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bSchools Information System is designed for a complex automation of educational and business processes in schools. Moreover the system provides information exchange between schools and governmental organizations that control educational processes and programs.Project Details Enterprise web system for task assignment, recording of transactions made by land agents and operational efficiency analysis.Project Details E‑time (entertainment time) — is a cross-platform application which enables you to check the schedule of more than 1,200 theaters and cinemas in one place and buy tickets safely inside the app. E‑time — your entertainment at the right time!Project Details Aristek has developed a system that manages almost everything, from storing petroleum products and transporting them to accounting and client relations management.Project Details Reliable scalable Electronic Health Record (EHR) platform for patients, doctors and healthcare providers. Safe place for health information exchange with quick access to essential patient’s information.Project Details Gas transportation management software provides online monitoring & controlling of gas transportation streams as well as contracts management A centralized platform for sports coaches and their clients with the opportunity to build, manage and analyze workouts. Coaches are able to build charged or free and searchable or non-searchable training programs.Project Details Multifunctional multilingual online store for diagnostics equipment of cars electronic systems.Project Details The online service offers a tool for creating forms and their timely completion control. The system contains the set of categorized templates of forms and automatically creates the new forms and sends the notifications of the necessity to complete the form.Project Details Comprehensive EMR (Electronic Medical Record) system for private group of clinics from the USA, specified on mental health. Easy and quick management of patient mental information and accompanying billing documents. System integrations with insurance companies and clearinghouses.Project Details Our customer wanted us to create an educational platform which would stay the connector of teachers and students, a progress tracking tool and the educational source to a large number of schools.Project Details Aristek developed a comprehensive ERP that automated all the business processes, improved employee productivity by 23% and decreased fuel expenditure by 31%.Project Details Aristek introduced augmented and virtual reality to a platform used by millions of students and teachers.Project Details Aristek helped develop an all-in-one solution for independent artists, designers, and other creators. The feature set gives the system a serious shot to challenge Shopify and BigCommerce in the art domain Aristek quickly put together a dedicated team and increased the customer’s development velocity by 20%.Project Details Aristek developed learning games that increased student engagement and simplify learning material assimilation by up to 30%.Project Details Aristek developed interactive 3D scenes that increase student engagement with the ratio of 20-30% as compared to the traditional approach.Project Details A comprehensive Enterprise Resource Planning (ERP) System that dramatically increased efficiency of a well-known US-based vehicle transportation company.Project Details Send Message

It seems simple enough — to get bigger, you need to eat more. But there’s more to it than that. We’ve seen too many people go through “bulk” phases that involve eating everything in sight — pizza, buffalo wings, alfredo pasta, cookies, chips, etc. — and call it their means to putting on significant muscle mass. Remember what we said yesterday about all calories not being equal? Well, it’s doubly true when you are eating many more of them. Just like we said before, it’s not always about the amount of calories you’re eating, but the quality. Even in a bulking phase, where you are definitely going to be upping your calorie intake, you want to make sure those calories are high-quality ones that serve a purpose, and not loads of empty calories that will only serve to add more undue body fat. Obviously, you’ll need more clean calories in your everyday diet to accomplish a muscle-mass gain that’s significant, and most of these should come from food — eggs, milk, lean meats, complex carbs, vegetables, etc. But you can easily add a large, quality-calorie increase (or two) a day with a good gainer shake.. MASS COMPLEX delivers your muscles the fuel needed for optimal growth. MASS COMPLEX also contains a natural digestive enzyme complex to improve digestion and assimilation, thus reducing bloating commonly associated with other weight gain supplements. Low in fat and sugar! MASS FUZION’s carbohydrate blend is designed not only to increase glycogen for improved amino acid delivery, but also to signal a big spike in insulin in the blood. This natural anabolic hormone, produced by the pancreas, is one of the most potent substances there is that can be naturally found in the body for muscle-building. MASS FUZION promotes insulin production post-workout with InsuCell-K, a formula designed to increase blood insulin levels. InsuCell-K also includes KreAlkalyn and Beta-Alanine, both shown to enhance muscle growth, strength and power in conjunction with each other.

TITLE: $SO(4,2)$ symmetry of the hydrogen atom QUESTION [7 upvotes]: The hydrogen atom with Hamiltonian obviously has $SO(3)$ symmetry since it just depends on the radius. $$ H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r}$$ This is generated by angular momentum $\mathbf{L} = \mathbf{r}\times \mathbf{p} $. In quantum mechanics class we learn there is $SO(4)$ symmetry due to the Runge-Lenz vector: $$ \mathbf{A}= \frac{1}{2m}(\mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p}) - k \frac{\mathbf{r}}{r}$$ Even classical symmetry like gravity has this kind of symmetry. I remember reading one time there is even greater symmetry for hydrogen atom. Possibly $SO(4,2)$ as in this article by Hagen Kleinert. Has anyone heard of this? How can one see that the Hydrogen atom has $SO(4)$ symmetry? REPLY [10 votes]: $SO(4,2)$ is called the full dynamical group of the Kepler (or Hydrogen atom problem). The $SO(4)$ , $SO(3,2)$ and $SO(4,1)$ subgroups of $SO(4,2)$ are called partial dynamical groups. Unlike symmetry groups which commute with the Hamiltonian, dynamical groups do not. They have the following properties: The system's phase space is a coadjoint orbit of the group, or equivalently, The system's Hilbert space is spanned by an irreducible representation of the group. In many cases, although, it is not necessary, the Hamiltonian itself is a generator of the dynamical group. The equivalence of the points 1. and 2. above stems from the fact that in the case under study there is a correspondence between coadjoint orbits and irreducible representations. The partial dynamical groups span only part of the Hydrogen atom spectrum through their irreducible representations: An $SO(4)$ irreducible representation spans the state vectors corresponding to a single energy shell of a bound state (fixed (and quantized) $n$ and varying $l$, $m$) . An $SO(3,2)$ irreducible representation spans the full continuous spectrum and an irreducible representation of $SO(4,1)$ spans the full bound spectrum. $SO(4,2)$ is the smallest group whose irreducible representations span both the continuous and the discrete spectrum. The use of dynamical group representations reduces the problem of finding the Hamiltonian spectrum to an algebraic problem, instead of a solution of differential equations.

\section{Function Fields}\label{sec:ff} In this section we give the analogue statements over function fields of the theorems obtained in the previous sections. For this section we let $\kappa$ be an algebraically closed field of characteristic zero. Let $\calC$ be a non-singular projective curve defined over $\kappa$ and let $K = \kappa(\calC)$ denote its function field. We refer to \cite[Section 7.2]{RTW} for the basic definitions of heights and proximity functions in the function field setting. We recall the definition of algebraic hyperbolicity. \begin{definition}\label{def:alg_hyp} Let $(X,D)$ be a pair of a non-singular projective variety $X$ defined over $\kappa$ and a normal crossing divisor $D$ on $X$. We say that $(X,D)$ is \emph{algebraically hyperbolic} if there exists an ample line bundle $\calL$ on $X$ and a positive constant $\alpha$ such that, for every non-singular projective curve $\calC$ and every morphism $\varphi: \calC \to X$ the following holds: \begin{equation}\label{eq:alg_hyp} \deg \varphi^* \calL \leq \alpha \cdot \left( 2g(\calC) - 2 + N^{[1]}_\varphi(D)\right), \end{equation} where $N^{[1]}_\varphi(D)$ is the cardinality of the support of $\varphi^*(D)$. We say that $(X,D)$ is \emph{pseudo algebraically hyperbolic} if there exists a proper closed subvariety $Z$ of $X$ such that \eqref{eq:alg_hyp} holds for every morphism $\varphi: \calC \to X$ such that $\varphi(\calC)$ is not contained in $Z$. \end{definition} We can now rephrase Theorems \ref{IP'}, \ref{proposition7} and \ref{Theorem7}. \begin{theorem}\label{th:ff1.3} Let $n\ge 2$, $r\ge 2n+1$ and $D_0,D_1, \dots, D_{r} $ be hypersurfaces in general position on $\mathbb P^n$ defined over $\kappa$. Let $\pi: X\to \mathbb P^n$ be the blowup long the union of subschemes $D_i\cap D_0$, $1\le i\le r$, and let $\widetilde D_i$ be the strict transform of $D_i$. Let $D=\widetilde D_1+\cdots+\widetilde D_r$. Then $ X\setminus D$ is algebraically pseudo-hyperbolic. \end{theorem} \begin{theorem}\label{th:ff1.4} Let $n\ge 2$ and $H_1,\hdots,H_{2n}$ be $2n$ hyperplanes in general position on $\mathbb P^n$ defined over $\kappa$. Choose $n+1$ points $P_i$, $1\le i\le n+1$ such that $P_i\in H_i$, $1\le i\le n+1$, and $P_i\notin H_j$ if $i\ne j$ for $1\le j\le 2n$. Let $\pi: X\to\mathbb P^n$ be the blowup of the $n+1$ points $P_i$, $1\le i\le n+1$, and let $D\subset X$ be the strict transform of $H_1+\cdots+H_{2n}$. Then $X\setminus D$ is algebraically pseudo-hyperbolic. \end{theorem} \begin{theorem}\label{th:ff1.5} Let $n\ge 2$, $q\ge 3n$ be an integer; for every index $i\in \mathbb Z/q\mathbb Z$, let $H_i$ be a hyperplane in $\mathbb P^n$ defined over $k$. Suppose that $H_i$'s are in general position. Let for each index $i\in \mathbb Z/q\mathbb Z$, $P_i$ be the intersection point $\cap_{j=0}^{n-1} H_{i+j}$. Let $\pi: X\to \mathbb P^n$ be the blow-up over the points $P_1,\hdots, P_q$ and let $\widetilde H_i\subset X$ be the corresponding strict transform of $H_i$ and let $D=\widetilde H_1+\cdots+\widetilde H_q$. Then $ X\setminus D$ is algebraically pseudo-hyperbolic. \end{theorem} We remark that, even if we stated the results in the so-called \emph{split case}, our proofs carry over almost verbatim to the non-split case as well. As in the analytic setting, the proofs of the above statements follow the same lines of the proof of our arithmetic results and the same strategy as in our previous paper \cite{RTW} with two modifications. On one hand we can use the results in Section \ref{sec:analytic} instead of \cite[Theorem 8.3 B]{levin_annal} for the case in which $2g(\calC) - 2 + N^{[1]}_\varphi(D)\leq 0$. On the other hand we replace the use of Theorem \ref{Ru-Vojta} with the following analogue that uses a version of the Schmidt subspace theorem over function fields recently obtained in \cite[Theorem 15]{GSW}. In particular this gives a better control on the exceptional set. \begin{theorem} \label{thm:ffconst} Let $X\subset \PP^m$ be a projective variety over $\kappa$ of dimension $n$, let $D_1,\cdots,D_q$ be effective Cartier divisors intersecting properly on $X$, and let $\calL$ be a big line sheaf. Then for any $\epsilon>0$, there exist constants $c_1$ and $c_2$, independent of the curve $\calC$ and the set $S$, and a finite collection of hypersurfaces $\mathcal Z$ (over $\kappa$) in $\PP^m$ of degree at most $c_2$ such that for any map $x=[x_0:\cdots:x_m] :\calC\to X$, where $x_i\in K$, outside the augmented base locus of $\calL$ we have either \begin{align*} \sum_{i=1}^q \beta_{\mathcal L, D_i} m_{D_i,S}(x)\le (1+\epsilon) h_{\calL}(x)+c_1 \max\left\{ 1, 2g(\calC)-2+|S|\right\}, \end{align*} or the image of $x$ is contained in $\mathcal Z$ . \end{theorem} \begin{proof} The proof is similar to the first part of the proof of \cite[Theorem 7.6]{RTW}. We will follow its argument and notation and only indicate the modification. Let $\epsilon>0$ be given. Since $\calL$ is a big line sheaf, there is a constant $c$ such that $\sum_{i=1}^qh_{D_i}(x)\le c h_{\calL}(x)$ for all $x\in X(K)$ outside the augmented base locus $B$ of $\calL$. By the properties of the local heights, together with the fact that $m_{D_i,S}\le h_{D_i}+O(1)$, we can choose $\beta_i\in\Q$ for all $i$ such that \[ \sum_{i=1}^q (\beta_{\mathcal L, D_i}-\beta_i) m_{D_i,S}(x)\le \frac{\epsilon}{2}h_{\calL}(x) \] for all $x\in X\setminus B(K)$. Therefore, we can assume that $\beta_{\mathcal L, D_i}=\beta_i\in\Q$ for all $i$ and also that $\beta_i\ne 0$ for each $i$. From now on we will assume that the point $x \in X(K)$ does \emph{not} lie on $B$. Choose positive integers $N$ and $b$ such that \begin{equation}\label{aut_choices} \left( 1 + \frac nb \right) \max_{1\le i\le q} \frac{\beta_i Nh^0(X, \calL^N) } {\sum_{m\ge1} h^0(X, \calL^N(-mD_i))} < 1 + \epsilon\;. \end{equation} Then, using \cite[Theorem 7.5]{RTW} with the same notation, we obtain \begin{equation}\label{eq:weilbase_ff} \begin{split} &\frac b{b+n} \left( \min_{1\leq i \leq q} \sum_{m\ge 1} \frac{h^0(\calL^N(-mD_i))}{\beta_i} \right) \sum_{i=1}^q \beta_i \lambda_{D_i,\mathfrak p} (x) \\ &\qquad\le \max_{1\le i\le T_1} \lambda_{\mathcal B_i,\mathfrak p} (x)+ O(1)= \max_{1\le i\le T_1} \sum_{j\in J_i} \lambda_{s_j,\mathfrak p} (x) + O(1). \end{split} \end{equation} Let $M=h^0(X,\calL^N)$, let the set $\{\phi_1,\dots,\phi_M\}$ be a basis of the vector space $H^0( X, \calL^N)$, and let \begin{align}\label{Phi} \Phi=[\phi_1,\dots,\phi_M]:X \dashrightarrow \PP^{M-1}(\kappa) \end{align} be the corresponding rational map. By \cite[Theorem 15]{GSW}, there exists a finite collection of linear subspaces $\mathcal R$ over $\kappa$ such that, whenever $\Phi\circ x$ is not in $\mathcal R$, we have the following \begin{equation}\label{eq:schmidt2_ff} \sum_{\mathfrak p\in S}\max_J \sum_{j\in J} \lambda_{s_j,\mathfrak p}(x) \le M \, h_{\calL^N}(x) +\frac{M(M-1)}2(2g-2+|S|), \end{equation} here the maximum is taken over all subsets $J$ of $\{1,\dots,T_2\}$ for which the sections $s_j$, $j\in J$, are linearly independent (with the same notation as in the proof of \cite[Theorem 7.1]{RTW}). We first consider when $\phi_1,\dots,\phi_M$ are linearly independent over $\kappa$. Combining \eqref{eq:weilbase_ff} and \eqref{eq:schmidt2_ff} gives $$\sum_{i=1}^q \beta_i m_{D_i,S}(x) \leq \left( 1 + \frac nb \right) \max_{1\le i\le q} \frac{\beta_i } {\sum_{m\ge1} h^0(\calL^N(-mD_i))} \, M\, h_{\calL^N}(x) +c_1'(2g-2+|S|) + O(1),$$ where $c_1'=\frac{M(M-1)}2. $ Using \eqref{aut_choices} and the fact that $h_{\calL^N}(x)=Nh_{\calL}(x)$, we have \begin{equation*} \sum_{i=1}^q \beta_i m_{D_i,S }(x) \leq \left( 1+\epsilon \right)h_{\calL}(x)+c_1'(2g-2+|S|) + O(1), \end{equation*} which implies the first case of the Theorem. To conclude we note that, if $\Phi\circ x$ is in one of the linear subspace of $\mathcal R$ over $\kappa$ in $\PP^{M-1}$, then $a_1\phi_1(x) + \dots + a_M \phi_M(x) = 0$, where $H=\{a_1z_1 + \dots + a_M z_M = 0\}$ is one of the hyperplanes (over $\kappa$) in $\PP^{M-1}$ coming from $\mathcal R$. On the other hand, since $\phi_1,\dots,\phi_M$ is a basis of $H^0(X,\calL^N)$, it follows that $\Phi(X)$ is not contained in $H$, hence $x(\calC)$ is contained in is the hypersurface coming from $a_1\phi_1 + \dots + a_M \phi_M = 0$ in $\PP^{m}$ (as $X\subset \mathbb P^m$) whose degree is bounded independently of $\calC$ and $x$ as wanted. Moreover, since $\mathcal R$ is a finite collection of linear subspaces over $\kappa$ in $\PP^{M-1}$, the number of $H$ is finite and hence the number of hypersurfaces obtained above is also finite. \end{proof} \nocite{}

TITLE: Why is the hint needed in Exercise 3.5.12 in Tao's Analysis 1 QUESTION [0 upvotes]: I did see this post and this post but I still don't really get it. The question is Let f : N×N → N be a function, and let c be a natural number. Show that there exists a function a : N → N such that a(0) = c and a(n++) = f(n, a(n)) for all n ∈ N, and furthermore that this function is unique. The hint given is Show inductively that for every natural number M ∈ N, there exists a unique function a : {n ∈ N : n ≤ M} → N such that aM(0) = c and aM(n+1) = f(n, a(n)) for all n ∈ N such that n<M.) My question is, why can't we just define a function a: N → N that obeys these properties. Jut let a(0)=c and a(n+1)=f(n,a(n)) Then a is defined for every natural number, and is unique. REPLY [0 votes]: $\def\NN{\mathbb{N}}$The point of the exercise is that one cannot define a function $f:\NN\to\NN$ by saying that $f(1)=1$ and $f(n+1)=2f(n)$ for all $n\in\NN$. This is NOT a definition of a function, but a property of a function. What one can do is to show that there exists a function which has that property and, moreover, that there is exactly one. The hint gives you the standard argument used to do this. Stackexchange does not let us use a font as big and as bold as one would need to properñy emphasize the fact that NOT above… Giving a property of something does not define anything in general. For example, it does not work to say let $x$ be the real number such that $x^2+1=0$ for there is no such real number. Similarly, it does not work so say let $f:\NN\to\NN$ be the function such that $f(0)=4$ and $f((n-5)^2)=n+1$ for all $n\in\NN$ for there is no such function. So when one wants to say, for example, let $f:\NN\to\NN$ be the function such that $f(1)=1$ and $f(n+1)=2f(n)$ for all $n\in\NN$ one has to prove that there is such a function. (I'll write $[a,b]$ for intervals in the integers here) Now suppose that you have a positive integer $N$ and you know that there is a function $h:[1,N]\to\NN$ such that $h(1)=1$ and $h(n+1)=2h(n)$ for all $n\in[1,N-1]$. I claim that then there is also a function $g:[1,N+1]\to\NN$ such that $g(1)=1$ and $g(n+1)=2g(n)$ for all $n\in[1,N]$, and moreover the restriction of $g$ to $[1,N]$ is equal to the function $h$. Indeed, we can define this function directly: we consider the set $$\{(i,h(i)):i\in[1,N]\}\cup\{(N+1,2h(N))\},$$ prove that it is a function $[1,N+1]\to\NN$ and check that it has the required property. In this way, by induction (one has to start it, I'll let you do that), one shows that for all $N\in\NN$ there exists a function $h_N:[1,N]\to\NN$ such that $h_N(1)=1$ and $h_N(n+1)=2h_N(n)$ for all $n\in [1,N-1]$ and that moreover these functions have the property that whenever $n$ and $m$ are elements of $\NN$ such that $n\leq m$ we have $h_m(n)=h_n(n)$. Now we can consider the set $$\{(n,h_n(n)):n\in\NN\},$$ check that it is a function $f:\NN\to\NN$ and that it has the property we want, namely, $f(1)=1$ and $f(n+1)=2f(n)$ for all $n\in\NN$. The key fact that makes all this work is that in constructing the functions $h_n$ and the function $f$ I did not wave my hands at any point: I simply used the definition of what a function is, and the operations of set theory that construct sets from sets — in fact, I only used unions and pairings. One can view a phrase like define the function $f:\NN\to\NN$ so that $f(1)=1$ and $f(n+1)=2f(n)$ for all $n\in\NN$ as a definition of a function through an algorithm that computes it. This does not eliminate the need to prove the exercise in Tao's book. Indeed, it is the fact that the claims of the exercise are true what give sense to defining a function through an algorithm to compute it. At some point one has to define what it means «to give an algorithm that defines a function» and it is that exercise, essentially, what one usually does. There is a whole theory, that of denotational semantics, that deals with giving sense to this sort of «definitions». And, trust me, it is not an easy task.

Smooth Hound Smith, a self-described foot stomping Americana roots duo, will perform at the historic Astra Theatre on Saturday, May 12, 2018. Doors will open at 6:30 p.m., with a yet-to-be-determined opening act beginning at 7:30 p.m. The East Nashville duo utilizes a combination of foot percussion, fuzzed out finger-picked guitar patterns, and tight harmonies to deliver their own unique style of folky, garage-infused rhythm and blues, soul, and rock n’ roll music. They have performed at Bonnaroo (2015) and their music has been featured on CMT’s show Nashville, MTV’s The Real World, and the Esquire Network. Their second full-length album, Sweet Tennessee Honey, was released in 2016 and features appearances by Natalie Maines (Dixie Chicks), Sarah Jarosz, and Jano Rix (The Wood Brothers). Seating for the Smooth Hound Smith performance will be general admission. Tickets are $17.95 and are on sale now at or by the direct link. The performance is presented by Next Act, Inc., and will be the second performance of their 2018 season. The first performance, the grand re-opening, is set for April 14, 2018, will feature Will Kimbrough (and band) and Patrick Sweany (solo acoustic), and sold out in seven days. CLICK HERE. Sorry, comments are closed for this post.

TITLE: The length of non perfect binary 1 error correcting codes QUESTION [4 upvotes]: I am interested in the best known number of code words in binary 1 error correcting codes of length $n$. I am aware of the Hamming code when $n=2^r-1$, but i would like to get lower bounds for other $n$ too. My motivation is completely theoretical and is related to graph theory. REPLY [7 votes]: Here is a table of the best known (linear and non-linear) binary codes for distance 3, for $n \leq 512\,$. Distance 3 is equivalent to being able to correct one error. The table only gives you the number of codewords, but the references given in the table will tell you how to construct the codes themselves. The best known codes for $n$ not in this table can be found by taking the code for the next larger $n$ in the table, repeatedly choosing a coordinate, and retaining only those codewords which have $0$'s (or $1$'s) in that position. You can delete $\ell$ coordinates and retain a $2^{-\ell}$ fraction of the codewords. REPLY [0 votes]: Demazure (in Cours d'algèbre) writes: Let $C$ a binary $t$-error correcting code of length $n$ and let $N$ the number of elements of $C$. Then $N(n + 1) \leqslant 2^n$. Moreover, the following conditions are equivalent: $C$ is perfect $N(n + 1) = 2^n$ there exists $r > 1$ such that $n = 2^r - 1$ and $N = 2^{n - r}$. Proof. Each ball of radius 1 has $n + 1$ elements (its center plus $n$ elements around). All $N$ balls centered in the codewords are disjoint (because $C$ is 1-error correcting) so they cover $N(n + 1)$ elements among the $2^n$ possible words. The equality characterizes the perfect codes, for which $N$ divides $2^n$.

NARK Kenya party leader Martha Karua has spill the beans of what transpired in the 2017 general election. Karua said that Jubilee party from its top led by Uhuru Kenyatta were added votes irregularly to help them win the seats. Here are some information from the Iron lady. Martha Karua says she was rigged out of Kirinyaga Gubernatorial and that Uhuru tried to buy her silence with a CS post that she turned down 😁#KaruaOnPunchline pic.twitter.com/2uCiKhswGd — Kenya West© (@KinyanBoy) August 11, 2019 Martha Karua: There was widespread cheating and rigging in 2017 elections #KaruaOnPunchline pic.twitter.com/mxMKjbrfLj — Lord Abraham Mutai (@ItsMutai) August 11, 2019 Anne Kiguta: What options do you have now? Martha Karua: There will be a complaint against a member of the bench to JSC. She knows herself. There is a criminal complaint against my video evidence in Kerugoya court. #KaruaOnPunchline pic.twitter.com/5a17oHU11X — K24 TV (@K24Tv) August 11, 2019 She has been fighting to unseat Governor Anne Waiguru from Kirinyaga County’s top seat. So far, the courts have said not. @MarthaKarua : The Judiciary said the six months within which to determine my petition should be heard had finished. #KaruaOnPunchline @AnneKiguta pic.twitter.com/tgxDiYXWyV — K24 TV (@K24Tv) August 11, 2019 Karua: There is no clause in the entire Constitution that talks about a remitted petition. It does not talk about when a petition is remitted by a higher court for hearing. #KaruaOnPunchline @AnneKiguta @MarthaKarua pic.twitter.com/5ApCwCHGdB — K24 TV (@K24Tv) August 11, 2019 Martha Karua has held different positions as far as the fight against corruption is concerned. He resigned after 2007 but worked from 2005 under an administration that was slow in the fight against corruption. Those in government and are opposed to the handshake and BBI should ask Karua what she did when in similar situation during Kibaki's tenure. Resign!#KaruaOnPunchline — Daniel Mwaniki (@DanDanielMwanik) August 11, 2019 Anonymous says The truth has always been there and the election thief should know that his regime is a thieving regime founded to thieving. Anonymous says She never said Uhuru rigged, even in kirinyaga she was unable to prove it. Anonymous says She never said Uhuru rigged, even in kirinyaga she was unable to prove it. These days losers refuse to accept results M O says True , let her feels the pain . Anonymous says.

NEW YORK (AP) - The Southeastern Conference isn't No. 1 in one category: revenues. Forbes released a list Wednesday of the estimated annual revenues for the top conferences in college sports (). The SEC, boasting the reigning national champions in football and men's basketball, is fourth. The Big Ten is No. 1 with revenues of $310 million - $250 million of that coming from TV. The Pacific-12 is second at $303 million and the Atlantic Coast Conference third at $293 million. The SEC had revenues of $270 million. Forbes' financial data came from the NCAA, BCS, conferences, news reports and estimates.

Extended family members like grandparents, uncles, aunts, and cousins can be a great support for children. However, these connections tend to get lost when complex family legal matters arise. Here’s what you should know about making your case for the legal right to visitation to a child with whom you have a familial relationship. Understanding Past Cases of Visitation for Extended Family In the past, grandparents could be denied access to their grandchild if the parents wished it so. Evaluations of how this would impact the child’s well being were often never considered, and even loving, supportive grandparents had difficulty getting legal support if the child’s parents wanted to restrict visitation. Unfortunately, this was often used as a tool for manipulation. Understanding Modern Cases of Visitation for Extended Family Currently, the United States has passed legislation that gives grandparents — and sometimes other third parties like uncles, aunts, and older siblings — the right to petition a Florida court for visitation rights. This can be done with or without the consent of the child’s parents. While visitation is not automatically awarded under these laws, it allows third parties to independently bring visitation cases before a judge for a decision. Under What Circumstances Are Grandparents Awarded Visitation? The most common way a grandparent is awarded visitation is after the child’s parents get a divorce or if one parent dies. Sometimes, a grandparent or other third party may be granted legal visitation rights if a parent becomes incarcerated, if the child was born to parents who are unmarried, or if the child previously lived with them. When child abuse, domestic violence, or substance abuse are involved, grandparents who can provide stability for the child are likely to be awarded visitation, even outside of the above circumstances. Steps to Take to Petition for Visitation Rights To start the process of obtaining legal visitation rights to your grandchild, ask your local court for a Petition for Grandparent Visitation form. Fill it out as accurately as possible and submit it to the appropriate office for filing. You’ll be given a hearing date where a judge will go over your case and make a determination on whether visitation should or should not be awarded. If your petition is contested by the parent(s) of the child, the legal process may become more complicated. Reach Out to a Tampa Family Attorney Now Don’t wait to get the help of an experienced family lawyer if you’re the grandparent or extended family member of a child who you’d like to have the legal right to visit with.

Washington D.C., Jun 9, 2020 / 02:00 pm (CNA).- Federal health officials say they hope the resolution of a disability rights case will help ensure hospital patients are deprived of necessary support during treatment, or left to die alone. “We’ve heard too many heart-wrenching stories of people literally dying alone during this crisis,” Roger Severino, head of the HHS Office of Civil Rights told reporters on a conference call on Tuesday. Respecting public health concerns through visitation restrictions should be balanced with other critical needs, Severino said, such as disabled patients having access to support persons and all patients having “access to clergy in their last moments.” The HHS, he said, has already issued guidance that “there should be access to persons in those end-of-life situations.” In a bulletin published on March 28, OCR instructed hospitals and other health care providers to respect “requests for religious accommodations in treatment and access to clergy or faith practices as practicable.” During the call on June 9, HHS announced it had resolved a complaint with the state of Connecticut regarding a 73 year-old woman with aphasia, a condition which limits a person’s ability to communicate, who was admitted to Hartford Hospital without her support person on April 19. Her daughter, Susan Fandacone, told reporters that her mother survived a brain aneurysm 11 years ago, but suffered from short-term memory loss and had lost her voice. When family members rushed her to the hospital, concerned that she had sepsis, they were told that they could not enter with her due to visitor limitations put in place to prevent the spread of the new coronavirus. With her mother’s short-term memory loss and inability to speak, Fandacone said she and her family feared for her mother’s well-being without an advocate to assist her. “She had no opportunity to be able to advocate for herself in any way,” Fandacone said. When “she started to fight for her life,” Fandacone said, “what they chose to do was to tie her down and to sedate her.” Disability rights advocates filed a complaint with the HHS Office of Civil Rights (OCR), alleging that the state limits on hospital visitations during the new coronavirus pandemic did not accommodate the needs of persons with disabilities. Without the presence of support persons to act as an advocate, the groups said, those with disabilities would not have equal access to the health care they needed; with support persons essential for informed consent procedures and communication with doctors and nurses. Worse, Severino told reporters Tuesday, narrowly-tailored visitation policies mean persons with disabilities could be left to die alone—an unacceptable situation. As part of the settlement, Connecticut’s acting health commissioner Deidre Gifford issued an executive order on Monday amending the policy. The order allows for persons with disabilities at short-term hospitals and outpatient clinics, dialysis units, and surgical facilities to have a designated support person with them, so long as that person is asymptomatic or has not tested positive for the coronavirus. Severino said the development was “a big step forward to making sure that people with disabilities are not left alone, and are not left to fend for themselves when reasonable modifications can be.

For 10 years, the Sugaw Creek Recreation Center has fielded a team in the Mecklenburg County Park and Recreation Department’s youth track and field program. With 124 participants this year, the Sugaw Creek Jaguars is one of the largest teams in the county. The Win With Integrity program holds countywide track and field meets from March through May. Considered a recreational-level program, Win With Integrity is open to youths age 4-18 years old and offers a number of running, jumping and throwing events. Once the Win With Integrity season was over, however, some Sugaw Creek athletes were interested in continuing their season on a more competitive level. As a result, head coach Cynthia Smith-Perkins developed an extension of the Jaguars program. For the first time, Sugaw Creek is supporting athletes in USATF and AAU events, allowing them to compete against some of the top runners, jumpers and throwers in the nation. On Aug. 2, Imunique Archie, Kendal McDougald and Cheyla Scott will finish their events at the 2015 USATF National Junior Olympic Track and Field Championships in Jacksonville, Fla. Imunique is a 13-year-old long jumper and middle sprinter. Kendal, 10, is a long-distance runner, and Cheyla, a tall, slender 9-year-old, specializes in the triathlon, a combination of running, jumping and throwing events. “It’s good to keep the kids together,” said Tanya Filmore, a Jaguars and Swift coach and Cheyla Scott’s mother. “We used to have to pick other (competitive) programs. Now we can keep our relays together. The bond is already built.” The Sugaw Creek team, known as the Carolina Swift, is new enough that the athletes continue to run with the status they held last year. Imunique and Kendal compete as unattached, and Cheyla runs for the Charlotte Flights, another local track and field program. All three athletes recently performed well at the Region 3 meet in Hampton, Va., on July 9-12. Imunique, who lives in the Spring Park neighborhood, placed second in the long jump in the 13- to 14-year-old division. “The secret to being a good long jumper is running fast and jumping as far as you can and reaching as far as you can,” said Imunique, a rising eighth-grader at Oaklawn Language Academy. “I’m a fast runner and I have long legs.” Cheyla won the 9- to 10-year-old girls’ triathlon, an event comprised of the 200-meter dash, long jump and shot put. Of the three events, Cheyla won the 200 meters and the high jump and finished second in the shot put. In the open events, Cheyla won the 100 meters, finished second in the high jump and fourth in the 400 meters. She did not compete in the open shot put. Entering the Junior Olympic Nationals, Cheyla was the national points leader in the triathlon. Last year, she achieved all-American status in the high jump, and in 2013 she won national titles in the 100 and 200 meters. “Running is the best, because I’m good at that,” said Cheyla. “But I’m good at high jump, too. I started the triathlon so I can show people I do track and field.” Kendal, who became interested in running long distances to fend off boredom at his brother’s football practices at the University City YMCA, placed fourth in the 800 meters and 1,500 meters. His Jaguar/Swift coaches taught him the proper technique for breathing during a long race, and Kendal’s strategy takes over from there. “I have to pace myself,” he said. “I don’t go too fast in the beginning. (But) I try to not let anyone get too far ahead.” Cheyla and her brother Peyton, another Swift member, also will be competing in the AAU Junior Olympic Nationals in Norfolk, Va., through Aug. 8. Joe Habina is a freelance writer: joehabina@gmail.com.

