is proven anywhere, but my guess is that, at least for the "local" version one can approach it using some sort of geometric optics construction. -For possible references (I haven't actually finished reading either, so they may not contain what you want), maybe you want to look at Michael Beals' book on propagation of singularities (sorry, the title escapes me at the moment) or Rauch's notes on Hyperbolic PDEs and Geometric Optics which I think you can find floating around on the internet.<|endoftext|> -TITLE: On successive regular cardinals with no ladders -QUESTION [5 upvotes]: Definition: Let $\kappa$ be an $\aleph$ cardinal, we say that $\langle f_\alpha\colon\alpha\to\kappa\mid\alpha<\kappa^+\rangle$ is a ladder if every $f_\alpha$ is injective. -Equivalently this is the range of a choice function from every injection of $\alpha$ into $|\alpha|$ (for $\alpha<\kappa$ we can always assume the identity is taken). Such ladder implies automatically that $\kappa^+$ is regular, since the union of $\kappa$ enumerated sets of size $\kappa$ is at most of size $\kappa$. (Recall that for well ordered sets $\kappa\times\kappa=\kappa$ even without the axiom of choice) -For example, then, if $\omega_1$ is singular then there is no such ladder of countable ordinals, since this would imply that $\aleph_1=\aleph_0^+$ is regular, therefore the existence of ladders for every successor ordinal is not provable in ZF alone. -However if we assume the existence of an inaccessible cardinal we can have the situation where $\aleph_1$ is indeed regular but there is no ladder avail. Indeed this is a necessary requirement since in such situation $\omega_1$ is inaccessible in $L$. -Both the forcing which makes $\aleph_1$ singular and the one which makes it regular without a ladder are essentially the same: collapse a limit cardinal to $\aleph_1$ and take symmetry model which ensures that no ladder exists, while setting $\aleph_1$ to have the cofinality of collapsed cardinal, i.e. singular or regular (if it was inaccessible). -Question I: Can we do this trick by replacing $\aleph_1$ by $\aleph_\alpha$ for any non-limit ordinal? So for example, $\aleph_5$ would be singular or regular without a ladder. -Question II: We do not need an inaccessible cardinal to have $\omega_1$ singular, nor $\omega_2$ singular. However if we want them both to be singular it already implies $0^\#$ exists, and requires Woodin cardinals. -Suppose $\aleph_1$ and $\aleph_2$ are both regular, and neither has a ladder. Can we do that "just" from the existence of two inaccessible cardinals, or would such phenomenon imply that some very large cardinals are playing in the background? - -REPLY [2 votes]: Both questions have a position answer. Which, in some sense, indicate that the non-existence of club sequences is somehow a weak property, relatively speaking. -For the first question this is quite trivial and really just requires a straightforward checking of the definitions. The point here being that if we start with a model of $\sf ZFC$ and symmetrically collapse a limit cardinal $\kappa$ to be $\aleph_{\alpha+1}$, then every set of ordinals is given by a bounded collapse, and in particular no ladder system can exist. - -As for the second question, If we have two inaccessible cardinals, $\kappa<\lambda$ and we force with the product of the obvious symmetric extensions, $\operatorname{Col}(\omega,<\kappa)\times\operatorname{Col}(\kappa,<\lambda)$. Because $\kappa<\lambda$, this can be seen also as a two-step iteration first collapsing $\lambda$ and then $\kappa$, and we can show that every subset of $\kappa$ was added by the collapse of $\kappa$, and every unbounded subset of $\lambda$ was added by the collapse of $\lambda$, so in either case, after the symmetries worked their magic, the above argument shows again that no ladders exist on $\aleph_1$ or $\aleph_2$. -(Note that if $\kappa$ is a singular cardinal the situation is different, because there is no reasonable way to change the successor of a singular cardinal without using large cardinals. So we may have $\lambda$ singular, but at least $\kappa$ needs to be regular.)<|endoftext|> -TITLE: Branches of the Fibonacci Word Tree -QUESTION [17 upvotes]: The Fibonacci word starts from $0$ subject to the rules $0 \mapsto 1, 1 \mapsto 01$ (or some variant thereof). The come from cutting sequences of the torus of a line of golden ratio slope. It is a 1D version of the Penrose Tiling. -...1010110101101101011010110110101101101011010110110101101... - -The Fibonacci word is has minimal complexity above a periodic word -- there are n+1 subwords of length n, making it a Sturmian word. - -5 subwords of length 4: "0101", "0110", "1010", "1011", "1101" -8 subwords of length 7: "0101101", "0110101", "0110110", "1010110", "1011010", "1011011", -"1101011", "1101101" - -As an experiment, I sorted the subwords of length n alphabetically The words are arranged in an infinite tree, where each word is descendant of its subword. -As a shorthand, I only placed the last letter of each word on each diagonal. Edges represent inclusion... drawn so any path from the top left corner "." appears in the Fibonacci word. Each Fibonacci subword corresponds to a path. -What is the structure of this tree? Is there any regularity to the location of the branches? -The complement of the tree becomes a tesselation of the Euclidean plane by "ribbon $\infty$-ominos", which is amusing. -.-1-1-0-1-1-0-1-0-1-1-0 -| | | -0 0-1-1 0-1-1-0-1-1-0 -| | | | -1-1 0 0-1-1-0-1 0-1 -| | | | | -0 0 1-1 0-1-1 0-1 -| | | | -1 1-1 0-1-1 0-1 -| | | | -1 0 0-1-1 0-1 -| | | | -0 1-1 0 1-1 -| | | -1-1 0 1-1 -| | | -0 0 1-1 -| | | -1 1 0 -| | -1 0 -| -0 - -REPLY [3 votes]: I was having difficulty seeing Nathaniel Shar's (interesting!) pattern. -This might be easier to interpret: -      - -(I thought it too intrusive to add this directly to his answer.)<|endoftext|> -TITLE: Bound the error in estimating a relative totient function -QUESTION [13 upvotes]: Let $n=p_1^{e_1}\cdots p_k^{e_k}$ be an integer with $k$ prime factors. We know that the number of integers less than $n$ and coprime to it is - $$\Phi(n)=n-\sum_i\frac n{p_i}+\sum_{i \lt j}\frac n{p_ip_j}-\cdots+(-1)^{k}\frac n{n}=nr$$ where $r=\prod(1-\frac 1{p_i})$. -For any positive integer $x$, the number of integers less than or equal to $x$ and relatively prime to $n$ is given exactly by the alternating sum $$\Phi(n,x)=x-\sum_i\lfloor{x/{p_i}}\rfloor+\sum_{i \lt j}\lfloor x/{p_ip_j}\rfloor-\cdots+(-1)^{k}\lfloor x/{n}\rfloor$$ - -How large and how small can the error $\Phi(n,x)-xr$ be? Can the absolute value of the error exceed $k$? - -Certainly the error can be no more (in absolute value) than $2^k$, probably it is easy to show that it could not be more than the middle binomial coefficient $\binom k{\lfloor k/2 \rfloor}$. I can see that it could get (almost) as large as $ k $ by imitating the following example: -The integers $1122659, 2245319, 4490639, 8981279, 17962559, 35925119, 71850239$ are a Cunningham chain as is $2,5,11,23,47$ in that each member is prime and one more than twice the previous one. If $n$ is the product of the $7$ primes from the first chain and $x=1122659\cdot64-1=71850175$ then all seven terms $ x/{p_i} $ are just a bit less than an integer so, without the rounding down to an integer, the estimate $rx$ will too small by about $7$ (the other terms are quite small). Of course it is not known for sure that there are arbitrary length chains. Maybe a similar idea could get an error of order $2k$ or $k^2.$ -later Thanks for the answers. I give one of my own below explaining that actually the best we could hope for is $2^{k-1}$ and then exhibiting a construction of D. H. Lehmer which attains $(1-2/q)2^{k-1}$ for arbitrary $q$. This is pretty much a result of the other answers, but I thought it was worth showing off the construction (which is not immediately clear from the article). - -REPLY [6 votes]: In fact the absolute value could not be greater than $2^{k-1}.$ Let $\{z\}=z-\lfloor z \rfloor$ denote the fractional part of $z$, then $$\Phi(n,x)-xr=0-\sum_i\{x/{p_i}\}+\sum_{i \lt j}\{ x/{p_ip_j}\}-\cdots+(-1)^{k}\{x/{n}\}$$ is a sum of $2^k$ terms, each between $0$ and $1$, half added and half subtracted. There is no reason to expect the first $k$ primes to give the optimal gap, and in fact they do not. For that matter, the question would be as interesting to me if the $p_i$ are simply relatively prime integers. The references given by Alan Haynes are apt and lead one to look at the article The distribution of totatives by D. H. Lehmer. Even there the key example is a bit hard to work out. If $q$ is any integer and $p_1\lt\dots\lt p_k$ are $k$ primes all of the form $p_i=c_iq-1$ then for $n=p_1p_2\cdots p_k$ there is a number $1 \lt x \lt n$ for which $|\Phi(n,x)-xr| \gt 2^{k-1} \frac{q-2}{q}.$ Let $C=\prod c_i$ It is esy to see that $$n=Cq^k-(\sum_iC/c_i)q^{k-1}+(\sum_{i\lt j}C/c_ic_j)q^{k-2}-\cdots+(-1)^k.$$ A similar expression holds for any product of several of the $p_i.$ The essentially unique extremal $x$ is $$\frac{n +(-1)^kp_1}{q}-1.$$ At this point I will simply illustrate with $k=4.$ Then -$$n=c_{1}c_{2}c_{3}c_{4}{q}^{4}- \left( c_{1}c_{2}c_{3}+c_{1}c_{2}c_{4}+c_{1}c_{3}c_{4}+c_{2}c_{3}c_{4} \right) {q}^{3} $$$$ + \left( c_{1}c_{2}+c_{1}c_{3}+c_{1}c_{4}+c_{2}c_{3}+c_{2}c_{4}+c_{3}c_{4} \right) {q}^{2}- \left( c_{1}+c_{2}+c_{3}+c_{4} \right) q+1$$ -and $$x=c_{1}c_{2}c_{3}c_{4}{q}^{3}- \left( c_{1}c_{2}c_{3}+c_{1}c_{2}c_{4}+c_{1}c_{3}c_{4}+c_{2}c_{3}c_{4} \right) {q}^{2} $$$$ + \left( c_{1}c_{2}+c_{1}c_{3}+c_{1}c_{4}+c_{2}c_{3}+c_{2}c_{4}+c_{3}c_{4} \right) {q}- \left( c_{2}+c_{3}+c_{4} \right)-1 $$ So $$x/p_1p_2p_3p_4-\lfloor x/p_1p_2p_3p_4\rfloor =x/n \approx 1/q$$ while, each term $$x/p_i-\lfloor x/p_i\rfloor=1-\frac{c_i-c_1+1}{c_iq-1} \approx 1-1/q $$ and each term $$x/(p_ip_j)-\lfloor x/(p_ip_j)\rfloor=\frac{c_ic_jq-(c_i+c_j)+c_1-1}{c_ic_jq^2-(c_i+c_j)q+1}\approx 1/q.$$ The terms of one sign are very about $1/q$ and those of the other are about $1-1/q.$ this accounts for the entire summation being close to $(1-2/q)2^{k-1}$ -The other extreme is at $n-1-x$.<|endoftext|> -TITLE: Weyl's law on asymptotic of Laplacian vs Hilbert's theorem on degree of a projective variety -QUESTION [14 upvotes]: Hi, -Let $M$ be a compact Riemannian manifold of dimension $n$. Define -the integer-valued function $N(k)$ to be the number of eigenvalues of the Laplacian on $M$ -which are less than or equal to $k$. Weyl's law states that the function $N(k)$ has asymptotic: -$$N(k) = C_n Vol(M) k^n + O(k^{n-1}),$$ -for some explicit constant C_n depending only on n. -Now suppose $M$ is a smooth complex projective subvariety of complex dimension d in ${\mathbb C}P^r$ equipped with the Fubini-Study metric from projective space. -Hilbert's theorem states that the Hilbert function $H(k)$ of $M$ has asymptotic -$$H(k) = (1/d!)Vol(M) k^d + O(k^{d-1}),$$ where $Vol(M)$ is the volume of $M$ with respect to -Fubini-Study metric which in turn is equal to degree of M as a subvariety of projective space. -Noting that $2d$ is real dimension of $M$, the two functions $H(k^2)$ and $N(k)$ have very similar asymptotic (given by dimension and volume). -Question: for smooth projective subvarieties of projective space with Fubini-Study metric, is there a connection between Hilbert function and number of eigenvalues of Laplacian? or the similarity in asymptotic is a coincidence? - -REPLY [13 votes]: There should be a relationship like this, because the heat function proof of Hirzebruch-Riemmann-Roch uses a much more refined equality between hilbert functions and Laplacian spectra. Let me see if I can put this together. -Let $M$ be a $d$-dimensional smooth projective variety. Let $L^k$ be the line bundle on $M$ obtained by restricting $\mathcal{O}(k)$ from projective space. Equip $M$ with the Fubini-Study metric. We have the complex of vector spaces: -$$0 \to C^{\infty}(L^k) \overset{\bar{\partial}}{\longrightarrow} C^{\infty}(L^k \otimes \Omega^{0,1}) \overset{\bar{\partial}}{\longrightarrow} \cdots \overset{\bar{\partial}}{\longrightarrow} C^{\infty}(L^k \otimes \Omega^{0,d}) \to 0. \quad (\dagger)$$ -Here $\Omega^{0,q}$ is the vector bundle of $(0,q)$-differential forms and $C^{\infty}(\mathrm{vector \ bundle})$ means $C^{\infty}$ sections of that vector bundle. We can use the two metrics on $L^k$ and on $X$ to define an inner products on $C^{\infty}(L^k \otimes \Omega^{0,q})$. Let $\bar{\partial}^{\ast}$ be the adjoint to $\bar{\partial}$ and let $\Delta_{L^k} = \bar{\partial} \bar{\partial}^{\ast} + \bar{\partial}^{\ast} \bar{\partial}$. This is not the ordinary Laplacian -- it acts on sections of $L^k$ tensored with differential forms rather than acting on differential forms. All of this is pretty standard : See Wells Differential Geometry on Complex Manifolds or Voisin's Complex Algebraic Geometry and Hodge Theory for background on this kind of construction. -Since $\bar{\partial}$ commutes with $\Delta_{L^k}$, the sequence $(\dagger)$ splits up into finite dimensional sequences for every eigenspace of $\Delta_{L^k}$. (There is some deep analysis with Sobolev spaces necessary to make this precise.) Let $(\dagger)_{\lambda}$ be the corresponding sequence of $\lambda$ eigensequences. Since $\Delta_{L^k}$ is positive semidefinite, you only get terms with $\lambda \geq 0$. -For $\lambda >0$, we have the homotopy $\mathrm{Id} = (1/\lambda) (\bar{\partial} \bar{\partial}^{\ast} + \bar{\partial}^{\ast} \bar{\partial})$, so $(\dagger)_{\lambda}$ is exact for $\lambda>0$. On $(\dagger)_0$, we have $\bar{\partial}=0$ so the sequence is trivial. For $k$ large, it turns out that only the first term has a nontrivial $0$-eigenspace. That eigenspace is the kernel of $\bar{\partial}$ acting on $C^{\infty}(L^k)$, which is to say, the holomorphic sections of $L^k$. So its dimension is the Hilbert function. More generally, for all $k$, the alternating sum of the dimensions of the $0$-eigenspaces is the Hilbert polynomial. -Gather up the eigenvalues into a generating function: -$$\theta(L^k, q, t) = \sum e^{-\lambda t} \dim{\LARGE (}\lambda\mathrm{-eigenspace \ of\ } C^{\infty}(L^k \otimes \Omega^{0,q}) {\LARGE )}.$$ -Then the above argument shows that -$$h(k) = \sum_q (-1)^q \theta(L^k, q, t). \quad (\S)$$ -Note that $(\S)$ holds for all $t$. -Now, in the height function proof, one engages in a detailed analysis of the asymptopics of $\theta(L^k, q, t)$ as $t \to 0$. When $t \to 0$, all of the $e^{-\lambda t}$ terms go to $1$, so $\theta$ blows up, and the rate at which it blows up depends on the growth rate of the eigenvalues of $\Delta_{L^k}$. One gets an asymptopic formula that looks like $\Theta(t) = a_{2d} t^{-d}+a_{2d-1} t^{-d+1/2} + \cdots + a_0 + O(t^{1/2})$. Plugging into $(\S)$ and comparing constant terms, one gets the following formula, which is Hirzebruch-Riemann-Roch slightly specialized to our setting: -$$h(k) = \int_M \frac{ e^{k \omega} \prod \alpha_i}{\prod (1-e^{- \alpha_i})}.$$ -Here $\omega$ is the Fubini-Study form, $\alpha_i$ are the Chern roots of $T^{\ast} M$ and we are using the convention that you ignore all terms not in top degree when integrating. You are interested only in the leading power of $k$. We have $e^{k \omega} = 1+k \omega + k \omega^2/2+ \cdots + k^d \omega^d/d!$, where the sum stops because we have reached top degree, so the leading term is $\frac{k^d}{d!} \int_M \omega^d = \frac{k^d}{d!} \mathrm{Vol}(M)$, which is your left hand side. -Is there a way to get the large $k$ behavior of the $\theta$'s without getting the exact result? I don't know. Also, you would want to switch from $\Delta_{L^k}$ to $\Delta$ at some point. -I haven't seen anyone work this out, but I do remember there is a lemma which relates $\Delta$ and $\Delta_{L^k}$ for $k$ large and that Griffiths and Harris use it in their proof of Serre vanishing. I covered this proof in my Hodge theory course, so I am a little embarrassed not to remember the details! It should be in the notes for April 12. Anyway, possibly you can figure out how to finish the proof from here.<|endoftext|> -TITLE: Are Gaussian Processes more important than other stochastic processes? -QUESTION [6 upvotes]: I am doing a course at university and it deals with Gaussian Processes mainly. We use them for fitting data and prediction, machine learning, regression, classification. Is there any particular reason why Gaussian Processes stand out from the millions other random processes you could have? -I suppose that's a bit like asking does the Normal distribution stand out from the million other distributions you can have but...There is a website http://www.gaussianprocess.org/ for the ones interested. - -REPLY [6 votes]: A well-known probabilist once told me that the 3 main classes of stochastic processes are Gaussian, Markov, and martingales. -Martingales are definitely useful in finance and also with respect to other betting games, but it seems to me that they are essential mostly because we have such powerful tools to study them. These tools have often proved useful in much broader contexts, for example the martingale problem/diffusions-- but in my own humble opinion, Gaussian and Markov processes are more natural. -As for Markov processes- the Markov property is clearly very natural. But again IMHO, these processes are most useful in modeling short time behavior, before central limit behavior has kicked in. Although, again stationary distributions and long-time behavior of Markov processes absent Gaussianity are definitely well worth studying (and of course the fruitful study of the mixing of Markov chains connects short term to long term behavior) -Gaussian processes are mostly used to study correlations and dependence, and one might argue that the notions of independence and dependence are what really make probability different from other fields (every remark has its caveat, and the connection between correlations and positive definite functions via Bochner's thm allows analysts to get in this game). Since anything with finite variance converges in a sense to a Gaussian, the idea is that Gaussian processes allow us to study how dependence behaves in the long run (for infinite variance, one simply replaces Gaussian with stable).<|endoftext|> -TITLE: path of almost complex structure in the definition of heegaard floer homology -QUESTION [5 upvotes]: In order to define Heegaard Floer Homology for a connected, closed, oriented 3 manifold, we fix a generic path of nearly symmetric almost complex strucutre $J_s$ over $Sym^g(\Sigma)$. By transversality theorem we can say the moduli spaces $M_{J_s}$ are all smooth. Under certain cases we can choose the path to be constant path $Sym^g(j)$ and still make the moduli space smooth. -Rather than a path, can we choose only one nearly symmetric almost complec structure $J$ over $Sym^g(\Sigma)$ other than $Sym^g(j)$ to define the heegaard floer homology? -If we can not, it is because of we can not make the moduli spaces smooth or apply the Gromov compactness theorem? -Thanks in advance. - -REPLY [4 votes]: According to Proposition 3.9 of Ozsvath and Szabo's original paper there are indeed topological conditions one can put on the homotopy class of discs to ensure that one can take a single almost complex structure. Moreover, there's an explicit complex structure (coming from a symmetric product of a complex structure on the Heegaard surface) for which you can achieve transversality. -Edit: On a second (more careful) reading of your question I realise you already knew this. Apologies. More pertinently to your question, if you have a regular almost complex structure then (provided you're only interested in finitely many homotopy classes at a time) you should be able to perturb $J$ slightly and it will remain regular (for each homotopy class regularity is an open condition). Certainly as long as it's tame you'll never run into problems with Gromov compactness. -Disclaimer: I know very little about Heegaard-Floer so maybe someone more specialised can say something more helpful and direct. Instead let me say something more general about transversality for holomorphic discs. -In general in Floer theory you need domain-dependent almost complex structures to achieve transversality for holomorphic discs/spheres. -The problem is that when proving transversality you make perturbations to the almost complex structure and if the disc is multiply-covered in some region then you may end up having to make different perturbations at the same point in the ambient manifold (to which different points of the disc are mapped). Open Riemann surfaces are particularly bad in this way because different regions can have different covering multiplicities (just think of something like Milnor's doodle). -By contrast, a closed holomorphic curve has an underlying simple curve for which one can prove transversality. Of course, for closed spheres in Calabi-Yau 3-folds, the branched covers of a simple, regular sphere are themselves not transverse (they necessarily occur in high dimensional families by varying the branch-points) and you need domain-dependent almost complex structures (or abstract perturbations) to cope with this (e.g. to prove the Aspinwall-Morrison formula) -For discs there is also a theorem of Lazzarini (probably in one of these papers) which lets you decompose a disc into subdiscs which are multiple covers of simple discs and this is how people using monotone Floer theory usually cope with the problem. Of course if you have Maslov 0 discs you run into the same problems as for the Aspinwall-Morrizon formula, hence the monotonicity requirement.<|endoftext|> -TITLE: Hodge numbers in a family -QUESTION [15 upvotes]: Let $X \to Y$ be a smooth projective morphism of smooth varieties over $\mathbb{C}$. Then the fibers $X_y, y \in Y$ have locally constant Hodge numbers $H^q(X_y, \Omega^p_{X_y})$. Namely, one can argue that the Hodge numbers are upper semicontinuous because they represent the kernels of the $\overline{\partial}$-Laplacian, and when you wiggle an operator, the kernel only jumps. However, their sum gives the sum of the Betti numbers of the fibers (since each $X_y$ is smooth and projective), which is locally constant by Ehresmann's fibration theorem. -In what generality is this claim about Hodge numbers true? For instance, will upper semicontinuity still be true (I would like to apply the semicontinuity theorem but don't know how the $\Omega^p_{X_y}$ fit into a flat family) if the morphism is, say, only proper? Is any of this true in characteristic $p$ where we don't have Hodge theory? - -REPLY [16 votes]: Without flatness you have little chance for this to even get off the ground: Let $f:X\to Y$ be the blow up of a smooth (closed) point of $Y$. Then all fibers except one consist of a single point, while the special fiber is a $\mathbb P^n$. That will have non-zero Hodge numbers that the others can't even dream about. -So, assume $f$ is flat, then if $X$ is smooth, then $\Omega_X$ is still locally free and so it is flat over $Y$ and hence you get that $\dim H^q(X_y, (\Omega^p_X)_y)$ is semi-continuous. This works in any characteristic and it does not have anything to do with Hodge theory. The shortcoming of this is that you're actually not getting $\dim H^q(X_y, \Omega^p_{X_y})$ even in the smooth case, because for that you would need $\Omega_{X/Y}$, but that's not flat and in some sense not the right object to consider. -If $Y$ is also smooth, then $\Omega_Y$ is locally free and if $f$ is dominant, then you still have the short exact sequence -$$ -0\to f^*\Omega_Y\to \Omega_X\to \Omega_{X/Y}\to 0 -$$ -The trouble is that the sheaf on the right is not locally free, so when you take exterior products, then it gets kind of tricky. One possibility is to construct complexes that behave very similarly to $\Omega_{X/Y}^p$ with respect to $\Omega_X^p$ and $\Omega_Y^p$. This is done in this paper. The primary goal of the paper is not what you want and I am not sure that you actually get semi-continuity, but you can at least check out the construction. -A general version of that construction is in this paper. (Sorry, neither of them are on arXiv). -Also, for singular fibers $\Omega_{X_y}$ is not the "right" thing to look at. -One can look at the objects that come from the Deligne-Du Bois complex and do Hodge theory for singular varieties. Then the restriction becomes a little tricky, because those are objects in a derived category and restriction is not exact, so you need to do something else. Completion along the fiber gives the right thing, but I don't know if there is a semi-continuity theorem using completion instead of restriction. That might be an interesting question to contemplate. This is related to the paper I linked above. See the references in that for more details.<|endoftext|> -TITLE: Lie algebras with abelian Cartan subalgebras -QUESTION [12 upvotes]: The Cartan subalgebras of a reductive Lie algebra are abelian. - -Are there non-reductive Lie algebras with abelian Cartan subalgebras? - -In fact, the elements of a Cartan subalgebra of a reductive Lie algebra are semisimple, so a weaker question is: - -Are there non-reductive Lie algebras with abelian Cartan subalgebras all of whose elements are semisimple? - -N.B.: I asked this earlier at math.SE. - -REPLY [2 votes]: If $H$ is a Cartan subalgebra of a Lie algebra $L$, and $A$ is an associative commutative algebra, then $H \otimes A$ is a Cartan subalgebra of $L \otimes A$. Specializing this to the case $L$ classical simple, we get another positive answer to the first question (but not to the second one). This example can be, probably, varied, by taking subalgebras of $L \otimes A$, or adding "tails" of derivations, etc.<|endoftext|> -TITLE: Linearly independent vector fields -QUESTION [6 upvotes]: Let $X_1,\dots,X_n$ be complete vector fields on $\mathbb R^n$ and suppose that $(X_1(p),\dots,X_n(p))$ is a basis for all $p \in \mathbb R^n$. -Question: Is it possible to choose a cube $C$ around the origin of $\mathbb R^n$ such that there is for every $p \in C$ a piecewise smooth curve $\alpha \subset C$ which connects $p$ with $0$ where the smooth parts of the curve are given by the flows of the vector fields $\pm X_1,\dots,\pm X_n$? (With other words: is it possible to travel from $0$ to $p$ following only the integral curves of the given vector fields in a bounded domain?) -For $n=2$ this is pretty clear; w.l.o.g. $X_1=\partial/\partial x_1$ and the image of $x_1=0$ under the flow $X_2$ fills all of $\mathbb R^2$, since the flow lines of $X_2$ intersect $x_1=0$ transversally. Now it is easy to find a $C$ and $\alpha$ for a $p \in C$. But I can not generalize this for $n$ arbitrary. -Edit: Instead of demanding $C$ to be a cube, one could also ask if there is a open neighbourhood of the origin with the desired properties. - -REPLY [6 votes]: For a suitable nbd of the origin $U$, yes, even following the $n$ flows in a prescribed order. Assuming $X_1,\dots,X_n$ locally Lipschitz continuous, the corresponding flows $\phi_1(t,x),\dots, \phi_n(t,x)$ are $C^1$ maps (in the pair) so the map $\Phi$ -$$(t _ 1,\dots,t_ n)\mapsto \phi _ n (t _ n, \phi_{n-1}( t _ {n-1}, \dots \phi _1 (t_1, 0))\dots)$$ -is of class $C^1$ on a nbd of $0\in\mathbb{R}^n$. Since for $(t,x)\to 0$ in $ \mathbb{R} \times\mathbb{R}^n$, -$$ \phi_j(t,x)= x + tX_j(0) + o\big((t,x) \big) $$ -it follows that the differential of $\Phi$ at $0\in\mathbb{R}^n$ is the linear map -$$(t _ 1,\dots,t_ n)\mapsto t_1X_1(0)+\dots t_n X_n(0)$$ -which is invertible because by assumption $X_1(0),\dots, X_n(0)$ is a base. By the implicit function theorem, $\Phi$ is a local diffeo between a cube $Q$ and a nbd $U$ of $0$. The images of the broken lines in $Q$ with edges parallel to the axes is what you want and reach any point of $U$. -(edit) Also, given any connected open subset $\Omega\subset \mathbb{R}^n$, any two points $p$ and $q$ of $\Omega$ can be joined by a continuous curve made by finitely many flow-lines of the fields $X_1,\dots,X_n$, just because the set of points that may be reached this way starting from $p$ is a non-empty open and closed subset of $\Omega$, thanks to the preceding local fact. -However, there could be problems to reach some point if $\Omega$ is not open, e.g a closed cube. Take e.g. in $n=2$ and two constant fields $X_1:=(2,1)$, $X_2:=(1,2)$. Then, the vertex $(r,r)$ of the square $Q:=[−r,r]\times[−r,r]$ is easily joined to the origin by a two-edges broken flow-line, but the vertex $(r,−r)$ can't be reached by any such broken line in $Q,$ in any finite number of steps.<|endoftext|> -TITLE: Continuous and smooth Lie groupoid cohomology -QUESTION [6 upvotes]: In the paper by Weinstein and Xu: Extensions of symplectic groupoids and quantization, J. Reine Angew. Math. 417 (1991), there are two versions of Lie groupoid cohomology. The same differential $\delta$ can be in fact applied to globally smooth cochains or only to cochains which are globally continous and smooth in a neighbourhood of the diagonal. -If $\Gamma$ denotes the total space of the groupoid and $\Gamma^{(n)}$ the set of composable $n$-tuples then one can consider: -$C^n(\Gamma:\mathbb R)={\cal C}^\infty(\Gamma^{(n)},\mathbb R)$, -$C^n_{es}(\Gamma:\mathbb R)=\{\sigma:\Gamma^{(n)}\to\mathbb R, \sigma\quad \mathrm{smooth}\quad \mathrm{around}\quad \Delta^{(n)}\}$ -and of course one could consider just continuous cochains -$C^n_0(\Gamma:\mathbb R)={\cal C}(\Gamma^{(n)},\mathbb R)$ -In the same paper it is proven that the first two cohomologies are different by giving an example in which globally smooth 2-cohomology is 0 while $H^2_{es}(\Gamma;\mathbb R)$ is non zero (in fact in the example coefficients are in $\mathbb S^1$ but this should make no big difference). -Does anyone knows: - -If also continuous cohomology differs from the two previous ones -Other examples in which the cohomologies are different and/or equal -General conditions under which the cohomologies are known to be equal. - -REPLY [7 votes]: Sorry for answering so late, I just was pointed to this question. Before answering your questions in the case of a Lie group (i.e., one object) let me point out that the difference between $\mathbb{R}$-coefficients and $\mathbb{S}^1$-coefficients it not negligible but is somehow the whole heart of the story. - -The cohomology of the continuous cochains is equal to the cohomology of the smooth cochains. Moreover, there is yet another complex $C^n_{ec}$ of cochains that are continuous around $\Delta^{(n)}$. Then $H^n_{ec}$ equals $H^n_{es}$ (all this for coefficients in $\mathbb{R}$, $\mathbb{S}^1$ or more general tori). -The universal cover $\mathbb{Z}\to\mathbb{R}\to\mathbb{S}^1$ is described by a cocycle in $Z^2_{es}$ that cannot be smooth (otherwise $ \mathbb{R}\to\mathbb{S}^1$ would have a smooth global section), see also 3. -$H^n_{es}=H^n$ (in your notation form above) if either the Lie group $G$ or the coefficients are contractible (in particular the difference between $\mathbb{R}$ -and $\mathbb{S}^1$ plays a significant role). - -A reference for the equality of smooth and continuous cohomology is Hochschild, Mostow, "Cohomology of Lie groups" and for the more general case this paper, together with F. Wagemann. -I expect that the case of general Lie groupoids can be treated similarly, for instance I would expect that $H^n=H^n_{es}$ if the Lie groupoid $\Gamma$ is topologically trivial (i.e., its simplicial manifold $N\Gamma$ is contractible). However, the generalisation of our approach is not entirely straight forward.<|endoftext|> -TITLE: Module categories over symmetric/braided monoidal categories -QUESTION [6 upvotes]: Given an algebraically closed field $k$ and a finitely generated commutative $k$-algebra $A$, all simple modules over $A$ are 1-dimensional - -What is the analogous statement for symmetric monoidal $k$-linear categories? -What is the analogous statement for braided monoidal $k$-linear categories? - -We can assume the category is Abelian and the product functor is right exact in both variables -I expect something of the sort "any simple (in some sense) module category is equivalent to $Vect$", although I have no idea how braided and symmetric are different in this repsect - -REPLY [8 votes]: Consider C = Rep(G) for a finite group G. This is a symmetric tensor category. Any subgroup $H \subset G$ yields a module category Rep(H) with the action given by restricting and then tensoring. This module category is simple in any appropriate sense, and in no interesting sense is it "1-dimensional."<|endoftext|> -TITLE: When is a finite cw-complex a compact topological manifold? -QUESTION [17 upvotes]: I think the statement of the question is pretty straightforward. Given a finite $n$-dimensional CW complex, are there necessary and sufficient conditions for determining that it is also a compact $n$-dimensional topological manifold (possibly with boundary)? Most of the questions I have found are about the converse, "When is a topological manifold a CW-complex?" so I thought it would be useful to consider the other side of the picture. -First of all, we already have 2nd countable and Hausdorff so we only need to determine when it is locally Euclidean. One necessary condition is that every point must be contained in the closure of at least one $n$-cell. One the other hand, a point can't be in too many $n$-cells, as the wedge of spheres is not a manifold. But these criteria are clearly not sufficient (or very precise: what does "too many" mean exactly?). -Another necessary condition that I think of right away is that the space must satisfy Poincare duality (with $\mathbb{Z}/2$ coefficients). I'm not sure if this is practical at all, but maybe it is useful if you are working with an explicit cell-structure and a concrete description of the cellular chain complex. -After that, I don't know how to proceed. I am assuming this is a difficult problem, since Google searches haven't answered my question yet. One idea that I have is to try and come up with an "obstruction" to this, where cell-by-cell we determine if every point in that cell admits an $n$-dimensional Euclidean nhd, and we have that the CW-complex is a manifold iff this "obstruction," computed at the homology/cohomology level, vanishes. -Any suggestions (or references to a solution) will be appreciated. - -REPLY [12 votes]: I think an answer to a question close to yours can be found in [1] and in the survey in the first section of [2]. -There some necessary and sufficient conditions are given for a simplicial complex to be a topological manifold. In particular you can look at Theorem 3.5 of [1] or Theorem 1.12 of [2]. (They are basically the if and only if version of Adam's answer for simplicial complexes) -Using the terminology of [2] we can state: - -Definition: An $n$-dimensional homology manifold is a pure simplicial complex $X$ of dimension $n$ in which, for every face $\sigma$, the complex $\mbox{Link}(\sigma)$ has the same homology of a $(\mbox{dim} X − \mbox{dim}\sigma − 1)$ -sphere. -Theorem: An $n$-dimensional homology manifold $X$ is a topological $n$-manifold iff for every vertex $v$ of $X$ $\mbox{Link}(v)$ is simply connected. - -[1] J.W.Cannon, The recognition problem: What is a topological manifold?,Bull. Amer. Math. Soc. 84 (1978), 832-866 -[2] B. Benedetti, Smoothing discrete Morse theory, Ann. Sc. Norm. Super. Pisa Cl. Sci. (5), Vol. XVI (2016), 335-368. preprint here<|endoftext|> -TITLE: flatness condition for local noetherian ring without nilpotent elements -QUESTION [9 upvotes]: In SGA 1, chapter 4, there is a corollary describing a flatness condition for a module of finite type over a local noetherian integral ring. The corollary is number 4.4, and states: -Suppose that $A$ is a local noetherian integral ring with maximal ideal $I$ and residue field $k = A/I$ and field of fractions $K$. Let $M$ be a module of finite type over $A$. Then saying that $M$ is flat (note: SGA actually cites here a previous proposition with equivalent conditions of flatness which in this case is satisfied) is equivalent to saying that $M\otimes_A K$ and $M\otimes_A k$ are vector spaces of the same dimension. -There is a remark immediately following this corollary saying the reader is left to generalize this to the case where $A$ is only assumed to be a ring without nilpotent elements. How does one show this more general case? - -REPLY [4 votes]: For a generalization to arbitrary noetherian reduced rings, see EGA III, 7.6.9. Applied to a projective resolution $P_*\to M$ for a finitely generated $A$-module $M$, it says in particular that $d(x)\colon x\mapsto\dim_{k(x)}(M\otimes k(x))$ is semi-continuous on $\mathrm{Spec}(A)$, and $M$ is flat iff $d$ is locally constant. The statement in SGA is a special case since semi-continuity together with $\dim_K(M\otimes K)=\dim_k(M\otimes k)$ already implies that $d$ is constant. A more elementary statement can be found in Mumford's treatment of semi-continuity in his Abelian Varieties (Lemma 1 in section II.5).<|endoftext|> -TITLE: Puiseux series expansion for space curves? -QUESTION [6 upvotes]: This result is apparently well known and used by many people. -I am, however, quite frustrated that I cannot seem to find a proof that I can understand. -For plane algebraic curves, this is not too hard. -For an irreducible polynomial $F(x,y) \in \mathbb{C}[x,y]$ nonconstant in both $x$ and $y$ -with $F(0,0) = 0$, -one can formally expand $y$ as a Puiseux series in $x$ -$$ -y(x) = \sum_{i=0}^\infty y_i x^{\alpha_i} -$$ -The Puiseux theorem actually states that such a series converges (in some sense) near $x=0$. -Alternatively, we can construct Riemann surface over the point $(0,0)\in \mathbb{C}^2$ of the algebraic function $y(x)$ which has the normal representation as holomorphic element -$$ -\begin{eqnarray} -x &=& t^m \cr -y &=& \sum_{i=1}^\infty y_i t^i -\end{eqnarray} -$$ -With some abuse of notation, I think we can even say the two are the same. -I can find many sources, among which I like Walker's representation the best. -Now how about space algebraic curves (of one complex dimension) in $\mathbb{C}^n$? -For a polynomial system -$F(x_0,\ldots,x_n) = (f_1(x_0,\ldots,x_n),\ldots,f_n(x_0,\ldots,x_n))$ -where each $f_i \in \mathbb{C}[x_0,\ldots,x_n]$ -with $F(0,\ldots,0) = (0,\ldots,0)$, -and lies only on a one (complex) dimensional irreducible component of $V(F)$, -if we fix a place of the this algebraic curve centred at $(0,\ldots,0)$, -we should be able to find a parametrization -$$ -\begin{eqnarray} -x_0 &=& t^m \cr -x_k &=& \sum_{i=1}^\infty c_{k,i} t^i -\end{eqnarray} -$$ -with convergent power series. -Or equivalently, we could express $x_1,\ldots,x_n$ as Puiseux series in $x_0$ -that converge in certain sense. -I could only find "proofs" that reference Hironaka's resolution of singularity, -which I don't think I can understand any time soon. -I'm hoping to find a proof using only complex geometry or basic complex algebraic geometry. -In particular, I was thinking maybe I can repeatedly apply -Weierstrass preparation theorem together with Puiseux theorem, -however, I'm not quite sure how to continue after the first step. - -REPLY [2 votes]: If $(C,0)$ is a germ of complex curve in $\mathbb{C}^n$, then you can find coordinates $(z_1, z')$ and a polydisc $V=V_1\times V'$ centered in $0$ such that the canonical projection $V\ni (z_1,z')\mapsto \pi(z_1,z')=z_1\in V_1$ is a ramified covering when restricted to $C$ with $p$ sheets. Let $S$ be the ramification locus, then $S$ is an analytic set in $V_1$, that is, a discrete set of points. Therefore, upon taking a smaller $V_1$, we can think that $S=\{0\}$ or it is empty. -Let us suppose $S$ is not empty, otherwise the projection is a local biholomorphism with the unit disc and this gives the thesis with $z_1=t$, $z_k=f_k(t)$, $f_k\in\mathcal{O}(D -)$. -Now, $C^*=C\cap \pi^{-1}(V_1\setminus S)$ is a $p-$sheeted covering of $V_1\setminus S=D\setminus\{0\}$, therefore it is isomorphic to the standard $p-$sheeted covering: -$$D\setminus\{0\}\ni t\mapsto t^p\in D\setminus\{0\}.$$ -This means that there exists a map $g:D\setminus\{0\}\to C^*$ such that $\pi\circ g (t)=t^p$; this map extends clearly to a holomorphic bijection by setting $g(0)=0$ (it is continuous in $0$ and holomorphic outside). -Therefore, we have -$$z_1=t^p$$ -$$z_k=g_k(t)=\sum_{j=0}^\infty a_{kj}t^j.$$ -The existence of such a system of coordinates is a standard result in local complex geometry, sometimes called local parametrization theorem.<|endoftext|> -TITLE: Giant Rat of Sumatra singularity -QUESTION [16 upvotes]: I would be grateful for explanations of the issues raised in any -of these three questions, or pointers to the relevant literature -(now updated with answers): - - How did a particular singularity come to be known as The Giant Rat of Sumatra? - -Answer: Named by Bruce & Giblin after a Sherlock Holmes reference, -as explained by Michael Biro and Daniel Moskovitch. - - What is the generic polynomial form of this singularity? - -Answer: $f_a(x,y)=x y(x-y)(x-a y)$ for a parameter $a$. This from -Daniel Moskovitch's answer -to the MO question, -"What are some examples of colorful language in serious mathematics papers?" -Here is a plot for $a=\frac{1}{4}$. Despite its formidable name, the singularity -appears rather tame to the eye: -               - - - Is there some natural equivalence relation that classifies all the giant-rat singularities -into the same class (unlike Arnold's $\cal{A}$-equivalence, which in my [limited] understanding, -does not). - -Answer: On p.199, Bruce & Giblin say that the polynomial above "gives uncountably -many inequivalent types" (for different parameters $a$). They address my question of another -equivlance relation: "topological equivalence is too weak to provide a workable -theory... Instead one has to work with 'universal stratified equivalence'," a theory due -to Eduard Looijenga. They go on to say, "Even for the giant rat of Sumatra one has -little idea what these models are. The world is indeed not yet prepared for its story." :-) - -I ask this as someone largely ignorant of singularity theory. Thanks for enlightening me (and other MO participants)! - -REPLY [10 votes]: I'd just like to add something about the plot of the graph of the function $z=\mathrm{f}(x,y)$. -The term "singularity", in this context, does not refer to a function whose graph is singular. -The Giant Rat is a function germ $\mathrm{f} : (\mathbb{R}^2,0) \to (\mathbb{R},0)$. We're interested in the set of $(x,y) \in \mathbb{R}^2$, very close to $(0,0) \in \mathbb{R}^2$ which are sent to $0 \in \mathbb{R}$. In other words, what does -$$\mathrm{f}^{-1}(0) = \{(x,y) \in \mathbb{R}^2 : \mathrm{f}(x,y)=0\}$$ -look like in a small neighbourhood of the origin? -If $a \neq 0,1$ then $xy(x-y)(x-ay)=0$ are four lines all crossing at $(0,0)$. This is quite a nasty singularity. Consider, for a moment, the $D_4$ singularity $x^3-xy^2$. The zero-level set of this is three distinct real lines through the origin. Any function germ $\mathrm{g}:(\mathbb{R}^2,0) \to (\mathbb{R},0)$, whose degree three Taylor series is a cubic polynomial in $x$ and $y$ whose zero-level set is three distinct real lines through the origin is $\mathscr{R}$-equivalent to $x^3-xy^2$, i.e. there is a diffeomorphism $\mathrm{\phi} : (\mathbb{R}^2,0) \to (\mathbb{R}^2,0)$ for which -$$\mathrm{g} \circ \phi = x^3 - xy^2$$ -This isn't totally unexpected because any three concurrent lines can be taken to any other three concurrent lines via a linear transformation, meaning that any cubic polynomial whose zero-level set is three distinct real lines can be taken to $x^3 - xy^2$ by a linear transformation. -Going back to the Giant Rat, with $a \neq 0,1$, we have four distinct real lines through the origin. Four concurrent lines can be taken to four other concurrent lines by a projective transformation if, and only if, they have the same cross ratio. The Giant Rat is so nasty because there are an uncountable number of singularity types in the family $xy(x-y)(x-ay)$ which are not $\mathscr{R}$-equivalent. The zero-level sets $xy(x-y)(x-ay)=0$ are almost all different from each other in the sense that no diffeomorphism can take $xy(x-y)(x-ay)=0$ to $xy(x-y)(x-by)=0$, where $a \neq b$. The cross ratios must be equal for there to be any hope of them being $\mathscr{R}$-equivalent. Equal cross ratios is necessary but not sufficient.<|endoftext|> -TITLE: An easy way to to explain the equivalence definitions of tangent spaces? -QUESTION [14 upvotes]: In a short talk, I had to explain, to an audience with little knowledge in geometry or algebra, the three different ways one can define the tangent space $T_x M$ of a smooth manifold $M$ at a point $x \in M$ and more generally the tangent bundle $T M$: - -Using equivalent classes of smooth curves through $x$ -Using derivations near $x$ -Using cotangent vectors at $x$ - -Just by looking at the definition, it is not at all clear why they should all define the same object. I went through the proof, but judging from their reaction, it was not very meaningful. I wonder if there is any way I can let them "see", with just intuition, that the three definitions are, in certain sense, the same. - -REPLY [15 votes]: What all three definitions have in common is that they each try to capture the first order behavior of a smooth function on $M$. - -The derivative of a smooth function $f$ along a curve $\gamma$ with $\gamma(0) = p$ depends on $\gamma$ only insofar as it depends on $\gamma'(0)$, and indeed it recovers the directional derivative of $f$ at $p$ in the direction $\gamma'(0)$. The directional derivatives of $f$ determine the total derivative of $f$ which in turn determines the first order behavior of $f$ (more or less by the definition of the total derivative). -Since a derivation $D$ at $p$ sees only the values of a function $f$ and its derivatives at $p$ (not near $p$), we can replace $f$ by a polynomial by Taylor's theorem. By the Leibniz rule, $D(P)$ depends only on the linear part of a polynomial $P$ and hence $D(f)$ depends only on the first order part of $f$. -Recall that the cotangent bundle of $M$ at $p$ is the space $I/I^2$ where $I$ is the ideal in $C^\infty(M)$ consisting of functions $f$ such that $f(p) = 0$. If we imagine replacing $C^\infty(M)$ by a polynomial ring then $I$ represents the ideal of polynomials whose lowest order part has degree $1$ and $I^2$ is the ideal of polynomials whose lowest order part has degree $2$. In this case $I/I^2$ is naturally identified with the space of linear polynomials. Thus the cotangent bundle at $p$ is in a sense the space of "first order parts" of smooth functions on $M$. - -This intuition acutally allows us to be a little more explicit about how the relevant identifications are made. -It's very easy to go from 1 to 2: if $\gamma$ is a curve in $M$ with $\gamma(0) = p$ then $D(f) = (f \circ \gamma)'(0)$ is a point derivation at $p$ which depends only on the equivalence class of $\gamma$ in $T_p M$. -To go from 2 back to 1, let $f$ be a smooth function and let $f(x) \sim \sum_\alpha c_\alpha x^\alpha$ (multi-index notation) be its Taylor series in a coordinate system centered at $p$. Then for any derivation $D$ at $p$ we have $D(f) = c_1 D(x_1) + \ldots + c_n D(x_n)$ by the Leibniz rule, so $D$ corresponds to the tangent vector $(D(x_1), \ldots, D(x_n))$. -To go from 2 to 3, let $D$ be a derivation at $p$ and observe that $D(f) = 0$ for any $f \in I/I^2$ by Taylor's theorem and the Leibniz rule. Thus $D$ determines a linear functional in $(I/I^2)^*$. -Finally, to go from 3 back to 2, let $\ell \in (I/I^2)^*$ and define a point derivation by $D(f) = \ell(f - f(p) + I^2)$.<|endoftext|> -TITLE: Does a manifold which bounds always admit a free involution? -QUESTION [5 upvotes]: If a closed smooth manifold $M$ admits a smooth free involution $T$, then it bounds. In fact, the mapping cylinder of the quotient map $M \to M/T$ is the manifold whose boundary is $M$. -Is the converse true? If not, then could someone give an example of a closed smooth manifold which bounds but does not admit any free involution. - -REPLY [6 votes]: Consider $M = \mathbb{CP}^2\#\mathbb{CP}^2$. All of its Stiefel-Whitney numbers vanish, so $M$ is unorientedly nullcobordant; more generally, $X\# X$ always bounds. -We have $H^2(M; \mathbb{Z}) \cong H^2(\mathbb{CP}^2; \mathbb{Z})\oplus H^2(\mathbb{CP}^2; \mathbb{Z}) \cong \mathbb{Z}a\oplus \mathbb{Z}b$, and $a^2 = b^2$ is a generator for $H^4(M; \mathbb{Z})$ while $ab = 0$. Now let $f : M \to M$ be a continuous map, and suppose $f^*a = ma + nb$, then -$$f^*(a^2) = (f^*a)^2 = (ma + nb)^2 = m^2a^2 + n^2b^2 = (m^2 + n^2)a^2$$ so $\deg f \geq 0$. In particular, if $f : M \to M$ is a homeomorphism, it must be orientation preserving. -Suppose $f : M \to M$ is a free involution. Then $N := M/\mathbb{Z}_2$ is a compact connected four-manifold which is orientable because $f$ preserves orientation, so $b_0(N) = b_4(N) = 1$. Furthermore, $N$ has fundamental group $\pi_1(N) = \mathbb{Z}_2$, so $b_1(N) = 0$ and hence $b_3(N) = 0$ by Poincaré duality. As the Euler characteristic is multiplicative under coverings, we see that $\chi(N) = \frac{1}{2}\chi(M) = \frac{1}{2}(4) = 2$ so $b_2(N) = 0$. On the other hand, the signature is also multiplicative under coverings, so $\sigma(N) = \frac{1}{2}\sigma(M) = \frac{1}{2}(2) = 1$; this is impossible as $b_2(N) = 0$. Therefore $M$ does not admit a free involution. -More generally, this argument can be used to show that $k\mathbb{CP}^2$ does not admit a fixed-point free homeomorphism of finite order. Therefore $k\mathbb{CP}^2$ is not the covering space of any manifold other than itself.<|endoftext|> -TITLE: A quadratic form represents all primes except for the primes 2 and 11. -QUESTION [17 upvotes]: By computer calculations, I found the following conjecture that the quadratic form $4x^2 + 2xy + 3y^2 + 4w^2 + 2wz + 3z^2$ represents all primes except for the two primes 2 and 11. Is it possible to prove the conjecture? Or, are there results to attack the conjecture? - -REPLY [21 votes]: As GH suggests, here the relevant Eisenstein and cusp spaces -are small enough that everything can be done explicitly. -It's even a bit better than the dimensions $5+4$ suggest, -because our quadratic form is isodual, which puts its -theta series in an eigenspace for the Atkin-Lehner involution $w_{44}$. -The resulting formula is particularly nice for $n$ prime, -and immediately shows that every prime other than $2$ and $11$ -is represented, and indeed the number of representations is -proportional to the number of points modulo the prime of an -elliptic curve of conductor $11$. -Namely: let -$$ -E_2(q) = 1 - 24 \sum_{n=1}^\infty \frac{nq^n}{1-q^n}; -$$ -this is not a modular form, but for every factor $d|44$ the combination -$$ -\varepsilon^{(d)}_2(q) := d \cdot E_2(q^d) - \frac{44}{d} E_2(q^{44/d}) -$$ -is a weight-2 form for $\Gamma_0(44)$. Let -$$ -\phi(q) = q \prod_{n=1}^\infty \bigl( (1-q^n)(1-q^{11n}) \bigr)^2 -= q - 2 q^2 - q^3 + 2 q^4 + q^5 + 2 q^6 - 2 q^7 \cdots -$$ -be the unique eigen-cuspform for $\Gamma_0(11)$, associated to the -elliptic curve $E: y^2+y=x^3-x^2$ of discriminant $-11$. Then the -theta function $\sum_{n=0}^\infty r(n) q^n$ is -$$ --\frac - {\varepsilon^{(1)}_2(q) - \varepsilon^{(2)}_2(q) + \varepsilon^{(4)}_2(q)} - {30} -- \frac45\bigl(\phi(q)+3\phi(q^2)+4\phi(q^4)\bigr). -$$ -The coefficients are obtained by matching $q$-expansions to $O(q^{125})$, -which is more than enough to prove that two weight-$2$ forms on $\Gamma_0(44)$ -coincide. In particular, for the number of representations of a prime -$p$ other than $2$ and $11$ we have -$$ -r(p) = \frac45 (p + 1 - a_p) -$$ -which is positive because $p+1 - a_p$ is the number of points on $E \bmod p$ -(which is indeed divisible by $5$ because $E$ has a rational $5$-torsion point -$x=y=0$).<|endoftext|> -TITLE: Surreal exponentiation -- are the varying definitions equivalent? If not, is there agreement on which ones are better? -QUESTION [26 upvotes]: The surreal numbers are sometimes introduced as a place where crazy expressions like $(\omega^2+5\omega-13)^{1/3-2/\omega}+\pi$ (to use the nLab's example) make sense. The problem is, there seem to be varying definitions of the exponential in the surreal numbers and since I can't find any recent reference that covers them all I have little idea whether they're actually the same or not. (When I say "exponential", I mean either $e^x$ or the general $a^x$ for $a>0$; obviously one can go back and forth between these so long as $e^x$ is indeed a bijection from surreals to positive surreals.) -To wit: - -Harry Gonshor gives one definition in his "An Introduction to the Theory of Surreal Numbers". -Gonshor mentions an earlier unpublished definition due to Martin Kruskal; so does Conway in the 2nd edition of "On Numbers and Games". Neither actually state this definition, but it is strictly speaking possible for someone who's never seen it to verify equivalence with it, because Conway mentions that it is inverse to a particular definition of the logarithm, which he does not explicitly state but gives enough information to deduce. Gonshor seems to suggest in his text that his definition is equivalent to Kruskal's unpublished one, but on the other hand never seems to explicitly state so. -Norman Alling's "Foundations of Analysis over Surreal Number Fields" looks like it might contain another definition? I'm not too clear on what he's doing, honestly, though it looks like it's restricted to non-infinite surreals... -Wikipedia's page [old version, this definition was removed from Wikipedia soon after asking this question] gives a totally uncited definition for $2^x$. I have no idea where this might be originally from. I suppose one could substitute in other surreals for 2 to generalize this? -Or else one could take Wikipedia's definition and generalize it in the way one usually does when starting from $e^x$? (I should hope this agrees with definition 4!). - -Note that the operation $x\mapsto \omega^x$ commonly used in the surreals is not related; though it's exponential in some sense, it's not surjective onto the positive surreals, and so definitions of a general exponential shouldn't attempt to agree with it. And of course definitions 1, 2, and 4/5 above are surjective onto the positive surreals. (Or Wikipedia claims #4/5 is, anyway.) -Edit: To avoid confusion, in what follows, I'll write $\exp_\omega x$ instead of the usual $\omega^x$, and reserve the notation $\omega^x$ for whatever that happens to be in the notion of exponentiation under discussion. -So, does anyone know to what extent these are actually equivalent? If they're not equivalent, is there agreement on which ones are the "right" definitions? (It seems like all of them have the right properties! And while it seems to be agreed that the idea behind Kruskal's definition is bad, that doesn't mean necessarily the definition itself is.) Or could anyone point me to any recent book which might clear all this up, or at least the source of Wikipedia's definition? -(I had originally intended to ask other questions about surreal exponentiation before finding that I wasn't sure what it actually was. I am hoping that whatever references people can point me to will answer my other questions as well.) -Slight update: Definitions 4 doesn't seem to agree with definition 5 (nor definition 1, see below); it would seem that definition 4 would imply $3^\omega=\omega$, while definition 5 would imply $3^\omega>\omega$. This raises the problem in that one could make more definitions by using definition 4 to define $a^x$ for some fixed $a$, and then generalizing it to $b^x$ for all $b$ via definition 5, and depending on your choice of starting $a$ -- whether $e$, 2, or something else -- you'd get different definitions of $b^x$ out. An entire proper class of distinct "exponentiation" operations! Well, perhaps not, perhaps not all starting values of $a>1$ yield an onto function -- perhaps 2 is special and it's the only one that does, though that seems pretty unlikely, and barring that, this is pretty bad regardless. Also, definition 4 seems pretty suspect as the "right" definition for another reason: If we plug in two ordinals, it looks like it will agree with ordinary ordinal exponentiation. This both disagees with Gonshor's definition (which would imply $\omega^\omega>\exp_\omega \omega$) and is suspect on its own, because we shouldn't expect to get one of the ordinary ordinal operations out of this (we use natural addition and multiplication in the surreals, not ordinary addition and multiplication). If indeed we get an ordinal operation out of this at all -- it would appear that by Gonshor's definition, $\omega^\omega$ would not even be an ordinal, instead being equal to $\exp_\omega \exp_\omega (1+1/\omega)$. -Oops: Sorry, that shouldn't be ordinary exponentiation, but rather the analogue of it based on natural multiplication. Regardless, still disagrees, still smells bad. - -REPLY [26 votes]: There is only one official definition of surreal exponentiation in the literature, the one due to Martin Kruskal. It was rediscovered by Harry Gonshor (with hints from Kruskal) and incorporated into his book (An Introduction to the Theory of Surreal Numbers) where important results on surreal exponentiation that go beyond Kruskal's initial discoveries are also found. -Surreal exponentiation was later discussed by Lou van den Dries and myself in “Fields of Surreal Numbers and Exponentiation”, Fundamenta Mathematicae 167 (2001), No. 2, pp. 173-188; erratum, ibid. 168, No. 2 (2001), pp. 295-297. We showed: -If No(alpha) is the set of surreal number of tree-rank less than an epsilon number alpha, then No(alpha) is an elementary extension of the ordered field of reals with exp and an elementary substructure of the ordered field of surreals with Kruskal-Gonshor exponentiation. -Surreal exponentiation is also discussed on pages 31 and 32 of my just-published paper: “The Absolute Arithmetic Continuum and the Unification of All Numbers Great and Small”, The Bulletin of Symbolic Logic 1 (2012), pp. 1-45. -Norman Alling's treatment of expX (in terms of power series) only holds when X is infinitesimal and in those those cases it coincides with the Kruskal-Gonshor definition. The definition of surreal exponentiation in Wikipedia seems - dubious. In fact, a number of features of the Wikipeda article on surreal numbers are dated or misleading.