\begin{document} \begin{abstract} In this paper we describe the oriented Riemannian four-manifolds $M$ for which the Atiyah-Hitchin-Singer or Eells-Salamon almost complex structure on the twistor space ${\mathcal Z}$ of $M$ determines a harmonic map from ${\mathcal Z}$ into its twistor space. \vspace{0,1cm} \noindent 2010 {\it Mathematics Subject Classification}. Primary 53C43, Secondary 58E20, 53C28 \vspace{0,1cm} \noindent {\it Key words: twistor spaces, almost complex structures, harmonic maps} \end{abstract} \thispagestyle{empty} \maketitle \section{Introduction}\label{Intro} The twistor approach has been used for years for studying conformal geometry of four-manifolds by means of complex geometric methods, and in this way many important results have been obtained. Moreover, the twistor spaces endowed with the Atiyah-Hitchin-Singer and Eells-Salamon almost complex structures are interesting geometric objects in their own right whose geometric properties have been studied by many authors. In this paper we look at these structures from the point of view of variational theory. The motivation behind is the fact that if a Riemannian manifold admits an almost complex structure compatible with its metric, it possesses many such structures (cf., for example, \cite{D16, DHM15}). Thus, it is natural to seek criteria that distinguish some of these structures among all. One way to obtain such a criterion is to consider the compatible almost complex structures on a Riemannian manifold $(N,h)$ as sections of its twistor bundle ${\mathcal Z}$. The smooth manifold ${\mathcal Z}$ admits a natural Riemannian metric $h_1$ such that the projection map $({\mathcal Z},h_1)\to (N,h)$ is a Riemannian submersion with totally geodesic fibres. From this point of view, E. Calabi and H. Gluck \cite{CG} have proposed to consider as "the best" those compatible almost complex structures $J$ on $(N,h)$ whose image $J(N)$ in ${\mathcal Z}$ is of minimal volume. They have proved that the standard almost Hermitian structure on the $6$-sphere $S^6$, defined by means of the Cayley numbers, can be characterized by that property. Another criterion has been discussed by C. M. Wood \cite{W1,W2} who has suggested to single out the structures $J$ that are harmonic sections of the twistor bundle ${\mathcal Z}$, i.e. critical points of the energy functional under variations through sections of ${\mathcal Z}$. While the K\"ahler structures are absolute minima of the energy functional, there are many examples of non-K\"ahler structures, which are harmonic sections \cite{W1,W2}. Sufficient conditions for a compatible almost complex structure to be a minimizer of the energy functional and examples of non-K\"ahler minimizers have been given by G. Bor, L. Hern\'andez-Lamoneda and M. Salvai \cite{BLS}. Forgetting the bundle structure of ${\mathcal Z}$, we can also consider compatible almost complex structures that are critical points of the energy functional under variations through all maps $N\to{\mathcal Z}$. These structures are genuine harmonic maps from $(N,h)$ into $({\mathcal Z},h_1)$; we refer to \cite{EL} for basic facts about harmonic maps. The problem when a compatible almost complex structure on a four-dimensional Riemannian manifold is a harmonic map into its twistor space has been studied in \cite{DHM15} (see also \cite{D16}). If the base manifold $N$ is oriented, the twistor space ${\mathcal Z}$ has two connected components often called positive and negative twistor spaces of $(N,h)$; their sections are compatible almost complex structures yielding the orientation and, respectively, the opposite orientation of $N$. Setting $h_t=\pi^{\ast}h+th^v$, $t>0$, where $\pi:{\cal Z}\to N$ is the projection map and $h^v$ is the metric of the fibre, define a $1$-parameter family of Riemannian metrics on ${\cal Z}$ compatible with the almost complex structures $\cal J_1$ and $\cal J_2$ on ${\mathcal Z}$ introduced respectively by Atiyah-Hitchin-Singer \cite{AHS} and Eells-Salamon \cite{ES}. In \cite{DM02} we have found geometric conditions on an oriented four-dimensional Riemannian manifold under which the almost complex structures $\cal J_1$ and $\cal J_2$ on its negative twistor space $({\mathcal Z},h_t)$ are harmonic sections. \smallskip \noindent {\bf Theorem 1.}~ {\it Let $(M,g)$ be an oriented Riemannian $4$-manifold and let $({\cal Z},h_t)$ be its negative twistor space. Then: \noindent $(i)$ The Atiyah-Hitchin-Singer almost complex structure ${\cal J}_1$ on $({\cal Z},h_t)$ is a harmonic section if and only if $(M,g)$ is a self-dual manifold. \noindent $(ii)$ The Eells-Salamon almost complex structure ${\cal J}_2$ on $({\cal Z},h_t)$ is a harmonic section if and only if $(M,g)$ is a self-dual manifold with constant scalar curvature}. \smallskip By a theorem of Atiyah-Hitchin-Singer \cite{AHS} the self-duality of $(M,g)$ is a necessary and sufficient condition for the integrability of the almost complex structure ${\cal J}_1$. In contrast, the almost complex structure ${\cal J}_2$ is never integrable by a result of Eells-Salamon \cite{ES} but it is very useful for constructing harmonic maps. \smallskip The aim of the present paper is to find the four-manifolds for which the almost complex structures $\cal J_1$ and $\cal J_2$ are harmonic maps. More precisely, we prove the following \smallskip \noindent {\bf Theorem 2.}~ {\it Let ${\cal J}_1$ and ${\cal J}_2$ be the Atiyah-Hitchin-Singer and Eells-Salamon almost complex structures on the (negative) twistor space $({\cal Z},h_t)$ of an oriented Riemannian four-manifold $(M,g)$. Each ${\cal J}_k$ (k=1 or 2) is a harmonic map if and only if $(M,g)$ is either self-dual and Einstein, or is locally the product of an open interval in ${\mathbb R}$ and a $3$-dimensional Riemannian manifold of constant curvature}. \smallskip Note that any compact self-dual Einstein manifold with positive scalar curvature is isometric to the $4$-sphere $\mathbb S^4$ or the complex projective space $\mathbb {CP}^2$ with their standard metrics \cite{FK, H81} (see also \cite[Theorem 13.30]{Besse}). In the case of negative scalar curvature, a complete classification is not available yet and the only known compact examples are quotients of the unit ball in $\mathbb C^2$ with the metric of constant negative curvature or the Bergman metric. In contrast, there are many local examples of self-dual Einstein metrics with a prescribed sign of the scalar curvature (cf., e.g., \cite {Der, H95, Le1, Le2, Le3, Ped, Tod}). Note also that every Riemannian manifold that locally is the product of an open interval in ${\mathbb R}$ and a $3$-dimensional Riemannian manifold of constant curvature $c$ is locally conformally flat with constant scalar curvature $6c$. It is not Einstein unless $c=0$, i.e. Ricci flat. The proof of Theorem 2 is based on an explicit formula for the second fundamental form $\widetilde\nabla J_{\ast}$ of a compatible almost complex structure $J$ on a Riemannian manifold considered as a map from the manifold into its twistor space (Proposition 1). In particular, it follows from Theorem 1 mentioned above that if the vertical part of $\mathit{Trace}\widetilde\nabla {\cal J}_{k\,\ast}$ vanishes then the manifold $(M, J)$ is self-dual. This simplifies the formulas for the values of the horizonal part of $\mathit{Trace}\widetilde\nabla {\cal J}_{k\,\ast}$ at vertical and horizontal vectors (Lemmas 1 and 2). Using these formulas we show that the Ricci tensor of $(M,g)$ is parallel and three of its eigenvalues coincide. Thus either $(M,g)$ is Einstein or exactly three of the eigenvalues coincide. In the second case a result in \cite[Lemma 1]{DGM} (essentially due to LeBrun and Apostolov) implies that the simple eigenvalue vanishes, thus $(M,g)$ is locally the product of an interval in $\mathbb R$ and a $3$-manifold of constant curvature. Note also that if $(h_t,{\cal J}_1)$ is a K\"ahler structure, then ${\cal J}_1$ is a totally geodesic map. It is a result of Friedrich-Kurke \cite{FK} that $(h_t,{\cal J}_1)$ is K\"ahler exactly when the base manifold is self-dual and Einstein with positive scalar curvature $12/t$. The necessary and sufficient conditions for ${\cal J}_1$ and ${\cal J}_2$ to be totally geodesic maps will be discussed elsewhere. \smallskip \noindent {\bf Acknowledgment}. We would like to thank the referee whose remarks helped to improve the final version of the paper. \section{Preliminaries}\label{prel} \subsection{The manifold of compatible linear complex structures}\label{CLCS} Let $V$ be a real vector space of even dimension $n=2m$ endowed with an Euclidean metric $g$. Denote by $F(V)$ the set of all complex structures on $V$ compatible with the metric $g$, i.e. $g$-orthogonal. This set has the structure of an imbedded submanifold of the vector space $\mathfrak{so}(V)$ of skew-symmetric endomorphisms of $(V,g)$. The group $O(V)$ of orthogonal transformations of $(V,g)$ acts smoothly and transitively on the set $F(V)$ by conjugation. The isotropy subgroup at a fixed $J\in F(V)$ consists of the orthogonal transformations commuting with $J$. Therefore $F(V)$ can be identified with the homogeneous space $O(2m)/U(m)$. In particular, $dim\,F(V)=m^2-m$. Moreover, $F(V)$ has two connected components. If we fix an orientation on $V$, these components consist of all complex structures on $V$ compatible with the metric $g$ and inducing $\pm$ the orientation of $V$; each of them has the homogeneous representation $SO(2m)/U(m)$. The tangent space of $F(V)$ at a point $J$ consists of all endomorphisms $Q\in \mathfrak{so}(V)$ anti-commuting with $J$ and we have the decomposition \begin{equation}\label{decom} \mathfrak{so}(V)=T_JF(V)\oplus \{S\in\mathfrak{so}(V):~ SJ-JS=0\}. \end{equation} This decomposition is orthogonal with respect to the metric $G(A,B)=-\frac{1}{n}\mathit{Trace} AB$ of $\mathfrak{so}(V)$ (the factor $1/n$ is chosen so that every $J\in F(V)$ to have unit norm). The metric $G$ on $F(V)$ is compatible with the almost complex structure ${\cal J}$ defined by $$ {\cal J}Q=JQ~\mbox{for}~ Q\in T_JF(V). $$ Let $J\in F(V)$ and let $e_1,...,e_{2m}$ be an orthonormal basis of $V$ such that $Je_{2k-1}=e_{2k}$, $k=1,...,m$. Define skew-symmetric endomorphisms $S_{a,b}$, $a,b=1,...,2m$, of $V$ setting $$ S_{a,b}e_c=\displaystyle{\sqrt{\frac{n}{2}}}(\delta_{ac}e_b-\delta_{bc}e_a),\quad c=1,...,2m. $$ The maps $S_{a,b}$, $1\leq a<b\leq 2m$, constitute a $G$-orthonormal basis of $\mathfrak{so}(V)$. Set $$ \begin{array}{c} A_{r,s}=\frac{1}{\sqrt 2}(S_{2r-1,2s-1}-S_{2r,2s}),\quad B_{r,s}=\frac{1}{\sqrt 2}(S_{2r-1,2s}+S_{2r,2s-1}),\\[6pt] r=1,...,m-1,\>s=r+1,...,m. \end{array} $$ Then $\{A_{r,s},B_{r,s}\}$ is a $G$-orthonormal basis of $T_JF(V)$ with $B_{r,s}={\cal J}A_{r,s}$. Denote by $D$ the Levi-Civita connection of the metric $G$ on $F(V)$. Let $X,Y$ be vector fields on $F(V)$ considered as $\mathfrak{so}(V)$-valued functions on $\mathfrak{so}(V)$. By the Koszul formula, for every $J\in F(V)$, \begin{equation}\label{coder} (D_{X}Y)_J=\frac{1}{2}(Y'(J)(X_J)+J\circ Y'(J)(X_J)\circ J) \end{equation} where $Y'(J)\in Hom(\mathfrak{so}(V),\mathfrak{so}(V))$ is the derivative of the function $Y:\mathfrak{so}(V)\to \mathfrak{so}(V)$ at the point $J$. The latter formula easily implies that $(G,{\cal J})$ is a K\"ahler structure on $F(V)$. Note also that the metric $G$ is Einstein with scalar curvature $\frac{m}{2}(m-1)(m^2-m)$ (see, for example,\cite{D05}). \subsection {The four-dimensional case} Suppose that $dim\,V=4$. Then, as is well-known, each of the two connected components of $F(V)$ can be identified with the unit sphere $S^2$. It is often convenient to describe this identification in terms of the space $\Lambda^2V$. The metric $g$ of $V$ induces a metric on $\Lambda^2V$ given by $$ g(x_1\wedge x_2,x_3\wedge x_4)=\frac{1}{2}[g(x_1,x_3)g(x_2,x_4)-g(x_1,x_4)g(x_2,x_3)], $$ the factor $1/2$ being chosen in consistence with \cite{DM91,DM02}. Consider the isomorphisms $\mathfrak{so}(V)\cong \Lambda^2V$ sending $\varphi\in \mathfrak{so}(V)$ to the $2$-vector $\varphi^{\wedge}$ for which $$ 2g(\varphi^{\wedge},x\wedge y)=g(\varphi x,y),\quad x,y\in V. $$ This isomorphism is an isometry with respect to the metric $G$ on $\mathfrak{so}(V)$ and the metric $g$ on $\Lambda^2V$. Given $a\in \Lambda^2V$, the skew-symmetric endomorphism of $V$ corresponding to $a$ under the inverse isomorphism will be denoted by $K_a$. Fix an orientation on $V$ and denote by $F_{\pm}(V)$ the set of complex structures on $V$ compatible with the metric $g$ and inducing $\pm$ the orientation of $V$. The Hodge star operator defines an endomorphism $\ast$ of $\Lambda^2V$ with $\ast^2=Id$. Hence we have the decomposition $$ \Lambda^2V=\Lambda^2_{-}V\oplus\Lambda^2_{+}V $$ where $\Lambda^2_{\pm}V$ are the subspaces of $\Lambda^2V$ corresponding to the $(\pm 1)$-eigen-values of the operator $\ast$. Let $(e_1,e_2,e_3,e_4)$ be an oriented orthonormal basis of $V$. Set \begin{equation}\label{s-basis} s_1^{\pm}=e_1\wedge e_2 \pm e_3\wedge e_4, \quad s_2^{\pm}=e_1\wedge e_3\pm e_4\wedge e_2, \quad s_3^{\pm}=e_1\wedge e_4\pm e_2\wedge e_3. \end{equation} Then $(s_1^{\pm},s_2^{\pm},s_3^{\pm})$ is an orthonormal basis of $\Lambda^2_{\pm}V$. Note that this basis defines an orientation on $\Lambda^2_{\pm}V$, which does not depend on the choice of the basis $(e_1,e_2,e_3,e_4)$ (see, for example, \cite{D16}). We call this orientation "canonical". It is easy to see that the isomorphism $\varphi\to \varphi^{\wedge}$ identifies $F_{\pm}(V)$ with the unit sphere $S(\Lambda^2_{\pm}V)$ of the Euclidean vector space $(\Lambda^2_{\pm}V,g)$. Under this isomorphism, if $J\in F_{\pm}(V)$, the tangent space $T_JF(V)=T_JF_{\pm}(V)$ is identified with the orthogonal complement $({\Bbb R} J^{\wedge})^{\perp}$ of the space ${\Bbb R}J^{\wedge}$ in $\Lambda^2_{\pm}V$. Consider the $3$-dimensional Euclidean space $(\Lambda^2_{\pm}V,g)$ with its canonical orientation and denote by $\times$ the usual vector-cross product in it. Then if $a,b\in\Lambda^2_{\pm}V$, the isomorphism $\Lambda^2V\cong \mathfrak{so}(V)$ sends $a\times b$ to $\pm\frac{1}{2}[K_a,K_b]$. Thus, if $J\in F_{\pm}(V)$ and $Q\in T_JF(V)=T_JF_{\pm}(V)$, we have \begin{equation}\label{calJ} ({\cal J}Q)^{\wedge}=\pm (J^{\wedge}\times Q^{\wedge}). \end{equation} \subsection{The twistor space of an even-dimensional Riemannian manifold} Let $(N,g)$ be a Riemannian manifold of dimension $n=2m$. Denote by $\pi:{\cal Z}\to N$ the bundle over $N$ whose fibre at every point $p\in N$ consists of all compatible complex structures on the Euclidean vector space $(T_pN,g_p)$. This is the associated bundle $$ {\cal Z}=O(N)\times_{O(n)}F({\Bbb R^n}) $$ where $O(N)$ is the principal bundle of orthonormal frames on $N$ and $F({\Bbb R^n})$ is the manifold of complex structures on ${\Bbb R^n}$ compatible with its standard metric. The manifold ${\cal Z}$ is called the twistor space of $(N,g)$. The Levi-Civita connection of $(N,g)$ gives rise to a splitting ${\cal V}\oplus {\cal H}$ of the tangent bundle of any bundle associated to $O(N)$ into vertical and horizontal parts. This allows one to define a natural $1$-parameter family of Riemannain metrics $h_t$, $t>0$, on the manifold ${\cal Z}$ sometimes called "the canonical variation of the metric of $N$" (\cite[Chapter 9 G]{Besse}). For every $J\in {\cal Z}$, the horizontal subspace ${\cal H}_J$ of $T_J{\cal Z}$ is isomorphic via the differential $\pi_{\ast J}$ to the tangent space $T_{\pi(J)}N$ and the metric $h_t$ on ${\cal H}_J$ is the lift of the metric $g$ on $T_{\pi(J)}N$, $h_t|{\cal H}_J=\pi^{\ast} g$. The vertical subspace ${\cal V}_J$ of $T_J{\cal Z}$ is the tangent space at $J$ to the fibre of the bundle ${\cal Z}$ through $J$ and $h_t|{\cal V}_J$ is defined as $t$ times the metric $G$ of this fibre. Finally, the horizontal space ${\cal H}_J$ and the vertical space ${\cal V}_J$ are declared to be orthogonal. Then, by the Vilms theorem \cite{V}, the projection $\pi:({\cal Z},h_t)\to (N,g)$ is a Riemannian submersion with totally geodesic fibres (this can also be proved directly). The manifold ${\cal Z}$ admits two almost complex structures ${\cal J}_1$ and ${\cal J}_2$ defined in the case $dim\,N=4$ by Atiyah-Hitchin-Singer \cite{AHS} and Eells-Salamon \cite{ES}, respectively. On a vertical space ${\cal V}_J$, ${\cal J}_1$ is defined to be the complex structure ${\cal J}_J$ of the fibre through $J$, while ${\cal J}_2$ is defined as the conjugate complex structure, i.e. ${\cal J}_{2}|{\cal V}_J=-{\cal J}_J$. On a horizontal space ${\cal H}_J$, ${\cal J}_1$ and ${\cal J}_2$ are both defined to be the lift to ${\cal H}_J$ of the endomorphism $J$ of $T_{\pi(J)}N$. The almost complex structures ${\cal J}_1$ and ${\cal J}_2$ are compatible with each metric $h_t$. \smallskip Consider ${\cal Z}$ as a submanifold of the bundle $$\pi:A(TN)=O(N)\times_{O(n)}so(n)\to N$$ of skew-symmetric endomorphisms of $TN$. The inclusion of ${\cal Z}$ into $A(TN)$ is fibre-preserving and, for every $J\in{\cal Z}$, the horizontal subspace ${\cal H}_J$ of $T_J{\cal Z}$ coincides with the horizontal subspace of $T_JA(TN)$ since the inclusion of $F({\Bbb R^n})$ into $so(n)$ is $O(n)$-equivariant. The Levi-Civita connection of $(N,g)$ determines a connection on the bundle $A(TN)$, both denoted by $\nabla$, and the corresponding curvatures are related by $$ (R(X,Y)\varphi)(Z)=R(X,Y)\varphi(Z)-\varphi(R(X,Y)Z) $$ for $\varphi\in A(TN)$, $X,Y,Z\in TN$. The curvature operator ${\cal R}$ is the self-adjoint endomorphism of $\Lambda ^{2}TN$ defined by $$ g({\cal R}(X\land Y),Z\land T)=g(R(X,Y)Z,T),\quad X,Y,Z,T\in TN. $$ Let us note that we adopt the following definition for the curvature tensor $R$ : $R(X,Y)=\nabla_{[X,Y]}-[\nabla_{X},\nabla_{Y}]$. \smallskip Let $(U,x_1,...,x_n)$ be a local coordinate system of $N$ and $E_1,...,E_n$ an orthonormal frame of $TN$ on $U$. Define sections $S_{ij}, 1\leq i,j\leq n$, of $A(TN)$ by the formula \begin{equation}\label{Sij} S_{ij}E_l=\sqrt{\frac{n}{2}}(\delta_{il}E_j - \delta_{lj}E_i),\quad l=1,...,n. \end{equation} Then $S_{ij}, i<j,$ form an orthonormal frame of $A(TN)$ with respect to the metric $G(a,b)=\displaystyle{-\frac{1}{n}}\mathit{Trace}(a\circ b)\, ;a,b\in A(TN)$. Set $$\tilde x_{i}(a)=x_{i}\circ\pi(a),\, y_{jl}(a)=\sqrt{\frac{2}{n}}G(a,S_{jl}), j<l,$$ for $a\in A(TN)$. Then $(\tilde x_{i},y_{jl})$ is a local coordinate system of the manifold $A(TN)$. Setting $y_{lk}=-y_{kl}$ for $l\geq k$, we have $aE_j=\sum_{l=1}^n y_{jl}E_l$, j=1,...,n. For each vector field $$X=\textstyle{\sum\limits_{i=1}^n} X^{i}\frac{\partial}{\partial x_i}$$ on $U$, the horizontal lift $X^h$ on $\pi^{-1}(U)$ is given by \begin{equation}\label{eq 3.1} X^{h}=\textstyle{\sum\limits_{i=1}^n }(X^{i}\circ\pi)\frac{\partial}{\partial\tilde x_i}- \textstyle{\sum\limits_{j<l}\sum\limits_{p<q}} y_{pq}G(\nabla_{X}S_{pq},S_{jl})\circ\pi\frac{\partial}{\partial y_{jl}}. \end{equation} Let $a\in A(TN)$ and $p=\pi(a)$. Then (\ref{eq 3.1}) implies that, under the standard identification $T_{a}A(TN)\cong A(T_{p}N)$ (=the skew-symmetric endomorphisms of $(T_{p}N,g_p)$), we have \begin{equation}\label{Lie-2} [X^{h},Y^{h}]_{a}=[X,Y]^h_a + R(X,Y)a. \end{equation} \smallskip Farther we shall often make use of the isomorphism $A(TN)\cong\Lambda^{2}TN$ that assigns to each $a\in A(T_{p}N)$ the 2-vector $a^{\wedge}$ for which $$ 2g(a^{\wedge},X\wedge Y)=g(aX,Y),\quad X,Y\in T_{p}N, $$ the metric on $\Lambda^{2}TN$ being defined by $$ g(X_1\wedge X_2,X_3\wedge X_4)= \frac{1}{2}[g(X_1,X_3)g(X_2,X_4)-g(X_1,X_4)g(X_2,X_3)]. $$ \begin{lemma}\label{R[a,b]} {\rm (\cite{D05})} For every $a,b\in A(T_pN)$ and $X,Y\in T_pN$, we have \begin{equation}\label{Rw} G(R(X,Y)a,b)=\frac{2}{n}g(R([a,b]^{\wedge})X,Y). \end{equation} \end{lemma} {\bf Proof}. Let $E_1,...,E_n$ be an orthonormal basis of $T_pN$. Then $$ [a,b]=\frac{1}{2}\textstyle{\sum\limits_{i,j=1}^n} g([a,b]E_i,E_j)E_i\wedge E_j. $$ Therefore $$ \begin{array}{l} g(R([a,b]^{\wedge})X,Y)\\[6pt] =\frac{1}{2}\sum\limits_{i,j=1}^n g(R(X,Y)E_i,E_j)[g(abE_i,E_j)+g(aE_i,bE_j)]\\[6pt] =\frac{1}{2}\sum\limits_{i=1}^ng(R(X,Y)E_i,abE_i)\\[6pt] \hspace{2.5cm}+\frac{1}{2}\sum\limits_{i,j,k=1}^n g(R(X,Y)E_i,E_j)g(E_i,aE_k)g(E_j,bE_k)\\[6pt] =-\frac{1}{2}\sum\limits_{i=1}^n g(a(R(X,Y)E_i),bE_i)+\frac{1}{2}\sum\limits_{k=1}^n g(R(X,Y)aE_k,bE_k)\\[6pt] =\frac{n}{2}G(R(X,Y)a,b). \end{array} $$ For every $J\in{\cal Z}$, we identify the vertical space ${\cal V}_J$ with the subspace of $A(T_{\pi(J)}N)$ of skew-symmetric endomorphisms anti-commuting with $J$. Then, for every section $K$ of the twistor space ${\cal Z}$ near a point $p\in N$ and every $X\in T_pN$, the endomorphism $\nabla_{X}K$ of $T_pN$ belongs to the vertical space ${\cal V}_{K(p)}$. Lemma~\ref{R[a,b]} implies that \begin{equation}\label{r-r} h_t(R(X,Y)J,V)=\frac{2t}{n}g(R([J,V]^{\wedge})X,Y)= \frac{4t}{n}g(R((J\circ V)^{\wedge})X,Y). \end{equation} \smallskip Denote by $D$ the Levi-Civita connection of $({\cal Z},h_t)$. \begin{lemma}\label{LC} {\rm (\cite{D05,DM91})} If $X,Y$ are vector fields on $N$ and $V$ is a vertical vector field on ${\cal Z}$, then \begin{equation}\label{D-hh} (D_{X^h}Y^h)_{J}=(\nabla_{X}Y)^h_{J}+\frac{1}{2}R_{p}(X\wedge Y)J \end{equation} \begin{equation}\label{D-vh} (D_{V}X^h)_{J}={\cal H}(D_{X^h}V)_{J}=-\frac{2t}{n}(R_{p}((J\circ V_J)^{\wedge})X)_J^h \end{equation} where $J\in{\cal Z}$, $p=\pi(J)$, and ${\cal H}$ means "the horizontal component". \end{lemma} {\bf Proof}. Identity (\ref{D-hh}) follows from the Koszul formula for the Levi-Civita connection and (\ref{Lie-2}). Let $W$ be a vertical vector field on ${\cal Z}$. Then $$ h_t(D_{V}X^h,W)=-h_t(X^h,D_{V}W)=0 $$ since the fibres are totally geodesic submanifolds, so $D_{V}W$ is a vertical vector field. Therefore $D_{V}X^h$ is a horizontal vector field. Moreover, $[V,X^h]$ is a vertical vector field, hence $D_{V}X^h={\cal H}D_{X^h}V$. Thus $$ h_t(D_{V}X^h,Y^h)=h_t(D_{X^h}V,Y^h)=-h_t(V,D_{X^h}Y^h). $$ Now (\ref{D-vh}) follows from (\ref{D-hh}) and (\ref{r-r}). \section{The second fundamental form of an almost Hermitian structure as a map into the twistor space} Now let $J$ be an almost complex structure on the manifold $N$ compatible with the metric $g$. Then $J$ can be considered as a section of the bundle $\pi:{\cal Z}\to N$. Thus we have a map $J:(N,g)\to ({\cal Z},h_t)$ between Riemannian manifolds. Let $J^{\ast}T{\cal Z}\to N$ be the pull-back of the bundle $T{\cal Z}\to {\cal Z}$ under the map $J:N\to{\cal Z}$. Then we can consider the differential $J_{\ast}:TN\to T{\cal Z}$ as a section of the bundle $Hom(TN,J^{\ast}T{\cal Z})\to N$. Denote by $\widetilde D$ the connection on $J^{\ast}T{\cal Z}$ induced by the Levi-Civita connection $D$ on $T{\cal Z}$. The Levi Civita connection $\nabla$ on $TN$ and the connection $\widetilde D$ on $J^{\ast}T{\cal Z}$ induce a connection $\widetilde\nabla$ on the bundle $Hom(TN,J^{\ast}T{\cal Z})$. Recall that the second fundamental form of the map $J$ is, by definition, $$ \widetilde\nabla J_{\ast}. $$ The map $J: (N,g)\to ({\cal Z},h_t)$ is harmonic if and only if $$ \mathit{Trace}_{g}\widetilde\nabla J_{\ast}=0. $$ Recall also that the map $J: (N,g)\to ({\cal Z},h_t)$ is totally geodesic exactly when $\widetilde\nabla J_{\ast}=0$. Any (local) section $a$ of the bundle $A(TN)$ determines a (local) vertical vector field $\widetilde a$ defined by $$ {\widetilde a}_I=\frac{1}{2}(a(p)+I\circ a(p)\circ I),\quad p=\pi(I). $$ Thus if $aE_j=\sum\limits_{l=1}^n a_{jl}E_l$, $$ {\widetilde a}=\textstyle{\sum\limits_{j<l}}\widetilde{a}_{jl}\frac{\partial}{\partial y_{jl}} $$ where $$ \widetilde{a}_{jl}=\frac{1}{2}[a_{jl}\circ\pi+\textstyle{\sum\limits_{r,s=1}^n} y_{jr}(a_{rs}\circ\pi)y_{sl}] $$ The next lemma is "folklore". \begin{lemma}\label{Xh-a til} If $I\in{\cal Z}$ and $X$ is a vector field on a neighbourhood of the point $p=\pi(I)$, then $$ [X^h,\widetilde a]_I=(\widetilde{\nabla_{X}a})_I. $$ \end{lemma} {\bf Proof}. Take an orthonormal frame $E_1,...,E_n$ of $TN$ near the point $p$ such that $\nabla E_i|_p=0$, $i=1,...,n$. Let $(\tilde x_{i},y_{jl})$, $1\leq j<l\leq n$, be the local coordinates of $A(TN)$ defined by means of a local coordinate system $x$ of $N$ at $p$ and the frame $E_1,..,E_n$. Then, by (\ref{eq 3.1}), $$ [X^h,\frac{\partial}{\partial y_{jl}}]_I=0,\quad j,l=1,...,n, \quad X^{h}=\textstyle{\sum\limits_{i=1}^n} X^{i}(p)(\frac{\partial}{\partial\tilde x_i})_I. $$ It follows that $$ [X^h,\widetilde a]_I=\frac{1}{2}[X_p(a_{jl})+\textstyle{\sum\limits_{k,m=1}^n} y_{jk}(I)X_p(a_{km})y_{ml}(I)]=(\widetilde{\nabla_{X}a})_I $$ since $$(\nabla_{X_p}a)(E_i)=\textstyle{\sum\limits_{l=1}^n} X_p(a_{jl})(E_l)_p.$$ \smallskip \noindent {\bf Remark 1}. For every $I\in{\cal Z}$, we can find local sections $a_1,...,a_{m^2-m}$ of $A(TN)$ whose values at $p=\pi(I)$ constitute a basis of the vertical space ${\cal V}_I$ and such that $\nabla a_{\alpha}|_p=0$, $\alpha=1,...,m^2-m$. Let $\widetilde a_{\alpha}$ be the vertical vector fields determined by the sections $a_{\alpha}$. Lemma~\ref{Xh-a til} and the Koszul formula for the Levi-Civita connection imply that $h_t(D_{\widetilde a_{\alpha}}\widetilde a_{\beta}, X^h)_I=0$ for every $X\in T_pN$. Therefore, for every vertical vector fields $U$ and $V$, the covariant derivative ($D_{U}V)_I$ at $I$ is a vertical vector. It follows that the fibres of the twistor bundle are totally geodesic submanifolds. \smallskip Let $I\in{\cal Z}$ and let $U,V\in{\cal V}_I$. Take sections $a$ and $b$ of $A(TN)$ such that $a(p)=U$, $b(p)=V$ for $p=\pi(I)$. Let $\widetilde a$ and $\widetilde b$ be the vertical vector fields determine by the sections $a$ and $b$. Taking into account the fact that the fibre of ${\cal Z}$ through the point $I$ is a totally geodesic submanifold and applying formula (\ref{coder}) we get \begin{equation}\label{a-b-til} (D_{\widetilde a}\widetilde b)_I=\frac{1}{4}[UVI+IVU+I(UVI+IVU)I]=0. \end{equation} \begin{lemma}\label{v-frame} For every $p\in N$, there exists a $h_t$-orthonormal frame of vertical vector fields $\{V_{\alpha}:~\alpha=1,...,m^2-m\}$ such that \noindent $(1)$ $\quad (D_{V_{\alpha}}V_{\beta})_{J(p)}=0$,~~ $\alpha,\beta=1,...,m^2-m$. \noindent $(2)$ $\quad$ If $X$ is a vector field near the point $p$, $[X^h,V_{\alpha}]_{J(p)}=0$. \noindent $(3)$ $\quad$ $\nabla_{X_p}(V_{\alpha}\circ J)\perp {\cal V}_{J(p)}$ \end{lemma} {\bf Proof}. Let $E_1,...,E_n$ be an orthonormal frame of $TN$ in a neighbourhood $N$ of $p$ such that $J(E_{2k-1})_p=(E_{2k})_p$, $k=1,...,m$, and $\nabla E_l|_p=0$, $l=1,...,n$. Define sections $S_{ij}$,$1\leq i,j\leq n$ by (\ref{Sij}) and, as in Section~\ref{prel}, set $$ \begin{array}{c} A_{r,s}=\frac{1}{\sqrt 2}(S_{2r-1,2s-1}-S_{2r,2s}),\quad B_{r,s}=\frac{1}{\sqrt 2}(S_{2r-1,2s}+S_{2r,2s-1}),\\[6pt] r=1,...,m-1,\>s=r+1,...,m. \end{array} $$ Then $\{(A_{r,s})_p,(B_{r,s})_p\}$ is a $G$-orthonormal basis of the vertical space ${\cal V}_{J(p)}$ such that $(B_{r,s})_p=J(A_{r,s})_p$. Note also that $\nabla A_{r,s}|_p=\nabla B_{r,s}|_p=0$. Let $\widetilde A_{r,s}$ and $\widetilde B_{r,s}$ be the vertical vector fields on ${\cal Z}$ determined by the sections $A_{r,s}$ and $B_{r,s}$ of $A(TN)$. These vector fields constitute a frame of the vertical bundle ${\cal V}$ in a neighbourhood of the point $J(p)$. Consider $\widetilde A_{r,s}\circ J$ as a section of $A(TN)$. Then, if $X\in T_pN$, we have $$ \begin{array}{c} \nabla_{X_p}(\widetilde A_{r,s}\circ J)=\frac{1}{2}\{(\nabla_{X_p}J)\circ (A_{r,s})_p\circ J_p+J_p\circ (A_{r,s})\circ (\nabla_{X_p}J)\}\\[6pt] =\frac{1}{2}\{-\nabla_{X_p}J\circ J_p\circ (A_{r,s})_p+J_p\circ (A_{r,s})\circ (\nabla_{X_p}J)\}\\[6pt] =\frac{1}{2}[(B_{r,s})_p,\nabla_{X_p}J] \end{array} $$ The endomorphisms $(B_{r,s})_p$ and $\nabla_{X_p}J$ of $T_pN$ belong to ${\cal V}_{J(p)}$, so they anti-commute with $J(p)$, hence their commutator commutes with $J(p)$. Therefore, in view of (\ref{decom}), the commutator $[(B_{r,s})_p,\nabla_{X_p}J]$ is $G$-orthogonal to the vertical space at $J(p)$. Thus $$ \nabla_{X_p}(\widetilde A_{r,s}\circ J)\perp{\cal V}_{J(p)} $$ and similarly $\nabla_{X_p}(\widetilde B_{r,s}\circ J)\perp{\cal V}_{J(p)}$. It is convenient to denote the elements of the frame $\{\widetilde A_{r,s},\widetilde B_{r,s}\}$ by $\{\widetilde V_1,...,\widetilde V_{m^2-m}\}$. In this way we have a frame of vertical vector fields near the point $J(p)$ with the property $(3)$ of the lemma. Properties $(1)$ and $(2)$ are also satisfied by this frame according to (\ref{a-b-til}) and Lemma~\ref{Xh-a til}, respectively. In particular, $$ (\widetilde V_{\gamma})_{J(p)}(h_t(\widetilde V_{\alpha},\widetilde V_{\beta}))=0,\quad \alpha,\beta,\gamma=1,...,m^2-m. $$ Note also that, in view of (\ref{D-vh}), $$ {\cal V}(D_{X^h}\widetilde V_{\alpha})_{J(p)}=[X^h,\widetilde V_{\alpha}]_{J(p)}=0, $$ hence $$ X^h_{J(p)}(h_t(\widetilde V_{\alpha},\widetilde V_{\beta}))=0. $$ Now it is clear that the $h_t$-orthonormal frame $\{V_1,...,V_{m^2-m}\}$ obtained from $\{\widetilde V_1,...,\widetilde V_{m^2-m}\}$ by the Gram-Schmidt process has the properties stated in the lemma. \begin{prop}\label{covder-dif} For every $X,Y\in T_pN$, $p\in N$, $$ \begin{array}{c} \widetilde\nabla J_{\ast}(X,Y) =\displaystyle{\frac{1}{2}}{\cal V}(\nabla^{2}_{XY}J + \nabla^{2}_{YX}J)\\[8pt] -\displaystyle{\frac{2t}{n}}[(R((J\circ \nabla_{X}J)^{\wedge})Y)^h_{J(p)} +(R((J\circ\nabla_{Y}J)^{\wedge})X)^h_{J(p)}] \end{array} $$ where $\nabla^{2}_{XY}J=\nabla_X\nabla_Y J-\nabla_{\nabla_XY}J$ is the second covariant derivative of $J$. \end{prop} {\bf Proof}. Extend $X$ and $Y$ to vector fields in a neighbourhood of the point $p$. Let $V_1,...,V_{m^2-m}$ be a $h_t$-orthonormal frame of vertical vector fields with the properties $(1)$ - $(3)$ stated in Lemma~\ref{v-frame}. We have $$ J_{\ast}\circ Y=Y^h\circ J+\nabla_{Y}J=Y^h\circ J+\textstyle{\sum\limits_{\alpha=1}^{m^2-m}}h_t(\nabla_{Y}J,V_\alpha\circ J)(V_{\alpha}\circ J), $$ hence $$ \begin{array}{c} {\widetilde D}_{X}(J_{\ast}\circ Y)=(D_{J_{\ast}X}Y^h)\circ J+ \sum\limits_{\alpha=1}^{m^2-m}h_t(\nabla_{Y}J,V_{\alpha})(D_{J_{\ast}X}V_{\alpha})\circ J\\[8pt] +t\sum\limits_{\alpha=1}^{m^2-m}G(\nabla_X\nabla_{Y} J,V_{\alpha}\circ J)(V_{\alpha}\circ J) \end{array} $$ This, in view of Lemma~\ref{LC}, implies $$ \begin{array}{c} \widetilde{D}_{X_p}(J_{\ast}\circ Y)=(\nabla_{X}Y)^h_{J(p)}+\displaystyle{\frac{1}{2}}R(X\wedge Y)J(p)\\[8pt]-\displaystyle{\frac{2t}{n}}(R((J\circ\nabla_{X}J)^{\wedge})Y)^h_{J(p)} +t\sum\limits_{\alpha=1}^{m^2-m}G(\nabla_{X_p}\nabla_{Y}J,V_{\alpha}\circ J)_{p}V_{\alpha}(J(p))\\[8pt] -\displaystyle{\frac{2t}{n}}(R((J\circ\nabla_{Y}J)^{\wedge})X)^h_{J(p)}\\[8pt] =(\nabla_{X_p}Y)^h_{J(p)} +\displaystyle{\frac{1}{2}}{\cal V}(\nabla_{X_p}\nabla_{Y}J +\nabla_{Y_p}\nabla_{X}J) +\frac{1}{2}\nabla_{[X,Y]_p}J \\[8pt] -\displaystyle{\frac{2t}{n}}[R((J\circ\nabla_{X}J)^{\wedge})Y)^h_{J(p)} +(R((J\circ\nabla_{Y}J)^{\wedge})X)^h_{J(p)}]. \end{array} $$ It follows that $$ \begin{array}{c} \widetilde\nabla J_{\ast}(X,Y)=\widetilde D_{X_p}(J_{\ast}\circ Y)- (\nabla_{X}Y)^h_{\sigma}-\nabla_{\nabla_{X_p}Y} J\\[8pt] =\displaystyle{\frac{1}{2}}{\cal V}(\nabla_{X_p}\nabla_{Y}J-\nabla_{\nabla_{X_p}Y} J +\nabla_{Y_p}\nabla_{X}J--\nabla_{\nabla_{Y_p}X} J)\\[8pt] -\displaystyle{\frac{2t}{n}}[R((J\circ\nabla_{X}J)^{\wedge})Y)^h_{J(p)} +(R((J\circ\nabla_{Y}J)^{\wedge})X)^h_{J(p)}]. \end{array} $$ \begin{cotmb} If $(N,g,J)$ is K\"ahler, the map $J:(N,g)\to ({\cal Z},h_t)$ is a totally geodesic isometric imbedding. \end{cotmb} \noindent {\bf Remark 2}. By a result of C. Wood \cite{W1,W2}, $J$ is a harmonic almost complex structure, i.e. a harmonic section of the twistor space $({\cal Z},h_t)\to (N,g)$ if and only if $[J,\nabla^{\ast}\nabla J]=0$ where $\nabla^{\ast}\nabla$ is the rough Laplacian. This, in view of the decomposition (\ref{decom}), is equivalent to the condition that the vertical part of $\nabla^{\ast}\nabla J=-\mathit{Trace}\nabla^2 J$ vanishes. Thus, by Proposition~\ref{covder-dif}, $ J$ is a harmonic section if and only if $$ {\cal V} \mathit{Trace}\widetilde\nabla J_{\ast}=0. $$ \section{The Atiyah-Hitchin-Singer and Eells-Salamon almost complex structures as harmonic sections} Let $(M,g)$ be an oriented Riemannian manifold of dimension four. The twistor space of such a manifold has two connected components, which can be identified with the unit sphere subbundles ${\cal Z}_{\pm}$ of the bundles $\Lambda^2_{\pm}TM\to M$, the eigensubbundles of the bundle $\pi:\Lambda^2TM\to M$ corresponding to the eigenvalues $\pm 1$ of the Hodge star operator. The sections of ${\cal Z}_{\pm}$ are the almost complex structures on $M$ compatible with the metric and $\pm$-orientation of $M$. The spaces ${\cal Z}_{+}$ and ${\cal Z}_{-}$ are called the "positive" and the "negative" twistor space of $(M,g)$. The Levi-Civita connection $\nabla$ of $M$ preserves the bundles $\Lambda^2_{\pm}TM$, so it induces a metric connection on each of them denoted again by $\nabla$. The horizontal distribution of $\Lambda^2_{\pm}TM$ with respect to $\nabla$ is tangent to the twistor space ${\cal Z}_{\pm}$. Thus we have the decomposition $T{\cal Z}_{\pm}={\cal H}\oplus {\cal V}$ of the tangent bundle of ${\cal Z}_{\pm}$ into horizontal and vertical components. The vertical space ${\cal V}_{\tau}=\{V\in T_{\tau}{\cal Z}_{\pm}:~ \pi_{\ast}V=0\}$ at a point $\tau\in{\cal Z}$ is the tangent space to the fibre of ${\cal Z}_{\pm}$ through $\tau$. Considering $T_{\tau}{\cal Z}_{\pm}$ as a subspace of $T_{\tau}(\Lambda^2_{\pm}TM)$ (as we shall always do), ${\cal V}_{\tau}$ is the orthogonal complement of $\tau$ in $\Lambda^2_{\pm}T_{\pi(\tau)}M$. Given $a\in \Lambda^2TM$, define, as in Sec.~\ref{CLCS}, an endomorphism $K_a$ of $T_{\pi(a)}M$ by $$ g(K_aX,Y)=2g(a,X\wedge Y),\quad X,Y\in T_{\pi(a)}M. $$ For $\sigma\in{\cal Z}_{\pm}$, $K_{\sigma}$ is a complex structure on the vector space $T_{\pi(\sigma)}M$ compatible with the metric and $\pm$ the orientation. Denote by $\times$ the vector-cross product in the $3$-dimensional oriented Euclidean space $(\Lambda^2_{\pm}T_pM,g_p)$, $p\in M$. It is easy to show that if $a,b\in\Lambda^2_{\pm}TM$ \begin{equation}\label{KaKb} K_a\circ K_b=-g(a,b)Id \pm K_{a\times b}. \end{equation} This identity implies that for every vertical vector $V\in {\cal V}_{\sigma}$ and every $X,Y\in T_{\pi(\sigma)}M$ \begin{equation}\label{aux} g(V,X\wedge K_{\sigma}Y)=g(V,K_{\sigma}X\wedge Y)=g(\sigma\times V,X\wedge Y). \end{equation} Note also that, in view of (\ref{calJ}), the Atiyah-Hitchin-Singer and Eells-Salamon almost complex structures ${\cal J}_1$ and ${\cal J}_2$ at a point $\sigma\in {\cal Z}_{\pm}$ can be written as $$ \begin{array}{c} {\cal J}_kV=\pm(-1)^{k+1}\sigma\times V~~\mbox{for}~~V\in{\cal V}_{\sigma},\\[6pt] {\cal J}_kX^h_{\sigma}=K_{\sigma}X~~\mbox{for}~~X\in T_{\pi(\sigma)}M,\\[6pt] k=1,2. \end{array} $$ Denote by ${\cal B}:\Lambda^2TM\to \Lambda^2TM$ the endomorphism corresponding to the traceless Ricci tensor. If $s$ denotes the scalar curvature of $(M,g)$ and $\rho:TM\to TM$ the Ricci operator, $g(\rho(X),Y)=Ricci(X,Y)$, we have $$ {\cal B}(X\wedge Y)=\rho(X)\wedge Y+X\wedge\rho(Y)-\frac{s}{2}X\wedge Y. $$ Let ${\cal W}: \Lambda^2TM\to \Lambda^2TM$ be the endomorphism corresponding the Weyl conformal tensor. Denote the restriction of ${\cal W}$ to $\Lambda^2_{\pm}TM$ by ${\cal W}_{\pm}$, so ${\cal W}_{\pm}$ sends $\Lambda^2_{\pm}TM$ to $\Lambda^2_{\pm}TM$ and vanishes on $\Lambda^2_{\mp}TM$. It is well known that the curvature operator decomposes as (see e.g. \cite[Chapter 1 H]{Besse}) $$ {\cal R}=\frac{s}{6}Id+{\cal B}+{\cal W}_{+}+{\cal W}_{-}. $$ Note that this differ by the factor $1/2$ from \cite{Besse} because of the factor $1/2$ in our definition of the induced metric on $\Lambda^2TM$. The Riemannian manifold $(M,g)$ is Einstein exactly when ${\cal B}=0$. It is called self-dual (anti-self-dual) if ${\cal W}_{-}=0$ (resp. ${\cal W}_{+}=0$). By a well-known result of Atiyah-Hitchin-Singer \cite{AHS}, the almost complex structure ${\cal J}_1$ on ${\cal Z}_{-}$ (resp. ${\cal Z}_{+}$) is integrable (i.e. comes from a complex structure) if and only if $(M,g)$ is self-dual (resp. anti-self-dual). On the other hand the almost complex structure ${\cal J}_2$ is never integrable by a result of Eells-Salamon \cite{ES} but nevertheless it is very useful in harmonic map theory. \smallskip {\noindent}{\bf Convention}. In what follows the negative twistor space ${\cal Z}_{-}$ will be called simply "the twistor space" and will be denoted by ${\cal Z}$. Changing the orientation of $M$ interchanges the roles of $\Lambda^2_{+}TM$ and $\Lambda^2_{-}TM$, respectively of ${\cal Z}_{+}$ and ${\cal Z}_{-}$. But note that the Fubini-Study metric on ${\Bbb C}{\Bbb P}^2$ is self-dual and not anti-self-dual, so the structure ${\cal J}_1$ on the negative twistor space ${\cal Z}_{-}$ is integrable while on ${\cal Z}_{+}$ it is not. This is one of the reasons to prefer ${\cal Z}_{-}$ rather than ${\cal Z}_{+}$. \smallskip Remark 2, Proposition~\ref{covder-dif} and Theorem~1 imply \begin{cotmb}\label{hs} \noindent $(i)$ ${\cal V} \mathit{Trace}\widetilde\nabla {\cal J}_{1\,\ast}=0$ if and only if $(M,g)$ is self-dual. \noindent $(ii)$ ${\cal V}\mathit{Trace}\widetilde\nabla {\cal J}_{2\,\ast}=0$ if and only if $(M,g)$ is self-dual and with constant scalar curvature. \end{cotmb} \section{The Atiyah-Hitchin-Singer and Eells-Salamon almost complex structures as harmonic maps} In this section we prove Theorem 2, which is the main result of the paper. Note first that the almost complex structure ${\cal J}_k$, $k=1$ or $2$, is a harmonic map if and only if ${\cal V}\mathit{Trace}\widetilde\nabla {\cal J}_{k\,\ast}=0$ and ${\cal H}\mathit{Trace}\widetilde\nabla {\cal J}_{k\,\ast}=0$. By Corollary~\ref{hs} if the vertical part of $\mathit{Trace}\widetilde\nabla {\cal J}_{k\,\ast}$ vanishes, then the manifold $(M,g)$ is self dual. According to Proposition~\ref{covder-dif} ${\cal H}\mathit{Trace}\widetilde\nabla {\cal J}_{k\,\ast}=0$, $k=1,2$, if and only if for every $\sigma\in{\cal Z}$ and every $F\in T_{\sigma}{\cal Z}$ $$ \mathit{Trace}_{h_t}\,\{T_{\sigma}{\cal Z}\ni A\to h_t(R_{\cal Z}(({\cal J}_{k}\circ D_{A}{\cal J}_k)^{\wedge})A),F)\}=0. $$ Set for brevity $$ Tr_k(F)=\mathit{Trace}_{h_t}\,\{T_{\sigma}{\cal Z}\ni A\to h_t(R_{\cal Z}(({\cal J}_{k}\circ D_{A}{\cal J}_k)^{\wedge})A),F)\}. $$ The next two technical lemmas, giving explicit formulas for $Tr_k(F)$ in the self-dual case, will be proved in the next section. \begin{lemma}\label{tr-ver} Suppose that $(M,g)$ is self-dual. Then, if $\sigma\in{\cal Z}$ and $U\in{\cal V}_{\sigma}$, $$ Tr_k(U)=\frac{t}{4}g({\cal B}(U),{\cal B}(\sigma)), \> k=1,2. $$ \end{lemma} \begin{lemma}\label{tr-horr} Suppose that $(M,g)$ is self-dual. Then, if $X\in T_{p}M$, $p=\pi(\sigma)$, $$ \begin{array}{c} Tr_k(X^h_{\sigma})=[1+(-1)^k]\displaystyle{\frac{s(p)}{144}}X(s) +\displaystyle{\frac{1}{12}}(\frac{ts(p)}{6}-2)X(s)\\[6pt] +\mathit{Trace}_{h_t}\,\{{\cal V}_{\sigma}\ni V\to [\displaystyle{\frac{t}{8}}g((\nabla_{X}{\cal B})( V),{\cal B}(V))\\[6pt] \hspace{6.5cm}+(-1)^{k+1}\displaystyle{\frac{ts(p)}{24}}g(\delta{\cal B}(K_VX),V)]\}. \end{array} $$ \end{lemma} \smallskip \noindent {\bf Proof of Theorem~2}. Suppose that ${\cal J}_1$ or ${\cal J}_2$ is a harmonic map. By Corollary~\ref{hs}, $(M,g)$ is self-dual or self-dual with constant scalar curvature. Moreover, $Tr_k(U)=0$ for every vertical vector $U$ and $Tr_k(X^h)=0$ for every horizontal vector $X^h$, $k=1$ or $k=2$. Note that in both cases the first term in the expression for $Tr_k(X^h)$ given in Lemma~\ref{tr-horr} vanishes: $$ [1+(-1)^k]\displaystyle{\frac{s(p)}{144}}X(s)=0,\quad k=1,2. $$ By Lemma~\ref{tr-ver}, for every $p\in M$ and every orthonormal basis $v_1,v_2,v_3$ of $\Lambda^2_{-}T_pM$, $g({\cal B}(v_i),{\cal B}(v_j))=0$, $i,j=1,2,3$, $i\neq j$. This implies $g({\cal B}(v_i),{\cal B}(v_i))=g({\cal B}(v_j),{\cal B}(v_j))$, $i\neq j$. It follows that the function ${\cal Z}_p\ni\sigma\to ||{\cal B}(\sigma)||^2$ is constant on the fibre ${\cal Z}_p$ of ${\cal Z}$ at $p$. Thus we have a smooth function $f$ on $M$ such that $f(p)=||{\cal B}(\sigma)||^2$ for every $\sigma\in{\cal Z}_p$. It follows that \begin{equation}\label{Xf} X(f)=2g((\nabla_{X}{\cal B})(\sigma),{\cal B}(\sigma)) \end{equation} for every tangent vector $X\in T_pM$. Let $E_1,...,E_4$ be an oriented orthonormal basis of $T_pM$ consisting of eigenvectors of $\rho$. Denote by $\lambda_1,...,\lambda_4$ the corresponding eigenvalues. We have $\lambda_1+\lambda_2+\lambda_3+\lambda_4=s$ and \begin{equation}\label{cal B} {\cal B}(X\wedge Y)=\rho(X)\wedge Y+X\wedge\rho(Y)-\frac{s}{2}X\wedge Y. \end{equation} Define $s_i^{+}$ and $s_i=s_i^{-}$, $i=1,2,3$, as in (\ref{s-basis}) by means of the basis $E_1,...,E_4$. Then $$ \begin{array}{c} {\cal B}(s_1)=(\lambda_1+\lambda_2-\displaystyle{\frac{s}{2}})s_1^{+},\quad {\cal B}(s_2)=(\lambda_1+\lambda_3-\displaystyle{\frac{s}{2}})s_2^{+},\\[6pt] {\cal B}(s_3)=(\lambda_1+\lambda_4-\displaystyle{\frac{s}{2}})s_3^{+}. \end{array} $$ Therefore $||{\cal B}(\cdot)||^2=const$ on the fibre ${\cal Z}_p$ if and only if $$ |\lambda_1+\lambda_2-\displaystyle{\frac{s}{2}}|=|\lambda_1+\lambda_3-\displaystyle{\frac{s}{2}}| =|\lambda_1+\lambda_4-\displaystyle{\frac{s}{2}}|, $$ i.e. if and only if, at every point $p\in M$, three eigenvalues of ${\rho}$ coincide. Moreover, $$ 3f(p)=||{\cal B}(s_1)||^2+||{\cal B}(s_2)||^2+||{\cal B}(s_3)||^2=||\rho||^2-\frac{s^2(p)}{4}. $$ and, by (\ref{Xf}), it follows that $$ \mathit{Trace}_{h_t}\,\{{\cal V}_{\sigma}\ni V\to g((\nabla_{X}{\cal B})(V),{\cal B}(V))\}=\frac{1}{3}X(||\rho||^2)-\frac{s(p)X(s)}{6}. $$ Fix a tangent vector $X\in T_pM$ and denote by $P$ the symmetric bilinear form on $\Lambda^2_{-}T_pM$ corresponding to the quadratic form \begin{equation}\label{P} P(a,a)=\frac{ts(p)}{24}g(\delta{\cal B}(K_aX),a). \end{equation} Set $$ \psi=-(\frac{ts(p)}{144}+\frac{1}{6})X(s)+\frac{t}{24}X(||\rho||^2). $$ Then for every $\sigma\in{\cal Z}_p$ and every $V\in{\cal V}_{\sigma}$ with $||V||_g=1$ we have \begin{equation}\label{tr-last} Tr_k(X^h_{\sigma})=(-1)^{k+1}[\frac{1}{t}P(V,V)+\frac{1}{t}P(\sigma\times V,\sigma\times V)]+\psi. \end{equation} Let $\{s_1,s_2,s_3\}$ be an orthonormal basis of $\Lambda^2_{-}T_pM$. Take $$ \sigma=\frac{1}{\sqrt{y_1^2+y_2^2+y_3^2}}(y_1s_1+y_2s_2+y_3s_3) $$ for $(y_1,y_2,y_3)\in{\Bbb R}^3$ with $y_1\neq 0$. Set $$ V=\frac{1}{\sqrt{y_1^2+y_2^2}}(-y_2s_1+y_1s_2). $$ Then $$ \sigma\times V=\frac{1}{\sqrt{(y_1^2+y_2^2)(y_1^2+y_2^2+y_3^2)}}(-y_1y_3s_1-y_2y_3s_2+(y_1^2+y_2^2)s_3). $$ Now varying $(y_1,y_2,y_3)$ we see from (\ref{tr-last}) that the identity $Tr_k(X^h_{\sigma})=0$ implies $$ P(s_i,s_j)=0,\quad (-1)^{k+1}\frac{1}{t}[P(s_i,s_i)+P(s_j,s_j)]+\psi=0,\quad i,j=1,2,3,\>i\neq j. $$ Since $P(s_i,s_j)=0$, $i\neq j$, for every orthonormal basis, we have $P(s_i,s_i)=P(s_j,s_j)$. Suppose that $s(p)\neq 0$. Then, by the latter identity, $$ g(\delta{\cal B}(K_{s_i}X),s_i)=g(\delta{\cal B}(K_{s_j}X),s_j),\quad i,j=1,2,3. $$ Take an oriented orthonormal basis $E_1,...,E_4$ of $T_pM$ and, using it, define $s_i=s_i^{-}$, $i=1,2,3$. Then $g(\delta{\cal B}(K_{s_1}X),s_1)=g(\delta{\cal B}(K_{s_2}X),s_2)$ for every $X\in T_pM$. This, in view of (\ref{KaKb}), gives $$ -g(\delta{\cal B}(X),s_1)=g(\delta{\cal B}(K_{s_3}X),s_2),\quad X\in T_pM. $$ Applying the latter identity for the basis $E_3,E_4,E_1,E_2$ we get $$ g(\delta{\cal B}(X),s_1)=g(\delta{\cal B}(K_{s_3}X),s_2). $$ Hence $g(\delta{\cal B}(X),s_1)=0$. Similarly $g(\delta{\cal B}(X),s_2)=(\delta{\cal B}(X),s_3)=0$. Therefore for every $X\in T_pM$ and $a\in\Lambda^2_{-}T_pM$ $$ g(\delta{\cal B}(X),a)=0. $$ Then, by (\ref{P}), $P(a,a)=0$ for every $a\in\Lambda^2_{-}T_pM$. Thus, we see from (\ref{tr-last}) that the condition $Tr_k(X^h_{\sigma})=0$ for every $\sigma\in{\cal Z}$, $X\in T_{\pi(\sigma)}M$ is equivalent to the identities $$ g(\delta{\cal B}(X),\sigma)=0,\quad \psi=0. $$ Identity (\ref{cal B}) implies that for every $X\in T_pM$ and every orthonormal basis $E_1,..,E_4$ of $T_pM$ $$ \delta{\cal B}(X)=\delta\rho\wedge X-\textstyle{\sum\limits_{m=1}^4}[E_m\wedge(\nabla_{E_m}\rho)(X)-\frac{1}{2}E_m(s)E_m\wedge X]. $$ Therefore the identity $g(\delta{\cal B}(X),\sigma)=0$ is equivalent to $$ g(\delta\rho,K_{\sigma}X)+\textstyle{\sum\limits_{m=1}^4}g((\nabla_{E_m}\rho)(X),K_{\sigma}E_m)+\frac{1}{2}(K_{\sigma}X)(s)=0. $$ This is equivalent to \begin{equation}\label{newtr} \textstyle{\sum\limits_{m=1}^4}g((\nabla_{E_m}\rho)(K_{\sigma}E_m),X)=0 \end{equation} since $g(\delta\rho,Z)=-\frac{1}{2}Z(s)$ by the second Bianchi identity and the Ricci operator $\rho$ is $g$-symmetric. Let $r(X,Y)$ be the Ricci tensor and set $$ dr(X,Y,Z)=(\nabla_{Y}r)(Z,X)-(\nabla_{Z}r)(Y,X). $$ Thus $$ dr(X,Y,Z)=g((\nabla_{Y}\rho)(Z),X)-g((\nabla_{Z}\rho)(Y),X). $$ The left-hand side of (\ref{newtr}) clearly does not depend on the choice of the basis $(E_1,...,E_4)$. So, take an oriented orthonormal basis $(E_1,...,E_4)$ such that $E_2=K_{\sigma}E_1$ and $E_4=-K_{\sigma}E_3$. Then $$ dr(X,E_1,E_2)-dr(X,E_3,E_4)=\textstyle{\sum\limits_{m=1}^4}g((\nabla_{E_m}\rho)(K_{\sigma}E_m),X). $$ Denote by $W_{-}$ the $4$-tensor corresponding to the operator ${\cal W}_{-}$, $$ W_{-}(X,Y,Z,T)=g({\cal W}_{-}(X\wedge Y),Z\wedge T). $$ Then the second Bianchi identity implies $$ dr(X,E_1,E_2)-dr(X,E_3,E_4)=-2[\delta W_{-}(X,E_1,E_2)-\delta W_{-}(X,E_3,E_4)]. $$ Since $(M,g)$ is self-dual, we see from the latter identity that identity (\ref{newtr}) is always satisfied. The above identity shows also that \begin{equation}\label{dr=0} dr(X,\sigma)=0,\quad \sigma\in{\cal Z},\> X\in T_{\pi(\sigma)}M. \end{equation} Let $\lambda_1(p)\leq \lambda_2(p)\leq\lambda_3(p)\leq \lambda_4(p)$ be the eigenvalues of the symmetric operator $\rho_p:T_pM\to T_pM$ in the ascending order. It is well-known that the functions $\lambda_1,...,,\lambda_4$ are continuous (see, e.g. \cite[Chapter Two, \S 5.7 ]{K} or \cite[Chapter I, \S 3]{R}). We have seen that, at every point of $M$, at least three eigenvalues of the operator $\rho$ coincide. The set $U$ of points at which exactly three eigenvalues coincide is open by the continuity of $\lambda_1,...,\lambda_4$. For every $p\in U$ denote the simple eigenvalue of $\rho$ by $\lambda(p)$ and the triple eigenvalue by $\mu(p)$, so the spectrum of $\rho$ is $(\lambda,\mu,\mu,\mu)$ with $\lambda(p)\neq \mu(p)$ for every $p\in U$. As is well-known, the implicit function theorem implies that the function $\lambda$ is smooth. It is also well-known that, in a neighbourhood of every point $p$ of $U$, there is a (smooth) unit vector field $E_1$ which is an eigenvector of $\rho$ corresponding to $\lambda$. (for a proof see \cite[Chapter 9, Theorem 7]{L}). Fix $p\in U$ and choose local vector fields $E_2,E_3,E_4$ such that $(E_1,E_2,E_3,E_4)$ is an oriented orthonormal frame. Let $\alpha$ be the dual $1$-form to $E_1$, $\alpha(X)=g(E_1,X)$. Then $$ r(X,Y)=(\lambda-\mu)\alpha(X)\alpha(Y)-\mu g(X,Y) $$ in a neighbourhood of $p$. Note that the function $\mu=\frac{1}{3}(s-\lambda)$ is also smooth. Hence the identity $\delta r=-\frac{1}{2}ds$ reads as \begin{equation}\label{delta r} \begin{array}{c} -E_1(\lambda-\mu)\alpha(X)-X(\mu)+(\lambda-\mu)[\delta\alpha.\alpha(X)-(\nabla_{E_1}\alpha)(X)]\\[6pt] =-\frac{1}{2}[X(\lambda)+3X(\mu)],\quad X\in TU. \end{array} \end{equation} Let $s_i=s_i^{-}$, $i=1,2,3$, be defined by means of $E_1,...,E_4$. Taking into account that $(\nabla_{X}\alpha)(E_1)=0$, we easily see that the identities $dr(E_k,s_1)=0$, $k=1,2,3,4$, give \begin{equation}\label{dr-s} \begin{array}{c} (\lambda-\mu)[(\nabla_{E_1}\alpha)(E_2)-(\nabla_{E_3}\alpha)(E_4)+(\nabla_{E_4}\alpha)(E_3)]-E_2(\lambda)=0\\[6pt] (\lambda-\mu)(\nabla_{E_2}\alpha)(E_2)-E_1(\mu)=0,\quad (\lambda-\mu)(\nabla_{E_2}\alpha)(E_3)-E_4(\mu)=0\\[6pt] (\lambda-\mu)(\nabla_{E_2}\alpha)(E_4)+E_3(\mu)=0. \end{array} \end{equation} The identities obtained from the latter ones by cycle permutations of $E_2,E_3,E_4$ also hold as a consequence of the identities $dr(E_k,s_2)=0$ and $dr(E_k,s_3)=0$. Thus \begin{equation}\label{jj} (\lambda-\mu)(\nabla_{E_j}\alpha)(E_j)=E_1(\mu),\quad j=2,3,4. \end{equation} Hence \begin{equation}\label{delta-alpha} (\lambda-\mu)\delta\alpha=-3E_1(\mu) \end{equation} Moreover, we have $$ (\nabla_{E_3}\alpha)(E_4)=E_2(\mu),\quad (\nabla_{E_4}\alpha)(E_3)=-E_2(\mu) $$ and the first identity of (\ref{dr-s}) gives $$ (\lambda-\mu)(\nabla_{E_1}\alpha)(E_2)=E_2(\lambda)+2E_2(\mu). $$ On the other hand identity (\ref{delta r}) implies $$ (\lambda-\mu)(\nabla_{E_1}\alpha)(E_2)=\frac{1}{2}E_2(\lambda)+\frac{1}{2}E_2(\mu). $$ It follows that $$ 0=\frac{1}{2}E_2(\lambda)+\frac{3}{2}E_2(\mu)=\frac{1}{2}E_2(s), $$ so $E_2(s)=0$. Similarly, $E_3(s)=0$ and $E_4(s)=0$. Identity (\ref{delta r}) for $X=E_1$ together with (\ref{delta-alpha}) implies $0=E_1(\lambda)+3E_1(\mu)=E_1(s)$. It follows that the scalar curvature $s$ is locally constant on $U$. Then identity $\psi=0$ implies that $||\rho||^2$ is locally constant. Thus in a neighbourhood of every point $p\in U$, we have $\lambda+3\mu=a$ and $\lambda^2+3\mu^2=b^2$ where $a$ and $b$ are some constants. It follows that $$ \mu=\frac{1}{12}(3a\pm\sqrt{12b^2-3a^2}). $$ Note that $12b^2-3a^2\neq 0$ since otherwise we would have $\mu=\frac{1}{4}a$, hence $\lambda=a-3\mu=\frac{1}{4}a=\mu$, a contradiction. Since $\mu$ is continuous, we see that $\mu$ is constant, hence $\lambda$ is also constant. Then, by (\ref{jj}), $(\nabla_{E_j}\alpha)(E_j)=0$ for $j=2,3,4$ and the first equation of (\ref{dr-s}) gives $(\nabla_{E_1}\alpha)(E_2)=0$. Similarly $(\nabla_{E_1}\alpha)(E_3)=(\nabla_{E_1}\alpha)(E_4)=0$. Thus $(\nabla_{X}\alpha)(E_j)=0$ for every $X$ and $j=2,3,4$. This and the obvious identity $(\nabla_{X}\alpha)(E_1)=0$ imply that the $1$-form $\alpha$ is parallel. It follows that the restriction of the Ricci tensor to $U$ is parallel. In the interior of the closed set $M\setminus U$ the eigenvalues of the Ricci tensor coincide, hence the metric $g$ is Einstein on this open set. Therefore the scalar curvature $s$ is locally constant on $Int\,(M\setminus U)$ and the Ricci tensor is parallel on it. Thus the Ricci tensor is parallel on the open set $U\cup Int\,(M\setminus U)=M\setminus bU$, where $bU$ stands for the boundary of $U$. Since $M\setminus bU$ is dense in $M$ it follows the Ricci tensor is parallel on $M$. This implies that the eigenvalues $\lambda_1\leq...\leq \lambda_4$ of the Ricci tensor are constant. Thus either $M$ is Einstein or exactly three of the eigenvalues coincide. Since $(M,g)$ is self-dual, in the second case the simple eigenvalue $\lambda$ vanishes by \cite[Lemma 1]{DGM}. Therefore $M$ is locally the product of an interval in ${\Bbb R}$ and a $3$-dimensional manifold of constant curvature. Conversely, suppose that $(M,g)$ is self dual and either Einstein or locally is the product of an interval and a manifold of constant curvature. Then at least three of the eigenvalues of the Ricci tensor coincide which, as we have seen, implies that $||{\cal B}(\cdot)||^2=const$ on every fibre of ${\cal Z}$. It follows that $g({\cal B}(\sigma),{\cal B}({\tau}))=0$ for every $\sigma,\tau\in{\cal Z}$ with $g(\sigma,\tau)=0$. Therefore $Tr_k(U)=0$ for every vertical vector $U$, $k=1,2$, by Lemma~\ref{tr-ver}. Moreover, $Tr_k(X^h)=0$ by Lemma~\ref{tr-horr} since the scalar curvature is constant and $\nabla{\cal B}=0$. \smallskip \noindent {\bf Remark 3}. According to Theorems~1 and 2, the conditions under which ${\mathcal J_1}$ or ${\mathcal J_2}$ is a harmonic section or a harmonic map do not depend on the parameter $t$ of the metric $h_t$. Taking certain special values of $t$, we can obtain a metric $h_t$ with nice properties (cf., for example, \cite{DM91, DGM, M}). \section{Proofs of Lemmas~\ref{tr-ver} and \ref{tr-horr}} Denote by $R_{\cal Z}$ the curvature tensor of the Riemannian manifold $({\cal Z},h_t)$. \smallskip Let $\Omega_{k,t}(A,B)=h_t({\cal J}_kA,B)$ be the fundamental $2$-form of the almost Hermitian manifold $({\cal Z},h_t,{\cal J}_k)$, $k=1,2$. Then, for $A,B,C\in T_{\sigma}{\cal Z}$, $$ h_t({\cal J}_{k}\circ D_{A}{\cal J}_k)^{\wedge},B\wedge C)=-\frac{1}{2}h_t((D_{A}{\cal J}_k)(B),{\cal J}_kC)=-\frac{1}{2}(D_{A}\Omega_{k,t})(B,{\cal J}_kC). $$ \begin{lemma}\label{D-Omega}{\rm (\cite{M})} Let $\sigma\in{\cal Z}$ and $X,Y\in T_{\pi(\sigma)}M$, $V\in{\cal V}_{\sigma}$. Then $$ (D_{X^h_{\sigma}}\Omega_{k,t})(Y^h_{\sigma},V)=\frac{t}{2}[(-1)^kg({\cal R}(V),X\wedge Y)-g({\cal R}(\sigma\times V),X\wedge K_{\sigma}Y)]. $$ $$ (D_{V}\Omega_{k,t})(X^h_{\sigma},Y^h_{\sigma})=\frac{t}{2}g({\cal R}(\sigma\times V),X\wedge K_{\sigma}Y +K_{\sigma}X\wedge Y)+2g(V,X\wedge Y). $$ Moreover, $(D_{A}\Omega_{k,t})(B,C)=0$ when $A,B,C$ are three horizontal vectors at $\sigma$ or at least two of them are vertical. \end{lemma} \begin{cotmb}\label{JDJ} Let $\sigma\in{\cal Z}$, $X\in T_{\pi(\sigma)}M$, $U\in {\cal V}_{\sigma}$. If $E_1,...,E_4$ is an orthonormal basis of $T_{\pi(\sigma)}M$ and $V_1,V_2$ is a $h_t$-orthonormal basis of ${\cal V}_{\sigma}$, $$ \begin{array}{l} ({\cal J}_k\circ D_{X^h_{\sigma}}{\cal J}_k)^{\wedge}=-\displaystyle{\frac{1}{2}}\textstyle{\sum\limits_{i=1}^4\sum\limits_{l=1}^2}[g({\cal R}(\sigma\times V_l),X\wedge E_i)\\[6pt] \hspace{5.5cm} +(-1)^k g({\cal R}(V_l),X\wedge K_{\sigma}E_i)](E_i^h)_{\sigma}\wedge V_l \end{array} $$ $$ \begin{array}{l} ({\cal J}_k\circ D_{U}{\cal J}_k)^{\wedge}=\textstyle{\sum\limits_{1\leq i<j\leq 4}}[\displaystyle{\frac{t}{2}}g({\cal R}(\sigma\times U),E_i\wedge E_j-K_{\sigma}E_i\wedge K_{\sigma}E_j)\\[8pt] \hspace{6cm}-2g(U,E_i\wedge K_{\sigma}E_j)](E_i^h)_{\sigma}\wedge (E_j^h)_{\sigma}. \end{array} $$ \end{cotmb} The sectional curvature of the Riemannian manifold $({\cal Z},h_t)$ can be computed in terms of the curvature of the base manifold $M$ by means of the following formula. \begin{prop}\label{sec}{\rm (\cite{DM91})} Let $E,F\in T_{\sigma}{\cal Z}$ and $X=\pi_{\ast}E$, $Y=\pi_{\ast}F$, $V={\cal V}E$, $W={\cal V}F$. Then $$ \begin{array}{c} h_t(R_{\cal Z}(E,F)E,F)=g(R(X,Y)X,Y)\\[6pt] -tg((\nabla_{X}{\cal R})(X\wedge Y),\sigma\times W)+tg((\nabla_{Y}{\cal R})(X\wedge Y),\sigma\times V)\\[6pt] -3tg({\cal R}(\sigma),X\wedge Y)g(\sigma\times V,W)\\[6pt] -t^2g(R(\sigma\times V)X,R(\sigma\times W)Y)+\displaystyle{\frac{t^2}{4}}||R(\sigma\times W)X+R(\sigma\times V)Y||^2\\[6pt] -\displaystyle{\frac{3t}{4}}||R(X,Y)\sigma||^2+t(||V||^2||W||^2-g(V,W)^2). \end{array} $$ \end{prop} Using this formula, the well-known expression of the Levi-Civita curvature tensor by means of sectional curvatures and differential Bianchi identity one gets the following. \begin{cotmb}\label{RZ} Let $\sigma\in{\cal Z}$, $X,Y,Z,T\in T_{\pi(\sigma)}M$, and $U,V,W\in{\cal V}_{\sigma}$. Then $$ \begin{array}{c} h_t(R_{\cal Z}(X^h,Y^h)Z^h,T^h)_{\sigma}=g(R(X,Y)Z,T)\\[6pt] -\displaystyle\frac{3t}{12}[2g(R(X,Y)\sigma,R(Z,T)\sigma) -g(R(X,T)\sigma,R(Y,Z)\sigma)\\[6pt] +g(R(X,Z)\sigma,R(Y,T)\sigma)]. \end{array} $$ $$ \begin{array}{c} h_t(R_{\cal Z}(X^h,Y^h)Z^h,U)_{\sigma}=-\displaystyle{\frac{t}{2}g(\nabla_{Z}{\cal R}(X\wedge Y),\sigma\times U)}.\\[8pt] h_t(R_{\cal Z}(X^h,U)Y^h,V)_{\sigma}=\displaystyle{\frac{t^2}{4}g(R(\sigma\times V)X,R(\sigma\times U)Y)}\\[6pt] +\displaystyle{\frac{t}{2}g({\cal R}(\sigma),X\wedge Y)g(\sigma\times V,U)}.\\[8pt] h_t(R_{\cal Z}(X^h,Y^h)U,V)_{\sigma} =\displaystyle{\frac{t^2}{4}}[g(R(\sigma\times V)X,R(\sigma\times U)Y)\\[6pt] \hspace{6cm}-g(R(\sigma\times U)X,R(\sigma\times V)Y)]\\[6pt] +tg({\cal R}(\sigma),X\wedge Y)g(\sigma\times V,U). \end{array} $$ $$ h_t(R_{\cal Z}(X^h,U)V,W)=0. $$ \end{cotmb} We have stated in Lemma~\ref{tr-ver} that if $(M,g)$ is self-dual, $$ Tr_k(U)=\frac{t}{4}g({\cal B}(U),{\cal B}(\sigma))~\rm{for}~\rm{ every}~ U\in{\cal V}_{\sigma},\> \sigma\in{\cal Z}. $$ \noindent {\bf Proof of Lemma~\ref{tr-ver}}. Let $E_1,...,E_4$ be an orthonormal basis of $T_pM$, $p=\pi(\sigma)$, such that $E_2=K_{\sigma}E_1$, $E_4=-K_{\sigma}E_3$. Define $s_1=s_1^{-},s_2=s_2^{-},s_3=s_3^{-}$ via (\ref{s-basis}) by means of $E_1,...,E_4$, so that $\sigma=s_1$ and ${\cal V}_{\sigma}=span\{s_2,s_3\}$. Thus $V_1=\frac{1}{\sqrt t}s_2$, $V_2=\frac{1}{\sqrt t}s_3$ is a $h_t$-orthonormal basis of ${\cal V}_{\sigma}$. \smallskip By Corollary~\ref{JDJ}, for every $U\in {\cal V}_{\sigma}$ \begin{equation}\label{TrU} \begin{array}{c} Tr_k(U)=-\displaystyle{\frac{1}{2}}\textstyle{\sum\limits_{i,j=1}^4\sum\limits_{l=1}^2}[g({\cal R}(\sigma\times V_l),E_j\wedge E_i)+(-1)^kg({\cal R}(V_l),E_j\wedge K_{\sigma}E_i]\\[6pt] \times h_t(R_{\cal Z}(E_i^h,V_l)E_j^h,U)\\[6pt] +\displaystyle{\frac{t}{2}}\textstyle{\sum\limits_{l=1}^2}\{g({\cal R}(\sigma\times V_l),s_2)[h_t(R_{\cal Z}(E_1^h,E_3^h)V_l,U)-h_t(R_{\cal Z}(E_4^h,E_2^h)V_l,U)]\\[6pt] \hspace{1.5cm}+g({\cal R}(\sigma\times V_l),s_3)[h_t(R_{\cal Z}(E_1^h,E_4^h)V_l,U)-h_t(R_{\cal Z}(E_2^h,E_3^h)V_l,U)]\}\\[6pt] -2\textstyle{\sum\limits_{1\leq i<j\leq 4}\sum\limits_{l=1}^2}g(V_l,E_i\wedge K_{\sigma}E_j)h_t(R_{\cal Z}(E_i^h,E_j^h)V_l,U) \end{array} \end{equation} We show first that \begin{equation}\label{Isum} \begin{array}{c} \textstyle{\sum\limits_{i,j=1}^4\sum\limits_{l=1}^2}[g({\cal R}(\sigma\times V_l),E_j\wedge E_i)h_t(R_{\cal Z}(E_i^h,V_l)E_j^h,U)\\[8pt] =-\displaystyle{\frac{t}{2}}\mathit{Trace}_{h_t}\{{\cal V}_{\sigma}\ni V\to g({\cal R}(\sigma\times V),{\cal R}(\sigma))g(\sigma\times U,V)\}. \end{array} \end{equation} In order to prove this identity, we note that if $F\in T_{\sigma}{\cal Z}$, $V\in{\cal V}_{\sigma}$ and $a\in \Lambda^2 T_{\pi(\sigma)}M$, the algebraic Bianchi identity implies \begin{equation}\label{a} \begin{array}{c} \textstyle{\sum\limits_{i,j=1}^4} g(a,E_j\wedge E_i)h_t(R_{\cal Z}(E_i^h,V)E_j^h,F)\\[8pt] =-\displaystyle{\frac{1}{2}}\textstyle{\sum\limits_{i,j=1}^4}g(a,E_i\wedge E_j)h_t(R_{\cal Z}(E_i^h,E_j^h)V,F). \end{array} \end{equation} Using the latter identity and Corollary~\ref{RZ} we obtain $$ \begin{array}{c} \sum\limits_{i,j=1}^4\sum\limits_{l=1}^2[g({\cal R}(\sigma\times V_l),E_j\wedge E_i)h_t(R_{\cal Z}(E_i^h,V_l)E_j^h,U)\\[8pt] =-\displaystyle{\frac{t}{2}}\textstyle{\sum\limits_{l=1}^2}g({\cal R}(\sigma\times V_l),{\cal R}(\sigma))g(\sigma\times U,V_l)\\[8pt] =-\displaystyle{\frac{t}{2}}\mathit{Trace}_{h_t}\{{\cal V}_{\sigma}\ni V\to g({\cal R}(\sigma\times V),{\cal R}(\sigma))g(\sigma\times U,V)\}. \end{array} $$ Next, we claim that \begin{equation}\label{IIsum} \textstyle{\sum\limits_{i,j=1}^4\sum\limits_{l=1}^2} g({\cal R}(V_l),E_j\wedge K_{\sigma}E_i) h_t(R_{\cal Z}(E_i^h,V_l)E_j^h,U)=0. \end{equation} For every $V\in{\cal V}_{\sigma}$, we have \begin{equation}\label{2dsum} \begin{array}{c} \sum\limits_{i,j=1}^4g({\cal R}(V),E_j\wedge K_{\sigma}E_i)h_t(R_{\cal Z}(E_i^h,V)E_j^h,U)\\[8pt] =g({\cal R}(V),E_1\wedge E_2)[h_t(R_{\cal Z}(E_1^h,V)E_1^h,U)+h_t(R_{\cal Z}(E_2^h,V)E_2^h,U)]\\[6pt] -g({\cal R}(V),E_3\wedge E_4)[h_t(R_{\cal Z}(E_3^h,V)E_3^h,U)+h_t(R_{\cal Z}(E_4^h,V)E_4^h,U)]\\[6pt] +g({\cal R}(V),E_1\wedge E_3)[h_t(R_{\cal Z}(E_4^h,V)E_1^h,U)+h_t(R_{\cal Z}(E_2^h,V)E_3^h,U)]\\[6pt] +g({\cal R}(V),E_1\wedge E_4)[-h_t(R_{\cal Z}(E_3^h,V)E_1^h,U)+h_t(R_{\cal Z}(E_2^h,V)E_4^h,U)]\\[6pt] +g({\cal R}(V),E_2\wedge E_3)[h_t(R_{\cal Z}(E_4^h,V)E_2^h,U)-h_t(R_{\cal Z}(E_1^h,V)E_3^h,U)]\\[6pt] +g({\cal R}(V),E_4\wedge E_2)[h_t(R_{\cal Z}(E_3^h,V)E_2^h,U)+h_t(R_{\cal Z}(E_1^h,V)E_4^h,U)]\\[6pt] \end{array} \end{equation} Corollary~\ref{RZ} implies that $$ \begin{array}{c} h_t(R_{\cal Z}(E_4^h,V)E_1^h,U)+h_t(R_{\cal Z}(E_2^h,V)E_3^h,U)\\[8pt] =\displaystyle{\frac{t^2}{4}}[g(R(\sigma\times U)E_4,E_2)g({\cal R}(\sigma\times V),s_1)\\[6pt] \hspace{5cm} +g(R(\sigma\times V)E_1,E_3)g({\cal R}(\sigma\times U),s_1)]\\[6pt] -\displaystyle{\frac{t}{2}}g({\cal R}(\sigma),s_3)g(\sigma\times U,V) \end{array} $$ Since $(M,g)$ is self-dual, for every $\tau\in\Lambda^2_{-}T_{\pi(\sigma)}M$, $$ {\cal R}(\tau)=\frac{s}{6}\tau +{\cal B}(\tau) $$ where ${\cal B}(\tau)\in \Lambda^2_{+}T_{\pi(\sigma)}M$. Therefore $$ g({\cal R}(\sigma\times V),s_1)=g({\cal R}(\sigma\times V),\sigma)=0 $$ and $$ g({\cal R}(\sigma\times U),s_1)=0,\quad g({\cal R}(\sigma),s_3)=0. $$ Thus \begin{equation}\label{sim} h_t(R_{\cal Z}(E_4^h,V)E_1^h,U)+h_t(R_{\cal Z}(E_2^h,V)E_3^h,U)=0. \end{equation} Similarly \begin{equation}\label{3ple} \begin{array}{l} -h_t(R_{\cal Z}(E_3^h,V)E_1^h,U)+h_t(R_{\cal Z}(E_2^h,V)E_4^h,U)=0\\[8pt] h_t(R_{\cal Z}(E_4^h,V)E_2^h,U)-h_t(R_{\cal Z}(E_1^h,V)E_3^h,U)=0\\[8pt] h_t(R_{\cal Z}(E_3^h,V)E_2^h,U)+h_t(R_{\cal Z}(E_1^h,V)E_4^h,U)=0. \end{array} \end{equation} Moreover, a straightforward computation gives $$ \begin{array}{c} \textstyle{\sum\limits_{l=1}^2 }\{g({\cal R}(V_l),E_1\wedge E_2)[h_t(R_{\cal Z}(E_1^h,V_l)E_1^h,U)+h_t(R_{\cal Z}(E_2^h,V_l)E_2^h,U)]\\[6pt] -g({\cal R}(V_l),E_3\wedge E_4)[h_t(R_{\cal Z}(E_3^h,V_l)E_3^h,U)+h_t(R_{\cal Z}(E_4^h,V_l)E_4^h,U)]\}\\[8pt] =\displaystyle{\frac{t^2}{8}}\textstyle{\sum\limits_{l=1}^2}g({\cal R}(V_l),s_1^{+})g({\cal R}(\sigma\times U),s_1)g({\cal B}(\sigma\times V_l),s_1^{+})\\[6pt] =0. \end{array} $$ In view of (\ref{2dsum}), the latter identity, (\ref{sim}) and (\ref{3ple}) imply (\ref{IIsum}). \smallskip Using the algebraic Bianchi identity, we see from (\ref{3ple}) that $$ \begin{array}{c} h_t(R_{\cal Z}(E_1^h,E_3^h)V,U)-h_t(R_{\cal Z}(E_4^h,E_2^h)V,U)=0\\[6pt] h_t(R_{\cal Z}(E_1^h,E_4^h)V,U)-h_t(R_{\cal Z}(E_2^h,E_3^h)V,U)=0. \end{array} $$ Hence \begin{equation}\label{IIIsum} \begin{array}{c} \sum\limits_{l=1}^2\{g({\cal R}(\sigma\times V_l),s_2)[h_t(R_{\cal Z}(E_1^h,E_3^h)V_l,U)-h_t(R_{\cal Z}(E_4^h,E_2^h)V_l,U)]\\[6pt] \hspace{1cm}+g({\cal R}(\sigma\times V_l),s_3)[h_t(R_{\cal Z}(E_1^h,E_4^h)V_l,U)-h_t(R_{\cal Z}(E_2^h,E_3^h)V_l,U)]\}\\[6pt] =0. \end{array} \end{equation} Using (\ref{aux}) and Corollary~\ref{RZ}, we get $$ \begin{array}{c} \sum\limits_{1\leq i<j\leq 4}g(V,E_i\wedge K_{\sigma}E_j)h_t(R_{\cal Z}(E_i^h,E_j^h)V,U)\\[8pt] =\displaystyle{\frac{t^2}{4}}\textstyle{\sum\limits_{1\leq i<j\leq 4}}g(\sigma\times V,E_i\wedge E_j)[g(R(\sigma\times U)E_i,R(\sigma\times V)E_j)\\[6pt] \hspace{7cm}-g(R(\sigma\times V)E_i,R(\sigma\times U)E_j]\\[9pt] =\displaystyle{\frac{t^2}{8}}\textstyle{\sum\limits_{i=1}^4}g(R(\sigma\times U)E_i,R(\sigma\times V)K_{\sigma\times V}E_i) \end{array} $$ Therefore $$ \begin{array}{c} \textstyle{\sum\limits_{1\leq i<j\leq 4}\sum\limits_{l=1}^2}g(V_l,E_i\wedge K_{\sigma}E_j)h_t(R_{\cal Z}(E_i^h,E_j^h)V_l,U)\\[8pt] =\displaystyle{\frac{t}{8}}\textstyle{\sum\limits_{i,k=1}^4}[g(R(\sigma\times U)E_i,E_k)g(R(s_3)K_{s_3}E_i,E_k)\\[6pt] \hspace{5cm}+g(R(\sigma\times U)E_i,E_k)g(R(s_2)K_{s_2}E_i,E_k)]\\[8pt] =\displaystyle{\frac{t}{8}}[-g({\cal R}(\sigma\times U),\sigma)g({\cal R}(s_3),s_2)+g({\cal R}(\sigma\times U),s_2)g({\cal R}(s_3),s_1)\\[6pt] \hspace{1.2cm}+g({\cal R}(\sigma\times U),\sigma)g({\cal R}(s_2),s_3)-g({\cal R}(\sigma\times U),s_3)g({\cal R}(s_2),s_1)]\\[8pt] \end{array} $$ This, by virtue of the self-duality of $(M,g)$, gives \begin{equation}\label{IVsum} \textstyle{\sum\limits_{1\leq i<j\leq 4}\sum\limits_{l=1}^2}g(V_l,E_i\wedge K_{\sigma}E_j)h_t(R_{\cal Z}(E_i^h,E_j^h)V_l,U)=0. \end{equation} Identities (\ref{TrU}),(\ref{Isum}),(\ref{IIsum}), (\ref{IIIsum}) and (\ref{IVsum}) imply $$ Tr_k(U)=\frac{t}{4}\mathit{Trace}_{h_t}\{{\cal V}_{\sigma}\ni V\to g({\cal R}(\sigma\times V),{\cal R}(\sigma))g(\sigma\times U,V)\},\> k=1,2. $$ Now the lemma follows from the latter identity since $g({\cal R}(\tau),{\cal R}(\sigma))=g({\cal B}(\tau),{\cal B}(\sigma))$ for every $\tau, \sigma$ with $\tau\perp\sigma$. \medskip Recall that, according to Lemma~\ref{tr-horr}, if $(M,g)$ is self-dual $$ \begin{array}{c} Tr_k(X^h_{\sigma})=[1+(-1)^k]\displaystyle{\frac{s(p)}{144}}X(s) +\displaystyle{\frac{1}{12}}(\frac{ts(p)}{6}-2)X(s)\\[6pt] +\mathit{Trace}_{h_t}\,\{{\cal V}_{\sigma}\ni V\to [\displaystyle{\frac{t}{8}}g((\nabla_{X}{\cal B})( V),{\cal B}(V))\\[6pt] \hspace{6.5cm}+(-1)^{k+1}\displaystyle{\frac{ts(p)}{24}}g(\delta{\cal B}(K_VX),V)]\}. \end{array} $$ for $X\in T_{\pi(\sigma)}$, $\sigma\in{\cal Z}$. \smallskip \noindent {\bf Proof of Lemma~\ref{tr-horr}}. Let $s_1=s_1^{-},s_2=s_2^{-},s_3=s_3^{-}$ be the basis of $\Lambda^2_{-}T_pM$, $p=\pi(\sigma)$, defined by means of an oriented orthonormal basis $E_1,...,E_4$ of $T_pM$ such that $E_2=K_{\sigma}E_1$, $E_4=-K_{\sigma}E_3$. Set $V_1=\frac{1}{\sqrt t}s_2$, $V_2=\frac{1}{\sqrt t}s_3$. Then, by Corollary~\ref{JDJ}, $$ \begin{array}{c} Tr_k(X^h_{\sigma})=-\displaystyle{\frac{1}{2}}\textstyle{\sum\limits_{i,j=1}^4\sum\limits_{l=1}^2}[g({\cal R}(\sigma\times V_l),E_j\wedge E_i)+(-1)^kg({\cal R}(V_l),E_j\wedge K_{\sigma}E_i] \\[6pt] \times h_t(R_{\cal Z}(E_i^h,V_l)E_j^h,X^h)\\[6pt] +\sum\limits_{i<j}\sum\limits_{l=1}^2[\displaystyle{\frac{t}{2}}g({\cal R}(\sigma\times V_l),E_i\wedge E_j-K_{\sigma}E_i\wedge K_{\sigma}E_j)-2g(V_l,E_i\wedge K_{\sigma}E_j)]\\[6pt] \times h_t(R_{\cal Z}(E_i^h,E_j^h)V_l,X^h). \end{array} $$ Identity (\ref{a}) and Corollary~\ref{RZ} imply $$ \begin{array}{c} \sum\limits_{i,j=1}^4\sum\limits_{l=1}^2g({\cal R}(\sigma\times V_l),E_j\wedge E_i)h_t(R_{\cal Z}(E_i^h,V_l)E_j^h,X^h)_{\sigma}\\[6pt] =-\displaystyle{\frac{t}{4}}\textstyle{\sum\limits_{l=1}^2}g(\displaystyle{\frac{1}{6}}X(s)\sigma\times V_l+(\nabla_{X}{\cal B})(\sigma\times V_l),\displaystyle{\frac{1}{6}}s(p)\sigma\times V_l+{\cal B})(\sigma\times V_l)\\[6pt] =-\displaystyle{\frac{s(p)}{72}}X(s)-\displaystyle{\frac{t}{4}}\mathit{Trace}_{h_t}\{{\cal V}_{\sigma}\ni V\to g((\nabla_{X}{\cal B})(V),{\cal B}(V))\}, \end{array} $$ where the latter identity follows from the fact that $g((\nabla_{X}{\cal B})(a),b)=0$ for every $a,b\in\Lambda^2_{-}T_pM$ (since the operator ${\cal B}$ sends $\Lambda^2_{-}TM$ into $\Lambda^2_{+}TM$, and the connection $\nabla$ preserves the bundles $\Lambda^2_{\pm}TM$). Taking into account identity (\ref{aux}) and the fact that $$ E_i\wedge E_j-K_{\sigma}E_i\wedge K_{\sigma}E_j\in\Lambda^2_{-}T_{\pi(\sigma)}M, $$ we have $$ \begin{array}{c} \sum\limits_{i<j}\sum\limits_{l=1}^2[\displaystyle{\frac{t}{2}}g({\cal R}(\sigma\times V_l),E_i\wedge E_j-K_{\sigma}E_i\wedge K_{\sigma}E_j)-2g(V_l,E_i\wedge K_{\sigma}E_j)]\\[6pt] \times h_t(R_{\cal Z}(E_i^h,E_j^h)V_l,X^h)_{\sigma}\\[8pt] =(\displaystyle{\frac{ts(p)}{6}}-2)\textstyle{\sum\limits_{i<j}\sum\limits_{l=1}^2}g(\sigma\times V_l,E_i\wedge E_j)h_t(R_{\cal Z}(E_i^h,E_j^h)V_l,X^h)\\[8pt] =\displaystyle{\frac{t}{4}(\frac{ts(p)}{6}-2)}\textstyle{\sum\limits_{l=1}^2}g((\nabla_{X}{\cal R})(\sigma\times V_l),\sigma\times V_l)\\ =\displaystyle{\frac{1}{4}(\frac{ts(p)}{6}-2)\frac{X(s)}{3}}. \end{array} $$ Thus $$ \begin{array}{c} Tr_k(X^h_{\sigma})=(-1)^{k+1}\displaystyle{\frac{1}{2}}\textstyle{\sum\limits_{i,j=1}^4\sum\limits_{l=1}^2}g({\cal R}(V_l),E_j\wedge K_{\sigma}E_ih_t(R_{\cal Z}(E_i^h,V_l)E_j^h,X^h)_{\sigma}\\[8pt] +\displaystyle{\frac{s(p)}{144}}X(s) +\displaystyle{\frac{1}{12}(\frac{ts(p)}{6}}-2)X(s)\\[6pt] +\mathit{Trace}_{h_t}\,\{{\cal V}_{\sigma}\ni V\to \displaystyle{\frac{t}{8}}g((\nabla_{X}{\cal B})(V),{\cal B}(V))\}\\[6pt] \end{array} $$ In order to compute the first summand in the right-hand side of the latter identity, it is convenient to set $C_{ilj}=h_t(R_{\cal Z}(E_i^h,V_l)E_j^h,X^h)_{\sigma}$. Then $$\begin{array}{c} \textstyle{\sum\limits_{i,j=1}^4\sum\limits_{l=1}}g({\cal R}(V_l),E_j\wedge K_{\sigma}E_i)h_t(R_{\cal Z}(E_i^h,V_l)E_j^h,X^h)_{\sigma}\\[6pt] =\displaystyle{\frac{1}{2}}\textstyle{\sum\limits_{l=1}^2}[g({\cal R}(V_l),s_1^{+}+s_1)C_{1l1}-g({\cal R}(V_l),s_3^{+}-s_3)C_{1l3} +g({\cal R}(V_l),s_2^{+}-s_2)C_{1l4}\\[10pt] +g({\cal R}(V_l),s_1^{+}+s_1)C_{2l2}+g({\cal R}(V),s_2^{+}+s_2)C_{2l3}+g({\cal R}(V),s_3^{+}+s_3)C_{2l4}\\[10pt] -g({\cal R}(V),s_3^{+}+s_3)C_{3l1}+g({\cal R}(V),s_2^{+}-s_2)C_{3l2} -g({\cal R}(V),s_1^{+}-s_1)C_{3l3}\\[10pt] g({\cal R}(V),s_2^{+}+s_2)C_{4l1}+g({\cal R}(V),s_3^{+}-s_3)C_{4l2} -g({\cal R}(V),s_1^{+}-s_1)C_{4l4}]\\[6pt] =\displaystyle{\frac{s(p)}{12\sqrt t}}[(-C_{114}+C_{213}-C_{312}+C_{411}) +(C_{123}+C_{224}-C_{321}-C_{422})]\\[10pt] +\displaystyle{\frac{1}2}\textstyle{\sum\limits_{l=1}^2} [g({\cal B}(V_l),s_1^{+})(C_{1l1}+C_{2l2}-C_{3l3}-C_{4l4})\\[10pt] \hspace{1.1cm}+g({\cal B}(V_l),s_2^{+})(C_{1l4}+C_{2l3}+C_{3l2}+C_{4l1})\\[10pt] \hspace{1.5cm}+g({\cal B}(V_l),s_3^{+})(-C_{1l3}+C_{2l4}-C_{3l1}+C_{4l2})]. \end{array} $$ By Corollary~\ref{RZ} $$ \begin{array}{c} -C_{124}+C_{223}-C_{322}+C_{421}\\[6pt] =\displaystyle{\frac{\sqrt t}{2}\textstyle{\sum\limits_{i=1}^4}[-\frac{1}{12}E_i(s)g(E_i,X)+g((\nabla_{E_i}{\cal B})(K_{s_3}E_i\wedge X),s_3)]}. \end{array} $$ For every $i=1,...,4$, $K_{s_3}E_i\wedge X+E_i\wedge K_{s_3}X\in\Lambda^2_{-}T_pM$. Hence $$ g((\nabla_{E_i}{\cal B})(K_{s_3}E_i\wedge X+E_i\wedge K_{s_3}X),s_3)=0. $$ It follows that $$ -C_{124}+C_{223}-C_{322}+C_{421}=\displaystyle{\frac{\sqrt t}{2}[-\frac{1}{12}X(s)+g(\delta{\cal B}(K_{s_3}X),s_3)]}. $$ Similarly $$ \begin{array}{c} C_{133}+C_{234}-C_{331}-C_{432}\\[6pt] =\displaystyle{\frac{\sqrt t}{2}[-\frac{1}{12}X(s)+g(\delta{\cal B}(K_{s_2}X),s_2)]}. \end{array} $$ Hence $$ \begin{array}{c} (-C_{124}+C_{223}-C_{322}+C_{421}) +(C_{133}+C_{234}-C_{331}-C_{432})\\[8pt] =\displaystyle{\frac{\sqrt t}{2}}[-\frac{1}{6}X(s)+t\mathit{Trace}_{h_t}\,\{{\cal V}_{\sigma}\ni V\to g(\delta{\cal B}(K_VX),V)\}. \end{array} $$ Set for short $$ \begin{array}{c} \Sigma(E_1,...,E_4)=\textstyle{\sum\limits_{l=1}^2} [g({\cal B}(V_l),s_1^{+})(C_{1l1}+C_{2l2}-C_{3l3}-C_{4l4})\\[10pt] \hspace{0.8cm}+g({\cal B}(V_l),s_2^{+})(C_{1l4}+C_{2l3}+C_{3l2}+C_{4l1})\\[10pt] \hspace{1.2cm}+g({\cal B}(V_l),s_3^{+})(-C_{1l3}+C_{2l4}-C_{3l1}+C_{4l2})]. \end{array} $$ Under this notation, we have $$ \begin{array}{c} Tr_k(X^h_{\sigma})=[1+(-1)^k]\displaystyle{\frac{s(p)}{144}}X(s) +\displaystyle{\frac{1}{12}}(\frac{ts(p)}{6}-2)X(s)\\[6pt] +\mathit{Trace}_{h_t}\,\{{\cal V}_{\sigma}\ni V\to [\displaystyle{\frac{t}{8}}g((\nabla_{X}{\cal B})(V),{\cal B}(V))\\[6pt] \hspace{6.5cm}+(-1)^{k+1}\displaystyle{\frac{ts(p)}{24}}g(\delta{\cal B}(K_VX),V)]\}.\\[6pt] +(-1)^{k+1}\displaystyle{\frac{1}{2}}\Sigma(E_1,...,E_4). \end{array} $$ In particular, the sum $\Sigma(E_1,...,E_4)$ does not depend on the choice of the oriented orthonormal basis $E_1,...,E_4$ (clearly it does not depend on the choice of the $h_t$-orthonormal basis $V_1,V_2$ of ${\cal V}_{\sigma}$ as well). Since $$ \Sigma(E_3,E_4,E_1,E_2)=-\Sigma(E_1,E_2,E_3,E_4), $$ it follows that $$ \Sigma(E_1,E_2,E_3,E_4)=0. $$ This proves the lemma.

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second round preview: Bruins vs. Canadiens by amuir29 Email Posted: Thu May. 1, 2014 Updated: Sat Jun. 14, 2014 By Allan MuirThe second round begins, fittingly, with two storied Original Six franchises renewing their bitter, bruising rivalry for the 34th time. Montreal holds an overall edge of 24-9, but Boston, which won the Stanley Cup in 2011, has advanced from seven of their last 11 postseason encounters.Regular season recapsMontreal wins series, 3-1Dec. 5: Montreal 2, Boston 1Jan. 30: Montreal 4, Boston 1Mar. 12: Boston 4, Montreal 1Mar. 24: Montreal 2, Boston 1 (SO)Notable injuriesBruins: D Dennis Seidenberg (ACL/MCL, indefinite), D Adam McQuaid (strained quad, indefinite), F Chris Kelly (lower body, day-to-day), D Corey Potter (lower body, day-to-day)Canadiens: F Alex Galchenyuk (lower body, day-to-day), F Travis Moen (upper body, day-to-day)Keys to a Bruins victoryThe formula is simple for the Bruins: get pucks to the net. Boston has finished in the top-five for shots fired during each of the past four seasons and the B's like to create their opportunities in bulk, using their size and strength to generate second and third chances off the rebound. All eyes will be on Milan Lucic, but the difference makers could be Carl Soderberg and Justin Florek on the third line. Both gave the Red Wings fits with their net drive in the first round. Patrice Bergeron was world class against Detroit--there aren't 10 forwards better than him at this time of year--but David Krejci has to be more effective. After leading the playoffs in scoring during two of the past three seasons, he was held to just two assists by the tight checking Wings.Even without Dennis Seidenberg and Adam McQuaid, the B's have a depth advantage on a defense corps that was the best in the Eastern Conference. Their blueliners are poised, smart and make it very tough for the opposition to get good looks. Dougie Hamilton was a revelation in the first round with his poise and timing. Tory Krug can create offense off the rush at even strength or on the power play, and Kevan Millar is a big moment waiting to happen. He keeps it simple, but can impact a game with a big hit or timely block.Ultimately, though, it'll come down to goaltending. Tuukka Rask was terrific in his last two starts against the Habs, allowing just two goals on 58 shots, but he's struggled against them in the past. Rebound control will be key.CAZENEUVE: The 10 Greatest Moments in Bruins-Canadiens Postseason HistoryKeys to a Canadiens victoryNetminder Carey Price is up against the Vezina Trophy favorite, so he'll need to put a very average first-round performance (.904 save percentage) behind him. He can be better, as his Sochi gold medal and career bests in save percentage (.927) and GAA (2.32) attest, but if he falters, the Habs have a proven Bruins-killer in backup Peter Budaj. They have another one up front in the deadline acquisition of Thomas Vanek, who has 30 goals and 62 points in 55 career games against Boston. Playing on the top line alongside Max Pacioretty (39 goals during the regular season) and David Desharnais, Vanek could be the game-breaker that Montreal needs. Rene Bourque was barely recognizable in the first round. He tied for the team lead with three goals while playing a robust power forward game. Can he keep it up? No such questions for Brendan Gallagher, who led the Canadiens in scoring (five points) and number of Lightning players annoyed (all of 'em) in the first round. He'll be a factor.On the back end, Alexei Emelin brings a hint of menace that will keep Boston's forwards on high alert, and P.K. Subban can electrify with his play at both ends of the ice...if he maintains his discipline. That'll be an issue for this entire team. The Habs were shorthanded just seven times against Tampa Bay, but allowed two goals on those chances (a 71.4 percent kill rate). Meanwhile, Boston feasted on the power play vs. the Wings, scoring six goals on 16 opportunities for a league-leading 37.5 percent success rate.The pickBruins in 6. I don't know what it is about the B's, but there always seems to be a good reason to bet against them. In this case, their history against the Habs seems to work pretty well...but I just can't do it. It's time to give Boston its due, to recognize the depth, experience and system that made this team the league's best during the regular season. The Bruins give the opposition so little that, while they might lose a game or two, it's hard to see the Habs beating them four-out-of-seven. Look for Boston's strength down the middle to be the difference.Series schedule(All times Eastern; * if necessary)Thursday, May 1: 7:30 p.m. Montreal at Boston NBCSN, CBC, RDSSaturday, May 3: 12:30 p.m. Montreal at Boston NBC, CBC, RDSTuesday, May 6: 7 p.m. Boston at Montreal CBC, RDS, NBCSNCOMPLETE SECOND ROUND SCHEDULEMORE BREAKDOWNS: Blackhawks vs. Wild | Ducks vs. Kings | Rangers vs. Penguins Promoted Stories Comments