<|endoftext|> -TITLE: Question about hereditary $C^*$-algebra -QUESTION [11 upvotes]: Can anyone give me a relatively simple proof or Some reference for the following fact.(I know that there is a proof of this theorem in Gerard J. Murphy'book: "$C^*$-Algebras and Operator Theory", but I'm sure that there should be a simple proof of this. -Every hereditary C*-subalgebra of a simple $C^*$-algebra is also simple! -Maybe this is easy for someone, but it makes me confused for a long time. I am a novice! - -REPLY [15 votes]: Here is a direct argument which may not differ much in its essence from the one in the book that you mention: -Say $A$ is simple and $B$ is a closed hereditary subalgebra of $A$. This means that if $a\in A$ and $b_1,b_2\in B$ then $b_1ab_2\in B$. Let $x\in B$ be non-zero and let us show that it generates $B$ as a closed two sided ideal. Since $x$ generates $A$ as a closed two-sided ideal, the finite sums of elements of the form $axa'$, with $a,a'\in A$, form a dense subset in $A$. In particular, if $y\in B$ and $\epsilon>0$ then there exists an element of the form -$$ -\sum_{i=1}^{n} a_ixa'_i -$$ within a distance $\epsilon$ of $y$. The problem with this is that the $a_{i}$s and $a'_i$s are not in $B$. They are only in $A$. This is fixed using that in a C*-algebra one always has that $|c^*|^{1/n}c|c|^{1/n}\to c$ for any $c$ (alternatively, you can use an approximate unit for $B$). Then for $k$ large enough the element -$$ -\sum_{i=1}^{n} (|y^*|^{1/k}a_i|x^*|^{1/k}) x (|x|^{1/k}a'_i |y|^{1/k} ) -$$ -is also within a distance of $\epsilon$ of $y$.<|endoftext|> -TITLE: Are all manifolds affine? -QUESTION [14 upvotes]: There is a classical result which says that the assignment $$M \mapsto C^{\infty}\left(M\right)$$ is an embedding of the category of (paracompact Hausdorff) smooth manifolds into the opposite category of $\mathbb{R}$-algebras. However, all of the proofs I have seen first establish this for manifolds of the form $\mathbb{R}^n$ and then use Whitney's embedding theorem. This of course uses the paracompact Hausdorff condtion in an essential way. My question is, is this functor still an embedding if we don't impose paracompactness conditions on our smooth manifolds? If so, is there a nice proof of this fact? If not, can someone provide a simple counterexample? Thanks! - -REPLY [14 votes]: The functor is not an embedding if we remove the paracompactness assumption. -I will need some preliminary definitions. Let $R$ be the long ray, i.e., the topological space given by $\omega_1 \times [0, 1)$ equipped with the order topology induced by lexicographic order, and let $L$ be the long line obtained by gluing together two long rays. Topologically, we can think of $L$ as the colimit of spaces $L_\alpha$ where $\alpha$ is a countable ordinal and $L_\alpha$ is obtained by gluing together two rays $R_\alpha = \alpha \times [0, 1)$. -According to the Wikipedia article, not only does there exist a smooth ($C^\infty$) manifold structure on $L$, there exist infinitely many smooth ($C^{\infty}$) manifold structures on $L$ which extend a given $C^1$ manifold structure. On the other hand, there is just one smooth structure on the ordinary line which extends a given $C^1$ structure. We exploit these facts to show that $M \mapsto C^\infty(M)$ is not an embedding. -Let $L$ and $L'$ be distinct smooth structures which restrict to the same $C^1$ structure. If $C^\infty(-) = \hom(-, \mathbb{R})$ were an embedding on general Hausdorff not necessarily paracompact manifolds, then $L$ and $L'$ would be isomorphic if their smooth algebras are isomorphic. So it is enough to show that $C^\infty(L)$ and $C^\infty(L')$ are isomorphic. Now it is well-known that every continuous function on the long line is eventually constant (constant outside some bounded neighborhood). In that case, consider the algebra $C_\alpha$ of smooth functions on $L_\alpha$ which are eventually constant. For $\alpha \leq \beta$ there is an obvious extension $C_\alpha \to C_\beta$, and we have -$$C^\infty(L) = colim_\alpha C_\alpha$$ -Similarly, we may write $C^\infty(L') = colim_\alpha C^\prime_\alpha$. However, notice that the identity function $L \to L'$ restricts to a diffeomorphism $L_\alpha \to L^\prime_\alpha$, because each $L_\alpha$ is topologically an ordinary line, where we had observed there is just one smooth structure extending the given $C^1$ structure. The isomorphisms $C_\alpha \to C^\prime_\alpha$ induce an isomorphism $C^\infty(L) \to C^\infty(L')$, as desired.<|endoftext|> -TITLE: Alternate proofs of Quillen's theorem on formal group laws and MU -QUESTION [11 upvotes]: This is related to this posting: -complex cobordism from formal group laws? -but not entirely the same. I'd like to know if there are any proofs of Quillen's theorem that $\pi_* (MU)$ is the Lazard ring (home of the universal formal group law) other than Quillen's. Specifically, is it possible to prove the result without a deep understanding of the structure of $H^*(MU)$ as a module over the Steenrod algebra? - -REPLY [4 votes]: Jacob Lurie taught a course at Harvard in the spring of 2010 which included a slick proof of Quillen's theorem. I'm pretty sure Steenrod operations are not mentioned at all. The course page is here, and the notes which include this proof are in lecture 7. Also, the proof of Lazard's theorem (on the structure of the Lazard ring) in lectures 2 and 3 is very cool.<|endoftext|> -TITLE: Non-compact complex surfaces which are not Kähler -QUESTION [23 upvotes]: Not every complex manifold is a Kähler manifold (i.e. a manifold which can be equipped with a Kähler metric). All Riemann surfaces are Kähler, but in dimension two and above, at least for compact manifolds, there is a necessary topological condition (i.e. the odd Betti numbers are even). This condition is also sufficient in dimension two, but not in higher dimensions. Therefore the task of finding examples of compact complex manifolds which are not Kähler is reduced to topological considerations. -In the non-compact setting, we can also find such manifolds. For example, let $H$ be a Hopf surface, which is a compact complex surface which is not Kähler. Then for $k > 0$, $M_{k+2} = H\times\mathbb{C}^k$ is a non-compact complex manifold which is not Kähler - any submanifold of a Kähler manifold is Kähler, and $H$ is a submanifold of $M_{k+2}$. This generates examples in dimensions three and above. So I ask the following question: - -Does anyone know of some (easy) explicit examples of non-compact complex surfaces which are not Kähler? - -REPLY [11 votes]: From any compact non-Kähler surface $X$ remove a point $p$. You're left with a non-Kahler, non-compact surface. For the proof, see Théorème 2.3 in A. Lamari - Courants kählériens et surfaces compactes. First, by a theorem of Shiffman, a Kähler form on $X\setminus {p}$ extends as a closed positive current to all of $X$. Then, locally around $p$, the singularity at $p$ can be fixed by using convolutions to obtain a smooth Kähler form on $X$.<|endoftext|> -TITLE: The Problem about 2-coloring finite plane -QUESTION [6 upvotes]: Suppose we color a $X \times X$ finite plane by red and blue arbitrarily. How large does X need to be to guarantee a monochromatic combinatorial square $k \times k$ -1 0 1 0 1 1 -1 1 1 1 1 1 -1 0 1 0 1 1 -1 1 1 1 1 1 -The above figure shows a combinatorial $2 \times 2$ filled square filled by zeros. - -REPLY [9 votes]: The exact answer, $15$, to this question is the content of my paper with Shalom Eliahou: -Here a copy of the corresponding entry of Math-Review: -Bacher, Roland; Eliahou, Shalom -Extremal binary matrices without constant 2-squares -J. Comb. 1 (2010), no. 1, [ISSN 1097-959X on cover], 77–100. -05D10 (11B75) -Summary: "In this paper we solve, by computational means, an open problem of Erickson: Let $[n]=\{1,…,n\}$; what is the smallest integer $n_0$ such that, for every $n\ge n_0$ and every 2-coloring of the grid $[n]\times[n]$, there is a constant 2-square, i.e. a $2\times2$ subgrid $S=\{i,i+t\}\times\{j,j+t\}$ whose four points are colored the same? It has been shown recently that $13\le n_0\le\min(W(2,8),5\cdot2240)$, where $W(2,8)$ is the still unknown eighth classical van der Waerden number. We obtain here the exact value $n_0=15$. In the process, we display 2-colorings of $[13]\times{\bf Z}$ and $[14]\times[14]$ without constant 2-squares, and show that this is best possible.''<|endoftext|> -TITLE: Does $S$ being a free rank-$n$ $R$-algebra imply that $S/R$ is free rank $n-1$? -QUESTION [6 upvotes]: Suppose we have a (commutative) ring $R$ and an $R$-algebra $S$. Furthermore, suppose that $S\cong R^n$ as $R$-modules, that is, $S$ is free of rank $n$ as an $R$-module. Can we always choose $1$ to be a basis element for $S$? Equivalently, is it necessary that $S/R \cong R^{n-1}$? -If not, how about in the case that $R=\mathbb{Z}$? -This is true in the case $n=1$: if we have a ring homomorphism $\phi: R\to S$ and an $R$-module isomorphism $\psi: S\to R^1$, it's not hard to show that $\phi$ must also be an isomorphism, making $S/R\cong R^0$. -And it is true if $n=2$ and $R=\mathbb{Z}$: in that case it is known that $S\cong \mathbb{Z}[x]/p(x)$, where $p$ is some degree 2 monic polynomial, so that in particular $S$ is generated freely as a $\mathbb{Z}$-module by 1 and $x$. - -REPLY [2 votes]: For the first question, it is nevertheless true that if $S$ is an $R$-algebra (containing $1$) that is finitely generated and projective as an $R$-module, then $S \simeq R \oplus S/R$ as $R$-modules, i.e., $R$ is a direct summand of $S$. See Lemma 1.3 of http://math.dartmouth.edu/~jvoight/articles/lowrank-051714.pdf. (I'm sure there are other references, but I couldn't find one easily when I was writing this.) -I guess it is clear by now that the "equivalently" part is not true.<|endoftext|> -TITLE: flat maps of monoids which are not localizations -QUESTION [5 upvotes]: It is well known that a localization $S^{-1}R$ of a commutative ring $R$ is flat as a $R$-module. -Rather, I am looking for extensions of rings which share certain properties of localizations, like flatness, while inverting nothing. -More precisely, I would like to find classes of examples, if such exist, of maps of monoids $f:M\to N$ verifying - -$N^{\times} = \{1\}$, -$f$ is injective, but not surjective, -"$M$ is cofinal in $N$" in the sense that the augmentation induces an isomorphism $\Bbb{Z}[N]\otimes_{\Bbb{Z}[M]}\Bbb{Z}[1]\cong\Bbb{Z}[1]$ of abelian groups (note that $\Bbb{Z}[1]\cong\Bbb{Z}$ as abelian groups), - -for which $\Bbb{Z}[N]$ is flat as either a left or right $\Bbb{Z}[M]$-module. -The first two conditions listed above are simply to guarantee that $f$ does not invert any elements whatsoever, while not being an isomorphism. -For completeness, and in case it is easier, I am actually looking for examples of maps $f:M\to N$ for which $\text{Tor}^{\Bbb{Z}[M]}_\ast(\Bbb{Z}[N],\Bbb{Z}[N])$ is $\Bbb{Z}[N]$ concentrated in degree zero. -Remark: this condition is related to the forgetful functor from the positive derived category of $\Bbb{Z}[N]$-modules to the derived category of $\Bbb{Z}[M]$-modules being full and faithful. -Edit: I have removed the extraneous commutativity condition on the monoids. - -REPLY [2 votes]: This is not really an answer, but it is too long for a comment. Suppose that $f\colon M\to N$ is a homomorphism of monoids (not necessarily commutative). Let's work with right modules. -First of all condition (3) can be phrased as follows: construct a graph with vertex set $N$ and an edge between $n$ and $nm$ for all $n\in N$ and $m\in M$. Then (3) happens iff this graph is connected. -Next there is a classical condition coming from category/topos theory that guarantees flatness of $\mathbb ZN$ over $\mathbb ZM$. In fact, it is equivalent to flatness of $N$ as a right $M$-set, but this is a stronger condition. Namely, consider the category of elements $\mathcal C$ of $N$ viewed as a presheaf over $M$. So the objects are elements of $N$ and there is an arrow $(n,m)\colon nm\to n$ associated to each $n\in N$ and $m\in M$. Composition is $(n,m)(nm,m')=(n,mm')$. - - -Fact. If the category $\mathcal C$ is filtered one has that $\mathbb ZN$ is flat over $\mathbb ZM$.<|endoftext|> -TITLE: Curvatures of stable and unstable manifolds -QUESTION [7 upvotes]: Let $(M,g)$ be a closed Riemannian manifold and $f:M\to M$ be a $C^r$ ($r\ge2$) Anosov diffeomorphism, that is, there is a continuous hyperbolic splitting $TM=E^s\oplus E^u$ with respect to the Riemannian metric $g$ over $M$. -We know there exists stable and unstable foliations $\mathcal{W}^s$ and $\mathcal{W}^u$ tangent to $E^s$ and $E^u$. -Moreover, all leaves $W^s(x)$ and $W^u(x)$ are $C^r$ submanifolds for every $x\in M$. -So we can talk about the sectional/Ricci/scalar curvatures of these submanifolds (endow with the restricted metrics $g|_{W^s(x)}\text{ and } g|{W^u(x)}$). -I want to know if there are some results and references about this topic. -This should be an interesting question, and some results in decay of correlations do assume the sectional curvatures are bounded. -Maybe there are some results for - -more general classes of maps (say partially hyperbolic diffeo's), -more general submanifolds (say foliations with continuous tangent $E=T\mathcal{W}$). - -Thank you! - -Edit: the relation between the metric $g$ and the hyperbolic splitting $TM=E^s\oplus E^u$. -The Riemannian metric $g$ induces a norm $\|\cdot\|_x$ on each tangent space $T_xM$ by $\|v\|^2_x=g_x(v,v)$ for every $v\in T_xM$. -The map $f:M\to M$ is said to be hyperbolic if there exists a continuous splitting $TM=E^s\oplus E^u$ such that - -the splitting is $Df$-invariant: $D_xf(E^s_x)=E^s_{fx}$ and $D_xf(E^u_x)=E^u_{fx}$, -$E^s$ is uniformly contracting: $\|D_xf^n(v)\|_{f^nx}\le C\lambda^n_s\|v\|_x$ for every $v\in E^s_x$, -$E^u$ is uniformly expanding: $\|D_xf^n(v)\|_{f^nx}\ge C^{-1}\lambda^n_u\|v\|_x$ for every $v\in E^u_x$, - -for some uniform constant $C\ge1$ and $0<\lambda_s<1<\lambda_u$. -So the hyperbolicity of the map $f$ does depend on the choice of Riemannian metric. -But being hyperbolic is not sensitive to the choice of metric. For example if $f$ is hyperbolic with respect to $(M,g)$ and we have another Riemannian metric $h$ with $C_1\cdot g_x(v,v)\le h_x(v,v)\le C_2\cdot g_x(v,v)$, then $f$ is also hyperbolic with respect to $(M,h)$ (with a possible different $C$). -The upper bound of the curvatures of stable and unstable manifolds (if exists) may depend on the choice of metrics. But the property of having bounded curvature should be independent of the choice of metrics. - -REPLY [2 votes]: Maybe late, but I believe the following reference is relevant. Is from a paper from Bonatti and Viana which works even for partially hyperbolic attractors (of course the $C^2$-hypothesis is crutial): -See Lemma 3.1 here http://w3.impa.br/~viana/out/mcontracting.pdf<|endoftext|> -TITLE: Examples of algorithms that came from category theory? -QUESTION [20 upvotes]: Generating Compiler Optimizations from Proofs is a wonderful paper. The authors say that they were faced with the problem, got stuck, then tried reasoning about it using category theory. They took the obvious tack, isolated the new idea, designed the abstract algorithm, and applied it to their specific case. -What other examples like this do you know of, where category theory clearly had a role in producing a nontrivial algorithm? - -REPLY [16 votes]: I'm glad you liked the paper. I thought you might like to know that the algorithms for both of my PLDI papers were actually designed using a category-theoretic framework I made for existential types (half of which turned out to be opfibrations, but I didn't know about those at the time). Others have asked to see it, so I'm finally giving in and posting the work-in-progress on my page for Inferable Object-Oriented Typed Assembly Language, since that's the problem which prompted me to make the framework once I get stuck. Just look for "Inferable Existential Quantification". Hope you like it! -One thing I'll note while I'm at it is that, in my experience, "algorithmic" category theory (as opposed to "type-theoretic" and "semantic" category theory) is largely built on concrete category theory. After all, many algorithms deal with structured sets of some form. Thus the techniques available to an algorithm often depend on whether the concrete category is topological or algebraic rather than whether it is closed. There's lots of stuff showing what kind of structurings preserve things like existence of pushouts and pullbacks, but I think it would be really interesting (and helpful) to also identify what kind of structurings preserve constructability/decidability of pushouts and pullbacks. By knowing that, we could build algorithms for a domain just by showing how the domain can be built from a sequence of constructability-preserving structurings that consequently guarantee the presence of the particular structures required by the algorithm. If anyone happens to already know of such research, I'd be very interested to see it.<|endoftext|> -TITLE: Is the ring of quaternionic polynomials factorial? -QUESTION [15 upvotes]: Denote by $\mathbb{H}[x_1,\dots,x_n]$ the ring of polynomials in $n$ variables with quaternionic coefficients, where the variables commute with each other and with the coefficients. Two polynomials $P,Q\in \mathbb{H}[x_1,\dots,x_n]$ are similar, if $P=a Q b$ for some $a,b\in \mathbb{H}$. A ring $\mathbb{K}$ is factorial, if the equality $P_1\cdot\dots\cdot P_n=Q_1\cdot\dots\cdot Q_m$, where $P_1,\dots, P_n,Q_1,\dots,Q_m\in \mathbb{K}$ are irreducible (and noninvertible) elements, imply that $n=m$ and there is a permutation $s\in S_n$ such that $P_k$ is similar to $Q_{s(k)}$ for each $k=1,\dots,n$? -By [1, Theorem 1] and [2, Theorem 2.1] it follows that $\mathbb{H}[x]$ is factorial. - -Is the ring $\mathbb{H}[x,y]$ factorial? - -This question is a continuation of the following ones: -When the determinant of a 2x2 polynomial matrix is a square? -Pythagorean 5-tuples -[1] Oystein Ore, Theory of non-commutative polynomials, Annals of Math. (II) 34, 1933, 480-508. -[2] Graziano Gentili and Daniele C. Struppa, On the Multiplicity of Zeroes of Polynomials with Quaternionic Coefficients, Milan J. Math. 76 (2008), 15-25, DOI 10.1007/s00032-008-0093-0. - -REPLY [15 votes]: Just to remove the question from the 'Unanswered' list: -$(x-i)\cdot((x+i)(y+j)+1)=((y+j)(x+i)+1)\cdot(x-i)$, hence $\mathbb{H}[x,y]$ is not factorial.<|endoftext|> -TITLE: Representation theory of reductive groups in characteristic $p$ as a limit of the theories in characteristic $0$ -QUESTION [20 upvotes]: This question is out of plain curiosity. The first sentence of Deligne's -Les corps locaux de caractéristique $p$, limites de corps locaux de caractéristique $0$ (1984) reads (in rough translation) as follows : - -D. Kazhdan has introduced the - principle that the representation - theory of a reductive group over a - local field of prime characteristic - $p$ is the limit, as the ramification - index tends to infinity, of theories - over local fields of characteristic - $0$ with the same residue field. - -He says that according to Langlands' philosophy, one should expect the same phenomenon to occur on the galoisian side, and goes on to establish a precise equivalence of categories justifying this principle (and clarifying the earlier work of M. Krasner from the forties). -I'm mainly interested in this side of the story, but I'm curious as to where Kazhdan's principle in representation theory was first enunciated. What are the standard references in English or French explaining this principle ? - -REPLY [14 votes]: I think, although it's dated later than Deligne's paper that you mentioned, that the first written instance of Kazhdan's principle is in the paper "Representations of groups over close local fields", Journal d'Analyse Math\'ematique, vol. 47,1986, pp.175--179. -This is in the same journal issue as "Cuspidal Geometry of p-adic Groups" (by Kazhdan) and "Trace Paley-Wiener Theorem for Reductive p-adic Groups" (by Bernstein, Deligne, Kazhdan). The book "Representations of reductive groups over a local field" appeared in 1984, and according to the MathSciNet review of the article "Le 'Centre' de Bernstein", the Trace Paley-Wiener Theorem paper was already a preprint in 1984. -So it seems to me that Kazhdan's principle was probably "in the air" by 1984, but not written down by him until the "close local fields" article above. I second Jim Humphreys' suggestion to contact Kazhdan himself for less speculative history.<|endoftext|> -TITLE: Who first cared about singular points? -QUESTION [15 upvotes]: If you look at the cross $C\subset \mathbb A^2_k$ given by $xy=0$ in the affine plane over the field $k$, you see or compute that it is exceptional at $O=(0,0)$ for many (obviously not independent) reasons: -$\bullet$ The gradient of $xy$ vanishes at $ O$ . -$\bullet$ Two irreducible components pass through $O$. -$\bullet$ If $k=\mathbb C$, the complement of $O$ is disconnected. -$\bullet$ The tangent cone of $C$ at $O$ is not a line . -$\bullet$ The maximal ideal $(x,y)\subset \mathcal O_{C,O}$cannot be generated by just one element. -$\bullet$ The sheaf $\Omega_{C/k}$ is not locally free. -$\bullet$ The $k$-morphism $Spec (k[t]/\langle t^2\rangle) \to C$ given by $x=t,=y=t$ cannot be lifted to the overscheme $Spec (k[t]/\langle t^2\rangle) \subset Spec (k[t]/\langle t^3\rangle) $. -This exceptional character of $O$ is covered by several negative adjectives: non smooth,non-regular, non manifold-like , singular,... -Although I know that the purely algebraic condition for singularity (in terms of number generators of the maximal ideal of a local ring) is due to Zariski and that smoothness in terms of infinitesimal liftings is due to Grothendieck, I don't know the earlier history of the concept of singularity. -So my question is: -Who first considered explicitly the concept of singularity for varieties , why the interest and what was the definition? -Edit -First of all, thanks for the interesting comments. It is certainly plausible that Newton knew what a singularity was, but from what I read (very little) his preoccupation seems to have been classification of curves by degree. -I am curious about when he or others first wrote down the dichotomy between singular and non singular varieties , in analogy with Descartes's sharp distinction between mechanical (=transcendental) curves and geometric (=algebraic) curves ( see here) . -[By the way, if you know French you will be delighted by Descartes's old-fashioned but easily understandable language] - -REPLY [4 votes]: Gauss' Disquisitiones generales circa superficies curvas, read at the Royal Society in Goettingen on 8 Oct. 1827, contain many termini like punctis singularibus or singulis punctis. Already on the first page we read cuius singula puncta repraesentare. -Dirichlet's collected works, edited by Kronecker and Fuchs, contain often the phrase "singular cases", and on pag. 365 punctum singulare, (written around 1850). -Riemann's collected works, edited by Weber and Dedekind, contain the first mentioning of "singulärer Punkt" on page 389 that I know of in German. It is taken from a paper about linear differential equations of 1857. -The Earliest Known Uses of Some of the Words of Mathematics supply the following dates: -SINGULAR POINT appears in a paper by George Green published in 1828. The paper also contains the synonymous phrase "singular value" [James A. Landau]. -Singular point appears in 1836 in the second edition of Elements of the Differential Calculus by John Radford Young. -In An Elementary Treatise on Curves, Functions and Forces (1846), Benjamin Peirce writes, "Those points of a curve, which present any peculiarity as to curvature or discontinuity, are called singular points." - Additional remark -Hermann, a correspondent of Leibniz, wrote about singulis locis and singulis punctis in letters to Leibniz of 11 Jan. 1711 and Jun 1712, respectively. [C. I. Gerhardt (ed.): "Leibnizens mathematische Schriften", Halle (1859) p. 364 and 368] -And we should not forget that l'Hospital's famous theorem (1696) has been designed for singular points of functions only.<|endoftext|> -TITLE: Bound on graph domination number when min degree is 7 -QUESTION [5 upvotes]: I have a graph $G$ whose minimum vertex degree is $\delta=7$. -I am seeking an upper bound on the domination number $\gamma(G)$ -in terms of the number of vertices $n$ of $G$. -I found a paper by -Edwin Clark, Boris Shekhtman, Stephen Suen, and David Fisher, -"Upper bounds for the domination number of a graph," Congressus -Numerantium, 132:99–124, 1998 (CiteSeer link), -that implies a bound of about $0.31 n$ (more precisely, $(18286568 / 58640175)n$). -I was hoping for a smaller upper bound. Perhaps there have been advances since that paper? -Any pointers to relevant literature would be much appreciated. Thanks! - -REPLY [5 votes]: See this article, it provides the best bound till date: -Bujtás, Csilla; Klavžar, Sandi, Improved upper bounds on the domination number of graphs with minimum degree at least five, Graphs Comb. 32, No. 2, 511-519 (2016). ZBL1339.05280. - -Abstract -An algorithmic upper bound on the domination number $\gamma$ of graphs in terms of the order $n$ and the minimum degree $\delta$ is proved. It is demonstrated that the bound improves best previous bounds for any $5 \leq \delta \leq 50$. In particular, for $\delta =5$, Xing et al. (Graphs Comb. 22:127–143, 2006) proved that $\gamma \leq 5n/14 < 0.3572n$. This bound is improved to $0.3440 n$. For $\delta=6$, Clark et al. (Congr. Numer. 132:99–123, 1998) established $\gamma < 0.3377n$, while Biró et al. (Bull. Inst. Comb. Appl. 64:73–83, 2012) recently improved it to $\gamma < 0.3340n$. Here the bound is further improved to $\gamma < 0.3159n$. For $\delta=7$, the best earlier bound $0.3088n$ is improved to $\gamma<0.2927n$.<|endoftext|> -TITLE: Finding local patterns in a circular list -QUESTION [9 upvotes]: Consider a list $\boldsymbol{x}=x_0,x_1,\ldots,x_{n-1}$, which we consider to be circular by taking the subscripts modulo $n$. The entries in the list are distinct integers. -A local pattern is a Boolean expression $P(i)$ involving inequalities between the values $x_i,x_{i+1},\ldots,x_{i+k}$ for some constant $k$. For example, we might have $P(i)=(x_i\lt x_{i+1})\wedge(x_i\lt x_{i+2})$. "Finding $P~$ in $\boldsymbol{x}$" means -finding a value of $i$ for which $P(i)$ is true, or determining that there is no such $i$. -Trivially, we can find any local pattern in $\boldsymbol{x}$ in $O(n)$ time just by trying each $i$. Perhaps surprisingly, some nontrivial local patterns can be found in $O(\log n)$ time. Consider the pattern for a local minimum $P(i)=(x_i\gt x_{i+1})\wedge(x_{i+1}\lt x_{i+2})$. Choose any three distinct entries $x_i,x_j,x_k$, where $x_j$ is the least. Without loss of generality, we can assume that $0\le i\lt j\lt k\lt n$. Let $\ell~$ be the midpoint (rounded towards $j$) of the longest of the intervals $[i,j]$ and $[j,k]$. If $x_j\lt x_\ell$, $x_j$ is the least of $x_i,x_j,x_\ell$, while if $x_j\gt x_\ell$, $x_\ell$ is the least of $x_j,x_\ell,x_k$. It is easy to see that continuing in this fashion finds a local minimum in $O(\log n)$ steps. -The question is: which local patterns can be found in $O(\log n)$ steps? - -REPLY [2 votes]: Here is a solution to the exercise Brendan assigned in his comment to Emil's partial answer. An occurrence of the Boolean expression there can be found with a fairly straightforward bisection. -To begin, let's add one more inequality, namely $x_{i+1} \gt x_{i+2}$, so that we're now looking for a stretch of four consecutive terms where the largest occurs in the second position. (If $Q(i)$ implies $P(i)$ and you find $Q(i)$, then you've also found $P(i)$.) Now start by sampling two sets of four consecutive terms that are opposite one another on the circle, and look for the largest of these eight values. If this maximum is in the second position among its group of four, you're done. Otherwise, cut the circle so that the maximum is either the first term in the "right" group or one of the last two in the left group. (Note: You might actually already be done by this point anyway, in case the group that doesn't have the overall maximum has its own maximum in its second position, but I'm just going to ignore that and not bother to check on anything but the overall biggest number.) -Now bisect: Take another group of four consecutive terms roughly halfway between the two current groups. If the second term in this group is a new overall maximum, you're done. Otherwise, continue the bisection on the side that contains the maximum. In $O(\log n)$ steps, this will produce a stretch of consecutive terms containing the overall maximum (of the values sampled) with at least one smaller term to its immediate left and two smaller terms to its immediate right.<|endoftext|> -TITLE: Eilenberg-Mazur swindle for higher K groups -QUESTION [11 upvotes]: The Eilenberg-Mazur swindle shows that the Grothendieck group of an additive category with countable coproducts is trivial. The strategy is to observe that any "Euler characteristic" $\chi$ on such a category must be zero, because for any object $P$, we have $$P \oplus \bigoplus_{i=1}^\infty P \simeq \bigoplus_{i = 1}^\infty P$$ -which implies that $\chi(P) + \chi(Q) = \chi(Q)$ for $Q = \bigoplus_{i=1}^\infty P$. -Is there any analog for the higher $K_i$ (of, say, an exact category in Quillen's sense)? I don't know any simple way of thinking of the higher $K_i$ (e.g. via Euler characteristics), but it would be interesting if, say, the associated K-theory space somehow had to be contractible. - -REPLY [14 votes]: Yes, there is an analogue, see Weibel's book, chapter V, \S 1.9 (http://www.math.rutgers.edu/~weibel/Kbook.html). He calls an exact category $A$ "flasque" if there is a functor $\infty: A \to A$ such that $A \coprod \infty(A) \cong \infty (A)$ and proves that the Quillen K-theory space $K(A)$ of a flasque category $A$ is contractible. Your countable coproduct gives an example. Analogous statements hold more generally for Waldhausen categories and Waldhausen K-theory. -Another result that fits into this context is that if $R$ is a ring, the group of ALL automorphisms of $R^{\infty}$ is acylic (Mac Duff, de la Harpe, "Acyclic groups of automorphisms"). This is proven by an Eilenberg-swindle argument as well.<|endoftext|> -TITLE: Can roots of any polynomial be expressed using Eulerian function? -QUESTION [7 upvotes]: I encountered an interesting function which is called "Eulerian" by the Wolfram's MathWorld: -$$\phi(q)=\prod_{k=1}^{\infty} (1-q^{k})$$ -It is interesting because it is claimed that roots of any polynomial can be expressed in this function and elementary functions. Is this true and how the roots of arbitrary polynomial can be expressed? -P.S. In Mathematica this function is inplemented as QPochhammer[q] - -REPLY [17 votes]: This Euler function is essentially the same as Dedekind's eta function (Wikipedia, Mathworld). The usual use of the $\eta$ function is to express various modular forms. In particular, you should be able to rewrite Hermite's solution of the quintic by modular functions in terms of the $\eta$ function. -I don't know whether you can express roots of higher degree polynomials in terms of $\eta$, but I would guess not. Here are my two hazy arguments: - -Hilbert conjectured that the roots of a general sextic could not be expressed using functions of one variable. I am told that this conjecture appears in Über die Gleichung neunten Grades, Mathematische Annalen Volume 97, Number 1, 243-250; I have not read this article. Abhyankar proves a version of Hilbert's conjecture in this paper, which I discussed in my answer here. Unfortunately, to my limited understanding of Abhyankar's result, it is about algebraic functions of one variable, so it is not clear to me that it answers your question. -According to Wikipedia, Hermite's solution of the quintic by modular forms was finally generalized to equations of arbitrary degree by Umemura, using Siegel modular forms. Siegel modular forms are analytic functions of many variables. This suggests to me that a generalization using modular forms in a single variable was found unworkable. - -As you can tell by the hesitant style of this answer, I find the results on solving equations by transcendental functions rather hard to follow; they all seem to be written in very equation heavy nineteenth century style. Since these questions come up fairly often on MO, it would be great if someone could recommend a good survey which translates them into the modern language.<|endoftext|> -TITLE: Is there always an uncountable $\kappa$ such that $\kappa^{\aleph_0}=2^\kappa$? -QUESTION [9 upvotes]: The cardinal equation $\kappa^{\aleph_0}=2^\kappa$ is satisfied by $\kappa=\aleph_0$. -It is also satisfied by any $\kappa$ for which $MA(\kappa)$ holds. -Under $GCH$, the equation is satisfied by $\kappa$ if and only if $cof(\kappa)=\aleph_0$. -So my question is: - -Is it consistent with $ZFC$ that the - only solution to - $\kappa^{\aleph_0}=2^\kappa$ is - $\kappa=\aleph_0$? - -I´m sorry if this is too basic, but I just don´t see it. - -REPLY [13 votes]: No, it is provable in ZFC that there are many $\kappa$ for which $2^\kappa=\kappa^\omega$. For example, let $\kappa=\beth_\omega$, and the same argument will work with any strong limit cardinal of cofinality $\omega$. Observe that every subset of $\kappa=\beth_\omega$ is determined by the sequence of its intersections with all the $\beth_n$'s. Thus, $2^\kappa\leq \Pi_n P(\beth_n)\leq \Pi_n\beth_{n+1}\leq\kappa^\omega$. But on the other hand, $\kappa^\omega\leq\kappa^\kappa\leq (2^\kappa)^\kappa=2^\kappa$. So this is a case where $2^\kappa=\kappa^\omega$.<|endoftext|> -TITLE: Does smoothness descend along flat morphisms? -QUESTION [14 upvotes]: Suppose $f:X\to Y$ is a flat morphism of schemes. If $X$ is smooth at $x$, must $Y$ be smooth at $f(x)$? - -If $f$ is locally finitely presented, then it is open (using EGA IV 1.10.4), so after replacing $Y$ by $f(X)$, we can assume $f$ is faithfully flat. I'd be happy to understand even the case where $X$ and $Y$ are local: - -Suppose $R$ and $S$ are local rings and $R\to S$ is a local homomorphism with $S$ (faithfully) flat over $R$. If $S$ is regular, must $R$ be regular? - -Note that I'm not asking if smoothness is "flat local"; there are certainly flat morphisms from singular things to smooth things (e.g. $k[x,y]/(x^2-y^2)$ is flat over $k[x]$). The question is whether there are flat morphisms from smooth schemes which hit singular points. - -REPLY [6 votes]: The answer to the first question is yes and (probably you known) is a consequence of the second question. -Suppose $f : X\to Y$ is a flat morphism of schemes over $S$ with $X$ smooth. Let $y=f(x)$ and let $s$ be the image of $x$ in $S$. By the positive answer to the second question, the geometric fiber $Y_{\bar{s}}$ is regular. So it only remains to see that$Y$ is flat over $S$ at $y$. -Consider the homomorphisms of local rings -$$ O_{S,s} \to O_{Y, y}\to O_{X,x}$$ -The second is flat by hypothesis, hence faithfully flat (because we deal with local rings), and the composition is flat by the smoothness of $X\to S$. So the first one is flat (easily seen using definition of flatness).<|endoftext|> -TITLE: quotient of Burnside group -QUESTION [5 upvotes]: Let G be a group of exponent n and m-generated. -Suppose further that every finite quotient of the Burnside group B(m,n) can be occurred as a finite quotient of G. Can we say that G≅B(m,n) ? - -REPLY [13 votes]: The answer is no. Take $\widehat B$ to be the profinite completion of $B(m,n)$. Then every finite quotient of $B(m,n)$ is still a quotient of $\widehat B$. However, from Zelmanov's solution of the Restricted Burnside Problem $\widehat B$ is finite. -In other words, take the intersection of all the normal subgroups of finite index in $B(m,n)$. From Zelmanov's solution of the RBP there are only finite number of them. Thus, the intersection is of finite index and the quotient satisfies your request, but it is finite (and of course generally, $B(m,n)$ is infinite). -Edit: Indeed as I commented above, let $q$ be a prime not dividing $n$ and let $S$ be a monster in which every element has order $q$. Then $G=\widehat B \oplus S$ satsifies the requirements. Clearly, $G$ has the same finite quotients as $B(m,n)$ since $S$ is simple. Now, $S$ is generated by any two elements not in the same cyclic subgroup. So let $y_1,y_2$ be two such generators, each has order $q$, and let $x_1,\ldots,x_m$ be generators of $B(m,n)$. Then $(x_1,y_1),(x_2,y_2),(x_3,1), \ldots, (x_m,1)$ generate $G$ since $(x_1,y_1)^n,(x_2,y_2)^n$ generate $S$.<|endoftext|> -TITLE: Stiefel-Whitney classes of a projective space bundle -QUESTION [5 upvotes]: Hi! -Let $\gamma_1$ denote the twisted line bundle over $S^1$ and add a trivial $(2k-1)$-bundle $\mathbb{R}^{2k-1}$. Consider the projective bundle $P(\gamma_1 \oplus \mathbb{R}^{2k-1})$ over $S^1$. Is it true that the first Stiefel-Whitney class of this bundle is $q^*(w_1)$ and all other vanish? Here $q$ denotes the projection from the total space of the projective bundle to $S^1$ and $w_1$ is the first Stiefel-Whitney class of $\gamma_1$. -EDIT: The question is about the tangential Stiefel-Whitney classes of the total space of $P(\gamma_1 \oplus \mathbb{R}^{2k-1})$. -best regards - -REPLY [2 votes]: You are asking whether a loop in the $2k-1$ manifold $P_k=P(\gamma_1\oplus R^{2k-1})$ is orientation reversing if and only if its projection to the base $S^1$ has odd degree. -$q:P_k\to S^1$ is a fiber bundle with fiber $F_k=RP^{2k-1}$. Since $RP^{2k-1}$ is orientable, a loop in $P_k$ which projects to a nullhomotopic loop in $S^1$ is homotopic into the fiber, hence orientation preserving. Thus $w_1(P_k)$ is either $q^*(w_1)$ or $0$. -But it isn't zero, since it isn't for $k=1$ (the Klein bottle is not orientable), and the normal bundle of $P_1\subset P_k$ is trivial. -EDIT: for the higher SW classes, The inclusion $F_k\subset P_k$ takes the SW classes to those of $F_k$ since the normal bundle of $F_k$ is trivial. If I recall correctly, the total SW class of $RP^{2k-1}$ equals $(1+t)^{2k}$, where $t\in H^1(RP^{2k-1};Z/2)$ denotes the generator. So, for example $(1+t)^6=1+t^2 + t^4$ and so for $k=3$, $w_2(P_k)$ and $w_4(P_k)$ are non-trivial.<|endoftext|> -TITLE: Cohomological dimension of a homomorphism -QUESTION [11 upvotes]: Let $G$ and $\Gamma$ be discrete groups, and let $\phi\colon\thinspace G\to \Gamma$ be a homomorphism. -Define its cohomological dimension $\operatorname{cd}\phi$ to be the least integer $d$ such that $\phi^\ast\colon H^i(\Gamma;M)\to H^i(G;M)$ is the zero homomorphism for all $i>d$ and all $\Gamma$-modules $M$ (where $M$ is regarded as a $G$-module via $\phi$). -Given that cohomological dimension of groups is such a well-studied invariant, I would have expected to find references to this relative notion in the literature. Alas, I cannot. - -Are there any references considering cohomological dimension of homomorphisms? - -and more specifically - -Does anyone know an example of a surjective homomorphism $\phi$ as above for which -$$\operatorname{cd} \phi < \min\lbrace \operatorname{cd}G, \operatorname{cd} \Gamma \rbrace?$$ - -EDIT: Thanks to Tom and Ralph's answers, I have been able to prove the following precise statement: -Let -$$ 1\to A \to G \stackrel{\phi}{\to} \Gamma \to 1$$ -be a central extension, where $H_\ast(A)$ is free and of finite type, and $\Gamma$ is a duality group with $\operatorname{cd}\Gamma = n$. Then $\operatorname{cd}\phi = n$. -Proof. We will show that $0\neq \phi^\ast\colon\thinspace H^n(\Gamma;\mathbb{Z}\Gamma)\to H^n(G;\mathbb{Z}\Gamma)$. This follows from the Lyndon-Hochschild-Serre spectral sequence. Since the action of $\Gamma$ on $A$ is trivial, and $\mathbb{Z}\Gamma$ is a trivial $A$-module, the $E_2$ term has -$$H^p(\Gamma;H^q(A;\mathbb{Z}\Gamma))\cong H^p(\Gamma;H^q(A)\otimes\mathbb{Z}\Gamma)$$ -in the $(p,q)$-position. Since $\Gamma$ is a duality group, this is zero for $p\neq n$. Hence there are no non-trivial differentials, and the edge homomorphism -$$\phi^\ast\colon\thinspace H^n(\Gamma;\mathbb{Z}\Gamma) \to H^n(G;\mathbb{Z}\Gamma)$$ -is an isomorphism. $\Box$ -Tom's answer shows that either centrality or finite type is necessary in the above statement. I haven't accepted it yet because I'm hoping someone will give an example with $\operatorname{cd} G <\infty$. -Update: Dranishnikov and Kuanyshov have investigated cohomological dimensions of homomorphisms more fully in a recent preprint https://arxiv.org/abs/2203.03734, in particular in relation to the Lusternik--Schnirelmann category of the classifying map $B\phi: BG\to B\Gamma$. They provide simpler examples of surjective $\phi$ with $\operatorname{cd}\phi<\min\{\operatorname{cd}G,\operatorname{cd}\Gamma\}$, including examples where $G$ and $\Gamma$ are geometrically finite. They also show that this phenomenon can't occur when $G$ and $\Gamma$ are torsion-free nilpotent groups. - -REPLY [3 votes]: One more example refering to the second question: Let -$$\Gamma = \left\lbrace \left. \begin{pmatrix} -1 & \ast & \ast \newline & 1 & \ast \newline & & 1 -\end{pmatrix} \right\vert\ \ast \in \mathbb{Z} \right\rbrace$$ -be the group of integral upper triangular matrices with unit diagonal. $\Gamma$ fits into the non-split central extension $$0 \to \mathbb{Z} \to \Gamma \to \mathbb{Z}^2 \to 0$$ that corresponds to a generator $\epsilon \in H^2(\mathbb{Z}^2;\mathbb{Z}) = \mathbb{Z}$. -Claim 1: $cd(\Gamma) = 3$ -Since $\mathbb{Z}$ resp. $\mathbb{Z}^2$ has cd $1$ resp. $2$, the LHS spectral sequence $E_2^{ij} = H^i(\mathbb{Z}^2;H^j(\mathbb{Z};M))$ shows $cd(\Gamma) \le 3$. Moreover, by positional reasons $E_\infty^{2,1}=E_2^{2,1}$ and in particular $E_\infty^{2,1} = \mathbb{Z}$ for $M = \mathbb{Z}$. Hence $cd(\Gamma) = 3$. -Claim 2: The inflation map $\text{inf}: H^2(\mathbb{Z}^2;M) \to H^2(\Gamma;M)$ is zero. -Since the image of inflation is just $E_\infty^{2,0} = \text{coker}(d_2^{0,1})$, it is sufficient to show that $d_2^{0,1}$ is surjective. Since the action of $\Gamma$ on $M$ is induced by the action of $\mathbb{Z}^2$, it follows that $\mathbb{Z}$ acts trivially on $M$. Futhermore, $\mathbb{Z}^2$ acts trivially on $H^\ast(\mathbb{Z};M)$ because $\mathbb{Z}$ is central. Therefore $E_2^{0,1} = Hom(\mathbb{Z},M)$. -Let $\alpha \in Hom(\mathbb{Z},M)$. Then -$$d_2^{0,1}: Hom(\mathbb{Z},M) \to H^2(\mathbb{Z}^2;M)$$ -is given by $d_2(\alpha)= - \alpha^\ast(\epsilon)$ (well-known formula) where -$$\alpha^\ast: H^2(\mathbb{Z}^2;\mathbb{Z}) \to H^2(\mathbb{Z}^2;M)$$ is induced by $\alpha$ on the coefficients. By using a projective resolution or by Poincare duality one easily sees that -$$\alpha^\ast : \mathbb{Z} \to M/\lbrace gm-m \mid g \in \mathbb{Z}^2 \rbrace =: \bar{M}$$ -is just $\alpha$ composed with the natural projection. Identifying $Hom(\mathbb{Z},M) = M$ now shows that $d_2^{0,1}: M \to \bar{M}$ is the natural projection and hence surjective.<|endoftext|> -TITLE: Representations of GL(2, Q_p) and GL(2, Z_p) -QUESTION [5 upvotes]: The cuspidal representations of $GL_n(F)$ a non archimedean field $F$ with ring of integers $o$ can be classified by inducing irreducible representation from $Z GL_n(o)$. -The general question: - -How does the representation theory of $GL_2(F)$ and the representation theory of $GL_2(o)$ affect each other? - -Uri Onn has shown that the irreducible representations of $GL(2,o)$ will depend only upon the residue characteristic. - -Does the representation theory of $GL_n(F)$ only depend upon the residue characteristic? Is there a connection to the reduction steps from local field of characteristic zero to fields of characteristic $p$ (e.g. as in the context of the Fundamental lemma)? - -In Stasinki "Smooth Representations of $GL(2,o)$" the representations are partionated in to similarity classes of $2 \times 2$ matrices over the residue field (pg.4419) via applying Clifford theory. - -What irreducible representations in Stasinski are important for describing the cuspidal representation of $GL(2, F)$? - -A little more general, I know that the irreducible dual of $GL_n(o)$ is far from being classified, so to what extent is the irreducible dual of $GL(n,F)$ described in Bushnell-Kutzo "The admissible dual of GL(N) via compact open subgroups". - -Is the classification of the irreducible dual of $GL(n,F)$ known only modulo the representation theory $GL(n, o)$ or do we know all the representations of $GL(n,o)$ needed for the dual of $GL(n,F)$? - -If the description of the dual of $GL(n,F)$ is possible independently of the dual of $GL_n(o)$, there is a natural last question in the context of Silberger "$PGL_2$ over the $p$ adics", where he classifies the irreducibe of $PGL_2(o)$ via the restricting the irreducible of $PGL_2(F)$. - -Can we classify the representation of $GL_n(o)$ by restricting the irreducible representation of $GL_n(F)$? - -REPLY [2 votes]: Some of your questions have several non-equivalent interpretations so it's not so easy to give precise answers, but I will try to give at least partial answers to two of your questions. -How does the representation theory of $GL_2(F)$ and the representation theory of $GL_2(o)$ affect each other? -One possible answer is that every supercuspidal representation of $GL_2(F)$ has a unique type on $GL_2(o)$ (this is due to Henniart for $GL_2$ and Paskunas for $GL_n$). So there is an interesting one-to-one correspondence between supercuspidals and certain representations of $GL_2(o)$. -What irreducible representations of $GL_2(o)$ are important for describing the cuspidal representations of GL(2,F)? -In light of the above connection between supercuspidals and types on $GL_2(o)$, at the very least the representations of $GL_2(o)$ which are supercuspidal types are important. I think these are the ones which are cuspidal or which have an orbit with Eisenstein polynomial. -Note that this does not say anything about non-supercuspidal representations and their types on $GL_2(o)$ so is certainly not an exhaustive answer.<|endoftext|> -TITLE: Estimating the distance from a variety by evaluating the polynomial -QUESTION [9 upvotes]: I have the conjecture stated below, which would be very helpful for a cryptographic security proof I'm working on for several months now. At first glance, it seemed a pretty natural question to me, but after asking a lot of people and also searching the internet I'm still without a solution or any helpful hints. This is the conjecture: -Let any multivariate polynomial $f\in\mathbb R[X_1,\ldots,X_n]$ be given, such that the corresponding variety $V_f$ is not empty: $V_f=\{x\in\mathbb R^n\,|\,f(x)=0\}\neq\emptyset$. Then there exists some constant $c>0$, such that for all $x\in\mathbb R^n$ one can estimate the distance of $x$ from $V$ as follows: $c\cdot\min_{y\in V_f}\|x-y\|^{\deg f}\leq|f(x)|$. -In particular, I only need this to hold true if $\deg(f)\leq4$ and $x$ is only taken from a compact set $K\subset\mathbb R^n$. If the conjecture turns out true, it can be used for generalization of a classification result of mine in theoretic cryptography. Here are some further notes: - -Since all norms on $\mathbb R^n$ are equivalent, the assertion either holds true for every norm or for none. -If $K\cap V_f=\emptyset$, one can just set $c:=\min_{x\in K,y\in V_f}|f(x)|\cdot\|x-y\|^{-\deg f}$. So, the "interesting" values $x$ are close to $V$. -If the conjecture is not true, then there exists a counterexample with irreducible polynomial $f$. Given any reducible counterexample, say $f=g\cdot h$, we can estimate $\min_{y\in V_f}\|x-y\|^{\deg f}$ by $\min_{y\in V_g}\|x-y\|^{\deg g}\cdot\min_{y\in V_h}\|x-y\|^{\deg h}$ and it follows straightforwardly that $g$ or $h$ must be a counterexample, too. -If $n=1$, i.e. $f$ is univariate, the conjecture is true, since the only irreducible polynomials with non-empty variety are of the form $f(x)=ax+b$. Clearly, these are no counterexamples and thus, as mentioned above, there don't exist any univariate counterexamples at all. -I was able to prove the weak variant of the conjecture with $x\in K$ for the case that $\deg(f)\leq2$, but the proof is quite long; it can be found here. - -REPLY [7 votes]: You want this: http://en.wikipedia.org/wiki/%C5%81ojasiewicz_inequality -The exponent might not be the degree, though.<|endoftext|> -TITLE: Exceptional points for generalized north-eastern knight walks in a quarter plane -QUESTION [11 upvotes]: Given two coprime integers $a < b$ of different parities, only a finite number of -points in $\mathbb N^2$ cannot be reached by a walk in $\mathbb N^2$, starting at the origin -and using only steps of the form $(b,\pm a),(\pm a,b)$ (and thus making an acute angle -with the north-eastern vector $(1,1)$). -Is there a good upper bound on the number of such exceptional points? Is there a good -upper bound on the coordinate sum $x+y$ of such an exceptional point $(x,y)$? -(Remark: A naive proof that almost all points can be reached gives an upper bound which is probably very far from the true value.) -Added in order to comply with the request made by user 9072: -Proof that almost all points can be reached: We consider $v=a(b,a)+b(-a,b)=(0,a^2+b^2)$ and $w=(a,b)+(-a,b)=(0,2b)$. -Since $a,b$ are coprime and have different parities, $a^2+b^2$ and $2b$ are coprime. The set -$$\{k(a^2+b^2),k=0,\ldots,2b-1\}$$ -with largest element $(2b-1)(a^2+b^2)<4b^3$ contains thus representants of all classes of $\mathbb Z/(2b)\mathbb Z$. This shows that all elements $(0,y)$ with $y\geq 4b^3$ are of the form $\mathbb N v+\mathbb N w$. Using the obvious symmetry with respect to coordinates, the same holds for $(x,0)$ with $x\geq 4b^3$. We get thus everything -in $4b^3(1,1)+\mathbb N^2$. All non-reachable points are thus contained in the two strips of width $4b^3-1$ -parallel to coordinate axes. Let $(x,y)$ -with $x=\min(x,y)<4b^3$ be a point of the vertical strip. If -$y\geq \left\lceil\frac{4b^3}{a}\right\rceil b+4b^3$, then -$$(x,y)\in \mathbb N(-a,b)+(4b^3)(1,1)+\mathbb N^2$$ and we are done. -This gives the very crude upper bound $64b^7$ on the number of non-reachable points in $\mathbb N^2$. -Some computational data: -For $a=1$ and $b\in\{2,4,6,\ldots,16\}$ the number of missing elements is given by $$\frac{1}{12}(4b^5-3b^4+8b^3)$$ -and the largest coordinate occuring in missing points is given by $b^3-b$. -For $a=2$ and $b\in\{3,5,7,9,\ldots,17\}$ the number of missing elements is given by $$\frac{1}{12}(2b^5+3b^4-2b^3-24b^2+12b+9)$$ -and the largest coordinate occuring in missing points is given by -$$\frac{1}{2}(b^3+2b^2-9b)$$ -if $b>3$ and by $10$ for $b=3$ (the polynomial formula yields the wrong value $9$). -For $a=3$ and $b\in\{4,8,10,14,16,20,22,26\}$ the number of missing elements is given by -$$\begin{cases}\frac{1}{36}(4b^5+15b^4-8b^3-240b^2+160b+384)\qquad&b\equiv 1\pmod 3,\\ - \frac{1}{36}(4b^5+15b^4-8b^3-240b^2+184b+360)&b\equiv 2\pmod 3\end{cases}$$ -and the largest coordinate occuring in missing points is given by -$$\begin{cases}\frac{1}{3}(b^3+4b^2-17b)\qquad&b\equiv 1\pmod 3,\\ - \frac{1}{3}(b^3+4b^2-15b)&b\equiv 2\pmod 3.\end{cases}$$ -For $a=4$ and $b\in\{5,7,9,\ldots,21\}$ the number of missing elements is given by -$$\begin{cases}\frac{1}{12}(b^5+6b^4-b^3-162b^2+120b+540)\qquad&b\equiv 1\pmod 4,\\ - \frac{1}{12}(b^5+6b^4-b^3-162b^2+132b+528)&b\equiv 3\pmod 4\end{cases}$$ -and the largest coordinate occuring in missing points is given by -$$\frac{1}{4}(b^3+6b^2-27b).$$ -There seems to be some structure in these data: Fixing $a$, in all examples the number of missing elements seems to be of the form -$$\frac{1}{3a}b^5+\frac{3a-4}{4a}b^4+O(b^3)$$ -and the largest coordinate occuring in missing elements seems to be of the form -$$\frac{1}{a}b^3+\frac{2(a-1)}{a}b^2+O(b)\ .$$ - -REPLY [11 votes]: I would like to see a solution. Below are a few comments which were obvious once I thought of them. I guess they do allow a naive argument that the number of exceptional points is finite, but I wonder if the OP has a nicer one. -I'll start with the (to me) -non-obvious observation: It appears that the exceptional points are all nicely contained in a non-convex quadrilateral with two sides on the axes (of course) and the other two lines of slopes $-b/a$ and $-a/b$. For example $a=4,b=7$ gives 2032 exceptional points. We can reach all points $(0,z)$ and $(z,0).$ for $z \ge 110$ as well as for for $42$ smaller values of $z$ -0, 14, 28, 41, 42, 47, 49, 55, 56, 57, 61, 63, 65, 69, 70, 71, 74, 75, 77, 79, 80, 82, 83, 84, 85, 88, 89, 90, 91, 93, 94, 96, 97, 98, 99, 102, 103, 104, 105, 106, 107, 108 -Hence there are 68 exceptional points on each axis. -The points $(3+4j,112-7j)$ and $(112-7j,3+4j)$ for $0 \le j \le 16$ are also exceptional and they lie on two lines which cross between $(43,44)$ and $(44,43)$. Here are the $\mathbf{2032}$ exceptional points: - -more obvious observations -Recall that given positive co-prime integers $u \lt v$, the set $S=\{su+tv \mid s,t \ge 0\}$ contains exactly one member of each pair $\{m,uv-u-v+m\}$ so neither $uv-2u-v$ nor $uv-u-v$ is in $S$ but everything between them or larger is (because $0$ and $u$ are in $S$ but anything between them or smaller is not). - -Let the rank of point $(x,y)$ be $x+y$ so that there are $r+1$ points of rank $r$ in $\mathbb N^2.$ Then each move increases the rank by either $b+a$ or $b-a$. Hence half the ranks up to $(b+a)(b-a)-(b+a)-(b-a)=b^2-2b+a^2$ are completely empty. This allows some lower bound. Those comments stay true even if we are allowed to venture outside $\mathbb N^2$ and then come back in. So it might be worthwhile bounding the number of exceptional points in $\mathbb N^2$ under those relaxed rules (which allow us to reach the otherwise exceptional points such as $(b-a,b-a)$ ). -Under the strict rules we may assume that we reach the non-exceptional points by first using only the two vectors $(a,b)$ and $(b,a)$ followed by later use of only $(-a,b)$ and $(b,-a).$ -Using just the vectors $(a,b)$ and $(b,a)$ gives only points which both have a rank $k(a+b)$ and are inside the cone $C$ between the lines $ax=by$ and $bx=ay.$ The first time that the entire rank is filled (using only those two vectors) is for $k_0=s+t$ obtained from the unique positive solution to $sa+tv=(a-1)(b-1)$. So $k_0 \le b-1-\frac {b-1}a.$ -Once we have a full rank between $(ka,kb)$ to $(kb,ka)$ for each $k \ge k_0$ (if not sooner) we can shift using the other two vectors to get full ranks from $(ka+jb,kb-ja)$ to $(kb-ja,ka+jb)$ (choosing ($1 \le j \le kb/a$). With a bit of work we could describe an $r_0$ so that we are sure to have every rank for $r \ge r_0$ full from $(r,0)$ to $(0,r).