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\bibliographystyle{alpha} \maketitle \begin{abstract} We review briefly the existing vertex-operator-algebraic constructions of various tensor category structures on module categories for affine Lie algebras. We discuss the results that were first conjectured in the work of Moore and Seiberg and led us to the construction of the modular tensor category structure in the positive integral level case. Then we review the existing constructions and results in the following three cases: (i) the level plus the dual Coxeter number is not a nonnegative rational number, (ii) the level is a positive integer and (iii) the level is an admissible number. We also present several open problems. \end{abstract} \section{Introduction} In 1988, Moore and Seiberg \cite{MS1} \cite{MS2} discovered that rational conformal field theories have properties analogous to the properties of tensor categories satisfying additional conditions. The results of Moore and Seiberg were obtained based on the operator product expansion and modular invariance of chiral vertex operators (intertwining operators). Mathematically, these two fundamental properties had been deep and important conjectures for many years. But it was the Verlinde formula and modular tensor category structure in the case of affine Lie algebras at positive integral levels (the two most well-known consequences of these two conjectures mentioned above) that first attracted the attentions of mathematicians. To use the tensor category structures to study the representation theory of affine Lie algebras and to solve related problems in algebra, topology, mathematical physics and other related areas, the first mathematical problem is to give constructions of these tensor categories. Kazhdan and Lusztig \cite{KL1}--\cite{KL5} first constructed a rigid braided tensor category structure on a category $\mathcal{O}_{\ell}$ of modules for the affine Lie algebra $\hat{\mathfrak{g}}$ of a finite-dimensional complex simple Lie algebra $\mathfrak{g}$ in the case that the level $\ell$ plus the dual Coxeter number is not a nonnegative rational number. Then several works, including those of Beilinson, Mazur and Feigin \cite{BFM} and Lepowsky and the author \cite{HL-affine}, constructed a braided tensor category structure using different approaches on a category $\widetilde{O}_{\ell}$ of $\hat{\mathfrak{g}}$-modules in the case that the level $\ell$ is a positive integer. The modular tensor category structure on $\widetilde{O}_{\ell}$ in the case that $\ell$ is a positive integer was constructed by the author \cite{H-rigidity} as a special case of a general construction of the modular tensor category structures conjectured by Moore and Seiberg \cite{MS2} for vertex operator algebras corresponding to rational conformal field theories. After a gap was filled later in \cite{F3} using the Verlinde formula proved by Faltings \cite{Fa}, Teleman \cite{Te} and the author \cite{H-Verlinde}, Finkelberg's work \cite{F1}--\cite{F3} can also be reinterpreted as giving a construction of the modular tensor category structure on $\widetilde{O}_{\ell}$ except for a few cases, including the important case of $\mathfrak{g}=\mathfrak{e}_{8}$ and $\ell=2$. In the case that the level $\ell$ is admissible, Creutzig, Yang and the author \cite{CHY} constructed a braided tensor category structure on a semisimple category $\widetilde{O}_{\ell, {\rm \scriptsize ord}}$ of $\hat{\mathfrak{g}}$-modules. In the special case of $\mathfrak{g}=\mathfrak{sl}_{2}$, the rigidity of $\widetilde{O}_{\ell, {\rm \scriptsize ord}}$ was proved and the problem of determining when $\widetilde{O}_{\ell, {\rm \scriptsize ord}}$ is a modular tensor category was solved completely in \cite{CHY}. Recently, Creutzig proved in \cite{C} that when $\mathfrak{g}$ is simple-laced and the level $\ell$ is admissible, the braided tensor category structure on $\widetilde{O}_{\ell, {\rm \scriptsize ord}}$ is rigid and thus $\widetilde{O}_{\ell, {\rm \scriptsize ord}}$ is a ribbon tensor category. These constructions are all implicitly or explicitly vertex-operator-algebraic. In the case that $\mathfrak{g}=\mathfrak{sl}_{n}$ and the level is a positive integer, Kawahigashi, Longo and M\"{u}ger \cite{KLM} gave a construction of the modular tensor category structure on $\widetilde{O}_{\ell}$ using the approach of conformal nets. In all the cases discussed above, not only have the braided tensor category structures been constructed, but also the underlying vertex tensor category structures or modular vertex tensor category structures (see \cite{HL1}--\cite{HL5}, \cite{tensor4}, \cite{HL-affine}, \cite{HLZ1}--\cite{HLZ9}, \cite{H-rigidity}, \cite{Z}, \cite{H-Z-correction} and \cite{CHY}). We would like to emphasize that from a vertex tensor category, we obtain naturally a braided tensor category (see \cite{HL2} and \cite{HL6}) but in general, a braided tensor category alone is not enough to construct a vertex tensor category. This fact corresponds to the fact that from a two-dimensional conformal field theory, we obtain naturally a three-dimensional topological field theory but in general, a three-dimensional topological field theory alone is not enough to construct a two-dimensional conformal field theory. Vertex tensor categories (see \cite{HL2}), rigid vertex tensor categories (see \cite{H-rigidity}) or modular vertex tensor categories (see \cite{H-rigidity}) contain much more information than braided tensor categories, ribbon tensor categories or modular tensor categories, respectively. In this paper, we review briefly the existing vertex-operator-algebraic constructions of tensor category structures on module categories for affine Lie algebras. We also present several open problems in this paper. For a review on the approach of conformal nets, including in particular the work \cite{KLM}, we refer the reader to \cite{K}. We recall some basic definitions and constructions in the representation theory of affine Lie algebras in the next section. In Section 3, we review briefly the important work \cite{MS1} \cite{MS2} of Moore and Seiberg. The review of the existing vertex-operator-algebraic constructions of various tensor category structures on module categories for affine Lie algebras are given in Section 4. In Subsection 4.1, we recall some basic terms in the theory of tensor categories. In Subsections 4.2, 4.3 and 4.4, we review the existing constructions in the case that the level plus the dual Coxeter number is not a nonnegative rational number, in the case that the level is a positive integer and in the case that the level is an admissible number, respectively. The open problems in these cases are also given in these subsections. We state the open problem for the remaining case in Subsection 4.5. \paragraph{Acknowledgments} This paper is an expanded and revised version of the slides of the author's talk at the ``10th Seminar on Conformal Field Theory: A conference on Vertex Algebras and Related Topics'' held at the Research Institute for Mathematical Sciences, Kyoto University, Kyoto, April 23--27, 2018. The author is grateful to the organizers Tomoyuki Arakawa, Peter Fiebig, Nils Scheithauer, Katrin Wendland and Hiroshi Yamauchi for their invitation and to Nils Scheithauer for the financial support. The author would also like to thank Thomas Creutzig for comments. \section{Affine Lie algebras and modules} We recall the basic definitions and constructions in the representation theory of affine Lie algebras. Let $\mathfrak{g}$ be a finite-dimensional complex simple Lie algebra of rank $r$ and $(\cdot, \cdot)$ the invariant symmetric bilinear form on $\mathfrak{g}$. The affine Lie algebra $\hat{\mathfrak{g}}$ is the vector space $\mathfrak{g} \otimes \C[t, t^{-1}] \oplus \C{\bf k}$ equipped with the bracket operation \begin{align*} [a \otimes t^m, b \otimes t^n] &= [a, b]\otimes t^{m+n} + (a, b)m\delta_{m+n,0}{\bf k},\\ [a \otimes t^m, {\bf k}] &= 0, \end{align*} for $a, b \in \mathfrak{g}$ and $m, n \in \Z$. Let $\hat{\mathfrak{g}}_{\pm} = \mathfrak{g}\otimes t^{\pm 1}\C[t^{\pm 1}]$. Then $$\hat{\mathfrak{g}} = \hat{\mathfrak{g}}_{-} \oplus \mathfrak{g} \oplus \C{\bf k} \oplus \hat{\mathfrak{g}}_{+}.$$ If ${\bf k}$ acts as a complex number $\ell$ on a $\hat{\mathfrak{g}}$-module, then $\ell$ is called the level of this module. Let $h$ and $h^{\vee}$ be the Coxeter number and dual Coxeter number, respectively, of $\mathfrak{g}$. Let $M$ be a $\mathfrak{g}$-module and let $\ell \in \C$. Assume that M can be decomposed as a direct sum of generalized eigenspaces of the Casimir operator. This decomposition gives a $\C$-grading to $M$. In the case that $\ell+h^{\vee}\ne 0$, by using this decomposition and defining the (conformal) weight of the generalized eigenspace of the Casimir operator with eigenvalue $n$ to be $\frac{n}{2(\ell+h^{\vee})}$, we obtain a $\C$-grading of $M$ by (conformal) weights. Let $\hat{\mathfrak{g}}_{+}$ act on $M$ trivially and let $\mathbf{k}$ act as the scalar multiplication by $\ell$. Then $M$ becomes a $\mathfrak{g} \oplus \C{\bf k} \oplus \hat{\mathfrak{g}}_{+}$-module and we have a $\C$-graded induced $\hat{\mathfrak{g}}$-module $$\widehat{M}_{\ell} = U(\hat{\mathfrak{g}}) \otimes_{U(\mathfrak{g}\oplus \C\mathbf{k}\oplus \hat{\mathfrak{g}}_{+})}M,$$ where the $\C$-grading is given by the $\C$-grading on $M$ and the grading on $U(\hat{\mathfrak{g}})$ induced from the grading on $\hat{\mathfrak{g}}$. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$. For $\lambda\in \mathfrak{h}^{*}$, let $L(\lambda)$ be the irreducible highest weight $\mathfrak{g}$-module with the highest weight $\lambda$. We use $M(\ell, \lambda)$ to denote the $\hat{\mathfrak{g}}$-module $\widehat{L(\lambda)}_{\ell}$. Let $J(\ell, \lambda)$ be the maximal proper submodule of $M(\ell, \lambda)$ and $L(\ell, \lambda) = M(\ell, \lambda)/J(\ell, \lambda)$. Then $L(\ell, \lambda)$ is the unique irreducible graded $\hat{\mathfrak{g}}$-module such that ${\bf k}$ acts as $\ell$ and the space of all elements annihilated by $\hat{\mathfrak{g}}_{+}$ is isomorphic to the $\mathfrak{g}$-module $L(\lambda)$. Frenkel and Zhu \cite{FZ} gave both $M(\ell, 0)$ and $L(\ell, 0)$ natural structures of vertex operator algebras. Moreover, they gave in \cite{FZ} both $M(\ell, \lambda)$ and $L(\ell, \lambda)$ structures of $M(\ell, 0)$-modules and $L(\ell, \lambda)$ a structure of an $L(\ell, 0)$-module for dominant integral $\lambda$. For $\ell \in \C$ such that $\ell+h^{\vee}\not\in \Q_{\ge 0}$, let $\mathcal{O}_{\ell}$ be the category of all the $\hat{\mathfrak{g}}$-modules of level $\ell$ having a finite composition series all of whose irreducible subquotients are of the form $L(\ell, \lambda)$ for dominant integral $\lambda\in \mathfrak{h}^{*}$. For $\ell\in \Z_{+}$, let $\widetilde{\mathcal{O}}_{\ell}$ be the category of $\hat{\mathfrak{g}}$-modules of level $\ell$ that are isomorphic to direct sums of irreducible $\hat{\mathfrak{g}}$-modules of the form $L(\ell, \lambda)$ for dominant integral $\lambda\in \mathfrak{h}^{*}$ such that $(\lambda, \theta)\le \ell$, where $\theta$ is the highest root of $\mathfrak{g}$. Admissible modules for affine Lie algebras were studied first by Kac and Wakimoto \cite{KW1} \cite{KW2}. The level of these modules are called admissible numbers. Let $\ell$ be an admissible number, that is, $\ell+h^{\vee}=\frac{p}{q}$ for $p, q\in \Z_{+}$, $(p, q)=1$, $p\ge h^{\vee}$ if $(r^{\vee}, q)=1$ and $p\ge h$ if $(r^{\vee}, q)=r^{\vee}$, where $r^{\vee}$ is the "lacety" or lacing number of $\mathfrak{g}$, that is, the maximum number of edges in the Dynkin diagram of $\mathfrak{g}$. Let $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$ be the category of $\hat{\mathfrak{g}}$-modules of level $\ell$ that are isomorphic to direct sums of irreducible modules for the vertex operator algebra $L(\ell, 0)$. \section{Operator product expansion, modular invariance and Moore-Seiberg equations} The important work \cite{MS1} \cite{MS2} of Moore-Seiberg on two-dimensional conformal field theory in 1988 led to a conjecture: The category $\widetilde{O}_{\ell}$ for $\ell\in \Z_{+}$ has a natural structure of a modular tensor category in the sense of Turaev \cite{T1} \cite{T2}. This conjecture was proved by the author in \cite{H-rigidity} in 2005 as a special case of a general construction of modular tensor categories for vertex operator algebras satisfying natural existence-of-nondegenerate-invariant-bilinear form, positive-energy, finiteness and reductivity conditions. Since there are still a lot of misunderstandings about this work of Moore and Seiberg, here we clarify what were the main assumptions in \cite{MS1} and \cite{MS2} and what were in fact proved in \cite{MS1} and \cite{MS2}. In \cite{MS2}, Moore and Seiberg formulated two major conjectures on chiral rational conformal field theories: The operator product expansion of chiral vertex operators (see the end of Page 190 in \cite{MS2}) and the modular invariance of chiral vertex operators (see the beginning of the second paragraph on Page 200 in \cite{MS2}). In \cite{MS2}, no proofs or explanations or discussions were given on why the operator product expansion and modular invariance of the chiral vertex operators formulated in \cite{MS2} are true. These were the most fundamental hypotheses, not results, of this work of Moore and Seiberg. Mathematically, chiral vertex operators are called intertwining operators in \cite{FHL}. Conformal field theory can be viewed as the study of intertwining operators. The two major hypotheses above are clearly major mathematical conjectures which turned out to be more difficult to prove than the consequences derived from them in the work \cite{MS1} and \cite{MS2}. From these two conjectures, Moore and Seiberg \cite{MS1} \cite{MS2} derived a set of polynomial equations. The Verlinde conjecture and the Verlinde formula \cite{V} were shown in \cite{MS1} to be consequences of this set of equations but, since these two conjectures above were not proved in \cite{MS1} \cite{MS2}, they remained to be conjectures until they were proved by the author in \cite{H-Verlinde} in 2004. Moreover, Moore and Seiberg noticed in \cite{MS2} that their set of polynomial equations has properties analogous to those of tensor categories and there are also additional properties that are not satisfied by the usual examples of tensor categories. Later, Turaev \cite{T1} \cite{T2} formulated the precise notion of modular tensor category. From the set of Moore-Seiberg polynomial equations, it is not difficult to obtain an abstract modular tensor category in the sense of Turaev. But one cannot construct a modular tensor category structure on the category of modules from this set of Moore-Seiberg equations or this abstract modular tensor category alone. Since this set of Moore-Seiberg polynomial equations are consequences of the two conjectures on the operator product expansion and modular invariance of chiral vertex operators (intertwining operators) mentioned above, we see that these two conjectures led to a conjecture that the category of modules for a vertex operator algebra for a rational conformal field theory has a natural structure of modular tensor category. The vertex operator algebra associated to the category $\widetilde{\mathcal{O}}_{\ell}$ for a finite-dimensional complex simple Lie algebra $\mathfrak{g}$ and $\ell\in \Z_{+}$ was conjectured to be such a vertex operator algebra. These two conjectures on the operator product expansion and modular invariance of intertwining operators were proved by the author in 2002 \cite{H-diff-eqn} and 2003 \cite{H-modular}, respectively, for vertex operator algebras satisfying suitable positive-energy, finiteness and reductivity conditions. \section{Tensor category structures on suitable categories of modules for affine Lie algebras} In this section, we review the constructions of tensor category structures on suitable categories of modules for affine Lie algebras. \subsection{Tensor categories} Before we discuss the results on the constructions of tensor category structures on suitable categories of modules for affine Lie algebras, we recall briefly here some basic terms we use in the theory of tensor categories to avoid confusions. See \cite{T2} and \cite{BK} for details. For definitions and terms in the theory of vertex tensor categories, we refer the reader to \cite{HL2} and \cite{H-rigidity}. A tensor category is an abelian category with a tensor product bifunctor, a unit object, an associativity isomorphism, a left unit isomorphism and a right unit isomorphism such that the pentagon and triangle diagram are commutative. A braided tensor category is a tensor category with a braiding isomorphism such that two hexagon diagrams are commutative. A braided tensor category is rigid if every object has a two-sided dual object. A ribbon tensor category is a rigid braided tensor category with a twist satisfying the balancing axioms. A modular tensor category is a semisimple ribbon tensor category with finitely many simple (irreducible) objects such that the matrix of the Hopf link invariants is invertible. \subsection{The case of $\ell+h^{\vee}\not\in \Q_{\ge 0}$} In this case, we have the following major result of Kazhdan and Lusztig \cite{KL1}--\cite{KL5}: \begin{thm}[Kazhdan-Lusztig] Let $\ell\in \C$ such that $\ell+h^{\vee}\not\in \Q_{\ge 0}$. Then $\mathcal{O}_{\ell}$ has a natural rigid braided tensor category structure. Moreover, this rigid braided tensor category is equivalent to the rigid braided tensor category of finite-dimensional integrable modules for a quantum group constructed from $\mathfrak{g}$ at $q=e^{\frac{i\pi}{\ell+h^{\vee}}}$. \end{thm} This result was announced in 1991 \cite{KL1} and the detailed constructions and proofs were published in 1993 \cite{KL2} \cite{KL3} and 1994 \cite{KL4} \cite{KL5}. The construction of the rigid braided tensor category structure, especially the rigidity, in this work depends heavily on the results on the quantum group side. In particular, this construction cannot be adapted directly to give constructions for the module categories at other levels. In 2008, using the logarithmic generalization \cite{HLZ1}--\cite{HLZ9} by Lepowsky, Zhang and the author of the semisimple tensor category theory of Lepowsky and the author \cite{HL1}--\cite{HL6} and of the author \cite{tensor4} \cite{H-diff-eqn}, Zhang \cite{Z} gave a vertex-operator-algebraic construction of the braided tensor category structure in this case, with a mistake corrected by the author in 2017 \cite{H-Z-correction}. The braided tensor category obtained in this construction can be shown easily to be indeed equivalent to the one constructed in \cite{KL1}--\cite{KL5} (for example, the tensor product bifunctors are in fact the same in these two constructions). In this vertex-operator-algebraic construction, the main work is the proof of the associativity of logarithmic intertwining operators (logarithmic operator product expansion) and the construction of the associativity isomorphism. By using the work \cite{HLZ1}--\cite{HLZ9}, this work in fact constructed a vertex tensor category structure and the braided tensor category structure was obtained naturally from this vertex tensor category structure. But in \cite{Z}, only the braided tensor category structure was obtained and the rigidity is not proved. Thus we still have the following interesting problem: \begin{prob} Give a proof of the rigidity in this case in the framework of the logarithmic tensor category theory of Lepowsky, Zhang and the author \cite{HLZ1}--\cite{HLZ9}, without using the results for modules for the corresponding quantum group. \end{prob} \subsection{The case of $\ell\in \Z_{+}$} This $\ell\in \Z_{+}$ case is what the original work and conjectures of Moore and Seiberg \cite{MS1} \cite{MS2} were about. The construction of modular tensor category structures in this case has a quite interesting and complicated story. Recall from Section 3 that the work of Moore and Seiberg \cite{MS1} \cite{MS2} led to a conjecture that the category $\widetilde{O}_{\ell}$ has a natural structure of modular tensor category in the sense of Turaev \cite{T1} \cite{T2}. In 1997, Lepowsky and the author \cite{HL-affine} gave a construction of the braided tensor category structure on the category $\widetilde{O}_{\ell}$ when $\ell\in \Z_{+}$, using the semisimple tensor product bifunctor constructed by Lepowsky and the author \cite{HL1}--\cite{HL6} and the associativity isomorphism constructed by author \cite{tensor4} in the general setting of the category of modules for a vertex operator algebra satisfying suitable conditions. This work obtained the braided tensor category structure naturally from a vertex tensor category structure constructed in \cite{HL-affine} using \cite{HL3}--\cite{HL6} and \cite{tensor4}. In 2001, using the method developed in an early unpublished work \cite{BFM} of Beilinson-Feigin-Mazur in 1991, Bakalov and Kirillov Jr. also gave in a book \cite{BK} a construction of the braided tensor category structure on the category $\widetilde{O}_{\ell}$ in this case. Indeed Beilinson and other people knew how to construct the braided tensor category structure in mid 1990's. But Beilinson informed the author in private discussions in 1996 that he did not know how to prove the rigidity. Bakalov and Kirillov Jr. also did not give a proof of the rigidity and the nondegeneracy property in \cite{BK}, though these are stated in \cite{BK} as parts of the main theorem (Theorem 7.0.1 in \cite{BK}). . Since there are a lot of confusions, we quote the sentence about rigidity in \cite{BK} after the proof of Corollary 7.9.3 (which is a precise statement of Theorem 7.0.1 in the book): ``As a matter of fact, we have not yet proved the rigidity (recall that modular functor only defines weak rigidity); however, it can be shown that this category is indeed rigid.'' This is the only place in the whole book where the rigidity of this tensor category is explicitly mentioned. There is also no reference given for a proof or even a sketch of a proof of this rigidity. In 2012, Bakalov and Kirillov Jr. informed the author explicitly in private communications that they do not have a proof of the rigidity. It was until 2005 (in fact the author already gave a talk on the proof in 2004 in the the Erwin Schr\"{o}dinger Institute but the paper was posted in the arXiv in 2005), the rigidity and the nondegeneracy property were finally proved by the author in \cite{H-rigidity} as a special case of a proof for a vertex operator algebra satisfying natural existence-of-nondegenerate-invariant-bilinear form, positive-energy, finiteess and reductivity conditions. In particular, the conjecture mentioned above was proved by the author in 2005: \begin{thm}[\cite{H-rigidity}] Let $\ell\in \Z_{+}$. Then the category $\widetilde{\mathcal{O}}_{\ell}$ has a natural structure of a modular tensor category. \end{thm} The author's proof of this theorem was based on a formula used by author \cite{H-Verlinde} to derive the Verlinde formula. This was the first indication that the rigidity is in fact deeply related to the Verlinde formula which in turn is a consequence of the operator product expansion (associaitivity) and modular invariance of intertwining operators. Assuming the existence of the rigid braided tensor category structure on $\widetilde{\mathcal{O}}_{\ell}$, Finkelberg in his thesis \cite{F1} in 1993 and then in a revision \cite{F2} in 1996 gave a proof that this rigid braided tensor category is equivalent to a semisimple subquotient of a rigid braided tensor category of modules for a quantum group constructed from $\mathfrak{g}$, using the equivalence constructed by Kazhdan and Lusztig \cite{KL1}--\cite{KL5}. This work had been reinterpreted as giving a construction of the rigid braided tensor category structure on $\widetilde{\mathcal{O}}_{\ell}$. But in 2012, the author found a gap in Finkelberg's paper. Through Ostrik, the author informed Finkelberg this gap and explained that in \cite{H-Verlinde} the Verlinde formula is needed to prove the rigidity. Then instead of obtaining the Verlinde formula in the affine Lie algebra case as a major consequence, Finkelberg filled the gap in \cite{F3} by using this formula whose proofs had been given using other methods by Faltings \cite{Fa}, Teleman \cite{Te} and the author \cite{H-Verlinde}. Even after the correction in 2013, Finkelberg's proof of the rigidity is not complete. There are a few cases, including the important $\mathfrak{g}=\mathfrak{e}_{8}$ and $\ell=2$ case, that his method does not work. Finkelberg's equivalence between the modular tensor category $\widetilde{\mathcal{O}}_{\ell}$ and a semisimple subquotient of a rigid braided tensor category of modules for a quantum group is also not complete because of the same few cases, including the $\mathfrak{g}=\mathfrak{e}_{8}$ and $\ell=2$ case, in which his method does not work. Thus we have the following open problem: \begin{prob} Find a direct construction of this equivalence without using the equivalence given by Kazhdan-Lusztig so that this equivalence covers all the cases, including the important $\mathfrak{g}=\mathfrak{e}_{8}$ and $\ell=2$ case. \end{prob} \subsection{The admissible case} In the case that $\ell$ is an admissible number, Creutzig, Yang and the author proved in 2017 the following result: \begin{thm}[\cite{CHY}]\label{CHY1} Let $\ell$ be an admissible number. Then the category $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$ has a natural structure of a braided tensor category with a twist. \end{thm} This theorem was proved using the logarithmic tensor category theory of Lepowsky, Zhang and the author \cite{HLZ2}--\cite{HLZ9}, some results of Kazhdan-Lusztig \cite{KL2} and some results of Arakawa \cite{A}. The logarithmic tensor category theory reduces the construction of such a braided tensor category structure to the verification of several conditions. Since it used \cite{HLZ2}--\cite{HLZ9} to obtain the braided tensor category structure, this work also gave a construction of a vertex tensor category structure. This is a semisimple category. At first one might want to use the early tensor category theory for semisimple category of modules by Lepowsky and the author \cite{HL1}--\cite{HL6} and by the author \cite{tensor4} and \cite{H-diff-eqn}. The main result in \cite{tensor4} in this semisimple theory constructing the associativity isomorphism in this theory needs a convergence and extension property without logarithm. If generalized modules (not necessarily lower bounded) for the affine Lie algebra vertex operator algebras in this case are all complete reducible, a result in \cite{H-diff-eqn} can be applied to this case to conclude that the convergence and extension property without logarithm holds. But in this case, we do not have such a strong complete reducibility theorem and thus we cannot directly use this semisimple theory. Instead, we use the logarithmic generalization of the semisimple theory, even though our theory is semisimple. In this theory, we do not need to prove that there is no logarithm in the analytic extension of the product of intertwining operators. We construct the associativity isomorphism from logarithmic intertwining operators. Finally, since the modules in the category are all semisimple, the logarithmic intertwining operators are all ordinary. In particular, our theory still has no logarithm. Another condition that needs to be verified is that the category should be closed under a suitable tensor product operation. This condition is verified using a result of Arakawa in \cite{A}. The most subtle condition is the condition that suitable submodules in the dual space of the tensor product of two modules in $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$ should also be in $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$. The verification of this condition uses the author's modification in 2017 of one main result in \cite{HLZ8} that had been used to correct a mistake in Zhang's construction \cite{Z} in the case of $\ell+h^{\vee}\not\in \Q_{\ge 0}$. It also uses some results of Kazhdan-Lusztig \cite{KL2} and Arakawa \cite{A}. Theorem \ref{CHY1} gives only a braided tensor category structure with a twist to the semisimple category $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$. The natural question is whether this category is rigid and modular. In the case of $\mathfrak{g}=\mathfrak{sl}_{2}$, Creutzig, Yang and the author gave a complete answer in the same paper: \begin{thm}[\cite{CHY}]\label{CHY2} Let $\mathfrak{g}=\mathfrak{sl}_{2}$ and $\ell=-2+\frac{p}{q}$ with $p, q$ coprime positive integers. Then the braided tensor category $\mathcal O_{\ell,{\rm \scriptsize ord}}$ is a ribbon tensor category and is a modular tensor category if and only if $q$ is odd. \end{thm} The idea of the proof of this theorem is roughly speaking the following: First use the theory of vertex operator algebra extensions established by Kirillov Jr., Lepowsky and the author \cite{HKL} and by Creutzig, Kanade and McRae \cite{CKM} to show that this tensor category is braided equivalent to a full tensor subcategory of the braided tensor category of modules for a minimal Virasoro vertex operator algebra. But the braided tensor category of modules for the minimal Virasoro vertex operator algebra in fact has a natural modular tensor category structure constructed by the author \cite{H-rigidity} as a special case of a general construction for vertex operator algebras corresponding to rational conformal field theories. Then Theorem \ref{CHY2} is proved by using this modular tensor category structure and the braided equivalence mentioned above. For general complex semisimple $\mathfrak{g}$, conjectures have been given in the paper \cite{CHY}. Let $\ell$ an admissible number. Then in particular, $\ell=-h^{\vee}+\frac{p}{q}$ with coprime positive integers $p, q$. \begin{conj}\label{CHY-conj-1} The braided tensor category structure on $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$ is rigid and thus is a ribbon tensor category. \end{conj} In the case that $\mathfrak{g}$ is simple-laced, Conjecture \ref{CHY-conj-1} has been proved recently by Creutzig: \begin{thm}[\cite{C}] When $\mathfrak{g}$ is simple-laced, the braided tensor category structure on $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$ is rigid and thus is a ribbon tensor category. \end{thm} The idea of the proof of this result is similar to the proof in the case of $\mathfrak{sl}_{2}$ discussed above but is much harder because it uses the ribbon tensor category (actually a modular tensor category) of modules for a vertex operator algebra called the rational principle $W$-algebra of $\mathfrak{g}$, instead of the ribbon tensor category of modules for a minimal Virasoro vertex operator algebra. Arakawa proved in \cite{A2} that such a $W$-algebra satisfies the conditions needed in Theorem 4.6 in \cite{H-rigidity} and thus by Theorem 4.6 in \cite{H-rigidity}, the category of modules for this $W$-algebra has a natural structure of a modular tensor category. Using the work \cite{ACL} of Arakawa, Creutzig and Linshaw on principal $W$-algebras of simply-laced Lie algebras, the theory of vertex operator algebra extensions in \cite{HKL} and \cite{CKM} mentioned above and suitable fusion rule results proved by Creutzig in \cite{C}, Creutzig constructed in \cite{C} a fully faithful braided tensor functor from a subcategory of the modular tensor category of modules for this $W$-algebra to a certain braided tensor category with a twist. The rigidity of the tensor category of modules for the $W$-algebra gives the rigidity of this braided tensor category with a twist so that it becomes a ribbon tensor category. Since $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$ is different from this ribbon tensor category by just the actions of certain simple currents, $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$ is also rigid and thus a ribbon tensor category. \begin{conj}\label{CHY-conj-2} The ribbon tensor category structure on $\mathcal{O}_{\ell, {\rm \scriptsize ord}}$ is modular except for the following list: \begin{enumerate} \item $\mathfrak g \in \{\mathfrak{sl}_{2n}, \mathfrak{so}_{2n}, \mathfrak{e}_7, \mathfrak{sp}_n\}$ and $q$ even. \item $\mathfrak g = \mathfrak{so}_{4n+1}$ and $q=0 \mod 4$. \item $\mathfrak g = \mathfrak{so}_{4n+3}$ and $q=2 \mod 4$. \end{enumerate} \end{conj} These conjectures follow from the following conjecture on the equivalence of these braided tensor categories with the braided tensor categories coming from module categories for quantum groups constructed from the same finite-dimensional simple Lie algebra $\mathfrak{g}$: \begin{conj}\label{CHY-conj-3} The category $\mathcal O_{\ell, {\rm \scriptsize ord}}$ and the semi-simplification $\mathcal C_\ell(\mathfrak g)$ of the category of tilting modules for $U_q(\mathfrak g)$ are equivalent as braided tensor categories, where $q=e^{\frac{\pi i }{r^{\scriptsize \vee}(\ell+h^{\scriptsize \vee})}}$. \end{conj} Since the rigidity and modularity of the braided tensor category of tilting modules for $U_q(\mathfrak g)$ were established by Sawin in 2003 \cite{S}, we see that Conjectures \ref{CHY-conj-1} and \ref{CHY-conj-2} are indeed consequences of Conjecture \ref{CHY-conj-3}. The results and conjectures in \cite{CHY} are all about the subcategory $\mathcal O_{\ell, {\rm \scriptsize ord}}$ of $\mathcal{O}_{\ell}$. We still have the following open problem: \begin{prob} Let $\ell$ be an admissible number. Is there a braided tensor category structure on $\mathcal{O}_{\ell}$? If there is, is it rigid or even modular in a suitable sense? \end{prob} \subsection{The remaining case: an open problem} For the remaining case, we have only an open problem: \begin{prob} Let $\ell$ be a rational number larger than or equal to $-h^{\scriptsize \vee}$ but is not admissible. What tensor category structures can we construct? Are they rigid, semisimple or modular? \end{prob}

TITLE: A valuation question about the proof of Löwenheim–Skolem theorem. QUESTION [1 upvotes]: In the wiki page of Löwenheim–Skolem theorem, there is such a valuation in the proof of the downward part: $|N|=\kappa$ From $\left\vert F(A) \right\vert \leq \left\vert A \right\vert + \left\vert \sigma \right\vert + \aleph_0 \,$, we can get $\left\vert F^n(A) \right\vert \leq \left\vert A \right\vert + n\left\vert \sigma \right\vert + n\aleph_0 \,,$ for $n\in \Bbb{N}$. Expanding $F^\omega(A)$ as $\bigcup_{m\in \omega}F^m(A)$, we know that $|N|\leq |A| + \aleph_0 \cdot |\sigma|$. But how can we get that $|N|=\kappa$? Postscript: the full sketch of the proof (of the downward part) in wiki: For each first-order $\sigma \,$-formula $\varphi(y,x_{1}, \ldots, x_{n}) \,,$ the axiom of choice implies the existence of a function $f_{\varphi}: M^n\to M$ such that, for all $a_{1}, \ldots, a_{n} \in M$, either $M\models\varphi(f_{\varphi} (a_1, \dots, a_n), a_1, \dots, a_n)$ or $M\models\neg\exists y \varphi(y, a_1, \dots, a_n) \,.$ Applying the axiom of choice again we get a function from the first order formulas $\varphi$ to such functions $f_{\varphi} \,.$ The family of functions $f_{\varphi}$ gives rise to a preclosure operator $F \,$ on the power set of $M \,$ $F(A) = \{b \in M \mid b = f_{\varphi}(a_1, \dots, a_n); \, \varphi \in \sigma ; \, a_1, \dots, a_n \in A \}$ for $A \subseteq M \,.$ Iterating $F \,$ countably many times results in a closure operator $F^{\omega} \,.$ Taking an arbitrary subset $A \subseteq M$ such that $\left\vert A \right\vert = \kappa$, and having defined $N = F^{\omega}(A) \,,$ one can see that also $\left\vert N \right\vert = \kappa \,.$ $N \,$ is an elementary substructure of $M \,$ by the Tarski–Vaught test. The trick used in this proof is essentially due to Skolem, who introduced function symbols for the Skolem functions $f_{\varphi}$ into the language. One could also define the $f_{\varphi}$ as partial functions such that $f_{\varphi}$ is defined if and only if $M \models \exists y \varphi(y,a_1,\dots,a_n) \,.$ The only important point is that $F \,$ is a preclosure operator such that $F(A) \,$ contains a solution for every formula with parameters in $A \,$ which has a solution in $M \,$ and that $\left\vert F(A) \right\vert \leq \left\vert A \right\vert + \left\vert \sigma \right\vert + \aleph_0 \,.$ REPLY [2 votes]: Here $\kappa$ is an infinite cardinal with $\kappa\geq |\sigma|$, and $|A| = \kappa$, right? Since $A\subseteq N$, $\kappa \leq |N|$. On the other hand, $|N|\leq |A|+\aleph_0\cdot |\sigma| = \max(\kappa,\aleph_0,|\sigma|) = \kappa$.

MINNEAPOLIS — President Barack Obama traveled to the nation’s heartland Monday to press his case for tougher national gun laws, even as he appeared to acknowledge that expanded background checks on gun sales were far more likely to pass Congress than a ban on military-style assault weapons. In a city once called “Murderapolis” for its homicide rate in the 1990s, the president cited successful gun violence prevention efforts here renewed his call for Congress to pass a series of measures, including a ban on the manufacture and sale of new assault weapons, limits on high-capacity magazines and an expansion of the criminal background checks that currently covers only about 60 percent of gun sales. But he openly demonstrated different expectations for the measures as Washington wages a bitterly divisive debate over the role of guns in society. Obama, top lawmakers in Congress and gun control advocacy groups appear nervous about the political chances, noted supported broader background checks. The same survey found that 53 percent supported a ban on some semiautomatic weapons, and that 63 percent would support limits on magazines. Some advocates of tougher gun laws say that Obama and his allies in Congress should not give up on pushing for all three measures, regardless of the opposition that the measures are likely to face from the NRA and other gun rights groups. R. T. Rybak Jr., the mayor of Minneapolis, mocked politicians in Washington who are unwilling to support an assault weapons ban. “Oh, it’s not going to pass,” Rybak said. “Well, guess what? People are dying out here, and I’m not satisfied with the lame kind of response that we’ve gotten from some of the people in Washington who look at this like some kind of game.”. The city has developed programs aimed at rehabilitation for young people who have committed violent crimes. And its leaders are pushing for faster and more comprehensive state background checks for people buying guns. Recent article comments» View more comments»

Dias bons e ruins. 39 Funny Pictures for Today Dummies of the Year Memes Humor, Dark Jokes, Himym, Funny Comics, Harry Potter, Funny Images, Minions, Funny Things, Cool Things & Here are 33 hilarious Dad parenting pictures compiled by loldamn, those Dads just show off their & of taking care of babies when Mom is not home&

Older female relatives celebrated in new photowork for International Women’s Day. The photo collective ‘Not Finished Article’ has chosen to celebrate International Women’s Day by creating a new work especially for the Porty Light Box. Not Finished Article’s work ‘Exquisite…#MakeItHappen’ is a homage to significant older female members in the family. Comprising of three full-length portraits of personal significance to three members of the collective, and linked together by a purple figure of ‘8’ to mark the date, NFA hope the Portobello community will do their own remembering, celebrating of Grandmothers and Great Aunties on International Women’s day. About NFA Not. NFA meet regularly to share work in progress, attend events such as the Recontres Arles photography festival, and have more recently contributed collective works to 2014 Frames projections at CCA, GI Festival and Street Level Photoworks. Contact NFA: Tweet us @notfinishedart © Older female relatives celebrated in new photowork for International Women’s Day © 2015 Not Finished Article

Singer Dusty Springfield, whose smoky voice graced such ’60s hits as “You Don’t Have to Say You Love Me,” ‘Son of a Preacher Man,” “The Windmills of Your Mind” and “The Look of Love” (from the movie “Casino Royale”), has died in London. She had been battling breast cancer for years. Next month she was due to be inducted into the Rock Hall of Fame. Springfield was 59.

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3 years and 11 months “What would you do if I bought you a ring?” “I’d wear it.” “That’s it?” “I don’t know, what was I supposed to say?” “Well, the question was, what would you do if I asked you to marry me?” “Yeah, and the answer was: ‘I’d wear it.’” “What would everyone else think?” “The same thing we’re gonna think eventually. That we’re young and stupid and don’t know a thing about life. But that has nothing to do with right now, does it?”

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\begin{document} \title[Values of resurgences]{Extreme values of the resurgence for homogeneous ideals in polynomial rings} \author{B.\ Harbourne} \address{Department of Mathematics\\ University of Nebraska\\ Lincoln, NE 68588-0130 USA} \email{bharbourne1@unl.edu} \author{J.\ Kettinger} \address{Department of Mathematics\\ University of Nebraska\\ Lincoln, NE 68588-0130 USA} \email{jkettinger@huskers.unl.edu} \author{F.\ Zimmitti} \address{Department of Mathematics\\ University of Nebraska\\ Lincoln, NE 68588-0130 USA} \email{frank.zimmitti@huskers.unl.edu} \begin{abstract} We show that two ostensibly different versions of the asymptotic resurgence introduced by Guardo, Harbourne and Van Tuyl in 2013 are the same. We also show that the resurgence and asymptotic resurgence attain their maximal values simultaneously, if at all, which we apply to a conjecture of Grifo. For radical ideals of points, we show that the resurgence and asymptotic resurgence attain their minimal values simultaneously. In addition, we introduce an integral closure version of the resurgence and relate it to the other versions of the resurgence. In closing we provide various examples and raise some related questions, and we finish with some remarks about computing the resurgence. \end{abstract} \date{edited: May 11, 2020; compiled \today} \thanks{ {\bf Acknowledgements}: Harbourne was partially supported by Simons Foundation grant \#524858. We also wish to thank M. DiPasquale, E. Grifo, C. Huneke and A. Seceleanu, and M. Dumnicki, H. Tutaj-Gasi\'nska, T. Szemberg and J. Szpond, for helpful comments. } \keywords{resurgence, asymptotic resurgence, symbolic power, integral closure, fat points, polynomial ring, ideals, containment problem} \subjclass[2010]{Primary: 14C20, 13B22, 13M10; Secondary: 14N05, 13D40 } \maketitle \section{Introduction} This paper is motivated by wanting to better understand concepts, conjectures and methods introduced in \cite{BH1}, \cite{GHVT}, \cite{G}, \cite{DFMS} and \cite{DD} involving various approaches to the containment problem. It makes particular use of the groundbreaking methods of \cite{DFMS} and \cite{DD}. Let $\KK$ be a field and let $N\geq1$. Then $R=\KK[\PP^N]$ denotes the polynomial ring $\KK[\PP^N]=\KK[x_0,\ldots,x_N]$. Now let $(0)\neq I\subsetneq \KK[\PP^N]$ be a homogeneous ideal; thus $I=\oplus_{t\geq0} I_t$, where $I_t$ is the $\KK$ vector space span of all homogeneous polynomials of degree $t$ in $I$. The symbolic power $I^{(m)}$ is defined as $$I^{(m)}=R\cap (\cap_{P\in{\rm Ass}(R/I)}I^mR_P)$$ where the intersections take place in $\KK(\PP^N)$. While $I^r\subseteq I^{(m)}$ holds if and only if $m\leq r$, the containment problem of determining for which $m$ and $r$ the containment $I^{(m)}\subseteq I^r$ holds is much more subtle. If $h_I$ is the minimum of $N$ and the bigheight of $I$ (i.e., the maximum of the heights of associated primes of $I$), it is known that \begin{equation}\label{ELSHH} I^{(rh_I)}\subseteq I^r \end{equation} \cite{ELS, HoHu}, so given $r$, the issue is for which $m$ with $m<rh_I$ do we have $I^{(m)}\subseteq I^r$. The {\it resurgence} $\rho(I)$, introduced in \cite{BH1}, gives some notion of how small the ratio $m/r$ can be and still be sure to have $I^{(m)}\subseteq I^r$; specifically, $$\rho(I)=\sup\Big\{\frac{m}{r} : I^{(m)}\not\subseteq I^r\Big\}.$$ A case of particular interest is that of ideals of {\em fat points}. Given distinct points $p_1,\ldots,p_s\in\PP^N$ and nonnegative integers $m_i$ (not all 0), let $Z=m_1p_1+\cdots+m_sp_s$ denote the scheme (called a fat point scheme) defined by the ideal $$I(Z)=\cap_{i=1}^s(I(p_i)^{m_i})\subseteq \KK[\PP^N],$$ where $I(p_i)$ is the ideal generated by all homogeneous polynomials vanishing at $p_i$. Note that $I(Z)$ is always nontrivial (i.e., not $(0)$ nor $(1)$). Symbolic powers of $I(Z)$ take the form $I(Z)^{(m)}=I(mZ)=\cap_{i=1}^s(I(p_i)^{mm_i})$. We say $Z$ is {\it reduced} if $m_i$ is either 0 or 1 for each $i$ (i.e., if $I(Z)$ is a radical ideal). Subsequent to \cite{BH1}, two asymptotic notions of the resurgence were introduced by \cite{GHVT}. The first is $$\rho'(I)=\limsup_t\rho(I,t)=\lim_{t\to\infty}\rho(I,t),$$ where $\rho(I,t)=\sup\Big\{\frac{m}{r} : I^{(m)}\not\subseteq I^r, m\geq t, r\geq t\Big\}$. The second is $$\widehat{\rho}(I)=\sup\Big\{\frac{m}{r} : I^{(mt)}\not\subseteq I^{rt}, t\gg0\Big\}.$$ A useful new perspective on $\widehat{\rho}(I)$ is given by \cite[Corollary 4.14]{DFMS}, which shows that $$\widehat{\rho}(I)=\sup\Big\{\frac{m}{r} : I^{(m)}\not\subseteq \overline{I^{r}}\Big\},$$ where $\overline{I^{r}}$ is the integral closure of ${I^{r}}$ (defined below). Since we always have $I^r\subseteq \overline{I^r}$, this new perspective makes it clear that we always have $\widehat{\rho}(I)\leq\rho(I)$, and that we have $\widehat{\rho}(I)=\rho(I)$ if $I^r= \overline{I^r}$ for all $r\geq1$. Our major results are to show that $\widehat{\rho}(I) = \rho'(I)$ (Theorem \ref{MT1}), that $\widehat{\rho}(I)=h_I$ if and only if $\rho(I)=h_I$ (Theorem \ref{MT2}), and that $\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$ when $Z\subset\PP^N$ is a reduced scheme of points (Theorem \ref{MT3}) and for every fat point subscheme $Z\subset\PP^2$ (Corollary \ref{CorIntRho1}). We also introduce a new version of the resurgence, $\rho_{int}$, based on integral closure, and relate it to the original resurgence. We then discuss the relevance of our results to a conjecture of Grifo. Finally we provide some examples and raise some questions, and include a discussion of the computability of the resurgence. \subsection{Background}\label{bkgrnd} Let $I\subseteq \KK[\PP^N]$ be a nontrivial homogeneous ideal. Given the comments and definitions above we have (see \cite{GHVT}) \begin{equation}\label{ELSHHbounds} 1\leq\widehat{\rho}(I)\leq \rho'(I)\leq \rho(I)\leq h_I. \end{equation} By \cite{BH1} and \cite{GHVT} we also have \begin{equation}\label{BHbounds} \frac{\alpha(I)}{\widehat{\alpha}(I)}\leq\widehat{\rho}(I)\leq \rho'(I)\leq \rho(I), \end{equation} where $\alpha(I)$ is the least degree of a nonzero element of $I$ and $\widehat{\alpha}(I)$ is the {\it Waldschmidt constant}, defined as $$\widehat{\alpha}(I)=\lim_{m\to\infty}\frac{\alpha(I^{(m)})}{m}.$$ For a nontrivial fat point subscheme $Z\subseteq\PP^N$, by \cite{BH1} we have in addition \begin{equation}\label{BHbounds2} \rho(I(Z))\leq \frac{{\rm reg}(I(Z))}{\widehat{\alpha}(I(Z))}, \end{equation} where ${\rm reg}(I(Z))$ is the Castelnuovo-Mumford regularity of $I(Z)$. A version of resurgence can be defined with integral closure replacing symbolic powers. We pause to briefly discuss the concept of integral closure. Given an ideal $I\subseteq R=\KK[\PP^N]$, we recall (see \cite{HuSw}) that the {\it integral closure} $\overline{I}$ of $I$ consists of all elements $c\in R$ satisfying for some $n\geq1$ a polynomial equation $$c^n+a_1c^{n-1}+\cdots+a_n=0$$ where $a_j\in I^j$. We say $I$ is {\it integrally closed} if $I=\overline{I}$. We note that $\overline{I}$ is monomial (resp. homogeneous) if $I$ is \cite[Proposition 1.4.2]{HuSw} (resp. \cite[Corollary 5.2.3]{HuSw}). If $I^r$ is integrally closed for all $r\geq1$, we say $I$ is {\it normal}. For example, the ideal $I(p_i)$ of a point $p_i\in\PP^N$ is normal. Likewise, $M=(x_0,\ldots,x_N)$ is normal. This is because $M$ is a monomial prime ideal and $I(p_i)$ is also, up to choice of coordinates, but monomial primes are normal. (Apply the usual criterion for integral closure for monomial ideals, that the integral closure of a monomial ideal $I$ is the monomial ideal associated to the convex hull of the Newton polygon of $I$ \cite{HuSw}.) As further examples of integrally closed ideals, we note $I(Z)$ is integrally closed for all $Z$, as is $M^t\cap I(Z)$ for every $t$; this is because intersections of integrally closed ideals are integrally closed. Now assume $\alpha(I(Z))={\rm reg}(I(Z))$; then $(I(Z)^r)_t=(I(rZ))_t$ for $t\geq \alpha(I(Z)^r)$ (apply \cite[Lemma 2.3.3(c)]{BH1} using the fact that $\alpha(I(Z)^r)=r\alpha(I(Z))$). Thus we have $I(Z)^r=M^{r\alpha(I)}\cap I(rZ)$ and hence $I(Z)$ is normal, if $\alpha(I(Z))={\rm reg}(I(Z))$. Thus, when $\alpha(I(Z))={\rm reg}(I(Z))$, we have $\widehat{\rho}(I(Z)) =\rho(I(Z))$ by the normality, but in fact \eqref{BHbounds} and \eqref{BHbounds2} give us more, namely $$\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}=\widehat{\rho}(I(Z)) =\rho(I(Z)).$$ Now we define the integral closure resurgence. Given any nontrivial homogeneous ideal $I\subset\KK[\PP^N]$ but replacing symbolic power by integral closure in the definition of resurgence gives us the {\it integral closure resurgence}, $$\rho_{int}(I)=\sup\Big\{\frac{m}{r} : \overline{I^m}\not\subseteq I^r\Big\}.$$ If symbolic powers of $I$ are integrally closed (as is the case when $I$ is the ideal of a fat point subscheme), we have $\rho_{int}(I)\leq\rho(I)$, and if moreover $I^{(m)}=\overline{I^m}$ for all $m\geq1$, it follows that $\rho_{int}(I)=\rho(I)$ (see Theorem \ref{BSkoda}). A lower bound such as $\rho_{int}(I)\leq\rho(I)$ is of interest since for any $c>1$ (as the proof of Theorem \ref{BSkoda} shows) it is in principle a finite calculation (although not necessarily an easy one) to verify whether or not $\rho_{int}(I)\geq c$, and if so to compute $\rho_{int}(I)$ exactly. In a further analogy of $\rho_{int}$ with $\rho$, by Lemma \ref{rhoIntLem} we have $1\leq\rho_{int}(I)\leq N$, and, if $I=\overline{I}$, we have $\overline{I^m}\subseteq I^r$ for all $m\geq Nr$. \section{Main results} We begin with the result that started this paper. \begin{proposition} Let $Z$ be a fat point subscheme of $\PP^N$ with an integer $c$ such that $I(cZ)^t=I(ctZ)$ for all $t\geq 1$. Then $\widehat{\rho}(I(Z)) =\rho'(I(Z))$. \end{proposition} \begin{proof} Let $b$ be a rational such that $\widehat{\rho}(I(Z))<c/b$. Pick any integer $d>0$ such that $db$ is an integer. Since $\widehat{\rho}(I(Z))<c/b=cd/(db)$, we have by definition of $\widehat{\rho}(I(Z))$ that $I(cdtZ)\subseteq I(Z)^{dbt}$ for $t\gg0$, Note that $I(cdZ)^t=I(cZ)^{dt}=I(cdtZ)$ for all $t\geq1$. Hence by \cite[Theorem 1.2(3)]{GHVT} we have $\rho'(I(Z))\leq cdt/(dbt)=c/b$. Since this holds for all $b$ with $c/b>\widehat{\rho}(I(Z))$, we have $\widehat{\rho}(I(Z))\geq\rho'(I(Z))$. But by \cite[Theorem 1.2(1)]{GHVT} we have $\widehat{\rho}(I(Z))\leq\rho'(I(Z))$, hence $\widehat{\rho}(I(Z))=\rho'(I(Z))$. \end{proof} A much stronger result can be proved based on an argument in \cite{DD}. \begin{theorem}\label{MT1} Let $I$ be a nontrivial homogeneous ideal of $\KK[\PP^N]$. Then $\widehat{\rho}(I) =\rho'(I)$. \end{theorem} \begin{proof} If $\widehat{\rho}(I) =\rho(I)$, then \eqref{ELSHHbounds} gives $\widehat{\rho}(I) =\rho'(I)$, so assume $\widehat{\rho}(I) < \rho(I)$. Thus there is an $\epsilon>0$ such that $\widehat{\rho}(I) + \epsilon < \rho(I)$. As in the proof of \cite[Proposition 2.6]{DD} (which in turn is a consequence of \cite[Lemma 4.12]{DFMS}), we have $$\frac{s}{r+N} < \widehat{\rho}(I)$$ whenever $I^{(s)}\not\subseteq I^r$. This means there are only finitely many $s$ and $r$ for which $\widehat{\rho}(I)+\epsilon\leq s/r$ holds but $I^{(s)}\subseteq I^r$ fails. (This is because $\widehat{\rho}(I) + \epsilon \leq s/r$ and $\frac{s}{r+N} < \widehat{\rho}(I)$ implies $r(\widehat{\rho}(I)+\epsilon)\leq s< \widehat{\rho}(I)(r+N)$ and $r < N\widehat{\rho}(I)/\epsilon$.) Hence for all $s$ and $r$ sufficiently large which have $I^{(s)}$ not contained in $I^r$, we will have $s/r < \widehat{\rho}(I)+\epsilon$ and hence $\rho'(I) \leq \widehat{\rho}(I)+\epsilon$. This is true for every $\epsilon > 0$, and so we get $\rho'(I) \leq \widehat{\rho}(I)$. Since we already have $\rho'(I) \geq \widehat{\rho}(I)$, we conclude that $\rho'(I) = \widehat{\rho}(I)$. \end{proof} An alternate statement of the next result is that $\widehat{\rho}(I)=h_I$ if and only if $\rho(I)=h_I$. We have learned that Theorem \ref{MT2} was also obtained independently by DiPasquale and Drabkin, but not included in \cite{DD}. \begin{theorem}\label{MT2} Let $I$ be a nontrivial homogeneous ideal of $\KK[\PP^N]$. Then $\widehat{\rho}(I)<h_I$ if and only if $\rho(I)<h_I$. \end{theorem} \begin{proof} If $\rho(I) < h_I$, then we have $\widehat{\rho}(I) \leq \rho(I) < h_I$ by \eqref{ELSHHbounds}. So assume $\widehat{\rho}(I) < h_I$. If $\widehat{\rho}(I)=\rho(I)$, then $\rho(I) < h_I$. If $\widehat{\rho}(I) < \rho(I)$, then by \cite[Proposition 2.6]{DD}, $\rho(I)$ is the maximum of finitely many ratios $s/r$ with $I^{(s)}\not\subseteq I^r$. Thus $\rho(I) = h_I$ would imply that $s/r = h_I$, which contradicts the result of \cite{ELS, HoHu} that $I^{(m)}\subseteq I^r$ whenever $m\geq rh_I$. \end{proof} The other extreme is also of interest. Let $Z\subset\PP^N$ be a fat point subscheme. The next result, Theorem \ref{MT3}, shows that the question of when $\rho(I(Z))=1$ is related to two concepts: to analytic spreads (see \cite{HuSw}) and to symbolic defects (see \cite{GGSV}). We can define the {\it analytic spread} $\ell(I)$ of a homogeneous ideal $I\subseteq R=\KK[\PP^N]$ as being the minimum number of elements of $I$ such that, after localizing at the irrelevant ideal, the ideal $J$ they generate has $\overline{J}=\overline{I}$ \cite[Corollary 1.2.5, Proposition 8.3.7]{HuSw}. The analytic spread $\ell(I)$ of an ideal $I\subseteq R$ is at least the height of $I$, since $J$ and $I$ have the same minimal primes, and the minimal number of generators of an ideal is at least the height of its minimal prime of minimal height. By \cite[Proposition 5.1.6]{HuSw}, it is also at most the dimension of $R$ (which is $N+1$). For $I(Z)\subset \KK[\PP^N]$ we thus have $N\leq\ell(I(Z))\leq N+1$. We say an ideal $B\subseteq\KK[\PP^N]$ is a {\it complete intersection} if $B$ is generated by a regular sequence; equivalently, $B$ is a complete intersection if all associated primes of $B$ have height $c$, where $c$ is the minimal number of generators of $B$. If for a fat point subscheme $Z\subset\PP^N$ its ideal $I(Z)$ is a complete intersection (i.e., $I(Z)$ has $N$ generators), then $\ell(I(Z))=N$, but $\ell(I(Z))=N$ can occur even when $I(Z)$ is not a complete intersection (for example, $\ell(I(mZ))=N$ when $I(Z)$ is a complete intersection but $m>1$). The {\it symbolic defect} ${\rm sdefect}(I(Z),m)$ is the minimum number of generators of the module $I(mZ)/I(Z)^m$ \cite{GGSV}. Thus ${\rm sdefect}(I(Z),m)=0$ if and only if $I(mZ)=I(Z)^m$. If $I(Z)$ is a complete intersection, then ${\rm sdefect}(I(Z),m)=0$ for all $m\geq1$ since we have $I(mZ)=I(Z)^m$ (indeed, if $I$ is generated by a regular sequence, then $I^r=I^{(r)}$ for all $r\geq 0$ by \cite[Lemma 5, Appendix 6]{ZS}). When $N=2$ and $Z$ is reduced (i.e., $I(Z)$ is radical), \cite[Theorem 2.6]{GGSV} gives a converse: if ${\rm sdefect}(I(Z),m)=0$ for all $m\geq1$, then $I(Z)$ is a complete intersection (see also \cite[Remark 2.5]{CFGLMNSSV} and \cite[Theorem 2.8]{HU}). Our next result gives a number of equivalent conditions for ${\rm sdefect}(I(Z),m)=0$ for all $m\geq1$ for any reduced fat point subscheme $Z\subset \PP^N$, thereby extending from $N=2$ to all $N$ the result that $I(mZ)=I(Z)^m$ for all $m\geq1$ if and only if $I(Z)$ is a complete intersection. We thank Seceleanu and Huneke for the implications (d) $\Rightarrow$ (e), and, for $Z$ reduced, (e) $\Rightarrow$ (a). We do not know if (e) $\Rightarrow$ (a) holds for nonreduced fat point schemes $Z$ (i.e., when $I(Z)$ is not radical), but see Corollaries \ref{CorIntRho1} and \ref{CorIntRho2}, and also Examples \ref{Ex1} and \ref{Ex1b} and Question \ref{Q4b}. \begin{theorem}\label{MT3} Let $Z$ be a nontrivial fat point subscheme of $\PP^N$. Then each of the following criteria implies the next. \begin{enumerate} \item[(a)] $I(Z)^m=I(mZ)$ (i.e., ${\rm sdefect}(I(Z),m)=0$) for all $m\geq1$. \item[(b)] $\rho(I(Z))=1$. \item[(c)] $\widehat{\rho}(I(Z))=1$. \vskip.3\baselineskip \item[(d)] $\overline{I(Z)^m}=I(mZ)$ for all $m\geq1$. \item[(e)] The analytic spread of $I(Z)$ is $N$. \end{enumerate} Moreover, if $Z$ is reduced or $N=1$, then (e) implies (a). In fact, if $Z$ is reduced or $N=1$, then each of the conditions (a)-(e) is equivalent to $I(Z)$ being a complete intersection. \end{theorem} \begin{proof} That (a) implies (b) is clear (since $I(Z)^m\subseteq I(Z)^r$ if and only if $m\geq r$), and (b) implies (c) since $1\leq\widehat{\rho}(I(Z))\leq\rho(I(Z))$. Next (c) is equivalent to (d) by \cite[Corollary 4.16]{DFMS}. Now we show (d) implies (e). By \cite[Proposition 5.4.7]{HuSw}, we have that (d) implies $\ell(I(Z))\neq N+1$, hence $\ell(I(Z)) = N$. If $N=1$, then $I(Z)$ is principal, hence (a) always holds and $I(Z)$ is a complete intersection. Finally assume that $Z$ is reduced. Thus the primary components of $I(Z)$ are ideals of points of multiplicity 1, hence are complete intersections (i.e., $I(Z)$ is locally a complete intersection). By \cite{CN}, $\ell(I(Z)) = N$ implies that the localization $I(Z)_M$ of $I(Z)$ at the irrelevant ideal $M\subset R=\KK[\PP^N]$ is a complete intersection, and hence $(I(Z)^m)_M=(I(Z)_M)^m$ is saturated for all $m\geq1$, so $I(Z)^m$ itself is saturated so (a) holds. Moreover, since (e) implies that $I(Z)_M$ is a complete intersection when $Z$ is reduced, the number of generators of $I(Z)_M$ is $N=\dim I(Z)_M/I(Z)_MM_M=\dim I(Z)/I(Z)M$, hence $I(Z)$ also has $N$ generators, so is itself is a complete intersection and hence (a) holds. \end{proof} We can now give a characterization of those $Z$ for which $I(mZ)=I(Z)^m$ for all $m\geq1$. This characterization is not so interesting in itself, but it does raise the question of whether the normality hypothesis can be dropped; see Question \ref{Q4b}. Although $I(Z)$ is not always normal (see Example \ref{Ex2}), we do not know any examples with $\widehat{\rho}(I(Z))=1$ where $I(Z)$ is not normal. \begin{corollary}\label{MT3Cor} Let $Z\subset\PP^N$ be a fat point subscheme. Then $I(mZ)=I(Z)^m$ for all $m\geq1$ if and only if $\widehat{\rho}(I(Z))=1$ and $I(Z)$ is normal. \end{corollary} \begin{proof} Having $I(mZ)=I(Z)^m$ for all $m\geq1$ implies $\widehat{\rho}(I(Z))=1$ by Theorem \ref{MT3}, and it implies $I(Z)$ is normal since $I(Z)^m\subseteq \overline{I(Z)^m}\subseteq I(mZ)$. Conversely, $\widehat{\rho}(I(Z))=1$ implies $I(mZ)=\overline{I(Z)^m}$ for all $m\geq1$ by Theorem \ref{MT3} and normality implies $\overline{I(Z)^m}=I(Z)^m$ for all $m\geq1$, hence $I(mZ)=I(Z)^m$ holds for all $m\geq1$. \end{proof} We now consider $\rho_{int}(I)$. Note for a nontrivial homogeneous ideal $I\subset\KK[\PP^N]$ that an ideal independent version of \eqref{ELSHH} can be stated as $I^{(Nr)}\subseteq I^r$, and the corresponding bounds \eqref{ELSHHbounds} on $\rho(I)$ are $1\leq \rho(I)\leq N$. In analogy with this, we have the following lemma. (See Example \ref{Ex4} for an example showing that the assumption $I=\overline{I}$ is needed both for the second part of Lemma \ref{rhoIntLem} and for Theorem \ref{BSkoda}(b).) \begin{lemma}\label{rhoIntLem} Let $I\subset\KK[\PP^N]$ be a nontrivial homogeneous ideal. Then $1\leq \rho_{int}(I)\leq N$. Moreover, if $N>1$ and $t>1$, or if $I=\overline{I}$ and $t\geq1$, then $\overline{I^{Nt}}\subseteq I^t$. \end{lemma} \begin{proof} Since $I^t\subseteq \overline{I^{t}}$ but $I^t\not\subseteq I^{t+1}$, we see that $\overline{I^t}\not\subseteq I^{t+1}$, so $1\leq\rho_{int}(I)$. The Brian{\c c}on-Skoda Theorem \cite{HuSw} asserts $\overline{I^{t+N}}\subseteq I^t$ for each $t\geq1$. For $t,N\geq2$ we have $tN\geq t+N$, hence $I^{Nt}\subseteq I^{t+N}$ and thus $\overline{I^{Nt}}\subseteq\overline{I^{t+N}}\subseteq I^t$. This also tells us that if $\overline{I^m}\not\subseteq I^t$, we must have either $m<Nt$ (and so $m/t<N$) or $t$ or $N$ must be equal to 1. By Brian{\c c}on-Skoda, $\overline{I^m}\not\subseteq I^t$ implies $m<t+N$, so if $t=1$, then $m/t = m\leq N$, while if $N=1$, then we have $m<t+1$, so $m/t\leq t/t=1=N$. Thus in all cases we have $m/t\leq N$, hence $\rho_{int}(I)\leq N$. We already saw that $\overline{I^{Nt}}\subseteq I^t$ if $N,t>1$. Assume $N>1$ but $t=1$. Then $I^N\subseteq I$ so if $I=\overline{I}$ we have $\overline{I^{N}}\subseteq I$. Finally assume $N=1$ and $I\subset \KK[x_0,x_1]$ is a nontrivial homogeneous ideal. Thus $I=(G_1,\ldots,G_s)$ for some nonzero homogeneous generators $G_i$. Let $F$ be the greatest common divisor of the $G_i$, and for each $i$ let $H_iF=G_i$. Then $I=(F)Q$ where $Q=(H_1,\ldots,H_s)$. If $\deg H_i=0$ for some $i$, we have $Q=(1)$ and so $I=(F)$. If $\deg H_i>0$ for all $i$, then $Q$ is primary for $(x_0,x_1)$. If $I=(F)$, then $I$ is normal, so we have $\overline{I^t}=I^t$ for all $t\geq1$. So say $I=(F)Q$ where $Q$ is primary for $(x_0,x_1)$, hence $I^t=(F^t)Q^t$ for each $t\geq1$. Note that $I^t=\overline{I^t}$ if and only if $Q^t=\overline{Q^t}$. (Here's why. Assume $I^t=\overline{I^t}$. By \cite[Remark 1.3.2(2)]{HuSw}, we have $\overline{I^t:J}\subseteq \overline{I^t}:\overline{J}$ for any ideal $J$. Take $J=(F^r)$. Then $Q^t\subseteq \overline{Q^t}= \overline{I^t:J}\subseteq \overline{I^t}:\overline{J}=I^t:(F^r)=Q^t$. Alternatively, say $x\in \overline{Q^t}$. Then $x^n+a_1x^{n-1}+\cdots+a_n=0$ for some $n$ and some $a_i\in Q^{ti}$. Multiplying by $F^{tn}$ gives $(F^tx)^n+F^ta_1(F^tx)^{n-1}+\cdots+(F^t)^na_n=0$ so $F^tx\in \overline{(F^t)Q^t}=\overline{I^t}=I^t=(F^t)Q^t$, so $x\in Q^t$. Now assume $Q^t = \overline{Q^t}$. Say $x\in \overline{I^t}$. Then $x^n+F^ta_1x^{n-1}+\cdots+(F^t)^na_n=0$ for some $n$ with $a_i\in Q^{ti}$. Say $F^t=AP^m$ where $P$ is an irreducible factor of $F$ and $P$ does not divide $A$. Then $P$ divides $x$. Let $yP=x$. Dividing out gives $y^n+AP^{m-1}a_1y^{n-1}+\cdots+A^nP^{n(m-1)}a_n=0$. We can keep dividing out until we have $z^n+Aa_1z^{n-1}+\cdots+A^na_n=0$, where $zP^m=x$ and $z\in\overline{(A)Q^t}$. Continuing in this way, dividing out irreducible factors of $F^t$, we eventually see that we get an element $w$ with $wF^t=x$, where $w\in\overline{Q^t}=Q^t$, hence $x\in (F^t)Q^t=I^t$.) Now we claim that $Q$ is normal if and only if $Q=\overline{Q}$. Given this, if $I=\overline{I}$, then $Q=\overline{Q}$, hence $Q^t=\overline{Q^t}$ so $I^t=\overline{I^t}$ (i.e., $I$ is normal), which is what we needed to show. But note that $Q$ normal implies by definition that $Q=\overline{Q}$. Conversely, assume $Q=\overline{Q}$. It suffices to show $Q^t=\overline{Q^t}$ for each $t$. By \cite[Proposition 4.8]{AM} and \cite[Proposition 1.1.4]{HuSw}, it is enough to check this after localizing at $(x_0,x_1)$. But by results of Zariski, $Q=\overline{Q}$ implies $Q^t=\overline{Q^t}$ for each $t$ in the local case \cite[Theorem 14.4.4]{HuSw}. \end{proof} \begin{theorem}\label{BSkoda} Let $I\subset \KK[\PP^N]$ be a nontrivial homogeneous ideal, $N\geq1$. \begin{enumerate} \item[(a)] We have $$1\leq\rho_{int}(I)=\max\Big\{\big\{\frac{m}{r}: \overline{I^m}\not\subseteq I^r\big\}\cup\{1\}\Big\}.$$ \item[(b)] If $I=\overline{I}$ and $N>1$, then $\rho_{int}(I)<N$. \item[(c)] If $I^{(m)}=\overline{I^{(m)}}$ for all $m\geq1$ (as for example is the case for $I=I(Z)$ for a fat point subscheme $Z\subset\PP^N$), then $\rho_{int}(I)\leq\rho(I)$. \item[(d)] If $I^{(m)}=\overline{I^m}$ for all $m\geq1$, then $\rho_{int}(I)=\rho(I)$. \end{enumerate} \end{theorem} \begin{proof} (a) By Lemma \ref{rhoIntLem} we have $1\leq \rho_{int}(I)$. In order for $\rho_{int}(I) > 1$ there must be a pair of positive integers $(r_0,m_0)$ with both $m_0/r_0 > 1$ and $\overline{I^{m_0}}\not\subseteq I^{r_0}$ (and hence $m_0<r_0+N$ by Brian{\c c}on-Skoda). Set $c=m_0/r_0$. If $\rho_{int}(I)>c$, then as before we have $(r,m)$ with both $m/r > c$ and $\overline{I^m}\not\subseteq I^r$ (and hence $m<r+N$). But there are only finitely many pairs $(r,m)$ with $cr<m$ and $m<r+N$ (in particular, we have $r< N/(c-1)$ and $cr<m<r+N$). Thus either $\rho_{int}(I)=1$ or $\rho_{int}(I)=\max\{\frac{m}{r}: \overline{I^m}\not\subseteq I^r\}$, hence $\rho_{int}(I)=\max\Big\{\big\{\frac{m}{r}: \overline{I^m}\not\subseteq I^r\big\}\cup\{1\}\Big\}.$ (b) Now assume $I=\overline{I}$ and $N>1$. By Lemma \ref{rhoIntLem} we have $\overline{I^{Nr}}\subseteq I^r$ for $r\geq1$, so $\overline{I^m}\not\subseteq I^r$ implies $m/r<N$, hence $\rho_{int}(I)$, being either 1 or a maximum of values $m/r$ less than $N$, is less than $N$. (c) Here we assume $I^{(m)}=\overline{I^{(m)}}$ for all $m\geq1$. Then since $I^m\subseteq I^{(m)}$ we have $\overline{I^m}\subseteq I^{(m)}$, so $\overline{I^m}\not\subseteq I^r$ implies $I^{(m)}\not\subseteq I^r$, and hence $\rho_{int}(I)\leq\rho(I)$. (d) Assuming $I^{m}=\overline{I^{(m)}}$ for all $m\geq1$, we have $$\rho_{int}(I) =\sup\Big\{\frac{m}{r} : \overline{I^m}\not\subseteq I^r\Big\}= \sup\Big\{\frac{m}{r} : I^{(m)}\not\subseteq I^r\Big\}=\rho(I).$$ \end{proof} We do not know any examples with $\rho_{int}(I(Z))>1$. For $Z\subset\PP^2$, there are none, by the next result, which is an immediate consequence of \cite[Theorem 3.3]{AH1}. We thank Huneke for alerting us to this result. \begin{corollary}\label{CorIntRho3} Let $Z\subset\PP^N$ be a nontrivial fat point subscheme and let $I=I(Z)$. Then we have the following: \begin{enumerate} \item[(a)] $\overline{I^{N+m-1}}\subseteq I^m$ for $m\geq1$; \item[(b)] $\rho_{int}(I(Z))\leq N/2$ for $N\geq2$; and \item[(c)] $\rho_{int}(I(Z))=1$ for $N=1,2$. \end{enumerate} \end{corollary} \begin{proof} (a) Let $M=(x_0,\ldots,x_N)\subset \KK[\PP^N]$. Using $I$ as a reduction for $I$ and $\ell$ for the analytic spread of $I$, \cite[Theorem 3.3]{AH1} states (after localizing at the irrelevant ideal $M$) that $\overline{I^{\ell+m}}\subseteq I^{\ell-N+m+1}$ for $m\geq0$. But $N\leq \ell\leq N+1$, so for $\ell=N$ we have $\overline{I^{N+m}}\subseteq I^{m+1}$, while for $\ell=N+1$ we have $\overline{I^{N+m+1}}\subseteq I^{m+2}$, both for $m\geq0$. Either way, we have $\overline{I^{N+m-1}}\subseteq I^m$ for $m\geq1$, and hence $\overline{I^{N+m-1}}\subseteq I^m$ holds without localizing, since all of the ideals are homogeneous. (b) Assume $N\geq 2$. For $r=1$ we have $\overline{I^m}\not\subseteq I^r$ for all $m\geq 1$, so consider $r>1$. Then we have $\overline{I^m}\subseteq I^r$ for all $m\geq N+r-1$, so the fractions $m/r$ for which we have $\overline{I^m}\not\subseteq I^r$ are contained in the set $\{m/r : 1\leq m\leq N+r-2, r\geq2\}$. The supremum occurs for $m=N+r-2$ and $r=2$, hence the supremum is $N/2$, so $\rho_{int}(I(Z))\leq N/2$. (c) When $N=1$ we have $\rho_{int}(I(Z))=1$ from Lemma \ref{rhoIntLem}, and when $N=2$ we have $\rho_{int}(I(Z))=1$ from (b). \end{proof} If $\rho_{int}(I(Z))$ is always 1, then the next result would imply that $\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$. In particular, it shows that $\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$ for every fat point subscheme $Z\subset\PP^2$. \begin{corollary}\label{CorIntRho1} Let $Z\subset\PP^N$ be a nontrivial fat point subscheme. If $\widehat{\rho}(I(Z))=1$, then $\rho_{int}(I(Z))=\rho(I(Z))$, hence $\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$ when $N=2$. \end{corollary} \begin{proof} By Theorem \ref{MT3}, $\widehat{\rho}(I(Z))=1$ implies $I(mZ) = \overline{I(Z)^m}$, so $\rho_{int}(I(Z))=\rho(I(Z))$. Since $\rho_{int}(I(Z))=1$ when $N=2$ by Corollary \ref{CorIntRho3}, and since $\rho(I(Z))=1$ implies $\widehat{\rho}(I(Z))=1$, we have $\widehat{\rho}(I(Z))=1$ if and only if $\rho(I(Z))=1$ when $N=2$. \end{proof} We now recover a version of \cite[Corollary 4.17]{DFMS}. \begin{corollary}\label{CorIntRho2} Let $Z\subset\PP^N$ be a nontrivial fat point subscheme. Then $\rho(I(Z))=1$ if and only if $\widehat{\rho}(I(Z))=\rho_{int}(I(Z))=1$. \end{corollary} \begin{proof} The proof is immediate from $$1\leq\rho_{int}(I(Z))\leq\rho(I(Z)),$$ $$1\leq\widehat{\rho}(I(Z))\leq\rho(I(Z))$$ and Corollary \ref{CorIntRho1}. \end{proof} \begin{remark}\label{rhointRem} For a nontrivial homogeneous ideal $I\subset\KK[\PP^N]$, it is also of interest to define $$\rho^{int}(I)=\sup\Big\{\frac{m+1}{r} : \overline{I^m}\not\subseteq I^r\Big\}.$$ This is exactly what \cite{DFMS} denotes as $K(I)$. Clearly we have $\rho_{int}(I)\leq \rho^{int}(I)$, and by applying Theorem \ref{BSkoda} we see we have equality if and only if $\rho^{int}(I)=1$, in which case $I$ is normal. By \cite[Proposition 4.19]{DFMS} we have $\rho(I)\leq \widehat{\rho}(I)\rho^{int}(I)$. Thus we have $$\rho_{int}(I(Z))\leq \rho(I(Z))\leq \widehat{\rho}(I(Z))\rho^{int}(I(Z))$$ for every fat point subscheme $Z\subset\PP^N$. \end{remark} \section{Grifo's Conjecture} We now discuss Grifo's containment conjecture \cite[Conjecture 2.1]{G}. In our context it says the following (we note it is true and easy to prove for $N=1$). \begin{conjecture}\label{GrifoConj} Let $I\subseteq\KK[\PP^N]$ be a radical homogeneous ideal. Then $I^{(h_Ir-h_I+1)}\subseteq I^r$ for all $r\gg0$. \end{conjecture} Remark 2.7 of \cite{G} shows the conjecture holds for $I$ if $\rho(I)<h_I$ (whether $I$ is radical or not), or if $\rho'(I)<h_I$, and raises the question of whether it holds when $\widehat{\rho}(I)<h_I$. This was answered affirmatively by \cite[Proposition 2.11]{GHM} with a direct proof (we thank Grifo and Huneke for bringing this result to our attention). Our results also answer this question affirmatively, in two ways. By Theorem \ref{MT1}, $\widehat{\rho}(I)<h_I$ implies $\rho(I)'<h_I$, hence Conjecture \ref{GrifoConj} holds for $I$ by the results of \cite{G}. And by Theorem \ref{MT2}, $\widehat{\rho}(I)<h_I$ implies $\rho(I)<h_I$, so again Conjecture \ref{GrifoConj} holds for $I$ by the results of \cite{G}. \begin{remark}\label{GrifoCorRem} When $I=I(Z)$ for a fat point scheme $Z\subset\PP^N$ we have $h_I=N$. No examples of a fat point scheme $Z\subset\PP^N$ (radical or not) are known for which it is not true that $I(Z)^{(Nr-N+1)}\subseteq I(Z)^r$ for all $r\gg0$. By Remark 2.7 of \cite{G}, one approach to proving that $I(Z)^{(Nr-N+1)}\subseteq I(Z)^r$ for all $r\gg0$ holds for all $Z$ is to show $\rho(I(Z))<N$ whenever $N>1$. This raises the question of: for which $Z$ is it known that $\rho(I(Z))<N$? Let $Z=m_1p_1+\cdots+m_sp_s$. If ${\rm gcd}(m_1,\ldots,m_s)>1$, then $\rho(I(Z))<N$ by \cite[Proposition 2.1(2)]{TX}. Thus it is the cases with ${\rm gcd}(m_1,\ldots,m_s)=1$ that remain of interest. Another approach is to apply our Theorem \ref{MT2}: $\rho(I(Z))<N$ holds if $\widehat{\rho}(I(Z))<N$. Aiming to show $\widehat{\rho}(I(Z))<N$ has the advantage that the results of \cite{DFMS,DD} suggest that $\widehat{\rho}(I(Z))$ is more accessible computationally than is $\rho(I(Z))$. Another advantage is that in most cases where $\widehat{\rho}(I(Z))$ is known we have $\widehat{\rho}(I(Z))=\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}$ (but see Example \ref{Ex3}) and in all known cases we have $\widehat{\alpha}(I(Z))\geq \frac{\alpha(I(mZ))+N-1}{m+N-1}$ for all $m\geq 1$. Assuming both we have $$\widehat{\rho}(I(Z))=\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}\leq \frac{\alpha(I(Z))}{\frac{\alpha(I(Z))+N-1}{N}} = N\frac{\alpha(I(Z))}{\alpha(I(Z))+N-1}<N$$ and thus we would have $\rho(I(Z))<N$. \end{remark} The previous paragraph merits further discussion. First we recall \cite[Conjecture 2.1]{HaHu}, which if true would refine the containment $I(rNZ)\subseteq I(Z)^r$ of \cite{ELS, HoHu}: \begin{conjecture}\label{HaHu} Let $Z\subset\PP^N$ be a fat point scheme. Let $M=(x_0,\ldots,x_N)$. Then $$I(rNZ)\subseteq M^{r(N-1)}I(Z)^r$$ holds for all $r>0$. \end{conjecture} A further refinement of $I(rNZ)\subseteq I(Z)^r$ is that $I(r(m+N-1)Z)\subseteq I(mZ)^r$ \cite{ELS,HoHu}. This suggests a refinement of Conjecture \ref{HaHu} (cf. \cite[Question 4.2.3]{HaHu}), namely \begin{equation}\label{HaHuRefinement} I(r(m+N-1)Z)\subseteq M^{r(N-1)}I(mZ)^r. \end{equation} If \eqref{HaHuRefinement} were true, then the following conjecture would also be true: \begin{conjecture}\label{Chud} Let $Z\subset\PP^N$ be a fat point scheme. Then $$\widehat{\alpha}(I(Z))\geq \frac{\alpha(I(mZ))+N-1}{m+N-1}$$ for all $m\geq 1$. \end{conjecture} The proof that Conjecture \ref{HaHu} implies Conjecture \ref{Chud} when $m=1$ is given in \cite{HaHu}. The same argument shows that \eqref{HaHuRefinement} implies Conjecture \ref{Chud}. The first version of Conjecture \ref{Chud} was posed by Chudnovsky \cite{Ch} over the complex numbers for the case that $m=1$ and $Z$ is reduced. He also sketched a proof of his conjecture for $N=2$ which works over any algebraically closed field (see \cite{HaHu} for a proof). Conjecture \ref{Chud} assuming $Z$ reduced was posed by Demailly \cite{Dem}. The best result currently known is by Esnault and Viehweg \cite{EV}. It is over the complex numbers and says that if $Z$ is reduced with $N>1$, then $$\widehat{\alpha}(I(Z))\geq \frac{\alpha(I(mZ))+1}{m+N-1}$$ holds for all $m\geq 1$. Thus for reduced $Z$ over the complex numbers, taking $m=1$, we have $$\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}\leq \frac{\alpha(I(Z))}{\frac{\alpha(I(Z))+1}{N}} = N\frac{\alpha(I(Z))}{\alpha(I(Z))+1}<N.$$ Hence $\rho(I(Z))<N$ holds over the complex numbers whenever $Z$ is reduced and $\widehat{\rho}(I(Z)) = \frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}$. \section{Examples and questions}\label{ExSect} For this section assume $Z$ is a fat point subscheme of $\PP^N$ (so $I(Z)$ is a nontrivial ideal of $\KK[\PP^N]$). \begin{example}\label{Ex1} If $I(Z)$ is a complete intersection or a power thereof, then $I(mZ)=I(Z)^m$ for all $m\geq1$, hence $\rho(I(Z))=1$ by Theorem \ref{MT3}. Again by Theorem \ref{MT3}, if $Z$ is reduced, then $\rho(I(Z))=1$ if and only if $I(Z)$ is a complete intersection. However, when $Z$ is not reduced, $\rho(I(Z))=1$ does not imply $I(Z)$ is a complete intersection, or even a power of a complete intersection ideal (homogeneous or not), as is shown by Example \ref{verticesEx} and Remark \ref{FatPtsAndCI1}. (Additionally, let $p_1,p_2,p_3\in\PP^N$ be noncollinear points and let $Z=m_1p_1+m_2p_2+m_3p_3$ with $1\leq m_1\leq m_2\leq m_3$. If either $m_1+m_2\leq m_3$ or if $m_1+m_2+m_3$ is even, then by \cite[Theorem 2]{BZ} we have $I(mZ)=I(Z)^m$ for all $m\geq1$ and hence $\rho(I(Z))=1$. See \cite[Example 5.1]{BH2} for additional examples in $\PP^2$ of $Z$ for which all powers of $I(Z)$ are symbolic. By Remark \ref{FatPtsAndCI2}, in none of these cases is there a homogeneous complete intersection ideal $J$ such that $I(Z)=J^r$ for some $r\geq1$. See \cite[Proposition 3.5]{HaHu} for a criterion for $Z\subset\PP^2$ such that $I(mZ)=I(Z)^m$ for all $m\geq1$; this gives further examples for which $I(Z)$ is not a power of a homogeneous complete intersection.) \end{example} It is an interesting problem to clarify which fat point subschemes $Z\subset\PP^N$ have $I(mZ)=I(Z)^m$ for all $m\geq1$. We do not know any $Z$ for which the analytic spread $\ell(I(Z))=N$ but for which $I(mZ)=I(Z)^m$ fails for some $m\geq1$. Nor do we know any $Z$ for which either $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$ but for which $I(mZ)=I(Z)^m$ fails for some $m\geq1$. Thus we have the following question. \begin{question}\label{Q-NonreducedZCI} Is it true that all powers of $I(Z)$ are symbolic (i.e., $I(mZ)=I(Z)^m$ for all $m\geq1$) if the analytic spread of $I(Z)$ is $N$, or if $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$? \end{question} \begin{example}\label{Ex1b} It is worth noting here that there is a monomial ideal $I$ with $\rho(I)=\widehat{\rho}(I)=1$ where $I^m\subsetneq I^{(m)}$ for every $m>1$ (\cite[Remark 3.5]{DD}; we thank DiPasquale and Grifo for bringing this to our attention). Since the ideal $I$ given in \cite[Remark 3.5]{DD} is a radical monomial ideal and powers of monomial primes are primary, we have $I^{(m)}=\cap_{P\in{\rm Ass(I)}}P^m$. But monomial primes are normal, so $I^{(m)}$ is integrally closed. Thus $\overline{I^m}\subseteq I^{(m)}$, and since $\widehat{\rho}(I)=1$ we have by \cite[Corollary 4.16]{DFMS} that $I^{(m)}\subseteq \overline{I^m}$, so $I^{(m)}=\overline{I^m}$ for all $m\geq 1$. Thus $I^m$ is integrally closed only for $m=1$, but nonetheless $\rho_{int}(I)=\rho(I)=1$. \end{example} Answering Question \ref{Q-NonreducedZCI} is closely related to whether having $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$ gives a complete solution to the containment problem for $I(Z)$. If $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$, then by Theorem \ref{MT3} when $I(Z)$ is radical, we do have a complete solution to the containment problem for $I(Z)$: $I(mZ)\not\subseteq I(Z)^r$ for $m<r$ and otherwise we have $I(mZ)\subseteq I(Z)^r$. When $I(Z)$ is not radical, we do not know if having either $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$ solves the containment problem for $I(Z)$. For example, having $\rho(I(Z))=1$ means $\widehat{\rho}(I(Z))=1$ and it means $I(mZ)\not\subseteq I(Z)^r$ for $m<r$ and $I(mZ)\subseteq I(Z)^r$ for $m>r$, but we do not know for which $m\geq1$ that we have $I(mZ)\subseteq I(Z)^m$. We also do not know any examples with $\widehat{\rho}(I(Z))=1$ but $\rho(I(Z))>1$. This raises the following question. \begin{question}\label{Q4b} Does $\widehat{\rho}(I(Z))=1$ always imply $\rho(I(Z))=1$? \end{question} \begin{example}\label{Ex4} The assumption $I=\overline{I}$ is needed in both Lemma \ref{rhoIntLem} and Theorem \ref{BSkoda}(b). For example, take $t=1$ and any $N\geq1$. Let $I=(x_0^{N+1},x_1^{N+1},\ldots,x_N^{N+1})\subset\KK[\PP^N]$. Then $x_0^N\cdots x_N^N\in\overline{I^N}=(x_0,\ldots,x_N)^{N(N+1)}$ but $x_0^N\cdots x_N^N\not\in I$, so $\overline{I^{tN}}\not\subseteq I^t$ and $\rho_{int}(I)=N$. \end{example} Although $I(Z)$ is not always normal (see Example \ref{Ex2}) and hence $I(Z)^m=\overline{I(Z)^m}$ can fail, we do not have an example with $\overline{I(Z)^m}\not\subseteq I(Z)^r$ when $m>r$, so we do not know of any $Z$ for which $\rho_{int}(I(Z))\neq1$. If $\rho_{int}(I(Z))=1$ were always true, then by Corollary \ref{CorIntRho2} we would have that $\rho(I(Z))=1$ if and only if $\widehat{\rho}(I(Z))=1$, thus answering Question \ref{Q4b}. This raises the following question. \begin{question}\label{FinalQuest2} Is it ever true that $\rho_{int}(I(Z)) >1$? \end{question} The next example shows that $\rho_{int}(I(Z))=1$ does not force $\rho(I(Z))=1$ or $\widehat{\rho}(I(Z))=1$, even if $Z$ is reduced. In contrast, we know that (b) (and hence (a)) of Theorem \ref{MT3} implies $\rho_{int}(I(Z))=1$, but we do not know if any of the other criteria of Theorem \ref{MT3} imply $\rho_{int}(I(Z))=1$, unless $Z$ is reduced or $N=2$. \begin{example}\label{Ex5} Examples occur with $\rho_{int}(I(Z))=1$ but with $\widehat{\rho}(I(Z))>1$. Let $Z\subset\PP^N$ be a {\it star configuration}, meaning we have $s>N$ general hyperplanes, and $Z$ is the reduced scheme consisting of $\binom{s}{N}$ points, where each point is the intersection of $N$ of the $s$ hyperplanes; see \cite[Definition 3.8]{HaHu}. Then the ideal $I(Z)$ has $\alpha(I(Z))={\rm reg}(I(Z))=s+N-1$ \cite[Lemma 2.4.2]{BH1} and $\widehat{\alpha}(I(Z))=s/N$ \cite[Lemma 2.4.1]{BH1}. As noted in \S\ref{bkgrnd}, $\alpha(I(Z))={\rm reg}(I(Z))$ implies that $I(Z)$ is normal and hence $\rho_{int}(I(Z))=1$, but $\alpha(I(Z))={\rm reg}(I(Z))$ also implies that $\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}=\widehat{\rho}(I(Z))=\rho(I(Z))$, and in the case of a star configuration we have $\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))}=N(s-N+1)/s>1$ when $N>1$. However we do not know of any $Z$ with $\widehat{\rho}(I(Z))<\rho_{int}(I(Z))$. \end{example} \begin{question}\label{FinalQuest} Is it ever true that $\rho_{int}(I(Z)) > \widehat{\rho}(I(Z))$? Do any of (c), (d) or (e) of Theorem \ref{MT3} imply $\rho_{int}(I(Z))=1$, when $Z$ is not reduced and $N>2$? \end{question} \begin{example}\label{Ex2} When $\widehat{\rho}(I(Z)) >1$, it is known that $\widehat{\rho}(I(Z)) < \rho(I(Z))$ can occur. For example, let $I = (x(y^n-z^n),y(z^n-x^n),z(x^n-y^n))\subset \CC[x,y,z]=\CC[\PP^2]$, so $I=I(Z)$ where $Z$ is a certain set of $n^2+3$ points. Then by \cite[Theorem 2.1]{DHNSST}, we have $\widehat{\rho}(I(Z)) = \frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))} =(n+1)/n < 3/2=\rho(I(Z))$ for $n\geq3$. Moreover, since $\widehat{\rho}(I(Z)) < \rho(I(Z))$, it follows that $I(Z)$ cannot be normal. \end{example} \begin{example}\label{Ex3} Moreover, examples of $Z$ occur with $\frac{\alpha(I(Z))}{\widehat{\alpha}(I(Z))} < \widehat{\rho}(I(Z))$. The results of \cite{DFMS} suggest that this should occur, but up to now no explicit examples have been given. For one such explicit example (indeed, the first we are aware of), let $Z$ consist of 8 points in the plane, where 3 of the points (say $p_1,p_2,p_3$) are general (we may as well assume they are the coordinate vertices) and the other 5 (say $p_4,\ldots,p_8$) are on a general line $L$ (defined by a linear form $F$) and are general on that line. Then one can show that $\alpha(I(Z))=3$, $\widehat{\alpha}(I(Z))=5/2$ and $I(25sZ)\subsetneq I(Z)^{19s+1}$ for all $s\geq1$, and hence that $\alpha(I(Z))/\widehat{\alpha}(I(Z)) =6/5 < 25/19\leq \widehat{\rho}(I(Z))$. (We now sketch the justification of these claims. The key is that $(I(25sZ))_{65s}$ vanishes on $L$ with order $15s$, but $(I(Z)^{19s+1})_{65s}$ vanishes on $L$ with order $15s+3$, hence $(I(25sZ))_{65s}\not\subseteq (I(Z)^{19s+1})_{65s}$ and so $I(25sZ)\not\subseteq I(Z)^{19s+1}$. It is easy to check that $\alpha(I(Z))=3$. Using Bezout's Theorem, one can show that $(I(2tZ))_{5t}=(xyzF^2)^t\KK\subset\KK[\PP^2]=\KK[x,y,z]$ and hence (since $\dim ((I(mZ))_i)>0$ implies $\dim ((I(mZ))_j)>1$ for all $j>i$) that $\alpha(I(2tZ))=5t$. (In more detail, note that $(xyzF^2)^t\in (I(2tZ))_{5t}$. If $(I(2tZ))_{5t}$ contained anything more than scalar multiples of $(xyzF^2)^t$, then $(I(8tZ))_{20t}$ would contain more than scalar multiples of $(xyzF^2)^{4t}$, so it's enough to consider $(I(8tZ))_{20t}$. But by Bezout, we have \begin{align*} (I(8tZ))_{20t}&=F^{5t}((I(8t(p_1+p_2+p_3)+3t(p_4+\cdots+p_8)))_{15t})\\ &=(xyz)^tF^{5t}((I(3t(2(p_1+p_2+p_3)+(p_4+\cdots+p_8))))_{12t})\\ &=(xyz)^tF^{5t+1}((I((3t(2(p_1+p_2+p_3))+(3t-1)(p_4+\cdots+p_8))))_{12t-1})\\ &=(xyz)^{t+1}F^{5t+1}((I((3t-1)(2(p_1+p_2+p_3)+(p_4+\cdots+p_8))))_{12t-4})\\ &=\cdots=(xyz)^{4t}F^{8t}\KK.)\\ \end{align*} \vskip-\baselineskip \noindent It follows that $\widehat{\alpha}(I(Z))=5/2$. We also have by Bezout that $(I(25sZ))_{65s}=F^{15s}(I(5sZ'))_{50s}$, where $Z'$ is the fat point subscheme obtained by taking the three coordinate vertices with multiplicity 5 and the five points on $L$ with multiplicity 2. Moreover, one can check that $(I(Z'))_{10}$ has greatest common divisor 1 (i.e., it defines a linear system which is fixed component free); to see this, write elements of $(I(Z'))_{10}$ in two different ways using conics and lines, where the two ways have no factors in common. Thus $(I(5sZ'))_{50s}$ also defines a linear system that is fixed component free, so $(I(25sZ))_{65s}=F^{15s}(I(5sZ'))_{50s}$ has $15sL$ as the divisorial part of its base locus. But $(I(Z))_t$ has $L$ in its base locus for $t=3,4$ and is fixed component free for $t\geq5$. Now $(I(Z))^{19s+1}$ is spanned by products of $a$ homogeneous elements of $I(Z)$ of degree 5 or more and $b$ elements of degree 4 or less with $a+b=19s+1$. Such an element vanishes on $L$ to order at least $b$. To minimize $b$ we want $a$ as large as possible, and hence we want to take all $a$ elements to have degree 5 and as many of the $b$ elements as possible to have degree 3, and thus we have the inequality $5a+3b\leq 65s$. Solving $5a+3b\leq 65s$ given $a+b=19s+1$ gives $2a\leq 8s-3$ and hence, taking $a$ as large as possible we have $a=4s-2$, so $b=15s+3$. I.e., every element of $((I(Z))^{19s+1})_{65s}$ vanishes on $L$ to order at least $15s+3$.) We do not know if $\widehat{\rho}(I(Z))=25/19$, nor do we know the value of $\rho(I(Z))$. \end{example} \begin{question}\label{rhohatValue} For the example $Z$ in the previous paragraph, what are the exact values of $\widehat{\rho}(I(Z))$, $\rho(I(Z))$ and $\rho_{int}(I(Z))$? \end{question} As Remark \ref{GrifoCorRem} explains, if the next question has a negative answer, then Conjecture \ref{GrifoConj} holds for $I=I(Z)$ whenever $\widehat{\rho}(I(Z))=\alpha(I(Z))/\widehat{\alpha}(I(Z))$. \begin{question} Is it ever false that $$\widehat{\alpha}(I(Z))\geq \frac{\alpha(I(Z))+1}{N}?$$ \end{question} By \cite{DFMS}, $\widehat{\rho}(I(Z))$ is equal to the maximum of $v(I(Z))/\widehat{v}(I(Z))$ for valuations $v$ supported on $I(Z)$, and this maximum always occurs for what is known as a Rees valuation. Thus we have $$\frac{v(I(Z))}{\widehat{v}(I(Z))}\leq \widehat{\rho}(I(Z))\leq N,$$ hence $$\frac{v(I(Z))}{N}\leq \widehat{v}(I(Z)).$$ This raises the question of whether Chudnovsky-like bounds occur for valuations: \begin{question}\label{FinalQuest3} Is it always true for $N>1$ that $$\frac{v(I(Z))+1}{N}\leq \widehat{v}(I(Z))?$$ \end{question} If the answer is affirmative, then for some $v$ we would have $$\widehat{\rho}(I(Z))=\frac{v(I(Z))}{\widehat{v}(I(Z))} \leq \frac{v(I(Z))}{\frac{v(I(Z))+1}{N}}<N$$ which would confirm Grifo's Conjecture for $I(Z)$. \begin{example}\label{verticesEx} Consider $Z=p_1+\cdots+p_N+2p_{N+1}\subset\PP^N$ where the points $p_i$ are the coordinate vertices. Here we show that $I(Z)^m=I(mZ)$ for all $m\geq1$. For $N=1$, $I(Z)^m=I(mZ)$ holds for every $Z$ (since fat point ideals are principal), so assume $N>1$. (We note that the case $N=2$ is covered by the results of \cite{BZ}.) Since $I(Z)$ is a monomial ideal, we merely have to check for each monomial $f=x_0^{a_0}\cdots x_N^{a_N}\in I(mZ)$ that $f\in I(Z)^m$; i.e., that $f$ can is divisible by a product of $m$ monomials, each in $I(Z)$. But if we choose coordinates $x_i$ such that $x_j\neq0$ at $p_j$ for each $j$, then $f\in I(mZ)$ is equivalent to $a_1+\cdots+a_N\geq 2m$ and $a_0+a_1+\cdots+a_N-a_i\geq m$ for each $0<i\leq N$. Without loss of generality we may assume $a_1\leq a_2\leq \cdots\leq a_N$, so the inequalities $a_0+a_1+\cdots+a_N-a_i\geq m$ reduce to the single inequality $a_0+a_1+\cdots+a_{N-1}\geq m$. Let us set $b=a_1+\cdots+a_{N-1}$. Then $f\in I(mZ)$ if and only if $a_0+b\geq m$ and $b+a_N\geq 2m$, so we want to show $a_0+b\geq m$ and $b+a_N\geq 2m$ implies $f\in I(Z)^m$. First suppose $b\geq m$. We have two cases: $b+a_N=2m$ and $b+a_N>2m$. First assume $b+a_N=2m$, so $0<b/(N-1)\leq a_{N-1}\leq a_N\leq b$. Let $e_0=(e_{01},\ldots,e_{0N})=(a_1,\cdots,a_N)$ so $e_0$ is the exponent vector of $g_0=x_1^{e_{01}}\cdots x_N^{e_{0N}}=x_1^{a_1}\cdots x_N^{a_N}$. Let $b_0=b=e_{01}+\cdots+e_{0,N-1}$. Note that $g_0\in I(mZ)$ and $g_0$ divides $f$, so it's enough to show $g_0\in I(Z)^m$. Starting with $e_0$ and $b_0$, we will recursively define a sequence $e_i$ of exponent vectors $e_i=(e_{i1},\ldots,e_{iN})$ and nonnegative integers $b_i$ for $0\leq i\leq \omega$ with $b_i>0$ for $0\leq i<\omega$ and $b_\omega=0$, satisfying the following conditions: \begin{enumerate} \item[(i)] the entries of $e_i$ are nonnegative and nondecreasing; \item[(ii)] $b_i+e_{i,N}$ is even, where $b_i=e_{i1}+\cdots+e_{i,N-1}$; \item[(iii)] $b_i\geq e_{iN}$. \end{enumerate} For $e_0$ we have $b_0=b\geq m>0$. Moreover, conditions (i)-(iii) hold for $e_0$: (i) holds by assumption; (ii) holds since $b_0+e_{0,N}=b+a_N=2m$ by assumption; (iii) holds since we noted above that $a_N\leq b$, but $e_{0N}=a_N$ and $b_0=b$. Given that conditions (i)-(iii) hold for some $e_i$ with $b_i>0$, we now define $e_{i+1}=(e_{i+1,1},\ldots,e_{i+1,N})$ with $b_i>b_{i+1}$. In brief, $j_i$ and $k_i$ are chosen to be as large as possible such that $j_i<k_i$ and so that the entries of $e_{i+1}=(e_{i+1,1},\ldots,e_{i+1,N})$ are nondecreasing, where $e_{i+1,l}=e_{il}$ for all $l$ except that $e_{i+1,j_i}=e_{ij_i}-1$ and $e_{i+1,k_i}=e_{ik_i}-1$. More precisely, $j_i$ is the least index $t$ such that $e_{it}=e_{i,N-1}$. Then $k_i=N$ if $e_{iN}>e_{i,N-1}$ and $k_i=j_i+1$ if $e_{iN}=e_{i,N-1}$. Since $b_{i+1}$ is either $b_i-1$ or $b_i-2$ we have $b_i>b_{i+1}$. It is easy to check that the construction ensures that the entries of $e_{i+1}$ are nonnegative and nondecreasing, so (i) holds for $e_{i+1}$. Since $b_i+e_{i,N}$ is even and $b_{i+1}+e_{i+1,N}=b_i+e_{i,N}-2$, (ii) holds. For (iii), if $e_{i+1,N}=e_{iN}$, then $k_i<N$ so $0<e_{i,N-2} = e_{i,N-1}=e_{i,N}$, and we cannot have $e_{i,N-2}=1$ with $e_{i,N-3}=0$ since then the sum of the entries of $e_i$ would be odd. Thus either $e_{i,N-2}>1$ or $e_{i,N-3}>0$; either way $b_i\geq 2+e_{i,N-1}=2+e_{iN}$, so $b_{i+1}=b_i-2\geq e_{iN}=e_{i+1,N}$. If $e_{i+1,N}=e_{iN}-1$, then $b_{i+1}=b_i-1$, so $b_i\geq e_{iN}$ implies $b_{i+1}\geq e_{i+1,N}$ (and hence $b_{i+1}\geq0$). Let $g_i$ be the monomial whose exponent vector is $e_i$, so $g_i=x_1^{e_{i1}}\cdots x_N^{e_{iN}}$. We have $g_\omega=1$, since $0=b_\omega\geq e_{\omega N}$, so $e_\omega=(0,\ldots,0)$. If $i<\omega$, then $g_i=g_{i+1}x_{j_i}x_{k_i}$. Thus $g_0=(x_{j_0}x_{k_0})\cdots (x_{j_{\omega-1}}x_{k_{\omega-1}})$, but each factor $x_{j_i}x_{k_i}$ is in $I(Z)$, and since $\deg g_0=b_0+e_{0N}=2m$, we see that there are $m$ factors, so $g_0\in I(Z)^m$, as we wanted to show. We now consider case 2, $b+a_N>2m$ (still under the assumption that $b\geq m$). Starting with $(a_1,\ldots,a_{N-1},a_N)$, replace $a_N$ by the smallest integer $a$ satisfying $a\geq a_{N-1}$ and $b+a\geq 2m$. We then have that $g=x_0^{a_0}\cdots x_{N-1}^{a_{N-1}}x_N^a$ divides $f$ and is still in $I(mZ)$. Thus it is enough to show that $g\in I(Z)^m$. If $b+a=2m$, we are done by case 1, so assume $b+a>2m$. Then by construction we have $a=a_{N-1}$. Since $b\geq m$, we must have $a\leq m$. It is now not hard to find new exponents $0\leq a_1'\leq a_2'\leq \cdots\leq a_{N-1}'$ such that $a_i'\leq a_i$ and $b'+a=2m$, where $b'=a_1'+\cdots+a_{N-1}'$. Let $g_0=x_1^{a_1'}\cdots x_{N-1}^{a_{N-1}'}x_N^a$; then $g$ divides $f$ and by case 1 we have $g\in I(Z)^m$. We are left with considering the case that $b<m$. We have $a_0+b\geq m$ and $b+a_N\geq 2m$. Clearly we can reduce $a_0$ and $a_N$ so that $a_0+b = m$ and $b+a_N = 2m$ (since the associated monomial $g$ divides $f$ and still is in $I(mZ)$). So we may assume $a_0+b = m$ and $b+a_N = 2m$, in which case $f=(x_1^{a_1}\cdots x_{N-1}^{a_{N-1}}x_N^b)x_0^{a_0}x_N^{a_N-b} =(x_1^{a_1}\cdots x_{N-1}^{a_{N-1}}x_N^b)x_0^{m-b}x_N^{2(m-b)} =(x_1^{a_1}\cdots x_{N-1}^{a_{N-1}}x_N^b)(x_0x_N^2)^{m-b}$. But $x_0X_N^2\in I(Z)$ and $x_1^{a_1}\cdots x_{N-1}^{a_{N-1}}x_N^b$ is a product of $b$ factors of the form $x_ix_N$ for $1\leq i\leq N-1$, and each of these is in $I(Z)$. Thus $f$ is a product of $m=b+(m-b)$ elements of $I(Z)$, hence $f\in I(Z)^m$. \end{example} \begin{remark}\label{FatPtsAndCI1} Again consider $Z=p_1+\cdots+p_N+2p_{N+1}\subset\PP^N$ where the points $p_i$ are the coordinate vertices and $N>1$. Then there is no $N$-generated ideal $J$, homogeneous or not, such that $I(Z)=J^s$ for some $s\geq1$. Suppose there were; say $J=(F_1,\dots,F_N)$. The ideal $I(Z)$ defines the $N+1$ coordinate lines in affine $N+1$ space, all taken with multiplicity 1 except one taken with multiplicity 2. Since each $F_i$ vanishes on each coordinate line, none of the $F_i$ can have any terms which are a power of a single variable (i.e., every term involves a product of two or more variables). Therefore, none of the $F_i$ have terms of degree less than 2. Since $(F_1,\dots,F_N)^s=J^s=I(Z)$ and since $I(Z)$ has elements with terms of degree 2, we see that $s=1$ (otherwise the least degree of a term of an element of $J^s$ would be at least $2s\geq4$). But $s=1$ implies that $J=I(Z)$ is homogeneous, and an easy argument shows that the least number of homogeneous generators of a homogeneous ideal is the least number of generators possible, which in this case is $\binom{N}{2}+N>N$ (since, for a particular ordering of the points $p_i$ we have $I(Z)=(x_ix_j:i>0, j>0)+(x_0x_i^2:i>0)$). \end{remark} \begin{remark}\label{FatPtsAndCI2} Let $Z=m_1p_1+\cdots+m_rp_r\subset\PP^N$ with $N>1$, $m_i>0$ for all $i$ and $\KK[\PP^N]=\KK[x_0,\ldots,x_N]$. Here we show that there is an ideal $J$ generated by $N$ forms such that $I(Z)=J^m$ for some $m\geq1$ if and only if $m_1=\cdots=m_r=m$ where $I(p_1+\cdots+p_r)$ is generated by $N$ forms. The reverse implication is known \cite[Lemma 5, Appendix 6]{ZS}, so assume $I(Z)=J^m$ for some $m\geq1$ with forms $F_i$ such that $J=(F_1,\ldots,F_N)$. Choose coordinates $x_0,\ldots,x_N$ such that none of the points lies on $x_0=0$ and $p_1=(1,0,\ldots,0)$. Let $U_0$ be the affine neighborhood defined by $x_0\neq0$, and let $f_i(x_1,\ldots,x_N)=F_i(1,x_1,\ldots,x_N)$ be the polynomial obtained by setting $x_0$ to 1 in $F_i$. Then on $U_0$, $Z$ is defined by the ideal $I'(Z)$ obtained from $I(Z)$ by setting $x_0=1$ in each element of $I(Z)$, so $I'(Z)=(J')^m$ where $J'=(f_1,\ldots,f_N)$. Localizing at $P_1=I(p_1)=(x_1,\ldots,x_N)$ gives $(P_1)_{P_1}^{m_1}=(I'(Z))_{P_1}=(J')^m_{P_1}$ and modding out by $P_1^{m_1+1}$ gives graded isomorphisms $$\frac{P_1^{m_1}}{P_1^{m_1+1}}\cong \frac{(P_1)_{P_1}^{m_1}}{(P_1)_{P_1}^{m_1+1}}= \frac{(J')^m_{P_1}}{(P_1)_{P_1}^{m_1+1}}\cong \frac{(J')^m+P_1^{m_1+1}}{(P_1)^{m_1+1}}.$$ Thus each $f_i$ can have no terms of degree less than $d=m_1/m$ and there must be terms of degree exactly $d$, so $d$ is an integer, hence $m_1=dm$ and $m\leq m_1$. But the vector space dimension of $\frac{P_1^{m_1}}{P_1^{m_1+1}}$ is $\binom{m_1+N-1}{N-1}$, and the vector space dimension of $\frac{(J')^m+P_1^{m_1+1}}{(P_1)^{m_1+1}}$ is at most $\binom{m+N-1}{N-1}$, so we have $\binom{m_1+N-1}{N-1}\leq \binom{m+N-1}{N-1}$, which implies $m_1\leq m$, hence $m=m_1$ and $d=1$. With a change of coordinates, the same argument works for each point $p_i$, so $m_i=m$ for all $i$, and, at each point $p_i$, the linear terms of the $f_j$ must be linearly independent (otherwise we would have $\dim_\KK\frac{(J')^m+P_1^{m_1+1}}{(P_1)^{m_1+1}}<\binom{m+N-1}{N-1}$). Thus $J'=I(p_1+\cdots+p_r)$ on $U_0$, hence $J=I(p_1+\cdots+p_r)\subseteq\KK[\PP^N]$. \end{remark} \begin{question} In Remark \ref{FatPtsAndCI2}, do we need to assume a priori that $J$ is homogeneous? \end{question} \section{Computational estimates of resurgences} It might be possible to address some of the foregoing questions computationally. More generally, it is of interest to consider to what extent quantities like resurgences can be computed. \subsection{Denkert's thesis} In an unpublished part of her thesis \cite{Den}, Denkert gives an algorithm for computing $\rho(I(Z))$ arbitrarily accurately when the symbolic Rees algebra $$R_s(I(Z))=\oplus_t I(tZ)x^t\subseteq \KK[\PP^N][x]$$ is Noetherian (equivalently, when for some $a\geq 1$, all powers of $I(aZ)$ are symbolic \cite[Theorem 2.1]{HHT}). Let $I=I(Z)\subseteq \KK[\PP^N]$ be nontrivial and assume $I^{(at)} = (I^{(a)})^t$ for all $t > 0$. For each $s\geq1$, let $b_s$ be the largest $b$ such that $I^{(asN)}\subseteq I^{b}$ and let $\epsilon>0$. Since $b_s\geq as$, we have $\frac{asN}{b_s}-\frac{asN}{b_s+1}=\frac{asN}{b_s(b_s+1)}\leq\frac{N}{as+1}$. So, by picking $s\gg0$ and $\epsilon$ small, we can make $\frac{N}{as+1}+\epsilon$ arbitrarily small. Denkert's algorithm either computes $\rho(I)$ exactly or gives an estimate which is accurate to less than $\frac{N}{as+1}+\epsilon$. Assume we have picked $s$ and $\epsilon$. Let $A=asN$ and let $B=b_s$. Thus we have $$I^{(At)}=(I^{(A)})^t\subseteq (I^{B})^t=I^{Bt}$$ for all $t\geq1$, and for $m\geq At$ and $r\leq Bt$ we have $$I^{(m)}\subseteq I^{(At)}\subseteq I^{Bt}\subseteq I^r.$$ If $r\geq B\lceil\frac{A}{B\epsilon}\rceil$ and $\frac{m}{r}\geq\frac{A}{B}+\epsilon$, we now show for the least $t$ such that $r<Bt$ that $m\geq At$, and hence $I^{(m)}\subseteq I^r$. Indeed, we have $B(t-1)\leq r<Bt$, so $t-1=\lfloor\frac{r}{B}\rfloor\geq\lceil\frac{A}{B\epsilon}\rceil$. Thus $(t-1)B\epsilon\geq A$, which is equivalent to $(\frac{A}{B}+\epsilon)(t-1)B\geq At$, hence we have $m\geq r(\frac{A}{B}+\epsilon)\geq B(t-1)(\frac{A}{B}+\epsilon)\geq At$. In summary, we have $I^{(m)}\subseteq I^r$ for all $r\geq B\lceil\frac{A}{B\epsilon}\rceil$ and $\frac{m}{r}\geq\frac{A}{B}+\epsilon$, and we have $I^{(A)}\not\subseteq I^{B+1}$. Thus $\frac{A}{B+1}\leq \rho(I)$ and either $\rho(I)< \frac{A}{B}+\epsilon$ or $\rho(I)=m/r$ for some $r< B\lceil\frac{A}{B\epsilon}\rceil$ with $N>m/r\geq\frac{A}{B}+\epsilon$ (and there are only finitely many such $r$ and $m$). Moreover, since we have $I^{(m)}\subseteq I^r$ for all but finitely many $m$ and $r$ with $m/r\geq\frac{A}{B}+\epsilon$, we have $\widehat{\rho}(I) \leq \frac{A}{B}+\epsilon$ for each $\epsilon>0$, and hence we have $\widehat{\rho}(I) \leq \frac{A}{B}$. This also follows by Theorem 1.2(3) of \cite{GHVT}. \subsection{DiPasquale-Drabkin method} An alternate approach is based on \cite{DD}. Again assume the symbolic Rees algebra of a homogeneous ideal $I\subset\KK[\PP^N]$ is finitely generated. Then \cite{DD} shows that we can compute $\widehat{\rho}(I)$ exactly (assuming we know the Rees valuations). Let $\epsilon>0$. Then, as in the proof of Theorem \ref{MT1}, whenever $I^{(m)}\not\subseteq I^r$, we have $\frac{s}{r+N} < \widehat{\rho}(I)$. But there are only finitely many $r$ for which there is an $s$ such that both $\frac{s}{r+N} < \widehat{\rho}(I)$ and $\widehat{\rho}(I)+\epsilon \leq s/r$ hold. For each of these $s$ and $r$ we check if $I^{(s)}$ is not contained in $I^r$. For all such $r$ and $s$ which occur (if any), then $\rho(I)$ is the maximum of their ratios $s/r$. If none occur, then $\widehat{\rho}(I) \leq \rho(I) < \widehat{\rho}(I)+ \epsilon$. The estimates on $\rho(I)$ given by using \cite{DD} versus those of \cite{Den} are somewhat similar, but each starts with different input data. Also, \cite{DD} gives us $\widehat{\rho}(I)$ exactly (if we know the Rees valuations), whereas \cite{Den} just gives us an upper bound on $\widehat{\rho}(I)$.