$<|endoftext|> -TITLE: When is a cube of cofibrations are "lattice"? -QUESTION [6 upvotes]: Let $C$ be a category with cofibrations in the sense of (Waldhausen, Algebraic K-Theory of Spaces) and denote by $F_n(C)$ the category with cofibrations consisting of sequences of $n$ cofibrations $A_0 \rightarrowtail A_1 \rightarrowtail \dotsc \rightarrowtail A_n$ in $C$. A cofibration in $F_n(C)$ is a commutative ladder consisting of "lattices". See the attached PDF for a precise definition. -Waldhausen gives a beautiful graphical proof that there is an equivalence $F_n F_m C \cong F_m F_n C$ (Lemma 1.1.5). The reason is, basically, that an object in both categories is given by a rectangular array of cofibrations, such that each square is a lattice, which is a symmetric condition. Now, Waldhausen only indicates why this equivalence $F_n F_m \cong F_m F_n$ also preserves cofibrations, and is therefore exact. Namely, he claims that a cofibration in $F_n F_m C$ is a $3$-dimensional diagram satisfying similarly some symmetric condition, but he does not name it explicitly. So what is this condition? So basically the question is: When should we call a commutative cube of cofibrations a lattice? -The full question contains many diagrams (also a 3-dimensional one), which cannot be properly displayed with MO-Latex. Therefore I've decided to write the full question, also with some of my attempts so far, as a PDF. I hope that this is not unappropriate for MO. -EDIT: Eva Höning pointed out to me that a cube of fibrations is a lattice if every one of the six squares is a lattice, and the map from the pushout of the cube without its tip to the tip is a cofibration. And it is rather straight forward to see that both $F_n F_m$ and $F_m F_n$ have these cubes as cofibrations. In the first place I was not careful enough with the definitions to see that, but it is really easy. - -REPLY [8 votes]: The notion you are looking for is well-known in homotopy theory under the name Reedy cofibration, but for some reason this name doesn't show up in papers about Waldhausen categories, even though the concept is used all the time. -To keep things close to your question let's say that $J$ is a finite poset (in general it can be any direct category). For a diagram $X : J \to \mathcal{C}$ and $j \in J$ we define a latching object $L_j X$ as the colimit of the restriction of $X$ to the subposet $\lbrace i \in J \mid i < j \rbrace$. If $L_j X$ exists, then it comes with a canonical map $L_j X \to X_j$. We say that a diagram $X$ is Reedy cofibrant if all $L_j X$s exist and the canonical maps $L_j X \to X_j$ are cofibrations. More generally, a map of Reedy cofibrant diagrams $X \to Y$ is a Reedy cofibration if all the induced maps $X_j \sqcup_{L_j X} L_j Y \to Y_j$ are cofibrations. -You can easily verify that if $\mathcal{C}$ is a category with cofibrations, then so is the category of Reedy cofibrant diagrams $J \to \mathcal{C}$ (and Reedy cofibrations as cofibrations). The same holds for Waldhausen categories. -Now, $F_n \mathcal{C}$ is nothing else but the category of Reedy cofibrant diagrams $[n] \to \mathcal{C}$ and you can verify that both categories $F_n F_m \mathcal{C}$ and $F_m F_n \mathcal{C}$ can be identified with the category of Reedy cofibrant diagrams $[m] \times [n] \to \mathcal{C}$ (as categories with cofibrations), which fills in the gap in the proof. -Edit: I realized that to answer all your concerns raised in the comments, I would have to go through the basic theory of Reedy cofibrations and an MO answer is not a good place for this. Instead I will try to indicate what needs to be done and point you to a reference (unfortunately, I don't know a reference that discusses exactly what you need). -Initially, I thought that I would simplify things by restricting to finite posets, but this only resulted in obscuring an important part of the story. So let's go to general direct categories (by the way, "direct" is very different from "directed"). A small category $J$ is direct if there is a functor $\mathrm{deg} : J \to \mathbb{N}$ such that for any non-identity morphism $i \to j$ in $J$ we have $\mathrm{deg} i < \mathrm{deg} j$. Everything I said above goes through, you just have to modify the definition of latching objects, $L_j X$ is defined to be the colimit over $\partial J \downarrow j$ i.e. the full subcategory of the slice $J \downarrow j$ spanned by non-identity morphisms. -The key fact is that the colimit of a Reedy cofibrant diagram always exists, the proof can be found in http://arxiv.org/abs/math/0610009v4 (Theorem 9.3.5). If you go through the proof you will see that this means in particular that if you know that some diagram is Reedy cofibrant below certain degree $m$, then the latching objects in degree $m$ exist. This should help you with seeing why $L_j (X_{i,\bullet})$ from your comment exists. -If $I$ and $J$ are direct, then so is $I \times J$ (just take the sum of degrees). Using the remarks above you should be able to verify that a diagram $I \to \mathcal{C}^J$ is Reedy cofibrant if and only if the corresponding diagram $I \times J \to \mathcal{C}$ is Reedy cofibrant. Moreover a morphism $X \to Y$ in $\mathcal{C}^J$ is a Reedy cofibration between Reedy cofibrant diagrams if and only if the corresponding diagram in $\mathcal{C}^{J \times [1]}$ is Reedy cofibrant. Those things are hopefully sufficient to answer your questions.<|endoftext|> -TITLE: The K-theoretic Farrell-Jones conjecture for cat(0) groups -QUESTION [8 upvotes]: Is the fibered K-theoretic farrell-jones conjecture true for cat(0)-groups? - -REPLY [14 votes]: Yes. More precisely, both the L-theory and K-theory Farrell-Jones conjectures with coefficients in any additive category have recently been proved for CAT(0) groups: - -Arthur Bartels, Wolfgang Lück, “The Borel Conjecture for hyperbolic and CAT(0)-groups”, Annals of Mathematics 175 (2012), 631–689, http://dx.doi.org/10.4007/annals.2012.175.2.5 -Christian Wegner, “The K-theoretic Farrell-Jones conjecture for CAT(0)-groups”, http://arxiv.org/abs/1012.3349, to appear in Proceedings of the AMS. - -In Arthur Bartels, Holger Reich, “Coefficients for the Farrell-Jones Conjecture”, Advances in Mathematics 209:1 (2007), 337–362, it is proved that “with coefficients is stronger than fibered,” see Corollary 4.3 and Remark 4.4 on page 346. -[Edited to clarify the relation between the “with coefficients” and “fibered” versions of the Farrell-Jones conjectures.]<|endoftext|> -TITLE: Smooth variety contained in another smooth variety -QUESTION [11 upvotes]: I apologize if this is a trivial question. If $X$ is a smooth irreducible codimension two subvariety of projective space $\mathbb P^n$, then does there always exist a smooth irreducible codimension one subvariety $Y \subset \mathbb P^n$ with $X \subset Y$ ? - -REPLY [15 votes]: For $n = 3$ this is always true. It is also true when $X$ is a complete intersection. -Suppose $n ≥ 4$, and suppose that $Y$ exists; then by a well known result of Lefschetz $X$ is a hyperplane section of $Y$, and so $X$ is a complete intersection. Now, for $m = 4$ there are subvarieties of $\mathbb P^4$ that are not complete intersections (for example, the image of a generic projection $\mathbb P^2 \to \mathbb P^4$ of a quadratic Veronese embedding of $\mathbb P^2 \subseteq \mathbb P^5$), so the answer is negative for these. For $m ≥ 5$ the existence of codimension 2 subvarieties that are not complete intersections is a big open question; for $m ≥ 7$ it is a particular case of a conjecture of Hartshorne that these should not exist.<|endoftext|> -TITLE: question about derived subgroup -QUESTION [8 upvotes]: Let $G$ be a free group. Then $G/G^{(n)}$ ($G^{(n)}$ is the $n$th derived subgroup.) acts on $G^{(n)}/G^{(n+1)}$ by conjugation, which makes $G^{(n)}/G^{(n+1)}$ a $\mathbb{Z}[G/G^{(n)}]$-module. What can I say about this module? Namely, I wonder whether $G^{(n)}/G^{(n+1)}$ is a free $\mathbb{Z}[G/G^{(n)}]$-module. If I want to study about this subject, what would be a good reference? - -REPLY [2 votes]: $G/G^{(n)}$ is the free solvable group of class $n$. The module mentioned in your question is projective (a submodule of a free module of rank = rank of $G$). -You can read about free solvable groups in Hanna Neumann's "Varieties of groups", and in Karras-Magnus-Solitar "Combinatorial group theory". Among newer papers, see for example Romanovskiĭ, N. S. -Coproducts of rigid groups. Algebra Logika 49 (2010), no. 6, 803--818, 845--846, 848; translation in -Algebra Logic 49 (2011), no. 6, 539–550 and references there. See also Lichtman, A. I. -Automorphism groups of free soluble groups. J. Algebra 174 (1995), no. 1, 132–149. See also this paper, especially p. 8 where they define Magnus embeddings and Fox derivatives.<|endoftext|> -TITLE: Triviality of direct multiples of flat complex vector bundles -QUESTION [8 upvotes]: Atiyah Patodi and Singer [Spectral asymmetry and Riemannian geometry III] write that if $E$ is a complex flat bundle (non holomorphic, just smooth and complex) on a compact manifold $X$ (more generally a CW-complex) there must be some positive integer $m$ such that the direct sum -$E\oplus E\oplus\cdots\oplus E$ ($m$ times) is a trivial bundle. More precisely, they write on top of page 19, "The vector bundle $V_a$ defined by $a$ is flat so its real Chern classes -vanish, hence some multiple $kV_a$ is (unitarily) trivial". -Of course since the Chern character is an isomorphism from $K^\bullet(X)\otimes \mathbb{Q}$ to $H^{2\bullet}(X;\mathbb{Q})$, we know the bundle $E$ (or, more precely its K-theory class) is torsion in topological K-theory but this only tells us that $E\oplus E\oplus\cdots\oplus E$ is stably trivial. -I checked an impressive amount of literature on characteristic classes without finding a clue. -Several authors cite directly Atiyah who, however, does not prove the claim. -Other authors say that the above statement cannot be true. -Does someone know the answer ? - -REPLY [5 votes]: As you already said, if $E$ is flat of rank $r$, then $ch(E)=0$, meaning that the class $[E] \in K^0 (X)$ is torsion and therefore $m[E]=0 \in K^0(X)$ for some $m$. Pick $m$ such that $mr >> dim X$. -Now the stabilization map $BU(n) \to BU$ is roughly $2n$-connected. This means that if $V \to X$ a vector bundle of sufficiently high rank then $V$ is trivial iff it is stably trivial. Apply this observation to $E^{ \oplus m}$. Done.<|endoftext|> -TITLE: My first question - on Affine Schemes in Algebraic Geometry -QUESTION [11 upvotes]: If R is a commutative ring (with unit) then we have an affine scheme Spec(R) which is an object of the category of ringed topological spaces. Is there any way of characterising this object relative to the category of ringed topological spaces? The underlying space of an affine scheme is compact and the structure sheaf is a ring, but these statements hardly go any way towards characterising an affine scheme. I am not looking for an answer that is necessarily strictly tied to the structure of the category of ringed topological spaces - just something that is topological and/or about the algebraic structure of the structure sheaf. -A non-answer is: 'An affine scheme is a ringed topological space of the form SpecR for some cummutative ring R.' -Thanks for any pointers, Christopher - -REPLY [3 votes]: Maybe you will be interested in the thesis of Mel Hochster. He characterizes the image of Spec in Top. The article based on it is called Prime Ideal Structures in Commutative Rings and appeared in the Transactions of the AMS in 1969.<|endoftext|> -TITLE: the dual space of C(X) (X is noncompact metric space) -QUESTION [21 upvotes]: It is well known that when $X$ is a compact space (or locally compact space), the dual space of $C(X)=\{f |f: -X\rightarrow \mathbb{C} \text{ is continuous and bounded} \}$ is $M(X)$, the space of Radon measures with bounded variation. -However, according to my knowledge, there are few books which discuss the case when X is noncompact, for example a complete separable metric space. -Even for the simplest example, when taking $X=\mathbb{R}$, what does $(C(X))^{*}$ mean? -Any advice and reference will be much appreciated. - -REPLY [11 votes]: The problem of obtaining a useful generalisation of the Riesz representation theorem for non-compact spaces was addressed in the 50's by R.C. Buck, amongst others. It was clear that it was necessary to leave the context of Banach spaces for a nice theory. Buck introduced the so-called strict topology on the space of bounded, continuous functions on a locally compact space and showed that the dual is the space of bounded Radon measurs on the underlying space. This was generalised to the case of completely regular spaces in the 60's using the theory of mixed topologies or Saks spaces which had been developed by the Polish school. The most succinct definition of the resulting topology on the above space is that it is the finest locally convex topology which agrees with compact convergence on bounded sets. There is a relatively complete theory---in particular, the Riesz representation theorem holds in its natural form.<|endoftext|> -TITLE: Pinball on the infinite plane -QUESTION [25 upvotes]: Imagine pinball on the infinite plane, with every lattice -point $\mathbb{Z}^2$ a point pin. -The ball has radius $r < \frac{1}{2}$. -It starts just touching the origin pin, and shoots off at angle $\theta$ -w.r.t. the horizontal. -It reflects from pins in the natural manner: -           - - -What happens? -More specifically, - -Q. - For a given $r$, what is the measure of the set of angles $\theta$ - for which the pinball remains a finite distance from the origin - forever? - -Many other questions suggest themselves, but let me leave it at that basic question for now. -This problem seems superficially similar to -Polya's Orchard problem (e.g., explored in the MO -question, -"Efficient visibility blocker -s in Polya’s orchard problem"), -but the reflections produce complex interactions. -It is also similar to Pach's enchanted forest problem, mentioned -in the MO question, -"Trapped rays bouncing between two convex bodies", -but it seems simpler than that unsolved problem, due to the lattice regularity. -Has it been considered before, in some guise? -If so, pointers would be appreciated. Thanks! -Addendum. -My question can be rephrased in terms of the "Sinai billiard," -as Anthony Quas explains: I am asking for the fate of radial rays in the situation illustrated: - -REPLY [5 votes]: This is an infinite horizon Lorentz gas. Ergodicity in the extended space has been shown for this model, but comparatively recently: D. Szasz and T. Varju, J. Stat. Phys. 129 59-80 (2007). But as noted already the set of initial conditions has zero measure, so this is not sufficient in itself. The initial conditions are a smooth one dimensional set in the full two dimensional collision space, so one would expect that if the set of orbits in the collision space with a bound $r$ has dimension $d(r)>1$, the desired set will have dimension -$d(r)-1$, approaching unity as $r\to\infty$.<|endoftext|> -TITLE: Faithful transitive actions by large groups on small sets -QUESTION [5 upvotes]: How large is the largest transitive subgroup of $S_n$ other than itself and $A_n$? In particular, does its size grow at least exponentially in $n$? - -REPLY [6 votes]: Re-edited to include Derek Holt's observation: This question seems to depend on the smallest prime divisor $p$ of $n.$ I am not sure how to proceed if $n =p.$ For example, if $n =p$ and (apart from $S_{p}$ and $A_{p}$) every doubly transitive permutation group of degree $p$ is solvable, then $S_p$ has no transitive subgroup of order greater than $p(p-1),$ other than $A_{p}$. Using the classification of finite simple groups, it seems likely that there are infinitely many such primes,(but this is not entirely straightforward, since it depends whether or not $p$ can be represented by certain cyclotomic polynomials). If, however, $n$ is not prime, then $p \leq \sqrt{n}$ and $S_n$ has the large transitive subgroup $ S_{\frac{n}{p}}\wr S_{p}$<|endoftext|> -TITLE: Uncountable family of infinite subsets with pairwise finite intersections -QUESTION [8 upvotes]: I am searching for a constructive proof of the following fact: If $X$ is an infinite set, there exists an uncountable family $(X_\alpha)_{\alpha \in A}$ of infinite subsets of $X$ such that $X_\alpha \cap X_\beta$ is finite whenever $\alpha \neq \beta$. The way I know how to prove this statement is as follows. -First, it suffices to prove the case when $X$ is countable. Thus we can choose a bijection between $X$ and $\mathbb{Q} \cap [0,1]$. To save notation we can tacitly assume that $X = \mathbb{Q} \cap [0,1]$. -Let the index set be $A = [0,1] \setminus X$, i.e. all the irrationals in $[0,1]$. For each $\alpha \in A$, choose a sequence $(x_{\alpha 1},x_{\alpha 2},\dots)$ of elements of $X$ such that $x_{\alpha n} \to \alpha$ as $n \to \infty$, and let $X_\alpha = \{ x_{\alpha_n} \mid n \in \mathbb{N} \}$. -Since $\alpha$ is irrational, the sequence $(x_{\alpha n})$ cannot be eventually constant, so $X_\alpha$ is infinite. And if $\alpha \neq \beta$ then the sequences $(x_{\alpha n})$ and $(x_{\beta n})$ can have only finitely many terms in common since they have different limits, so $X_\alpha \cap X_\beta$ is finite. -Is it possible to do this in a more constructive way? I know very little about set theory and logic, so I apologize if this question is too elementary. Also, I wasn't sure about any relevant tags other than set-theory, so please feel free to add appropriate tags. -Edit: to clarify, I didn't have a clear notion of what I meant by "constructive" here. What I didn't like about the proof I gave above was that it required a choice of sequence of rationals converging to each irrational. The answers so far all address this concern adequately. - -REPLY [12 votes]: Here is a transcription of Todd Eisworth's answer into constructive mathematics. His construction is more or less constructive, he just uneccessarily says things which ruin the constructivity such as: - -In constructive mathematics only decidable subsets of $\mathbb{N}$ have characteristic functions. But we can avoid the problem by speaking only about the characteristic functions. -Not every countable set can be put in bijective correspondence with the natural numbers. This can be avoided, because we don't need such a bijection, just an injection from the natural numbers into the set (which is the definition of "infinite"). -In constructive mathematics cardinals don't work too well, so it is better not to mention $2^{\aleph_0}$. This can be avoidded by talking explicitly about a set being uncountable (i.e., there is no surjection from $\mathbb{N}$ onto the set). - -And here is the proof: -The set $2^\ast$ of finite sequences of $0$'s and $1$'s is in bijective correspondence with $\mathbb{N}$, therefore it clearly suffices to find an uncountable collection of subsets of $2^\ast$ such that any two of them have only a finite intersection. Once such a collection is found, it can be embeded into $X$. -The set $2^\mathbb{N}$ is uncountable by the usual Cantor's diagonal argument (which is constructive!). Given any $f \in 2^\mathbb{N}$, let $S_f \subseteq 2^\ast$ be the set of finite prefixes of $f$. The assignment $f \mapsto S_f$ is injective, therefore the $S_f$'s form an uncountable family. If $f$ and $g$ are distinct, then there is a smallest $n \in \mathbb{N}$ such that $f(n) \neq g(n)$, hence $S_f \cap S_g$ is finite, as it contains precisely the first $n$ prefixes of $f$. - -REPLY [3 votes]: There are several constructions, some of which are quite nice and visual, and which you are therefore more likely to consider "constructive". These examples were shown to me by Imre Leader. - -Your example: take your infinite set to be $\mathbf{Q}$ and look at approximations to reals. -Take your infinite set to be the nodes of an infinite binary tree, and take your sets to be the paths from the root to infinity. Every two such paths will go separate ways after finitely many steps. -Take your infinite set to be the quarter-lattice $\mathbf{N}^2$, and take your sets, indexed by $\theta\in[0,\pi/2)$, to consist of the lattice points between $y=x \tan(\theta)$ and $y=x\tan(\theta) + 1$. Any two such slivers have finite intersection. - -Complementary question: Find an uncountable totally ordered chain of subsets of $\mathbf{N}$.<|endoftext|> -TITLE: Higher categories in logic -QUESTION [8 upvotes]: I've read somewhere (probably in the nlab) that higher category theory has application in logic. -By the way since now the only applications of higher category theory I've seen are in homotopy theory and mathematical physics, so I was wondering if anyone could give me some example of applications of higher categorical methods in logic (any reference would be appreciated). - -REPLY [11 votes]: To elaborate on Peter Arndt's answer a bit: indeed, considering terms as 1-cells and rewriting rules as 2-cells, you can indeed obtain a rather productive higher categorical view on various constructions of rewriting theory. Morally, this would mean that cofibrant replacements in various model categories (of modules, or associative algebras) can be described in a rather pleasant geometric way, - computing syzygies becomes building higher-dimensional cells of a cell complex from the low-dimensional ones. There are several good references to master this kind of viewpoint, best processed when viewed together. Those are the paper of Craig Squier "Word problems and a homological finiteness condition for monoids" from which all sort of started, when it comes to theoretical computer science, at least (this paper re-discovered in a somewhat different context results from a paper of David Anick "On the homology of associative algebras"), the paper of Jean-Louis Loday "Homotopical syzygies", and indeed, recent papers of the French computer science school, such as Yves Lafont, Francois Metayer, Yves Guiraud, Philippe Malbos and many others.<|endoftext|> -TITLE: When is a Topological pushout also a Smooth pushout? -QUESTION [6 upvotes]: I feel like this problem has not been solved, but I'm interested in knowing any results on it. More specifically, I mean: -Let $B\stackrel{f}{\leftarrow} A \stackrel{g}{\rightarrow} C$ be a diagram of Smooth manifolds and Smooth functions, and let $B\stackrel{k}{\rightarrow} D \stackrel{l}{\leftarrow} C$ be the Topological pushout (that is, $D$ obtains a Set structure as the quotient of $B \coprod C$ by the images of $f$ and $g$, and receives the weak Topology induced by $k$ and $l$). GOAL: Determine necessary and sufficient conditions for the Topological manifold $D$ to receive a natural Smooth manifold structure making the diagram a Smooth pushout. -To make it seem less daunting, break it into steps: -1) When is $D$ a Topological manifold (that is, inherits a Topological atlas from the diagram)? -2) When does $D$ inherit Smooth transition functions? (or whatever your favorite definition of "smooth structure" is) -2') If $D$ does not inherit a Smooth structure from the diagram, can we give it one? (I believe this is just Hirsch-Mazur) -3) Can Smooth functions be defined piecewise? -I'm certain that #1 is most difficult. Using $A\subset\mathbb{R}=B=C$ and $f,g$ inclusion maps, $A=0$ gives a point with 4 arms (not locally Euclidean) and $A=\mathbb{R}\setminus0$ gives the line with two origins (not Hausdorff). I find it difficult to find concise conditions that rule out those two situations. My only idea is to assume that everything is also cellular and try to come up with an obstruction class somehow (hence this question: When is a finite cw-complex a compact topological manifold?) -I have a few ideas and results for #2. For example, attaching manifolds along a submanifold (as defined in Kosinski's "Differential Manifolds") can be expressed with a smooth pushout diagram, which I proved by making a more general sufficient condition for "passing" a Smooth structure from two (or $n$) Smooth manifolds onto a Topological manifold. My intuition says that the criteria for #3 will be almost the same as for #2 -I'm interested in seeing what else is out there right now, and Google/arXiv searches aren't turning anything up. I'm half-expecting to see the usual "This is basically Surgery Theory" response.... - -REPLY [4 votes]: I think you should look at -MR2342857 (2008i:58016) -Pradines, Jean -In Ehresmann's footsteps: from group geometries to groupoid geometries. (English summary) Geometry and topology of manifolds, 87–157, -Banach Center Publ., 76, Polish Acad. Sci., Warsaw, 2007. -(arXiv:0711.1608 ) which discusses these issues. -Compare also with -arXiv:0807.1704 [pdf, ps, other] - Convenient Categories of Smooth Spaces - John C. Baez, Alexander E. Hoffnung<|endoftext|> -TITLE: Weak convergence implying norm convergence -QUESTION [5 upvotes]: A surprising (to me) consequence of Hahn-Banach is that when a sequence converges weakly then there is another sequence made of (finite) convex combination which converges in norm (to the same element). -In particular, this has a consequence in the context of $\ell^p$ spaces (the space of $p$-summable sequences, with $1 < p < \infty$). Assume there is a sequence of elements $\lbrace x_n \rbrace$ (with $x_n \in \ell^r$ for all $n$ and $1 < r < p$) which converges weakly (in the sense of $\ell^p$) to an element $x \in \ell^r$. Under the extra hypothesis that $\|x_n\|_{\ell^r} \leq K$ (for some $K \in \mathbb{R}$), the above theorem implies that there will be another sequence (given by finite convex combinations) which converges in $\ell^r$. -$\textbf{Question:}$ Suppose a sequence $\lbrace y_n \rbrace$ converges in $\ell^p$ norm to $y$, where $y_n$ and $y \in \ell^r$ (again with $1 < r < p < \infty$). Let $Y = \overline{ \textrm{Span} \lbrace y_n \rbrace }^{\ell^r}$ be the closure of the space generated by the $y_n$s in $\ell^r$. Does $y \in Y$? Since this is probably false (but I would be happy with a counter-example), are there conditions (weaker than boundedness of the sequence) which may allow to reach the same conclusion? - -REPLY [7 votes]: No. In $\ell_r \oplus_r \ell_r$ let $y_n = (z_n, x_n)$ where $x_n$ are disjoint, $\|x_n\|_r=4^n$, $\|x_n\|_p \to 0$, $\|z_n\|_r = 1$, and $\|z_n-y\|_r \to 0$ for some non zero $y$.<|endoftext|> -TITLE: Non-isomorphic groups with the same oriented Cayley graph -QUESTION [17 upvotes]: There are many examples of two non-isomorphic groups with the same Cayley graph. If the graph is non-oriented, asking for the generating set to be minimal does not make the task much harder. However, I was unable to answer the following -$\textbf{Question:}$ Say a set $S$ inside a group is "mag" if it is minimal (for the inclusion) among sets that generate the group ($\textit{i.e.}$ such that the words in $S \cup S^{-1}$ give the full group). Find two non-isomorphic groups $G$ and $H$ such that the oriented Cayley graphs of $G$ and $H$ with respect to $S \subset G$ and $T \subset H$ are isomorphic, given that $S$ and $T$ are "mag". -Apologies in advance if the answer is well-known... -PS: Here by Cayley graph of $G$ w.r.t. $S$, I mean the graph whose vertices are $G$ and $(g,h)$ is an edge if and only if $\exists s \in S, g =hs$. (There are two conventions, but it matters little for the question.) The point being that if $(g,h)$ is an edge then $(h,g)$ need not be one. -PPS: given minimality of $S$ and finiteness of the group, $S$ will usually be anti-symetric. -Here are examples. Take $G = \mathbb{Z}$ and $ H= (\mathbb{Z}/2 \mathbb{Z}) *(\mathbb{Z}/2 \mathbb{Z})$. For $H$, picking $ \lbrace a,b \rbrace$ the resulting Cayley graph is going to be infinite unoriented path ("two-way" infinite street). For $G$, you could pick $S =\lbrace 1\rbrace$ or $S = \lbrace -1\rbrace$ but not $S = \lbrace -1, 1\rbrace$. In either case, the resulting (oriented) Cayley graph is going to be a infinite unorientd path ("one-way" infinite street). -There might be two questions: for finite groups and infinite groups. I suspect one could find them more easily by looking at infinite groups. - -REPLY [5 votes]: If we consider the dihedral group of order 12, $G = \langle a, b \mid a^2 = b^2 = (ab)^6 = e \rangle$, then the Cayley graph corresponding to $\{ a, b \}$ is the cyclic graph on 12 vertices with edges labeled alternately by $a$ and $b$. We may then consider $H_1 = G \times \mathbb Z/2 \mathbb Z$ and $H_2 = G \rtimes \mathbb Z/2 \mathbb Z$ where the copies of $\mathbb Z/ 2 \mathbb Z$ are generated by $c$ and $d$ respectively, and where $d$ acts by $d a d = b$, $d b d = a$. Then the Cayley graph of $H_1$ with respect to $\{ a, b, c \}$ consists of two copies of cyclic graphs of order 12 with edges labeled by $c$ connecting the two graphs. Since $d$ just interchanges $a$ and $b$ we have that the Cayley graph of $H_2$ with respect to $\{ a, b, d \}$ can be obtained from the Cayley graph of $H_1$ by just relabeling the second copy of the cyclic graph, so that the two Cayley graphs will be isomorphic. -Unfortunately, the generating set for $H_2$ is not minimal since by the definition of $d$ we have $b \in \langle a, d \rangle$. This can be fixed however by taking two automorphisms $c$ and $d$ which are complicated enough so that this doesn't occur. -Specifically, we can let $c$ act on $G$ by $cac = ababa$, and $cbc = babab$, and we can let $d$ act on $G$ by applying $c$ and then interchanging $a$ and $b$, i.e., $dad = babab$, and $dbd = ababa$. It's not hard to see that these indeed define order two automorphisms of $G$, and for the same reason as above we have that the Cayley graphs of $H_1$ and $H_2$ will be isomorphic. -It is also not hard to check that we now have $| \langle a, c \rangle | = | \langle b, c \rangle | = 12 < 24$ and $| \langle a, d \rangle | = | \langle a, d \rangle | = 8 < 24$ so that the generating sets are now minimal. Moreover, the groups $H_1$ and $H_2$ will not be isomorphic, this can be seen for instance by counting the number of elements of order 2, (I counted 15 for $H_1$ and 9 for $H_2$, but I've omitted the tedious details).<|endoftext|> -TITLE: Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ? -QUESTION [52 upvotes]: The function $\Gamma(s)$ does not have zeros, but $\Gamma(s)\pm \Gamma(1-s)$ does. -Ignoring the real solutions for now and assuming $s \in \mathbb{C}$ then: -$\Gamma(s)-\Gamma(1-s)$ yields zeros at: -$\frac12 \pm 2.70269111740240387016556585336 i$ - $\frac12 \pm 5.05334476784919736779735104686 i$ - $\frac12 \pm 6.82188969510663531320292827393 i$ - $\frac12 \pm 8.37303293891455628139008877004 i$ - $\frac12 \pm 9.79770751746885191388078483695 i$ - $\frac12 \pm 11.1361746342106720656243966380 i$ - $\frac12 \pm 12.4106273718343980402685363665 i$ -$\dots$ -and -$\Gamma(s)+\Gamma(1-s)$ gives zeros at: -$\frac12 \pm 4.01094805906156869492043027819 i$ - $\frac12 \pm 5.97476992595365858561703252235 i$ - $\frac12 \pm 7.61704024553573658642606787126 i$ - $\frac12 \pm 9.09805003388841581320246381948 i$ - $\frac12 \pm 10.4760650707765536619292369200 i$ - $\frac12 \pm 11.7804020877663106830617193188 i$ - $\frac12 \pm 13.0283749883477570386353012761 i$ -$\dots$ -By multiplication, both functions can be combined into: $\Gamma(s)^2 - \Gamma(1-s)^2$ -After playing with the domain of $s$ and inspecting the associated 3D output charts, I now dare to conjecture that all 'complex' zeros of this function must have a real part of $\frac12$. -Has this been proven? If not, appreciate any thoughts on possible approaches. -Thanks! - -REPLY [7 votes]: Just to point out there are very good approximation to complex zeros off your line of -$$ \Gamma(s)-\Gamma(1-s) \qquad(1)$$ -At $\rho \approx -1.69711183621729718874218687438 - 0.305228379993226071272967719419 i$ (1) appears to vanish while $\Gamma(\rho) \approx 1.4648039 + 0.3642699441 i$ -Root finding with better precision converges to $\rho$ while (1) still appear to vanish in both sage and gp/pari (modulo bugs). -Checked to precision $5000$ digits and (1) still appears to vanish. -Here is $\rho$ with $100$ digits of precision: --1.697111836217297188742186874382163077146364585981726518217373889827452772242797069678994954785699956 - 0.3052283799932260712729677194188512919331197338088909477524842921187943642970297308885952936796125572*I - - -... for $ \Gamma(s)+\Gamma(1-s)$ approximation of zero appears $\rho \approx -0.60940537628997711023 - 0.82913081575572747216 i$ checked to $5000$ digits of precision. -With 100 digits: - -0.6094053762899771102337308158313839002012166649163876907688596366808893391382113824494098816671945331 - 0.8291308157557274721587141536678087800797120641344787653174391388417832472543392187032283839972409848*I - -Edit In comments juan suggested using x-ray to investigate the zeros. -The primary reference for x-ray I found is X-Ray of Riemann zeta-function, J. Arias-de-Reyna -AFAICT x-ray are the plots of Re(f(s))=0 and Im(f(s))=0. The zeros are the intersection. -The x-ray and juan's comments suggest the above quadruples of zeros are indeed zeros off $\frac12$ and possibly there are no more complex zeros zeros off the line. -Here is the x-ray of $ \Gamma(s)-\Gamma(1-s)$. Blue is $\Re(\Gamma(s)-\Gamma(1-s))=0$ and red is $\Im(\Gamma(s)-\Gamma(1-s))=0$. - -(source) - - -(source)<|endoftext|> -TITLE: The semidihedral group of order 16 and ko -QUESTION [12 upvotes]: Let $\mathcal{A}(1)$ denote the subalgebra of the $\mathrm{mod}\ 2$ Steenrod algebra generated by $\mathrm{Sq}^1$ and $\mathrm{Sq}^2$. -The cohomology with $\mathbf{F}_2$ coefficients of the semidihedral group -$$SD_{16} = \langle g, h \mid g^8 = h^2 = 1, hgh = g^3\rangle$$ -of order $16$ is isomorphic to $\mathrm{Ext}^\ast_{\mathcal{A}(1)}(\mathbf{F}_2, \mathbf{F}_2)$. Is there a topological explanation for this isomorphism? - -REPLY [11 votes]: Not a 'topological' explanation, but A(1) is the 8 dimensional member of the family of semi-dihedral algebras, whose members of dimension $2^n$ for $n > 3$ are the mod 2 group rings of the semidihedral group of that order. Their cohomology ring is insensitive to the differences; that difference is reflected in the order of the higher Bockstein from H^3 to H^4. -Crawley-Boevey completely analyzed the structure of the modules over these algebras in -Crawley-Boevey, W. W. Functorial filtrations. III. Semidihedral algebras. J. London Math. Soc. (2) 40 (1989), no. 1, 31–39. -where he applies his results from Functorial Filtrations I and II to determine all the representations of "semidihedral algebras" $A_m = k\langle a,b | a^3 = b^2 = 0, a^2 = (ba)^mb\rangle$ (CAVEAT: when $k$ is a field with more than 2 elements). -There is one of these of $k$-dimension $4n-1$ for each $n \geq 2$. -The 7 dimensional one is $A(1)$ modulo its socle ($a=Sq^2$ and $b=Sq^1$), and the $2^n-1$ dimensional one is -the group ring of the semidihedral group modulo its socle for larger $n$. -Quotient by the socle is sensible here, since $A(1)$ and group rings are Frobenius, so that free modules are injective, hence split off. Modules with no free summand are all pulled back from the quotient by the socle. -There is a more recent (and I believe simpler) account of the representation theory as well, but I am embarrassed to admit that I cannot remember the author. -Since Crawley-Boevey's work actually classifies the modules over $A(1) \otimes GF_4$, I have always thought it would be entertaining to do the Galois descent to $A(1)$.<|endoftext|> -TITLE: Generalizing the Fundamental Theorem of Symmetric Polynomials -QUESTION [10 upvotes]: The fundamental theorem of symmetric polynomials tells us that the ring $\mathbb{Z}[x_1,\ldots,x_n]^{S_n}$ of symmetric polynomials in $n$ variables is generated (without relations) by the elementary symmetric polynomials $e_i(\bar{x})$, for $i$ between $1$ and $n$. I'm looking for a reference in the literature for a similar theorem in more variables, which should look something like this: -Consider the action of $S_n$ on $\mathbb{Z}[x_1,\ldots,x_n,y_1,\ldots,y_n]=\mathbb{Z}[\bar{x},\bar{y}]$ given by permuting the $x_i$ and the $y_i$ simultaneously. The fixed subring $\mathbb{Z}[\bar{x},\bar{y}]^{S_n}$ is generated by the elementary symmetric polynomials $e_i(\bar{m})$, where $m=m(x,y)$ is a monomial. (For example, if $m(x,y)=x^2y$, then $e_1(\bar{m}) = x_1^2y_1 + x_2^2y_2 + \ldots$.) -As an example, consider the $S_2$-invariant polynomial $(x_1+y_1)(x_2+y_2)$. It can be written as $(x_1+x_2)(y_1+y_2) - (x_1 y_1 + x_2 y_2) + (x_1x_2) + (y_1y_2)$, i.e. $e_1(\bar x)e_1(\bar y) - e_1(\overline{xy}) + e_2(\bar{x}) + e_2(\bar y)$. -I'd also be interested to know what the relations are in such a presentation of $\mathbb{Z}[\bar{x},\bar{y}]^{S_n}$. Certainly we can do better by excluding the monomials $x^m$ and $y^m$ for $m\geq 2$, as each such $e_i(\bar x^m)$ is already covered by the ordinary fundamental theorem. There also seem to be a handful of other relations around $i=n$, such as the observation that $e_n(\overline{xy}) = e_n(\bar x)e_n(\bar y)$, and possibly others. - -REPLY [3 votes]: Another systematic study of multisymmetric polynomials is in the thesis of Emmanuel Briand -available in .ps format at: -http://emmanuel.jean.briand.free.fr/publications/polms/ -see also his other publications at: -http://emmanuel.jean.briand.free.fr/publications/ -Although not well-known the study of multisymmetric functions is very old and related -to the search for explicit formulas for multidimensional resultants. -Among the classics who worked on this: Poisson, Schlafli, Brill, Gordan, Junker. -A good account of some of this old work is in the book by Faa di Bruno -"Theorie general de l'elimination", p. 109--114 and 129--131. -Another little-known reference is the beginning of Lecture Notes in Math no. 896 by Bernard Angeniol -"Familles de Cycles Algebriques - Schema de Chow".<|endoftext|> -TITLE: tangent bundles of orbifolds -QUESTION [9 upvotes]: Hello, -I want to take a unit $n$-ball $B_n$ and quotient it by some subgroup $G \subset \mathbb{Z}_2^n$. -Here is a sketch of the construction: take the obvious inscribed $n$-hypercube of $B_n$; we can associate the hypercube's vertices with the $2^n$ bitstrings of length $n$, and take some sub-vector space (thought of as a subgroup) $G$ of $\mathbb{Z}_2^n$. We can then think of $G$ as identifying the vertices of the hypercube and it is fairly obvious that we can then quotient $B_n$ by $G$ to get some quotient space $B_n/G$. -Now, I'm interested in constructing a "tangent bundle" for this quotient which is not quite a manifold and then calculate / construct the Stiefel-Whitney classes for this bundle. I use quotes since I'm not sure about how to do this, or if it is well-defined. - -are the quotients $B_n/G$ for these "well-behaved looking" $G$ well-understood as topological spaces? In particular, do we know their cohomologies? I can see $H^1(B_n/G) = G$, but that's about it for now =) It looks like equivariant cohomology would come into play, but my topology skills are elementary and I don't have a feeling of what is "easy" or "hard" in equivariant cohomology. -what is the correct notion of "tangent bundle" for this kind of orbifold, if any? $B_n$ itself is a nice $n$-manifold with boundary, so it seems like I want to look at something like $TB_n/G \rightarrow B_n/G$, but I am in alien waters and I feel I should be careful about singularities. -A natural followup question: if so, we can formally define Stiefel-Whitney classes (and spin structures). Are its Stiefel-Whitney classes well-understood? - -Some googling has led me to two papers: - -http://arxiv.org/abs/math/0409136 -http://arxiv.org/abs/1004.1979 - -and I'm trying to read them now, but advice from experts would be great. =) - -REPLY [5 votes]: As far as (co)homology is concerned, you may replace the orbifold quotient of a manifold $M$ by a finite group $G$ by the so-called Borel construction $(M\times EG)/G$. The "tangent bundle" of the orbifold is then a usual vector bundle over the space $(M\times EG)/G$, and you can define all the usual characteristic classes. -I should say: -- $EG$ is a contractible $G$-space with free action. -- The total space of the "tangent bundle" is $(TM\times EG)/G$.<|endoftext|> -TITLE: Example of a manifold which is not a homogeneous space of any Lie group -QUESTION [74 upvotes]: Every manifold that I ever met in a differential geometry class was a homogeneous space: spheres, tori, Grassmannians, flag manifolds, Stiefel manifolds, etc. What is an example of a connected smooth manifold which is not a homogeneous space of any Lie group? -The only candidates for examples I can come up with are two-dimensional compact surfaces of genus at least two. They don't seem to be obviously homogeneous, but I don't know how to prove that they are not. And if there are two-dimensional examples then there should be tons of higher-dimensional ones. -The question can be trivially rephrased by asking for a manifold which does not carry a transitive action of a Lie group. Of course, the diffeomorphism group of a connected manifold acts transitively, but this is an infinite-dimensional group and so doesn't count as a Lie group for my purposes. -Orientable examples would be nice, but nonorientable would be ok too. - -REPLY [10 votes]: I would think that many examples from 3-manifold theory would work. Take any compact, oriented, irreducible 3-manifold $M$ whose torus decomposition is nontrivial and has at least one hyperbolic piece. Such examples at least have no locally homogeneous Riemannian metric, as a consequence of Thurston's analysis of the 8 geometries of 3-manifold theory. A specific example of this sort can be obtained from a hyperbolic knot complement in $S^3$ by deleting an open solid torus neighborhood of the knot and doubling across the resulting 2-torus boundary; the doubling torus produces a characteristic $Z^2$ subgroup of $\pi_1(M)$. These examples have universal cover homeomorphic to $R^3$, and so they have trivial $\pi_2$. By the homotopy exact sequence there would be a quotient group $\pi_1(G) / \pi_1(H)$ identified with a subgroup of $\pi_1(G/H)$ whose quotient set is $\pi_0(H)$. Perhaps, in order to get a proof, one can analyze this situation by considering the intersection of the $Z^2$ subgroup with the $\pi_1(G) / \pi_1(H)$ subgroup.<|endoftext|> -TITLE: Reference request: Dickman, On the frequency of numbers containing prime factors -QUESTION [5 upvotes]: I've been trying without success to find the paper -Dickman, Karl, "On the frequency of numbers containing prime factors of a certain relative magnitude." Ark Mal., Astronomi och Physik, 22A (10), 1930. I checked at Rice U. and it was not available. -Actually,I would be satisfied with an exposition of the information it contains, in some notes or another paper. This is not extremely important, I'm just curious to learn how the Dickman function is derived. -Does anyone have any ideas? -Thanks, -Tom - -REPLY [4 votes]: Answering from the far future! I present some literature on this topic: -1) K. Dickman in his original paper of 1930 gave an heuristic argument that can be found in pages 382-383 of The art of computer programming, volume 2 (third edition) by Knuth. -2) V. Ramaswami made the argument rigorous in his 1949 paper On the number of positive integers less than $x$ and free of prime divisors greater than $x^c$. -See also the review MR0031958 for links to "weaker" papers (according to the reviewer) of Chowla and Vijayaraghavan and of Buhštab published around the same time, plus further work of Ramaswami on the question. -3) The second part of the AMS memoir Numbers with small prime factors, and the least $k$th power non-residue by K.K. Norton (1971) surveys Dickman's function. -4) As already mentioned by Dimitris Koukoulopoulos, the paper Integers without large prime factors by Hildebrand and Tenenbaum (1993) is a survey which actualizes that of Norton. -5) Integers without large prime factors: from Ramanujan to de Brujin by P. Moree (2014) is a small recent survey highlighting the contributions of de Brujin to the question. -6) The very recent paper Exact asymptotics of positive solutions to Dickman equation by Diblík and Medina (2018) improves the well-known asymptotic formulas for Dickman's function. -7) Also very recent are tight estimates for effective counts of smooth numbers due to Lichtman and Pomerance (2018), published in Explicit estimates for the distribution of numbers free of large prime factors. -8) A modern textbook reference for Dickman's function theorem is Theorem 7.2 of Multiplicative number theory I: classical theory (2006) by Montgomery and Vaughan.<|endoftext|> -TITLE: Intuition behind generic points in a scheme -QUESTION [24 upvotes]: In a scheme, each point is a generic point of its closure. In particular each closed point is a generic point of itself (the set containing it only), but that's perhaps of little interest. A point that's not closed, is probably more interesting, and there are no such thing in ordinary varieties. -What I have been wondering about is that there must be a good reason they are called generic points. Here are what I have got so far: - -A non-closed generic point is not closed, so it cannot be cut out from the scheme by any polynomial equations in any affine patch, and thus it does not posses any extra algebraic property that's not shared by others. -A non-closed generic point is not a specialization of the scheme. - -Are these correct? If not, what's the right intuition? Also, can the following statement be make more precise using the language of generic points? - -A degree $n$ and a generic degree $m$ algebraic curves intersect at $n \cdot m$ distinct points in $\mathbb{P}^2$ (the planar Bezout's theorem) -Common solutions of $n$ polynomial systems in $n$ variables with generic complex coefficients in $\mathbb{C}^\ast$ are all isolated. (corollary of the Cheater's homotopy theorem) -The number of common isolated solutions of $n$ polynomial systems in $n$ variables with generic complex coefficients in $\mathbb{C}^\ast$ equals the mixed volume of the Newton polytopes of the system. (Bernshtein's theorem) -Generic points on a nonreduced scheme have isosingular structure (i.e., at all such points the local ring fail to be reduced in exactly the same way). - -In each case I am familiar with their original meaning of the word generic, but I'm wondering if we can state the genericity conditions using the concept of generic points of schemes. - -REPLY [6 votes]: My favorite view on generic points is found in Mumford's book: Complex Projective Varieties, on page 2. It goes as roughly as follows: -Definition: Let $k \subset \mathbb{C}$ be a subfield of the complex numbers and $V$ an affine complex variety. A point $x \in V$ is $k$-generic if every polynomial with values in $k$ that vanishes on $x$, vanishes on all of $V$. -Proposition: If $\mathbb{C}/k$ has infinite transcendental degree, then every variety $V$ has a $k$-generic point. -Proof: Extend $k$ by all coefficients of a finite set of equations for $V$. Note that $\mathbb{C}/k$ has still infinite transcendental degree. But now $V$ becomes a variety over $k$ in a canonical way. The function field $L$ of $V$ is an $k$-extension of finite transcendental degree, and therefore can be embedded into $\mathbb{C}$. The images of the coordinate function $X_i \in L$ in $\mathbb{C}$ give the coordinates of a $k$-generic point. -For the relation to the generic point $\eta \in V$ from scheme theory note the following: - -If we define a $k$-Zariski topology on $V(\mathbb{C})$ defined by polynomials over $k$, a $k$-generic point $x$ will be non-closed with closure $V$. -The field $K(x)/k$ generated by the coefficients of $x$, is canonically isomorphic to the function field $L=K(V)$, which is the residue field of the generic point $K(\eta)$. - -From the abstract perspective the function field $L=K(\eta)$ is as good as $K(x)$, if not better, because it does not depend on choices. Moreover, all $k$-linear algebraic operations cant tell a difference between $\eta$ and $x$.<|endoftext|> -TITLE: Meeting management -QUESTION [6 upvotes]: A friend wants to have ten meetings of six people every day for five days with no pair of people meeting twice. Is this possible? It appears to be a question about maximal decomposition of a complete graph on 60 vertices into sets of 10 disjoint complete graphs on 6 vertices such that no edge is used twice. Does anyone know a useful theorem or, better, a way of constructing an example? - -REPLY [4 votes]: Look at the edit history for an old idea (if you wish). -Here is a schedule which runs for 8 days. Each row is an agenda for a day. It consists of 60 digits in 6 groups of 10 (for readability). Note that I will usually count days, people and meetings each day starting from 0 ( except that the schedule below starts with day 1.) Day 2 begins 0123456777.8777 which tells us that people 7,8,9,11,12, and 13 make up meeting 7 that day. You may confirm that no two of those six people is ever in a meeting together on another day. -0001000234.5167513847.9497368284.1239662875.3198562431.7649587529 -0123456777.8777465293.8508631463.0984282401.9598350211.5946032961 -0123456465.2738999699.9630784282.4017578350.2115746032.7615803418 -0123456724.0185873502.1999499928.6157034172.6846528375.0763146308 -0123456157.4603276158.0341826799.9299950863.1463078428.2401757835 -0123456630.7842824017.5783502115.7460399919.9934182674.6527385086 -0123456276.1580341826.7465273850.8631463078.4299909997.8350211574 -0123456783.5021157460.3276158034.1826746527.3850863149.9979998240 -The method of generation is systematic (I thought it might work at least for 5 days and was pleased to get 8). I only sketch it here, still it took less time to do (with Maple) than it does to describe. -Begin with the field of $9$ elements which could be thought of as $\mathbb{Z}_3[i]$, the Gaussian integers $\mod{3}$. Code it somehow, I used $0 \rightarrow 0$ and $(1+i)^{j-1} \rightarrow j$ for $1 \le j \le 8$. -Using the method of Kevin Costello, one can construct a 9 day schedule of meetings for 63 people which has 9 meetings of 7 people each day (and no pair is ever in the same meeting twice.) What we will do is take 80 of those 81 meetings, reduce each one to six people and arrange the 80 meetings into eight days each with ten meetings. Call a meeting big if it has seven people and good once it is reduced to six people. -On day $0$, meeting $j$ has people $7j,7j+1, \cdots,7j+6$ so the ninth meeting (meeting $8$) is $56,57,58,59,60,61,62$ Go through days 1 through 8 and erase people 60,61,62 (note that no two of them are ever in the same meeting after day 0). At this stage each of those eight days has nine meetings, three are good and six are big. We wish to remove one person from each big meeting to make them good and use the removed people for a tenth good meeting. To have a chance of doing this appropriately, we look for a meeting from day 0 which has only one member already in a good meeting. If we manage to do this then the other six people are sure to be distributed one in each big meeting. Use them to make a tenth good meeting. -An example may help: Here is the schedule for day 4 after people $60,61,62$ are removed (note that I have sorted the people in each meeting and sorted the list of meetings according to least person in the meeting) -$\small [0, 10, 18, 26, 34, 50, 58], [1, 11, 20, 21, 31, 37, 54], [2, 8, 19, 28, 39, 45], -[3, 16, 27, 35, 47, 53, 57], [4, 9, 24, 36, 42, 55],$ $\small [5, 13, 17, 22, 32, 44, 49],[6, 25, 30, 40, 43, 52, 56], [7, 15, 33, 38, 48, 51], [12, 14, 23, 29, 41, 46, 59]$ -The eighteen people who are already in good meetings are: -2 4 * 7 8 9 * 15 19 *24 *28 33 *36 38 39 *42 45 48 51 55 so the only day 0 meeting which only includes one of them is [21,22,23,24,25,26,27]. Removing person 24 from that group and the other six people from their groups yelds -$\small [0, 10, 18, 34, 50, 58], [7, 15, 33, 38, 48, 51], [12, 14, 29, 41, 46, 59], - [1, 11, 20, 31, 37, 54], [2, 8, 19, 28, 39, 45],$ -$\small [3, 16, 35, 47, 53, 57], [4, 9, 24, 36, 42, 55], [5, 13, 17, 32, 44, 49], [6, 30, 40, 43, 52, 56], [21, 22, 23, 25, 26, 27]$ -0123456724.0185873502.1999499928.6157034172.6846528375.0763146308 is the resulting code appearing as line 4 above. The first meeting explains the digit 0 in positions 0,10,18,34,50,58. -As it turns out, each day from 1 to 8 is compatible with exactly one meeting from day 0 and vice versa, perhaps there is an explanation for this, but I am satisfied that it happened this once. -Reviewing this I see that I could have pushed the construction by Kevin to get $10$ days with $90$ meetings (I think.) It does not seem that this would allow 10 days of meetings with $60$ people but I have not carefully checked (nor do I plan to.)<|endoftext|> -TITLE: Sections measure zero imply set is measure zero? -QUESTION [6 upvotes]: I have a subset $B\subset\mathbb{R}^n\times\mathbb{R}^m$ that I want to show has measure zero. I know that the sections $B^x = \{y : (x,y)\in B\}$ all have measure zero. I do not know if $B$ is measurable. Is this enough to conclude that $B$ is a measurable set with measure zero? -EDIT : I would like to say that when randomly selecting $a\in\mathbb{R}^n\times\mathbb{R}^m$, I am almost always not in $B$. Since I know that $B^x = \{y : (x,y)\in B\}$ all have measure zero, it seems like a reasonable conclusion. I don't know enough probability or measure theory to put this in a rigorous way, so any suggestions would be great. - -REPLY [13 votes]: Since the Sierpinski article is in French an uses slightly old-fashioned notation, let me sketch a proof of the result. -Theorem. There is a function $f:\mathbb R\to\mathbb R$ whose graph is not a measurable subset of $\mathbb R^2$. -Proof. We first show that a set $A\subseteq\mathbb R$ of size $<2^{\aleph_0}$ cannot have a complement of measure zero (in fact, if such a set is measurable, then it is of measure zero, but we don't need this). -To see this, we enlarge $A$ so that it becomes a subgroup of $(\mathbb R,+)$ that is still of size $<2^{\aleph_0}$. -Now choose $x\in\mathbb R\setminus A$. -The coset $A+x$ is disjoint from $A$. Hence $\mathbb R$ is the union of $\mathbb R\setminus A$ and $\mathbb R\setminus(A+x)$. -It follows that not both of these sets can be of measure zero. -However, if $A$ is measurable, they have the same measure. So the complement of $A$ is not of measure zero. -Now let $(B_\alpha)_{\alpha<2^{\aleph_0}}$ be an enumeration of all Borel subsets of $\mathbb R^2$ of measure zero. -For each $\alpha<2^{\aleph_0}$ choose a pair $(x_\alpha,y_\alpha)\in\mathbb R^2$ such that -$x_\alpha\not\in\{x_\beta:\beta<\alpha\}$ and $(x_\alpha,y_\alpha)\not\in B_\alpha$. -This is possible since for each $\alpha$ the set $$((\mathbb R\setminus -\{x_\beta:\beta<\alpha\})\times\mathbb R)\setminus B_\alpha$$ is not of measure zero by the remark at the beginning of the proof. -Now $\{(x_\alpha,y_\alpha):\alpha<2^{\aleph_0}\}$ is a partial function from $\mathbb R$ to $\mathbb R$ that is not contained in any measure zero Borel subset of the plane. -It follows that this partial function is not of measure zero. -Extend the function to any total function $f:\mathbb R\to\mathbb R$. -The extended function still is not of measure zero. -But since the sections are singletons, the graph of this function cannot be measurable by Fubini's theorem. \qed<|endoftext|> -TITLE: indecomposable modules of the symmetric group -QUESTION [8 upvotes]: Quite simply: - -What is known about the indecomposable modules of the symmetric group $S_n$ in positive characteristic? - -Since the answer is likely to be "very little", perhaps I should rather ask - -What is known about the indecomposable modules of $S_4$ in characteristic $2$ ? - -References appreciated. -Thanks! -Pierre - -REPLY [10 votes]: Well you certainly asked the right question. If the Sylow subgroup is cyclic then there is a nice way to write down all the (finitely many) indecomposable modules, see for example the last chapter of Alperin's "Local Representation Theory" or 6.5 in Benson's "Representations and Cohomology." If $p$ is odd and one is not in the cyclic defect case then the representation type is wild, and classifying the indecomposable modules is hopeless in some precise sense that it is at least as difficult as for a polynomial algebra on two noncommuting variables. -In between these two is something called "tame" representation type, which happens for symmetric groups only for $p=2$ and $n=4, 5$, where there is a dihedral Sylow. (Or also for larger $n$ but but blocks of weight $2$). However by work of Jost it is known that all these blocks are Morita equivalent to either the principal block of $S_4$ or $S_5$. The main reference for this situation is Erdmann's LNM 1428 although there is a lot of work since. See also for example the paper "Representation type of Hecke algebras of type A" by Erdmann-Nakano that gives the quiver and relations for the principal blocks of $S_4$ and $S_5$ in characteristic two. -As an aside, even in the wild type case I find it an interesting question to ask if one can classify indecomposable, self-dual modules that have Specht filtrations.<|endoftext|> -TITLE: Are injective quasi-coherent modules acyclic? -QUESTION [6 upvotes]: Let $X$ be a scheme and $F$ be an injective object of $\mathrm{Qcoh}(X)$. Is it true that $F$ is acyclic with respect to the usual sheaf cohomology? -For noetherian schemes $X$ this is well-known; then $F$ even turns out to be flasque. I don't really care for pathological schemes, but I would like to know if it's true for quasi-compact quasi-separated schemes. -If $X$ is affine, then it is also well-known (no matter if $X$ is noetherian or not), because then actually every quasi-coherent module is acyclic. Remark that even on an affine scheme, whose underlying topological space is noetherian, there are injective quasi-coherent modules which are not flasque (SGA 6, Exp. II, App. I); but of course this does not influence the answer. -The background is that I would like to define cohomology within $\mathrm{Qcoh}(X)$, without using the category of (not necessarily quasi-coherent) $\mathcal{O}_X$-modules or even all sheaves on $X$. This works because $\mathrm{Qcoh}(X)$ is a Grothendieck category (without any assumptions on $X$), thus has enough injective objects. This cohomology would turn out to be the usual sheaf cohomology (i.e. computed in $\mathrm{Sh}(X)$) if and only if injective objects are acyclic with respect to the usual sheaf cohomology. -EDIT: a-fortiori answers the question affirmatively if $X$ is quasi-compact and semi-separated. Is there any chance to get the result also when $X$ is just assumed to be quasi-compact and quasi-separated? I've already convinced myself that the proof cannot be translated verbatim. - -REPLY [5 votes]: The case of quasi-compact semi-separated schemes is treated in the references given in the comments above.