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\begin{abstract} Techniques from higher categories and higher-dimensional rewriting are becoming increasingly important for understanding the finer, computational properties of higher algebraic theories that arise, among other fields, in quantum computation. These theories have often the property of containing simpler sub-theories, whose interaction is regulated in a limited number of ways, which reveals a topological substrate when pictured by string diagrams. By exploring the double nature of computads as presentations of higher algebraic theories, and combinatorial descriptions of ``directed spaces'', we develop a basic language of directed topology for the compositional study of algebraic theories. We present constructions of computads, all with clear analogues in standard topology, that capture in great generality such notions as homomorphisms and actions, and the interactions of monoids and comonoids that lead to the theory of Frobenius algebras and of bialgebras. After a number of examples, we describe how a fragment of the ZX calculus can be reconstructed in this framework. \end{abstract} \section{Introduction} A traditional presentation of an algebraic theory consists of a number of generating operations, together with a number of equations that they satisfy. If we are concerned with computational aspects of the presentation --- looking for normalisation procedures, for instance --- it is commonplace to replace equations with directed \emph{rewrite rules}. Then, in the analysis of critical pairs and confluences of a rewrite system, we are led to consider relations between different sequences of rewrites, which can in turn be relaxed to ``rewrites of rewrites'', and so on, leading into \emph{higher-dimensional rewriting theory} \cite{mimram2014towards}. From this perspective, the dichotomy between generators and relations in a presentation is resolved: they both become generators of a higher-dimensional algebraic theory, only differing in dimension. The natural setting for higher-dimensional rewriting is \emph{higher category theory}, where, besides the objects (0-cells) and morphisms (1-cells) of basic category theory, there can also be $n$-cells between $(n-1)$-cells, for any $n > 0$. The use of terminology borrowed from topology is not coincidental: there is a sense in which the ``directed $n$-cells'' of higher categories behave like topological $n$-cells. This is exemplified by the successful application of methods from homology theory in the study of rewriting systems, based on this analogy \cite{lafont2007algebra, lafont2009polygraphic}; however, it is perhaps best pictured through the use of \emph{string diagrams} \cite{selinger2011survey, hinze2016equational} (or, more recently, surface diagrams \cite{dunn2016surface}) for reasoning about higher categories. In one especially relevant application, string diagrams have emerged as a strong contender for a high-level, native syntax for quantum programming \cite{zeng2015abstract, coecke2015picturing}, whose highly symmetrical semantics --- in pure, finite-dimensional quantum theory, all processes are reversible, and inputs can be turned into outputs and vice versa \cite{selinger2011finite} --- require a quite unusual amount of interplay between algebraic and coalgebraic structures. Better understanding the computational properties of theories such as the ZX calculus \cite{coecke2008interacting, backens2014zx} and its refinement, the ZW calculus \cite{hadzihasanovic2015diagrammatic}, is pivotal in making them viable for the efficient design of quantum algorithms and protocols. Although these theories include, as a whole, a relatively large number of axioms, they contain a number of simpler sub-theories, whose interactions are regulated by the axioms in fairly predictable ways: something is a homomorphism of something else, something is an action on something else... Beyond these motivating examples, such a factorisation seems to be a property of many theories: as a simple case, think of $*$-monoids, which can be seen as an interaction of the theory of monoids and the theory of involutions. So we asked ourselves the question: \begin{itemize} \item Is there a way to study algebraic theories \emph{compositionally}, so that one can derive properties of the larger theory from its components, and the few ways in which they are allowed to interact? \end{itemize} There has already been, in fact, an attempt to develop a compositional algebra, through Lack's ``composing PROPs'' framework \cite{lack2004composing}. In this setting, a presentation of a fragment of the ZX calculus --- the theory of \emph{interacting bialgebras} --- was successfully constructed from the theories of monoids and of comonoids \cite{bonchi2014interacting}. There are, nevertheless, two downsides to this approach, relative to our objectives. Firstly, composition relies on the choice of a ``distributive law'', which conceptually amounts to stating what the normal form for operations of the resulting theory should be. Thus, we can already tell what the resulting theory will \emph{globally} look like, which subtracts something from its heuristic value, especially when we only have an algebraic presentation at hand. In fact, ``composing PROPs'' is mostly useful to derive axioms when concrete models of the component theories are available, also suggesting a ``concrete'' way of composing them. Secondly, it is a \emph{flat} composition, in that it works in a strictly 2-categorical framework, and fails to account for any of the topological properties of the interactions. For instance, two specular distributive laws for monoids and comonoids lead, respectively, to the theory of special Frobenius algebras, and to the theory of bialgebras. Both the theory of monoids and the theory of comonoids are \emph{planar} --- none of the axioms require the swapping of inputs or outputs of operations --- and so is the theory of Frobenius algebras; hence, the interaction leading to Frobenius algebras is not supposed to change the dimension of generators. On the other hand, the bialgebra law --- a part of the theory of bialgebras --- is \emph{not} planar, and is in fact best represented by a string diagram in 3 dimensions, where the monoid part and the comonoid part belong to different, orthogonal planes, as in the following picture. \begin{equation*} \input{diagrams/bialgebra_3d_bis.tikz} \end{equation*} So the interaction leading to bialgebras should be of a different, dimension-increasing sort. In this paper, we try to lay the groundwork for an alternative approach, and make a case for the following assertions: \begin{enumerate} \item that there exists a way of studying algebraic theories compositionally, with a small number of basic constructions corresponding to the most frequent interactions; \item that the language for compositional algebra is a kind of combinatorial \emph{directed topology}, all interactions having clear analogues in standard topology. \end{enumerate} In Section \ref{sec:computads}, we briefly present our technical framework of choice, the theory of \emph{computads} or polygraphs, and build a basic vocabulary of directed topology in this context. In Section \ref{sec:basic}, we use these tools to construct presentations of some basic theories, such as the theory of monoids, from even simpler ones. In Section \ref{sec:homo}, we show how two kinds of interaction capture the notion of homomorphism and of (co)action. Finally, in Section \ref{sec:smash}, we introduce the dimension-changing operation which enables us to obtain the theory of bialgebras (and of commutative monoids) from the theory of monoids. We conclude by describing a partial reconstruction ``from scratch'' of the theory of interacting bialgebras, and discuss some of the many possible further directions of this project.

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Jon with Nana, Pappy, Mom, and Dad; Connecticut, summer 2009 I'm afraid I've been giving America short shrift on this blog recently. It makes sense, I guess, since I was just in London for 12 days and I'll be flying back to England in five weeks for my wedding, but I'm still feeling a little guilty about it. Maybe part of the reason I haven't been so gung-ho about this #AmericanSummer is because I don't have a real vacation planned. I mean, my bachelorette weekend away on the Eastern Shore will be amazing and yes, I have two trips to the UK this summer, but, honestly, they're more like working holidays. I did so much wedding stuff (and professional things as well) while I was in London recently and I'm sure that the days leading up to my wedding will be full, too. Anyway, all of that is very grown-up, you know? But last night we had our first massive thunderstorm of the summer in DC and today my parents head up to my maternal grandparents' house in Connecticut and, because of that, I'm feeling a great pull to the summers of my childhood. We'd pile into the car (a boxy Volvo station wagon, obviously; I take great delight in the cliché that my family was in the '90s) and drive up I-95, the way back stuffed with duffle bags and kid-friendly snacks. Snickers, our yellow lab, would take turns standing on either my sister or me so he could stick his head out the window to catch the air flying by, and Sarah and I would shriek when his drool dripped onto our bare arms. We'd have to pull over at rest stops at least twice for pee breaks, making the 6 hour journey last about 8, and I only ever remember getting Nathan's Hot Dogs for lunch. At various points, Mom and/or Dad would yell at us to stop fighting, and, when we were all getting along, we'd play word games and look for license plates from far away states. When we finally pulled onto the dirt road that led to Nana and Pappy's house, the relief from all was palpable. Writing out all of that makes me feel like I should go apologize to my parents. Hang on a second... Sarah and I spent a fantastic weekend in the country with our grandparents last summer, and it's actually where Jon really got to know Mom and Dad and met Nana and Pappy back in August 2009, so it's got wonderful adult memories associated with it, too. I'd love to go back next summer with Jon and Charlie and the whole mischpocha... dress / cardigan / sunglasses / shoes linking up with {long distance loving} and with Found Love, Now What? and We Took The Road Less Traveled That picture needs more sunscreen. ;) you are so right! fixed :) Much better. My pale paleness thanks you. Cuuute outfit. And it amuses me that a collage of an American summer must involve bug spray and citronella candles. Except those darn mosquitoes are not funny at all. I get to return to the family cottage of all my childhood summers next week... can't wait! Lovely family pic! Thanks for sharing! Whenever I get to see you and your sister, I cannot help but to think that you're like your father, and she's like your mother! Tú te pareces a tu padre, y ella se parece a tu madre. Am I right? Gracias, enjoy your weekend! María um OBVIOUSLY. my letters home from summer camp when I was a kid would always include the latest counts of my mosquito bites. ew. This outfit is SO east coast. There's no way you'd catch me in a cardigan right now! ;) it gets cool in the evenings! ah! usually people say the other way around but you're right, I definitely have Dad's smile there :) I would love a big summer thunderstorm, sounds so relaxing somehow! Absolutely love that outfit, I'm a huge fan of cardigans. They're perfect! And Off Spray is, of course, a must. I grew up spending my summer's on a lake in Wisconsin, so I know plenty about how bad mosquito bites can get! So happy that I found you through the blog hop! Beautiful photograph! I am in love with that dress! My family used to holiday all over the UK when I was younger - it's only as we got older that my parents decided to stay closer to home. How they managed 12 hour road trips to Cornwall from Yorkshire with three kids under the age of 10, I will never know.... Maybe I should go apologise/thank them too, haha. soweliveforever.com That dress is perfect! I need it! it would have been if I hadn't had two scarydycat dogs freaking out! Oh man YES! I went to summer camp in the woods in Minnesota, where I learned that the state bird is the mosquito :P thanks, Andrew! it was a great holiday. sometimes you do find pretty things from blog giveaways! that's how I found this :) oh my goodness - I have NEVER met anyone in the UK who would do a 12 hour road trip! everyone says that anything longer than 4 or 5 hours is too much for one drive. you're practically American! haha First, I love that cardigan. Second, your recap of your drive up to CT (where do they live?) makes me think of our drives to Florida from CT. You must have stopped in New Jersey for those Nathan's hot dogs, since they're at just about every single rest stop! And I have always wanted a boxy Volvo wagon. Serious dream car. :) i love that tank top tress and i love how you included sunblock and off for your american summer post :) hope you enjoy your summer trips! :) rhea @ I love that your American Summer involves OFF and citronella candles! Brilliant!! I can see the road trip memories you've conjured perfectly, you've painted them so well. Now I'm nostalgic for my childhood and the Taconic State Parkway :) Ooooh that makes me feel strangely proud! Haha.

Back before the Golden Age of Television, there really wasn’t enough compelling content on TV. People would have to suffer through channel surfing across TLC reality shows and infomercials for hours on end, disappointed all the way through. Nowadays, the problem has been turned on its head. There are far too many well-made television shows than anyone could ever watch in a single lifetime. Hold on to your remotes, though, because long list of must-watch TV shows is about to get longer. Time to focus your attention on Korean TV shows or K-dramas that have taken the world by storm. Once you turn your Netflix on to these shows, you might find yourself binge watching for hours on end. Be it a romantic drama or a period piece, similar elements can be found across the K-drama spectrum. They are known for their engaging story arcs, well-written narratives, pulpy relationships, and, very famously, extremely attractive lead actors. As an added bonus, K-dramas like to tie their narratives up in a nice little bow all in one season over a digestible number of episodes. It will never jump the shark at an awkward point in the story, unlike American TV shows. There are a lot of good shows to look through, and it may be a bit overwhelming to have to see it all. Here are some of the best one-season Korean dramas to give you a nice place to start with. 1. Strong Girl Do Bong-soon Do Bong-soon (played by Park Bo-young) belongs to a bloodline of women blessed with superhuman strength. She dreams of being able to star as the main character of a video game that she would make herself. She also seeks to become an elegant and delicate woman to catch the attention of her crush, police officer In Guk-doo (played by Ji Soo). Bong-soon is hired as a security guard to the CEO of Ainsoft, a gaming company, Ahn Min-hyuk (played by Park Hyung-sik). In contrast to Guk-doo, Min-hyuk has a complete disregard for rules, more than a little spoiled, extremely playful, and has a particular dislike for police officers. Min-hyuk began to look for a bodyguard after being stalked and receiving anonymous threats. After watching Bong-soon fight off a group of men who were threatening an old driver of an elementary school bus, Min-hyuk sought to hire her immediately. At the same time, people have been kidnapped in Dobong-dong, where Bong-soon lives. After the kidnapper targets Bong-soon’s best friend, she begins to hunt the culprit. She slowly learns to gain control over her strength to use to help those around her, with help and training from Min-hyuk. Their relationship gradually develops into something more throughout the season. 2. Oh My Ghost Oh My Ghost features a an extremely timid woman with low self-esteem named Na Bong-sun (played by Park Bo-young). Because of her personality, she has no close friends of her own and tends to be very lonely. She works in Sun Restaurant as an assistant chef, where she is constantly reprimanded for a variety of reasons. She also occasionally sees ghosts, thanks to her grandmother, who is a shaman. One day, Bong-sun gets possessed by a lustful virgin ghost named Shin Soon-ae (played by Kim Seul-gi). Soon-ae’s life was cut short and she did not get to experience much romance. She therefore believes that this is what keeps her on Earth. Soon-ae has been possessing various women to seduce as many men as she can so she can finally lose her virginity, which she believes will allow her to move on to the afterlife. She finds the perfect vessel for her romantic misadventures in the shy Bong-sun. Bong-sun’s boss is arrogant star chef Kang Sun-woo (played by Jo Jung-suk), whom she secretly has a crush on. Sun-woo hasn’t dated anyone since he got his heart broken by his college friend Lee So-hyung (played by Park Jung-ah), who is now a TV producer. After being possessed, Bong-sun’s personality turns around, coming from timid and shy to dynamic and confident. Because of this, she finally catches his eye. Meanwhile, the mystery surrounding Soon-ae’s death involves Sun-woo’s brother-in-law, a kind police officer, Choi Sung-jae (Lim Ju-hwan), who may not be what he seems. 3. Boys Over Flowers Based on a Japanese shōjo manga series of the same name, Boys Over Flowers is about a regular girl from a working-class family who attends a prestigious academy and is thus considered an outcast by the school’s population of students from wealthy families. She catches the attention of the school’s wealthiest group of students, commonly known as the F4, after its leader notices that she is the only girl in the entire school who does not fight for his attention. Her confidence is constantly tested by his bullying, which changes into a budding relationship over time. Boys Over Flowers stars Ku Hye-sun as the head-strong lead woman Geum Jan-di, and Lee Min-ho and Kim Hyun-Joong as the lead males and members of the F4. The series’ 25 episodes aired in Korea and all over Asia, where it was extremely popular. The base storyline of arrogant, popular, handsome boy falls for a confident, slightly stubborn regular girl who initially have a strained or complicated relationship with each other is all too familiar, but viewers from all over the world love it all the same. Boys Over Flowers received high viewership ratings wherever it aired and is now watchable through Netflix. 4. Romance is a Bonus Book Romance is a Bonus Book features Lee Jong-suk as Cha Eun-ho, a bestselling author and a book publishing company’s senior editor. As a child, he suffered an accident from which he was saved by Kang Dan-i (played by Lee Na-young). They become close childhood friends. Now, Dan-i is a broke and unemployed single mother after a successful career as a copywriter. In her attempts to find a job, she lies about her credentials and background and receives a job in the publishing company that Eun-ho works in. The both encounter professional and personal issues while exploring their feelings for each other. 5. Black Black follows a grim reaper named Grim Reaper 444 who possesses the body of a detective named Han Moo-gang (played by Song Seung-heon). Moo-gang is a detective who is investigating the 20-year-old serial murder cold case which included the death of his brother. Investigating with him is Kang Ha-ram (played by Go Ara), a clairvoyant who can foretell the nature and time of the deaths of people. Ha-ram decided to partner up with Moo-gang in an attempt to prevent his death, which she foretold. While on duty, Moo-gang is shot dead, after which he rises back to life after being possessed by Grim Reaper 444. In contrast with Moo-gang’s warm personality, Grim Reaper 444 is cold and quick to anger. Grim Reaper 444 partners with Ha-ram to find a runaway grim reaper, given that grim reapers who possess human bodies can be hard to find. Their relationship begins to more complicated as Grim Reaper 444 falls in love with Ha-ram, which is forbidden in the laws of the grim reapers. Read also: Best Korean Movies Of All Time 6. Stranger Stranger is a highly successful and award-winning Korean legal thriller. It stars Cho Seung-woo as Hwang Shi-Mok, a prosecutor who was left emotionless after a brain surgery performed on him as a child. He is cold towards others, extremely rational, and has a strong sense of right and wrong. He is partnered with Detective Han Yeo-jin (played by Bae Doo-na), an energetic detective who remains to be the only female detective on their team. Together, their team attempt to unravel a web of corruption and cover-ups that spans across the police, business, and political sectors. Stranger deviates from the norm of K-dramas by being less focused on the romance between the lead characters in exchange for the monumental task that they have ahead of them. There is still romance involved of course, with Yeo-jin obviously being highly fond of Shi-Mok and another team member with a big crush on the prosecutor, but the show’s main draw is its intelligently designed heroes, villains, and plots. The villains appeared to be just a few steps ahead of the heroes at every turn, and use everything they can to throw the heroes off the chase. Apart from praising the protagonists’ acting and the plot, critics also lauded the well written and acted villains of the story. The villains do not fit the trope of evil for evil’s sake. They are remarkably human, with personalities and motivations that concretely support the actions that they take throughout the story. In the New York Times’ list of Best Shows of 2017, Stranger was listed among the best international shows. It won several awards in Korea, notably getting the Baeksang Arts Awards’ Grand Prize for Television. It premiered in Korean on the tvN network and in Netflix simultaneously on June 10, 2017, with Netflix reportedly purchasing each of Stranger’s 16 episodes at $200,000 per episode. 7. Cain and Abel As is obvious by the name, this series is heavily inspired by the biblical story of Cain and Abel. The Korean series Cain and Abel is So Jib-sub and Shin Hyun-joon as brothers Lee Cho-in and Lee Seon-woo. Cho-in is a successful doctor, while Seon-woo harbors jealousy and ill will against his brother, as he believes that Cho-in took everything he has ever valued in his life, the love and attention of their father, the love of his life, and being a better doctor. It relies heavily on traditional drama elements like betrayal, sickness, affairs, disownment, and memory loss, among others to create a titan of a twisting drama to watch. Best Korean Dramas On Netflix? Ready to get started on these Korean TV shows? Tune in to Netflix and experience a whole new genre of shows never seen before. This article was written by a freelancer. Related articles: Top 10 Best K-Pop Boy Groups Of All Time Top 10 Best K-Pop Girl Groups Of All Time 13 Shows Like Black Mirror You Can Watch Today Top 10 Best Korean Dramas Of All Time Games Like Overcooked 2 – Five Alternatives To Check Out The Best Zombie Films Or Shows On Netflix Today Top Shows Like Arrow You Can Watch On Netflix Top 8 Shows like Brooklyn Nine-Nine Best Netflix Superhero Shows To Watch Today

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The Soho Hotel, housing 150 refugees mainly from Syria and Iraq, is one of two SolidarityNow buildings in Athens supported by UNHCR and funded by the EU. NurPhoto/ Press Association. All rights reserved.Southern European countries and more specifically Greece and Italy have faced rising flows of asylum seekers and irregular migrants since 2013, and particularly during the 2015-2016 period. Indeed no-one predicted the dramatic escalation of irregular maritime arrivals during 2015, lasting till March 2016 along the Turkey-Greece corridor, and to this day along the Libya-Italy corridor. Major challenges that such arrivals bring to light include the need for internal EU solidarity in sharing responsibility, the ‘burden’ of processing asylum seeking applications, and eventually of integrating those admitted as refugees. The Emergency Relocation Mechanism launched first in May 2015 and further expanded in September 2015[1] has played an important role as an EU policy response, seeking to enforce the more equitable sharing of responsibility for asylum-seekers among member-states, and thereby taking some of the pressure off the frontline states. The plan foresaw 160,000 relocation places in total, to be implemented over a two-year period (September 2015 to September 2017), more specifically assigning 66,400 places to people to be relocated from Greece and 39,000 from Italy to other EU countries. Relocation is selective and applies only to those nationalities whose applications for international protection (refugee status or subsidiary protection) have over a 75 per cent success rate on the basis of quarterly Eurostat data. Thus, Syrians have steadily been eligible for relocation while for instance Afghans, Iraqis or Eritreans have moved over and under this threshold. The 2015 challenge In order to understand how the emergency relocation quotas came about, one needs to consider the size of the flows reaching Greece via Turkey and Italy via Libya in 2015. During 2015, Greece received through its sea border with Turkey over 800,000 people while the flows continued unabated during the first three months of 2016 when a further 140,000 people crossed the narrow straits that divide the Turkish coasts from the Greek islands in the Aegean sea. The top countries of origin of irregular maritime arrivals to Greece were Syria, with over 50% of the arrivals, Afghanistan (25%- 30%) and Iraq (nearly over 10% and nearly 20% in recent months). These were populations clearly in need of international protection. In Italy, arrivals have been rising since 2014 with over 150,000 in 2015, approximately 181,000 in 2016 and nearly 60,000 already in the first half of 2017. The nationality composition differs significantly compared to Greece with Nigerians, Eritreans, Sudanese, Somalis and Gambians being among the five largest nationality groups. Among these clearly Eritreans, Sudanese and Somalis are people fleeing violence and persecution in their own country and hence the flows involve a high percentage of asylum seekers. The emergency relocation scheme was both innovative in terms of asylum policy and ambitious in its conception. It showed the strong political will on the part of the President of the European Commission to enforce burden sharing. Naturally its implementation was not free of challenges. Indeed the first implementation issues stemmed from the actual pre-registration and processing of people upon first reception in Greece and Italy. They included a certain reluctance on the part of the destination member-states and the emergence of what has been dubbed ‘shopping lists’ of people for relocation (for example, vulnerable groups such as unaccompanied minors, single mothers, victims of trafficking, or highly-educated persons). Table 1: State of Play of Emergency Relocation Mechanism, 13 October 2017 Source: European Commission, periodic reports available here. The European Commission has also described as problematic the inadequate or unjustified grounds for rejecting relocation requests.[2] However, the Greek Asylum Service, for instance, has explained how following up on rejections of relocation is difficult, because of the relocation scheme’s overall design, given that member-states’ sovereign right to refuse to relocate asylum seekers cannot be challenged (AIRE, ECRE report). Naturally the whole issue has been further complicated by fears of terrorist infiltration of asylum seekers, particularly from Syria. Responding to the legal action taken by Slovakia and Hungary, the Court of Justice of the European Union has upheld that the emergency relocation quotas are in line with EU law. While this decision is important at the political level, it probably did not do much to change the implementation level where problems and delays have persisted.[3] As of 13 October 2017, there had been approximately 48,000 requests for relocation lodged by Greece and Italy and just over 30,000 people were actually relocated (see table 1 and 2). Table 2: State of Play of Emergency Relocation Mechanism, 28 Sep. 2017 Source: accessed on 17 October 2017. Ιt is important to clarify that when it comes to Greece, relocation is available only to people who entered Greece between 16 September 2015 and 19 March 2016. While this has been unpopular among people on the move at first, it has since been reconsidered, as the alternative is to remain in Greece with few prospects for employment. Despite its implementation problems the emergency relocation quotas have been an important policy innovation. They have been a first step towards reconsidering the Dublin Regulation and the first safe country principle. They have introduced a new way for direct sharing of responsibility for asylum-seeking among member-states. While presented as an emergency solution, the quotas are likely to become a more long-term, permanent mechanism in the impending reform of the EU Asylum system, now that the main criteria – size of the country, GDP, and number of refugees already present – have been established. This is finally taking up suggestions from researchers that had actually been circulating for at least two years. Reaching the ear of the European Commission The question arises whether the relocation quotas emerged as a result of evidence-based policy making and researcher-policymaker dialogue, or whether they are the result of a political move taken in the heat of the moment, under pressure from an unfolding crisis. One might argue that all these elements combined. The idea had been circulating among academic and policy circles about asylum seeker relocation quotas. It was in November 2014 that Jesús Fernández-Huerta Moraga and Hillel Rapoport presented a model for tradable refugee admission quotas[4] in a possible reform of the EU asylum policy while the same authors had also elaborated a proposal on tradable immigration quotas[5]. Other authors had considered the issue of responsibility sharing and the imbalances created by the Dublin system and the first safe country principle since it found itself operating in an uneven implementation context[6]. However, the question arises whether such proposals reached as far as European Commission headquarters because the Commission was looking for scientific evidence on the best possible approach, or whether the emergency, political will and perhaps civil society mobilisation was more instrumental in this? Indeed our answer rather favours the latter reasons over and above the effects of scientific evidence. True, the European Commission leadership was open to scientific evidence because of the situation on the ground and the need to address it through radical and effective measures. But the backing of the German political leadership in admitting refugees and showing solidarity was also crucial to the decision. The emergence of a spontaneous transnational civil solidarity movement not only in the frontline countries, Greece and Italy, but also in Austria, Germany and other countries where local people greeted asylum seekers at train stations offering food, clothes and hospitality of course played an important part too. In other words, the moment was ripe and the political will was there. ‘Bad blood’ However, one also has to consider the downside of the emergency relocation quotas’ measure. Indeed the quotas, as explained above, were strongly contested by Central Eastern European member states and were almost forced upon them. This has created ‘bad blood’ for the current and future Dublin reform negotiations where a permanent relocation mechanism is a red flag for many of these countries. Indeed the decision of the ECJ marks an important development, but consensus needs to be built further around this measure of direct asylum responsibility sharing. Naturally here an important part can again be played by civil society and by experts who circulate the idea, if they explain what it entails and continue to build support for it. Overall it is clear that political leadership and political will are paramount for research findings and proposals to be taken on board and transformed into policy solutions. [1] Council Decision (EU) 2015/1601 of 22 September 2015 establishing provisional measures in the area of international protection for the benefit of Italy and Greece. [2] European Commission 18, May 2016 available here. [3] For a full commentary see here. [4] Jesús Fernández-Huertas Moraga and Hillel Rapoport (2014) Tradable Refugee-Admission Quotas and EU Asylum Policy IZA DP No. 8683 November 2014, available here. [5] Fernandez Huertas Moraga, J. and Rapoport, H. (2014) Tradable Immigration Quotas, Journal of Public Economics, Volume 115, July 2014, Pages 94-108. [6] Bovens, L. Chatkupt, C. Smead, L. (2011) Measuring common standards and equal responsibility sharing in EU asylum outcome data, European Union Politics 13(1) 70–93. Read more Get our weekly email CommentsWe encourage anyone to comment, please consult the oD commenting guidelines if you have any questions.