<|endoftext|> -TITLE: Non-residually finite matrix groups -QUESTION [20 upvotes]: By Malcev's theorem, every finitely generated linear group is residually finite (RF). -On the other hand, say, the group of rational numbers is linear, but is not residually finite. Thus, one has to impose some restrictions on linear groups to avoid this example. Discreteness sounds like a reasonable hypothesis. It is a nice and easy exercise to show that -every discrete subgroup of $PSL(2, {\mathbb R})$ is residually finite. -Question 1. (This question came from Fanny Kassel) Are there non-RF discrete subgroups of -$PSL(2, {\mathbb C})$? -On the other hand, almost surely, there are no simple discrete infinite subgroups $\Gamma$ of rank 1 Lie groups. (Take a high power of a hyperbolic element in $\Gamma$, then its normal closure in $\Gamma$ should have infinite index in $\Gamma$.) This argument fails however in the higher rank case because of the Margulis' normal subgroups theorem for higher rank lattices. -Question 2. Are there infinite simple discrete subgroups of $SL(n, {\mathbb R})$? -It is hard to imagine that such things could exist, but I see no way to rule them out... - -REPLY [2 votes]: The answer to question 2 is no [edit :] when the group is finitely generated. It follows from Malceev's result. In the case of finitely presented, here is a simple proof. -Proof : Let $G$ be a group defined by a finite number $n$ of generators and a finite number of relations. -Let $k$ be a field. Pick an element $g$ in $G$. -The set of morphisms $\varphi:G\to GL_m(k)$ such that $\varphi(g)\ne 1$ can be seen as an algebraic subvariety of $M_m(k)^n$. In fact this variety is defined over $\mathbf Z$. -So if this variety admits a point over a field $k$ of characteristic $0$, then it has to have points over infinitely many finite fields. -Thus if the group is simple and infinite, the variety is empty.<|endoftext|> -TITLE: Do disjoint unions and fiber products commute? -QUESTION [11 upvotes]: Do disjoint unions and fiber products commute? -In other words, is the following statement true? -Statement: Let $C$ be a category with (infinite) coproducts and fiber products. Let {$U_{i}$} be a family of objects in $C$, and denote the coproduct of them by $U = \coprod_{i}U_{i}$. Moreover, let $U_{i} \to X$ and $Y\to X$ be morphisms in $C$, and $U\to X$ be the morphism induced by the universality of coproduct. Then, $U\times_{X}Y \cong \coprod_{i}(U_{i}\times_{X}Y)$. -If $C$ is the category of schemes, this statement will be true. This is because fiber products of schemes are constructed locally at first, and glued together. -However, I could not prove this by using universality (i.e. in categorical settings). -My questions are: - -Is the above statement true? If so, then how can one prove it? - -If the statement is false, what kind of counter example exists? - -If the statement is false, then, please change the statement replacing "coproducts" by "disjoint unions". Is the NEW statement true? - - -Here, disjoint union of $U_{i}$'s means coproduct $U=\coprod_{i}U_{i}$ satisfying that the fiber products $U_{i}\times_{U}U_{j}$ are the strict initial object if $i \neq j$. Here, strict initial object means initial object $\phi$ such that for any object $X$, the set of morphisms $Hom(X, \phi)$ is the empty set if $X$ is not isomorphic to $\phi$. (This is the generalization of empty set in the category of sets or schemes.) -Later -Counterexamples for the first statement exist (e.g. the category of pointed sets or the opposite of the category of sets). -However, those are not for the refined statement in my question 3. -Does anybody have ideas for it? - -REPLY [9 votes]: The more general question is whether pullbacks and colimits commute. This is true if the category $\mathcal C$ is locally cartesian closed, i.e. every slice category $\mathcal C /c$ of objects over a given object $c$ of $\mathcal C$ is cartesian closed. For then pullback has a right adjoint. -These ideas are of interest in the topological case, though even if $\mathcal C$ is a convenient category of spaces, the categories $\mathcal C/c$ are not always cartesian closed. -Nonetheless, the use of such categories was initiated by R. Thom and continued by Peter Booth, see for example: -Booth, Peter I. -A unified treatment of some basic problems in homotopy theory. -Bull. Amer. Math. Soc. 79 (1973), 331–336. -and his related papers: these ideas were also taken up by Ioan James under the term "Fibrewise topology", see -Crabb, Michael; James, Ioan; -Fibrewise homotopy theory. -Springer Monographs in Mathematics. Springer-Verlag London, Ltd., London, 1998. -There is a paper -Johnstone, P. T. -On a topological topos. -Proc. London Math. Soc. (3) 38 (1979), no. 2, 237–271. -giving a locally cartesian closed category, in fact a topos, with sequential spaces as a reflective subcategory, but this has not yet been used in algebraic opology, to my knowledge. -August 19, 2014 A doctoral thesis in this area, "Topos Theoretic Methods in General Topology" by Hamed Harasani, Bangor 1988. is available here.<|endoftext|> -TITLE: Injective dimension of graded-injective modules -QUESTION [13 upvotes]: In "Existence theorems..." Van den Bergh proposes the following "pleasant excercise in homological algebra": - -Let $A$ be a connected graded noetherian $k$-algebra (that is, $\mathbb N$-graded with $A_0 = k$ a field). A graded module $I$ which is injective in the category of left graded modules has injective dimension at most $1$ in the category of left modules. - -There is a proof of this fact for commutative graded algebras in this paper by Fossum and Foxby, but I don't really see how to transfer this to the non-commutative setting. Can anyone provide any pointers or a reference to a proof? Thanks in advance. -PS: In a later paper, Yekutieli and Zhang state that the only proof they know of this fact is "quite involved", which eased my anxiety at being unable to solve a pleasant excercise in the area I'm supposed to be PhD-ing in... - -REPLY [3 votes]: Here is a proof of Van den Bergh's theorem, just posted on the arxiv: -http://arxiv.org/abs/1407.5916 . -My proof is pretty much self contained, and very detailed. The proof uses the same strategy as the original proof of Van den Bergh (which was very similar to the one given above by Ralph), namely passing through the Rees ring. But it is somewehat more conceptual, in that it isolates the precise "reason" for the dimension jump. This is possible since I work with derived categories (and not with the more cumbersome method of spectral sequences).<|endoftext|> -TITLE: Conformal structure does not see conical singularities -QUESTION [7 upvotes]: the conformal structure does not see the conical singularities of a polyhedral surface. - -This is a quote from the Preface of Quantum Triangulations (eds.: Carfora, Marzuoli). -The sentiment is attributed to Marc Troyanov, in a 1991 Trans. AMS paper, -"Prescribing Curvature on Compact Surfaces with Conical Singularities," -that I cannot access easily. -Can anyone provide a brief explication or unpacking of the meaning of this intriguing -statement? I'd appreciate it—thanks! - -REPLY [4 votes]: Well the question was posted 6 years ago...nevertheless. The reason why the conformal structure does not see the conical singularities of a polyhedral surface is clear if one writes the metric in a neighborhood of a singular point in conformal (isothermal) coordinates. It takes the form $ds^2 = |z|^{2\beta}|dz|^2$ (if the cone angle is $\theta = 2\pi(\beta+1)$). -Another argument is that the capacity of a conical singularity is zero, therefore a complex atlas defined from the conformal structure on the complement of the singular set of the polyhedral metric extends to the singular points by Riemann's theorem on removable singularities.<|endoftext|> -TITLE: Is there an integral version of Faltings' isomorphism in p-adic Hodge theory between etale and Hodge cohomologies -QUESTION [7 upvotes]: Let $K$ be a $p$-adic field, that is a complete discrete valuation ring of characteristic $0$ with a perfect residue field $k$ of characteristic $p > 0$ (to simplify one could also take $K$ to be a finite extension of $\mathbb{Q}_p$). Let $\mathbb{C}_K$ be the completion of a (fixed) algebraic closure $\overline{K}$ of $K$. Then one of Faltings' theorems in $p$-adic Hodge theory says for any smooth proper $K$-scheme $X$ there is a natural isomorphism of $\mathbb{C}_K$-semilinear $\mathrm{Gal}(\overline{K}/K) = \mathrm{Gal}(\mathbb{C}_K/K)$-representations -$\mathbb{C}_K \otimes_{\mathbb{Q}_p} H^n_{et}(X_{\overline{K}}, \mathbb{Q}_p) \cong \bigoplus_{q\in\mathbb{Z}} (\mathbb{C}_K(-q) \otimes_K H^{n - q}(X, \Omega_{X/K}^q)).$ -Here $X_{\overline{K}}$ is the base change of $X$ to the algebraic closure, whereas $\mathbb{C}_K(s)$ stands for the usual $s$-th order Tate twist by the cyclotomic character describing the action of the absolute Galois group on the $p$-power roots of unity. -My question is: is there an integral version of the above isomorphism? Let me be more precise and explain what I mean by this: let $\mathcal{X}$ be a smooth proper scheme over the valuation ring $\mathcal{O}_K$ of $K$ and let $\mathcal{O}_{\overline{K}}$ be the valuation ring of the algebraic closure. Is there an isomorphism similar to the one above relating, say, $H^n_{et}(\mathcal{X}\times_{\mathcal{O}_K} \mathcal{O}_{\overline{K}}, \mathbb{Z}_p)$ and the $H^{n - q}(\mathcal{X}, \Omega^q_{\mathcal{X}/\mathcal{O}_K})$'s? - -REPLY [11 votes]: Dear Kestutis, -your question is integral in two ways: first of all, you would like to consider schemes over a whole DVR instead of the generic fiber only. Secondly, you would like to have a comparison theorem between two $\mathcal{O}_\overline{K}$-modules and not only between $K$-vector spaces. The two are tightly connected since $H_\text{dR}(\mathcal{X})\otimes K=H_\text{dR}(X)$, if $X$ denotes the generic fiber of $\mathcal{X}$. -As for the first question we have the so-called Crystalline Conjecture proven by Faltings and many others. It says that if $\mathcal{X}$ is proper and smooth over $\mathcal{O}_K$ (I keep your notations) with generic fiber $X$ and special fiber $\overline{\mathcal{X}}$, then -$$ -B_\text{cris}\otimes_{K_0}H_\text{cris}^m(\overline{\mathcal{X}})\cong B_\text{cris}\otimes_{\mathbb{Q}_p}H_\text{et}(X_\overline{K},\mathbb{Q}_p) -$$ -where $B_\text{cris}$ is Fontaine's ring of periods. If you extend scalars to $B_\text{dR}\supset B_\text{cris}$ you get back the comparison isomorphism that you quote because $B_\text{dR}$ admits a filtration whose associated graded ring is $\bigoplus_{q\in\mathbb{Z}}\mathbb{C}_K(q)$; and because of a basic result by Berthelot and Berthelot-Ogus saying that crystalline cohomology (which is the group $H_\text{cris}(\overline{\mathcal{X}})$ in my equation above and is a cohomology theory of the special fiber with coefficients in $W(k)$, the Witt vectors of the residue field $k$ of $\mathcal{O}_K$) is isomorphic to de Rham cohomology after you tensor both with $\text{Frac}(W(k))$. You could give a look at -1) J.-M. Fontaine, Représentations $p$-adiques semi-stables, in Périodes $p$-adiques, Astérisque 223, Section 6.1 for a discussion about the Crystalline Conjecture -2) P. Berthelot and A. Ogus, Notes on crystalline cohomology, Princeton University Press, Chapter 7 (the very last theorem) for the isomorphism $H_\text{cris}\cong H_\text{dR}$ -Let me also remark that much of the above can be generalized assuming only that $\mathcal{X}$ be semistable instead of smooth (by Tsuji). -The second question is more subtle. As I mentioned, already the isomorphism $H_\text{cris}(\overline{\mathcal{X}})=H_\text{dR}(\mathcal{X})$ is false in general: it is true if $K/\mathbb{Q}_p$ (or $\mathcal{O}_K/\mathbb{Z}_p$, equivalently) has absolute ramification index $e\leq p-1$. In this case, at least, we dispose of an interpretation of the de Rham cohomology of $\mathcal{X}$ in terms of crystalline cohomology. We also have an "integral structure" $A_\text{cris}\subseteq B_\text{cris}$ and one can hope to have something like -$$ -A_\text{cris}\otimes_{W(k)}H_\text{cris}^m(\overline{\mathcal{X}})\stackrel{?}{\cong} A_\text{cris}\otimes_{\mathbb{Z}_p}Hˆm_\text{et}(\mathcal{X}_{\mathcal{O}_\overline{K}},\mathbb{Z}_p) -$$ -The problem is that $A_\text{cris}$ (and even $A_\text{cris}[1/p]$!) is much smaller than $B_\text{cris}$: roughly speaking, it cannot detect "negative Hodge-Tate weights". -[EDIT] As Keerthi Madapusi Pera and Matthew Emerton commented, in some special cases the above isomorphism holds true. They give precise references in the comments.<|endoftext|> -TITLE: Recovering a measure from its moments -QUESTION [6 upvotes]: Suppose we are given moments of a measure on the interval [0,1]. Is there some practical way to recover the measure itself? I am particularly interested in the case where the measure density is given by some smooth positive function, and I would like to determine the behavior of this function near x=0. - -REPLY [3 votes]: An answer on your question is given in the following book: -[ST] J. A. Shohat and J. D. Tamarkin "The problem of moments". -See [ST], page 90 and what follows. In particular, the case of an absolutely continuous measure is considered on page 95.<|endoftext|> -TITLE: Is $H^i(X,F)$ finitely generated over $\Gamma(O_X)$ if $F$ is coherent? -QUESTION [8 upvotes]: Suppose $\mathcal{X}$ is a smooth quasi-projective variety over $\mathbb{C}$ (I apologize if these hypotheses have little to do with the question at hand). Let $\mathcal{F}$ be a coherent sheaf on $\mathcal{X}$. Is $H^i(X,\mathcal{F})$ finitely generated over $\Gamma(O_X)$ if $\mathcal{F}$ is coherent ? This statement is simple enough that I probably would have heard it if it were true. -If it makes a difference, the statement that I really would like to understand is whether $\Gamma(O_X) \to Ext^i(\mathcal{G},\mathcal{G})$ is module finite for any coherent sheaf. -Most of the schemes that I am "friendly with" are either projective or affine. The statements are correct in those particular cases, so probably I just need to learn more examples... Thanks! - -REPLY [17 votes]: This is false even for $\mathrm H^0$: take $X$ to be $\mathbb A^2 \smallsetminus \{0\}$, and as $F$ the structure sheaf of $L \smallsetminus \{0\}$, where $L$ is a line through $0$.<|endoftext|> -TITLE: Connections between various generalized algebraic geometries (Toen-Vaquié, Durov, Diers, Lurie)? -QUESTION [52 upvotes]: As far as I know, there are four possible ways to generalize algebraic geometry by 'simply' replacing the basic category of rings with something similar but more general: -$\bullet$ In the approach by Toen-Vaquié we fix a nice symmetric monoidal category $C$, also called a relative context. An affine scheme is defined to be an algebra object in $C$, and an arbitrary scheme is a certain presheaf on affine schemes. We optain the category $\mathrm{Sch}(C)$ of schemes relative to $C$. -$\bullet$ In Durov's theory a generalized ring is an algebraic monad which is commutative in a certain sense. Then affine schemes are defined to be the spectra of generalized rings and arbitrary schemes are optained by gluing. This results in the category $\mathrm{genSch}$. -$\bullet$ In his book Categories of commutative algebras Yves Diers considers Zariski categories, which seem to axiomatize familiar properties of categories of commutative algebras. If $\mathcal{A}$ is such a Zariski category, then one can develope commutative algebra internal to $\mathcal{A}$, construct affine schemes and then by gluing also schemes as usual. We optain the category $\mathrm{Sch}(\mathcal{A})$. -$\bullet$ In derived algebraic geometry one replaces the category of rings with the category of simplicial rings (but I don't really know enough about that, yet). -My question is: What are the connections between these 'generalized algebraic geometries'? -Fortunately there is a map of $\mathbb{F}_1$-land which draws connections between all these various approaches to schemes over $\mathbb{F}_1$. For example monoid schemes à la Deitmar/Kato are in the intersection of Toen-Vaquiè and Durov. Note, however, that the theories mentioned above are far more general. -Specifically, one might ask the following questions: Is the category of generalized rings a Zariski category and is Durov's theory (say, with the unary localization theory) a special case of the one by Yves Diers? What is the relationship between Toen-Vaquié schemes relative to the symmetric monoidal category of simplicial rings and derived schemes? If $C$ is a relative context, is then the category of algebra objects in $C$ a Zariski category and do the corresponding schemes coincide? Probably not because Diers never mentions monoids as an example, but perhaps it's the other way round? Of course, many more questions are out there ... -Probably I'm not the first one with this question, therefore I've also put the "reference-request" tag. It would be great if there is some paper like "Mapping AG-land". - -REPLY [5 votes]: In 2015 appeared one more very interesting work by Connes and Consani; their Absolute algebra and Segal’s $Γ$-rings (au dessous de $\overline{\operatorname{Spec}(\mathbb Z)}$) also contains the view of interconnections between previous approaches to the subject.<|endoftext|> -TITLE: 0 eigenvalue for a symmetric tridiagonal matrix -QUESTION [9 upvotes]: Let $T\in \mathbb{R}^{n\times n}$ be a symmetric tridiagonal matrix having the off--diagonal entries equal to -1. The diagonal entries are all positive, $a_i>0$, $i=\overline{1,n}$, and there exist $j$ and $k$, $j\neq k$ such that $a_j=a_k\leq 1$. $a_j$ and $a_k$ are the smallest diagonal entries. -I'm interested under what supplemental conditions can such a matrix have the smallest eigenvalue equal to 0? - -REPLY [10 votes]: To simplify things a little, I describe conditions under which the smallest eigenvalue is strictly positive. These can be adjusted to get equality to zero. -Necessary and sufficient conditions for positive definiteness of the tridiagonal matrix in question are described below. -Definition (Chain Sequence). A sequence $\lbrace x_k \rbrace_{k > 0}$ is a chain sequence if there exists another sequence $\lbrace y_k \rbrace_{k\ge 0}$ such that -\begin{equation*} - x_k = y_k(1-y_{k-1}), -\end{equation*} -where $y_0 \in [0,1)$ and $y_k \in (0,1)$ for $k > 0$. -By the Wall-Wetzel Theorem, your tridiagonal matrix is positive definite if and only if -\begin{equation*} - \left\lbrace \frac{1}{a_ka_{k+1}} \right\rbrace_{k=1}^{n-1} -\end{equation*} -is a chain sequence. -Example. In particular, if the entries of the matrix satisfy, -\begin{equation*} - 0 < \frac{1}{a_ka_{k+1}} < \frac{1}{4\cos^2\left(\frac{\pi}{n+1}\right)},\quad k=1,\ldots,n-1, -\end{equation*} -then it is positive definite. - -For additional information and details about this material, please see: - -M. Andelic, and C. M. Da Fonesca. Sufficient conditions for positive definiteness of tridiagonal matrices revisited. (2010).<|endoftext|> -TITLE: Examples for "nice" Boolean algebras that are not complete or not atomic -QUESTION [9 upvotes]: A Boolean algebra may, or may not, be complete (i.e, any set of elements has a sup and an inf) or atomic (i.e., every element is a sup of some set of atoms). -Boolean Algebras that are complete as well as atomic (also called CABAs) are of course precisely those that are isomorphic to some power set (equipped with the obvious choices for the operations), or equivalently stated, those that form a category dually equivalent to $Set$. -The category of all Boolean algbras, however, is well-known to be equivalent to the category of Stone spaces (compact totally disconnected Hausdorff spaces) with continuous morphisms. Thus, for a Boolean algebra (of infinite cardinality), it is a very special case to be complete and atomic. My question is: - -What are nice examples for Boolean Algebras that are not complete or not atomic? - -Please understand that I do not look for any kind of example (so the emphasis lies on the word "nice"). I am, for instance, aware that looking at free BAs would lead to such an example, and I also know the classic example of the BA that is formed by all finite and confinite sets of integers. Also, as mentioned above, I know how the Stone Duality transforms Stone spaces into Boolean algebras, so please don't simply say "the clopen subsets of a that-and-that Stone-Space form a Boolean algebra". -I admit that nice is a somewhat vague notion. What I mean are Boolean Algebras that arise naturally (except those I have already mentioned) and are of special interest for some reason (yes, I know that this formulation is not vague at all). - -REPLY [3 votes]: The 3-volume "Handbook of Boolean Algebras" abounds with examples of Boolean algebras, but of course only few of them are "nice". - -You may have a look at Koppelberg's chapter 6 in volume 1, "Special classes of BAs", and also at several chapters in volume 3 describing some of these classes in more detail. This may challenge or sharpen your idea of "nice". -Generalizing your question one could also ask for nice constructions of Boolean algebras (products, free products, quotients, iterated quotients, interval algebras, completions, etc); some of them will yield nice algebras. -For example, dividing any BA $B$ by the ideal $At(B)$ generated by the atoms will yield a new BA $B':=B/At(B)$; this new algebra $B'$ will often be atomless (e.g. if you start with a power set BA, as mentioned by Andreas Blass). If not, you can continue this process of dividing, even transfinitely often. -Monk's chapter 12 in volume 2 describes several methods for getting "many" BAs with particular properties. For example, starting with any linear order $L$, the interval algebra $I(L)$ is defined as the subalgebra of $P(L)$ (the power set algebra) generated by the half open intervals $[a,b)$ in $L$. Sufficiently different/nice linear orders will give nonisomorphic/nice BAs. -For $L=\mathbb N$ you will get the (highly atomic) finite-cofinite subalgebra of $P(\mathbb N)$, and for $L=\mathbb Q$ you will get the countable atomless BA, mentioned by Joel Hamkins.<|endoftext|> -TITLE: Quick computation of the Pontryagin dual group of torus -QUESTION [6 upvotes]: I'm looking for a quick way to compute the Pontryagin dual group of the n-dimensional torus $\mathbb{T}^n$ (with $\mathbb{T} := \mathbb{R} / \mathbb{Z}$). The only way I know is from "Dikran Dikranjan - Introduction to Topological Groups" and is very tedious. -I thought so: $\mathbb{T}^n$ is compact then $\widehat{\mathbb{T}^n}$ has the discrete topology. The set $E$ of all the functions -$\mathbb{T}^n \to \mathbb{C} : x \mapsto e^{2 \pi a \cdot x}$ -with $a \in \mathbb{Z}^d$ is a subset of $\widehat{\mathbb{T}^d}$ which separates the points of $\mathbb{T}^d$ (I mean: for -any $x,y \in \mathbb{T}^d$ with $x \neq y$ exists some $\chi \in E$ such that $\chi(x) \neq \chi(y)$. -Now I was hoping that Stone-Weierstass theorem could help me to prove that $E$ is dense in $\widehat{\mathbb{T}^d}$ and so that $\widehat{\mathbb{T}^d} = E$, because $\widehat{\mathbb{T}^d}$ is discrete. However I don't know how to apply Stone-Weierstass theorem in this case. -I thank you for any ideas! - -REPLY [3 votes]: If your purpose of calculating the dual of the torus is to have the Fourier inversion formula, I just want to add the remark that you don't need to go through all the machinery of locally compact topological groups in order to prove the basic results of Fourier analysis. -For the circle, one needs only note that the distribution -$u(x) = \sum_{n \in {\mathbb Z}} e^{2 \pi i n x}$ -on the torus ${\mathbb R}/{\mathbb Z}$ is invariant under multiplication by $e^{2 \pi i x}$, and is therefore a multiple of the delta-function $\delta(x)$ -- integrating against the smooth function $1$, you see that, in fact, $u(x) = \delta(x)$. (This move is simply how you usually calculate the "sum of a geometric series".) Substituting into the general formula $f(x) = \int_{{\mathbb R}/{\mathbb Z}} f(y) \delta(x - y) dy$ gives the Fourier inversion formula. -The above is really just for a conceptual point of view, since it's unnecessary to quote distribution theory for this purpose. To give a direct proof, suppose $f(x)$ is smooth; after translating, it is enough to show that $f(0) = \sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_n \int f(x) e^{- 2 \pi i n x} dx$ (which converges absolutely because you can integrate by parts enough times to prove decay of $\hat{f}(n)$). We can assume without loss of generality that $f(0) = 0$ by subtracting the constant $f(0)$ (this is how we know "the constant" in the inversion formula is correct and corresponds to testing against $1$ in the previous argument). -In the case $f(0) = 0$, we can write $f(x) = (e^{2 \pi i x} - 1) g(x)$ for some smooth function $g(x)$ (which proves the previous claim that if $e^{2 \pi i x} u = u$ then $u(x) = C \delta(x)$). But then, -$\sum_{n \in {\mathbb Z}} \hat{f}(n) = \sum_{n \in {\mathbb Z}} (\hat{g}(n - 1) - \hat{g}(n))$ -but this is $0$ because $\hat{g}$ is integrable. (In particular, assuming $f$ to be $C^\infty$ was certainly unnecessary, and the Fourier inversion formula holds pointwise for functions in a certain class.) -If you like to think in terms of Stone Weierstrass, you can view this argument is saying that the map $Tf$ which takes a Fourier transform and then performs the inverse Fourier transform is the identity. It relies on the fact that $f(x) = (e^{2 \pi i x} - e^{2 \pi i x_0}) g(x)$ whenever $f$ vanishes at $x_0$ which is a strong sense in which trigonometric polynomials separate points. The conclusion that $T$ is a multiple of the identity then follows from $T$ being linear not only over ${\mathbb C}$ but also over the algebra of trigonometric polynomials. -Another remark about distribution theory here is that for the argument of KConrad above, the equation $\gamma'(x + s) = \gamma'(s) \gamma(x)$'' already makes sense when $\gamma$ is a distribution (in particular, if $\gamma$ is continuous). The way it makes sense is by integrating by parts against a test function, which is related to the integration performed the accepted proof. One can use a very similar argument to KConrad's and show that the only distributions solving $\gamma(x + y) = \gamma(x)\gamma(y)$ are exponentials as I'll (almost) show below. -Finally, observe that the fact that the Fourier inversion formula holds for $L^2$ functions implies in particular that all the characters have been accounted for. After all, if you had any other character, it would be orthogonal to all the exponential functions and hence have $\hat{\gamma} = 0$. However, this argument requires not only that the Fourier transform is an isometry on smooth functions in $L^2$ (a consequence of the inversion formula), but one also has to argue that $L^2$ functions can be approximated by smooth functions, which is typically done by mollification $f(x) \approx \int f(x + h) \eta(h) dh$ for a smooth function $\eta$ of integral $\int \eta(h) dh = 1$ and small support. The same kinds of mollifications or similar calculations are used in proofs of Stone Weierstrass theorem; by a similar reasoning, once you have enough characters to separate points, you know you have a complete set. -But, as KConrad showed with $\eta$ the characteristic function of a short interval, when $f = \gamma$ is a character, mollification shows $\int \gamma(x + h) \eta(h) dh = \gamma(x) \int \gamma(h) \eta(h) dh$, which implies in particular that $\gamma(x)$ is smooth because $\int_{{\mathbb R}/{\mathbb Z}} \gamma(x +h) \eta(h) dh = \int_{{\mathbb R}/{\mathbb Z}} \gamma(h) \eta(h - x) dh$ is always smooth (even when $\gamma$ is just a distribution). -But then once you have established that $\gamma$ is smooth, the same equation $\gamma(x) \gamma(y) = \gamma(x+y)$ which proves the orthogonality of characters can also be differentiated classically and evaluated at $y = 0$, and then (as KConrad noted) you just need to solve the ODE that defines the exponential functions. So I'll end here now that we have "come full circle".<|endoftext|> -TITLE: Manin's lectures on algebraic geometry -QUESTION [14 upvotes]: In 1966-1968 Yuri Manin gave a 2 year lecture course in algebraic geometry at Moscow State University. The course starts from scratch and culminates in the proof of the Riemann-Roch theorem following SGA6 with some simplifications. The course was subsequently published by the Moscow University Press in two parts. Part 2 contained the proof of the Riemann-Roch theorem and Part 1 the preliminary material. (Part 2 also appeared as a paper in the Russian Mathematical Surveys.) -However, the version of Part 1 that I have has only chapters 1 to 15, and in Part 2 Manin makes references to chapters 25 and 22 of Part 1. This is why I would like to ask whether any more of Part 1 exists, apart from chapters 1-15, and if so, whether it has been made publicly available. -Thank you. - -REPLY [5 votes]: Hi Dmitry, -I am interested in the note but I cannot find your email address. -I would rather make this as a comment, but my reputation point is still below 50. -If you read this posting, please send your note to jazzpiano3@gmail.com. -I would appreciate it.<|endoftext|> -TITLE: What can we learn about an elementary embedding from the image of the ordinals? -QUESTION [9 upvotes]: If $j : V \rightarrow M$ is an elementary embedding, what can we learn in $M$ from $j''ORD$? That is, what is $M[j''ORD]$? -In particular, -Is it $M[j''ORD]$ equal to all of $V$? -If not, do we get a model intermediate between $M$ and $V$? If $\kappa$ is the critical point of $j$, is $\kappa$ still a large cardinal in $M[j''ORD]$? I am thinking of $j$ arising from a measure on $\kappa$, but I'm also interested in the more general situation. -My thinking goes like this: If we have the image of all of $V$, we can reconstruct $V$ itself by taking the Mostowski collapse of $j''V$ (and $j$ is the inverse of the Mostowski collapse). In $M[j''ORD]$, let's consider the class $W$ of sets with rank in $j''ORD$. Does the Mostowski collapse of $W$ yield all of $V$, and if not, what's missing? -EDIT: formatting, clarification of question. - -REPLY [8 votes]: Nice question, Jonas! -Yes, in the case that $V$ satisfies ZFC and $M\subset V$, then indeed $M[j''\text{Ord}]=V$. To see this, consider first the case of a set of ordinals $A\subset\theta$ in $V$. Notice that from $j''\theta$ we may reconstruct $j\upharpoonright\theta$. Further, $j(A)$ is in $M$, and from $j(A)$ and $j\upharpoonright\theta$ we may easily reconstruct $A$ itself. So every set of ordinals in $V$ is in $M[j''\text{Ord}]$. If $V$ satisfies ZFC, then this suffices, since every set is coded by a set of ordinals. Namely, (as you know) if $a$ is any set, then $\langle \text{TC}(\{a\}),{\in}\rangle\cong\langle\theta,E\rangle$ for some cardinal $\theta$ and some binary relation $E$ on $\theta$, and then using a Gödel pairing function to code $E$ as a single set $A\subset\theta$. So for any set $a$, we find $E$ and then $A$, and by the previous argument $A$ is in $M[j''\text{Ord}]$, and so also $E$ and hence $a$ itself is there. Thus, $M[j''\text{Ord}]=V$, as desired. -The argument relies on the axiom of choice, and in the most general case, I believe this is required.<|endoftext|> -TITLE: A followup on non-homogeneous spaces. -QUESTION [5 upvotes]: This question asks for an example of a manifold which is not a homogeneous space of any Lie Group, and many examples are given in the answers. However: is there a an example known with a metric of positive sectional curvature (this, apparently, is a question asked by Marcel Berger many years ago, so was open then, but I have no idea of its current status...)? - -REPLY [4 votes]: Yes, some quotients of the 3-sphere are examples. Namely, view $S^3$ as the group of quaternions of unit one. Consider two finite subgroups $F_1,F_2$ of $S^3$, with $F_1$ not abelian, and $F_2$ cyclic of odd prime order $p$ not dividing the order of $F_1$. So $F_1\times F_2$ acts freely on the sphere by $(g,h)\cdot z=gzh^{-1}$. I claim that the quotient $X=F_1\backslash S^3/F_2$ is not homeomorphic to a homogeneous space of any connected Lie group $G$. -Assume the contrary. First, by a result of Montgomery (Proc AMS 1950), any maximal compact subgroup of $G$ acts transitively (it's a general fact, using only that $X$ has finite fundamental group). So we can assume that $G$ is compact. Then a result of Montgomery-Samelson (Annals 1943) tells us that some simple $S$ subgroup acts transitively. (The result concerns actions on spheres, and is applied by considering the action of a finite covering of $G$ on $S^3$.) The isometry group of a 3-dimensional Riemannian manifold is at most 6-dimensional, and the only simple compact Lie group of dimension at most 6 is $SO(3)$ (and its universal covering). Thus $X=S^3/F$ for some finite subgroup $F$ of $S^3$, isomorphic to $F_1\times F_2$. But in $S^3$, the centralizer of any nontrivial element of odd order is abelian. This is a contradiction. -These examples are not simply connected, of course. It seems that the other answers deal with the much more delicate simply connected case.<|endoftext|> -TITLE: Easy and Hard problems in Mathematics -QUESTION [5 upvotes]: Modified question: -I would like to know some examples of problems in Mathematics, for pedagogical purposes, which do not involve difficult techiques to solve the problem but with a change of context turns them into monstrous-unimaginably difficult to solve problems. -By changing the context I mean, by changing one class of objects in the problem to a related -class of objects. For example, from directed graph to undirected graph or Zygmund class to -Log-lipshitz class. By changing a 'less-than problem' to 'greater-than problem'. From 2-case -problem to 3-case problem. There are plenty of such examples in Theoretical Computer Science or Computational Complexity theory. I need some examples in Mathematics. Lot of examples fall in this category but I am looking for only extreme examples like the ones I stated below. Since, this question is asked for pedagogical purpose it would be interesting if there is a story behind the problem. -Examples of problems: - -Linear Programming to integer linear programming -2-coloring to 3-coloring -Eulerian graph to Hamiltonian graph -Undirected graph case to directed graph case in Shannon's switching game -2-SAT to 3-SAT - -One thought which motivated me to pose this question is: what if Konigsberg problem has been -formulated as a vertex problem. Would Leonard Euler get inspried to create graph theory? No -doubt, history speaks differently as Konigsberg problem is stated in terms of edges. Not only -Euler solved this problem but created a branch of mathematics out of it! And I am not sure what turn of events would have taken place had the problem been posed in terms of vertices. -IMHO, there are look-alike easy problems and hard problems coexisting but it is the easy problems which saved mathematicians day and hard ones which gave them incentive to work harder. -Some pointers for hardness of problem: problems which need sophisticated tools, techniques which diverge from the routine ones, radical thinking or bold ideas to solve the them, like Poincare conjecture. Or, those problems which do not have adequate tools yet to attempt them, like (NP=P?). -I would appreciate any answers in this direction. Thank you in advance. - -REPLY [6 votes]: Finding the $\ell_p$ operator norm of a matrix is easy for $p=1,2,\infty$, but it is NP-hard for any other $p$ (see this discussion, for example).<|endoftext|> -TITLE: Full isometry groups of Stiefel and Grassmann manifolds -QUESTION [7 upvotes]: Hi, -I'm looking for a reference for the full isometry groups of the -(i) complex Stiefel manifolds $U(m)/U(m-l)$, either for the Euclidean metric (i.e. identifying it with orthonormal $m \times l$-matrices and using the inner product $\operatorname{tr}(X^\ast Y)$) or for the induced quotient metric (where we use on $U(m)$ the Euclidean metric), and of the -(ii) real and / or complex (doesn't matter for me now) Grassmann manifolds with the induced quotient metric from the Euclidean metric on O(n), resp. U(n). -It is easy to see that left, resp. right multiplication with unitary / orthogonal matrices are isometries, so I'm interested whether there are more isometries and especially how to show that there are no more. -Thanks. -Asked that question already on Math.SE and got no answer: https://math.stackexchange.com/q/112175/7110 - -REPLY [4 votes]: As long as connected groups of isometries are concerned, Grassmann manifolds are symmetric spaces, -so the identity component of its isometry group -is $G$ in its symmetric presentation $G/H$ ($G$ connected) as a homogeneous space, namely, $SO(n)$ for $n$ odd and $SO(n)/\mathbf Z_2$ for $n$ even in the real case, -and $PU(n)=SU(n)/\mathbf Z_n$ in the complex case. -(Note that $U(n)$ acts on the left on the Grassmannian with a $U(1)$-kernel (its center), so the effectivized group is the projectivization -$PU(n)$. Moreover the center of $U(n)$ meets -$SU(n)\subset U(n)$ along its center, which consists of $\omega I$ where $\omega$ is an $n$-th root of unit.) -Further, Cartan described the full isometry groups of symmetric spaces, and an explicit result is easy to figure out in the case of Grassmann manifolds. -I do not remember now, but you can find Cartan's description in the book of O. Loos on symmetric spaces, the second volume. I tend to agree with Ryan when he writes that in the case of Grassmann manifolds, the full isometry group should be -$G\times N_G(H)$. -About Stiefel manifolds: with the metric you describe, they are normal homogeneous spaces -$G/H$, i. e. have the metric induced from a -bi-invariant Riemannian metric on $G$. There is a recent paper by S. Reggiani with a very effective way of computing the identity component of isometry groups of normal homogeneous spaces -in here. -Added: I looked up Loos, "Symmetric spaces, II", Theorem 4.4 and the ensuing Table 10 on page 156 for the full isometry group of the real and complex Grassmannians. If I understand correctly, indeed in the case of complex Grassmannians -$SU(n)/S(U(p)\times U(n-p))$, every isometry comes from left multiplication by elements from $SU(n)$ except for two cases: an isometry induced by complex conjugation; and mapping a $p$-plane to its orthogonal complement in case $n=2p\geq4$. In the case of -real unoriented Grassmannians $SO(n)/S(O(p)\times O(n-p))$, every isometry comes -from left multiplication by an element of $O(n)$ except for: mapping a $p$-plane to its orthogonal complement in case $n=2p\geq4$; the symmetric group $S_3$ in case $n=2p=8$, -coming from outer automorphisms of $\mathfrak{so}(8)$.<|endoftext|> -TITLE: Is Sheafification Functor Exact? -QUESTION [11 upvotes]: I know that sheafification functor from the category of abelian presheaves on $C$ to the category of abelian sheaves on $C$. Here, $C$ is a category with Grothendieck pretopology. -My question is: -How about the sheafification functor from the category of presheaves of "sets" on $C$ to the category of sheaves of "sets" on $C$? -Is this an exact functor? (i.e. preserving finite limits and finite colimits?) -If so, how can one prove it? -In fact, I want to know whether sheafification functor preserves cartesian products or not. -Please give me any advice. - -REPLY [11 votes]: Martin has probably answered everything Hiro meant to ask, but, since "whether sheafification functor preserves cartesian products or not" didn't explicitly say finite products, let me add that sheafification will not in general preserve infinite products. Intuitively, the reason is that a section of a product of sheafifications is a family of locally defined sections of the original presheaves, and there might not be a single covering over which all those local sections are simultaneously available.<|endoftext|> -TITLE: Spreading out flat morphisms of schemes -QUESTION [5 upvotes]: In EGA IV, Chapter 8, projective systems of schemes (and morphisms between them) are considered. Let $(S_{\lambda})_{\lambda \in L}$ be a projective system of schemes and let $S$ be the projective limit of this system. -The first version of my question I think assumes that you are somewhat familiar with this part of EGA. The second version is a special case of the first version, and doesn't require familiarity with EGA IV.8. - -In Théorème 8.10.5, it is stated that for every property P in a given list of properties (starting with: isomorphism, monomorphism, immersion...), and under some additional conditions that I won't mention right now, if a morphism $f : X \rightarrow Y$ of $S$-schemes with property P can be "spread out" to a morphism $f_{\alpha} : X_{\alpha} \rightarrow Y_{\alpha}$ of $S_{\alpha}$-schemes for some $\alpha \in L$, then we can choose $\alpha$ in such a way that $f_{\alpha}$ has property P. The list of properties P given in EGA for which the Théorème works, however, does not contain flatness. Is there a reason for this? I.e., is there a counter-example for the statement of the theorem if we include flatness in the list? If so, are there any reasonable extra assumptions that we can put into the theorem so that it also works for flatness? -This is the situation I am most interested in. Suppose that $R$ is a commutative ring, then let $R_i$ be the directed system of its finitely generated subrings. Put $S_i := \textrm{Spec}(R_i)$, so that $(S_i)_{i \in I}$ is a projective system of affine schemes. I have a morphism $f : X \rightarrow S$ that is proper, flat and of finite presentation. Does $f$ necessarily arise as the base-change of some $f_i : X_i \rightarrow S_i$ that is also proper, flat and of finite presentation? - -REPLY [5 votes]: I think what you are looking for is in the book, only later: see EGA IV, (11.2.6).<|endoftext|> -TITLE: Higher "Cartan-Eilenberg" Resolutions -QUESTION [11 upvotes]: I am a number theory graduate student learning a bit of homological algebra, and I am curious about higher complexes in abelian categories. I apologize if my post is slightly vague as I am not an expert in this area. I will use capital roman letters to denote objects or complexes, but the usage is clearly stated. -Motivation -To motivate my question, let us start from a single object $M$ in an Abelian category $\mathcal{C}$. If $\mathcal{C}$ has enough projectives, we can form a projective resolution $P\to M$ of $M$, and apply a right exact additive functor $F:\mathcal{C}\to\mathcal{D}$ to $P$, and calculuate homology. Here $\mathcal{D}$ is just some other abelian category. This will give us the derived functors of $F$, and is a standard and well-known construction. -Of course, it doesn't stop there. In the same abelian category $\mathcal{C}$ with enough projectives, any chain complex $M$ also has a (left) Cartan-Eilenberg resolution $P\to M$. Recall $P$ is an upper half plane double complex and the map $P\to M$ is just a chain map $P_{\bullet,0}\to M_\bullet$. Finally, $P$ is required to satisfy some axioms making it into a sort of 2-dimensional version of a projective resolution. I won't go into detail because this is also fairly standard. -The point is that we can also apply a right-exact functor $F$ to this double complex $P$ and take the the homology of the total direct-sum complex of $P$ (if it exists!); that is $H_i(Tot^\oplus(FP))$, to get the hyperderived functors of $F$. -The Question -It seems as though there is a natural generalization. One can easily define an $n$-complex in an analogous fashion to $2$-complexes. Higher dimensional complexes don't really show up much as far as I can tell, although I believe in Cartan-Eilenberg a $4$-complex is used somewhere (sorry, I don't have the book with me!). -So I suppose my question is: - -Suppose $\mathcal{C}$ is an abelian category with enough projectives. Is it true that for any $n$, an $n$-complex $M$ has some appropriate higher Cartan-Eilenberg resolution (which would be an $n+1$-complex)? - -Appropriate means that if $P\to M$ is this hypothetical higher Cartan-Eilenberg resolution, then applying a right exact additive functor $F$ to $P$ and taking the homology of the total direct-sum (if it exists) complex gives the "correct" notion of $n$-hyperderived functors. -Comments -I have searched the literature for this concept but I could not find anything relevant. I am thinking that there are two possibilities (a) yes, higher Cartan-Eilenberg resolutions exists and are interesting, or (b) yes, higher Cartan-Eilenberg resolutions exist but don't capture any new information and so are not that interesting. I'd be a bit surprised if they don't exist but I do not have enough experience in homological algebra to understand the bigger picture here. -Also, we could have phrased this question in terms of injectives and (right) Cartan-Eilenberg resolutions. -Thanks - -REPLY [3 votes]: The process can certainly be iterated as explained by Marc (see also Weibel, Homological Algebra, 1.2.5. Moreover cf. 1.2.3, 2.2.2 for the fact that the category of chain complexes over an abelian category with enough projectives is again an abelian category with enough projectives). -However, it seems to me that it isn't often used. A reason might be, that in many (most ?) cases one isn't interested in a double complex (or higher dimensional analogs) itself but in the (co)homology of of its total complex (the definition of hyperderived functors in your question is an example for this point of view). -However, in this case there is no need to jump into a higher dimension to define a projective resolution. For, there is an alternative definition for the projective resolution of a chain complex that is - in my opinion - much more elegant and easier to work with than with Cartan-Eilenberg's definition: - -A projective resolution of the chain complex $C$ is a complex $P$ of projectives together with a quasi-isomorphism $f: P \to C$ (i.e. $f$ is a chain map such that $H_n(f): H_n(P) \to H_n(C)$ is an isomorphism for all $n$). - -Note that such a $P$ is in general no projective object in the category of chain complexes, but it yields the same hyperderived functors, hypercohomology spectral sequences, etc. For a textbook reference of this definition see for example - -McCleary, A User's Guide to Spectral Sequences (before Theorem 12.12) -Benson, Representations and Cohomology I, Definition 2.7.4<|endoftext|> -TITLE: Numbers with known irrationality measures? -QUESTION [15 upvotes]: For a given real number $x$, let $R_x$ be the set of real numbers $r$ such that the inequality -$$\displaystyle \left| x - \frac{p}{q} \right| < \frac{1}{q^r}$$ -has at most finitely many solutions with integers $p,q$. Define the irrationality measure of $x$, say $\mu(x)$, to be the infimum of $R_x$. -It is known that if $x$ is algebraic and not rational, then $\mu(x)$ is 2, by Roth's Theorem. It is trivial that if $x$ is rational, then $\mu(x) = 1$. I believe it is also known that all real numbers except a set of measure 0 has irrationality measure of 2, but I am unsure of the reference. -For some known transcendental numbers, upper bounds for $\mu$ are known. For example, we know that $\mu(\pi) < 7.6063$ (Salikhov, V. Kh. "On the Irrationality Measure of ." Usp. Mat. Nauk 63, 163-164, 2008. English transl. in Russ. Math. Surv 63, 570-572, 2008.) -Are there any general results concerning a set of transcendental numbers $x$ with $\mu(x) = 2$? Are there any known, 'interesting' numbers (expressible in well-known functions or constants) $x$ with $\mu(x) = 2$? - -REPLY [11 votes]: Yes, there are uncountably many "explicit" real numbers that are (i) badly approximable and (ii) transcendental and (iii) have easy-to-write-down binary expansions. See my paper with van der Poorten, Folded Continued Fractions, J. Number Theory 40 (1992), 237-250. I'm surprised Gerry Myerson didn't remember that!<|endoftext|> -TITLE: How can I see that the slice of a presheaf category is equivalent to the presheaf category of the category of elements? -QUESTION [5 upvotes]: Let $\mathcal{C}$ be a small category and $P$ a presheaf on $\mathcal{C}$. Then there is an equivalence $$\widehat{\mathcal{C}} / P \cong \widehat{\int_{\mathcal{C}}} P$$ Now there are (at least) two ways to see this (from what I've managed to ascertain.) You might derive the equivalence as it were 2-categorically using the universal property of $\widehat{\mathcal{C}}$ which is the equivalence $\textbf{Cat}/\widehat{\mathcal{C}} = \textbf{Func }(\widehat{\mathcal{C}}) \cong \textbf{DFib}(\mathcal{C})$ where $A$ is any small category and $\textbf{DFib}$ are the discrete fibrations over $\mathcal{C}$. You do this by showing that any functor $A \rightarrow \widehat{\mathcal{C}}/P$ corresponds to a functor $A \rightarrow \widehat{\mathcal{C}}$ and a natural transformation $A \rightarrow 1 \overset{P}{\rightarrow} \mathcal{C}$ (where $1$ is terminal in $\textbf{Cat}$) and hence that the corresponding morphism of fibrations corresponds to a discrete fibration of ${\int_{\mathcal{C}}}P$ and hence that -$$\textbf{Func }(\widehat{\mathcal{C}}/P) \cong \textbf{DFib}(\int_{\mathcal{C}}P) \cong \textbf{Func }(\widehat{\int_{\mathcal{C}}P}) $$ -which by universality gives the desired equivalence. Now I have to admit I'm a bit sketchy on the details of this, as it uses many non-obvious properties of fibrations. A more detailed version of this approach was given as a partial answer to this question. -I am more interested in the more elementary approach, i.e. establishing an explicit equivalence between the two categories, so that I can understand, morally, why this equivalence ought to hold. The only natural functor I've been able to find between the two is given firstly by factorizing $$\int_{\mathcal{C}}P \overset{\pi_P}{\rightarrow} \mathcal{C} \overset{y}{\rightarrow} \widehat{\mathcal{C}}$$ through the Yoneda embedding to $\widehat{\int_{\mathcal{C}}P}$ and the canonical colimit functor $L : \widehat{\int_{\mathcal{C}}P} \rightarrow \widehat{\mathcal{C}}$ given since $y \circ \pi_P$ is a functor to a cocomplete category, and then getting to $\widehat{\mathcal{C}}/P$ by pullback $p$ along $P \rightarrow 1$. So the end result is a functor $$ p \circ L: \widehat{\int_{\mathcal{C}}P} \rightarrow \widehat{\mathcal{C}}/P$$ Does this functor provide the desired equivalence? There seems to be significant loss of information at $p$, but on the other hand $L$ does have the property that it makes the two respective Yonneda embeddings commute (i.e. $L \circ y = y \circ \pi_P$), so morally it seems that the information lost at $p$ should be the same amount of information lost by the projection $\pi_P : \int_{\mathcal{C}}P \rightarrow \mathcal{C}$ (which ought to make $p \circ L$ full and faithful; and I think essential surjectivity is straightforward.) -But I also get the sense I am chatting nonsense. If this functor does nothing of the sort then the question, I suppose, is what is the most elementary functor that provides this equivalence? -Any helpful comments will be greatly appreciated both by myself and my bloodshot eyes. - -REPLY [2 votes]: Sorry for digging up a decade-old post. I post this answer because I find this more conceptual. -Recall that a small full subcategory $\mathcal{A}\subset \mathcal{B}$ of a locally small category is said to be dense if $b=\operatorname{colim}_{b\to a\in \mathcal{A}\downarrow b}a$ for every $b\in \mathcal{B}$. Now say that a dense subcategory $\mathcal{A}\subset \mathcal{B}$ is nice if for each $a\in \mathcal{A}$, the functor $\mathcal{B}(b,-):B\to\mathsf{Set}$ preserves colimits. By the pointwise formula for left Kan extensions, we can verify that if $\mathcal{A}\subset \mathcal{B}$ is nice, then $B$ is a free cocompletion of $\mathcal{A}$. So it suffices to verify that $\mathcal{C}/P\subset \widehat{\mathcal{C}}/P$ is nice, but this is immediate from the density theorem and the Yoneda lemma.<|endoftext|> -TITLE: How to find Colin Day's PhD Thesis -QUESTION [9 upvotes]: A week or so ago, I was saddened to read Jim Stasheff's post on the AlgTop mailing list, announcing the passing of Colin Day, after a long bout with cancer. -I was thinking of reading Colin Day's PhD thesis in his memory, and trying to understand it (the title certainly sounds interesting); but I found it to be difficult to access. The reference is: - -Day, Colin -A topological construction of Vassiliev style invariants of links. -Thesis (Ph.D.)--The University of North Carolina at Chapel Hill. 1993. 99 pp. - -The standard course of action (I think) would be to write to the UNC library and to pay for access; but I'm wondering whether his PhD thesis, or a good summary of it, or anything related is freely available anywhere. -Does anyone know how Colin Day's work can be easily accessed? - -REPLY [15 votes]: I appreciate the interest shown. It apparently is on microfiche. -I am in contact with the UNC math library -and hope to have access I can share. -Update: It is now available at -http://hans.math.upenn.edu/~jds/ -scroll down to Colin... -Please let me know if you access it or if you have trouble accessing it.<|endoftext|> -TITLE: Parametric families for large torsion subgroups of elliptic curves -QUESTION [9 upvotes]: The following are two facts about $\mathbb{Z}/9\mathbb{Z}$, $\mathbb{Z}/10\mathbb{Z}$, $\mathbb{Z}/12\mathbb{Z}$, -$\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/8\mathbb{Z}$. -(a) According to Andrej Dujella's web-site "High rank elliptic curves with prescribed torsion", nobody has yet produced a parametric family of -elliptic curves, over $\mathbb{Q}(t)$, with these groups as the torsion subgroup and the rank of the curves at least $1$. All the other possible torsion -subgroups have such families. -(b) According to Noam Elkies' preprint "Three lectures on elliptic surfaces and curves of high rank", these are the only groups, -in the list of elliptic curve (over $\mathbb{Q}$) torsion possibilities, that do NOT appear as the torsion for elliptic $K3$ surfaces. -It seems that these observations ought to be related. -Is this wishful thinking, obvious, or somewhere between? Any observations, references etc. welcome. - -REPLY [7 votes]: Here's the short answer: -Observation (b) is, in my understanding, Elkies' explanation as to why the techniques he used to break most of the records on Dujella's list cannot be directly applied for these maximal torsion subgroups. -Observation (a) follows from observation (b) and the fact that Elkies techniques are the best available for finding elliptic curves with large rank over $\Bbb Q(t)$. This isn't a mathematically precise statement, but check the scoreboard. -End of Short Answer -I'm personally interested in trying to find elliptic curves over $\Bbb Q(t)$ with these torsion subgroups and positive rank. It would help with increasing the rank records over $\Bbb Q$ via specialization, and there might also be some applications to the elliptic curve method (ECM) of factorization (See Atkin and Morain's Paper). -I have a long standing personal obsession with the $\Bbb Z / 2 \times \Bbb Z / 8$ case. There the universal elliptic curve is: -$$E_{\Bbb Z / 2 \times \Bbb Z / 8} : y^2 = x(x + (2t)^4)(x + (t^2 - 1)^4)$$ -which has discriminant degree is 48. Elkies pointed out to me that it's a quadratic base change from a K3 surface, namely the universal elliptic curve with an 8-torsion point. -A general result of Alice Silverberg's ensures that the only $\Bbb Q(t)$-rational points of this elliptic curve are the 16 torsion points. If you want to find elliptic curves over $\Bbb Q(t)$ with 16-torsion points and positive rank, then you need to find rational curves on the associated surface other than the singular fibers and the 16 curves coming from the torsion sections. -No one has been able to find such a curve on this surface. I've asked around wether or not such a curve should exist, I've gotten one no, one yes, and nothing else very definitive. -My guess is that such curves do exist, but I don't have a very good idea where to start looking other than staring at a lot of data concerning rank 2 curves over $\Bbb Q$ with 16 torsion points and hoping to find a pattern. -Most of the same information applies for the other 3 torsion subgroups mentioned, but I don't know the universal elliptic curves by heart, and I don't already have a massive amount of data computed regarding specializations. Of course one can find models for them in Kubert's table. Interestingly Dujella attributes the records over $\Bbb Q(t)$ for these torsion subgroups to Kubert. It might be more appropriate to attribute some of them Fricke, or Levi at the latest.