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The Sword of Welleran and Other Stories/The Lord of Cities THE LORD OF CITIES I came one day upon a road that wandered so aimlessly that it was suited to my mood, so I followed it, and it led me presently among deep woods. Somewhere in the midst of them Autumn held his court, sitting wreathed with gorgeous garlands; and it was the day before his annual festival of the Dance of Leaves, the courtly festival upon which hungry Winter rushes mob-like, and there arise the furious cries of the North Wind triumphing, and all the splendour and grace of the woods is gone, and Autumn flees away, discrowned and forgotten, and never again returns. Other Autumns arise, other Autumns, and fall before other Winters. A road led away to the left, but my road went straight on. The road to the left had a trodden appearance; there were wheel tracks on it, and it seemed the correct way to take. It looked as if no one could have any business with the road that led straight on and up the hill. Therefore I went straight on and up the hill; and here and there on the road grew blades of grass undisturbed in the repose and hush that the road had earned from going up and down the world; for you can go by this road, as you can go by all roads, to London, to Lincoln, to the North of Scotland, to the West of Wales, and to Wrellisford where roads end. Presently the woods ended, and I came to the open fields and at the same moment to the top of the hill, and saw the high places of Somerset and the downs of Wilts spread out along the horizon. Suddenly I saw underneath me the village of Wrellisford, with no sound in its street but the voice of the Wrellis roaring as he tumbled over a weir above the village. So I followed my road down over the crest of the hill, and the road became more languid as I descended, and less and less concerned with the cares of a highway. Here a spring broke out in the middle of it, and here another. The road never heeded. A stream ran right across it, still it straggled on. Suddenly it gave up the minimum property that a road should possess, and, renouncing its connection with High Streets, its lineage of Piccadilly, shrank to one side and became an unpretentious footpath. Then it led me to the old bridge over the stream, and thus I came to Wrellisford, and found after travelling in many lands a village with no wheel tracks in its street. On the other side of the bridge, my friend the road struggled a few yards up a grassy slope, and there ceased. Over all the village hung a great stillness, with the roar of the Wrellis cutting right across it, and there came occasionally the bark of a dog that kept watch over the broken stillness and over the sanctity of that untravelled road. That terrible and wasting fever that, unlike so many plagues, comes not from the East but from the West, the fever of hurry, had not come here—only the Wrellis hurried on his eternal quest, but it was a calm and placid hurry that gave one time for song. It was in the early afternoon, and nobody was about. Either they worked beyond the mysterious valley that nursed Wrellisford and hid it from the world, or else they secluded themselves within their old-time houses that were roofed with tiles of stone. I sat down upon the old stone bridge and watched the Wrellis, who seemed to me to be the only traveller that came from far away into this village where roads end, and passed on beyond it. And yet the Wrellis comes singing out of eternity, and tarries for a very little while in the village where roads end, and passes on into eternity again; and so surely do all that dwell in Wrellisford. I wondered as I leaned upon the bridge in what place the Wrellis would first find the sea, whether as he wound idly through meadows on his long quest he would suddenly behold him, and, leaping down over some rocky cliff, take to him at once the message of the hills. Or whether, widening slowly into some grand and tidal estuary, he would take his waste of waters to the sea and the might of the river should meet with the might of the waves, like to two Emperors clad in gleaming mail meeting midway between two hosts of war; and the little Wrellis would become a haven for returning ships and a setting-out place for adventurous men. A little beyond the bridge there stood an old mill with a ruined roof, and a small branch of the Wrellis rushed through its emptiness shouting, like a boy playing alone in a corridor of some desolate house. The mill-wheel was gone, but there lay there still great bars and wheels and cogs, the bones of some dead industry. I know not what industry was once lord in that house, I know not what retinue of workers mourns him now; I only know who is lord there to-day in all those empty chambers. For as soon as I entered, I saw a whole wall draped with his marvellous black tapestry, without price because inimitable and too delicate to pass from hand to hand among merchants. I looked at the wonderful complexity of its infinite threads, my finger sank into it for more than an inch without feeling the touch; so black it was and so carefully wrought, sombrely covering the whole of the wall, that it might have been worked to commemorate the deaths of all that ever lived there, as indeed it was. I looked through a hole in the wall into an inner chamber where a worn-out driving band went among many wheels, and there this priceless inimitable stuff not merely clothed the walls but hung from bars and ceiling in beautiful draperies, in marvellous festoons. Nothing was ugly in this desolate house, for the busy artist's soul of its present lord had beautified everything in its desolation. It was the unmistakable work of the spider, in whose house I was, and the house was utterly desolate but for him, and silent but for the roar of the Wrellis and the shout of the little stream. Then I turned homewards; and as I went up and over the hill and lost the sight of the village, I saw the road whiten and harden and gradually broaden out till the tracks of wheels appeared; and it went afar to take the young men of Wrellisford into the wide ways of the earth—to the new West and the mysterious East, and into the troubled South. And that night, when the house was still and sleep was far off, hushing hamlets and giving ease to cities, my fancy wandered up that aimless road and came suddenly to Wrellisford. And it seemed to me that the travelling of so many people for so many years between Wrellisford and John o' Groat's, talking to one another as they went or muttering alone, had given the road a voice. And it seemed to me that night that the road spoke to the river by Wrellisford bridge, speaking with the voice of many pilgrims. And the road said to the river: "I rest here. How is it with you?" And the river, who is always speaking, said: "I rest nowhere from doing the Work of the World. I carry the murmur of inner lands to the sea, and to the abysses voices of the hills." "It is I," said the road, "that do the Work of the World, and take from city to city the rumour of each. There is nothing higher than Man and the making of cities. What do you do for Man?" And the river said: "Beauty and song are higher than Man. I carry the news seaward of the first song of the thrush after the furious retreat of winter northward, and the first timid anemone learns from me that she is safe and that spring has truly come. Oh but the song of all the birds in spring is more beautiful than Man, and the first coming of the hyacinth more delectable than his face! When spring is fallen upon the days of summer, I carry away with mournful joy at night petal by petal the rhododendron's bloom. No lit procession of purple kings is nigh so fair as that. No beautiful death of well-beloved men hath such a glory of forlornness. And I bear far away the pink and white petals of the apple-blossom's youth when the laborious time comes for his work in the world and for the bearing of apples. And I am robed each day and every night anew with the beauty of heaven, and I make lovely visions of the trees. But Man! What is Man? In the ancient parliament of the elder hills, when the grey ones speak together, they say nought of Man, but concern themselves only with their brethren the stars. Or when they wrap themselves in purple cloaks at evening, they lament some old irreparable wrong, or, uttering some mountain hymn, all mourn the set of sun." "Your beauty," said the road, "and the beauty of the sky, and of the rhododendron blossom and of spring, live only in the mind of Man, and except in the mind of Man the mountains have no voices. Nothing is beautiful that has not been seen by Man's eye. Or if your rhododendron blossom was beautiful for a moment, it soon withered and was drowned, and spring soon passes away; beauty can only live on in the mind of Man. I bring thought into the mind of Man swiftly from distant places every day. I know the Telegraph—I know him well; he and I have walked for hundreds of miles together. There is no work in the world except for Man and the making of his cities. I take wares to and fro from city to city." "My little stream in the field there," said the river, "used to make wares in that house for awhile once." "Ah," said the road, "I remember, but I brought cheaper ones from distant cities. Nothing is of any importance but making cities for Man." "I know so little about him," said the river, "but I have a great deal of work to do—I have all this water to send down to the sea; and then to-morrow or next day all the leaves of Autumn will be coming this way. It will be very beautiful. The sea is a very, very wonderful place. I know all about it; I have heard shepherd boys singing of it, and some times before a storm the gulls come up. It is a place all blue and shining and full of pearls, and has in it coral islands and isles of spice, and storms and galleons and the bones of Drake. The sea is much greater than Man. When I come to the sea, he will know that I have worked well for him. But I must hurry, for I have much to do. This bridge delays me a little; some day I will carry it away." "Oh, you must not do that," said the road. "Oh, not for a long time," said the river. "Some centuries perhaps—and I have much to do besides. There is my song to sing, for instance, and that alone is more beautiful than any noise that Man makes." "All work is for Man," said the road, "and for the building of cities. There is no beauty or romance or mystery in the sea except for the men that sail abroad upon it, and for those that stay at home and dream of them. As for your song, it rings night and morning, year in, year out, in the ears of men that are born in Wrellisford; at night it is part of their dreams, at morning it is the voice of day, and so it becomes part of their souls. But the song is not beautiful in itself. I take these men with your song in their souls up over the edge of the valley and a long way off beyond, and I am a strong and dusty road up there, and they go with your song in their souls and turn it into music and gladden cities. But nothing is the Work of the World except work for Man." "I wish I was quite sure about the Work of the World," said the stream; "I wish I knew for certain for whom we work. I feel almost sure that it is for the sea. He is very great and beautiful. I think that there can be no greater master than the sea. I think that some day he may be so full of romance and mystery and sound of sheep bells and murmur of mist-hidden hills, which we streams shall have brought him, that there will be no more music or beauty left in the world, and all the world will end; and perhaps the streams shall gather at the last, we all together, to the sea. Or perhaps the sea will give us at the last unto each one his own again, giving back all that he has garnered in the years—the little petals of the apple-blossom and the mourned ones of the rhododendron, and our old visions of the trees and sky; so many memories have left the hills. But who may say? For who knows the tides of the sea?" "Be; to-morrow I go to see if he be still at his post. For me Babylon was built, and rocky Tyre; and still men build my cities! All the Work of the World is the making of cities, and all of them I inherit."

Tipping Points 2017: Finding Energy-Climate BalanceJune 21-22, 2017 About the Forum Climate and Energy for The Top Leaders Under Forty: Tipping Points gathers the top leaders under forty years old from around the world to discuss one of the most pressing challenges the Millennial generation is inheriting: how to balance mitigating and adapting to climate change with the need for secure and reliable energy to fuel our future. Competitively selected from around the world, delegates include elected officials, senior advisors to heads-of-state, rising executives at top global companies, entrepreneurs and young CEOs, internationally renowned artists, civil society actors, journalists, academics, and other leaders at the cutting edge of global trends all under the age of forty. A Global & Bipartisan Conversation Between Past, Present, and Future Leaders: Tipping Points is a bipartisan and cross-generational dialogue about the future, featuring a lineup of senior speakers alongside young leaders that represent the full spectrum of views on energy and climate from both sides of the Atlantic, including former US Secretary of Energy Ernest Moniz, former EPA Administrator Gov. Christie Todd Whitman, Chairman and CEO of the Libra Group George Logothetis, CEO of Meridiam Thierry Deau, and Campaign Director for Greenpeace USA Leila Deen, among many others. Inspired, Interactive & Unconventional: Tipping Points promises an inspired and unconventional experience featuring panel discussions, TED-style talks, film screenings, and the co-painting by delegates of a mural that will hang prominently in Washington, DC following the event to provide a sustained call-to-action to solve this global issue. A Global Leadership Transition: Millennials are the largest generational cohort on Earth and already occupy executive roles in key institutions and firms around the world. With each passing year, they acquire more power to shape the future. As this leadership transition unfolds, Tipping Points aims to cement itself as the preeminent gathering for top decision makers in the Millennial generation. Organizing Partners: Tipping Points is an annual gathering organized by the Millennium Leadership Program (MLP), the Atlantic Council’s center for rising leaders. The 2017 Tipping Points conference, focused on climate change and the global energy transition, is organized in partnership with the Ecologic Institute, a German institution for applied interdisciplinary environmental research and policy analysis, as part of a joint initiative, the Emerging Leaders in Energy and Environmental Policy Network (ELEEP –), with seed funding from the European Commission and conference support from corporate and foundation donations.

[ltr.[/ltr] [ltr.[/ltr] [ltr]Upper Resistance Line: It takes at least two reaction highs to form the upper resistance line, ideally three. Each reaction high should be lower than the previous highs.[/ltr] [ltr]Lower Support Line: At least two reaction lows are required to form the lower support line. Each reaction low should be lower than the previous lows.[/ltr] [ltr.[/ltr] [ltr.[/ltr] [ltr]Volume: While volume is not particularly important on rising wedges, it is an essential ingredient to confirm a falling wedge breakout. Without an expansion of volume, the breakout will lack conviction and be vulnerable to failure.[]

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Gov. Scott Walker’s office announced this week that Department of Corrections (DOC) Secretary Jon Litscher is retiring from the agency. Walker has appointed DOC Deputy Secretary Cathy Jess to replace Litscher beginning June 11. Litscher previously served as DOC secretary from 1999-2003 and returned in 2016. He has also served as the secretary of the Department of Employment Relations, executive secretary of the Higher Educational Aids Board, and superintendent of the Cambria-Friesland School District. Jess has served as DOC deputy secretary since 2016 and was previously administrator of the Division of Adult Institutions. She has also served as a correctional officer at the Wisconsin Resource Center, warden at Dodge Correctional Institution and Wisconsin Women’s Correctional System, and deputy warden at the Waupun and Oshkosh correctional institutions.

TITLE: Vector Space or not QUESTION [0 upvotes]: Show that the functions $( c_1 + c_2 \sin^2x + c_3 \cos^2x)$ form a vector space. Find the basis. My solution: The set of all linear combinations of $\{1, \sin^2x, \cos^2x\}$ has been given. So we can write any equation $ c_1 + c_2 \sin^2x + c_3 \cos^2x$ in the form $ c_1 \sin^2x + c_1 \cos^2x + c_2 \sin^2x + c_3 \cos^2x$ Then we can rewrite this as $ a \sin^2x + b \cos^2x$ Assuming this is a vector space, I can say the basis is $\{\sin^2x, \cos^2x\}$ and the dimension is 2. The condition for it to be a vector space is $ (c + d)(x + y) = cx + dx + cy + dy$ where x, y are the vectors and c, d are constants. If I put $\{\sin^2x, \cos^2x\}$ as x and y then the above equation simply becomes $c + d$. That is not a proof of being a vector space. How do I actually use this prove that this is a vector space ? REPLY [3 votes]: I am assuming that you are working over the field $\Bbb R$. Let $V$ be your set of functions. If $c_1+c_2\sin^2+c_3\cos^2,d_1+d_2\sin^2+d_3\cos^2\in V$, then their sum is$$c_1+d_1+(c_2+d_2)\sin^2+(c_3+d_3)\cos^2\in V.$$Also, if $\lambda\in\Bbb R$, then$$\lambda(c_1+c_2\sin^2+c_3\cos^2)=\lambda c_1+\lambda c_2\sin^2+\lambda c_3\sin^2\in V.$$So, since $V\ne\emptyset$ (for instance, the null function is in $V$), $V$ is a vector space. And a basis of $V$ (not the basis) is $\{\cos^2,\sin^2\}$, since it spans $V$ and it is linearly independent. So, you have indeed $\dim V=2$.

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This announcement came on my Facebook Wall and wanted to pass this on to you. Denise Hanney of Spa Expectations Corporation is ordering 10 now and 10 during October which is Breast Cancer Awareness Month. Read below and connect to something important. Read below and connect to something important. AnnieMae's Cheese Cake For those who may not know Anniemae's Cheesecakes & Moore is named after my Grandmother who lost her battle with breast cancer, during breast cancer awareness month and 3 days before my birthday in October 2004. Not only did it claim the life of my grandmother but also my great-grandmother in 2001. Since then I have become an advocate for breast cancer awareness. To do my part I found a great organization Cancer Cares which offered me grief counseling in dealing with their deaths and to get a full understanding about breast cancer. So on September 25th, my Mom's Birthday (who passed in 2006) I am launching our Sho Fly Cares Collection. We have made Pink Ribbon Ballerina Slippers with the Pink Ribbon emblem on them. A slipper will be placed on one of our large Cupcakes and presented in a pink box. Proceeds from the sale of each cupcake will go to Cancercare.Org. Please join us in the fight against breast cancer. I know first hand how devastating this disease is and can be. Please help support an organization that helps those still battling breast cancer. Each large cupcake & Shoe is $6 Warmly Danielle L. Moore Confectionist 1 comment: I just love your newsletters. They are very informative.

Legislative act to cut July raises, 14th salary The government is preparing a legislative act that will suspend businesses’ obligation to raise their basic salaries by 2.6 percent as of July 2012, in accordance with the collective labor contract agreed by employers and labor unions. Kathimerini has learned that the same act is also going to include the incorporation of about half of the Christmas, Easter and summer bonuses into the monthly salaries, effectively signalling the end of the so-called 14th salary. The government will use this method to bypass the parliamentary procedure and avoid having to come to an agreement with unions and employers. In doing so, they would satisfy one of the demands of the European Union and the International Monetary Fund, which were set out in a letter to the Labor Ministry this week. In their letter, the troika representatives note that while the 13th and 14th salaries have been abolished in the public sector, they still apply to the private sector as well as to state corporations that operate infrastructure networks, such as the Public Power Corporation and the Athens and Thessaloniki Water Companies.

{\bf Problem.} Let $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1.$ What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$? {\bf Level.} Level 5 {\bf Type.} Intermediate Algebra {\bf Solution.} We have that \[g(x^{12}) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1.\]Note that \[(x - 1)g(x) = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1) = x^6 - 1.\]Also, \begin{align*} g(x^{12}) - 6 &= (x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1) - 6 \\ &= (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1). \end{align*}We can write \[(x^{60} - 1) = (x^6 - 1)(x^{54} + x^{48} + x^{42} + \dots + x^6 + 1).\]In the same way, $x^{48} - 1,$ $x^{36} - 1,$ $x^{24} - 1,$ and $x^{12} - 1$ are all multiples of $x^6 - 1,$ so they are multiples of $g(x).$ We have shown that $g(x^{12}) - 6$ is a multiple of $g(x),$ so the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$ is $\boxed{6}.$

TITLE: Why does photon have only two possible eigenvalues of helicity? QUESTION [13 upvotes]: Photon is a spin-1 particle. Were it massive, its spin projected along some direction would be either 1, -1, or 0. But photons can only be in an eigenstate of $S_z$ with eigenvalue $\pm 1$ (z as the momentum direction). I know this results from the transverse nature of EM waves, but how to derive this from the internal symmetry of photons? I read that the internal spacetime symmetry of massive particles are $O(3)$, and massless particles $E(2)$. But I can't find any references describing how $E(2)$ precludes the existence of photons with helicity 0. REPLY [6 votes]: It derives not from the internal symmetry itself but from the fact that it is a gauge symmetry. Your symmetry group assignments are not those of the symmetry group but of the little group of the representation. If you assume in addition that the representation is irreducible, you end up in the massless case (with little group ISO(2)=E(2)) with a helicity representation, which picks up from a vector representation only the transversal part, corresponding to a gauge symmetry. Because of reflection symmetry (parity), there are two helicity degrees of freedom. Under the connected part of the Poincare group, this splits into two irreducible representations of fixed helicity, corresponding left and right circular polarization. This is described in full detail in Section 5.9. of the quantum field theory book (Part I) by Weinberg. In particular, the 2-valuedness (rather than the 3-valuedness) of the helicity is discussed after (5.9.16).

Watch these one and a half year old twin girls learn the names of different fruits and veggies by helping dad unpack the groceries. Does it get any cuter than this? Filmed in San Francisco, California. silenttd65,097 viewsNEW Newsflare30,291 viewsNEW rumblestaff301,791 views corybart1,601 views TYarber371,263 views heathermichelle550711,371 views Ztoaplay31,653 views ZhemJZ1,022 views trarau4,477 views BabyBlueEyes5,163 views

TITLE: Suppose an isolated north pole exists. Then would it surely move along magnetic field line of a bar magnet as written in my book? QUESTION [0 upvotes]: My book says that magnetic field lines due to a bar magnet are the path along which a unit north pole would move (if it existed). Will it surely move along the field line? I mean the two poles of magnet will exert forces whose resultant will always be tangential to field line then which force will act as centripetal force? REPLY [0 votes]: A hypothetical isolated magnetic pole (often called a magnetic monopole) would feel a force in the direction of the magnetic field lines, and so its acceleration vector would be tangent to a field line at all times. However, the trajectory traced out by such a monopole does not necessarily follow a field line. If it did follow a field line, then its velocity (rather than its acceleration) would be tangent to a field line at all times. Note that the above statements apply equally as well to electric charges and their paths relative to electric field lines. For example, a "light" negative charge such as an electron can orbit a "heavy" positive charge such as a proton in a circular orbit; in this instance, the trajectory does not follow the field lines. Similarly, a "light" north pole could orbit a "heavy" south pole in a circular orbit that does not follow a magnetic field line.

Vlo Studios is an innovative agency creating visual, digital and social communications for fashion, beauty, lifestyle and luxury brands. It is founded by Vera Lobik and situated in the vibrant city of Rotterdam. With experience in content creation, social media strategy and management I offer a unique creative approach. Through powerful visuals, appealing design and compelling stories, I elevate brand’s social media and offer relevance to consumers. People don’t buy products. They buy a feeling, a story and values that they associate with. As a brand it is important to get your brand’s story and values out there. As tangible as possible. My passion is to get your story clear and for all to see. I pour your brand’s unique values into strong imagery and stories for every specific social media channel. Then I make sure as many people as possible engage with your content. Let them truly see you, and they will love you. Vera Lobik – founder/creative director I majored in Fashion & Management at the Amsterdam Fashion Institute and during this period, I realized I have a big passion for beautiful visuals and social media. I’ve continued to learn more while working for fashion brands such as Zoe Karssen, Suitsupply and Suistudio. I was Head of Social Media of Suistudio and as a digital native, continue to stay up-to-date with all the developments in social media. Now, I want to use this experience to assist other brands with creating a consistent online presence and reaching their target audience. Let me support you with content creation and social media.

New Delhi, Feb 4 (IANS) Delhi Chief Minister Arvind Kejriwal on Monday wrote to Union Minister Nitin Gadkari demanding that journalists should get exemption from toll tax all over the country. In a letter to the Union Minister for Road, Transport and National Highways, Kejriwal said many journalists have made the request. The Chief Minister said: “Some journalists from Haryana met me and requested that they should be exempted from paying toll tax in Haryana. “I think this demand of the journalists is very genuine and must be accepted. I think and thus request you to please exempt journalists all over the country from paying toll tax.” (IANS)

TITLE: Why the polynomials with real coefficient and having degree >3 is not a vector space? QUESTION [1 upvotes]: In my textbook it is written that- all polynomials with real coefficient and having degree $≤3$ is a vector space. Does it also means that $>3$ are not vector space? REPLY [3 votes]: Let $P_{n>3}$ be the set of polynomials with degree larger than 3. Then obviously, it does not contain an element that plays the role of zero vector. It is also not closed under addition. On the other hand, $P_{n\leq3}$ is a vector space because it contains the polynomial $P(X) = 0$ (zero vector), and it is closed under scalar multiplication and addition (taking a linear combination of 2 polynomials of degree 3 will never have degree larger than 3). EDIT1: you seem to be struggling with the question: 'Why is the degree of $P(x) = 0$ zero?' Let me try to answer that question for you. For any constant $c \in \mathbb{R}$, we can write $P(X) = c = cX^0$. Hence, the highest power of $X$ we see here is $0$. Since $0$ is such a constant, it makes sense to say that $P(X) = 0$ has degree $0$. There is actually a bit ambiguity here, since we can also write, for example $P(X) = 0*X^3$, so that's why we chose to define the degree of the zero polynomial as $0$. Some authors say the degree of this polynomial is $-1$ or $- \infty$, but for these kinds of exercises it is best that you define it as being $0$ to not confuse yourself with the linear algebra concepts. EDIT2: Now you ask why $P_{n>3}$ does not have a zero polynomial. The easy answer is, that if it would have one, it would be $P(X) = 0$! In a vector space (or more generally a group), a subspace (or subgroup) always shares the same neutral element as the mother vector space, which here, clearly is impossible.

. thanks Sınıf Öğretmenleri Obummer should give lots of love to you, teleprompter. He'd be lost without you! Kinda off topic, but - Has anyone heard if there has been a great increase in teleprompter orders? You'd think that is one company that would be doing a booming business - everyone wants to be a Swivelhead like Obama, right? So we were talking - That'd be a funny comedy sketch: Democrats Aping Obama! Have ordinary people using teleprompters during everyday life. Going to the bank, getting a traffic ticket, having sex... Can you pull this off? True Romance Dunno if you guys saw the above movie,as it was perfect for a VD date. Your gal,expected one thing,judging that book by it's cover,and got quite another. Having no shortage of Machine Gun fire and other guy movie 'twists',it also contained Dennis Hopper and Christopher Walken in a stunning 5 minute scene most will never forget.That scene made the movie.Some will know which one. Today,tho it is a less faithful form of romance,that has folks fighting AIDS in Africa.This morning I caught an AP article pimping a device for "painless" male Circumscisions. I have my doubts. From the article.........("Experts explore ways to Circumcise Men"). "The main problem I can foresee with this is actually persuading men to sign up for it.........." "Well yes,one imagines talking them into it might be a problem . Most men cannot imagine sufficient foot / cost incentives down at the' Lumber' yard. Others ,once adult ,refuse all negotiations on the matter. This is not some IRS mileage deduction we hope to incentivize,or foot cost analysis from yesterdays adventures in True Romance Land. Some lesser fellows might demand rates per centimeter,and justly so, as not all are equiped w uh,Annacondae. Rather than the " Ring " above promising pain free pruning., I would suggest a Wedding Ring instead. That will shrink things up. (Cross posted at the Barnyard,(me.com)). Lotsa LOLs, TOTUS! Good entry! Rose, there is a very astute video done by "The Onion" of The Dear Reader using TOTUS at the dinner table...our host no doubt knows more about this. As for myself, I wouldn't mind the everyday Jane having a portable TOTUS. It might help those challenged with foot in mouth disease. BTW, TOTUS, I noticed LOTUS wrote a Valentine for you the other day on her blog. Are you really married to your job? If so, why the unexplained absences? Does The Dear Reader put you through so much overtime that you just, work, recharge, work, recharge? XXOO,(and you know I don't hand those out lightly) Madame DeFarge [TOTUS]"made her a Valentine's Day card out of a bunch of scrap paper ... and ... aluminum foil and sparkly stuff ... ." LOL, Totus, that wasn't Mitchell's card -- it was her gift, a new purse (matches most of her outfits). No, seriously, it was Dope's applied physics project for his Ph.D. from M.I.T.. # # # # WILL HE GET A PASSING GRADE?! Wroooong question, buddy! He will get the Ph.D. That's affirmative. ************* Word (finally got a good one!): umster What one Secret Serviceman says to another every morning... "Here comes the umster." Dope: Uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuh, Hi. Mrs. Butterworth humiliates disabled kid. The TSA made a disabled kid take off his LEG BRACES before he could board a plane bound for Orlando and Disney World. Our doughy chief of Homeland Security really knows no bounds, does she? Well, Chris, AFTER ALL, he was probably just another of those "___ retards" who might compete in [giggle] Special Olympics. Mrs. Butterworth most likely said in her smarmiest voice, "The system worked." Really inspires confidence in Dope & Co. Security, Inc., doesn't it? Their X-ray machine can't even tell the difference between leg braces and dangerous weapons. Dope & Co. agent: Uuh, what's dat in da violin case? Mooslim Terrorist [SMILE]: Why, that is my.... "arm brace." D&C: Why'd you put it in your violin case? MT: Easier to carry. D&C: Where's da violin? MT: In the case. D&C: I don't see it. MT: It's very small. D&C: Duh.... why do you have such a tiny violin? MT: Well, I only play a little. D&C: Okay. Move along. Next! ******************************** Next day's headline: ISOLATED OVERSEAS CONTINGENCY OPERATIVE BLOWS UP PLANE WITH "ARM BRACE" New York -- Apparently dissatisfied with the service, a lone man...... Hey everyone! Did you all see the new Fox News Poll that said that a majority would NOT vote to re-elect Obama to a 2nd Term? Oh...wait...sorry. It was a CNN poll. Bleak turns to Bleaker... Bwah, ha, ha, ha, haaaaaaaaa! Via Mark Steyn's site (steynonline.com), I just read the Philadelphia Inquirer story about the horrendous treatment of the above-mentioned 4-YEAR-OLD boy required to not only remove his leg braces, but forced to walk without them to satisfy the TSA agent. And, yup. The little guy is developmentally delayed. One of Rahm's "____ retards." His parents just want to put this behind them. I wish some Philly-area attorney would help them SUE THOSE ___ Obama & Co. Security JERKS! Here's the URL for the article's page: yoohoo, bettyann -- you are missed. Barak (who's insane) Obama...Mmmm...mmm...mm. Just had to get that out there. Thank you for commenting, Chris. I was beginning to try to write a parody song in my head, but had gotten stuck. It's going something like: Where have all the FOTAE gone? Long time passing . . . . Where have all the FOTAE gone? Long time ago-o. Where have all the FOTAE gone -- Gone to ________, everyone -- Oh how will we ever know? Oh how will we -- ever know? -- So - that's about as creative as I can get (Needs your assistance, TWW --). I don't know where you all have gone! TOTUS, will you be in Denver today helping that LOOSER Bennet? And then off you go to Sin City for reid. The mayor of vegas doesn't want you there. Tell your master to stay in DC and get some work done! Where have all the FOTAE gone ? No need looking at the Barnyard aero,nobody there but a chicken or two. Not even the Ghost of Mary Travers haunts the place,presently. Maybe they are at the MOTUS mirror site tho,as a few FOTS have taken roost there. Some FOTAE flew the Coop here when TP Binged on 220,forever,it seemed,and I fancy, forgot and yes,failed his faithful FOTS fans. Yes,he has a great excuse,being way busy working for POTUS as he does,and having little time for us here.Even typing at light speed. Fresh roots are springing from here tho,and spring chicken sites have sprung from Hen and Rooster Romances also having something to cluck about.All grew weary of waiting for the new 'Postage' to arrive.Also most were somewhere between weary and wary of rumored pay per spew USPS style postage cost increases. Remember Team Teleprompter Commenteers,when you were in 11th grade,and your Paramour kept pestering you to 'prove your love' ? You may need to'do it'again here,and there (at me.com),too,eventually.By Clucking up. Not just Chicken Feed here and there,websites need EYEBALLS to survive,and like zombies, BRAINS,which are nearly as much fun to pick as noses. As our literatae,if not the Twitterattae,will recall from Pope,I believe............ 'You can pick your friends,you can pick your nose,but you cannot pick your friend's noses'. Sounds like him but maybe it was that other Pope who was a noteworthy Card,frequently using his hand buzzer on his Bishops when they were receiving a blessing . What a Cardinal and what a fountain of wisdom that man was and we FOTS are a fortunate few to hear his words again today. Your welcome ,Of CORPSE. Thanks for that gag Tea.(TWW). You other Dear Prudents out there,do what you can to drag fingertip to keyboard again. Just so we don't have to send out the search dogs.We are worried about you all. Even the Telepromter who is again MIA,and again presumed drunk on power of an inappropriate voltage. Somone flip his breaker before he slips into permanent ' OFF ' time. [Chris] "Barak (who's insane)" :D LOL, takes after his Muslim brother, So Damn Insane. Ha, HAAAA! ************************* Aero, cool song. You don't need my or anybody's help. Yeah, the mass exodus of the FOTS has saddened me, too. TOTUS' neglect to post has caused most of it, I'm sure, "But," I think to myself, "why didn't they at least say, 'Good bye?'" I don't mean everyone, but there were several cool posters who showed up regularly and just -POOF!-- dis-a-ppeared. Sigh. Well, barring sudden death, if I abandon ship, I will be sure to post a final good bye. Take heart, though, there are several regulars who are still "here," but are posting much less frequently. I'm glad YOU are here. ************************************** Sunflower, too! Hooray! [Sunflower] "The mayor of Vegas doesn't want you there." Uh, huh! WAY TO GO Mayor! That is SO COOL that he dissed the Fraud from Abroad and refused to meet with him. Wish the republicans would do the same re: Hopey the Clown's Feb. 25th public relations stunt. Tiger Woods “Turning Japanese” Tiger pulls a Toyota. Wants to apologize to those who would buy the products he would like to keep endorsing. The Vapor’s Turning Japanese. Someone sent me an Emailed photo of you last night TP,since they know I am a fan. It was taken at the elementary school as BG was enlightening the kids w talk of Bush's secret racist earthquake bomb and the children's torture facility at GITMO. I think that came off of your left screen. It is so odd seeing you in tandem so to speak,but when speaking here,(occasionally) you speak w only one voice,two screens or not. None of that yin and yang,left and right stuff from your ego, >>and<< alter ego screens. Here you speak exclusively from the right side,and I think those servers serving it. (Oh,BTW,how is the service from those waiting to serve in the Marine Corpse? Are they good waiters?) As the midterms approach,your left screen will feel neglected,as the President will favor script to his right.The centrist kind. This is a traditional election ploy as expected as is the usual question asked as to whether one was better off than before,or after,the election. Bet that question gets all the ink that Edward's mistress-e's pregnancy did,as it reflects poorly upon,ahem,Democratocracy. Most have noticed by now,the PRESS CORPSE,doesn't notice that sort of thing. It is an embarrassment to their party, the Won's one. Gee I hate to interrupt guys ,but someone alert the media,AP,and all the MSM that we are ON TO THEM. As noted above Obama has read mostly LEFT screen Teleprompter printout lately,in that Jeckel and Hyde twin and evil altertwwin configuration of his . And yet when he writes here he writes right. Could that left screen,(parts Manchurian), be an AP plant? When reading left is TP an AP_LIE_ANCE Machine?Or just as he claims simply inanimate? Inquiring minds at the National Enquirer want to know.The MSM Press Corpse already does. Will Obama apologize to the world for all the Olympic medals we're winning? Sounds an awful lot like 'Colonialism' of precious metals to me by our 'evil' country. Hi, Preptile. LOL, Chris, D'oh!bama, World Champion Bower, [having only watched one of the awards ceremonies] thinks the Winter Olympics is just the Preliminaries for the Ultimate Bowing event next summer. [He's never watched much TV -- when your brain's been fried from cocaine, TV just irritates you.] Giggly: So, Mister O, you been watching the Winter Olympics? D'oh!: Naaa. Are you kidding? No one can out-bow The One [pulls from under his collar his Golden Chain Award he got for bowing to King Saud] -- I bow when it COUNTS. I'll watch the REAL Olympics, next summer. That crap up in Canada, "I'm not sure how you say this in [Canadian]," is just a bunch of "typical white person[s]" acting "stupidly". Gig: [smirk, fingers fluttering] Still mad about not bringing home the Olympic bacon to Chicago, eh? D'oh!: [GLARE -- chin JUT] I didn't sit in REVrund Wright's church for TWENTY YEARS for nuthin'! I know that "Winter Olympics" is just "white people with blue eyes" ["my man" Lula of Brazil, 2008] and JEEEWWWWZZZ bossin' us half-black folkssss around. I'll be at the SUMMER Olympics where "white will do what's right AS IN GET OUT OF THE WAY SUCKAH and the brown will stay around" (or somethin' like that) [speaker at Fraud's inauguration 1/09]. Oh, I had put that enlightened piece of literary prose out of my mind, TWW, and you drug it back!!! "brown stick aroun...yella is mella...better red than dead" (Did I get that last one wrong...?) LOL, Susan, your bodacious quotationning reminded me of a bumper sticker we posted in our garage in 2004, "Flush the 2 Johns." You got that last one right, given that you are (sort of, heh, heh) quoting a lib-tool. As for you and me, "BETTER DEAD THAN RED." How's that head cold or whatever was troubling you? I hope that you are feeling much better. Say, speaking of putting horrible things out of our minds........ was that who I think it is who just ran past the window?!!!! Naaaa. He's miles away...... isn't he? He wouldn't traipse clear down to Georgia, not in this weather..... not....... NAKED!!!!! Bwah, ha, ha, ha, haaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa! Say, since the TOTUS Cafe is pretty much empty.... the above reminds me of a funny-scary time I had up on a local mountain. A time when using logic backfired. A girlfriend and I back packed up to spend the night on Mount ____. On our way up, we passed three rowdy semi-creepy guys heading down. The sun had already set as we arrived at our camp "site" and we hurried to set up our tent and start a little campfire. We chatted and laughed happily as we sat by our fire eating dinner until........... . We heard something. Something or..... aaaack!... someOne was creeping through the underbrush, twigs breaking underfoot... . Me: Oh, well [hearty chuckle], it's probably just a deer. If it's a bear or something, our fire will scare it away. Friend: [anxiously peering into the blackness surrounding us] Yeah, ha, ha. Me: There's no way it's those creepy guys we passed on our way up. They wouldn't bother coming clear all the way back up here. Naa, no one would hike 2 hours up that trail just to attack us -- [felt happy for 15 seconds] -- except A MANIAC!!!! We didn't sleep well that night. Don't worry, Susan, Fwank's safely ensconced in his little shack up in Maine. His slobber can't reach you from there. Almost, but you are a good 500 feet beyond his record flying spittle. Wouldn't hurt, though, to go make sure that front door is locked. And pull the blinds. Well, that's enough of THAT, huh? Gas expands to fill the space available. Shrug. Got this off GRETAWIRE.Guess we can change the words to his mantra now... Yes, he can, mmm mmm mmm Maybe, he can mmm mmm mmm No, he hasn't, mmm mmm mmm Maybe, he could have, mmm mmm mmm didn't figure he would, mmm mmm mmm No, he can't, mmm mmm mmm . TWW-(or Tea and P calls you) What's with the camping story! That was scarey! (Johnnie I'm on the 1st step.) I could hear the snapping twigs and rustling leaves as the creeps crept closer! Oh, ha ha! Fwank is in Maine with his bff and his bff's weed! Ha ha ha. When I laugh I cough.... Baaaaaaaad bad cough...husband's so bad he had chest xray...but doc said "no problem"! That was good news. Tonite I will see if wine helps. Nothing else does. How long's the cough been going on, Susan? Possibly whooping cough (pertussis)? My daughter (then I) had that a few years ago -- and I wouldn't even wish that on BHO. It's awful!! and if one has been taking antibiotics before being tested for w.c., it will show a false negative. And we'd both had immunizations for pertussis some time before -- guess too long before. Just asking -- daughter's doctor treated her for all sorts of things before sending her elsewhere, without even thinking of whooping cough. Thanks for sticking around, all you faithful FOTS -- All so kind -- and humorous! Thanks Aero! It'd been 2-3 weeks for me, almost 6 weeks for husband...Did the Google,as W would say, on whooping cough...have no fever, cough not as intense as a whooping crane, Candada goose, or dog with kennel cough (which if you've heard is really like the Cananda goose!) so will have another dose of Mucinex or wine. Speaking of Canadians, watching USA v Canada hockey... USA not good enough to win, but you gotta hope as score is USA up 3-2 at the end of 2nd! Holy Cow, Mabel! Well, USA WAS good enough to win! A heart-stopping 5-3 victory over the 'better' Canadians. In truth, Canada was better. You build great teams from the goalie out, and Miller bested Brodeur last night. That empty-net goal was best one in the history of EVER. Chris- You said it all! Miller, altho he plays for an NHL team I do not care for, was super! That last flury of shots Canada took made me dizzy and I want to re-watch as it was nerve racking hockey! Great game, great game! HURRAH FOR THE USA! Glad you guys posted about the U.S. v. Canada match, otherwise, I would have never known. Glad you were entertained, Susan and Aero -- those kinds of experiences are fun to laugh about...... afterward. :D Sunflower, neat and TRUE concise summary of "Little Zero's" year of pretending to be POTUS. Re: calling me "Tea," I don't know why Mr. P. calls me that. It kind of bugs me, but whatever. Shrug. I prefer TWW or TruthWillWin or something along those lines. GET WELL SOON, Susan and Mr. Susan! I recently came across your blog and have been reading along. I thought I would leave my first comment. I don't know what to say except that I have enjoyed reading. Nice blog. I will keep visiting this blog very often. Lucy We don't know what to say either, robbie-Lucy! So jump in! {TWW-thanks for the health wishes. And as for the appellation- I took it as a 'term of endearment' and thought it sweet!} Thank you for this very useful information. sohbet odalari cheap ed hardy cheap ugg boots puma chaussures hommes Nike Air Rift Ed Hardy clothing cheap cell phones wholesale sikis izle sikis izle erotik resim sapık salak videolar

TITLE: Algebraic vs. Integral Closure of a Ring QUESTION [10 upvotes]: Let $R\subseteq S$ be a ring extension. It is true that the set of elements of $S$ that are are integral over $R$ (i.e. satisfy a monic polynomial equation over $R$) is a subring of $S$. Can anyone provide an example showing that the set of elements of $S$ that are merely algebraic over $R$ (i.e. satisfy any polynomial equation over $R$) is not necessarily a subring of $S$? Thanks. EDIT: (April 2016) Thanks to Pavel Čoupek for the solution. I have incorporated this result and much more into a homework assignment. See here for the full assignment with solutions: http://www.math.miami.edu/~armstrong/762sp16/762hw3sol.pdf REPLY [10 votes]: At least in the integral domain case, the "algebraic closure" forms a subring as well: Assume $R \subseteq S$ is an extension of integral domains. Denote by $K, F$ the fraction fields of $R, S$ respectively. Then we have inclusions $R \subseteq S \subseteq F$, $R \subseteq K \subseteq F$. Denote by $B$ the algebraic closure of $K$ in $F$, and by $A$ the "algebraic closure of $R$ in $S$" (i.e. all roots in $S$ of all nonzero polynomials in $R$). The claim is that $B \cap S=A$. The inclusion $A \subseteq B \cap S$ is obvious because of the inclusion $R[x]\subseteq K[x]$. Take $a \in B \cap S$. Then $f(a)=0$ for some nonzero $f \in K[x]$. Then there is $g \in R[x]$ such that $g(a)=0$ - just multiply all the coefficients of $f$ by their common denominator. Hence $a \in A$, so the inclusion $B \cap S \subseteq A$ holds as well. Hence, we expressd $A$ as an intersection of two subrings of $F$ - thus, $A$ is a ring as well (subring of $S$).