<|endoftext|> -TITLE: Continuous selections from sums of compact sets -QUESTION [8 upvotes]: This question is somehow related to the last open problem from Grothendieck's thesis about completeness of locally convex inductive limit. However, a particular case of the problem boils down to a very concrete question in Banach spaces: For two compact and absolutely convex sets $K_1, K_2$ the (Minkowski-) sum $K_1+K_2$ just consists of all sums $x_1+x_2$ with $x_j \in K_j$. One might ask whether one can choose the $x_j$ depending continuously on the sum, that is - -Are there continuous decomposition maps $f_j: K_1+K_2 \to K_j$ with $f_1(z)+f_2(z)=z$ for all $z\in K_1+K_2$? - -The answer to this question is negative, as I learned from Petr Holicky: -Let $e_n$ be the unit vectors of $\ell_2(\mathbb N_0)$, $K_1=\overline{\Gamma(e_0+\frac{1}{n} e_n: n\in\mathbb N)}$ (where $\Gamma$ denotes the absolutly convex hull) and $K_2 = \overline{ \Gamma(-e_0+\frac{1}{n} e_n: n\in\mathbb N)}$. -In this example one has $K_1 \subseteq 3 K_2$ and there is a trivial continuous selection -$f_j: K_1+K_2 \to C K_j$ with $f_1(z)+f_2(z)=z$ for $C=3$ (namely, $f_1(z)=z$, $f_2(z)=0$). -Now the question becomes: - -Is there a universal constant $C>0$ such that for all absolutely convex compact sets $K_1,K_2$ in a Banach space there are continuous decomposition maps $f_j: K_1+K_2 \to C K_j$ with $f_1(z)+f_2(z)=z$ for all $z\in K_1+K_2$? - -A close look at Holicky's example in an article "(LB)-spaces of vector-valued continuous functions", Bull. Lond. Math. Soc. 40 (2008), no. 3, 505–515 showed that necessarily $C\ge 2$. - -REPLY [4 votes]: No, there does not exist any such universal constant C. -I'll build up a counterexample inductively. First, suppose that we have the following. - -(i) Let $K_1,K_2$ be compact and absolutely convex subsets of Hilbert space $H$, and $C\ge0$ be a constant such that there do not exist continuous functions $f_j\colon K_1+K_2\to CK_j$ such that $f_1(x)+f_2(x)=x$. - -For example, as long as $K_1+K_2$ contains at least one nonzero element then they satisfy (i) for any constant $C < 1$. I'll show that we can construct new compact and absolutely convex sets $\tilde K_1,\tilde K_2$ for which (i) holds with $C$ replaced by $C+2$. Applying this process inductively shows that there exist sets $K_1,K_2$ satisfying (i) for any fixed $C \ge 0$. -Let $K_1,K_2,C$ be as in (i). By embedding $H$ in a larger Hilbert space if necessary, we can assume that we have unit vectors $u_{mn},v_{1mnk},v_{2mnk}$ ($m,n,k=1,2,\ldots$), all of which are mutually orthogonal and orthogonal to $K_1+K_2$. Let $(x_{1,1},x_{2,1}),(x_{1,2},x_{2,2}),(x_{1,3},x_{2,3}),\ldots$ be a sequence dense in $K_1\times K_2$. Set $w_{jmnk}=x_{jm}+(-1)^j\frac1{mn}u_{mn}+\frac1{mnk}v_{jmnk}$. -Define the compact sets $\tilde K_j$ by -$$ -\tilde K_j=\overline{\Gamma\left(w_{jmnk}\colon m,n,k=1,2,\ldots\right)}. -$$ -Suppose that $f_j\colon\tilde K_1+\tilde K_2\to(C+2)\tilde K_j$ are continuous functions satisfying $f_1(x)+f_2(x)=x$. I'll show that this contradicts (i). -For $x$ in $H$, define $\theta_j(x)$ to be the infimum of $c\in\mathbb{R}_+$ such that $x\in c\tilde K_j$. Set $w_{jmn}=x_{j,m}+(-1)^j\frac1{mn}u_{mn}=\lim_{k\to\infty}w_{jmnk}$. Any $x$ in the subspace generated by $\tilde K_j$ can be decomposed as -$$ -\begin{align} -x = \lambda_j\tilde x+\sum_{m,n}\lambda_{jmn}w_{jmn}+\sum_{m,n,k}\lambda_{j,m,n,k}w_{jmnk}&&{\rm(1)} -\end{align} -$$ -for some $\tilde x\in K_j$. -The real coefficients $\lambda_{jmn},\lambda_{jmnk}$ are uniquely determined, and I will regard them as functions of $x$. Also, we can choose $\lambda_j$ to be nonnegative and as small as possible, in which case it is also uniquely determined. It can be seen that -$$ -\theta_j(x)=\lambda_j+\sum_{m,n}\vert\lambda_{jmn}\vert+\sum_{m,n,k}\vert\lambda_{jmnk}\vert. -$$ -Let $\pi\colon H\to H$ be the orthogonal projection onto the closed subspace generated by $K_1+K_2$. Then $K_j=\pi(\tilde K_j)$. -We have $\pi f_1(x)+\pi f_2(x)=x$ for all $x\in K_1+K_2$ so, by (i) and continuity of $f_j$, there exists $\bar m\in\mathbb{N}$ and $j\in\lbrace 1,2\rbrace$ such that $\pi f_j(x_{1\bar m}+x_{2\bar m})\not\in CK_j$. Wlog, suppose that $j=1$. Then, $f_1(x_{1\bar m}+x_{2\bar m})\not\in C\tilde K_1$ or, equivalently, $\theta_j(f_1(x_{1\bar m}+x_{2\bar m})) > C$. Set $y_j=f_j(x_{1\bar m}+x_{2\bar m})$. -Also, set $\epsilon=\theta_1(y_1)-C > 0$. Then fix $\bar n$ large enough that $\vert\lambda_{1\bar m\bar n}(y_1)\vert < \epsilon/2$. -Since $w_{1mnk}+w_{2mnk}\to w_{1mn}+w_{2mn}=x_{1m}+x_{2m}$ as $k\to\infty$, continuity of $f_j$ allows us to write -$$ -f_j(w_{1\bar m\bar nk}+w_{2\bar m\bar nk})=w_{j\bar m\bar nk}+y_j-w_{j\bar m\bar n}+(-1)^jr_k -$$ -where $r_k\to0$, and is in the subspace generated by $\tilde K_1\cap \tilde K_2$. -Note that the coefficients $\lambda_1,\lambda_{1mn},\lambda_{1mnk}$ in expansion (1) are all the same for $y_1-w_{1\bar m\bar n}$ as for $y_1$, except for $\lambda_{1\bar m\bar n}$ which satisfies -$$ -\vert\lambda_{1\bar m\bar n}(y_1-w_{1\bar m\bar n})\vert\ge1-\vert\lambda_{1\bar m\bar n}(y_1)\vert > 1-\epsilon/2 > 1+\vert\lambda_{1\bar m\bar n}(y_1)\vert-\epsilon. -$$ -So, -$$ -\theta_1(y_1-w_{1\bar m\bar n}) > 1+\theta_1(y_1)-\epsilon = C+1. -$$ -As $r_k\to0$ we have $\theta_1(y_1-w_{1\bar m\bar n}-r_k) > C+1$ for large $k$. Similarly, the coefficients in expansion (1) are the same for $w_{1\bar m\bar nk}+y_1-w_{1\bar m\bar n}-r_k$ as for $y_1-w_{1\bar m\bar n}-r_k$, except $\lambda_{1\bar m\bar n k}$. Also, $r_k$ is in the subspace generated by $\tilde K_1\cap\tilde K_2$ and, hence, $\lambda_{1\bar m\bar n k}(r_k)=0$. Therefore, -$$ -\begin{align} -&\vert\lambda_{1\bar m\bar n k}(w_{1\bar m\bar nk}+y_1-w_{1\bar m\bar n}-r_k)\vert-\vert\lambda_{1\bar m\bar n k}(y_1-w_{1\bar m\bar n}-r_k)\vert\cr -&\qquad\ge 1-2\vert\lambda_{1\bar m\bar n k}(y_1-w_{1\bar m\bar n})\vert\to1 -\end{align} -$$ -as $k\to\infty$. -So, -$$ -\begin{align} -\liminf_{k\to\infty}\theta_1(w_{1\bar m\bar nk}+y_1-w_{1\bar m\bar n}-r_k)&\ge\liminf_{k\to\infty}\theta_1(y_1-w_{1\bar m\bar n}-r_k)+1\cr -& > C+2. -\end{align} -$$ -So, $f_1(w_{1\bar m\bar n k}+w_{2\bar m\bar n k})\not\in(C+2)\tilde K_1$ for large $k$, giving the required contradiction.<|endoftext|> -TITLE: Orbits of exterior products -QUESTION [6 upvotes]: In linear algebra one learns a lot of normal forms (Which I want to think of as a classification of the orbits of a group action on some set). For example if $V$ is a $k$-vectorspace $GL(V)$ acts on $V$ with exactly two orbits - the orbit of $0$ and the other orbit. -Now what happens if we let $GL(V)$ act diagonally on -$V^{\wedge n}$ or $V^{\otimes n}$? Can one give a normal form for this group action or at least find the number of orbits ? -The motivation for this question comes from surgery theory. The number of isomorphism types of fake $n$-tori (for $n\ge 5$) is given by the number of orbits of $GL_n(\mathbb{F}_2)$ acting on $(\mathbb{F}_2^n)^{\wedge n-3}$. - -REPLY [5 votes]: To the best of my knowledge, already for $n=3$ case (which probably is close, if not identical, to $n=\dim(V)-3$ because of the duality argument) the situation is far from clear for $\dim(V)$ starting from $10$ or so (the complexity somewhat depends on the ground field). Some hints on that are in papers like [1], [2], [3], and many others which you find in Mathscinet or such asking for "classification of trivectors". -Of course, for $n=2$ the classification is straightforward.<|endoftext|> -TITLE: Does every feasible partial order relation on the natural numbers extend to a feasible linear order relation? -QUESTION [13 upvotes]: It is well known that every partial order on a set can be extended -to a linear order on that set. That is, for every partial order -$\lhd$ on a set $X$, there is a linear order $\prec$ on $X$ such -that ${\lhd}\subseteq\prec$, meaning $a\lhd b\implies a\prec b$. In the -most general case, with uncountable $X$, one appeals to Zorn's -lemma. For relations on $\mathbb{N}$, however, the linearization -process is effective. My question concerns the relative complexity -of such linearizations $\prec$ in comparison with the original -order $\lhd$. (See also Jirka Hanika's -related question on reducing minimal element search for partial orders to that of linear orders in the -case of finite orders.) -Question. Does every polynomial time decidable partial order -relation $\lhd$ on $\mathbb{N}$ extend to a polynomial time -decidable linear order relation on $\mathbb{N}$? -I expect not. I think there will be a polynomial time decidable -partial order relation on $\mathbb{N}$ that does not extend to any -polynomial time linear order. The reason I believe so is that the -natural method of constructing a linear order extending a given -partial order seems to proceed fundamentally in series rather than -parallel, in the sense that one gradually linearizes increasing -portions of the partial order, but one must keep track of what one -did earlier, in order not to conflict with later decisions. But -with a polynomial time construction, one cannot afford to inspect -the earlier parts of the linearization. -Meanwhile, every computable partial order $\lhd$ on $\mathbb{N}$ -does extend to a computable linear order $\prec$ on $\mathbb{N}$, and this is what I meant when I said that linearizaton is effective, by -the following procedure: at stage $n$, we know how the numbers up -to $n$ relate with respect to $\lhd$ and we have built the desired -relation $\prec$ on the numbers below $n$. The new number $n$ -divides this linear order into those that are $\lhd$-below $n$, -incomparable to $n$, and $\lhd$-above $n$. We may now proceed to -linearize $n$ into the order by placing $n$ as high as possible, -say, above all the nodes so far to which it is incomparable, if -any. This recursive procedure produces a linear order extending -$\lhd$. -The point for the question is that this algorithm seems to require -an exponential increase in the time complexity, since on a given -input one must construct the relation on all nodes up to that node -before knowing what to do, and this takes exponential time. Indeed, -it isn't even clear whether one should be able to find a -linearization in the class NP. -Question. Does every polynomial time partial order extend to -an NP linear order? -I expect not, since even a polynomial size certificate seems -insufficient in general to track the linearization construction, -which has exponential size. -Finally, let me point out that the analogue at the level of c.e. -orders attains the negative result. -Theorem. There is a c.e. partial order $\lhd$ on -$\mathbb{N}$ that does not extend to any c.e. linear order on -$\mathbb{N}$. -Proof. Part of the point is that every c.e. linear order is -actually decidable. Let $A,B\subset\mathbb{N}$ be computable -inseparable c.e. sets, meaning that they are disjoint c.e. sets and -there is no decidable set containing $A$ and disjoint from $B$. -Define the partial order $\lhd$ by placing every element of $A$ -below $0$ and $0$ below every element of $B$, but otherwise -elements are incomparable. This is a c.e. relation, since given two -numbers $a$ and $b$, we can say how they are related once they are -enumerated into $A$ or $B$, and otherwise they are unrelated. But -if the relation extends to a linear order $\prec$ on $\mathbb{N}$, -then the set of $\prec$-predecessors of $0$ will be a computable -separation of $A$ and $B$, a contradiction. QED -Can one similarly employ a polytime version of inseparability to -answer the main question? -There are similarly many analogues of the main question in terms of other complexity classes. Please answer if you have interesting positive or negative results for any of them. - -REPLY [5 votes]: A tiny step forward can be done by observing that some complexity classes, such as $EXP$, are fortunately closed under exponential slowdown. -Every exponential time decidable partial order relation $\triangleleft$ on $N$ extends to an exponential time decidable linear order relation on $N$. -The proof is directly based on the construction described in the question. -Denote the restriction of $\triangleleft$ to $\{0,1,...,n-1\}$ as $\triangleleft_n$. Construct a corresponding sequence of linear orders $\prec_n$, each also on $\{0,1,...,n-1\}$. -Start from $\prec_0=\triangleleft_0=\emptyset$ and continue recursively so that -each $\prec_n$ extends both $\triangleleft_n$ and $\prec_{n-1}$ by defining, for $x -TITLE: Contraction mapping with no fixed point -QUESTION [5 upvotes]: I am interested in constructing the following "counter-example" to the Banach's fixed point theorem. -Let $K=$ {$ g\in L_1: \|g\|=1, g(\cdot)\ge0 $}. - Clearly, $K$ is not a compact and $K$ is not closed. -My A question is: is it possible to construct a [edit] nonexpansive mapping $f: K\to K$ with no fixed point? ( i.e. a mapping $f$ such that [edit] for all $x\neq y\in K$ one has $\|f(x)-f(y)\| < \|x-y\|$. ) - -REPLY [4 votes]: There need not be a fixed point. First note that by composing with a conditional expectation onto the closed span of indicator functions of disjoint sets it is sufficient to build an example on $W:=\{x\in \ell_1 : x_i \ge 0, \sum x_i =1\}$. Given $x\in W$, define $y=Tx \in W$ by $y_1=0$, $y_2 = x_1/2$, and, for $k\ge 2$, $y_{k+1} = y_k/2 + x_{k+1}/2$. It is obvious that $T$ is nonexpansive. The inequality $\|Tx-Ty\|<\|x-y\|$ when $x\not= y$ follows from the fact that if $x\not= y$ there are coordinates $i$ and $j$ s.t. $x_iy_j$.<|endoftext|> -TITLE: Dualizable classifying spaces -QUESTION [7 upvotes]: If $G$ is a finitely generated free group, then its classifying space $B G$ can be presented as a finite CW complex (a finite bouquet of circles), and therefore is Spanier-Whitehead dualizable. Are there any other discrete groups $G$ with the property that $B G$ is Spanier-Whitehead dualizable? - -REPLY [2 votes]: This example is undoubtedly covered by John Klein's general response, but is somewhat illustrative in its own right. -By the Cartan-Hadamard theorem, the universal cover of every connected, non-positively curved manifold $M$ is homeomorphic to $\mathbb{R}^n$. So if $\Gamma = \pi_1(M)$ is the fundamental group of $M$, then $M = \mathbb{R}^n / \Gamma = B \Gamma$ is the classifying space of $\Gamma$. If $M$ is itself compact, it is dualisable. -One tends to consider the case when $\Gamma$ is a discrete subgroup of $Isom(\mathbb{H}^n)$, the group of isometries of hyperbolic space. If $\Gamma$ acts freely and cocompactly on $\mathbb{H}^n$, then $B\Gamma$ is stably dualisable.<|endoftext|> -TITLE: If graph is tree what can be said about its adjacency matrix ? -QUESTION [8 upvotes]: Question If graph is tree what can be said about its adjacency matrix ? And vice versa ? -Especially I am interested in case when graph is bipartite graph. -Such graphs are related to error-correction codes (see e.g. Adjacency matrices of graphs as parity check matrices of error correcting codes ). -If they are trees belief propagation is known to produce exact results. - -REPLY [4 votes]: A graph is bipartite iff the odd powers of the adjacency matrix have all 0's on the diagonal. So this implies that the sum of the $i$-th powers of the eigenvalues is 0 for each odd $i$. Since the adjacency matrix is symmetric, it has real eigenvalues. Thus, the eigenvalues are real numbers $\lambda_1,\dots ,\lambda_n $ with $\sum_{j=1}^n \lambda_j^i = 0$ for each odd $i$. -I am guessing this probably should mean that the nonzero eigenvalues come in pairs of equal magnitude and opposite sign. I wonder if there is a good trick for efficiently proving this sort of thing -- about collections of real numbers satisfying such an infinite family of relations? (If so, I don't know this trick.) -${\bf Edit:}$ Douglas Zare proved my above conjecture as a comment, so it is true for bipartite graphs that the nonzero eigenvalues of the adjacency matrix come in pairs of equal magnitude and opposite sign.<|endoftext|> -TITLE: When does the doubling map preserve measure ? -QUESTION [8 upvotes]: Let $(G,+)$ be a locally compact (abelian) group endowed with its natural Haar measure. The doubling-map $T_2 : g \in G \mapsto g + g $ defines an endomorphism of $G$. - -Is there some (natural) condition on $G$ which ensures that the map $T_2$ is measure-preserving? - -When $G$ is finite, it is of course the same as asking $G$ to be of odd order (i.e. that there's no $2$-torsion). But $T_2$ may fail even when $G$ has no $2$-torsion, as shows $G = \mathbb{R}$. The group of $2$-adic integers $\mathbb{Z}_2$ (with its natural topology) shows that lack of 2-torsion and compactness are not sufficient either. -It is well-known that $T_2$ is measure-preserving when $G = \mathbb{T} = \mathbb{R}/ \mathbb{Z}$, or even when $G = \mathbb{T}^d$ (despite the $2$-torsion). I believe this is still true for any compact connected (abelian) group (I think connectedness would prevent pathologies such as $\mathbb{Z}_2$) but I have no convincing argument so far. - -REPLY [6 votes]: For an endomorphism $\rho$ to preserve Haar measure on a compact group $G$, it is necessary and sufficient that $\rho$ be surjective. One can easily check that the image $\rho^*(m)$ of Haar measure $m$ under $\rho$ is a translation-invariant probability measure as long as $\rho$ is surjective. If $\rho$ is not surjective, the image $H=\rho(G)$ is a compact subgroup, either of positive finite index or infinite index. In either case, $\rho^{-1}(H)$ will not have the same Haar measure as $H.$ -When $G$ is compact and connected, the doubling map is surjective (one way to see this is to observe that the dual $\widehat{G}$ is torsion-free, so the doubling map induces an injective map on the dual). -More generally, if the dual of the compact group $G$ lacks $2$-torsion, then the doubling map will be surjective, and so will preserve Haar measure. -Edit: The following argument and conclusion are in error, as Yves' comment shows. - -The structure theorem for locally compact abelian groups reduces the general question to two cases: (1) $G$ is compact, and (2) $G$ is discrete. Fix a compact group $G$ where the doubling map preserves Haar measure, and let $\rho:G\to G$ be the doubling map. By the structure theorem, $G$ has an open subgroup $W$ isomorphic to $K \times \mathbb R^n$ for some nonnegative integer $n,$ where $K$ is a compact group. Since the doubling map does not preserve Haar measure on $\mathbb R$, we must have $n=0.$ If $\rho$ preserves Haar measure on $G$ then the induced map $\tilde{\rho}:G/W\to G/W$ preserves Haar measure on $G/W,$ so $\tilde{\rho}$ is injective. -Since $\tilde{\rho}$ is injective, we have $\rho^{-1}(W)=W.$ Thus $\rho(W)$ is a subgroup of $W$, and by the argument in the first paragraph, we necessarily have $\rho(W)=W.$ -We have (edit: not) shown: -The doubling map on G preserves Haar measure if and only if $G$ has a compact open subgroup $W$ such that: -(i) The doubling map on $G/W$ is injective (i.e. $G/W$ lacks 2-torsion). -(ii) $\widehat{W}$ lacks $2$-torsion.<|endoftext|> -TITLE: Nonisotopic homotopy equivalent Morse functions -QUESTION [15 upvotes]: One can cut a manifold up along the critical levels of a Morse function and deduce something about the topology. In particular the critical points (and the connecting gradient flowlines) define a chain complex which can be used to compute homology. -A minimal Morse function on a compact manifold is a Morse function which has exactly enough critical points to generate the homology (e.g. perfect if the homology is torsion free). The existence of a minimal Morse function on a simply-connected h-cobordism between simply-connected manifolds of high enough dimension is equivalent to the h-cobordism theorem. -Two Morse functions are called 'homotopy equivalent' if there is a diffeomorphism isotopic to the identity sending critical levels to critical levels. I don't know if this is the best terminology, but it seems to be the one people use. -Matsumoto proved that on a simply-connected manifold any two minimal Morse functions which are homotopy equivalent are isotopic through Morse functions. This seems to be a beefed up version of the argument which proves the h-cobordism theorem. -Of course, in the non-simply-connected case one expects something different. On page 45 of the English translation of Sharko's book 'Functions on manifolds: algebraic and topological aspects' (Translations of Mathematical Monographs, AMS 1993), just after Corollary 2.3 he claims that there are examples of homotopy equivalent but nonisotopic Morse functions on non-simply-connected manifolds. -My question is: can anyone give me an example of a pair of homotopy equivalent but nonisotopic Morse functions on a non-simply-connected manifold? What about minimal ones? -The reference Sharko gives there seems to have nothing to do with this (it points to a paper of Heller, 'Homological resolutions of complexes with operators', Ann. Math. 1954) so I assume the bibliographic numbering is flawed. If I check the adjacent and promising-looking reference (Hatcher and Wagoner's 'Pseudoisotopies of compact manifolds', Asterisque Volume 6, 1973) I find that I would need to know more Cerf theory to work out what this example might look like. -Any ideas would be very welcome! - -REPLY [8 votes]: I no longer think 3-dimensional lens spaces are a productive strategy. What you need is to have a manifold $M$ as a level-set of the Morse function and you need a non-trivial diffeomorphism of $M$ to be pseudo-isotopic to the identity. -The idea is that roughly, between any two consecutive critical levels of your Morse function (modulo the degeneracies on the ends) your flow is giving you a pseudo-isotopy diffeomorphism of $M$ -- by that I mean a diffeomorphism of $M \times [0,1]$ which is the identity on $M \times \{0\}$. -So provided $M$ admits non-trivial pseudo-isotopy diffeomorphisms you've got a head-start. Cerf's theorem says pseudo-isotopy implies isotopy for simply connected manifolds. So I suppose that's a tool Matsumoto uses. In particular, 2-manifolds don't admit any non-trivial pseudo-isotopy diffeomorphisms so that's likely not a productive route to go. -I think there's sort of a tautological answer to your question. Let $M$ be compact manifold. The manifold I'm going to give you is $M \times [0,1]$. And I'm going to give you two Morse functions on it with no critical points. The first one is $f(p,t) = t$, where $(p,t) \in M \times [0,1]$. Let $\phi : M \times [0,1] \to M \times [0,1]$ be a non-trivial pseudo-isotopy diffeomorphism, then the 2nd function is $f \circ \phi$. This is sort of a cheezy non-constructive "example". -By design, these are homotopy-equivalent but non-isotopic Morse functions. The critical point set of both is empty. That such functions $\phi$ exist when $M$ is not simply-connected, I'm not sure who first noticed this. But I think there are examples in the Hatcher-Wagoner reference you cite, when $M = (S^1)^n$ for $n \geq 7$. Hatcher's The Second Obstruction for pseudo-isotopies (1972) mentions that $\pi_0$ of the pseudo-isotopy group for $M$ is non-trivial whenever $M$ is not simply connected and the dimension is at least seven. -I think if you want an example on a closed manifold you'll have to work a fair bit more but I suspect capping off $(S^1)^n \times [0,1]$ with copies of $D^2 \times (S^1)^{n-1}$ on both sides should work with the Hatcher-Wagoner classes.<|endoftext|> -TITLE: What is the theory of local rings and local ring homomorphisms? -QUESTION [31 upvotes]: It is well-known that the category of local rings and ring homomorphisms admits an axiomatisation in coherent logic. Explicitly, it is the coherent theory over the signature $0, 1, -, +, \times$ with the usual axioms for rings, plus the axioms -$$0 = 1 \vdash \bot$$ -$$\top \vdash (\exists b . \; a \times b = 1) \lor (\exists b . \; (1 - a) \times b = 1)$$ -See, for example, [Sheaves in Geometry and Logic, Ch. VIII, §6]. Unfortunately, because homomorphisms are only required to commute with the various things in the signature, the homomorphisms here are just ring homomorphisms and need not be local. It appears to me that the neatest way to fix this is to introduce a unary relation symbol $(\quad) \in \mathfrak{m}$, with the intention that $\mathfrak{m}$ is interpreted as the unique maximal ideal of the local ring. Then, by the usual rules for homomorphisms of models, a homomorphism $R \to R'$ must map elements of $\mathfrak{m}$ to elements of $\mathfrak{m}'$. But is there a way to axiomatise the theory so that - -we get a coherent, or at least geometric theory, and -the category of models in $\textbf{Set}$ is indeed the category of local rings and local ring homomorphisms, and -the structure sheaf homomorphism $f^\ast \mathscr{O}_{Y} \to \mathscr{O}_{X}$ of morphism of locally ringed spaces $X \to Y$ is a homomorphism in the category of models for this theory? - -Ideally, we'd like to define $\mathfrak{m}$ to be the subsheaf of nowhere invertible sections defined by -$$\{ s \in \mathscr{O} : \nexists t . \; s \times t = 1 \}$$ -but unfortunately $\nexists t . \; s \times t = 1$ is not a geometric formula. (The formula $\forall t . \; s \times t \ne 1$ is equivalent to the previous one but has the same defect.) We can salvage one half of the biimplication as the geometric sequent -$$(a \in \mathfrak{m}) \land (\exists b . \; a \times b = 1) \vdash \bot$$ -which merely expresses the requirement that "$a$ is not in $\mathfrak{m}$ if $a$ is invertible", but we also need to express the requirement that "$a$ is in $\mathfrak{m}$ if $a$ is not invertible". One possibility is the following: -$$\top \vdash (\exists b . \; a \times b = 1) \lor (a \in \mathfrak{m})$$ -These two axioms appear to give the correct characterisation of $\mathfrak{m}$ in intuitionistic first order logic: it is easy to derive from these axioms that -$$a \in \mathfrak{m} \dashv \vdash \nexists b . \; a \times b = 1$$ -so the interpretation of $\mathfrak{m}$ is completely determined by the axioms, at least in a topos. -But does every local ring object (in the sense of the first paragraph) admit an $\mathfrak{m}$ satisfying these axioms? The answer appears to be no, for the reason that these axioms assert that a every section of a sheaf of a local ring admits an open cover of the space by open sets on which the restriction is either invertible or nowhere invertible – and this is certainly not true in the contexts of interest. Can this idea be rescued with a more clever approach? - -REPLY [5 votes]: The following should give some insight to this question, which was more or less the inspiring example of Diers' work on locally multipresentable categories. This involves the "multi" versions of usual universal constructions involved in Gabriel-Ulmer duality. -Locally (finitely) multipresentable categories are (finitely) accessible categories with multi-colimits - or equivalently, (finitely) accessible categories with connected limits commuting with (finitely) filtered colimits. -It is known that any locally finitely multipresentable category can be axiomatized by a small finite-limit/coproduct sketch, equivalently by a disjunctive geometric theory - which often happens to be finitely disjunctive, as in the example below. -In particular, Diers proposed in his thesis a process to construct locally finitely multipresentable categories as follows. If you start with a locally (finitely) multipresentable category $\mathcal{A}$ - hence in particular with a locally (finitely) presentable category - and a choice of a small family $ \Gamma $ of cones of finitely presented maps, then you can take the category $\mathcal{A}_\Gamma$ of objects that are injective relatively to cones in $\Gamma$, and morphisms that are right orthogonal to each arrows involved in the cones of $\Gamma$ (in general arrows in the cones of $\Gamma$ are chosen in a left class in an orthogonal factorization system, so the morphisms in $\mathcal{A}_\Gamma$ are in particular in the corresponding right class). -Then $\mathcal{A}_\Gamma$ can be shown to be locally multipresentable itself, and we have a (finitely) accessible functor $ \mathcal{A}_\Gamma \hookrightarrow \mathcal{A}$ which is a right multi-adjoint and only has to be relatively full and faithful, but not necessarily full. Multireflexiveness involves a kind of small object argument returning a factorization of morphisms with local codomain. -In the example of local rings, take as $\Gamma$ as consisting of the single cone made of the following two finitely presented localizations of the free ring on one generator as in Zariski topology -$ \mathbb{Z}[X] \twoheadrightarrow \mathbb{Z}[X,Y]/(XY-1)$ and $\mathbb{Z}[X] \twoheadrightarrow \mathbb{Z}[X,Y]/((X-1)Y-1)$ -Then as expected the objects of $CRing_\Gamma$ are the local rings, but moreover conservativity of a ring homomorphisms just has to be tested relatively to either one of those localizations. Hence $CRing_\Gamma$ is the category with the good choice of morphisms. In this case relative fullness comes from the fact that conservative morphisms are a right class in the (localization, conservative) factorization system in $CRing$ and hence have right cancellability. And the result above says that $CRing_\Gamma = LocRing^{Cons} $ is a locally finitely multipresentable category. Hence there must be a small finite-limit/coproduct sketch axiomatizing it, and hence a disjunctive theory in some convenient signature (though I must admit I don't know which one). -However I don't think that the finite limit part of this sketch is the same as the finite limit sketch of commutative rings, because there is a result by Adamek-Rosicky suggesting that the category of models of a finite limit-coproduct sketch should be a full multireflective subcategory of the category of models of the underlying finite limit sketch. -There is also an interesting paper from Johnstone about disjunctive theories and their link with Diers multipresentability that may be of interest relatively to this question.<|endoftext|> -TITLE: How to mentor an exceptional high school student? -QUESTION [16 upvotes]: I have a unique and, quite truthfully, humbling opportunity. The parents of an exceptionally talented high school freshman have reached out to me and asked if I might be able to help. -This kid is seriously good; he came to our state high school math contest and blew away the competition. He is active in a local Math Circle and is in extremely capable hands, and his parents inform me that he is extremely active in the Art of Problem Solving and the Worldwide Online Olympiad Training program. -I think my role, insofar as I could help, would be to introduce him to advanced topics and/or research. I am thinking about suggesting to him that he read Apostol's intro book on analytic number theory, or Stillwell's Naive Lie Theory; he might also enjoy some serious combinatorics and/or learning about the partition function. (Or, ... ?) -Does anyone have any suggestions for helping such a student? -Thank you! -Frank - -REPLY [12 votes]: And don't forget to play ball with him every once in a while.<|endoftext|> -TITLE: Is the pairing between contours and functions perfect (modulo the kernel given by Stokes' theorem)? -QUESTION [10 upvotes]: Let $s: \mathbb C^n \to \mathbb C$ be a homogeneous degree-$d$ polynomial which is nonsingular (in the sense that the hypersurface it defines in $\mathbb{CP}^{n-1}$ is smooth; equivalently the discriminant does not vanish). Write $\Omega$ for the canonical holomorphic $n$-form $\mathrm d z_1\cdots\mathrm d z_n$ on $\mathbb C^n$. I am interested in integrals against the measure $\exp(s)\Omega$. -Specifically, one can ask about integrals of the form $\int_\gamma f\exp(s)\Omega$ for some contour $\gamma$, where $f$ is a polynomial. -By a "contour" I mean a boundary-free (but not compact) real-$n$-dimensional submanifold of $\mathbb C^n$. For the integral to converge, you'd like exponential decay at the ends of $\gamma$; i.e. at the ends of $\gamma$ we should have $\Re(s)\to -\infty$. By Stokes' Theorem, small perturbations of $\gamma$ don't matter, and neither do compact cycles. So all that matters in $\gamma$ is the class it represents in the relative homology group $\mathrm H_n(\mathbb C^n,\lbrace \Re(s)\lt 0\rbrace)$. If $n\gt 1$, by the long exact sequence for relative homology this group is equal to $\mathrm H_{n-1}(\lbrace \Re(s)\lt 0\rbrace)$. (When $n=1$ it is the quotient of this by $\mathrm H_0(\mathbb C^n)$ which is one-dimensional.) -Stoke's theorem also implies that integrals of total derivatives (of functions that vanish at infinity) vanish. Namely (provided $\gamma$ is as above), if $f$ is of the form $\frac{\partial g}{\partial z_i} + \frac{\partial s}{\partial z_i}g$ for some $i=1,\dots,n$ and some polynomial $g$, then $\int_\gamma f\exp(s)\Omega = 0$. - -Thus the integral determines a pairing of the form - $$ \textstyle \mathrm H_n\bigl(\mathbb C^n,\lbrace \Re(s)\lt 0\rbrace\bigr) \otimes \bigl(\mathbb C[z_1,\dots,z_n] / \sum_i (\text{image of } \frac{\partial}{\partial z_i} + \frac{\partial s}{\partial z_i})\bigr) \to \mathbb C $$ - My question is whether this pairing is perfect. If it matters, I am primarily interested in the case when $s$ is generic. - -The answer is (somewhat trivially) "yes" when $n=1$ or when $d=2$. When $n=2$ I convinced myself that the answer is yes for "diagonal" $s(z_1,z_2) = z_1^d + z_2^d$; at least, we calculated the dimension for $d=3$, and a similar argument should work for higher $d$. At least some of the people I have asked have given the prediction that the answer is "no" in general. -For generic $s$ I have complete control over the "algebraic side": I can give an explicit basis for the quotient and for this basis and the monomial basis explicit formulas for the map. An easy part of this is to see that this "algebraic" piece is $(d-1)^n$-dimensional. -On the other hand, I'm not very good at algebraic topology and algebraic geometry, and the homology group is definitely inaccessible to pure algebra and is rather a piece of real algebraic geometry. After talking with a number of folks at Berkeley, we've been unable to calculate even the dimension of the relative homology group, but maybe there were tricks we didn't think of. Or maybe there's an a priori reason why the pairing is perfect, and then I would have a calculation of this "real topology" side. - -REPLY [5 votes]: This does not address the perfectness of the pairing, but just the dimension calculation on the topology side. -Call your hypersurface $H(d,n-2)$. Introduce one more variable $z_{n+1}$ and consider the polynomial $s(z_1,\dots ,z_n)+z_{n+1}^d$. It defines a smooth hypersurface $H(d,n-1)$ in $P(n)$ whose intersection with $P(n-1)$ is $H(d,n-2)$. -Your set of all $z\in \mathbb C^n$ where $s(z)$ has negative real part is the product of an open disk with the set where $s(z)=-1$. (Choose a branch of the $d$th root function in that half-plane.) And the latter set is the difference $H(d,n-1)-H(d,n-2)$. So the question is about the homology of this difference set. -The homology, in fact the diffeomorphism type, of $H(d,n-2)$ is independent of the choice of polynomial, since the space of all nonsingular homogeneous complex polynomials of degree $d$ in $n$ variables is connected. So we can use the simplest polynomial $\Sigma_j z_j^d$ if we want. -The relative homology $H_j(P(n+1),H(d,n))$ is trivial for $j\le n$ by a theorem of Lefschetz that can be proved using Morse theory. Thus the homology of $H(d,n)$ is isomorphic to that of complex projective space in dimensions $0$ through $n-1$. The same is true in dimensions $n+1$ through $2n$ by duality, since $H(d,n)$ is an oriented $2n$-manifold (although the inclusion $H\to P(n+1)$ is not an isomorphism in these higher dimensions, just mod torsion). So basically the homology of $H(d,n)$ looks like that of $P(n)$ plus something extra in the middle dimension. -You can calculate the rank of this extra bit by calculating the Euler characteristic of $H(d,n)$. For this let's use that simplest equation. $H(d,n)$ has an obvious map to $P(n)$, forgetting the last coordinate. This map is $1:1$ over $H(d,n-1)$ and otherwise $d:1$. Using this you can inductively find the Euler characteristic of $H(d,n)$ and thus the rank of the extra homology in the middle. -The homology of the difference $H(d,n)-H(d,n-1)$ can now be found by a long exact sequence, using also a Thom isomorphism $H_j(H(d,n),H(d,n)-H(d,n-1))=H_{j-2}(H(d,n-1)$. Except for $H_0=\mathbb Z$, it is all in dimension $n$ and of rank $(d-1)^{n+1}$. -EDIT Actually the rank calculation is easier with the affine variety than the projective: Write $V(d,n)=H(d,n)-H(d,n-1)$. Then $V(d,n)$ is a $d$-fold branched cover of affine $n$-space, unbranched except over $V(d,n-1)$ where it is totally branched. If we write $e(d,n)$ for the Euler number of $V(d,n)$, it follows that $e(d,n)-1=(1-d)(e(d,n-1)-1)$.<|endoftext|> -TITLE: What prerequisites do I need to read the book Ricci Flow and the Poincare Conjecture, published by CMI -QUESTION [14 upvotes]: As mentioned in the title, I want to understand the proof of Poincare Conjecture by Perelman, what prerequisites do I need? - -REPLY [7 votes]: My humble advice for learning about Ricci flow generally, after obtaining some background in Riemannian geometry, would be to start with a book which gets you to important results quickly. An excellent book is the one by Peter Topping. (The only typo I observed there is the one regarding backwards uniqueness, which is now due to Brett Kotschwar.) After that, there are excellent books on the differentiable spherical space form theorem by Brendle and Andrews--Hopper; see also the original papers of Boehm--Wilking and Brendle--Schoen. -What is irreplaceable is to read and to reread the original works by the masters, i.e., Hamilton and Perelman. A collection of Ricci flow papers, mostly by Hamilton, is edited by H.D. Cao, etal.; this is a convenient place to get Hamilton's papers in one place. Perelman's papers are on arXiv. There are a number of excellent expositions of their work (focused on Perelman's work), which actually go beyond expositions and include various degrees of original work, namely in alphabetical order: Bessieres--Besson--Boileau--Maillot--Porti, Cao--Zhu, Kleiner--Lott, and Morgan--Tian. -The above remarks only pertain to Riemannian Ricci flow.<|endoftext|> -TITLE: Geometric realization of Hochschild complex -QUESTION [10 upvotes]: Let $A$ be a commutative $\mathbb{C}$-algebra, and consider $C_{\bullet}(A,A)$ the simplicial Hochschild homology module of $A$ with respect to itself (i.e. $C_{n}(A,A)=A^{\otimes (n+1)}$). This is a simplicial commutative $\mathbb{C}$-algebra and we can take its $Spec$ levelwise, to get a cosimplicial $\mathbb{C}$-scheme $X_A :=Spec(C_{\bullet}(A,A))$. If $A$ is of finite type over $\mathbb{C}$, then we may take levelwise the associated topological space for the analytic topology, and get a cosimplicial topological space, denoted $X_{A}^{top}$. My question is whether somebody knows what the geometric realization of $X_{A}^{top}$ looks like (e.g. any relation to the topological free loop space of $(Spec(A))^{top}$?). - -REPLY [3 votes]: Sam Gunningham had the right idea, but I would like to fix it up for the construction of the homotopy pullback that I am aware of. -Alternatively one can compute $HH(A;A)$ as $\mathrm{Tor}^{A^e}(A,A)$ where $A^e$ is the enveloping algebra of $A$, $A\otimes A^{op}$, since your $A$ is commutative $A^{e}\cong A\otimes A$. These Tor groups can be calculated via the simplicial complex which in degree $n$ is $A\otimes (A^{e})^{\otimes n}\otimes A\cong A^{\otimes {2n+2}}$. As a side note the geometric realization of this complex gives a topological commutative ring (Because $A$ is commutative) which models $THH^{HZ}(HA;HA)$ via the Eilenberg-Maclane functor. Applying your contravariant functor to spaces to the simplicial algebra we obtain a cosimplicial space modeling the homotopy pullback $Spec(A)^{top}\times_{Spec(A)^{top}\times Spec(A)^{top}}\times Spec(A)^{top}$ (see 3.3 of http://www.math.uiuc.edu/~reldred2/tot-primer.pdf ) which as Sam pointed out, models the free loop space.<|endoftext|> -TITLE: Bounding the commutator [A,B] in terms of the numerical radius -QUESTION [17 upvotes]: Given a norm $N$ over ${\bf M}_n(\mathbb C)$, it is a natural question to find the best constant $C_N$ such that -$$N([A,B])\le C_N N(A)N(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$ -The answer is known at least in the following cases: - -the operator norm $\|A\|_2=\sup\frac{\|Ax\|_2}{\|x\|_2}$ where the norm over $\mathbb C^n$ is the standard Hermitian $\|x\|_2^2=\sum_j|x_j|^2$. Then -$$\|[A,B]\|_2\le2\|A\|_2\|B\|_2$$ -is optimal for $n\ge2$. -the Frobenius norm $\|A\|^2_F=\sum_{i,j}|a_{ij}|^2$. Then a theorem by Böttcher & Wentzel (2008) tells us that -$$\|[A,B]\|_F\le\sqrt2\|A\|_F\|B\|_F,$$ -and again this is optimal. - - -I have a third norm in mind, yet of a different nature: the numerical radius - $$r(A)=\sup_{x\ne0}\frac{|x^*Ax|}{\|x\|^2}.$$ - This is the smallest radius of a disk $D(0;r)$ containing the numerical range (or Hausdorffian) of the matrix. - What is the optimal constant $C_{nr}$ such that $r([A,B])\le C_{nr}r(A)r(B)$ for all $A,B$ in ${\bf M}_n(\mathbb C)$ ? - -Let me point out that $r$ is not submultiplicative. We have at best $r(MN)\le 4r(M)r(N)$, which gives by the triangle inequality $r([A,B])\le8r(A)r(B)$, but this is certainly not optimal. However, it is a super-stable norm, in the sense that $r(M^k)\le r(M)^k$ for every $k\ge1$. -This question naturally extends to $n$-commutators, in the spirit of my previous question Standard polynomials applied to matrices . -Edit. See below Piotr Migdal's answer and my adaptation of it. It gives $C_{nr}=4$. - -REPLY [6 votes]: The answer by Piotr Migdal can be modified to give the accurate inequality -$$r([A,B])\le4r(A)r(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$ -The only new argument is that for every matrix $M$, there exists an angle $\theta$ such that $r(M)=\|{\rm Re}(e^{-i\theta}M)\|_2.$ -Actually, we do have -$$r(M)=\sup_\alpha\|{\rm Re}(e^{-i\alpha}M)\|_2.$$ -Hereabove, the real part is defined as ${\rm Re} N=\frac12(N+\bar N^T)$. Notice that I employ the notation $\|\cdot\|$ (operator norm) which coincides with $\sigma_1$. -Let us apply this to $M=[A,B]$. With $\theta$ as above, let us decompose $e^{-i\theta}A=A_{\theta h}+iA_{\theta a}$. Then let us proceed as Piotr did: -$$r([A,B])=\|{\rm Re}[e^{-i\theta}A,B]\| = \|i[A_{\theta h},B_a]+i[A_{\theta a},B_h]\|\le2(\|A_{\theta h}\|\cdot\|B_a\|+\|A_{\theta a}\|\cdot\|B_h\|),$$ -which gives -$$r([A,B])\le4r(e^{-i\theta}A)r(B).$$ -Hence the result.<|endoftext|> -TITLE: Rank of the absolute-value matrix $|M|$ vs. rank of $M$ -QUESTION [6 upvotes]: Let $M$ be a real matrix of rank $r$ (and let us set $M=UV^T$, with $U,V^T\in\mathbb{R}^{n\times r}$, to fix the notation). -Let $|M|$ be the matrix obtained by taking the absolute value of each entry of $M$. Clearly $\operatorname{rk} |M|$ can be much smaller than $r$ --- take for instance Hadamard matrices. -However, what about the other direction? Is there a way to bound $\operatorname{rk} |M|$ from above in terms of $r$? -If I take random $U$ and $V$ with $n=200$, $r=2$, numerically $\operatorname{rk} |M|$ seems to be between 120 and 150 --- so definitely not as low as $r$ but also suspiciously far from being full-rank. - -REPLY [6 votes]: I do not think much/anything can be done. -Let us leave the simple special cases of rank $M$ equal $0$ or $1$ aside. -So, an example of a $n$ times $n$ rank two matrix $M$ such that the rank of $|M|$ is full: -Take the two vectors $e=(1, \dots, 1)$ and $u = (0, -1, -2, \dots, -(n-1))$. -Consider the matrix $M$ formed by $e$ and $je + u$ for $j=0, \dots n-2$. -The absolute value $|M|$ has full rank since the line for any $j\ge 1$ is $je+u + v_j$ where -$v_j = (2 \max(0, (i-1)-j) )_i$.` and thus has exactly the first $j+1$ coordinates equal to $0$. -So, we get $e$, and $u= - |u|$ and all the $v_j$ for $j=1, \dots, n-2$ in the spanned space, and these $n$ are independent. -Variations of this should give (all?) kinds of intermideate phenomena. -(Edit: slight change and explanation; perhaps the orginal would also work but the present example seems clearer and was the 'real' original, which I thought I should modify while typing for some dubious reasons. Sorry for the edit-noise.)<|endoftext|> -TITLE: Sum of Squares of Normal distributions -QUESTION [8 upvotes]: Given $X_i \sim \mathcal{N}(\mu_i,\sigma_i^2)$, for $i = 1,\dots,n$. How does one find the distribution of $D = \sum_{i=1}^n X_i^2$? In the case that all the standard deviations are the same (i.e. $\sigma_i = \sigma_1$ for all $i$), the random variable $D/\sigma_1^2$ has a noncentral chi-squared distribution. But when I don't have this simplifying assumption I don't know what to do. Is there a well known distribution for this? - -REPLY [6 votes]: All you could conceivably want to know about the subject (and many things you might not) are in Mathai + Provost, Quadratic Forms in Random Variables.<|endoftext|> -TITLE: Algorithm to find all (up to isomorphism) perfect matchings of quartic plane graphs -QUESTION [5 upvotes]: I need to find all (up to isomorphism) perfect matchings of some quartic plane graphs. I haven't found any specific algorithm to give me all the perfect matchings. Does anybody know about such an algorithm or any results that could be useful when implementing such an algorithm? At the moment I can only think of a branch-and-bound approach. -I don't really expect there to be an algorithm for this specific case, but I thought I'd mention all properties of the graphs. Maybe there are results for plane graphs, planar graphs or quartic graphs. - -REPLY [3 votes]: I want to mention that you can exploit your favorite symbolic algebra package to do this job for smallish graphs. I've used this method successfully for graphs with 20-30 vertices. -The FKT algorithm expresses the number of matchings of a planar graph as a pfaffian, where each nonzero entry of the matrix is $\pm 1$, corresponding to an edge $e$ of the graph. Write down the same matrix but replace $\pm 1$ by $\pm x(e)$, where $x(e)$ is a formal variable corresponding to the edge $e$, and ask your favorite computer algebra package to take the determinant. The result will be $\sum_M \prod_{e \in M} x(e)$, where the sum is over all matchings $M$. So you can look at the output and immediately read off your matchings. -This is most useful when the graph is bipartite, as the Pfaffian then simplifies to a determinant, and where the graph is small and/or symmetric enough to carry out the orientation part of the algorithm by hand.<|endoftext|> -TITLE: How would Hilbert and Weber think about the Langlands programme? -QUESTION [29 upvotes]: Explanations to a general mathematical audience about the Langlands programme often advertise it as "non-abelian class field theory". They usually begin as follows: a modern style formulation of classical class field theory is to say that for a global field $K$, the Artin map defines an isomorphism from the group of connected components of the idele class group to the Galois group $\operatorname{Gal}(K^{ab}|K)$. Pushing this even further, we see that we have a canonical identification of characters of the idele class group with characters of the absolute Galois group $\operatorname{Gal}(\bar{K}|K)$. -Then people usually go on to say that this should extend to a correspondence between a certain class of $n$-dimensional Galois representations and a certain class of representations of $\operatorname{GL}_n(\mathbb{A}_K)$ (where $\mathbb{A}_K$ denotes the adeles of $K$), and very soon they have disappeared into (to me) far off realms. -While it should be clear from my description that I have no clue whatsoever concerning the Langlands programme, I know a little bit about global class field theory in its traditional formulation. That is, I understand it as a means to describe and classify abelian extensions of $K$ with prescribed ramifications, with the Artin map giving an isomorphism from a ray ideal class group of $K$ (say) to the Galois group of the corresponding ray class field over $K$. -So, my question is: - -Do there exist results in the global Langlands programme which give us back some down-to-earth, may be ideal-theoretic, insights about number field extensions? And the same question for yet open questions in the global Langlands programme: would their answers give us some sort of "classical" information? - -REPLY [12 votes]: This question deserves an expert answer such as this one by Emerton, but allow me to offer an outsider's perspective. The following remarks are taken from my expository article arXiv:1007.4426. -First recall that the proportion of primes $p$ for which $T^2+1$ has -no roots (resp. two distinct roots) in $\mathbf{F}_p$ is $1/2$ (resp. $1/2$), -and that the proportion of $p$ for which $T^3-T-1$ has no roots -(resp. exactly one root, resp. three distinct roots) in $\mathbf{F}_p$ is -$1/3$ (resp. $1/2$, resp. $1/6$). -What is the analogue of the foregoing for the number of roots $N_p(f)$ of -$f=S^2+S-T^3+T^2$ in $\mathbf{F}_p$? A theorem of Hasse implies that $a_p=p-N_p(f)$ lies in the -interval $[-2\sqrt p,+2\sqrt p]$, so $a_p/2\sqrt p$ lies in $[-1,+1]$. What -is the proportion of primes $p$ for which $a_p/2\sqrt p$ lies in a given -interval $I\subset[-1,+1]$? It was predicted by Sato (on numerical grounds) -and Tate (on theoretical grounds), not just for this $f$ but for all -$f\in\mathbf{Z}[S,T]$ defining an "elliptic curve without complex multiplications", -that the proportion of such $p$ is equal to the area -$$ -{2\over\pi}\int_{I}\sqrt{1-x^2}\;dx. -$$ -of the portion of the unit semicircle projecting onto $I$. The Sato-Tate -conjecture for elliptic curves over $\mathbf{Q}$ was settled in 2008 by -Clozel, Harris, Shepherd-Barron and Taylor. -There is an analogue for "higher weights". Let $c_n$ (for $n>0$) be the -coefficient of $q^n$ in the formal product -$$ -\eta_{1^{24}}= -q\prod_{k=1}^{+\infty}(1-q^{k})^{24}=0+1.q^1+\sum_{n>1}c_nq^n. -$$ -In 1916, Ramanujan had made some deep conjectures about these -$c_n$; some of them, such as $c_{mm'}=c_mc_{m'}$ if $\gcd(m,m')=1$ and -$$ -c_{p^r}=c_{p^{r-1}}c_p-p^{11}c_{p^{r-2}} -$$ -for $r>1$ and primes $p$, which can be more succintly expressed as the -identity -$$ -\sum_{n>0}c_nn^{-s}=\prod_p{1\over 1-c_p.p^{-s}+p^{11}.p^{-2s}} -$$ -when the real part of $s$ is $>(12+1)/2$, were proved by Mordell in -1917. The last of Ramanujan's conjectures was proved by Deligne -only in the 1970s: for every prime $p$, the number -$t_p=c_p/2p^{11/2}$ lies in the interval $[-1,+1]$. -All these properties of the $c_n$ follow from the fact that the corresponding -function $F(\tau)=\sum_{n>0}c_ne^{2i\pi\tau.n}$ of a complex variable $\tau=x+iy$ ($y>0$) in -$\mathfrak{H}$ is a "primitive eigenform of weight $12$ and level $1$" (which -basically amounts to the identity $F(-1/\tau)=\tau^{12}F(\tau)$). -(Incidentally, Ramanujan had also conjectured some congruences satisfied by -the $c_p$ modulo $2^{11}$, $3^7$, $5^3$, $7$, $23$ and $691$, such as -$c_p\equiv1+p^{11}\pmod{691}$ for every prime $p$; they were at the origin of -Serre's modularity conjecture recently proved by Khare-Wintenberger and Kisin.) -We may therefore ask how these $t_p=c_p/2p^{11/2}$ are distributed: for -example are there as many primes $p$ with $t_p\in[-1,0]$ as with -$t_p\in[0,+1]$? Sato and Tate predicted in the 1960s that the precise -proportion of primes $p$ for which $t_p\in I$, for given interval $I\subset[-1,+1]$, is -$$ -{2\over\pi}\int_{I}\sqrt{1-x^2}\;dx. -$$ -This is expressed by saying that the $t_p=c_p/2p^{11/2}$ are -equidistributed in the interval $[-1,+1]$ with respect to the measure -$(2/\pi)\sqrt{1-x^2}\;dx$. Recently Barnet-Lamb, Geraghty, Harris and Taylor -have proved that such is indeed the case. -Their main theorem implies many such equidistribution results, including the -one recalled above for the elliptic curve $S^2+S-T^3+T^2=0$; for an -introduction to such density theorems, see -Taylor's review article Reciprocity laws and density theorems.<|endoftext|> -TITLE: Generators of the derived category -QUESTION [5 upvotes]: For a ring $R$, which is a finite-dimensional algebra over a field, the category of finite-dimensional, projective, right $R$-modules, $\mathcal{P}_R$ is generated by the indecomposable projective modules, in the sense that every object is isomorphic to a direct sum of them. Is there a similar statement for the bounded-above derived category $\mathcal{D}^-_R=\mathcal{K}^-\left(\mathcal{P}_R\right)$? I want to say that $\mathcal{D}^-_R$ is generated by projective resolutions of the irreducible modules. Is that right? - -REPLY [2 votes]: In a triangulated or dg-category, this is not the usual notion of "generators." One definition is that there is no smaller triangulated subcategory containing the objects. In this sense, the indecomposable projectives do generate the homotopy category over all projectives. This is also true for the projective resolutions of the simples; since the algebra is finite dimensional, there's a finite iterated cone of the resolutions of the simples which is quasi-isomorphic to any indecomposable projective, even if the algebra doesn't have finite global dimension.<|endoftext|> -TITLE: Is S^2 x S^4 a complex manifold? -QUESTION [35 upvotes]: As observed by Calabi a long time ago, the manifold $S^2\times S^4$ admits an almost-complex structure (obtained by embedding it in $\mathbb{R}^7$ and using the octonionic product), which however is not integrable. -Is it known whether $S^2\times S^4$ admits an integrable complex structure? -A few remarks. - -This is stated as an open problem in Calabi's paper, but perhaps it has been solved in the meantime? -This is similar to the case of $S^6$, which is still open (see this question). -One can also ask the same question for $\Sigma\times S^4$ for $\Sigma$ any compact Riemann surface -It seems that some people believe that every almost-complex manifold of real dimension $6$ or more admits an integrable complex structure (see this other question). -Even more generally (and this is obviously still open), one can ask about an arbitrary finite product of even dimensional spheres (excluding $S^0$). It is known that this is almost complex iff the only factors that appear are $S^2, S^6$ and $S^2\times S^4$. - If one allows connected sums, then for example $(S^2\times S^4)\#2(S^3\times S^3)$ is a complex manifold, and in fact it has complex structures with trivial canonical bundles (see for example here and here). - -REPLY [7 votes]: This is still an open problem. See this paper for some progress, which was prompted by this MO question.<|endoftext|> -TITLE: Symmetric distribution of maj over des in pattern avoidance classes -QUESTION [9 upvotes]: It appears from computation to be the case (and would prove at least one clause of a conjecture advanced by Bruce Sagan and collaborators in a recent preprint) that in some pattern avoidance classes of permutations, the distribution of the major index is symmetric among permutations with a given descent number. For instance, $$\sum_{S_6(1234)} q^{maj(\sigma)}t^{des(\sigma)} = \dots + (10 q^6 + 35 q^7 + 66 q^8 + 80 q^9 + 66 q^{10} + 35 q^{11} + 10 q^{12}) t^3 + \dots.$$ -As you can see, the coefficient of $t^3$ is a symmetric polynomial. -This is not the case for all avoidance classes: for instance, $$\sum_{S_6(2134)} q^{maj(\sigma)}t^{des(\sigma)} = \dots + (4 q^6 + 21 q^7 + 42 q^8 + 61 q^9 + 56 q^{10} + 35 q^{11} + 10 q^{12}) t^3 + \dots.$$ -Before I start hammering at this, I was wondering if this was known to be the case for any particular avoidance classes, and if so, which. Since I have not even found any papers that seem to deal with the subject, pointers to one you know of would also be gratefully received. - -REPLY [11 votes]: Let $w=a_1 a_2\cdots a_n\in S_n$. Set $w' =n+1-a_n,n+1-a_{n-1},\dots,n+1-a_1$. Then (1) $\mathrm{des}(w)=\mathrm{des}(w')$, (2) $\mathrm{maj}(w)+\mathrm{maj}(w')=dn$, where $d=\mathrm{des}(w)$, and (3) $\mathrm{is}(w)=\mathrm{is}(w')$, where $\mathrm{is}(w)$ denotes the length of the longest increasing subsequence of $w$. This implies the suggested symmetry property for $S_n(12\cdots p)$. Since the map $w\mapsto w'$ also preserves the length of the longest decreasing subsequence (in fact, it preserves the RSK insertion tableau), the symmetry property also holds for $S_n(12\cdots p, q\cdots 21)$.<|endoftext|> -TITLE: Finite map from quasi-projective variety -QUESTION [7 upvotes]: Suppose I have a quasiprojective variety $X$ and a finite surjective map -$$f: X \rightarrow Y$$ -to a scheme $Y$. Is it true that $Y$ is quasiprojective as well? It seems like the answer could be no, but I don't know enough examples of non-projective schemes. - -REPLY [11 votes]: It is true if $Y$ is normal. Let me explain why. -I will say that a variety has the Chevalley-Kleiman (CK) property if every finite subset is contained in an affine open. By Corollary 48 of Kollar's "Quotients by finite equivalence relations" arXiv:0812.3608, if $f:X\to Y$ is finite and surjective, then if $X$ has the CK property, $Y$ has it too. -Now, it is clear that a quasi-projective variety has the CK property, and a normal variety with the CK property is quasi-projective by Corollary 2 of "Quasi-projectivity of normal varieties" arXiv:1112.0975. -Putting this together, we see that if $f:X\to Y$ is a finite surjective map of varieties with $X$ quasi-projective and $Y$ normal, then $Y$ is quasi-projective.<|endoftext|> -TITLE: Diophantine elements in SU(2) -QUESTION [7 upvotes]: Following notions from [1], call a set of elements $g_1, \dots, g_k \in G = SU(2)$ Diophantine if it satisfies the following property: there exists a constant $D$ such that for every word $W_m$ of length $m$ in $g_1, \dots, g_k$, if $W_m \neq \pm e$ (identity element), then $\Vert W_m \pm e\Vert \geq D^{-m}$. It can be proved that elements with algebraic entries are Diophantine. -Bourgain and Gambrud prove that a certain property (spectral gap) holds for $g_1, \dots g_k$ that: a) generate a free subgroup of $SU(2)$, b) are Diophantine. The first property is generic in measure. The second is not known to be generic in measure, but sets of Diophantine elements are dense in $G^k$ (since all rational-entry elements are). Does it follow that set of elements that a) generate free subgroup, b) are Diophantine, is dense in $G^{k}$? -Maybe it's almost obvious, since the set of elements generating a free subgroup is a complement of a countable union of proper subvarieties, but somehow one must rule out the possibility that being Diophantine and freeness are "anticorrelated" (a priori, we only know that elements with algebraic elements are Diophantine, and this is a countable set). Or maybe in fact Diophantine elements form a much larger set? -[1] Jean Bourgain, Alex Gamburd, "On the spectral gap for finitely-generated subgroups of SU(2)" - -REPLY [3 votes]: Tao's argument essentially does the job. However, I would start with a cocompact arithmetic lattice -$\Gamma$ in $SL(2, {\mathbb R})$ (which is, hence, not commensurable, up to conjugation, to $SL(2, {\mathbb Z})$). Then every Galois conjugation $\sigma(\Gamma)$ of $\Gamma$ will be a subgroup of $SU(2)$ (essentially, by the definition of arithmeticity). The group $\sigma(\Gamma)$ is isomorphic to $\Gamma$. Thus, every free subgroup $F$ in $\Gamma$ yields a free subgroup $F':=\sigma(F)$ of $SU(2)$ whose matrix entries are contained in a fixed number field $K$. Thus, $F'$ will be dense in $SU(2)$. In particular, every $k$ elements $g_1,...,g_k\in SU(2)$ can be approximated (arbitrarily well) by elements $f_1,...,f_k\in F'$. The elements $f_1,...,f_k$ will generate a free subgroup $F''$ in $SU(2)$ and the homomorphisms $\psi: F_k\to F''$ sending free generators $x_i$ to the elements $f_i$, will approximate the given homomorphism $\phi: F_k\to SU(2)$ sending $x_i$ to $g_i$. This answers your original question. Note that, a priori, rank of $\psi(F_k)$ is less than $k$. -One can do better and replace $f_i\in F$ by elements $h_i\in F$ so that the elements $h_i\in F$ generate a free subgroup of rank $k$ and the homomorphisms $\eta: x_i\to \sigma(h_i)$ still approximate $\phi$. You would need the following: - -If $H\subset F$ is a finitely-generated subgroup of infinite index then every element $g\in SU(2)$ can be approximated arbitrarily well by elements $\sigma(f)$, where $f\in F$ is such that the cyclic group $\langle f\rangle$ generated by $f$ intersects $H$ only at the identity. -Let $H\subset F$ be a finitely-generated subgroup, $\hat{h}\in F$ is such that group $\langle \hat{h}\rangle$ intersects $H$ only at the identity. Then for all -sufficiently high powers $h:=(\hat{h})^m$, the group generated by $H$ and $h$ is the free product $H \star {\mathbb Z}$ and has infinite index in $F$. - -Given this, you construct elements $h_i\in F$ inductively, taking $\hat{h}_{i}$ in $F- \langle h_1,...,h_{i-1}\rangle$ (so that $\sigma(h_i)$ is close to $g_i\in SU(2)$) and then replace this $\hat{h}_i$ with its high power $h_i$ so that $\sigma(h_i)$ is still close to $g_i$ and the group generated by $h_1,...,h_i$ is free of rank $i$.<|endoftext|> -TITLE: Triangulation of Surfaces without Jordan-Schoenflies -QUESTION [6 upvotes]: Does anyone know of a proof of the fact that any 2-manifold can be triangulated that does not use the Jordan-Curve Theorem or the Jordan-Schoenflies Theorem? Thanks for your help - -REPLY [14 votes]: Take the proof that any compact smooth manifold admits triangulations, and set the dimension to two. -The idea goes like this: - -Embed your surface (or $n$-manifold) in $\mathbb R^5$ ($\mathbb R^{2n+1}$ in general). -Triangulate $\mathbb R^5$, and make the surface transverse to the triangulation. If the surface does not intersect each simplex in a locally linear manner, subdivide the triangulation and repeat this step until it does. -The pull-back of the triangulation to the surface is a decomposition into convex polyhedra. A subdivision turns this into a triangulation. - -Paraphrasing Allen Hatcher: -If you're interested in topological surfaces, the paper -A.J.S. Hamilton, The triangulation of 3-manifolds, Oxford Quart. J. Math. 27 (1976), 63-70 -takes the Kirby-Siebenmann machinery and scales it down to 3 dimensions where it becomes somewhat simpler, so one can prove existence and uniqueness of triangulations of 3-manifolds using only standard PL techniques, such as results of Waldhausen. Presumably the same approach would work for surfaces. Since the method works in 3 dimensions it can't be using the topological Shoenflies theorem since this fails in 3 dimensions. On the other hand, it would use some PL (or smooth) surface theory so it wouldn't be entirely "from scratch". -edit: Allen wrote this argument up in a recent paper. See this thread for details https://mathoverflow.net/a/151760/353<|endoftext|> -TITLE: What is the "positive part" of the unit ball in $M_n(R)$ ? -QUESTION [12 upvotes]: In ${\bf M}_n(\mathbb R)$, let us consider the usual operator norm -$$\|A\|=\sup\frac{\|Ax\|}{\|x\|},$$ -where $\|x\|$ is the Euclidian norm. -The closed unit ball $B$ is the set of contractions (in the terminology used by operator theorists). It is a convex compact subset of ${\bf M}_n(\mathbb R)$. By Krein-Milman (finite dimensional case), it is the convex hull of its subset ${\rm ext}(B)$ of extremal points. It turns out that ${\rm ext}(B)$ is the orthogonal group ${\bf O}_n(\mathbb R)$. -Now, remember that ${\bf O}_n(\mathbb R)$ has two connected components, a positive one ${\bf SO}_n(\mathbb R)$ and a negative one ${\bf O}_n^-(\mathbb R)$. - -What is the convex hull of ${\bf SO}_n(\mathbb R)$ ? - -Clearly, it is a compact convex subset, included in $B$. It is a strict subset of $B$, because it does not meet ${\bf O}_n^-(\mathbb R)$. The title refers to the "positive part" of $B$, but this could be inappropriate, in the sense that it could meet the convex hull of ${\bf O}_n^-(\mathbb R)$ non-trivially. -Remark also that this convex hull is invariant under multiplication at right or left by an element of ${\bf SO}_n(\mathbb R)$. Therefore it would be enough to decide which diagonal matrices ${\rm diag}(a_1,\ldots,a_n)$ with $|a_1|\le a_2\le\cdots\le a_n$ it contains. -When $n=2$, ${\bf SO}_n(\mathbb R)$ is a circle and its convex hull is a disk, obviously a much smaller set (even from the dimensional point of view) than $B$. - -REPLY [10 votes]: I'm a bit late in answering this. But in case there is still interest, please have a look at: -Saunderson, Parrilo, Willsky. Semidefinite descriptions of the convex hull of rotation matrices, SIAM J. Optimization, 25(3), 1314-1343, 2015. -This paper provides explicit spectrahedral representations for conv $SO(n)$ (Theorem 1.3).<|endoftext|> -TITLE: Is there anything special about the Riemann surface $y^2 = x(x^{10}+11x^5-1)$? -QUESTION [25 upvotes]: I stumbled upon the fact that the Bolza surface can be obtained as the locus of the equation, -$y^2 = x^5-x$ -Its automorphism group has the highest order for genus $2$, namely $48$. I recognized $x^5-x$ as a polynomial invariant of the octahedron. (In fact, the Bolza surface is connected to the octahedron.) -If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface, -$y^2 = x(x^{10}+11x^5-1)$ -have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.) - -POSTSCRIPT: -My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define, -$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$ -and, -$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$ -then, -$w^5-10aw^3+45a^2w-a^2 = 0$ -for arbitrary r. This in fact is the Brioschi quintic form which the general quintic can be reduced into. Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a polynomial invariant for the octahedron. -Other than in formulas using Ramanujan's continued fractions, I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special. As Elkies wonderfully showed, it turns out that it is. - -REPLY [38 votes]: Yes, this Riemann surface, call it $C: y^2 = x^{11}-11x^6-x$, -is quite special: not only does it have the maximal number of automorphisms -for a hyperelliptic surface of genus $5$, but it is a modular curve -in at least two ways, both of which exhibit its full automorphism group. -One is a classical (elliptic) modular curve of level $10$, intermediate between -$X(5)$ and $X(10)$, with $[C:X(5)] = 2$ (the hyperelliptic map) and -$[X(10):C] = 3$ (a cyclic cover); this modular curve parametrizes -elliptic curves $E$ with full level-$5$ structure and odd ${\rm Gal}(E[2])$, -or equivalently full level-$5$ structure and square $j(E)-12^3$. -Explicitly, $E$ has Weierstrass equation $Y^2 = X^3 - A(x)X/48 + B(x)/864$ -where $A(x) = x^{20} + 228x^{15} + 494x^{10} - 228x^5 + 1$ and -$$ -x^{30} - 522x^{25} - 10005x^{20} - 10005x^{10} + 522x^5 + 1 -$$ -are polynomials with roots at the $20$- and $30$-point orbits of $A_5$. -We have $A^3 - B^2 = 12^3 (x^{11}-11x^6-x)^5$, -so $j - 12^3 = B^2/(x^{11}-11x^6-x)^5$. The corresponding congruence -subgroup $\Gamma$ of ${\rm SL}_2({\bf Z})$ is the index-$2$ subgroup of -$\Gamma(5)$ consisting of matrices that reduce mod $2$ to the index-$2$ -subgroup of ${\rm SL}_2({\bf Z}/2{\bf Z})$, with $[\Gamma : \Gamma(10)] = 3$. -This $\Gamma$ is normal in ${\rm SL}_2({\bf Z})$, and the quotient group -is ${\rm Aut}(C)$. -Another modular approach to $C$ is via the $(2,3,10)$ triangle group, -call it $G^*$, which appears in class VIII of the nineteen commensurability -classes tabulated in - -Takeuchi, K.: Commensurability classes of arithmetic triangle groups, J. Fac. Sci. Univ. Tokyo 24 (1977), 201-212. - -According to Takeuchi's table, $G^*$ is the normalizer of the unit-norm -group $G_1$ of a maximal order in a quaternion algebra over -${\bf Q}(\sqrt 5)$ ramified over one real place and the prime $(\sqrt 5)$. -Moreover $G_1$ is the $(3,3,5)$ triangle group, contained in $G^*$ with -index $2$. Let $G_5$ be the normal subgroup of $G_1$ consisting of -units congruent to $1 \bmod (\sqrt 5)$. Then -$G^*/G_5 \cong \lbrace \pm 1 \rbrace \times A_5$, and -the quotient of the upper half plane $\cal H$ by $G_5$ has genus $5$, -so must be our $C$. Moreover, ${\cal H} / G_5$ has no elliptic points, -so this identifies the image of the fundamental group $\pi_1(C)$ in -${\rm Aut}{\cal H} = {\rm SL}_2({\bf R})$ with an arithmetic congruence group. -P.S. Roy Smith already noted that if we allow also non-hyperelliptic -Riemann surfaces then the maximal number of automorphisms for genus $5$ -is not $120$ but $192$. An explicit model for a Riemann surface $S$ -with $192$ automorphisms is the intersection of three quadrics -$$ -y^2 = x_0 x_1, -\phantom{and} -{y'}^2 = x_0^2 - x_1^2, -\phantom{and} -{y''}^2 = x_0^2 + x_1^2 -$$ -in ${\bf P}^4$. Then $(x_0:x_1:y:y':y'') \mapsto (x_0:x_1)$ -gives a normal cover $S \rightarrow {\bf P}^1$ with Galois group -$N = ({\bf Z}/2{\bf Z})^3$ acting by arbitrary sign changes on -$y,y',y''$, ramified above the vertices of a regular octahedron, -with each of $x_0 x_1, x_0^2 - x_1^2, x_0^2 + x_1^2$ vanishing -on an opposite pair of vertices. I claim that there is an exact sequence -$1 \rightarrow N \rightarrow {\rm Aut}(S) \rightarrow S_4 \rightarrow 1$, -so in particular $\#({\rm Aut}(S)) = 2^3 4! = 192$. Indeed let -$G$ be the subgroup of ${\rm Aut}(S)$ that stabilizes the span of -$\lbrace x_0, x_1 \rbrace$. Then $G$ contains $N$ as the kernel of -a homomorphism $G \rightarrow {\rm Aut}({\bf P}^1)$ given by the -action on $(x_0:x_1)$. The image is contained in the -group $S_4$ of rotations of the octahedron, and indeed equals $S_4$ -because any rotation permutes the three opposite pairs of vertices -and thus lifts to ${\rm Aut}(S)$. Therefore ${\rm Aut}(S)$ contains -a group $G$ of order $2^3 4! = 192$, and by the Hurwitz bound -this must be the full group of automorphisms, QED<|endoftext|> -TITLE: Relation between Metalanguage and Object Language -QUESTION [7 upvotes]: first,I think we can avoid set theory to bulid the first order logic , by the operation of the finite string.but I have The following questions: -How does "meta-logic" work. I don't really know this stuff yet, but from what I can see right now, meta-logic proves things about formal languages and logics in general. But does it use some logic to do so? Like if I want to prove that two formal languages are equivalent in some respect, aren't I presupposing a "background" formal language? And won't my choice of a "background" (meta) language affect what I can and can't demonstrate? For example, what logic was Godel using when he proved his famous theorems? Was it a bivalent one? A three valued logic? etc -In short,I'm still not sure how reasoning about all possible formal languages work. For example, suppose I say something of the form "for all formal theories, F, if F has property X, then F must have property Y". If I wanted to prove something like that, how does such very general reasoning work? What I mean is that in such a proof, what kind of logic would be employed (for example, would it be a two valued logic?), and does the choice of logic affect the outcome? Do logicians agree on some kind of meta-meta logic, which they use to reason about absolutely everything? Or do they just choose their favorite one? -if metalogic is just predicate logic,It seems circular to me! we build the theory of predicate logic by using predicate logic?For example, in proving some theorem in the object language we seem to assume that it is already correct (in the metalanguage). Or defining some connective in the object language, we use that connective in the metalanguage to do so. It's like they're saying "Alright guys! We are going to prove a bunch of stuff about logic! Oh, by the way, you have to take all this stuff we are about to prove for granted, but don't worry, that's just the "metalanguage"." Something about this seems wrong to me. Maybe I have misunderstood? - -REPLY [6 votes]: If I want to prove that two formal languages are equivalent in some respect, aren't I presupposing a "background" formal language? - -Yes -- but the distinction between object language and meta language can be studied carefully. This is an important part of proof theory, as well as (in a more modern context) of the theory of programming languages. Let me quote from Olivier Danvy's entry on "Self-interpreter" in the appendix to Jean-Yves Girard's Locus Solum: - -Overall, a computer system is constructed inductively as a (finite) tower of interpreters, from the micro-code all the way up to the graphical user interface. Compilers and partial evaluators were invented to collapse interpretive levels because too many levels make a computer system impracticably slow. The concept of meta levels therefore is forced on computer scientists: I cannot make my program work, but maybe the bug is in the compiler? Or is it in the compiler that compiled the compiler? Maybe the misbehaviour is due to a system upgrade? Do we need to reboot? and so on. Most of the time, this kind of conceptual regression is daunting even though it is rooted in the history of the system at hand, and thus necessarily finite. - -In this view (in contrast to the views expressed in some of the other answers), the meta language does not have any special status: it is distinguished from the object language by its role rather than by its character. In particular, a meta language $L_1$, used to interpret some object language $L_2$, may itself be the object language of some interpretation in $L_0$. -On the other hand, often both mathematicians and computer scientists are interested in keeping the meta language as minimalistic as possible, simpler than the object language in some sense. A paradigmatic example is Gentzen's cut-elimination argument, which proves the consistency of first-order Peano arithmetic (PA). If the meta language for this proof is taken to be ZFC, then the result seems vacuous, since ZFC already includes PA (indeed is much more complicated than PA). However, in fact only a very small logical fragment of ZFC is needed to formalize the proof, namely PRA + $\epsilon_0$ (primitive recursive arithmetic plus transfinite induction up to $\epsilon_0$). Although PRA + $\epsilon_0$ is not included in PA (so that Gentzen's theorem does not contradict Gödel's), neither is PA included in PRA + $\epsilon_0$, since the latter only allows (transfinite) induction on quantifier-free statements. This is what prevents Gentzen's argument from being "circular", and the sense in which it reduces a statement about the object language to a "simpler" meta language.<|endoftext|> -TITLE: Integration By Parts on Non-compact Manifolds -QUESTION [12 upvotes]: This is undoubtedly a very easy question, but perhaps there are some subtleties. Under what circumstances can we integrate by parts over a non-compact Riemannian manifold? I am aware that having bounded curvature is sufficient (is there a reference for this?), but can this be weakened? - -REPLY [16 votes]: I know of two ways to approach this in general. One common way is to exhaust the manifold with a sequence of compact domains with smooth boundary and show that when you integrate by parts on the compact domain, the two integrals converge and the boundary term converges to zero. -I, however, usually prefer the second approach, which is to use a sequence of cut-off functions, i.e. compactly supported smooth functions that converge to 1 on any compact domain. There is no boundary term, but, when you integrate by parts, you get an extra term involving the gradient of the cut-off function. You then show that this term converges to zero, given a suitable sequence of cut-off functions and suitable assumptions about the growth or decay rate of the volume form of the manifold and the integrand.<|endoftext|> -TITLE: Contour Integral with Gamma functions and 2F1 -QUESTION [6 upvotes]: Given the following contour integral -$$\frac{1}{2\pi j}\int^{c+j\infty}_{c-j\infty} \frac{\Gamma(-1+a+s)\Gamma(b+s)}{\Gamma(3+a-s)}\cos(-1+a+s)\, -{}_2F_1\Big(-1-a+s,-1+a+s;\frac{1}{2};z\Big) y^s\: \mathrm{d}s ,$$ -where $a,b,z,y \in \mathbb{R}$, as noticed there are two poles given by -$$P^{(1)}_k = 1-a-k \quad P^{(2)}_k=-b-k \quad\text{ where }\quad k=0,1,2,\dotsc,\infty.$$ -The questions are: - -What is the suitable contour to sum the residues and solve the integral? -Does the zero caused by $\Gamma(3+a-s)$ cancel any of the poles? - -REPLY [3 votes]: 1) I imagine this integral came from a Mellin transform approach to compute another integral. You would come up with an appropriate 'c' lying in the overlap region of definition of your two original Mellin transforms (e. g. one might be defined for $c > 1/2$ and the other for $0 < c < 1$, in which case you would choose $1/2 < c < 1$. -Then you need to estimate the behavior of your integrand for large $\Im(s) $, to determine which direction you could move the integration contour and pick up poles to develop a series for your integral. -2) Yes, the poles of the $\Gamma$ function in the denominator can cancel poles in the numerator, you just need to determine those $s$ where that will occur. This all depends on the relative values of $a$ and $b$. Also, the $\cos$ function could cancel poles. -I hope this is not too vague to be of help! -Regards, -Tom<|endoftext|> -TITLE: Galois representations attached to Maass form -QUESTION [17 upvotes]: So, how does one construct a galois representation from a Maass form? -For a modular cusp eigenform, I am familiar with the work of Eichler-Shimura, Deligne, Deligne-Serre, and realize these are different situations because of the involved geometry. I am also familiar with Maass's construction of Maass forms of weight zero from Hecke characters on real quadratic fields, so I can reverse this to answer a tiny bit of my question. There is also Langlands-Tunnell, which I am not familiar with. Finally, I realize that most Maass forms are not conjectured to be associated to galois representations. -Searching the web did not yield much. But I do want to ask an interesting precise question, so here it is: - -Is there an infinite family of Maass eigenforms, such that an irreducible galois representation of infinite image is constructed to each form, and these do not somehow arise from Maass's original construction or Langlands-Tunnell? - -If not, is there a conjectural association that has been checked (without proof) computationally? - -REPLY [17 votes]: Several remarks before answering your questions: (1) Langlands-Tunnell is a result in the other direction: from Galois representation to automorphic forms; it is therefore not relevant. (2) One expects to be able to attach Galois representations only to -certain types of Mass forms, those whose component at infinity in algebraic (in the automorphic representation settings) or equivalently, whose eigenvalue for the Laplacian is $1/4$. (3) this Galois representation is expected to take values in Gl${}_2(\mathbb{C})$, hence to have finite image. -So you ask: "how does one construct a galois representation from a Maass form?". -The answer is: one still doesn't know how to. It's one of the most striking open problem in the Langlands program. There was 25 years ago an announcement that this problem -has been essentially solved (with published articles), but it was soon after retracted: -see the two references given by Chandan in comments. -And for your displayed question about the infinite family, stripped of the reference to Langlands-Tunell and of the "infinite image" condition, the answer is no, as far as I know,<|endoftext|> -TITLE: A concept of dynamical coherence -QUESTION [8 upvotes]: I'm trying to make an overview of the study of partial hyperbolicity and there is an interesting concept of dynamical coherence which appears there. Some call it mild (see the Thesis of Pablo Carrasco, Compact Dynamical Foliations 2010), some call it strong and unnatural (see the work of Amy Wilkinson and Keith Burns Dynamical coherence, accessibility and center bunching). The definition which is the most common is that local cental-unstable $E^{cu}$ and center-stable $E^{cs}$ bundles integrate to foliations $W^{cu}$ and $W^{cs}$. -Let us suppose, that in a normally hyperbolic case, i.e. when we already have the $E^c$ that integrates to a foliation F, at which some diffeomorphism is hyperbolic. -My question is how the normally hyperbolic (i.e. partially hyperbolic on foliation) system could be dynamically incoherent and is this concept somewhat related to the concept of local product structure? -My question is, what is a simplest example of a normally hyperbolic foliation when $E^cu$ and $E^cs$ do not integrate to foliations? And how "often" does it happen in the world of normally hyperbolic foliations? -PS. Updated after a useful remark of Rafael Potrie, the definition of a dynamical coherence is now more precise. - -REPLY [4 votes]: In general, it is not known if a partially hyperbolic diffeomorphism should be dynamically coherent. -There are two obstructions for integrability of the center bundle. One is that the distributions are not integrable (Frobenius conditions fails) and the other one is that the distributions may lack of diferentiability (and so uniqueness of integrability may fail). -For the first obstruction, Wilkinson noted that in diffeomorphisms with high dimensional center (i.e. Anosov automorphisms on nilmanifolds) the bracket condition fails. -In the absolute partially hyperbolic setting, for diffeomorphisms of $T^3$ dynamical coherence has been obtained by Brin-Burago and Ivanov (http://www.pdmi.ras.ru/~svivanov/papers/coherence.pdf). This has been extended to nilmanifolds by Hammerlindl and Parwani (http://arxiv.org/abs/1103.3724, http://arxiv.org/abs/1001.1029). -For pointwise partially hyperbolic systems (a weaker condition), recent examples have been constructed by Rodriguez-Hertz, Rodriguez-Hertz and Ures showing that dynamical coherence may fail even in $T^3$. On the other hand, I have recently proved that if the partially hyperbolic diffeomorphism of $T^3$ is transitive (or volume preserving) then it must be dynamically coherent (see http://www.mat.puc-rio.br/edai/textos/potrie.pdf). -Local product structure I believe has something to do with plaque-expansiveness, which allows one to show robustness of dynamical coherence thanks to the work of Hirsch-Pugh and Shub, but in general it is not enough to show the existence of an invariant foliation tangent to the center.<|endoftext|> -TITLE: Finite number of cardinal in a model of $ZFC + \neg CH$ ? -QUESTION [5 upvotes]: Hello everybody, little question for logicians: -Considering $ZFC + \neg CH$, is it possible to construct a model $V \vDash ZFC + \neg CH$ -such that there exists -a finite number of cardinal between $\aleph_0$ and $2^{\aleph_0}$ ? -Same question with exactly one cardinal between $\aleph_0$ and $2^{\aleph_0}$ ? -Thanks. - -REPLY [11 votes]: Yes! Almost anything is possible. You can force over a model of ZFC + CH to create a new model where $2^{\aleph_0}$ is $\aleph_2$, for example, so that there is one cardinal between $\aleph_0$ and the continuum. The idea is to create new binary sequences, new real numbers, with a partial order (a notion of forcing) and allow a generic filter to make it coherent and take you to the new universe where the continuum is a new size. You could have 3 or 4 or any finite number of cardinals between $\aleph_0$ and $2^{\aleph_0}$ by adding new subsets of natural numbers. -See the discussion below about how the answer to the question "What size can the continuum be?" is due to Cohen, Solovay, and Easton. Also, see in the comments how the continuum could reach as far up as $\aleph_{2^{\aleph_0}}$, so there are continuum many cardinals between omega and $2^{\aleph_0}$. Hamkins' paper on the Multiverse shows that the ability to force to create models which have a variety of sizes of the continuum settles the continuum hypothesis. You can read all about how to add new reals to create a new model in Thomas Jech's Set Theory or Kenneth Kunen's book on the same subject.<|endoftext|> -TITLE: What is the computational-complexity-theoretic analogue of computable inseparability? For example, if P is not NP, are there disjoint NP sets with no separation in P? -QUESTION [11 upvotes]: Disjoint sets $A$ and $B$ are computably inseparable, if there -is no computable separating set, a computable set $C$ containing $A$ and disjoint from $B$. The -existence of c.e. computably inseparable sets is a fundamental -phenomenon explaining and unifing many arguments in computability theory. -Perhaps the easiest example of a c.e. computably inseparable pair of sets is the following, -where $\varphi_e$ is the function computed by program $e$. -$$A=\{e\mid -\varphi_e(0)\downarrow =0\},\qquad B=\{e\mid -\varphi_e(0)\downarrow=1\}.$$ -To see this, suppose $C$ is decidable, $A\subset C$ and $B\cap -C=\emptyset$. Since $C$ is computable, we may design via the recursion theorem a program $e$ such that -$\varphi_e(0)\downarrow=1$ just in case $e\in C$, and otherwise -$\varphi_e(0)=0$, and this gives an immediate contradiction. Another computably inseparable pair is the set of theorems versus the set of negations of theorems of PA, or your favorite consistent theory containing arithmetic. -Question. What is the computational-complexity-theoretic -analogue of computable inseparability? -Specifically, if $P\neq NP$, then are there disjoint NP sets with -no separating set in $P$? -This plainly fails if $P=NP$. -Edit. Mark Sapir points out that if we assume $\text{P}\neq\text{NP}\cap\text{Co-NP}$, then there is are some easy examples, namely, any set $L\in\text{NP}\cap\text{Co-NP}\setminus P$ together with its complement. In light of this (since this particular example won't help me with my intended purpose), let me modify the particular question to: -Question. Under some standard complexity theory hypothesis, such as $\text{P}\neq \text{NP}\cap\text{Co-NP}$, are there disjoint sets in NP with no separation in $\text{NP}\cap\text{Co-NP}$? -And are there other analogues of the phenomenon with other complexity classes? - -REPLY [7 votes]: This should be a comment but is too long. -You may want to look at the literature on promise problems for NP-Disjoint pairs. They consider pairs (A,B) of disjoint NP-languages. One is to think of this as a promise that the words belonging to a language L belong to A and that none of the words in B belong to L. Then a P-language S with $A\subseteq S$ and $S\cap B=\emptyset$ is a polynomial time algorithm which on $A\cup B$ gets the correct answer and behaves arbitrarily on the remaining inputs. There seem to be a big literature on this. It seems if $P\neq UP$ (where UP is unambiguous polynomial time), there exist P-inseparable NP-sets. The assumption $P\neq UP$ is equivalent to the existence of a one-to-one worst-case one-way function. Apparently it is not known if $UP$ has complete problems. I am no expert on this.<|endoftext|> -TITLE: Can every nonempty set carry abelian group structure? -QUESTION [7 upvotes]: Possible Duplicate: -Does every non-empty set admit a group structure (in ZF)? -Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group? - -If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite? -If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality. -Apologies if it is obvious, my group theory knowledge is just insufficient. - -REPLY [14 votes]: If the axiom of choice holds, then this is an immediate consequence of the upward Lowenheim-Skolem theorem. Any first order theory in a finite language with an infinite model, such as the theory of the infinite cyclic group, admits models of every infinite cardinality. Thus, one can find infinite abelian groups of any given size satisfying exactly the same theory, in the language of group theory, as your favorite infinite abelian group. -Without the axiom of choice, there are sets admitting no group structure at all, as show in Ashutosh's answer to the question I mention in the comment.<|endoftext|> -TITLE: Length spectrum of spheres -QUESTION [6 upvotes]: It is known (thanks to Hingston, Bangert, Franks, Birckhoff, etc) that $(S^2, g)$ has lots of primitive closed geodesics for any Riemannian metric $g$ (Riemannian is crucial here, this is not true for Finsler metrics).The question is: does the length spectrum determine the metric? This is obviously a completely different question from the similar question for hyperbolic surfaces, where spectral methods are available. In particular, it is not even obvious that if closed geodesics are the same length, then $g$ is the (appropriately scaled) round metric. - -REPLY [7 votes]: As to the question of closed geodesics all of the same length, there are Zoll metrics, so the length spectrum doesn't characterize even the round metric. See e.g. Besse's "Manifolds all of whose geodesics are closed" Springer 1978, or recent work by LeBrun and Mason on twistor methods applied to Zoll metrics. -On the other hand (and extremally opposite case), for a suitably generic metric, there is the Duistermaat-Guillemin trace formula which implies that the length and Laplace spectra determine each other. But there are isospectral non isometric surfaces. I don't know if there are spherical examples (this seems plausible via Sunada -- or rather Buser -- construction), and even if Zoll metrics are isospectral (which I doubt). -Hope this helps. -EDIT : in fact Berger proved that the round 2-sphere is determined by its laplace spectrum inside smooth metrics, so Zoll's aren't all isospectral.<|endoftext|> -TITLE: The Gysin long exact sequence for the complement of the zero section of a line bundle over a (possibly) singular base -QUESTION [5 upvotes]: Let $pr:X\to Z$ be a line bundle of (possibly) singular varieties (that could be reducible) over a characteristic $p$ field ($p$ could be zero); $U$ is the complement to the zero section $Z\to X$. Then for any $l\neq p$, $n>0$, there should exist a Gysin long exact sequence for the (etale or singular) cohomology $\dots \to H^i(X,\mathbb{Z}/l^n\mathbb{Z}) \to H^i(U,\mathbb{Z}/l^n\mathbb{Z})\to H^{i-1}(Z,\mathbb{Z}/l^n\mathbb{Z}(-1))\to \dots$ that is functorial in $pr$. Is this true? -Note that one can compute the cohomology of $U$ as the hypercohomology $H^*(Z,Rpr'_\ast\mathbb{Z}/l^n\mathbb{Z}_U)$, where $pr': U\to Z$ is the corresponding prinicple $G_m$-bundle. Hence the problem is to verify that $R^1pr'_*\mathbb{Z}/l^n\mathbb{Z}_U\cong \mathbb{Z}/l^n\mathbb{Z}_Z(1)$. Certainly, there is such an isomorphism for the trivial $G_m$-bundle $U\cong G_m\times Z$. Since this isomorphism does not seem to depend on the choice of a trivialization, one can 'glue' 'global' functorial isomorphisms from these 'local' isomorphisms. Yet I would prefer to have a reference for this result (or for some similar one). -Cf. https://mathoverflow.net/questions/89171/on-the-cohomology-of-g-m-bundles-and-purity-for-singular-varieties - -REPLY [3 votes]: Comment promoted to answer: -The following is Corollaire 1.5 in SGA5, Exposé VII: - -Let $Z$ be any scheme, $p:X\to Z$ a rank $r$ vector bundle, $U=X-Z$, and $q:U\to Z$ the retsriction of $p$. Let $n$ be coprime to the residual characteristics of $Z$ and let $L$ be a sheaf of $\mathbb{Z}/n$-modules on $Z$. Then there is a long exact sequence -$$\to H^{\ast-2r}(Z,L(-r))\to H^\ast(Z,L)\to H^\ast(U,q^\ast L)\to H^{\ast+1-2r}(Z,L(-r))\to$$ - -That's how the corollary is stated but they also show that the sequence is naturally isomorphic to the more familar-looking sequence for cohomology with supports -$$\to H^{\ast}_Z(X,p^\ast L)\to H^\ast(X,p^\ast L)\to H^\ast(U,q^\ast L)\to H^{\ast+1}_Z(X,p^\ast L)\to$$ -You should definitely have a look at the rest of exposé VII in SGA5, which proves a lot of related "expected" theorems for étale cohomology with very mild hypotheses (compared to what can be found elsewhere).<|endoftext|> -TITLE: compressibility of Seifert surface after 0-surgery -QUESTION [15 upvotes]: Gabai's solution of the Property R conjecture shows that a minimal genus Seifert surface of a knot, capped off in the 0-framed surgery along that knot, is of minimal genus in its homology class. In particular, it is incompressible in the 0-surgered manifold. On the other hand, there may be incompressible Seifert surfaces for the knot that are not of minimal genus. (For example many pretzel knots bound incompressible surfaces of arbitrarily high genus.) Presumably, there may be such a (non-minimal genus) incompressible surface that becomes compressible in the 0-surgered manifold. Does anyone know an example of this phenomenon? - -REPLY [10 votes]: Revised answer: -I found a paper of Ozawa and Tsutsumi which -constructs examples of Seifert surfaces which compress -when capped off in the 0-framed surgery. -See also Tsutsumi for further examples. -In fact, their constructions give knots with -Seifert surfaces with -"accidental peripheral" curves. That is, -given a knot $K$ with exterior $E(K)$, and -Seifert surface $S\subset E(K)$, then $S$ -has an accidental peripheral curve if it -has an essential curve $c\subset S$ (not parallel to the -boundary of $S$) which is homotopic into $\partial E(K)$. -When one performs 0-framed surgery on $K$ and -caps off $S$ to get a closed non-separating -surface $\hat{S}\subset S^3_K(0)$, then the -curve $c\subset \hat{S}$ is essential in $\hat{S}$ -since it is in $S$ and $S$ has only one boundary component, but it will be homotopically trivial in $S^3_K(0)$, since as they point out the homotopy of $c$ to $\partial E(K)$ will be realized by an annulus with interior embedded in $E(K)-S$, and hence intersecting $\partial E(K)$ in a curve of slope $0$ which gets capped off to a disk in $S^3_K(0)$. -Here's an explicit example from Tsutsumi's paper: - -Previous answer: -(this was too long for a comment): -I don't know an answer, but I'll make a suggestion. If you have a knot with infinitely many Seifert surfaces, then there's a closed incompressible surface (it's a kind of limit measured lamination). http://www.ams.org/mathscinet-getitem?mr=2420023 -If you perform 0-framed surgery, and the Seifert surfaces remain incompressible when capped off, then this closed surface should also be incompressible in the Dehn filled manifold. So my suggestion is to find a knot with a closed incompressible surface in its complement which compresses in 0-framed surgery. The pretzel knot examples don't seem to work. One possibility is to stick a handlebody with this property in $S^3$, such as http://www.ams.org/mathscinet-getitem?mr=2032111. Then you would also need to check that summing a minimal genus Seifert surface with the closed surface infinitely many times remains incompressible. I'm not sure how to do this part.<|endoftext|> -TITLE: Induced character for non-injective homomorphisms -QUESTION [9 upvotes]: Any group homomorphism $\phi\colon H\to G$ gives rise to an induction/restriction adjunction between $G$-representations and $H$-representations: -$$ \hom_G(\phi_! M, N) \cong \hom_H(M, \phi^* N) $$ -However, it seems that most textbooks and web pages about representation theory inexplicably consider only the case when $\phi$ is injective, i.e. exhibits $H$ as a subgroup of $G$. In this case, there are formulas for the character of $\phi_! M$ in terms of the character of $M$: -$$ \chi_{\phi_!(M)}(g) = \frac{1}{|H|} \sum_{k\in G \atop k^{-1} g k \in H} \chi_M(k^{-1} g k) = \sum_{\text{cosets } k H \atop k^{-1} g k \in H} \chi_M(k^{-1} g k) . $$ -Can someone give a reference for versions of these formulas when $\phi$ is not injective? - -REPLY [5 votes]: Exercise 7.1 in Serre's Linear Representations of Finite Groups gives a formula (without proof) in the case where $\phi$ is surjective. It is probably straightforward to compose this formula with your formula for the injective case to get the general formula.<|endoftext|> -TITLE: Reference request for TQFT, functoriality -QUESTION [10 upvotes]: I am reading Turaev's blue book Quantum Invariants of Knots and 3-manifolds. -It is difficult for me to understand the proof of Theorem 1.9 in chapter 4, which says; -The function $(M, \partial_{-}M, \partial_{+}M) \mapsto \tau(M): T(\partial_{-}M) \to T(\partial_{+}M)$ extends the modular functor $T$ to a non-degenerate topological quantum field theory. -The proof of the functoriality is unclear for me. I tried to look at Turaev's papers but its harder to understand. Also I don't understand the proof of computation of annomalies (Theorem 4.3 on chapter 4). The method of the proof seems to extend the method of the proof of functoriality to 4-manifolds. -Could you suggest me a textbook or paper etc that explain these theorems or similar material? -Or could you show me more detailed proof of functoriality (and the computation of anomalies) here? - -REPLY [2 votes]: Turaev himself has reviewed this construction of a 3-dimensional TQFT in the article Quantum 3-Manifold Invariants (2006). As "further reading" he suggests the 2001 lecture notes by Bakalov and Kirillov, which are published by the AMS but can be freely downloaded in a non-final form from here. Chapters 4 and 5 address the Turaev-Reshetikhin construction. A quick look suggests that these lecture notes are quite gentle on the reader, and may be along the lines of what you are looking for. - -These lectures are devoted to the discussion of the relation between - tensor categories, modular functor, and 3D topological quantum field - theory. They were written as a textbook; all the results there are - known. Our only contribution is putting it all together, filling the - gaps, and simplifying some arguments.<|endoftext|> -TITLE: Sum of Gaussian binomial coefficients. -QUESTION [9 upvotes]: We all know that $\sum_{i=0}^{n}{n \choose i}=2^{n}$. Is there a similar result regarding the q-binomial coefficients? (a.k.a Gaussian binomial coefficients) - $\sum_{i=0}^{n}{n \choose i}_{q}=?$ - -REPLY [6 votes]: For Gaussian binomial coefficients we have -$$ -\sum_{k = 0}^n \binom nk_q = \sum_{m = 0}^\infty a_m q^m, -$$ -where -$$ -a_m = \sum_{\lambda\vdash m} \#\{k\in \mathbf Z_{\geq 0}\mid \lambda_1\leq n-k, \lambda'_1\leq k\}. -$$ -The notation $\lambda\vdash m$ signifies that $\lambda$ is an integer partition of $m$. -Also $\lambda_1$ is the first (largest) part of $\lambda$ and $\lambda'_1$ is the number of positive parts in $\lambda$. -This follows from the following well-known fact about Gaussian binomial coefficients: -$$ -\binom nk_q = \sum_\lambda q^{|\lambda|}, -$$ -the sum being over all partitions $\lambda$ where $\lambda_1\leq n-k$ and $\lambda'_1\leq k$ (see, for example Stanley's Enumerative Combinatorics, vol. 1, or Eq. (4) in my expository article titled Counting subspaces in a Finite Vector Space). - -REPLY [3 votes]: One more version - analog of $\sum_{i=0}^n(-1)^i\binom ni=0$: -$$ -\sum_{i=0}^n(-1)^i\binom ni_q=\begin{cases}0,&n=2k-1\\ -\prod_{j=1}^k(1-q^{2j-1}),&n=2k\end{cases} -$$<|endoftext|> -TITLE: Frobenius formula for the determinant -QUESTION [9 upvotes]: Is there a formula for the determinant of an induced representation, e.g. in the fashion of the Frobenius character formula. -I would hope for something: - -$$ det \; Ind_H^G \rho(g) = (-1)^\alpha \prod\limits_{\gamma, \gamma_1 \in G/H \atop \gamma^{-1}g\gamma_1 \in H} det \rho(\gamma^{-1}g\gamma_1).$$ - -If you have a reference or a quick proof, that would be most helpful. - -REPLY [8 votes]: The answer can be formulated in compact form using the usual transfer map $V_H^G\colon G\to H/H'$ or the (not-so-well-known) construction of tensor induction: Namely, - -$$ \det( { \rm Ind}_H^G\; \rho ) - = ( {\rm sign}_{[G:H]})^{\rho(1)} ((\det\rho)\circ V_H^G) - =( { \rm sign }_{ [G:H] })^{\rho(1)} (\det \rho)^{\otimes G}, $$ - -where ${\rm sign}_{[G:H]}$ is the permutation sign character of $G$ on the cosets of $H$. An exposition of tensor induction and a proof of this formula is contained in Curtis and Reiner, Methods of Representation Theory I (see Proposition 13.15). It seems this formula is due to Gallagher (MR0185017). -Of course, when evaluating at $g\in G$, this yields the last formula in Geoff Robinson's answer.<|endoftext|> -TITLE: Does the length spectrum determine the volume? -QUESTION [10 upvotes]: Is it true that the length spectrum of a compact Riemannian manifold determines its volume? -This question was inspired by the MO question Length spectrum of spheres . BS's answer recalls a theorem of Duistermaat and Guillemin (Duistermaat, J. J.; Guillemin, V. W. (1975), "The spectrum of positive elliptic operators and periodic bicharacteristics", Inventiones Mathematicae 29 (1): 39–79) that states that for generic metrics the length spectrum and the spectrum of the Laplacian determine each other. Since Weyl's asymptotic formula implies that the spectrum of the Laplacian determines the volume, it may seem that the length spectrum determines the volume. Is this right? -Addendum. -This question seems more useful interesting and useful if it is reformulated as follows: what does the set of lengths of periodic geodesics and/or its various analogues such as the length spectrum and the marked length spectrum say about the volume of a compact Riemannian or Finsler manifold? -Here are more concrete questions around this topic: -1. Let $M$ be a simply-connected, compact $n$-manifold. Does there exist a quantity $c(M) > 0$ such that for any Riemannian metric $g$ on $M$ the volume of $(M,g)$ is bounded below by $c(M)$ times the $n$-th power of the length of the shortest periodic geodesic of $(M,g)$? -This is a really hard open problem. It was solved by Croke for the $2$-sphere in -Area and the length of the shortest closed geodesic J. Differential Geom. Volume 27, Number 1 (1988), 1-21. The current record for $c(S^2)$ is held by Rotman: -The length of a shortest closed geodesic and the area of a -dimensional sphere Proc. Amer. Math. Soc. 134 (2006), 3041-3047. However, the conjectured optimal constant $1/2\sqrt{3}$ -is still a challenge. -2. Are there known (explicit?) examples of compact Riemannian manifolds with the same set of lengths of periodic geodesics (the length set) and different volume? -BS gave a reference (see his answer) where Huber shows that two compact hyperbolic surfaces with the same length spectrum have the same laplace spectrum and hence the same volume. For negatively curved, compact surfaces J.-P. Otal showed that the marked length spectrum determines the surface up to isometries. This is not true for Finsler metrics (they are much more flexible than Riemannian metrics), however: -3. If two negatively-curved, compact Finsler surfaces have the same marked spectrum, do they also have the same area? -However, posing these type of question is easy. I'm more interested in collecting some success stories. - -REPLY [5 votes]: This seems like a difficult question, even for closed hyperbolic manifolds. -Indeed Marcos Salvai, in -"On the Laplace and complex length spectra of locally symmetric spaces of negative curvature." -Math. Nachr. 239/240 (2002), plus erratum on his web page -proved that the complex length spectrum (with multiplicities) and the volume of a closed (oriented) hyperbolic manifold (real, complex, quaternionic or octonionic) determine the laplace spectrum (with multiplicities, even on forms) but cannot dispense with the volume. Hence even the complex length spectrum (length spectrum and holonomies) is not shown to determine volume. In the erratum, he seems to be confident that this can be repaired, but this is not published. -However, there is a recent preprint by Dubi Kelmer, which shows among other things, that the length spectrum determines the laplace spectrum (hence the volume) for compact hyperbolic manifolds (real, complex, quaternionic or octonionic). -The methods seem strongly Lie group representation-theoretic, not generalizing to non locally symmetric (or homogeneous) manifolds. -Interestingly, he leaves open the question wether the laplace spectrum determines the multiplicities in the length spectrum -(it is known that the length set is determined), whereas Salvai proves that the laplace-beltrami spectrum on forms determines the complex length spectrum, by following the proof by Gordon and Mao Math. Res. Lett. 1 (1994), no. 6, that it determines the length spectrum. -Complicated situation...<|endoftext|> -TITLE: Are weak and strong convergence of sequences not equivalent? -QUESTION [12 upvotes]: For some infinite-dimensional Banach spaces $E$, it is easy to find sequences $\langle x_i:i\in\mathbb N_0\rangle$ which converge to zero weakly but not in the norm topology, i.e. we have $\lim_{i\to\infty}y(x_i)=0$ for all $y\in E^*$ but $\lim_{i\to\infty}\|x_i\|\not=0$. For example, taking $E=c_0(\mathbb N_0)$ or $E=\ell^p(\mathbb N_0)$ with $ 1 < p < +\infty $, the sequence of standard unit vectors provides such an example. For $C([0,1])$, the sequence $\langle\langle (i+1)(1-t)t^{i+1}:0\le t\le 1\rangle:i\in\mathbb N_0\rangle$ is an example. However, I cannot find such an example for $\ell^1(\mathbb N_0)$. So I ask - Are weak and strong convergence of sequences equivalent in $\ell^1(\mathbb N_0)$? What about $L^1([0,1])$? Is there possibly a general result saying that in any infinite-dimensional Banach space weak and strong convergence of sequences are not equivalent? - -REPLY [21 votes]: Banach spaces where all weakly convergent sequences are norm convergent are said to have the Schur property. A classical Theorem of Schur says that $\ell^1(I)$ has the Schur property for every set $I$. $L^1([0,1])$ does not have it. I guess that you can find this in -P. Wojtaszczyk's "Banach Spaces for Analysts".<|endoftext|> -TITLE: How far can one get with the Gross-Siebert program? -QUESTION [15 upvotes]: The Gross-Siebert program is said to be an algebraic analog of the SYZ conjecture and they used toric degeneration to construct a mirror dual of Calabi-Yau varieties. It seems like the singular central fiber, as is mostly given by some combinatoric data, has some sort of "mirror dual" that's easily written down. -My question is, how far can one get with this program? Are the CY varieties arising this way "the" mirror dual of the original CY varieties or just one candidate of its mirror? What are the most recent progress on it? -Also, this seems very different from the homological mirror conjecture saying that the equivalence of the derived category on one and Fukaya category on another. Can anyone here connect the dots for me? - -REPLY [3 votes]: For the question how SYZ is related to HMS, I have recently posted a paper on the arXiv (http://arxiv.org/abs/1201.6454) which explains this connection as a sort of Fourier-Mukai transform. There I gave a local explanation. Global situation requires more machinery, but it is a similar picture.<|endoftext|> -TITLE: q-analog of the matrix exponential -QUESTION [5 upvotes]: I am a fan of the Matrix exponential $\exp(X)$, defined for any complex matrix $X$ by -\begin{equation*} - \exp(X) := \sum_{k \ge 0} \frac{X^k}{k!}. -\end{equation*} -I have a fleeting acquaintance with q-analog's (essentially, I know that they exist, but have almost no idea what use they serve, which is part of the reason why I am asking this question). -Thus, my question is - -Has the following q-analog of the matrix exponential - \begin{equation*} - \exp_q(X) := \sum_{k \ge 0} \frac{X^k}{[k]_q!}, -\end{equation*} - been studied previously? If so, in what context? - - -PS: More generally, the above question can be rephrased in terms of q-analogs of functions of matrices (which includes scalar, vector, and matrix valued functions). But I wanted to limit my focus to a concrete case of special interest to me. - -REPLY [8 votes]: It's more an example than a general answer. Details can be found here : http://arxiv.org/abs/math/0512500 -It is convenient to replace $q$ by $q^2$ in the formula that you gave. Doing so we have the following desirable identities: - -$\exp_q(x)\exp_{-q}(-x)=1$ -if $xy=q^2yx$, then $\exp_q(y)\exp_q(x)=\exp_q(x+y)$ - -Now if $\mathfrak g$ is a simple Lie algebra, $\Phi^+$ a choice of positive roots, $h_i,e_{\alpha},f_{\alpha}$ the generators of $U_q(\mathfrak g)$ associated to the Chevalley basis of $\mathfrak g$, $>$ a normal ordering on $\Phi^+$, and $q_{\alpha}=q^{(\alpha,\alpha)/2}$, then the R-matrix of $U_q(\mathfrak g)$ is given by -$$ -R=K \prod_{\alpha \in \Phi^+}^> R_{\alpha} -$$ -where -$$K=q^{\sum h_i \otimes h^i}$$ -and -$$R_{\alpha}=\exp_{q_{\alpha}^{-1}}((q_{\alpha}-q_{\alpha}^{-1})e_{\alpha} \otimes f_{\alpha})$$ -Here $q$ is either a generic complex number, or a variable, in which case we work over the field $\mathbb{Q}(q^{\frac12})$. -It is a universal formula, but of course you can specialize it to an element of $End(V\otimes V)$ for any $U_q(\mathfrak g)$-module $V$.<|endoftext|> -TITLE: Geometric meaning of L-genus -QUESTION [16 upvotes]: Is there any reasonable geometric meaning of the L-genus for smooth manifolds? or perhaps easier, for complex algebraic surfaces? -The question came up after a friend and I realized that we don't understand why one would expect to have such a formula for the signature in terms of Pontrjagin classes (i.e. the signature theorem). Any insights about this will be appreciated. - -REPLY [13 votes]: I take that you ask how in the world did Hirzebruch come up with the complicated expression of the L-genus? -The key fact behind this is that the signature is a genus, i.e. a ring morphism $\gamma:\Omega^\bullet_+\to\mathbb{R}$ from the oriented cobordism ring to the (ring of) reals. By Thom's work we deduce that a genus is determined by its values on $\mathbb{CP}^{2n}$ and thus by the generating series -$$r^\gamma(t)=1+\sum_{n\geq 1}\gamma(\mathbb{CP}^{2n})t^{2n}.$$ -In the case of signature we have -$$r^\gamma(t)=\frac{1}{1-t^2}.$$ -How does one go from this series to the the function $\frac{\xi}{\tanh \xi}$ that enters into the defintion of $L$? As a teaser, let me point out that -$$ \xi =\log \left( \frac{1}{1-t^2}\right) \Leftrightarrow t=\tanh \xi.$$ -The signature theorem follows from the above observations using a bit of algebraic combinatorics. For details see Chap. 7 of these lecture notes for a graduate course on this topic that I taught in 2008.<|endoftext|> -TITLE: Naive definition of surface area doesn't work? -QUESTION [12 upvotes]: A first stab at a definition of surface area might go like this: -Let S be a surface. Select finitely many points from S and make a bunch of triangles having these points as vertexes. Add up the areas of all of the triangles. This is a finite approximation to the surface area. Now to get the actual surface area, we increase the number of points so that they "densely cover" the surface (Maybe a good working definition would be that any open set of S should eventually contain some vertexes of triangles). -I seem to remember reading about a counterexample to this naive definition, but I can't find a reference. I believe there is even a natural looking polygonal approximation to the cylinder whose surface area diverges to infinity. Can anyone help me out? - -REPLY [25 votes]: Perhaps you are thinking of the Schwartz Lantern? -It converges to the cylinder in the Hausdorff metric but its area can be arranged -to head toward $\infty$. -It was mentioned in the earlier MO question, "Convergence of finite element method: counterexamples." -There is nice applet here showing the lantern rotating. -Here is an image from Conan Wu's blog:<|endoftext|> -TITLE: Why are some q-analogues more canonical than others? -QUESTION [15 upvotes]: It is striking that some q-analogs of functions, operators, identities and especially whole theorems seem quite "canonical", e.g. - -the factorial and the q-Gamma function -the basic hypergeometric series (at least $_{r+1}\Phi_r$ and $_{r}\Psi_r$) -q-Pi and the q-Wallis formula - -In a strict sense, q-analogs can of course not be canonical, as we might throw in almost everywhere powers of $q$ without changing the limit if $q\to1$. - -I mean canonical in the sense that these are the forms that require the least extra powers of $q$, or, more importantly, that other q-identities/theorems using them also tend to avoid such extra powers at best, making the formulae shorter. - - -Does it really make sense to call these (and certain other) q-analogs "canonical"? And if so, is there an explanation why some are much more canonical than others? - -(Or is there a better definition of canonical?) -The canonicity of the q-binomial coefficients is obviously accounted for by their relationship with linear subspaces. (So this is not an analytical criteria using the limit $q\to1$.) -What about the q-Gamma function? We may consider it canonical because of the (?!) q-analogue of the Bohr-Mollerup theorem proved by R. Askey, which states that for $0\lt q\lt 1$, the only logarithmically convex function satisfying $f(1)=1$ and $f(x+1)=\frac{q^x –1}{q–1}f(x)$ is the q-gamma function -$ \Gamma_q(z)=(1-q)^{1-x} \frac{(q\;;\; q)_{\infty}}{{(q^x;\; q) _\infty }}.$ -Also note that this formula looks at least as elegant as Euler's definition of $\Gamma(z)= \dfrac{1}{z} \prod\limits_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}.$ -On the other hand, one of the most basic identities, the recursion of binomial coefficients, has only an "asymmetric" q-analog and thus two of them: -${\genfrac[]00{n}{k}}_q=q^k{\genfrac[]00{n-1}{k}}_q+{\genfrac[]00{n-1}{k-1}}_q={\genfrac[]00{n-1}{k}}_q+q^{n-k}{\genfrac[]00{n-1}{k-1}}_q$. -For the classical orthogonal polynomials, it looks like there exist systematically "nice" q-analogs (see e.g. this survey), but it is not clear if there is a certain sense in which those can be considered canonical. Maybe for the Chebyshev polynomials, there is one "best" q-analog. - -Is there a reasonable way of considering certain q-analogs of orthogonal polynomials "canonical", e.g. their uniqueness w.r.t. to an appropriate criteria, as for most classical orthogonal polynomials? -Has a q-analog of a polynomial identity (e.g. involving binomial coefficients) more chances of being canonical if it has a combinatorial interpretation? - -For the q-derivative, there are at least two completely different approaches, both with their merits. So there is no use looking for canonicity there. -But nevertheless the next question: - -Are q-analogs conceptually similar to an extension from $\mathbb R$ to $\mathbb C$? - -By the latter I mean the following: -I wonder if generally speaking, the shift from an entity to its q-analog(s) can be likened, at least sometimes, to the shift of passing from $\mathbb R$ to $\mathbb C$, in the sense that some q-analogs provide a more complete picture than the entity itself (cp. for the $\mathbb R\to\mathbb C$ case the fundamental theorem of algebra or the meromorphic extension of the zeta function)? - -Many features in $\mathbb C$, e.g. the residue theorem, cannot be reduced to $\mathbb R$. Likewise for example, identities of "infinite q-polynomials" (i.e. of generating functions), cannot be taken to the limit $q\to1$. -In situations where there are several useful q-analogs, e.g. for the q-exponential function or the q-cosine, we might consider those corresponding to different panes of a Riemann surface like the one of $\sqrt{z}$ or $\ln z$. -And we can take it further: -The next step after $\mathbb R$ to $\mathbb C$ are the quaternions. -The next step after q-analogs are p,q-analogs. It looks like they haven't been studied a lot yet. - -Thank you for reading me so far. -Some of the questions may be rather subjective. As for the title, I had thought at first about "Why are most q-analogues canonical?" But then I figured that in the whole ocean of q-analogues, maybe there are just some "very" canonical islands, but for the vast majority the notion of canonicity is more or less fuzzy. Is that a feasible perception? -Even though some of these thoughts are somewhat philosopical, anyway, here goes. Looking forward to your input! - -REPLY [8 votes]: As I said above I have some difficulty to denote specific $q-$analogues as canonical. -Consider as example the Catalan numbers $\frac{1}{{n + 1}}{2n\choose n}$ . They have a simple generating function $f(z)$ which satisfies $f(z) = 1 + zf(z)^2 $ or equivalently $f(z) = \frac{{1 - \sqrt {1 - 4z} }}{{2z}}$ and they are characterized by the fact that all Hankel determinants are 1. -Which of the following simple $q - $analogues should be called "canonical"? -a) The polynomials $C_n (q)$ introduced by Carlitz with generating function $F(z) = 1 + zF(z)F(qz)$. They also have very simple Hankel determinants, but there is no known formula for the polynomials themselves. -b) The polynomials $\frac{1}{{[n + 1]}}{2n\brack n}$ . They have a simple formula but no simple formula for their generating function and no simple Hankel determinants. -c) The $q - $Catalan numbers $c_n (q)$ introduced by George Andrews. Their generating function $A(z)$ is a $q - $analogue of $ \frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$ Let $h(z)$ be the $q - $analogue of $\sqrt {1 + z}$ defined by $h(z)h(qz)=1+z$. Then $A(z)= \frac{1+q}{{4qz}}(1-h(-4qz))$. They have both simple formulas and a simple formula for the generating function and also simple Hankel determinants. But they are not polynomials in $q.$<|endoftext|> -TITLE: Reference Request: Non-Standard Models of PA -QUESTION [7 upvotes]: I am attempting to write an expository paper on non-standard models of PA that is accesible to students taking an introductory graduate course in mathematical logic (covering Godel's incompleteness theorems, the diagonalization lemma, models, etc.). In this paper I want to give an explanation of some results such as Tennanbaum's Theorem (there does not exist a countable recursive model of PA that is not isomorphic to the standard model). By, "give an explanation of", I mean to actually work through an explanation of the proof, some of the techniques involved, and the general overlap between techniques used in the proofs of Tennanbaum's theorem, some theorems proven by Rosser on extensions of PA, Robinson's overspill lemma, etc. (Note: I want to avoid digressing into an explanation of forcing if possible). -My question is, what books or online resources do you know of that would be useful for me? That is, do you happen to know of surveys of these topics that are around on arXiv or JSTOR? I have been digging through the mathematical logic section of arXiv for papers and I found a few that are useful, but I thought that some mathematicians/logicians on MO might know of some papers that give a just survey of the introductory results regarding non-standard models of PA. -Thank you! - -REPLY [2 votes]: Peter Smith has a pretty good handout on Tennenbaum's theorem that I found useful when learning that material. As others have mentioned, Richard Kaye's Models of Peano Arithmetic is the go-to reference work here. Kossak and Schmerl's The Structure of Models of Peano Arithmetic gives the state of the art, but you probably won't need this one.<|endoftext|> -TITLE: Symmetric tensor product of bosonic/fermionic Hilbert space -QUESTION [8 upvotes]: Consider two representation of the group $SU(n)$: $Sym^k(\mathbb{C}^n)$ and $\wedge^k\mathbb{C}^n$ ($k\leq n$) and take their symmetric tensor products: $Sym^2(Sym^k(\mathbb{C}^n))$, $Sym^2(\wedge^k\mathbb{C}^n)$. I have two questions concerning those representations: -Question 1: -It appears that $Sym^2(Sym^k(\mathbb{C}^n))$ and $Sym^2(\wedge^k\mathbb{C}^n)$ are multiplicity free (I checked this for low dimensional cases in Lie). Is that true in general? Is there an easy proof of this fact? -Question 2: -Which irreps of $SU(n)$ will appear in $Sym^2(Sym^k(\mathbb{C}^n))$ and $Sym^2(\wedge^k\mathbb{C}^n)$ respectively? In particular, which young diagrams of the symmetric group $S_{2k}$ will be present? - -REPLY [6 votes]: For question two, you are asking about a composition of two Schur functors, i.e. a plethysm. More specifically you want to know $h_2 \circ h_k$ and $h_2 \circ e_k$. These can be found in Example 9 in the section on plethysm in Symmetric functions and Hall polynomials: -$$ h_2 \circ h_k = \sum_{j \text{ even}} s_{(2k-j,j)}$$ -and -$$ h_2 \circ e_k = \sum_{j \text{ even}} s_{(k+j,k-j)^T}$$ -where $(\cdot)^T$ denotes the transposed Young diagram, and the sum is taken over those even $j$ that make the subscript a valid Young diagram. These translate to universal identities between representations, i.e. an isomorphism of functors. They express how the composed functors $\mathrm{Sym}^2(\mathrm{Sym}^k(-))$ and $\mathrm{Sym}^2(\bigwedge^k(-))$ can be written as direct sums of Schur functors. -When you apply a Schur functor to the defining representation of $\mathrm{SU}(n)$, the result is nonzero if and only if the corresponding partition has at most $n$ parts. Think about how the functor $\bigwedge^k$ vanishes on any vector space of dimension less than $k$. This explains the observation you make in a comment below, that not all terms in the second sum will appear if $n$ is small.<|endoftext|> -TITLE: Is there "Schur-Weyl duality" for infinite dimensional unitary group? -QUESTION [18 upvotes]: To what extent does the relation between the diagonal representation of $SU(n)$ in $(\mathbb{C}^n)^{\otimes k}$ and representations of the symmetric group $S_k$ remain valid when instead of the group $SU(n)$ we take the general unitary group $U(\mathcal{H})$ ($\mathcal{H}$ is some separable Hilbert space) and its diagonal representation in $\mathcal{H}^{\otimes k}$? In particular is it true that irreducible representations of of $U(\mathcal{H})$ in $\mathcal{H}^{\otimes k}$ correspond to young diagrams of $S_k$ just like in the finite dimensional case? Is it possible to get any irreducible representation of $U(\mathcal{H})$ by considering decompositions of $\mathcal{H}^{\otimes k}$ for sufficiently big $k$? - -REPLY [2 votes]: Thank you Anatoly and Tador for your answers - they pointed to the very useful references. Despite to what Anatoly Claims in his answer there is a very rich representation theory of $U(\mathcal{H})$ (not only for separable Hilbert spaces). I am not an expert in this field but as far as I can tell the paper by DOUG PICKRELL: http://www.ams.org/journals/proc/1988-102-02/S0002-9939-1988-0921009-X/S0002-9939-1988-0921009-X.pdf answers nearly all my questions. It stands that all separable representations (ie. representations in separable Hilbert space) of $U(\mathcal{H})$ ($\mathcal{H}$ - separable) decompose onto irreducible components that correspond to irreducibles that are taken "naivly" from some $\mathcal{H}^{\otimes k}$. It seams that this theory was known for $U(\infty)$ but in this paper author handled the disturbing "Galkin Algebra". I only wonder what happens to the determinant representation of $U(n)$ if we go to $\infty$ limit?<|endoftext|> -TITLE: Software for symbolic matrix calculus? -QUESTION [25 upvotes]: Is it possible to get widely available math software (Maple/Matlab/Mathematica, etc) to symbolically differentiate vector and scalar functions of matrices, returning the result in terms of the original matrices and vectors involved? I have in mind the simple sort of rules collected here for example. -On a few separate occasions I've scoured around the internet for such a thing and only turned up a bunch of incomplete threads of various vintage (like this or this or this). -So my main question is if I am missing the right keywords to find what is obvious to people who use this sort of functionality all the time, and what platform it is available on if so. -If in fact this sort of functionality is not available in any of the commonly used software my question is if this is because of some sort of practical obstruction I am not seeing or simply because the problems for which it would be useful are simple enough to be done by hand (which is what I've ended up doing after I spend 4 hours searching for the "easy" way). - -REPLY [21 votes]: Indeed, I was having the same problem. Hence, I implemented a matrix calculus toolbox myself. You can find it at www.matrixcalculus.org. It can compute vector and matrix derivatives and will return the result in terms of the original vectors and matrices involved.<|endoftext|> -TITLE: Kunen tree and Martin tree -QUESTION [7 upvotes]: Do we know under which conditions the Kunen tree (Recall the Kunen tree provides an analysis of the equivalence classes of functions $f: \omega_1 \to \omega_1$ with respect to the normal measure $W^1_1$ on $\omega_1$ under $AD$) is homogeneous, if it is? Do we know if and under which hypothesis the Martin tree (it is a generalization of the Kunen tree to all the projective hierarchy) is homogeneous? Any reference treating their homogeneity will be appreciated. Thx. - -REPLY [4 votes]: Let me answer my question which I stumbled on while looking at my profile. The answer is yes. Actually assuming $AD$, if $\kappa$ is less than the supremum of the Suslin cardinals then every tree $T$ on $\omega \times \kappa$ is weakly homogeneous (Martin and Woodin). -Also assuming that we have a Woodin $\delta$, a tree $T$ on $\omega \times \alpha$, $\alpha$ some ordinal, then there is a $\kappa < \delta$ such that in the generic extension $V[G]$ where $G$ is generic for $Col(\omega, \kappa)$, $T$ is $< \delta$ weakly homogeneous. -One can modify the assumption and just assume that there is a pair $T$ and $U$ of $\delta^+$ absolutely complementing trees.<|endoftext|> -TITLE: Can anyone explain to me what is an assembly map? -QUESTION [16 upvotes]: Or can you give me a good place to read about things related to assembly map, besides wikipedia? I am specially interested in the case of algebraic K-theory. Would appreciated if you could provide examples. - -REPLY [3 votes]: You can find lots of definitions in Hambleton and Pedersen's paper "Identifying assembly maps in K- and L-theory " at http://www.maths.ed.ac.uk/~aar/papers/idensurg.pdf -For a torsion-free group $G$, the assembly map $H_n(BG;K\mathbb{Z})\rightarrow K_n(\mathbb{Z}G))$ is induced by the contraction $EG\rightarrow pt$. In General, the algebraic $K$-theory is difficult to compute. The Farrell-Jones conjecture (claiming the assembly map is an isomorphism for each $n$) says that the algebraic $K$-theory of group rings can be computed by a homology theory, which is easier to deal with.<|endoftext|> -TITLE: The most general context of Mather's Cube Theorems -QUESTION [12 upvotes]: Quite simply, I'd like to know what is the broadest or most natural context in which either (or both) of Mather's cube theorems hold. If you like, this may mean any of - -What properties of $Top$ or $Top^*$ are essential to the proofs? -(where) are model/homotopical categories verifying Mather's theorems studied as such in the literature? -Are there more examples known verifying Mather's theorems? - -I ask because Mather's proof strikes me as fairly gritty and seems to rely on explicit cellular constructions. - -For reference, the cube theorems concern a cubical diagram whose faces commute up to homotopy in a coherent way, and assert - -If one pair of opposite faces are homotopy push-outs and the two remaining faces adjecent the source vertex are homotopy pull-backs, then the final two faces are also homotopy pull-backs -If two pairs of opposite faces are homotopy pull-backs, and the remaining face adjacent the target vertex is a homotopy push-out, then the remaining face is a homotopy push-out. - -REPLY [17 votes]: Let $\mathcal{X}$ be an $\infty$-category (i.e., a homotopy theory) which admits small homotopy colimits, a set of small generators, and has the property that homotopy colimits in $\mathcal{X}$ commute with homotopy pullback. Then $\mathcal{X}$ satisfies the Mather cube theorem if and only if $\mathcal{X}$ is an $\infty$-topos: that is, it can be described as a left exact localization of an $\infty$-category of presheaves of spaces. (I learned this from Charles Rezk). Such homotopy theories are studied extensively in my book "Higher Topos Theory" (see in particular Proposition 6.1.3.10 and the remark which follows it).<|endoftext|> -TITLE: Mapping from a finite index subgroup onto the whole group -QUESTION [22 upvotes]: Dear All, -here is the question: -Does there exist a finitely generated group $G$ with a proper subgroup $H$ of finite index, and an (onto) homomorphism $\phi:G\to G$ such that $\phi(H)=G$? -My guess is "no", for the following reason (and this is basically where the question came from): in Semigroup Theory there is a notion of Rees index -- for a subsemigroup $T$ in a semigroup $S$, the Rees index is just $|S\setminus T|$. The thing is that group index and Rees index share the same features: say for almost all classical finiteness conditions $\mathcal{P}$, which make sense both for groups and semigroups, the passage of $\mathcal{P}$ to sub- or supergroups of finite index holds if and only if this passage holds for sub- or supersemigroups of finite Rees index. There are also some other cases of analogy between the indices. Now, the question from the post is "no" for Rees index in the semigroup case, so I wonder if the same is true for the groups. -Also, I beleive the answer to the question may shed some light on self-similar groups. - -REPLY [11 votes]: Here is a variation on Henry's nice argument which uses Malcev's theorem. Let $N$ be the intersection of all finite index normal subgroups of $G$. Clearly $\phi(N)\subseteq N$ because a surjective endomorphism takes finite index normal subgroups to finite index normal subgroups. Thus $\phi$ induces a proper endomorphism of the finitely generated residually finite group $G/N$. By Malcev's theorem that f.g. residually finite groups are Hopfian, it follows $\phi$ induces an automorphism, which means $\ker \phi$ is contained in $N$. But then since each finite index subgroup contains a finite index normal subgroup, we have $\ker \phi\subseteq H$, which is a contradiction as Henry points out since in that case one would have $[G:H]=[\phi(G):\phi(H)]$.<|endoftext|> -TITLE: Parabolic induction GL(n,Zp) -QUESTION [6 upvotes]: Let $P$ be a parabolic subgroup of $GL(n)$ with Levi decomposition $P =MN$, where $N$ is the unipotent radical. -Let $\pi$ be an irreducible representation of $M(\mathbf{Z}_p)$ inflated to $P(\mathbf{Z}_p)$, - -how does $ Ind_{P(\mathbf{Z}_p)}^{GL_n(\mathbf{Z}_p)} \pi$ - decompose? - -It would be sufficient for me to know the result in the simplest case, where $P$ is a Borel subgroup - -REPLY [6 votes]: As has been noted, the answer is known for $\mathrm{GL}_2$. For $n>2$ there are certain cases where $\rho:=\mathrm{Ind}_{P(\mathbb{Z}/p^r)}^{\mathrm{GL}_n(\mathbb{Z}/p^r)}\pi$ is irreducible (see for example the result of Hill referred to in the question Parabolic induction for GL(2,Z/pn)). -For $\mathrm{GL}_3$ there are two papers by Campbell and Nevins (http://arxiv.org/pdf/0710.3261v1.pdf and http://arxiv.org/pdf/0710.3263.pdf) which study the decomposition of representations of the form $\rho$. In particular the authors almost achieve a decomposition of $\mathrm{Ind}_{B(\mathcal{O}_r)}^{\mathrm{GL}_3(\mathcal{O}_r)}\mathbf{1}$, but it turns out there are irreducible consituents which depend on the residue characteristic $p$, and these are hard to pin down explicitly in a uniform way. -Contrary to what's stated in the question, the simplest case is not when $P$ is a Borel but when it is a maximal parabolic subgroup of $\mathrm{GL}_n$. In this case one can find an explicit set of representatives for $P(\mathcal{O}_r)\backslash \mathrm{GL}_n(\mathcal{O}_r)/P(\mathcal{O}_r)$, independent of $p$, and consequently a complete decomposition of $\mathrm{Ind}_{P(\mathcal{O}_r)}^{\mathrm{GL}_n(\mathcal{O}_r)}\mathbf{1}$. This was done by Hill in "On the nilpotent representations of $\mathrm{GL}_n(\mathcal{O})$", and is also studied in a more general context in Onn & Bader (http://arxiv.org/abs/math/0404408). -In general, the problem of decomposing $\mathrm{Ind}_{B(\mathcal{O}_r)}^{\mathrm{GL}_n(\mathcal{O}_r)}\mathbf{1}$ will in some way or another involve a description of $B(\mathcal{O}_r)\backslash \mathrm{GL}_n(\mathcal{O}_r)/B(\mathcal{O}_r)$, and this is claimed to be a wild problem in Onn, Prasad & Vaserstein (http://arxiv.org/pdf/math/0506094.pdf), although Prasad seems to express a small degree of reservation in Bruhat decomposition for G(R), R local ring or R=Z/p^r. -Looking at the results for $\mathrm{GL}_3$ it seems very likely that the problem of decomposing a general representation of the form $\rho$ is hopelessly complicated. As mentioned above there are partial results, and this is probably the most one can expect in general. -The "Iwahori decomposition plus the Bruhat decomposition over the residue field" approach alluded to in one of the comments seems unlikely to me to lead to a complete solution since it would reasonably be independent of $p$ while the decomposition problem in general probably depends on $p$ (because the spaces of double cosets do).<|endoftext|> -TITLE: Extremal points of the unit ball of M_n(R) ... -QUESTION [18 upvotes]: The unit ball of ${\bf M}_n(\mathbb R)$ is a compact convex subset. As such, it is (Krein-Milman theorem) the convex envelop of its extremal points. So far, so good; but the unit ball depends of the choice of a norm. There are a few natural choices: - -the Schur-Frobenius (Hilbert-Schmidt) norm $\|X\|_F=\sqrt{{\rm Tr}X^TX}$. This is the standard Euclidian norm over ${\bf M}_n(\mathbb R)\sim\mathbb R^{n^2}$. The extremal points form the unit sphere. -the operator norm $\|X\|_2$ associated with the Euclidian norm over $\mathbb R^n$. The extremal points form the orthogonal group ${\bf O}_n(\mathbb R)$. See a related question. -the operator norm $\|X\|_1$ associated with the $\ell_1$-norm over $\mathbb R^n$. The extremal points are the sign-permutation matrices. There are only $2^nn!$ of them. -the numerical radius $r(X)=\sup|y^*Xy|$, where the supremum is taken over all unit complex vectors (real vectors are not enough). It is the smallest radius of a disk $D(0;r)$ containing the numerical range (Hausdorffian) of $X$. So, this my question: - - -Let $B_{\rm nr}$ be the unit ball when we endow the $n\times n$-matrices with the norm $r$. What are the extremal points of $B_{\rm nr}$ ? - -Edit. To answer Geoff's comment. If $A$ is a normal matrix, then $r(A)=\rho(A)$ (the numerical range is the convex envelop of the spectrum in this case). But if $A$ is not normal, one usually have (not always) $r(A)>\rho(A)$. Thus not all matrices with spectral radius equal to $1$ belong to $B_{\rm nr}$. And even if $A$ is normal and $\rho(A)=1$, it is not extremal unless all eigenvalues have unit modulus. -Re-edit. After thinking to the case $n=2$, I have the opinion that considering the unit ball for $r$ in ${\bf M}_n(\mathbb R)$ is unnecessarily complicated. It would be more reasonable to work in ${\bf M}_n(\mathbb C)$ instead. - -REPLY [5 votes]: After working a little bit, I solved the $2\times 2$ case. Surprisingly, ${\rm ext}(B_{\rm nr})$ is not closed. It consists of (recall that if $n=2$, the numerical range is an ellipse $E_A$ together with its interior; the ellipse may degenerate to a segment or a point) - -homotheties $zI_2$ with $|z|=1$, -matrices $A$ whose ellipse $E_A$ is tangent interiorly to $\bf S^1$ at two points $w,z$ such that $w+z\ne0$, -nilpotent matrices $N$ with $\|N\|_F=2$. This is the case where $E_N=\bf S^1$, -normal matrices with distinct eigenvalues $w,z$ such that $w+z\ne0$ and $|w|=|z|=1$. - -It is the condition $w+z\ne0$ which prevents ${\rm ext}(B_{\rm nr})$ from being closed. -The second class above may be seen as the general one, the other ones being limits of it. -See the details in the draft.<|endoftext|> -TITLE: Self-avoiding walk on $\mathbb{Z}$ -QUESTION [15 upvotes]: This one is an unanswered question in Math.SE. I've posted it here because I think it deserves more attention. - -How many sequences $\{a_n\}$ exist satisfying: - a) $a_1=0$ - b) $\forall k\ge1 $ either $a_{k+1}=a_k+k$ or $a_{k+1}=a_k-k$ - c) $a_i \neq a_j$ whenever $i \neq j$ - d) $\mathbb{Z}=\{a_i\}_{i > 0}$ - -Are the two below alternating sequences the only solutions? - -$a_{2k}=k$, $a_{2k+1}=-k$ -$a_{2k}=-k$, $a_{2k+1}=k$ - -Also, it is known that Recamán's sequence satisfies a) $a_1=1$ and b), c) as above - -REPLY [11 votes]: There are many other solutions. As explained in Douglas Zare's comments, the idea is to choose a cell and to make the step large enough in order to visit it. -Here are the details (which are best followed with pen and paper...). Suppose that at some time, the convex hull of the integers which are already covered is the interval $[a,b]$. We also assume that we are at one end of the interval, say at $b$, and that we are ready to walk with step $s$. We want to visit a given integer $x \in (a,b)$. We want to do this using only cells at the right of $b$, and in such a way that with one more move, we may reach a cell situated at the left of $a$. By induction, it will then be possible to cover $\mathbf{Z}$. I will write $+$ for moving to the right and $-$ for moving to the left. -The proof consists of the following reductions. - -We may assume $s > \frac{b-a}{2}$ and $x < \frac{a+b}{2}$. Indeed, starting at $b$ with stepsize $s$, do $++(-+)^{s-1}$. We land at $b+3s$ and are ready to walk with step $3s$. The diameter $L=b-a$ and the step $s$ have changed to $(L+3s,3s)$. Thus the ratio $r=L/s$ has changed to $(L+3s)/(3s) = 1+r/3$. Iterating the process, we can make $r$ arbitrarily close to the fixed point $3/2$, in particular we can make $r<2$ which means $s>L/2$ as requested. It is also clear that iterating the process we can make $(x-a)/L$ arbitrary small, in particular we may assume $x<\frac{a+b}{2}$. -We may assume $b-x \equiv 1 \pmod{3}$. Indeed, we may always achieve this during the first reduction, by doing $++(-+)^{s-2}$ or $++(-+)^{s-3}$ instead of $++(-+)^{s-1}$ (which has the effect to land in $b+3s-1$ or $b+3s-2$ instead of $b+3s$). -Let $b-x=3k+1$ with $k \geq 0$. Do $+(+-)^k -$ and you are at $x$, and then do one more step to the left. This works because by assumption $3k < L < 2s$, so we use only cells at the right of $b$, and because when we are at $x$ we have step $s+2k+2>x-a$ which enables to escape at the left of $a$.<|endoftext|> -TITLE: Verbal subgroups of free groups and the corresponding automorphisms -QUESTION [8 upvotes]: Let $F_n$ be a free group of finite rank $n$ and let $V$ be a verbal subgroup of $F_n.$ In a "nice" situation one could expect that -$(*):$ the group $\mathrm{Aut}_V(F_n) \le \mathrm{Aut}(F_n)$ whose elements fix each element of $F_n$ mod $V$ acts transitively on the set of all primitive elements from the coset $xV$ of a primitive element $x$ of $F_n.$ -I have two questions related to the situation above. -1) Is the property $(*)$ true for the verbal subgroups $V=V([x_1,\ldots,x_m])=\gamma_m(F_n),$ i.e. for terms of the lower cenral series of $F_n?$ (Asked this question on math.stackexchange, got no answers). -Most likely, the answer to 1) is affirmative, and might be obtained from the classic description of the generators of the group $\mathrm{IA}(F_n)$ of IA-automorphisms of $F_n$ due to Magnus. I just don't see how, and being not versed in German I am not able to read the original Magnus' proof. Any comments on the plan etc. of Magmus' proof are welcome. -2) Are there examples of verbal subgroups $V$ for which the property $(*)$ is not true? - -REPLY [5 votes]: (This started out as a comment, but it grew too long) -My guess is that this is a very hard problem, though I'd be happy to be proven wrong. But I can say something about Magnus's theorem on generators for $IA_n$. I know of three proofs of this result. - -Magnus's original proof. This is based on the short exact sequence -$$1 \longrightarrow IA_n \longrightarrow Aut(F_n) \longrightarrow GL(n,\mathbb{Z}) \longrightarrow 1.$$ -Magnus constructed a finite presentation $\langle S | R \rangle$ for $GL(n,\mathbb{Z})$ (you can find a modern exposition of this in Milnor's book on algebraic k-theory). The elements $S$ can be lifted in a natural way to a generating set $\tilde{S}$ for $Aut(F_n)$. This implies that the lifts $\tilde{R}$ of the relations are normal generators for $IA_n$. Magnus then proved with a brute-force computation that the subgroup generated by his finite generating set was normal and contained $\tilde{R}$, which implies that his finite set is a generating set. -In the paper -M. Bestvina, K.-U. Bux and D. Margalit, Dimension of the Torelli group for Out(Fn), Invent. Math. 170 (2007), no. 1, 1–32 -there is a topological proof (using the action of $IA_n$ on auter space) that Magnus's normal generating set is a normal generating set. This paper also contains a brief description of the calculation needed to go from this to the finite generating set. -In my paper "The complex of partial bases for F_n and finite generation of the Torelli subgroup of Aut(F_n)" with Matt Day we gave yet another topological proof that Magnus's normal generating set is a normal generating set using a space analogous to the curve complex. We also have a somewhat more complete sketch of the calculation needed to go from this to the finite generating set. - -EDIT : I forgot to point out that in the paper described in the third item above, we also prove that the desired result holds for $V$ equal to the commutator subgroup. I'm pretty sure that this is the only special case of the desired result that appears in the literature.<|endoftext|> -TITLE: Categorical interpretation of quasi-compact quasi-separated schemes -QUESTION [19 upvotes]: Let $X$ be a scheme. Consider the global section functor $\Gamma : \mathrm{Qcoh}(X) \to \mathrm{Ab}$. It is well-known that $\Gamma$ preserves filtered colimits if $X$ is assumed to be quasi-compact and quasi-separated. In other words, $\mathcal{O}_X$ is a compact object. Even more, it can be shown (using the results of EGA I, 6.9) that compact objects are precisely the quasi-coherent modules which are locally of finite presentation. If we define a locally finitely presentable tensor category to be a tensor category $C$, whose underlying category is locally finitely presentable and such that a) $1_C$ is compact, b) the compact objects are closed under $\otimes_C$, then $\mathrm{Qcoh}(X)$ turns out to be such a category. -Does the converse also hold? This would be a categorical interpretation of this typical finiteness condition in algebraic geometry. -Question. When $\Gamma$ preserves filtered colimits, does it follow that $X$ is quasi-compact and quasi-separated? If not, what if we even assume that $\mathrm{Qcoh}(X)$ is a locally finitely presentable tensor category? -There are other natural conditions which might imply that $X$ is qc+qs, for example that even all cohomology functors $H^i : \mathrm{Qcoh}(X) \to \mathrm{Ab}$ preserve filtered colimits. Besides there are relative versions of this question: When $f : X \to Y$ is a morphism of schemes, then $f$ is quasi-compact and quasi-separated iff $f_\*$ or $f^\*$ has which property? -EDIT: Let us spell out what it means that $\Gamma$ preserves direct sums: Let $M_i$ be (infinitely many) quasi-coherent modules on $X$, $s_i \in \Gamma(M_i)$ global sections, and $X = \cup_j X_j$ an arbitrary open covering, then the following holds: If for every $j$ we have $s_i |_{X_j} = 0$ for almost all $i$, then $s_i = 0$ for almost all $i$. So the bounds for the $i$ with $s_i \neq 0$ on each $X_j$ are bounded. Why should this happen for a scheme which is not quasi-compact? But in order to prove this, we would have to construct appropriate quasi-coherent modules, and this can only be done my transfinite methods if $X$ is general (see for example here). - -REPLY [11 votes]: Let me show that if $\Gamma$ preserves filtered colimits, then $X$ is quasicompact. (At the moment I don't know about 'quasiseparated'; but, as Martin points out, I only use the injectivity of $\varinjlim\circ \Gamma \to \Gamma\circ \varinjlim$ for filtered inductive systems). -Assume $X$ is not quasicompact. Then there is a filtered decreasing family $(Y_i)$ of nonempty closed subschemes of $X$, with empty intersection. The structure sheaves $\mathcal{O}_{Y_i}$ form an inductive system with colimit zero, but the unit section of any $Y_{i_0}$ is an element of $\varinjlim_i \Gamma(\mathcal{O}_{Y_i})$ which is nonzero because each $Y_i$ is nonempty.<|endoftext|> -TITLE: Simple discrete subgroups of Lie groups -QUESTION [28 upvotes]: Upon Ian Agol's suggestion, I separated this question from the one on non-residual finiteness in -Non-residually finite matrix groups -Question. Are there infinitely generated simple discrete subgroups of $SL(n, {\mathbb R})$, $n\ge 3$? -Note: Infinitely generated means not admitting a finite generating set. -Background: By Malcev's theorem, every finitely generated matrix group is residually finite. In particular, it is either finite or non-simple. On the other hand, there are (say, countable) simple matrix groups like $PSL(n, {\mathbb Q})$. The question is what happens in the case of discrete infinitely generated groups. On one hand, arguments similar to Malcev's appear to fail since there are discrete non-residually finite subgroups of $SO(3,1)$, see Agol's examples in Non-residually finite matrix groups -Moreover, as Ian observes, one can construct groups like this which have no nontrivial finite quotients. However, an easy ping-pong argument shows the following: -Lemma. There are no simple discrete infinite subgroups $\Gamma$ in rank 1 Lie groups. -Proof. The statement is clear if $\Gamma$ is free, so suppose that it is not. Take a high power of a hyperbolic element in $\Gamma$, then its normal closure in $\Gamma$ will be free and, hence, a proper normal subgroup of $\Gamma$. If $\Gamma$ has no hyperbolic elements, then it is virtually nilpotent and, hence, not simple. -This argument, however, will fail miserably for discrete subgroups in Lie groups of rank $\ge 2$ since the latter contain irreducible lattices $\Gamma$. Such lattices (by Margulis' normal subgroups theorem) contain no infinite normal subgroups of infinite index. Of course, Margulis' argument will not work for infinitely generated groups, but the point is that any proof of non-simplicity of discrete subgroups $\Gamma$ of higher rank Lie groups would have to use the fact that $\Gamma$ is not a lattice. (The only way to exploit this fact that I can see at this point is that an infinitely generated group $\Gamma$ does not have Property (T), but it is unclear to me how to use this even to get started on proving non-simplicity.) -So, maybe there are simple infinite discrete subgroups of higher rank Lie groups $G$. The problem is that there are not that many ways to construct discrete infinitely generated subgroups in such $G$'s. I know only of two reasonably general constructions: -(1). Some form of ping-pong (used, for instance, by Margulis and Soifer to construct maximal subgroups of matrix groups). However, the resulting groups will be free products, so never simple. -Addendum: I am not claiming here that maximal subgroups $H$ (in a f.g. group $G$) constructed by Margulis and Soifer are not free products, only that the key step in the construction of $H$ is a free product $F$ (actually, a free group in the main paper of Margulis and Soifer). Then the subgroup $H\subset G$ is a maximal proper subgroup of $G$ containing $F$: This last step of the proof is nothing by Zorn's lemma and does not contain any constructive information. (All what we know about $H$ is that it is maximal and has infinite index in $G$.) In particular, $H$ itself is useless as far as constructing simple subgroups of matrix groups. Incidentally, it is an open problem, going back to Prasad and Tits: Does, say, $SL(n, {\mathbb Z}), n\ge 3$, contain a maximal subgroup which is not (virtually) a nontrivial free product. Margulis and Soifer prove that, for $n\ge 4$, one can get a maximal subgroup $H$ containing a free product of ${\mathbb Z}^2$ and of a free group. -(2). Vinberg's theorem on discreteness of groups generated by linear reflections in faces of convex "polyhedral" cones (such cones would have to have infinitely many faces in our situation, so polyhedrality has to be taken with the grain of salt). Of course, Coxeter groups one obtains as the result are not simple, but the only obvious normal subgroup is of index 2, so maybe one can obtain a virtually simple group (i.e., a group containing a simple subgroup of finite index) in this fashion. Could Coxeter groups with finite-dimensional Davis complex be virtually simple? Hard to imagine, but, at this point, this seems to be the best hope for an example of a discrete infinite simple matrix group. -Update: Actually, one can show that (infinite) discrete Coxeter subgroups of Lie groups are never virtually simple. Even more, any discrete subgroup $\Gamma$ of $SL(n, {\mathbb R})$, which contains an abstract rank 1 diagonalizable element, is never virtually simple (same proof as in the case of rank 1 Lie groups works). Here abstract rank 1 element of an (abstract) group $\Gamma$ is an infinite order element $\gamma$ so that for any $k\ne 0$ the centralizer of $\gamma^k$ in $\Gamma$ is virtually cyclic with uniform bound in "virtually." So, maybe one can show that all discrete Zariski dense subgroups of $SL(n, {\mathbb R})$ without abstract rank 1 elements, are lattices by running some Margulis-type argument... - -REPLY [2 votes]: This is not an answer; but Margulis and Soifer construct maximal subgroups of lattices $\Gamma$ . Contrary to what Misha says, they are not free products (nor necessarily free groups), but they are maximal subgroups which CONTAIN free subgroups which map onto every finite quotient of the lattice $\Gamma$. It is entirely possible that maximal subgroups are simple (I do not know a proof, one way or another). -Aakumadula<|endoftext|> -TITLE: Existence of Rational Orthogonal Matrices -QUESTION [9 upvotes]: Question: -Let $A\in\mathbb{R}^{n \times n}$ be an orthogonal matrix and let $\varepsilon>0$. Then does there exist a rational orthogonal matrix $B\in\mathbb{R}^{n \times n}$ such that $\|A-B\|<\varepsilon$? -Definitions: - -A matrix $A\in\mathbb{R}^{n \times n}$ is an orthogonal matrix if $A^T=A^{-1}$ -A matrix $A\in\mathbb{R}^{n \times n}$ is a rational matrix if every entry of it is rational. - -REPLY [23 votes]: Yes. It is a theorem of Cayley that the mapping $S \rightarrow (S-I)^{-1}(S+1)$ gives a correspondence between the set of $n\times n$ skew-symmetric matrices over $\mathbb{Q}$ and the set of $n\times n$ orthogonal matrices which do not have one as an eigenvalue. Since the mapping is nice, and rational skew-symmetric matrices are dense in the set of all skew-symmetric matrices, you have your result. For more, see the very nice paper by Liebeck and Osborne - -REPLY [13 votes]: Sure. Consider matrices which fix $n-2$ of the standard basis vectors and describe a rotation in the plane spanned by the last two about an angle $\theta$ such that $\sin \theta, \cos \theta$ are both rational; these are dense in all such rotations, and all such rotations generate the orthogonal group, so the corresponding products (all of which are rational) are dense in the orthogonal group. - -REPLY [6 votes]: I should say yes. For this, I shall use the fact that in the unit sphere $\mathbb S^{d-1}$, the set of rational vectors is dense. I shall proceed by induction over $n$. -So let $A\in {\bf O}_n(\mathbb R)$ be given. Let $\vec v_1$ be its first column, an element of ${\mathbb S}^{n-1}$. We can choose a rational unit vector $\vec w_1$ arbitrarily close to $\vec v_1$. The first step is to construct a rational orthogonal matrix $B$ with first column $\vec w_1$. To this end we choose inductively rational unit vectors $\vec w_2,\ldots,\vec w_n$. This is possible because at each step, we may take a rational unit vector in the unit sphere of a "rational" subspace. Here, a subspace $F$ is rational if it admits a rational basis. -Now, let us form $A_1=B^{-1}A$. This is a orthogonal matrix, whose first column is arbitrarily close to $\vec e_1$. Hence its first line is close to $(1,0,\ldots,0)$ as well. Thus -$$A_1\sim\begin{pmatrix} 1 & 0^T \\\\ 0 & R \end{pmatrix}.$$ -The matrix $R$ is arbitrarily close to ${\bf O}_{n-1}({\mathbb R})$. By the induction hypothesis, there exists a rational orthogonal matrix $Q$ arbitraly close to $R$. Then -$$B\begin{pmatrix} 1 & 0^T \\\\ 0 & Q \end{pmatrix}$$ -is arbitrarily close to $A$.<|endoftext|> -TITLE: Second homology group of free nilpotent p-group -QUESTION [11 upvotes]: Let $F_n$ be a free group on $n$ generators. Fix a prime $p$. Let $\gamma_k^p(F_n)$ be the mod $p$ lower central series, i.e. the inductively defined series -$$\gamma_0^p(F_n) = F_n \quad \text{and} \quad \gamma_{k+1}^p(F_n) = (\gamma_{k}^p(F_n))^p [F_n,\gamma_k^p(F_n)].$$ -Observe that the quotients $\gamma_{k}^p(F_n) / \gamma_{k+1}^p(F_n)$ are abelian $p$-groups. Moreover, the quotients $N_n^p := F_n / \gamma_{k}^p(F_n)$ are $p$-groups of nilpotency class $k$. They are universal with this property -- if $G$ is a $p$-group of nilpotency class $k$ and $g_1,\ldots,g_n \in G$, then there is a unique homomorphism $N_n^p \rightarrow G$ taking the generators of $N_n^p$ to the $g_i$. -Question : What are $H_2(N^p_n;\mathbb{Z})$ and $H_2(N^p_n;\mathbb{F}_p)$? - -REPLY [2 votes]: For brevity, I denote by $F$ the free group on $n$ generators, $\lambda_k$ the $k$th terme of the $p$-lower series of $F$, and $N_k$ the relatively free group $F/\lambda_k$. -Also I use the following 5- terme exact sequence (see Generators of sections of free groups): -$$0 \longrightarrow H_2(N_k;\mathbb{Z}/p) \longrightarrow \lambda_k/[F,\lambda_k]{\lambda_k}^p \longrightarrow H_1(F;\mathbb{Z}/p) \longrightarrow H_1(N_k;\mathbb{Z}/p) \longrightarrow 0.$$ -Now since $N_k$ is minimally $n$-generated $p$-group, it follows from $ H_1(N_k;\mathbb{Z}/p) = N_k/[N_k,N_k]{N_k}^p$ that $ H_1(N_k;\mathbb{Z}/p) \cong (\mathbb{Z}/p)^n$, and also we have $H_1(F;\mathbb{Z}/p) = F/[F,F]F^p \cong (\mathbb{Z}/p)^n$. Now the above exact sequence reduces to -$$0 \longrightarrow H_2(N_k;\mathbb{Z}/p) \longrightarrow \lambda_k/[F,\lambda_k]{\lambda_k}^p \longrightarrow 0$$ -so we have $H_2(N_k;\mathbb{Z}/p) \cong \lambda_k/\lambda_{k+1}$, and this is completely determined by the rank $r_k$ of the elementary abalian $p$-group $\lambda_k/\lambda_{k+1}$. -The integer $r_k$ can be determined using Lie methods, see for instance Corollary 18 in "The automorphism group of a finite p-group is almost always a p-group" Journal of Algebra, Volume 312 (2007) G. T. Helleloid and U. Martin. In which it is proved that -$$r_k = \sum_{i=1}^{k} 1/i (\sum_{j/i} \mu(i/j) n^j)$$ where $\mu(t)$ denotes the Mobius function.<|endoftext|> -TITLE: Prime numbers in arithmetic progressions : uniformity with respect to the modulus -QUESTION [10 upvotes]: Most of the proofs of Dirichlet's theorem on primes in arithmetic progressions actually give a Mertens-like theorem, and then the (weaker) statement - -Chebyshev-like bound : if $(a,q) = 1$ then -$$\sum_{\substack{n \leq X \\ n \equiv a \mod q}} \Lambda(n) \gg_q \frac{X}{\varphi(q)} $$ - (the factor $\varphi (q)$ is introduced here only for cosmetic reasons) - -There are basically two ways in which this could be strenghtened : - -for fixed $q$, in the $X$-aspect : this amounts to replace the $\gg $ above by $\sim$, which is exactly the prime number theorem in arithmetic progressions. -in the $q$-aspect : one asks for explicit $\epsilon$ (depending on $q$) satisfying -$$\sum_{\substack{n \leq X \\ n \equiv a \mod q}} \Lambda(n) \geq \epsilon \frac{X}{\varphi(q)} $$ - -If complex analysis is allowed, the Siegel-Walfisz solves both problems and gives $\epsilon = 1 - o(1)$ in the range $q \ll \left( \log X \right) ^A $ (for any $A>0$). But I'm especially interested in elementary methods (with the usual meaning of the word "elementary" in this context). Following step by step Dirichlet's proof (or at least one of its modern variants), I managed to prove that -$$ \epsilon = e^{- C \varphi(q) \left( \log q \right)^9} $$ -is admissible. Apart from the unimportant $\log$ factors, I haven't improved this yet. Hence my questions : - -What is the best (known) lower bound on $\epsilon$ that one can reach by elementary methods ? -What is the wider allowed range for $q$ that one gets from the elementary proofs of the prime number in arithmetic progressions ? - -References are welcome, I've found none so far. -EDIT : In view of the comments and answers below, I conclude that what I'm asking for is not as classic I thought it was. I summarize the state of the question : - -there are (relatively easy and) elementary proofs of Siegel's theorem, but deducing from it a Siegel-Walfisz theorem seems to require complex (or Fourier) analysis. -No elementary proof of Linnik's theorem exists in the literature, but Micah Milinovich suggests below that A. Granville could have further information on this subject. It may be worth contacting him. - -EDIT 2 : My questions are essentially solved by the "anonymous's" comment below. There indeed exists an elementary proof of Linnik's theorem (or at least a proof avoiding most of the complex analysis machinery in original Linnik's proof). The last use of complex analysis originated results lies in the explicit use of $\Psi(X,q,a) = \frac{X-\frac{X^{\beta}}{\beta}}{\varphi(q)} + \text{Error term}$ (according to Andrew Granville, this seems to be fixed, but details are not clear -Thanks for your help ! - -REPLY [4 votes]: Have a look at equation (5) in -Sur un théorème de Mertens in Manuscripta Mathematica 108 (2002), pages 495--513. -by myself (O. Ramaré). You can find the paper on my web site. I get $\epsilon_q = \exp(-c\sqrt{q}(\log q)^2)$ in an elementary manner. Better would improve on the Siegel zero location. Sorry, but the paper is in french :) -I hope that helps! Olivier<|endoftext|> -TITLE: A Realization Problem for Character Tables -QUESTION [10 upvotes]: Given an $n\times n$ table of complex numbers, are there known sufficient conditions for the table to be the character table of a finite group? Representation theory gives plenty of necessary conditions, but I can't imagine they'd be enough in general. - -REPLY [6 votes]: You should look up an older article by Stephen Gagola, Jr., but read some of the arguments skeptically (as I did a long time ago when exploring this question in a graduate introduction to finite group representations): -Gagola, Stephen M., Jr.(1-KNTS) -Formal character tables. -Michigan Math. J. 33 (1986), no. 1, 3–10. -I'm not sure whether more interesting results are known by now, but it's a difficult problem which has been around for a long time. I think it's fair to describe the problem as essentially open, though inevitable in this subject.<|endoftext|> -TITLE: Endomorphisms rings of elliptic curves and congruences of $j$ -QUESTION [12 upvotes]: Let $p$ be a prime number, $K/\mathbf{Q}_p$ a finite extension, with integers $O_K$, valuation ideal $\mathfrak{p}$, and residue field $k_\mathfrak{p}$. Let $E$ be an elliptic curve over $K$ with good reduction $E_\mathfrak{p}$ over $k_\mathfrak{p}$. -If $\ell$ is a prime $\neq p$, then $T_\ell(E)$ is identified with $T_\ell(E_\mathfrak{p})$ in a natural way, by the good reduction of $E$. As it turns out such a Galois representation is determined, up to isomorphism, by the characteristic polynomial $f_{E_\mathfrak{p}}(x)=x^2-a_{E_\mathfrak{p}}x+|k_\mathfrak{p}|$ associated to $E_\mathfrak{p}$ and by $j_E$ mod $\mathfrak{p}=j_{E_\mathfrak{p}}$ UNLESS we are in the following (very) ${\it special}$ case: -$p\equiv 3$ mod $4$; $|k_\mathfrak{p}|=p^{2m+1}$; $a_{E_\mathfrak{p}}=0$; $\ell=2$; and $j_E\equiv 1728$ mod $\mathfrak{p}$. -If the first three conditions hold, then $E_\mathfrak{p}$ is supersingular and its endomorphisms ring over $k_\mathfrak{p}$ is "only" isomorphic to an order in $\mathbf{Q}(\sqrt{-p})$ containing $\sqrt{-p}$, and thus isomorphic to either $\mathbf{Z}[\sqrt{-p}]$ or to $\mathbf{Z}[(1+\sqrt{-p})/2]$. The second case occurs precisely when all the two torsion is defined over $k_\mathfrak{p}$, the first case when $E_\mathfrak{p}[2]$ has only two $k_\mathfrak{p}$-points. Both cases do arise and give rise to non-isomorphic $T_2(E_\mathfrak{p})$. -Essentially by Deuring's Lifting Lemma one can decide which of the two possibilities occurs by looking at the $j$-invariant of $E_\mathfrak{p}$ UNLESS this is equal to 1728. The point is that if $j_{E_\mathfrak{p}}\neq 1728$ then the two $k_\mathfrak{p}$-forms of $E_\mathfrak{p}\otimes_{k_\mathfrak{p}}\bar k_\mathfrak{p}$ lying in the $k_\mathfrak{p}$-isogeny class $a_{E_\mathfrak{p}}=0$ have the same ring of endomorphisms over $k_\mathfrak{p}$, out of the two possibilities listed above. The opposite being true when $j_{E_\mathfrak{p}}=1728$ (this fact is very related to the analysis of the mod $p$ reduction of Hilbert Class Polynomials associated to discriminants $-p$ and $-4p$ done by Gross and Elkies (cf. $\S 2$, Proposition, in Elkies' "The existence of infinitely many supersingular primes for every elliptic curve over $\mathbf{Q}$", Inventiones 89 (1987))). -In other words, in the special case the pair $(f_{E_{\mathfrak{p}}}(x), j_E$ mod $\mathfrak{p})$ does ${\it not}$ determine $T_2(E_\mathfrak{p})$. -Here is the question then: in the special case can we determine what is the endomorphism ring of $E_\mathfrak{p}$ (and hence $T_2(E)$) from congruences of $j_E$ mod a higher power of $\mathfrak{p}$ (or of $p$)? -It is not even clear to me whether this should be possible, let alone what power of $p$ we would need to tell one case from the other. The hope behind this is that the $j$-invariant of $E$ be "close" to that of the CM lift of $E_\mathfrak{p}$ and of its endomorphisms ring over $k_\mathfrak{p}$. Thanks. -PS: I do not know if the question above has anything to do with Is there a "classical" proof of this $j$-value congruence? -[EDIT: I realize that for clarity of exposition I should have probably recalled that $T_\ell(E_\mathfrak{p})$ for $\ell\neq p$, in the above notation, is a free ${\rm End}(E_\mathfrak{p})\otimes\mathbf{Z}_\ell$-module of rank one. Therefore, roughly, the knowledge of either of the two is equivalent to that of the other] - -REPLY [2 votes]: For simplicity assume $p \neq 3$. Given an elliptic curve $E$ in Weierstrass form with one two-torsion point defined over its field of definition, the other 2-torsion points are defined over the field of definition if and only if the discriminant $g_2^3 - 27 g_3^2$ is a perfect square. (Clear from properties of the discriminant of a cubic). -So we need to determine whether the discriminant is a perfect square mod $\mathfrak p$. -We have $$j = 1728 \frac{g_2^3}{ g_2^3 - 27 g_3^2} = 1728 + \frac{ (216g_3)^2}{ g_2^3 - 27 g_3^2} $$ -so $j-1728$ is a perfect square divided by the discriminant. -To determine whether the discriminant is a perfect square mod $\mathfrak p$, it is sufficient to evaluate the $\mathfrak p$-adic leading term of $j$, then divide by any perfect square with equal $\mathfrak p$-adic valuation, obtaining a residue mod $\mathfrak p$, which is either a quadrati residue or nonresidue. In the first case the order is maximal and the second case it is nonmaximal. -This works unless $j$ is exactly 1728, in which case it is clear that the $j$-invariant is not sufficient, as then there are multiple curves with distinct $2$-torsion but the same $j$-invariant.<|endoftext|> -TITLE: Applications of the "almost commuting" theorem of H. Lin -QUESTION [11 upvotes]: H. Lin proved that "almost commuting" hermitian matrices are "nearly commuting." To be more precise, Lin showed that given $\epsilon > 0$ there exists a $\delta > 0$ such that if $A, B \in M_N$ are self-adjoint, with $|| AB - BA || < \delta$, and $\|A\|, \|B\|\le 1$, then there exists $X, Y \in M_N$ with $XY = YX$ such that $ || A - X || + || B - Y || < \epsilon$. Here, $|| . ||$ is the usual operator norm and $\delta = \delta(\epsilon)$ does not depend on the dimension N. -Does anyone know of any applications of this theorem? - -REPLY [15 votes]: Lin's theorem shows the existence of a localized basis for the low-energy space in models of non-interacting fermions on a finite lattice on a disk. This was observed by Matt Hastings. - -See "Topology and phases in fermionic systems" in the Journal of Statistical Mechanics: Theory and Experiment, 2008, L01001, especially the next to last paragraph. - -Other norms come up in applications. It can be hard to figure what norm is relevant in engineering and science papers, but it is often clearly not the operator norm. Generically speaking, approximating two commuting hermitian matrices by commuting hermitian matrices is equivalent to joint approximate diagonalization of those matrices. - -A famous algorithm in blind source separation if by Cardoso and Souloumiac, "Jacobi angles for simultaneous diagonalization" in SIAM Journal on Matrix Analysis and Applications, vol 17, no. 1, 161--164, 1996. - -The Cardoso and Souloumiac algorithm was used later in computational quantum chemistry, as in Francois Gygi, Jean-Luc Fattebert and Eric Schwegler, "Computation of Maximally Localized Wannier Functions using a simultaneous diagonalization algorithm," -Computer Physics Communications, Volume 155, Issue 1, 1 September-15 September 2003, 1--6. - -Jon von Neumann considered almost commuting operators. See "Proof of the ergodic theorem and the H-theorem in quantum mechanics," The European Physical Journal, 1--37, 2010 and the commentary that accompanied the translation, "Long-time behavior of macroscopic quantum systems, by Goldstein, S. and Lebowitz, J.L. and Tumulka, R. and Zanghi, N., The European Physical Journal H, 1--28, 2010.<|endoftext|> -TITLE: Direct proof that the centralizer of $GL(V)$ acting on $V^{\otimes n}$ is spanned by $S_n$ -QUESTION [23 upvotes]: Let $V$ be a finite dimensional vector space over a field of characteristic zero. Let $A$ be the space of maps in $\mathrm{End}(V^{\otimes n})$ which commute with the natural $GL(V)$ action. Clearly, any permutation of the tensor factors is in $A$. I am looking for an elementary proof that these permutations span $A$. -If $\dim V \geq n$, there is a very simple proof. Take $e_1$, $e_2$, ..., $e_n$ in $V$ linearly independent and let $\alpha \in A$. Then $\alpha(e_1 \otimes e_2 \otimes \cdots \otimes e_n)$ must be a $t_1 t_2 \cdots t_n$ eigenvector for the action of the matrix $\mathrm{diag}(t_1, t_2, \ldots )$ in $GL(V)$. So $\alpha(e_1 \otimes \cdots \otimes e_n) = \sum_{\sigma \in S_n} c_{\sigma} e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)}$ for some constants $c_{\sigma}$. It is then straightforward to show that $\alpha$ is given by the corresponding linear combination of permutations. -I feel like there should be an elementary, if not very well motivated, extension of the above argument for the case where $\dim V < n$, but I'm not finding it. -Motivation: I'm planning a course on the combinatorial side of $GL_N$ representation theory -- symmetric polynomials, jdt, RSK and, if I can pull it off, some more modern things like honeycombs and crystals. Since it will be advertised as a combinatorics course, I want to prove a few key results that give the dictionary between combinatorics and representation theory, and then do all the rest on the combinatorial side. Based on the lectures I have outlined so far, I think this will be one of the few key results. -The standard proof is to show that the centralizer of $k[S_n]$ is spanned by $GL(V)$, and then apply the double centralizer theorem. Although the double centralizer theorem (at least, over $\mathbb{C}$) doesn't formally involve anything I won't be covering, I think it is pretty hard to present it to people who aren't extremely happy with the representation theory of semi-simple algebras. So I am looking for an alternate route. - -REPLY [5 votes]: I'm going to write up Mark Wildon's proof as I understand it. As in the standard proof, we start by showing the Key Lemma that the centralizer of $k[S_n]$ is linearly spanned by $GL(V)$. Decompose $V^{\otimes n}$ into $S_n$-irreps, and let $\alpha$ be an endomorphism of $V^{\otimes n}$ commuting with $GL(V)$. For each irrep $U$ of $S_n$, let $U_1$, ..., $U_a$ be the occurrences of $U$ in $V^{\otimes n}$. -For any $U_i$, consider the endomorphism of $V^{\otimes n}$ which acts by $1$ on $U_i$ and on $0$ on all of the other summands of $V^{\otimes n}$. This commutes with $k[S_n]$ so, by the Key Lemma it is a linear combination of maps in $GL(V)$. Hence $\alpha$ commutes with it, which means that $\alpha$ takes $U_i$ to $U_i$ by some map $\alpha_i$. -Consider the endomorphism of $V^{\otimes n}$ which takes $U_i$ to $U_j$ by an $S_n$-equivariant endomorphism and acts by $0$ on every other summand of $V^{\otimes n}$. This commutes with $k[S_n]$ so, by the Key Lemma it is a linear combination of maps in $GL(V)$. Hence $\alpha$ commutes with it, which means that $\alpha_i = \alpha_j$. (Abusing equals to mean "is taken to the other along the isomorphism $U_i \to U_j$, which is unique up to scalar".) Write $\alpha(U)$ for the common value of $\alpha_1$, $\alpha_2$, ..., $\alpha_a$. -There are now two ways to finish the proof. -Standard Argument: By Maschke and Artin-Wedderburn, there is an element in $k[S_n]$ which acts on each irrep $U$ by $\alpha(U)$. This element of $k[S_n]$ induces $\alpha$. -Mark Wildon's Argument: Let $V \subset W$. We will show that we can extended $\alpha$ to an endomorphism $\beta$ of $W^{\otimes n}$ which commutes with $GL(W)$. Decompose $W^{\otimes n}$ into $S_n$ irreps, so that the previous decomposition of $V^{\otimes n}$ occurs as a subset of the summands. Let the occurrences of $U$ be $U_1 \oplus U_2 \cdots \oplus U_a \oplus \cdots \oplus U_b$. Define a linear map $\beta:W^{\otimes n}\to W^{\otimes n}$ to act on all of the $U_i$ by $\alpha(U)$ or, if $a=0$ so that $\alpha(U)$ is undefined, define $\beta$ to act on the $U_i$ by $0$. -We claim that $\beta$ commutes with $GL(W)$. Proof: Any element of $GL(W)$ commutes with $k[S_n]$. So (by Schur's lemma), it can only map $U_i$ to a linear combination of other $U_j$'s, and the component of $\alpha$ mapping $U_i$ to $U_j$ is a scalar multiple of the standard isomorphism. Clearly, $\beta$ commutes with any map of this form. -Now, by my argument in the original post, take $\dim W \geq n$ to see that $\beta$ is induced by an element of $S_n$. Then $\alpha$ is also induced by this element of $S_n$.<|endoftext|> -TITLE: An iterated tensor product integral -QUESTION [7 upvotes]: In "Differential equations driven by rough paths" (Terry Lyons, et al) section 1.4.2 it's claimed that the symmetric part of the tensor: -$\int_{0 \le u_1 \le \cdots \le u_j \le t} \mathrm{d}X_{u_1} \otimes \cdots \otimes \mathrm{d}X_{u_j}$ -is exactly $\frac{1}{j!}(X_t - X_0)^{\otimes j}$. It is assumed that $X:[0,T] \to V$ is a Lipschitz path in some finite dimensional vector space $V$. -Is there a simple way to establish this fact? Also I would be grateful for any reference on this type of integrals involving tensors since I've never encountered them before. -Further background -Integration with respect to $X$ is defined for any continuous $f: [0,T] \to L(V,W)$ ($W$ a finite dimensional vector space and $L(V,W)$ the space of linear maps from $V$ to $W$) as: -$\int f_s \mathrm{d}X_s = \lim \sum_{i = 0}^{r -1} f_{t_i}(X_{t_{i+1}} - X_{t_i})$ -where the limit is taken over partitions $t_0 = 0 \le t_1 \le \cdots \le t_r = T$ whose largest interval is decreasing to $0$ in length. The result belongs to the space $W$. -Hence, to make the above iterated integral fit into the definition (and give a result in $V^{\otimes j}$) it seems to me that one must consider each element $w_1 \otimes \cdots \otimes w_i \in V^{\otimes i}$ as an element of $L(V, V^{\otimes i+1})$ via the map $v \mapsto w_1 \otimes \cdots \otimes w_i \otimes v$. However after doing this I still have trouble establishing the identity. -As an example, with the above interpretation I get: -$\int_{0 \le s \le t \le T} \mathrm{d}X_s \otimes \mathrm{d}X_t = \frac{1}{2}T^2 v_1\otimes v_1 + \frac{2}{3}T^3 v_1\otimes v_2 + \frac{1}{3}T^3 v_2 \otimes v_1 + \frac{1}{2}T^4 v_2 \otimes v_2$ -for $X_t = t v_1 + t^2 v_2$ in a $2$-dimensional vector space with basis $(v_1,v_2)$. This example shows that the result of the integral itself may be non-symmetric. - -REPLY [7 votes]: Let's just view the function $\dot{X}(t) = \frac{d}{dt} X_t$ as a bounded function taking values in the vector space $V$. The notation $dX_{u_1}$ means $\dot{X}(u_1) du_1$ Then $dX_{u_1} \otimes \cdots \otimes dX_{u_k} = \dot{X}(u_1) \otimes \cdots \otimes \dot{X}(u_k) du_1 \cdots du_k$, should be regarded as a bounded function (or measure) on $[0,t]^k$ taking values in the vector space $V^{\otimes k}$. -Because the operation of projecting to a symmetric part is a linear operation from $V^{\otimes k}$ to itself, you can take it inside of the integral. Let's call this symmetric part $dX_{u_1} \cdots dX_{u_k}$. I'll use the symbol $u \cdot v$ to denote the symmetric product of of $u, v \in V$ -- that is, $u \cdot u \cdot u = u \otimes u \otimes u$, and the general product is defined by the polarization identity. For example $u \cdot v = ( u \otimes v + v \otimes u) / 2$. The resulting multiplication is commutative. -Therefore your integral is equal to -$\int_{0 \leq u_1 \leq \ldots \leq u_k \leq t} dX_{u_1} \cdots dX_{u_k}$ -Because the symmetric product is commutative, for any permutation $\sigma$ of $\{ 1 \ldots k\}$ you get the same value by integrating any of the permuted regions -$ \int_{0 \leq u_1 \leq \ldots \leq u_k \leq t} dX_{u_1} \cdots dX_{u_k} = \int_{0 \leq u_{\sigma(1)} \leq \ldots \leq u_{\sigma(k)} \leq t} dX_{u_1} \cdots dX_{u_k}$ -Observe also that the region $0 \leq u_1 \leq \ldots \leq u_k \leq t$ is a fundamental domain for the action of $S_k$ on the cube $[0,t]$ -- that is, you can get the whole cube by permuting the variables $u_i$, and none of these regions overlap. Therefore, since there are $k!$ such permutations, we sum over permutations to conclude -$ k! \int_{0 \leq u_1 \leq \ldots \leq u_k \leq t} dX_{u_1} \cdots dX_{u_k} = \int_{[0,t]^k} dX_{u_1} \cdots dX_{u_k}$ -But, by Fubini and the multilinearity of the symmetric product, the integral on the right is just $(X_t - X_0)^k$. -I should remark that this same proof in an even simpler setting also produces the $\frac{1}{k!}$ that appears when you prove the Taylor expansion through iterative use of the fundamental theorem of calculus.<|endoftext|> -TITLE: For which fields is the inverse Galois problem known? -QUESTION [8 upvotes]: The inverse Galois problem is known for (or in Jarden's and Fried's terminology, the following fields are universally admissible) function fields over henselian fields (like $\mathbb{Q}_p(x)$); function fields over large fields (like $\mathbb{C}(x)$); and large Hilbertian fields (conjecturally $\mathbb{Q}^{ab}$, although I'm not certain that any field is known to be in this category). -Clarification: -A large field $K$ (a.k.a. an ample field) is a field such that if $V$ is a variety of dimension $\geq 1$ over $K$ with at least one smooth $K$-rational point, then it has infinitely many smooth $K$-rational points. For example any algebraically closed field is large. -A Hilbertian field is more difficult to explain, but it suffices to say that any number field and any function field (over any field) is Hilbertian. -My question is: -Is there a proof (not a conjecture) that there exists a field $K$ which is neither a function field over a henselian field, nor a function field over a large field, nor a large Hilbertian field, such that the inverse Galois problem is true over that field? (i.e. that every finite group is realizable as a Galois group over that field) - -REPLY [8 votes]: You should find what you want in the following article by Jochen Koenigsmann: The regular inverse Galois problem over non-large fields. J. Europ. Math. Soc., 6(4):425–434, 2004.<|endoftext|> -TITLE: When is the category of pro-objects a homotopy category? -QUESTION [15 upvotes]: For a category $C$, there is a category Pro-$C$ whose objects are cofiltered diagrams $I \to C$ and whose morphisms are given by -$$ -{\rm Hom}(\{x_s\},\{y_t\}) = \varprojlim_t\ \varinjlim_s\ {\rm Hom}(x_s,y_t). -$$ -Generally, this category is fairly hard to work with. This is especially true because several types of maps of pro-objects are defined in terms of the existence of a representing map of diagrams with certain properties, and it can be very difficult to rectify several distinct properties at once. -One way to describe the category of pro-objects is by inverting morphisms. Specifically, we can form a more restricted category of pairs $(I,F)$ of a cofiltered index category $I$ and a diagram $F: I \to C$, with morphisms defined as pairs of a functor and a natural transformation of diagrams. Certain maps of cofiltered diagrams become isomorphisms of pro-objects (the most important ones being reindexing along a final subcategory). Inverting them gives us the pro-category. -In some sense, this automatically provides us with a "category with weak equivalences", but it's intrinsically very large and it's not necessarily clear if the "homotopy theory" is tractable. -Are there any circumstances under which the category of diagrams in $C$ automatically has the structure of a model category, with weak equivalences being pro-isomorphisms? In these cases, does the category Pro-$C$ have an interesting homotopy theory or are the mapping spaces essentially discrete? -Obviously being complete and cocomplete is going to be an obstacle to this kind of structure. Failing that, is there any further possibility of gaining control on the homotopy theory? -Having said all this, I've been a little bit vague about what I mean by a "map of diagrams" because I'd be open to the idea of having slightly restricted classes of maps in the definition. - -REPLY [9 votes]: There are some answers to this way back. There is a lovely answer to several of your questions in -D. A. Edwards and H. M. Hastings, 1976, ˇCech and Steenrod homotopy -theories with applications to geometric topology , volume 542 of Lecture -Notes in Maths , Springer-Verlag. -The category of prosimplicial sets has a model category structure that corresponds to a geometrically defined notion of strong shape theory (i.e. a homotopy coherent version of Borsuk's shape theory). Edwards and Hastings extended a result of Chapman and showed this model category theory also to be a form of proper homotopy theory. (There is also some discussion of this in my article: -T. Porter, 1995, Proper homotopy theory , in Handbook of Algebraic -Topology , 127–167, North-Holland, Amsterdam. ) -The story does not end there. Because of the connection with étale homotopy theory (Artin and Mazur), there was a revival of interest in pro-categories in the last few years and there is a good discussion in -H. Fausk and D. Isaksen, Model structures on pro-categories , Homology, -Homotopy and Applications, 9, (2007), 367 – 398. -I suggest that you also look at others of Dan Isaksen's papers on this area as they answer more of the quetions that you have asked. -On another point that you mention, the rectification process for properties is reasonably well understood due to what is known as the reindexing lemma (the simplest case is in Artin and Mazur's lecture notes but there are much fuller versions some of which are discussed in another of Isaksen's papers -D. C. Isaksen, Completions of pro-spaces , Math. Z., 250, (2005), 113 – -143. ) -If you read these papers carefully you will come to the conclusion that certain problems are still not fully understood especially when pro-finite simplicial sets are concerned, and the applications of those beasties are again very important so that is a good area to explore!!! -(See also work by Quick (Profinite homotopy theory , Documenta Mathematica, 13, -(2008), 585–612.) and Pridham (Pro-algebraic homotopy types , Proc. Lond. Math. Soc. -(3), 97, (2008), 273 – 338. ) They show some of the more recent stuff on this with some good applications. There are copies on the ArXiv.)<|endoftext|> -TITLE: Is there a way of canonically labelling permutation groups? -QUESTION [21 upvotes]: When working with large numbers of graphs, a canonical labelling routine is essential as, after the initial cost of canonically labelling each graph, it permits isomorphism checks to be replaced with identity checks. -Currently I am working with large numbers of transitive permutation groups; the degree is small (40ish) but the size of group can be large, and my programs are gumming up due to having to perform large numbers of - not exactly isomorphism - but conjugacy checks. (I need to check each pair of groups to see if they are conjugate in the symmetric group.) -This process would be considerably easier if there was a known way of finding a canonical conjugate of the incoming permutation groups... - -REPLY [2 votes]: In GAP, for transitive permutation groups of degree at most 30 one can use TransitiveIdentification(G) from the package TransGrp by Alexander Hulpke. This gives the index of the (unique!) permutation group in the library that is conjugate to the given permutation group and it is apparently quite fast. See section 13 of -Hulpke, Alexander, Constructing transitive permutation groups., J. Symb. Comput. 39, No. 1, 1-30 (2005). ZBL1131.20003 MR2168238.<|endoftext|> -TITLE: 3rd homotopy group of a compact Simple Lie Group -QUESTION [12 upvotes]: Suppose $G$ is a compact simple Lie group with Lie algebra $\mathfrak g$. Then we know that $\pi_3(G)=Z$. Now suppose that $H_\alpha$ is a co-root vector in correspondence with a root $\alpha$. So it means that there are $X_\alpha$ and $Y_\alpha$ such that $span${$H_\alpha, X_\alpha, Y_\alpha$} is a sub-Lie algebra of $\mathfrak g$ isomorphic to $\mathfrak{su}(2)$. It induces a map of Lie groups $\phi:SU(2) \to G$. I'm wondering what's the image of this map as an element of $\pi_3(G)$ in terms of $G$. - -REPLY [15 votes]: This number is called the index of the map $\phi: SU(2)\to G$. It can be defined for any homomorphism $\phi:H\to G$ where $H$ is simple. Algebraically it can be computed as follows. Since $\mathfrak h$ is simple the restriction of the Killing form of $\mathfrak g$ to $\mathfrak h$ is a constant multiple of the Killing form of $\mathfrak h$. That constant is the index of $\phi$. In the specific case you are asking about for a simple root $\alpha$ the index can also be written as $\frac{(\alpha_{max},\alpha_{max})}{(\alpha,\alpha)}$ where $\alpha_{max}$ is the longest simple root of $\mathfrak g$. Note that from the classification of compact simple Lie groups this can only be equal to 1,2 or 3. -See Onishchik, "Topology of transitive transformation groups", §3.10 and §17.2 for details on this.<|endoftext|> -TITLE: Erdős-Szekeres for first differences -QUESTION [22 upvotes]: The classical Erdős-Szekeres theorem says that any sequence of $n^2+1$ -real numbers contains a monotonic $(n+1)$-term subsequence. Suppose, however, -that we want to find a subsequence which is not necessarily -monotonic itself, but has the sequence of its first differences monotonic. -How long has the original sequence to be to ensure that such an $(n+1)$-term -subsequence can be found? - -For positive integer $n$, what is the smallest integer $N=N(n)$ such that every - $N$-element sequence of real numbers contains an $n$-element subsequence - $(a_1,\ldots,a_n)$ with either $a_2-a_1\le\dotsb\le a_n-a_{n-1}$, or - $a_2-a_1\ge\dotsb\ge a_n-a_{n-1}$? - -Trivially, we have $N(1)=1$, $N(2)=2$, and $N(3)=3$. However, I do not know -the value of $N(4)$. - -The state of the art as of 05.03.12. The nice argument of Boris Bukh (below) shows that $N(n)$ is exponential in $n$. Still, this does not completely settle the problem. -Updated 16.03.12: the absolutely beautiful, must-upvote solution by Sergey Norin is posted below. - -REPLY [21 votes]: For brevity let me call a sequence with non-decreasing first differences convex, and a sequence with non-increasing first differences concave. -Let $M(r,s)$ denote the minimum integer $N$ so that every $N$-element sequence of real numbers contains a convex subsequence of length $r+1$ or a concave subsequence of length $s+1$. Below I attempt to show that - -$$M(r,s)=\binom{r+s-2}{r-1}+1.$$ - -The lower bound: Let $[m]$ denote the set $\{1,\ldots,m\}$. -For $S \subseteq [r+s-2]$, let $x_S := \sum_{i \in S}3^i$. Consider the sequence $(x_S\;|\; S \subseteq [r+s-2], |S|=s-1)$ with elements in increasing order. It has $\binom{r+s-2}{r-1}$ elements. We will show that this sequence contains no convex subsequence of length $r+1$ and no concave subsequence of length $s+1$. -Let $x_{S_1},x_{S_2}, \ldots,x_{S_n}$ be a convex subsequence. Let $d_i$ be the maximum element of $S_{i+1} \setminus S_{i}$. Then $ 3^{d_i}/2 < x_{S_{i+1}}-x_{S_i}< 3^{d_i+1}/2.$ It follows that $(d_1,d_2,\ldots,d_{n-1})$ is a strictly increasing sequence, and that $d_i \not \in S_1$ for $i \in [n-1]$. Therefore $n \leq r$ as desired. -The proof showing that there is no concave subsequence of length $s+1$ is symmetric. (One can replace $x_{S_i}$ by $x_{[r+s-2]}-x_{S_i}$.) -The upper bound: The goal is to imitate the elegant pigeonhole argument of Seidenberg for Erdős-Szekeres theorem. (See e.g. Wikipedia.) -Let $N = \binom{r+s-2}{r-1}+1$. Let ${\bf a}=(a_1,\ldots, a_N)$ be a sequence. -For $i \in [N]$ and $k \in [r-1]$ let $\alpha_i(k)$ be the minimum real number so that a $(k+1)$-term convex subsequence of ${\bf a}$ ending with $a_i$ has $\alpha_i(k)$ as the difference of the last two terms. Set $\alpha_i(k)=+\infty$ if no such subsequence exists. -Note that for fixed $i$ the sequence $\alpha_i(\cdot)$ is non-decreasing. -Analogously, for $k \in [s-1]$ we define $\beta_i(k)$ to be the maximum real number so that a $(k+1)$-term concave subsequence of ${\bf a}$ ending with $a_i$ has $\beta_i(k)$ as the difference of the last two terms. Set $\beta_i(k)=-\infty$ if no such subsequence exists. -The sequence $\beta_i(\cdot)$ is non-increasing. -For given $i$ arrange the elements of the multiset $\{\alpha_i(1),\ldots, \alpha_i(r-1),\beta_i(1),\ldots,\beta_i(s-1)\}$ in increasing order, with alphas preceding betas, when the values are the same. Now we consider the resulting sequence as a sequence of $r+s-2$ symbols each of which is either $\alpha$ or $\beta$, ignoring the indexing. Call this sequence $\bf \chi_i$. For example, we always have -$${\bf \chi_1}=(\beta, \ldots,\beta, \alpha, \ldots, \alpha),$$ -$${\bf \chi_2}=(\beta,\ldots,\beta, \alpha, \beta, \alpha, \ldots, \alpha),$$ -and ${\bf \chi_3}$ depends on $a_1$, $a_2$ and $a_3$. -There are $N-1$ possible sequences and by -pigeonhole principle we have ${\bf \chi_i}={\bf \chi_j}$ for some $1 \leq i < j \leq N$. Let $z=a_j-a_i$. Let $r'$ be chosen to be maximum so that $\alpha_{i}(r') \leq z$, and let $r'=0$ if no such $r'$ exists. Let $s'$ be chosen to be maximum so that $\beta_{i}(s') \geq z$, and let $s'=0$ if no such $s'$ exists. Note that $\alpha_{j}(r'+1) \leq z$ and $\beta_{j}(s'+1) \geq z$. If $r'=r-1$ or $s'=s-1$ then $\bf{a}$ contains a convex subsequence of length $r+1$ or a concave subsequence of length $s+1$ as desired. Otherwise, $r'+1$ alphas precede $s'+1$ betas in $\chi_j$, -while in $\chi_i$ after the first $r'+1$ alphas we encounter only $\leq s'$ betas. -This contradiction finishes the proof.<|endoftext|> -TITLE: Orthogonality in non-inner product spaces -QUESTION [10 upvotes]: I have come across a notion of orthogonality of two vectors in a normed space not necessarily inner product space. Two vectors $x$ and $y$ in a normed space are said to be orthogonal (represented $x\perp y$) if $||x||\leq ||x+\alpha y||,$ for every $\alpha,$ a scalar. -1) What is the rational behind the definition above? I guess, it has got something to do with minimum overlap between $x$ and $y$. -2) Is this unique generalization of the concept of orthogonality from inner product spaces? -Thank you. - -REPLY [4 votes]: Concerning your follow-up question (iii) there is the following very nice result: For Birkhoff-James orthogonality -it is easy to find examples where $y\perp x$ but -$\left\|x\right\|/\left\|x+\alpha y\right\| > 1$ for some real $\alpha$, and so -natural to investigate the largest such value -$\left\|x\right\|/\left\|x+\alpha y\right\|$ over $X$. -In "R. L. Thele, Some results on the radial projection in Banach spaces. -Proc. Amer. Math. Soc., 42(2):484--486", it is it is shown that this quantity -is exactly the Lipshitz constant for the radial projection onto the unit ball -in this norm.<|endoftext|> -TITLE: Groups quasi-isometric to reducible nonuniform lattices -QUESTION [7 upvotes]: It is known that a finitely group $G$ is quasi-isometric to a nonuniform irreducible lattice $\Lambda$ in a semisimple Lie group if and only if $G$ and -$\Lambda$ are commensurable (see references in this survey of Farb). - Question. What is known about groups quasi-isometric to reducible nonuniform lattices in semisimple Lie groups? -As usual in this business "semisimple" means "noncompact, connected, semisimple, with finite center". - -REPLY [6 votes]: Here is a partial answer: Suppose $\Gamma = \Gamma_1 \times \dots \times \Gamma_n$ and all the $\Gamma_i$ are irreducible lattices in $G_i$, where each $G_i$ has real rank at least two. -It has been a long time, and I do not remember all the details, but I think it may be true that any quasi-isometry from a product of such lattices $\Gamma_1 \times \dots \times \Gamma_n$ to itself preserves the factors (up to permutation). -I am looking at Lemma 10.3 of my paper in JAMS from 1998 http://www.math.uchicago.edu/~eskin/sl3z.ps. It is stated for irreducible lattices, but that does seem to be used in the proof. Of course I could be missing something. -If self quasi-isometries are indeed factor preserving, then one has the same classification as for irreducible lattices. -One more comment: the reason my proof fails when you have a factor $\Gamma_i$ in a real rank one group $G_i$ is that I quote Lubotsky-Mozes-Raghunathan which does not work in that case.<|endoftext|> -TITLE: Infinite mathematics as non-standard finite mathematics? -QUESTION [24 upvotes]: I have in mind something like the following: - -Start with some suitable version of "finite" mathematics. Some possibilities might be maybe ZFC with a suitable anti-infinity axiom, the topos $\mathbf{FinSet}$, Peano arithmetic, Turing machines... something whose objects are suitably "finite". -Then, posit the existence of both a standard and a non-standard model. -Now, in this setting, where we have access both to a standard model and a non-standard extension, use the non-standard objects as proxies for infinite objects (e.g. maybe some sort of set theory that has a set of natural numbers), and develop ordinary mathematics this way. - -Has anybody worked on such a thing? Does anyone know of references of it being done? Or suggestions that it can't work out? -(P.S. I wasn't sure how to tag this....) - -Edit: After more thought and reviewing the answers thus far, I think I can state an example of the sort of thing i was imagining. Define a first-order theory with two types $T_1$ and $T_2$, two binary relation symbols $\in_1, \in_2$ (one for each sort), and a map $\tau : T_1 \to T_2$ satisfying: - -$(T_1, \in_1)$ satisfies the axioms of finite set theory -$(T_2, \in_2)$ satisfies the axioms of finite set theory -$\tau$ is injective -$\tau$ is not surjective -$\tau$ satisfies an axiom schema that says it's an elementary embedding - -and the question is to what extent we can develop infinite set theory in this theory. - -REPLY [6 votes]: You might want to look into the work of Vopenka and his collaborators on what they call alternative set theory. The formal theory looks a bit strange, since it allows proper classes to be subclasses of sets, but one interpretation of the theory is that the sets are the internal sets of a nonstandard model of finite-set-theory while the classes are arbitrary (not necessarily internal) subsets. There's a small book, "Mathematics in the Alternative Set Theory," by Vopenka, that should serve as a good introduction to the subject.<|endoftext|> -TITLE: Rationality of algebraic groups -QUESTION [5 upvotes]: The Cayley parametrization of $O(n),$ as in my answer to this question makes one wonder: which algebraic groups are actually rational? I am sure this is very well understood, just not by me... - -REPLY [6 votes]: A reductive group $G$ over a ground field $k$ need not be a rational variety over $k$ (although the group $G_{/\ell}$ obtained by scalar extension is a rational $\ell$ variety, where $\ell$ is an algebraic closure of $k$). Maybe a good place to look concerning these matters is Merkurjev's 1996 Publ. Math. IHES paper "R-equivalence and rationality problem for semisimple adjoint classical algebraic groups". -In the intro to that paper, Merkurjev makes some interesting observations: - -Chevalley showed in the 50s that there are algebraic tori over local fields which are not rational varieties. -algebraic tori over dimension <= 2 are always rational varieties; thus, any reductive -group of rank <=2 is rational. (Chevalley proved that the variety of maximal tori in a connected group is a rational variety, so the rationality of $G$ boils down to the rationality of some maximal torus of $G$). -any semisimple group over a number field which is a counterexample to weak approximation is an example of a non-rational group. - -In Merkurjev's paper, you will find explicit examples of groups $G$ which are not stably rational -- a $k$-variety $X$ is stably rational if $X \times \operatorname{Aff}^d$ is a rational $k$-variety for some $d \ge 0$. -One way to show that $G$ is not stably rational (and hence not rational) is to show that -the "group of $R$-equivalence classes" for $G$ is non-trivial; -this group of $R$-equivalence classes was introduced by Manin. Merkurjev's paper is devoted to the computation of the group of $R$-equivalence classes for semisimple, adjoint, classical groups.<|endoftext|> -TITLE: Stone-Čech compactification of $\mathbb R$ -QUESTION [9 upvotes]: Let $\beta X$ - is a Stone-Čech compactification of $X$. $I=(-1,1)$ - is an interval of the real line. Is it true that $\beta \mathbb R\setminus I = \beta(\mathbb R\setminus I)$? In other words, it means that a finite interval does not affect on the "compactification of infinity". -Update: -Great thanks for realized calculations. - -REPLY [2 votes]: More generally: if $X$ is normal and $A$ is closed in $X$ then, by the Tietze-Urysohn theorem, the closure in $\beta X$ of $A$ is $\beta A$. In the example above $X=\mathbb{R}$ and $A=\mathbb{R} \setminus (-1,1)$. -As the closure of $(-1,1)$ in $\beta\mathbb{R}$ is just $[-1,1]$ the desired equality follows.<|endoftext|> -TITLE: laplace equation on manifolds with boundary -QUESTION [6 upvotes]: in aubin's book on page 104 theorem 4.7 there is the theorem: Let $(M,g)$ be a compact $C^{\infty}$ Riemannian manifold. There exists a weak solution $\varphi \in H_{1}$ of $\Delta \varphi = f $ if and only if $\int f dvol = 0$. The solution is unique up to a constant. If $f \in C^{r + \alpha}$ ($r \geq 0$ a integer or $r=+\infty$, $0 < \alpha < 1$), then $\varphi \in C^{r+2+\alpha}$. -in this theorem the manifolds do not have boundary. -My question: are there similar results with riemannian manifolds with boundary ? -hope for answers. -william - -REPLY [8 votes]: Similar results hold for manifolds with boundary, but you need to include boundary conditions. The most common boundary conditions in the case of Laplacian are the Dirichlet and the Neumann conditions. -The only tricky part in the case with boundary is regularity along the boundary. (In the regularirty results you need to include assumptions on the regularity of the boundary data. Two good references are - -Gilbarg & Trudinger: Elliptic partial differential equations of second order - -or - -C. Morrey: Multiple integrals in the calculus of variations.<|endoftext|> -TITLE: About the definition of quasi-equivalent dg-categories -QUESTION [5 upvotes]: In the article Grothendieck Ring of pretriangulated categories by Bondal-Larsen-Lunts, two dg-categories $\mathcal A$ and $\mathcal B$ are called quasi-equivalent if there is a chain of dg-categories and quasi-equivalences -$\mathcal A \leftarrow \mathcal C_1 \rightarrow \cdots \leftarrow \mathcal C_n \rightarrow \mathcal B$. -Another possibility, I think, would be to call $\mathcal A$ and $\mathcal B$ quasi-equivalent if they are isomorphic in the homotopy category $\mathrm{Ho}(\mathbf{dgcat})$. Clearly if $\mathcal A$ and $\mathcal B$ are quasi-equivalent as in the definition of Bondal-Larsen-Lunts, they are also isomorphic in $\mathrm{Ho}(\mathbf{dgcat})$. Is the converse true? What is the "better" definition of quasi-equivalent dg-categories? -Thanks in advance! - -REPLY [5 votes]: The converse is true. -You can see this for example as follows: first since DG-categories admit a model structure each morpism $A \to B$ in Ho(dgCat) can be realized as a 3-step span like this -$A \leftarrow A' \to B' \leftarrow B$ (here $A'$ is a cofibrant replacement and $B'$ is a fibrant replacement). Then such a morpism is an isomorphism in the homotopy category be the 2-out-of-3 property if and only if the middle morphism $A'\to B'$ is a weak equivalence. Thus each isomorphism in the homotopy category comes from a Zig-Zag of weak equivalences (and this argument is valid in each model category).<|endoftext|> -TITLE: Commutator of closed subgroups -QUESTION [10 upvotes]: Suppose we have a simply-connected Lie group $G$. Let $G_1$ and $G_2$ be two closed and connected subgroups of $G$. Is it true that the commutator $[G_1,G_2]$ is a closed subgroup of $G$? - -REPLY [7 votes]: No. Let's make an example in which both $G_1$ and $G_2$ are one-dimensional. -Start by choosing $x$ and $y$ in $\frak{sl}_2(\mathbb R)$ such that the subgroup determined by $[x,y]$ is a circle (the square of this matrix has negative trace) but the subgroups generated by $x$ and by $y$ are isomorphic to $\mathbb R$ (their squares have positive trace). -Now in $\frak{sl}_2(\mathbb R)\times \frak{sl}_2(\mathbb R)$ consider the elements $(x,x)$ and $(y,cy)$, where $c$ is some irrational number. These give closed noncompact one-dimensional subgroups $G_1$ and $G_2$, but their commutator gives a dense line in a torus. -$SL_2(\mathbb R)\times SL_2(\mathbb R)$ is not simply connected, so embed it in $SL_4(\mathbb R)$ and take that, or rather its double cover, to be $G$.<|endoftext|> -TITLE: Maps between sets and coalgebras -QUESTION [6 upvotes]: Hi, -To every set $X$ there corresponds a group-like coalgebra $kX$, with basis $X$. "Grouplike" means that there is a basis $X$ with $\epsilon(x)=1$ and $\Delta(x)=x\otimes x$ for all $x\in X$. -First, somewhat vague, question: is there a standard way of proving that a coalgebra $C$ is group-like? I have a linearly independent subset, but can't prove yet that it's a basis. -Specifically, I'm interested in the following. For $C,D$ coalgebras, there is a "measuring coalgebra" $(C,D)$, see Sweedler, Hopf algebras, chapter VII. If $X,Y$ are sets, and $kX,kY$ are their coalgebras, then "obviously" one should have $(kX,kY)=k Map(X,Y)$. However, it doesn't seem to follow easily from the definitions. -Recall that $(C,D)$ is defined as follows. Consider the cofree coalgebra on $U$ on $Hom_k(C,D)$; it admits an evaluation map $U\otimes C\to D$, simply written $u(c)$. Define then $(C,D)$ as the maximal subcoalgebra of $U$ such that $\epsilon(u(c))=\epsilon(u)\epsilon(c)$ and $\Delta(u(c))=\Delta(u)(\Delta(c))$. -Letting $TC,TD$ denote the tensor algebras ovec $C,D$ respectively, $U$ naturally sits inside $Hom_{graded}(TC,TD)$. Then, if $f:X\to Y$ is a map, there corresponds the graded homomorphism $f(x_1\otimes...\otimes x_n)=f(x_1)\otimes...\otimes f(x_n)$ from $TC$ to $TD$, which is group-like. What is to be shown is that every homomorphism in $(C,D)$ is a linear combination of such $f$'s. - -REPLY [5 votes]: Edit: The answer I gave before reached a conclusion opposite to the one that grok and I came to in discussion later, offline. It turns out the original answer was correct up until the very end where I did some unfortunate handwaving; these lines have been corrected. In short, I am happy to say that grok's original surmise was entirely correct. -Note: we are working throughout with the category of cocommutative coalgebras over a field $k$ (from here on I will omit the words "cocomutative" and "commutative" as tacitly understood). This is a cartesian closed category whose cartesian product is given by $\otimes_k$; essentially by definition, the measuring coalgebra $M(C, D)$ plays the role of the exponential or internal hom, which we also denote by $D^C$. - -The natural coalgebra map $k(Y^X) \to M(k(X), k(Y))$ is in fact an isomorphism for all sets $X$ and $Y$. If $X$ has finite cardinality $n$, this is easy to see as follows: the functor -$$k(-): Set \to CocommCoalg$$ -preserves arbitrary coproducts (in fact, arbitrary colimits), and also finite products because $k(-)$ takes cartesian products to tensor products, and tensor products $C \otimes_k D$ give cartesian products in $CocommCoalg$. In particular, $k(-)$ preserves $n$-fold cartesian products $Y^n = Y \times \ldots \times Y$, and we have a series of canonical isomorphisms -$$k(Y^n) \cong k(Y)^n \cong (k(Y)^k)^n \cong k(Y)^{\sum^n k} \cong k(Y)^{k(n)}$$ -as desired. (The second isomorphism comes about because $k$ is the terminal coalgebra. Throughout we are using exponential notation for the measuring coalgebra, and using formal categorical properties of exponentials.) -I do not know whether this line of argument extends to infinite $n$. But a longer-winded analysis will give us this conclusion anyway, as follows. -First, here is a categorical way to think of coalgebras. By the fundamental theorem of coalgebras, every coalgebra is the union of its finite-dimensional subcoalgebras. As a consequence, one can show the category of coalgebras is locally finitely presentable, hence equivalent to the category of left exact functors to $Set$ from the category opposite to $Coalg_{fd}$, the category of finite-dimensional coalgebras. That opposite category is (equivalent to) the category $Alg_{fd}$ of finite-dimensional algebras. The equivalence -$$Coalg \to Lex(Alg_{fd}, Set)$$ -sends a coalgebra $C$ to the functor that takes a finite-dimensional algebra $A$ to $Coalg(A^\ast, C)$, the set of coalgebra morphisms from the dual $A^\ast$ to $C$. -Given coalgebras $C$ and $D$, the measuring coalgebra $D^C = M(C, D)$ is the coalgebra that represents the left-exact functor $A \mapsto Coalg(A^\ast \otimes C, D)$. -Now let $C = k(X)$, $D = k(Y)$. Suppose for the moment that $Y$ is finite. Then the coalgebra $k(Y)$ represents the left exact functor on $Alg_{fd}$ given by -$$A \mapsto Coalg(A^\ast, k(Y)) = Alg(k^Y, A)$$ -where $k^Y$ is the algebra product of $Y$ copies of $k$. An algebra map $k^Y \to A$ picks out $|Y|$ many mutually orthogonal idempotents in $A$ which sum to $1$. So $k(Y)$ represents the functor that takes $A$ to the set of functions $e: Y \to A$ where the $e_y$ are mutually orthogonal idempotents summing to $1$. -For $Y$ infinite, the coalgebra $k(Y)$ is the union or filtered colimit of $k(Y_i)$ where $Y_i$ ranges over finite subsets of $Y$. Consequently, $k(Y)$ represents the functor which takes $A$ to the set of functions $e: Y \to A$ of finite support, again where the $e_y$ are mutually orthogonal idempotents summing to $1$. For short, let me call such functions "distributions", although "probability distribution" might be more accurate. (Of course, the phrase "of finite support" is somewhat redundant since $A$ is finite-dimensional; the point is that there is no uniform bound on the size of the support.) -Now let us turn our attention to $M(k(X), k(Y))$. It represents the functor -$$A \mapsto Coalg(A^\ast \otimes k(X), k(Y))$$ -where $A^\ast \otimes k(X)$ is a coalgebra coproduct of $|X|$ copies of $A^\ast$. By some abstract nonsense, we see this functor is identified with -$$A \mapsto Coalg(A^\ast, k(Y))^X$$ -where the right side is an $X$-indexed product of copies of $Coalg(A^\ast, k(Y))$. This just gives us $X$-tuples of distributions $e: Y \to A$. -In this language, the natural map -$$\omega: Dist(Y^X, A) \to Dist(Y, A)^X$$ -takes a distribution $\{e_\phi\}$, where $\phi$ ranges over maps $X \to Y$, to an $X$-tuple whose component at $x$ is the distribution $e^x: Y \to A$ defined by -$$e^x_y = \sum_{\phi: \phi(x) = y} e_\phi$$ -The map $\omega$ is invertible: given an $X$-tuple of distributions $\{e^x_{-}: Y \to A\}$ indexed by $x \in X$, define a distribution $e_{-}: Y^X \to A$ which takes a function $\phi: X \to Y$ to -$$e_\phi = \prod_{x \in X} e^{x}_{\phi(x)}$$ -(This may look like an infinite product, but it's okay since each finite-dimensional algebra $A$ has only finitely many distinct idempotents, and idempotency plus commutativity allows us to coalesce infinitely many factors of the same idempotent into just one.) It is not hard to see that the $e_\phi$ are mutually orthogonal and sum to $1$, and that this rule indeed gives the inverse to $\omega$, as desired.<|endoftext|> -TITLE: Why are lacunary series so badly behaved? -QUESTION [27 upvotes]: Hi! -I just came across the Ostroski-Hadamard gap theorem, and while I can understand the proofs as well as the principle that the series $\sum_{n=0}^\infty z^{2^n}$ ought to have a singularity at every $2^n$-th root of unity for every $n$, I feel I'm missing some intuition into what exactly is going on. -Specifically, there is certainly the intuition that the faster a power series' coefficients decrease, the larger the radius of convergence will be - say, comparing the geometric series with the exponential power series. When contrasted with lacunary series, this seems to fail: the coefficients seem to be increasingly "smaller", at least in an average sense, but the function becomes terribly ill-behaved. (One could try and argue that in the Cesàro sense the coefficients do tend to zero: if $\sum_{n=0}^\infty z^{2^n}=\sum_{k=0}^\infty a_k z^k$, then $\frac{1}{n}\sum_{k=0}^n a_k\approx\frac{\lfloor\log_2(n)\rfloor}{n}\rightarrow0$ as $n\rightarrow\infty$. On the other hand, the power series $\sum_{k=0}^\infty \frac{z^k}{k}$, while having the same radius of convergence, can easily, if non-uniquely, be analytically extended to the whole complex plane; I'd expect the same of any series of the form $\sum_{k=0}^\infty \frac{\log(k)}{k}z^k$.) -Can anyone share some insight? - -REPLY [7 votes]: The intuition is simple. Consider the example $f(z)=\sum_{n=0}^\infty z^{2^n}.$ -This function satisfies the functional equation $f(z^2)=z^2+f(z)$. On the positive ray, -we evidently have $f(r)\to\infty$ as $r\to 1$, and the functional equation shows that -the same must happen on all rays $\{ re^{i\theta}:01$. This is an implementation of the general principle: a lacunary series behaves in the same way in all directions. So if the radius of convergence is $R<\infty$ then -all points $Re^{i\theta}$ must be singular. - -The final form of this theorem is due to E. Fabry, and it is called "Fabry's gap theorem", which implies, for example that $\sum_{n=0}^\infty -z^{n^2}$ is singular at every boundary point of the circle of convergence. This gap theorem is in turn is a very special case of "Fabry's General Theorem". -The best source for all of this is the German book of L. Bieberbach, Analytische Fortsetzung, Springer 1955, except that it is somewhat out of date. He describes the story of Fabry's General Theorem and related results in great detail. -For a modern exposition of Fabry's theorems in English, I recommend my papers - - -MR2595767 Eremenko, Alexandre, Densities in Fabry's theorem. Illinois J. Math. 52 (2008), no. 4, 1277–1290, and -MR2431054 Eremenko, Alexandre, A version of Fabry's theorem for power series with regularly varying coefficients. Proc. Amer. Math. Soc. 136 (2008), no. 12, 4389–4394. -Remark. Alexandr Ostrowski was 5 years old when Fabry published his general theorem. So it is unclear why the Wikipedia author calls it "Hadamard-Ostrowski". -Remark 2. A complex analyst will not describe such behavior as "badly behaved". Anyway, this behavior is typical for analytic functions, both in the sense of Baire category and in the sense of measure.<|endoftext|> -TITLE: A Kunneth formula for the etale cohomology of the product of ('simple') varieties over not (necessarily) algebraically closed field -QUESTION [5 upvotes]: If $X$ and $Y$ are varieties over an algebraically closed field, then in the corresponding derived category of complexes we have $RH_{et}(X\times Y,\mathbb{Z}/l^n\mathbb{Z})\cong RH_{et}(X,\mathbb{Z}/l^n \mathbb{Z})\otimes RH_{et}(Y,\mathbb{Z}/l^n\mathbb{Z})$. Where can I found a proof of this fact whose 'idea' would be clear (I don't want to assume that $X$ and $Y$ are smooth, and I don't want to consider cohomology with compact support)? Does the Leray spectral sequence help here? In order to compute the stalks of the corresponding (higher) direct images, one has to consider $Y$ being a strictly henselian scheme; how can this be done? -And what happens when the base field is no longer algebraically closed? In particular, I would like to understand completely the case when $X$ is singular, but $Y$ is just $G_m$ (an affine line minus one point). This particular case is related with the question The Gysin long exact sequence for the complement of the zero section of a line bundle over a (possibly) singular base - -REPLY [11 votes]: The only reference I know for Künneth theorems in this generality are SGA4 and SGA4.5. Specifically, SGA4 has theorems for étale cohomology with proper support (which hold in ridiculous generality), but SGA4.5's "Théorèmes de finitude en cohomologie $l$-adique" allows you to get the same results for étale cohomology, provided that the schemes are of finite type over a field (no other assumptions are needed!). The basic Künneth theorem itself is given as Corollaire 1.11 in that exposé. It is stated for schemes of finite type over a separably closed field, but if I'm not mistaken the proof works over any field whatsoever if you replace global sections by pushforwards to the étale topos of the base field. So this only gives you an equivalence between chain complexes with an action of the Galois group. The actual "étale cochains" are the invariants of those, and I don't know how to compute the invariants of a tensor product. But at least this reduces the problem to one in group cohomology. -ADDED: Here's a different approach. I think it may be true that for $X$ and $Y$ schemes of finite type over a field $k$, the cohomology of say $\mathbb{Z}/l^n$ over $X\times_kY$ can be computed as the $\mathbb{Z}/l^n$-cohomology of the homotopy pullback of étale homotopy types $Et(X)\times_{Et(k)}Et(Y)$. My argument is currently very WRONG though. The idea is that (1) $Et(X)$ should be the homotopy fixed points of $Et(\bar X)$ (where $\bar X=X\times_k\bar k$), (2) the result is obviously true over $\bar k$ (by the Künneth theorem), and (3) homotopy fixed points commute with homotopy pullbacks.The usefulness of this is limited since it is not easy to compute the cohomology of a homotopy pullback, but this would give a "closed formula" answer only in terms of $X$, $Y$, and $k$. -ADDED 2: I gave this a bit more thought and it obviously doesn't work. $Et(X)$ is the homotopy orbits of $Et(\bar X)$, and again I don't know how to compute the homotopy orbits of a pullback. The pullback formula is still true if $Y$ is proper and smooth over $k$. In this case you have a fiber sequence of $l$-completed étale homotopy types: -$$Et(\bar Y)\to Et(X\times_kY)\to Et(X)$$ -which you can compare with the fiber sequence -$$Et(\bar Y)\to Et(X)\times_{Et(k)}Et(Y)\to Et(X)$$ -so that $Et(X\times_kY)$ and $Et(X)\times_{Et(k)}Et(Y)$ are $\mathbb{Z}/l^n$-cohomologically equivalent. These fiber sequences are established in this paper by Friedlander (Corollary 4.8). I wonder if the properness assumption can be removed using SGA 4.5. -ADDED 3: OK, I'm pretty sure the above fiber sequence argument will work for any $X$ and $Y$ of finite type over a field. The only reason Friedlander considers proper and smooth maps is to apply a corollary of the proper/smooth base change theorems from SGA4, exposé XVI, which I think works without that assumption when over a field, thanks to SGA4.5. I need this result myself so I will try to check it thoroughly.<|endoftext|> -TITLE: Mean value property with fixed radius -QUESTION [7 upvotes]: Let $f$ be a continuous function defined on $\mathbb{R^n}$. It is well known that both the spherical mean value property (MVP) of $f$, i.e. -$$f(x)=\frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)}f,\ \forall x\in\mathbb{R^n}, r>0$$ -and the ball MVP, i.e. -$$f(x)=\frac{1}{|B(x,r)|}\int_{B(x,r)}f,\ \forall x\in\mathbb{R^n},r>0$$ -imply that $f$ is harmonic. -Note that in the definitions we require the redius $r$ to run over all the positive numbers. Out of curiosity I tried to find non-harmonic functions which satysfy the MVPs only for $r=1$. I did some search and found a remarkable fact called Delsarte's two-radius theorem saying that the spherical MVP with two fixed radii is enough to imply harmonicity of $f$. But for the $1$-radius MVP I haven't found any statement. -In the case $n=1$ examples have been found nicely in this M.SE post. But it is still unclear to me how to construct similar examples in higher dimensions. Any comments would be appreciated! - -REPLY [3 votes]: It is a theorem of Hansen and Nadirashvili: W. Hansen, N. Nadirashvili, "Littlewood's one circle problem" J. London Math. Soc. , 50 (1994) pp. 349–360 that one radius (which is allowed to be a function of the point) is not enough for the spherical MVP for functions defined in domains in $\mathbb{C}$ -- this is still open in dimension > 2, while it IS enough for the ball MVP (W. Hansen, N. Nadirashvili, "A converse to the mean value theorem for harmonic functions" Acta Math. , 171 (1993) pp. 139–163), in every dimension. If instead of a domain we take all of $\mathbb{C}$ the answer is YES, and the proof is elementary, see: -W. Hansen, A strong version of Liouville’s theorem, Am. Math. Mon. 115 (7) (2008) 583–595. -EDIT -To answer @George's comment: I was actually misled by Hansen's statement in a later paper. What he actually seems to show is that a positive bounded function satisfying the mean-value property for a single $r(x)$ at every $x,$ where $r(x) < | x| + M,$ for some $M$ is constant.<|endoftext|> -TITLE: Why is $\mathbb{Q}$-factoriality not local in the étale topology? -QUESTION [10 upvotes]: I was reading Kollár and Mori's book today and stumbled on the following passage: -"The $\mathbb{Q}$-factoriality assumption is a very natural one if we start with smooth varieties, and it makes many proofs easier. On the other hand it is a rather unstable condition in general. It is not local in the Euclidean (or étale) topology, and it is very hard to keep track of when we pass from a variety to a divisor in an inductive proof." -(For those curious this is in section 3.7) -My question is, what are some examples that show why $\mathbb{Q}$-factoriality is not local in the étale topology? - -REPLY [4 votes]: A different set of examples, closer to Kollár--Mori, comes from projective geometry over $\mathbb C$. For a projective variety $X$ over $\mathbb C$, $\mathbb Q$-factoriality is a global topological property: it depends on the prime divisors lying on $X$, rather than just on the local analytic type of its singular points, the issue being that local $\mathbb Q$-Weil but not $\mathbb Q$-Cartier divisors may fail to glue to a global non-$\mathbb Q$-Cartier divisor. For example, for a quartic threefold $X=X_4\subset {\mathbb P}^4$, a small number of ordinary double points do not effect the factoriality, even though the threefold ordinary double point is manifestly non-factorial. See this paper with a nice introduction which has a fuller discussion, explicit examples and further references.<|endoftext|> -TITLE: theorems equivalent to the parallel postulate -QUESTION [6 upvotes]: Is there a good survey article listing all the theorems of Euclidean geometry that are equivalent to the parallel postulate? - -REPLY [3 votes]: Doug Chatham's answer is the best I received; see Theorem 23.7 of George E. Martin's "The Foundations of Geometry and the Non-Euclidean Plane".<|endoftext|> -TITLE: A strange sum over bipartite graphs -QUESTION [21 upvotes]: While mucking around with some generating functions related to enumeration of regular bipartite graphs, I stumbled across the following cutie. I wonder if anyone has seen it before, and/or if anyone sees a nice interpretation. The sum is over all simple bipartite graphs $G$ with $n$ vertices on each side, the product is over all $2n$ vertices $v$ of $G$ and $d_G(v)$ is the degree of vertex $v$ in graph $G$. -$$\sum_{G\subseteq K_{n,n}} ~ \prod_{v\in V(G)} (n-2d_G(v)) = 2^{n^2}n!~.\kern 3cm (1)$$ -Addition 1, proof. Consider $n^2$ commuting indeterminates $\{x_{ij}\}_{i,j=1\ldots n}$. The polynomial -$$P(\boldsymbol{x}) = \prod_{i=1}^n \sum_{j=1}^n x_{i,j} \times \prod_{j=1}^n \sum_{i=1}^n x_{i,j}.$$ -has terms with each variable having power 0, 1 or 2. Consider the total coefficient $C~$ of the terms with only even powers. One way is to sum each variable over $\pm 1$, which makes the terms we don't want cancel out and the others get multiplied by $2^{n^2}$. This gives (1), interpreting $G$ as the bipartite graph whose edges are the variables with value $-1$. Alternatively, note that each term has total degree $2n$ and the only possible such term with even degrees is $\prod_{i=1}^n x_{i,\sigma(i)}^2$ for some permutation $\sigma$. This shows $C=n!~$. -Addition 2, generalisation. Define the numbers -$$\rho(n,k,d) = \sum_{j\ge 0} (-1)^j \binom{d}{j} \binom{n-d}{k-j}.$$ -Let $k_1,\ldots,k_{2n}$ be a sequence of nonnegative integers. Then -$$\sum_{G\subseteq K_{n,n}} ~ \prod_{v\in V(G)} \rho(n,k_v,d_G(v)) - = 2^{n^2} B(\boldsymbol{k}), $$ -where $B(\boldsymbol{k})$ is the number of simple bipartite graphs with vertex $v$ having degree $k_v$ for all $v$. The case (1) follows from $\rho(n,1,d)=n-2d$. The proof of the general case is similar. - -REPLY [13 votes]: I'm not sure if this is the right interpretation or not...it may really just be another way of encoding the generating function argument. Let $H$ be a random bipartite graph where every edge appears independently with probability $1/2$. Then the left hand side is -$$2^{n^2} E \left(\prod_v f(v) \right),$$ -where $f(v)$ is equal to $\sum_u x(u,v)$ and $x(u,v)$ is $1$ if an edge is not present, $-1$ if an edge is present. Expanding out the product and using linearity of expectation, we can write this as -$$2^{n^2} \sum_{\sigma} E \left(\prod_{v} x(v,\sigma(v))\right)$$ -Where $\sigma$ consists of all mappings taking each vertex to a vertex on the opposite side. -Any $\sigma$ for which some edge $(v,\sigma(v))$ appears only once has $0$ expectation due to independence. The $\sigma$ for which every edge appears for both of its endpoints correspond to matchings between the left and right side, of which there are $n!$. (This last observation corresponds to the fact that the expected square of the permanent of an $n \times n$ random Bernoulli matrix is $n!$...I think it goes at least back to Turan).<|endoftext|> -TITLE: Entropy conjecture for distributions over $\mathbb{Z}_n$ -QUESTION [8 upvotes]: Suppose we have two independent random variables $X$ (with distribution $p_X$) and $Y$ (with distribution $p_Y$) which take values in the cyclic group $\mathbb{Z}_n$. Let $Z = X +Y$, where the addition is done modulo $n$. Distribution of $Z$ is given by -$$p_Z = p_X \circledast p_Y$$ where $\circledast$ stands for cyclic convolution. Suppose we were interested in minimising $H(Z)$ while keeping $H(X)$ and $H(Y)$ fixed. I also impose an additional constraint that $p_X(0) > 0$ and $p_Y(0) > 0$. The conjecture is as follows- - -The $p_X$ and $p_Y$ which minimise $H(Z)$ have to be supported on the smallest possible subgroup $G$ of $Z_n$ that satisfies $$\log_2(|G|) \geq \max(H(X),H(Y))$$ - -Some Remarks: - -The constraint that $p_X(0) > 0$ and $p_Y(0) > 0$ is essentially to rule out $X$ and $Y$ being supported on some coset of $G$. One can even assume $p_X(0)$ and $p_Y(0)$ are the maximum probabilities in the distributions $p_X$ and $p_Y$ since we can always "rotate" the distributions and still keep $H(Z)$ the same. -The intuition behind this is that we have the trivial lower bound -$$H(Z) \geq \max(H(X),H(Y)$$ -and when $H(Y) \leq H(X) = \log_2 |G|$, this lower bound is actually attained. Also intuitively the more concentrated a distribution, the lesser its entropy. So forcing $X$ and $Y$ to be supported in $G$ also ensures $Z$ is supported only on $G$, which is "small". -I'm not sure but I think one can also pose this as a norm-extremisation problem instead of entropy by using the fact $$H(p_X) = \lim_{p \to 1}\frac{1}{1-p}\log_2 \sum p_X(i)^p$$ -I unfortunately have very little empirical evidence. An exhaustive search of all possible "entropy fixing" distributions of $X$ and $Y$ goes out of hand very fast. My codes run only till $n=5$ in reasonable time, and $4$ being the only composite number till then, I've verified it only for $4$. Since things are much more cleaner for $2^n$ than for an arbitrary composite number, maybe its true only for $\mathbb{Z}_{2^n}$. - -REPLY [2 votes]: The conjecture is wrong! It wasn't as complicated as I thought it was. -A simple counter example is over $\mathbb{Z}_6$. Consider $H(X) = 1$ and $H(y) = 1 +\epsilon$ where $\epsilon$ is very small. Now the conjecture would imply $X$ and $Y$ should both be supported on $\{0,2,4\}$ giving $H(X+Y) \approx 1.3326$. But consider $$p_X = [0.5, 0, 0, 0.5, 0, 0]$$ and $$p_Y = [0.5 - \delta, \frac{\delta}{2}, \frac{\delta}{2},0.5 - \delta, \frac{\delta}{2}, \frac{\delta}{2}]$$ for a suitable $\delta$ which adjusts the entropy to be $1+ \epsilon$. -$$p_X \circledast p_Y = p_Y $$ -so we have $H(X+Y) = 1 + \epsilon < 1.3326$. Thus disproved.<|endoftext|> -TITLE: Can the Sylow p-subgroup of a finite group of Lie type be cyclic? -QUESTION [6 upvotes]: Let $G$ be a finite group of Lie type in characteristic $p$. When is the Sylow $p$-subgroup of $G$ cyclic? - -REPLY [12 votes]: What is meant by "finite group of Lie type" needs to be made precise. But at least the simple groups of Lie type in characteristic $p$ with a cyclic -Sylow $p$-subgroup are easy to specify: these are the groups $\text{PSL}(2,p)$ with $p>3$ along with one twisted group usually denoted $^2 \text{G}_2(3)'$ with $p=3$ (which is isomorphic to $\text{SL}(2,8)$). Of course there are also some closely related non-simple groups of Lie type including a few very small groups with $p=2$ -This is summarized on page 74 of my 2005 Cambridge Univ. Press book Modular Representations of Finite Groups of Lie Type along with what I hope are sufficient references to the scattered literature. -P.S. Whether or not a finite group has a cyclic Sylow subgroup (for some prime) usually comes up in two contexts: blocks with a cyclic defect group (Brauer, Dade) and finite representation type for finite dimensional algebras including group algebras. Are there other motivations?<|endoftext|> -TITLE: When is the norm of all positive operators on an ordered Banach space determined by their values on the positive cone? -QUESTION [8 upvotes]: I'm trying to investigate the interplay between the norm and cone of positive elements in ordered Banach spaces. In particular, I would like a nice characterization of when the norm of a positive operator can be obtained as a supremum of norms over all norm 1 positive -elements (see 1. below). It is easy to show that this always holds for Banach lattices, but gets tricky when trying to prove this more generally. -To be a bit more detailed, let $X$ be a Banach space with a cone (or wedge) of 'positive' elements, denoted by $X_{+}$; $X$ is then called an ordered Banach space and $x\leq y$ means $y-x\in X_{+}$. I will call $X_{+}$ generating if $X=X_{+}-X_{-}$; proper if $X_{+}\cap(-X_{+})=\{0\}$; and normal if $0\leq x\leq y$ implies $\|x\|\leq\|y\|$. A bounded linear operator $T:X\to X$ is called positive, if $x\geq 0$ implies $Tx\geq 0$. An ordered Banach space with the property that every positive operator satisfies $$\|T\|=\sup\{\|Tx\|:x\in X_{+},\|x\|=1\}$$ I will say has the positive operator property, i.e., the norms of positive operators are completely determined by their behavior on the cone. -My question is: - -Let $X$ be an ordered Banach space - with a closed, proper, generating, - normal cone. Is there a characterization of the positive operator property in terms of the cone-norm interaction? - -Thus far I have found a few sufficient conditions: - -For all $x\in X$ there exist $X\ni x_{1},x_{2}\geq0$ such that $x=x_{1}-x_{2}$ and $\|x_{j}\|\leq\|x\|$ for $j=1,2$. -For all $x\in X$ and $X\ni a,b\geq0$, $-a\leq x\leq b$ implies $\|x\|\leq\max \{ \|a\|,\|b\| \}$. - -together imply the positive operator property. -Another sufficient condition is having the property that for any $x\in X$ it holds that $\|x\| = \inf \{\|z_+ + z_-\|:z_\pm \geq 0;x=z_+ -z_-\}$. Being a Banach lattice is sufficient to have this property (by invoking the property $\|x\|=\||x|\|$), but seems to be a bit more general since $\mathbb{R}^3$ with the Euclidean norm and the `ice-cream cone' $\{(x_1,x_2,x_3):x_1\geq (x_2^2+x_3^2)^{1/2} \}$ (which is not a lattice) also has this property (so being a lattice is not necessary for the positive operator property). -I've been unable to prove that any of these conditions are necessary. The last one had been my best bet so far, but I'm beginning to doubt the existence of a nice list of conditions on the norm and cone that together are necessary and sufficient. -Any suggestions, counterexamples or pointers to the literature that anyone -may have will be greatly appreciated. -UPDATE: Cleaned it up a bit, and posted some new information I found since first posting. I found no mention in any literature to this question. I also asked around a bit with no hits from people who are a bit in the know, so it seems to be wide open. Gets the 'open-problem' tag. - -REPLY [7 votes]: Here are examples to show that neither of the two conditions 1. and 2. that you give as jointly sufficient can just be dropped. -Take $X$ to be two-dimensional real space with the pointwise order. Norm it by taking the closed unit ball to be the convex hull of -$\{(-3,3),(0,2),(2,0),(3,-3),(0,-2),(-2,0),(-3,3)\}$. On the positive cone, $\|(x,y)\|=(x+y)/2$. -Clearly, $(-1,-1)\le (-1,1)\le (1,1)$ and $\|(-1,-1)\|=\|(1,1)\|=1$ whilst $\|(-1,1)\|=1/3$. -Any attempt to write $(-1,1)=u-v$ with $u$ and $v$ positive will have to have $\|u\|,\|v\|\ge 1/2$, so your condition 1 fails. Condition 2 does hold. -Define $T(x,y)=(y,y)$, which is certainly positive. For $(x,y)\ge 0$, $\|T(x,y)\|=y\le 2\|(x,y)\|$. -On the other hand $\|T(-1,1)\|=\|(1,1)\|=1$ whilst $\|(-1,1)\|=1/3$ so that $\|T\|\ge 3$. -For the second example, take $Y$ to be two-dimensional real space with the pointwise order and normed by $\|(x,y)\|=\max\{\|(x,y)\|_\infty, |x-y|\}$. -Take $Z$ to be two-dimensional real space with the pointwise order and supremum norm. For $(x,y)\ge (0,0)$ the two norms are equal. - $Y$ has your property 1 but not 2. The identity map from $Z$ into $Y$ has norm at least 2 as $\|(-1,1)\|=2$ whilst $\|(-1,1)\|_\infty=1$. -On the positive cone all norms of images are equal. To get an example from a space into itself, take $X=Y\oplus Z$ and the operator -$(y,z)\mapsto (z,0)$. \ No newline at end of file