TITLE: An ellipse with major axis $4$ and minor axis $2$ touches both the coordinate axes. Locus of its Center and Focus is? QUESTION [2 upvotes]: An ellipse with major axis $4$ and minor axis $2$ touches both the coordinate axes. Locus of its Center and Focus is? My Approach: For locus of Center. Since it is touching the coordinate axes, the coordinate axes will act as tangents making angle of $90^{\circ}$, so origin will lies on Director Circle. Center of director circle will be same as center of Ellipse. Let center of Ellipse be $(h,k)$, so the equation of director circle will be $$(x-h)^2+(y-k)^2=(\text{semi major axis})^2+ (\text{semi minor axis})^2,$$ $$i.e.\;\;\;(x-h)^2+(y-k)^2=(2)^2+ (1)^2.$$ Because it passes through the origin, $(0-h)^2+(0-k)^2=(2)^2+ (1)^2$ $\implies $ $h^2+k^2=5$ $\implies $ locus of center of Ellipse is $x^2+y^2=5$. For Locus of Focus: I assumed focus as $(x_1,y_1)$ but i cannot proceed further. I know one property, that product of distance of tangents from Foci is constant and equal to square of semi-minor axis and lies on auxiliary circle, but that leads me nowhere. Note There are many solutions available on internet, but they all did this with the same method of taking axis of ellipse parallel to coordinate axis. I don't want to do using that method. I want it for slanted ellipse as shown in attached image. REPLY [0 votes]: The method used in the other answer was pretty neat, but I found another approach using the property you mentioned. Assume one of the foci to be $(k,h)$, and the angle made by the ellipse's major axis with the $x$-axis to be $\alpha$. We know the distance between two foci to be $2ae$, where $a=2$ and $e=\frac{\sqrt{3}}{2},$ so the other foci will be: $$\big(k+2\sqrt{3}\cos(\alpha),\;h+2\sqrt{3}\sin(\alpha)\big).$$ Using the property that you mentioned of ellipse that the product of the length of the perpendicular segments from the foci on any tangent to the ellipse is $b^2$, we get: $$k(k+2\sqrt{3}\cos(\alpha))=h(h+2\sqrt{3}\sin(\alpha))=b^2=1,$$ as the perpendicular from two foci on tangent $x=0$ is $k$ and $k+2\sqrt{3}\cos(\alpha)$. Thus we get $$\cos(\alpha)=\frac{1-k^2}{2k\sqrt3} $$ and $$ \sin(\alpha)=\frac{1-h^2}{2h\sqrt3}.$$ On squaring and adding, we get $$ h^2+k^2+\frac{1}{k^2}+\frac{1}{h^2}=16.$$ Replacing $(k,h)$ with $(x,y)$: $$ x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=16.$$

\begin{document} \title[continuous piecewise linear differential systems] {LIMIT CYCLES BIFURCATING FROM A PERIOD ANNULUS\\ IN CONTINUOUS PIECEWISE LINEAR DIFFERENTIAL \\ SYSTEMS WITH THREE ZONES} \let\thefootnote\relax\footnotetext{Corresponding author Claudio Pessoa: Departamento de Matem\'{a}tica, IBILCE--UNESP, 15054--000 S. J. Rio Preto, S\~ao Paulo, Brazil. email: pessoa@ibilce.unesp.br} \author[M. F. S. Lima, C. Pessoa and W. F. Pereira] {Maur\'{\i}cio Firmino Silva Lima$^{1}$, Claudio Pessoa$^{2}$ and Weber F. Pereira$^{2}$} \address{$^1$ Centro de Matem\'atica Computa\c c\~ao e Cogni\c c\~ao, Universidade Federal do ABC\\ Santo Andr\'e, S\~ao Paulo, 09210-170, Brazil} \email{mauricio.lima@ufabc.edu.br} \address{$^2$ Departamento de Matem\'{a}tica, IBILCE--UNESP \\ J. Rio Preto, S\~ao Paulo, 15054--000, Brazil} \email{pessoa@ibilce.unesp.br}\email{weberf@ibilce.unesp.br} \subjclass[2010]{34A36, 34C29, 37G15} \keywords{piecewise linear vector fields, Poincar\'e map, limit cycles, center, focus} \maketitle \begin{abstract} We study a class of planar continuous piecewise linear vector fields with three zones. Using the Poincar\'{e} map and some techniques for proving the existence of limit cycles for smooth differential systems, we prove that this class admits at least two limit cycles that appear by perturbations of a period annulus. Moreover, we describe the bifurcation of the limit cycles for this class through two examples of two-parameter families of piecewise linear vector fields with three zones. \end{abstract} \section{Introduction} One of the most important and studied problem in the qualitative theory of differential systems, in particular for planar systems, is the maximum number, stability and position of limit cycles, see for instance \cite{Yan} and \cite{Zou}. The famous second part of the 16th Hilbert problem proposed in the $1900^\prime$s and that have been studied for a long time addresses the problem of the maximum number and position of limit cycles restricted to the planar polynomial differential equations. In the piecewise continuous context this problem has been studied by many authors, see for instance \cite{Freire}, \cite{Hogan}, \cite{Llibre3} and \cite{van}. This class of system is very important, mainly by their numerous applications, where they arise in a natural way, for instance in control theory \cite{Lefschetz} and \cite{Narendra}, design of electric circuits \cite{Chua}, neurobiology \cite{FitzHugh} and \cite{Nagumo}, etc. In this context, specially for piecewise linear differential systems, many works have been developed. Most of them obtaining results on the existence and uniqueness of limit cycles for systems with only one curve of discontinuity. For systems with more then one curve of discontinuity not many works are available and, more important than this, recently (see \cite{Llibre4}) an example with more then one limit cycle could be obtained for a special class of Liénard piecewise linear differential system with two curves of discontinuity. As far as we know the paper \cite{Llibre4} is, until this moment, the only place where at least two limit cycles was obtained in this context. It is important to observe that the characterization of all possible cases with two (or maybe more than two) limit cycles is far from being completely solved. In this paper we give a contribution in this direction where we provide a family of piecewise linear differential systems with at least two limit cycles. We observe that the bifurcation that give rise to the second limit cycle is very close to the one that appear in \cite{Llibre4}, namely, one limit cycle visiting the three zones and the second limit cycle visiting two zones and that bifurcates of a period annulus. We observe that in \cite{Llibre4} this period annulus is obtained of a center and the second limit cycle appears when this center becomes a focus stable/unstable under parameter changes while, in the present paper, this region is obtained from two foci of different stability with eigenvalues of the same modulus and the second limit cycle appears when these foci have attracting/repelling with different magnitude. However, in both cases, the piecewise linear differential systems have a center. In \cite{Llibre4} this center is located in the central region while in the present work the center is supposed to be in the left region. In order to set the problem, consider the plane $\mathbb R^2$ divided in three closed regions $R_-,$ $R_o$ and $R_+$ which frontier are given by two parallel straight lines $L_-$ and $L_+$ symmetric with respect to the origin such that $(0,0)\in R_o$ and the regions $R_-$ and $R_+$ have as boundary the respective straight lines $L_-$ and $L_+$. In this paper, we study the existence of limit cycles for the family of differential systems \begin{equation} {\bf x'}=\left\{\begin{array}{ll} A_-{\bf x}+B_- & {\bf x}\in R_-, \\ A_o{\bf x}+B_o & {\bf x}\in R_o, \\ A_+{\bf x}+B_+ & {\bf x}\in R_+, \end{array} \right.\label{system 1} \end{equation} that are continuous piecewise linear differential systems with tree zones where ${\bf x}=(x,y)\in\mathbb{R}^2$, $A_i\in\mathcal{M}_2(\mathbb{R}),$ $B_i\in\mathbb{R}^2,\;i\in\{-,o,+\}$, and ${\bf x'}=\dfrac{d{\bf x}}{dt}$ with $t$ the time. Let $X_i=A_i{\bf x}+B_i$, $i\in \{-, o, +\}$, the linear vector fields given in \eqref{system 1}. Denote by $X$ the continuous piecewise linear vector field associated to system \eqref{system 1}, i.e. ${\bf x'}=X({\bf x})$, where ${\bf x}=(x,y)\in\mathbb{R}^2$. Therefore $X\mid_{R_i}=X_i\mid_{R_i}$, $i\in \{-, o, +\}$, i.e., $X$ restrict to each one of these zones are linear systems with constant coefficients that are glued continuously at the common boundary. Our goal is improve the results of the papers \cite{Mauricio} and \cite{ClauWeMau} considering the case not covered in these papers and where at least two limit cycles can be present. Thus, as in \cite{Mauricio} and \cite{ClauWeMau}, we assume that $X_o$ has an equilibrium of focus type and the equilibria of $X_-$ and $X_+$ are a center or a focus. From the next result we can refine these assumptions a bit more, but before we need some notation. We say that the vector field $X_i$ has a \textit{real equilibrium} ${\bf x}^*$ in $R_i$ with $i\in\{-,o,+\}$ if ${\bf x}^*$ is an equilibrium of $X_i$ and ${\bf x}^*\in R_i.$ Otherwise, we will say that $X_i$ has a \textit{virtual equilibrium} ${\bf x}^*$ in $R_i^c$ if ${\bf x}^*$ is an equilibrium of $X_i$ and ${\bf x}^*\in R_i^c$, where $R_i^c$ denotes the complementary of $R_i$ in $\mathbb{R}^2$. From now on for $i\in\{-,o,+\}$ we denote by $d_i$ the determinant of the matrix $A_i$ and by $t_i$ its trace. Also we denote by ${\mbox Int}(\Gamma)$ the open region limited by a closed Jordan curve $\Gamma.$ The next result is an immediate consequence of the Green's Formula (see also Proposition 3 of \cite{Llibre2}). \begin{proposition} If system {\rm (\ref{system 1})} has a simple invariant closed curve $\Gamma$ then $$\int\!\!\!\int_{\mbox{ Int}_-(\Gamma)}t_-dxdy+\int\!\!\!\int_{\mbox{ Int}_o(\Gamma)}t_odxdy+ \int\!\!\!\int_{\mbox{ Int}_+(\Gamma)}t_+dxdy =t_-S_-+t_oS_o+t_+S_+=0,$$ where $\mbox{ Int}_i(\Gamma)=\mbox{ Int}(\Gamma)\cap R_i$ and $S_i=\mbox{ area}(\mbox{Int}_i(\Gamma))$ with $i\in\{-,o,+\}.$\label{prop green} \end{proposition} From this result if $X_o$ has a focus and either $X_-$ or $X_+$ has a center then a necessary condition for the existence of such $\Gamma$ is that $\Gamma$ visit at least the two zones having a real or virtual focus with different stability. Then, from now on, we assume the following hypothesis: \begin{table}[h] \begin{tabular}{ll} (H1) & $X_o$ has a focus. \\ \\ (H2) & The others equilibria of $X_-$ and $X_+$ are a center and a focus with \\ & different stability with respect to the focus of $X_o$. \end{tabular} \end{table} As usual a {\it limit cycle} $\mathcal{C}$ of (\ref{system 1}) is an isolated periodic orbit of (\ref{system 1}) in the set of all periodic orbits of (\ref{system 1}). We say that $\mathcal{C}$ is {\it hyperbolic} if the integral of the divergence of the system along it is different from zero, for more details see for instance \cite{dumortier}. The main result of this paper is the following. \begin{theorem} \label{the_paper2:01} Assume that system {\rm (\ref{system 1})} satisfies assumptions {\rm (H1)}, {\rm (H2)} and $X_o$ has a real focus at the boundary of $R_o$. If the focus of $X_o$ belongs to $L_+$ (respectively $L_-$) and $X_+$ (respectively $X_-$) also has a focus at the same point of $L_+$ (respectively $L_-$) and both the foci give rise to a center for system \eqref{system 1}, then at least two limit cycles can appear by small perturbations of the parameters of system \eqref{system 1}. \end{theorem} \begin{remark} Explicit conditions for the existence of at least two limit cycle in terms of the parameters of system {\rm (\ref{system 1})} are techniques, depend of some ingredients defined in the paper, and are summarized in Proposition \ref{teo two limit cycle} of Section \ref{sec Lim b_2< -1}. \end{remark} The paper is organized as follows. In Section \ref{sec Norm Form} we obtain a normal form to system \eqref{system 1} that simplifies the computations. In Section \ref{sec Poin Map} we study the behavior of the Poincar\'e maps associated to system \eqref{system 1} that will be used to study the problem and at the Section \ref{sec Lim b_2< -1} we prove Theorem \ref{the_paper2:01}. Finally, in Section \ref{sec examples} we give two examples that resume the bifurcations of limit cycles to system \eqref{system 1} under hypothesis (H1) and (H2). \section{Normal Form} \label{sec Norm Form} The following result, proved in \cite{Mauricio}, give us a convenient normal form to write system (\ref{system 1}) with the number of parameters reduced, and so it will be useful to obtain the Poincar\'{e} return map and its derivatives. \begin{lemma} Suppose that system (\ref{system 1}) is such that $d_o>0.$ Then there exists a linear change of coordinates that writes system {\rm (\ref{system 1})} into the form $\dot{{\bf x}}=X({\bf x}),$ with $L_-=\{x=-1\},$ $L_+=\{x=1\},$ $R_-=\{(x,y)\in\mathbb{R}^2;\;x\leq-1\},$ $R_o=\{(x,y)\in\mathbb{R}^2;\;-1\leq x\leq1\},$ $R_+=\{(x,y)\in\mathbb{R}^2;\;x\geq1\}$ and \begin{equation} X({\bf x})=\left\{\begin{array}{ll} A_-{\bf x}+B_- & {\bf x}\in R_-, \\ A_o{\bf x}+B_o & {\bf x}\in R_o, \\ A_+{\bf x}+B_+ & {\bf x}\in R_+, \end{array}\right.\label{base system nf} \end{equation} where $A_-=\left(\begin{array}{cc} a_{11} & -1 \\ 1-b_2+d_2 & a_1 \end{array}\right),$ $B_-=\left(\begin{array}{c} a_{11} \\ d_2 \end{array}\right),$ $A_o=\left(\begin{array}{cc} 0 & -1 \\ 1 & a_1 \end{array}\right),$ $B_o=\left(\begin{array}{c} 0 \\ b_2 \end{array}\right),$ $A_+=\left(\begin{array}{cc} c_{11} & -1 \\ 1+b_2-f_2 & a_1 \end{array}\right)$ and $B_+=\left(\begin{array}{c} -c_{11} \\ f_2 \end{array}\right).$ The dot denotes derivative with respect to a new time $s.$\label{nf} \end{lemma} We call {\it contact point} to a point of a straight line where the vector field is tangent to it. The next lemma, which proof can be found in \cite{ClauWeMau}, characterizes the equilibria of the vector fields $X_i$, $i\in \{-, o, +\}$, in real or virtual with respect the sign of the parameter $b_2$ of the normal form \eqref{base system nf}. \begin{lemma}\label{equilibrium} In the coordinates given by Lemma \ref{nf}, there is a unique contact point of system {\rm (\ref{base system nf})} with each one of the straight lines $L_-$ and $L_+$. These points are respectively $p_-=(-1,0)$ and $p_+=(1,0).$ Moreover under the assumptions {\rm (H1)} and {\rm (H2)}, we have \begin{itemize} \item[(i)] if $b_2<-1$, the equilibrium points of $X_-$ and $X_o$ are virtual and the equilibrium point of $X_+$ is real; \item[(ii)] if $b_2=-1$, $X_o$ and $X_+$ have an equilibrium point at $(1,0)$, and $X_-$ has a virtual equilibrium point; \item[(iii)] if $|b_2|<1$, the equilibrium points of $X_-$ and $X_+$ are virtual and the equilibrium point of $X_o$ is real; \item[(iv)] if $b_2=1$, $X_-$ and $X_o$ have an equilibrium point at $(-1,0)$, and $X_+$ has a virtual equilibrium point; \item[(v)] if $b_2>1$, the equilibrium points of $X_o$ and $X_+$ are virtual and the equilibrium point of $X_-$ is real. \end{itemize} \end{lemma} From Lemmas \ref{nf} and \ref{equilibrium} we can divide the problem of study the existence of limit cycles for system \eqref{system 1} with hypothesis (H1) and (H2) in the cases given in the table below. \medskip \begin{table}[h] \scalefont{0.9} \caption{All the cases under the hypothesis (H1) and (H2).} {\begin{tabular}{|c|c|c|c|c|c|} \hline & Case $b_2<-1$ & Case $b_2=-1$ & Case $|b_2|<1$ & $b_2=1$ & $b_2>1$ \\ \hline Equilibrium of $X_-$ & Virtual & Virtual & Virtual & $(-1,0)$ & Real \\ \hline Equilibrium of $X_o$ & Virtual & $(1,0)$ & Real & $(-1,0)$ & Virtual \\ \hline Equilibrium of $X_+$ & Real & $(1,0)$ & Virtual & Virtual & Virtual \\ \hline \end{tabular}} \end{table} The Case $|b_2|<1$ has been studied in \cite{Mauricio}, in this paper it is showed that under the hypothesis (H1) and (H2) and additionally assuming that $X_o$ has a real focus in the interior of $R_o$, system \eqref{system 1} has a unique limit cycle. Some another cases have been studied in \cite{ClauWeMau}, where it is showed that under the hypothesis (H1) and (H2) and additionally assuming that \begin{itemize} \item[(a)] $X_o$ has a virtual focus and $X_+$ (respectively $X_-$) has a real center, system {\rm (\ref{system 1})} has a unique limit cycle, which is hyperbolic; \item[(b)] $X_o$ has a real focus at the boundary of $R_o$, system {\rm (\ref{system 1})} has a unique limit cycle, which is hyperbolic. Except when the focus of $X_o$ belongs to $L_+$ (respectively $L_-$) and $X_+$ (respectively $X_-$) also has a focus and both foci give rise to a center for system \eqref{system 1}. In this case system \eqref{system 1} has no limit cycles. \end{itemize} Note that, the Cases $b_2>1$ and $b_2=1$ are equivalent, by a rotation through of an angle of $180^o$, to the Cases $b_2<-1$ and $b_2=-1$, respectively. Therefore, from above results, remains only the case $b_2<-1$ with $X_+$ having a real focus to be studied. Here, Theorem \ref{the_paper2:01} shows that we have at least two limit cycles in this case. A complete study of the case $b_2<-1$, when $X_+$ has a real focus, it is still an open problem. \section{Poincar\'{e} Return Map} \label{sec Poin Map} In this section we will rewrite the problem of finding limit cycles that visit the three zones $R_i$, $i\in\{-, o, +\}$ or even only two of them in terms of finding the fixed points of an appropriated Poincar\'{e} return map. This Poincar\'{e} return map will be the composition of four or two different Poincar\'{e} maps, according to the number of zones. As in \cite{ClauWeMau}, these Poincar\'{e} maps will be defined in the transversal sections $L_\pm^{O}=\{(\pm1,y);\,y\geq0\}$ and $L_\pm^{I}=\{(\pm1,y);\,y\leq0\}.$ In order to study the qualitative behavior of each one of these maps, we will do a convenient parameterization in the transversal sections $L_\pm^{O}$ and $L_\pm^{I}$ assuming $b_2\neq \pm 1$. We parametrize $L_-^{O}$ by the parameter $c$ defined as follows. Let $p_-=(-1,0)$ be the contact point of $X_-$ with $L_-$ and $\dot{p}_-=X_-(p_-)=(0,b_2-1).$ Given, $p\in L_-^{O}$ we take $c\geq 0$ as the unique non-negative real satisfying $p=p_--c \dot{p}_-.$ Similarly we parametrize $L_-^{I}$ by the parameter $d$, i.e. given $p\in L_-^{I}$ we take $d\geq 0$ as the unique non-negative real satisfying $p=p_-+d \dot{p}_-$. In a very close way we parametrize $L_+^{O}$ by the parameter $b$ as follows. Let $p_+=(1,0)$ be the contact point of $X_+$ with $L_+$ and $\dot{p}_+=X_+(p_+)=(0, b_2 + 1).$ Given, $p\in L_+^{O}$ we take $b\geq 0$ as the unique non-negative real stisfying $p=p_+-b \dot{p}_+.$ In a similar way we parametrize $L_+^{I}$ by the parameter $a,$ i.e. given $p\in L_+^{I}$ we take $a\geq 0$ as the unique non-negative real satisfying $p=p_++a \dot{p}_+$. {For} study the limit cycles of system (\ref{base system nf}) that visit the three zones $R_i$, $i\in\{-, o, +\}$, the Poincar\'{e} return map $\Pi$ is defined on $L_-^O$ and its fixed points correspond to these limit cycles. This map involves all the vector fields $X_i$, $i\in \{-, o, +\}$, and has the form \[ \Pi=\bar{\pi}_o\circ\pi_+\circ \pi_o\circ\pi_-, \] where the Poincar\'{e} maps $\bar{\pi}_o$, $\pi_+$, $\pi_o$, and $\pi_-$ will be defined as follows. The vector field $X_-$ points toward the region $R_-$ in $L_-^O$ while in $L_-^{I}$ the vector field points toward the region $R_o.$ Then we can define a Poincar\'{e} map $\Pi_-$ from $L_-^{O}$ to $L_-^{I}$ by $\Pi_-(p)=q$ as being the first return map in forward time of the flow of $X_-$ to $L_-,$ i.e. if $\varphi_-(s,p)$ is the solution of $\dot{\bf x}=A_-{\bf x}+B_-$ such that $\varphi_-(0,p)=p$ and $p\in L_-^{O},$ then $q=\varphi_-(s,p)\in L_-^{I}$ with $s\geq0$. Note that $\Pi_-(p_-)=p_-.$ Now, using the parametrizations of $L_\pm^O$ and $L_\pm^I$ previously defined, given $p\in L_-^{O}$ and $q\in L_-^{I}$ there exist unique $c\geq0$ and $d\geq0$ such that $p=p_--c\dot{p}_-$ and $q=p_-+d\dot{p}_-.$ So $\Pi_-$ induces a mapping $\pi_-$ given by $\pi_-({c})=d.$ Observe that the qualitative behavior of $\Pi_-$ is equivalent to the qualitative behavior of $\pi_-.$ Analogously we can define the Poincar\'{e} maps $\Pi_+$ from $L_+^I$ to $L_+^O$, through the flow of $X_+$, $\Pi_o$ from $L_-^I$ to $L_+^I$ and $\bar{\Pi}_o$ from $L_+^O$ to $L_-^O$, bouth through of the flow of $X_o$ and, by the parametrization defined to this transversal sections, we obtain the correspondents induced Poincar\'{e} maps $\pi_+$, $\pi_o$ and $\bar{\pi}_o$ as above. The maps $\bar{\pi}_o$, $\pi_+$, $\pi_o$, and $\pi_-$ are invariant under linear change of coordinates and translation, see Proposition 4.3.7 in \cite{Llibre1}. Then, in the Case $b_2<-1$, to compute the maps $\pi_\pm$ we can suppose that the equilibria of $X_\pm$ are at the origin and that the matrices $A_\pm$ are given in their real Jordan normal form. In what follows, we will consider the maps $\bar{\pi}_o$, $\pi_+$, $\pi_o$, and $\pi_-$ instead of $\bar{\Pi}_o$, $\Pi_+$, $\Pi_o$, and $\Pi_-$. We will study the qualitative behavior of each one of these maps separately in order to understand the global behavior of the general Poincar\'{e} return map. The next lemma will be useful in the study of the Poincar\'{e} return map associated to system (\ref{base system nf}). The proof of this lemma can be found in \cite{Llibre1} Lemma 4.4.10 and also in \cite{Mauricio}. \begin{lemma} Consider the function $\varphi:\mathbb{R}^2\rightarrow\mathbb{R}$ given by $\varphi(x,y)=1-e^{xy}(\cos x-y\sin x).$ The qualitative behavior of $\varphi(x,y_o)$ in $(-\infty,2\pi)$ is represented in Figure \ref{grafico1} when $y_o>0,$ and in Figure \ref{grafico2} when $y_o<0.$ \label{basicfunction} \end{lemma} \begin{figure}[!htb] \begin{minipage}[b]{0.45\linewidth} \centering \includegraphics[width=4cm,height=5cm,angle=-90]{grafico1.eps} \caption{Function $\varphi(x,y_o)$ for $y_o>0$.} \label{grafico1} \end{minipage} \hspace{0.5cm} \begin{minipage}[b]{0.45\linewidth} \centering \includegraphics[width=4cm,height=5cm,angle=-90]{grafico2.eps} \caption{Function $\varphi(x,y_o)$ for $y_o<0$.} \label{grafico2} \end{minipage} \end{figure} \begin{proposition}\label{prop mau 1} Consider the vector field $X_-$ in $R_-$ with a virtual center or focus equilibrium and such that $t_- \geq 0.$ Let $\pi_-$ be the map associated to the Poincar\'{e} map $\Pi_-:L_-^O\rightarrow L_-^I$ defined by the flow of the linear system $\dot{{\bf x}}=A_-{\bf x}+B_-.$ \begin{itemize} \item[a] If $t_->0$, then the map $\pi_{-}$ is such that $\pi_{-}:[0,\infty)\rightarrow[0,\infty),$ $\pi_{-}(0)=0,$ $\displaystyle\lim_{c\rightarrow\infty}\pi_{-}({c})=+\infty$ and $\pi_{-}(c)>c$ in $(0,\infty).$ \begin{itemize} \item[(a.1)] If $c\in(0,\infty)$, then $(\pi_{-})'(c)=\dfrac{c}{\pi_-(c)}e^{2\gamma_-\tau_-}.$ Moreover $(\pi_{-})'(c)>1$ and $\displaystyle\lim_{c\rightarrow0}(\pi_-)'(c)=1.$ \item[(a.2)] If $c\in(0,\infty)$, then $(\pi_{-})''(c)>0.$ \item[(a.3)] The straight line $d=e^{\gamma_-\pi}c-t_-(1+e^{\gamma_-\pi})/d_-$ in the plane $(c,d)$ is an asymptote of the graph of $\pi_-$ when $c$ tends to $+\infty$ where $\gamma_-=t_-/\sqrt{4d_--t_-^2}.$ So $\displaystyle\lim_{c\rightarrow\infty}(\pi_-)'(c)=e^{\gamma_-\pi}.$ \end{itemize} \item[(b)] If $t_-=0$, then $\pi_{-}$ is the identity map in $[0,\infty).$ \end{itemize} \end{proposition} \begin{proof} See Proposition 3.2 of \cite{Mauricio}. \end{proof} \begin{proposition}\label{prop.8} Consider the vector field $X_+$ in $R_+$ with a real focus equilibrium and such that $t_+>0.$ Let $\pi_+$ be the map associated to the Poincar\'{e} map $\Pi_+:L_+^I\rightarrow L_+^O$ defined by the flow of the linear system $\dot{{\bf x}}=A_+{\bf x}+B_+.$ \begin{itemize} \item[(a)] Then the maps $\pi_{+}$ is such that $\pi_{+}:[0,\infty)\rightarrow[b^\ast_+,\infty),$ $\pi_+(0)=b^\ast_+ >0$, $\displaystyle\lim_{a\rightarrow\infty}\pi_{+}(a)=+\infty$ and $\pi_{+}(a)>a$ in $(0,\infty).$ \begin{itemize} \item[(a.1)] If $a\in(0,\infty)$, then $(\pi_{+})'(a)=\dfrac{a}{\pi_+(a)}e^{2\gamma_+\tau_+}.$ Moreover $(\pi_{+})'(a)>0$ and $\displaystyle\lim_{a\rightarrow0}(\pi_+)'(a)=0.$ \item[(a.2)] If $a\in(0,\infty)$, then $(\pi_{+})''(a)>0.$ \item[(a.3)] The straight line $b=e^{\gamma_+\pi}a+t_+(1+e^{\gamma_+\pi})/d_+$ in the plane $(a,b)$ is an asymptote of the graph of $\pi_+$ when $a$ tends to $+\infty$ where $\gamma_+=t_+/\sqrt{4d_+-t_+^2}.$ So $\displaystyle\lim_{a\rightarrow\infty}(\pi_+)'(a)=e^{\gamma_+\pi}.$ \end{itemize} \end{itemize} \label{prop p1} \end{proposition} \begin{proof} Let $p_+$ be the contact point of the flow with $L_+$ and $p$ and $q$ such that $\Pi_+(p)=q$. As $q$ is in the orbit of $p$ in the forward time we have that $q=\varphi(s,p)$ with $s\geq0.$ Moreover for computing the map $\pi_+$ we can suppose that the real equilibrium is at the origin and that matrix $A_+$ is in its real Jordan normal form. Let $p_+^*$ be the contact point $p_+$ in the coordinates in which $A_+$ is in its real Jordan normal form and the real equilibrium of $X_+$ is at the origin. We denote by $\dot{p}_+^*=X_+(p_+^*).$ So we can write $$q=\varphi_+(s,p)=e^{A_+s}p.$$ As $p=p_+^*+a\dot{p}_+^*$ and $q=p_+^*-\pi_+(a)\dot{p}_+^*$, we obtain $$p_+^*-\pi_+(a)\dot{p}_+^*=e^{A_+s}(p_+^*+a\dot{p}_+^*).$$ Now using the fact that $\dot{p}_+^*=A_+p_+^*$, we have \begin{equation} (Id-\pi_+(a)A_+)p_+^*=e^{A_+s}(Id+aA_+)p_+^*,\label{equation pi-} \end{equation} where $a\geq0,$ $\pi_+(a)\geq0,$ $s\geq0$ and the matrix $A_+$ is given by $A_+=\left(\begin{array}{cc} \alpha_+ & -\beta_+ \\ \beta_+ & \alpha_+ \end{array}\right)$ with $\alpha_+=\dfrac{t_+}{2}.$ Since $p_+^*\neq(0,0)$, we obtain from equation (\ref{equation pi-}), that $b=\pi_+(a)$ is defined by the system \begin{equation} \begin{array}{l} \!\!\!\!1-b\alpha_+\!\!=e^{\alpha_+s}(\cos(\beta_+s)+a[\alpha_+\cos(\beta_+s)-\beta_+\sin(\beta_+s)]), \\ \\ \!\!\!\!b\beta_+\!\!= -e^{\alpha_+s}(\sin(\beta_+s)+a[\alpha_+\sin(\beta_+s)+\beta_+\cos(\beta_+s)]), \end{array}\label{system sol} \end{equation} and the inequalities $a\geq0,\,b\geq0$ and $s\geq0.$ Let $\pi_+(0)=b^\ast_+$, we have that $b^\ast_+ >0$ (see Figure \ref{figg1}). \begin{figure}[h!] \begin{center} \begin{overpic}[width=5cm,height=4.5cm]{fig_prop11_2.eps} \put(58,71){\tiny $b_+^*$} \put(6,-6){\tiny $L_-$} \put(64,-6){\tiny $L_+$} \end{overpic} \end{center} \caption{The flow of the three zones vector fields with a real focus equilibrium and $t_+>0$.} \label{figg1} \end{figure} Moreover if $a=a_o,\,b=b_o$ and $s=s_o$ is a solution of (\ref{system sol}), then $s_o$ is the flight time between the points $p=p_+^*+a\dot{p}_+^*$ and $q=p_+^*-b\dot{p}_+^*.$ Define $\tau_+=\beta_+s$ and $\gamma_+=\alpha_+/\beta_+.$ Solving system (\ref{system sol}) with respect to $\tau_+$ we obtain the following parametric equations for $\pi_+(a)=b,$ \begin{equation} \begin{array}{rr} a(\tau_+)=-\dfrac{\beta_+e^{-\gamma_+\tau_+}}{d_+\sin\tau_+}\varphi(\tau_+,\gamma_+) & \mbox{ and } \\ & \\ b(\tau_+)=-\dfrac{\beta_+e^{\gamma_+\tau_+}}{d_+\sin\tau_+}\varphi(\tau_+,-\gamma_+), & \\ \end{array}\label{parametricsolution} \end{equation} where $\varphi$ is the function described in Lemma \ref{basicfunction}. Since $A_+$ is given in its real Jordan normal form, $\tau_+$ is the angle covered by the solution during the flight time $s$. Hence we conclude that $\tau_+ \in (\pi,\tau^\ast)$, where $\tau^\ast<2\pi$. Note that $\tau^\ast$ is the angle covered by the solution during the flight time $s^\ast$, i.e. $e^{s^\ast A}p_+^\ast=\Pi_+(p_+^\ast)$. Now since $\displaystyle\lim_{\tau_+\rightarrow\pi^+}a(\tau_+)=+\infty$ and $\displaystyle\lim_{\tau_+\rightarrow\pi^+}b(\tau_+)=+\infty$ it follows that the domain of definition of $\pi_+$ is $[0,+\infty)$ and $\displaystyle\lim_{a\rightarrow\infty}\pi_+(a)=+\infty.$ Moreover, when $\tau_+\in(\pi,\tau^\ast)$ we have $$b(\tau_+)-a(\tau_+)=-\dfrac{2\beta_+}{d_+\sin\tau_+}(\sinh(\gamma_+\tau_+)-\gamma_+\sin\tau_+).$$ Since $\sinh(\gamma_+\tau_+)>\gamma_+\sin\tau_+$ when $\tau_+\in(\pi,\tau^\ast),$ we conclude from the expression above that $b(\tau_+)>a(\tau_+)$ if $\tau_+\in(\pi,\tau^\ast).$ Therefore $\pi_+(a)>a$ in $(0,+\infty).$ So statement (a) is proved. Derivating (\ref{parametricsolution}) with respect to $\tau_+$ it follows that $$ \dfrac{da}{d\tau_+}=-\dfrac{\beta_+}{d_+\sin^2\tau_+}\varphi(\tau_+,-\gamma_+) $$ and $$ \dfrac{db}{d\tau_+}=-\dfrac{\beta_+}{d_+\sin^2\tau_+}\varphi(\tau_+,\gamma_+).$$ Thus $(\pi_+)'(a)=\dfrac{\varphi(\tau_+,\gamma_+)}{\varphi(\tau_+,-\gamma_+)}=\dfrac{a}{b}e^{2\gamma_+\tau_+}$ and $\displaystyle\lim_{a\rightarrow0}(\pi_+)'(a)=0$, because $\displaystyle\lim_{a\rightarrow0} b=\displaystyle\lim_{a\rightarrow0} \pi_+(a)=b^\ast_+\neq 0$. Therefore substatement (a.1) is proved. Now we observe that $$\begin{array}{c} (\pi_+)''(a)=\dfrac{d}{d\tau_+}\left(\dfrac{db}{da}\right)\dfrac{1}{\frac{da}{d\tau_+}}= \\ - \dfrac{2d_+(1+\gamma_+^2)\sin^3\tau_+}{\beta_+\varphi(\tau_+,-\gamma_+)^3}(\sinh(\gamma_+\tau_+)-\gamma_+\sin\tau_+)>0. \\ \end{array}$$ Therefore substatement (a.2) follows. From expression (\ref{parametricsolution}) it follows that $$ \lim_{a\rightarrow\infty}\dfrac{\pi_+(a)}{a} =\displaystyle\lim_{\tau_+\rightarrow\pi}\dfrac{b(\tau_+)}{a(\tau_+)}= \lim_{\tau_+\rightarrow\pi}e^{2\gamma_+\tau_+}\dfrac{\varphi(\tau_+,-\gamma_+)}{\varphi(\tau_+,\gamma_+)} =e^{\gamma_+\pi}. $$ On the other hand by applying the L'H\^{o}ptal's rule it is easy to check that $\displaystyle\lim_{a\rightarrow+\infty}(\pi_+(a)-e^{\gamma_+\pi}a)=t_+(1+e^{\gamma_+ \pi})/d_+$, which implies that the straight line $b=e^{\gamma_+\pi}a+t_+(1+e^{\gamma_+ \pi})/d_+$ is an asymptote of the graph of $\pi_+(a)$ and we obtain substatement (a.3). \end{proof} \begin{proposition}\label{prop.9} Consider the vector field $X_+$ in $R_+$ with a real focus equilibrium and such that $t_+<0.$ Let $\pi_+$ be the map associated to the Poincar\'{e} map $\Pi_+:D_+^*\subset L_+^I\rightarrow L_+^O$ defined by the flow of the linear system $\dot{{\bf x}}=A_+{\bf x}+B_+$, where $D_+^*$ is a subset of $L_+^I$ which the mapping $\Pi_+$ is well defined. \begin{itemize} \item[(a)] Then the maps $\pi_{+}$ is such that $\pi_{+}:[a^\ast_+,\infty)\rightarrow[0,\infty),$ $\pi_+(a^\ast_+)=0$, $\displaystyle\lim_{a\rightarrow\infty}\pi_{+}(a)=+\infty$ and $\pi_{+}(a)<a$ in $(a^\ast_+,\infty).$ \begin{itemize} \item[(a.1)] If $a\in(a^\ast_+,\infty)$, then $(\pi_{+})'(a)=\dfrac{a}{\pi_+(a)}e^{2\gamma_+\tau_+}.$ Moreover $(\pi_{+})'(a)>0$ and $\displaystyle\lim_{a\rightarrow a^\ast_+}(\pi_+)'(a)=+\infty.$ \item[(a.2)] If $a\in(a^\ast_+,\infty)$, then $(\pi_{+})''(a)<0.$ \item[(a.3)] The straight line $b=e^{\gamma_+\pi}a+t_+(1+e^{\gamma_+\pi})/d_+$ in the plane $(a,b)$ is an asymptote of the graph of $\pi_+$ when $a$ tends to $+\infty$ where $\gamma_+=t_+/\sqrt{4d_+-t_+^2}.$ So $\displaystyle\lim_{a\rightarrow\infty}(\pi_+)'(a)=e^{\gamma_+\pi}.$ \end{itemize} \end{itemize}\label{prop p1} \end{proposition} \begin{proof} The proof follows in a similar way to the proof of Proposition \ref{prop.8}. \end{proof} \begin{proposition}\label{prop.10} Consider the vector field $X_o$ in $R_o$ with a virtual focus equilibrium. Let $\pi_o$ be the map associated to the Poincar\'{e} map $\Pi_o: L_-^I\rightarrow L_+^I$ defined by the flow of the linear system $\dot{{\bf x}}=A_o{\bf x}+B_o$ from the straight line $L_-$ to the straight line $L_+$. \begin{itemize} \item[(a)] Then the map $\pi_o$ is such that $\pi_o:[0,\infty)\rightarrow[a^*_o,\infty),$ $a^*_o> 0$ with $\pi_o(0)=a^*_o$ and $\displaystyle\lim_{d\rightarrow\infty}\pi_o(d)=+\infty.$ \item[(b)] If $d\in[0,\infty)$, then $\pi_o'(d)=\left(\dfrac{1-b_2}{1+b_2}\right)^2 e^{2\gamma_o\tau_o}\dfrac{d}{\pi_o(d)},$ with $\tau_o\rightarrow0$ when $d\rightarrow\infty$ and $\displaystyle\lim_{b\rightarrow\infty}\pi_o'(d)=-\dfrac{1-b_2}{1+b2}$. \end{itemize} \end{proposition} \begin{proof} See Proposition 3.3 of \cite{ClauWeMau}. \end{proof} \begin{proposition} \label{prop.11} Consider the vector field $X_o$ in $R_o$ with a virtual focus equilibrium. Let $\bar{\pi}_o$ be the map associated to the Poincar\'{e} map $\bar{\Pi}_o:\bar{D}_o^*\subset L_+^O\rightarrow L_-^O$ defined by the flow of the linear system $\dot{{\bf x}}=A_o{\bf x}+B_o$ from the straight line $L_+$ to the straight line $L_-$, where $\bar{D}_o^*$ is a subset of $L_+^O$ in which the mapping $\bar{\Pi}_o$ is well defined. \begin{itemize} \item[(a)] Then the map $\bar{\pi}_o$ is such that $\bar{\pi}_o:[b^*_o,\infty)\rightarrow[0,\infty),$ $b^*_o> 0$ with $\bar{\pi}_o(b^*_o)=0$ and $\displaystyle\lim_{b\rightarrow\infty}\bar{\pi}_o(b)=+\infty.$ \item[(b)] If $b\in(b^*_o,\infty)$, then $\bar{\pi}_o'(b)=\left(\dfrac{b_2+1}{b_2-1}\right)^2 e^{2\gamma_o\bar{\tau}_o}\dfrac{b}{\bar{\pi}_o(b)},$ with $\bar{\tau}_o\rightarrow0$ when $b\rightarrow\infty$, $\displaystyle\lim_{b\rightarrow\infty}\bar{\pi}_o'(b)=\dfrac{b_2+1}{b_2-1}$ and $\displaystyle\lim_{b\rightarrow b^*_o}\bar{\pi}_o'(b)=\infty$. \end{itemize} \end{proposition} \begin{proof} See Proposition 3.3 of \cite{ClauWeMau}. \end{proof} \section{Limit Cycles when $b_2<-1$ and $X_+$ Has a Real Focus} \label{sec Lim b_2< -1} In this section we will assume that $b_2\leq-1$ and $X_+$ has a real focus and so, by Lemma \ref{equilibrium}\;{(i)-(ii)} and hypothesis (H1) and (H2), $X_-$ has a virtual center and $X_o$ has either a virtual focus (when $b_2<-1$) or a real focus at $(1,0)$ (when $b_2=-1$). As the previuous sections, we denote by $\gamma_i=\dfrac{\alpha_i}{\beta_i},\,i\in\{-,o,+\}.$ Thus $\gamma_-=0$ and $\gamma_+\gamma_o<0.$ By proof of Proposition 12 from \cite{ClauWeMau}, if $b_2=-1$ and $\gamma_++\gamma_o=0$ the foci of $X_+$ and $X_o$ give rise to a center at the point $(1,0)$. Moreover, we have a bounded period annulus which the border is the equilibriun point $(1,0)$ and the periodic orbit tangent to $L_-$ at point $(-1,0)$, see Figure \ref{figura1a}. \begin{figure}[h!] \begin{center} \begin{overpic}[width=5.5cm,height=3.5cm]{figura1a.eps} \put(12,-4){\tiny$L_-$} \put(72,-4){\tiny $L_+$} \end{overpic} \end{center} \caption{Period annulus which the border is the equilibrium point $(1,0)$ and the orbit by the point $(-1,0)$. } \label{figura1a} \end{figure} Note that for $b_2<-1$ with $|b_2+1|$ small enough, the focus of $X_+$ is real, $a_+^*<a_o^*$ and $a_+^*<b_o^*$, where $a_o^*=\pi_o(0)$, $b_o^*=\bar\pi_o^{-1}(0)$ and $a_+^*=\pi_+(0)$. Hence, the orbits of the periodic annuls are broken and give us the four possible phase portraits described in the Figures \ref{figura8ab} and \ref{figura9ab}, with $a_+^o=\pi_+^{-1}(b_o^*)$. \begin{figure}[!htb] \begin{minipage}[b]{0.45\linewidth} \centering \begin{overpic}[width=5.5cm,height=3.5cm]{figura8a} \put(17,30){\tiny $0$} \put(78,12){\tiny $a_o^*$} \put(71,15){\tiny $a_o^+$} \put(69,21){\tiny $a_+^*$} \put(78,56){\tiny $b_o^*$} \put(13,-3){\tiny $L_-$} \put(74,-3){\tiny $L_+$} \put(38,-15){ Case $(a)$} \end{overpic} \end{minipage} \hspace{0.5cm} \begin{minipage}[b]{0.45\linewidth} \centering \begin{overpic}[width=5.5cm,height=3.5cm]{figura8b} \put(16.5,31){\tiny $0$} \put(77.5,13){\tiny $a_o^*$} \put(70,7){\tiny $a_o^+$} \put(70,20){\tiny $a_+^*$} \put(77,56){\tiny $b_o^*$} \put(13,-3){\tiny $L_-$} \put(74,-3){\tiny $L_+$} \put(38,-15){ Case $(b)$} \end{overpic} \end{minipage} \vspace{1cm} \caption{Phase portraits when the periodic annulus is broken and $\gamma_+<0$.} \label{figura8ab} \end{figure} \begin{figure}[!htb] \begin{minipage}[b]{0.45\linewidth} \centering \begin{overpic}[width=5.5cm,height=3.5cm]{figura9a} \put(16.5,30){\tiny $0$} \put(78,10.5){\tiny $a_o^*$} \put(71.5,14.5){\tiny $a_o^+$} \put(71.5,43.5){\tiny $a_+^*$} \put(78,56){\tiny $b_o^*$} \put(13,-3){\tiny $L_-$} \put(74,-3){\tiny $L_+$} \put(38,-15){ Case $(a)$} \end{overpic} \end{minipage} \hspace{0.5cm} \begin{minipage}[b]{0.45\linewidth} \centering \begin{overpic}[width=5.5cm,height=3.5cm]{figura9b} \put(16.5,31){\tiny $0$} \put(78.5,12.5){\tiny $a_o^*$} \put(71,7){\tiny $a_o^+$} \put(72,44){\tiny $a_+^*$} \put(79,56){\tiny $b_o^*$} \put(13,-3){\tiny $L_-$} \put(75,-3){\tiny $L_+$} \put(38,-15){ Case $(b)$} \end{overpic} \end{minipage} \vspace{1cm} \caption{Phase portraits when the periodic annulus is broken and $\gamma_+>0$.} \label{figura9ab} \end{figure} We will study the periodic orbits of period annulus that persist by small perturbations of the parameters of system {\rm (\ref{base system nf})}. More precisely, we are interested in the limit cycles that bifurcate from the period annulus. For this, we will consider the Poincar\'e maps defined in two and three zones respectively. Firstly, we begin studying the sign of the displacement function \[ a_o^*-a_o^+. \] Note that, for $b=-1$ and $\gamma_++\gamma_o=0$, $a_o^*-a_o^+=0$. \begin{lemma} Let $\varphi_o(s, q)=(\varphi^1_o(s, q), \varphi^2_o(s, q))$ be the flow of $X_o$ by the point $q$. Then $\Pi_o(0) = \varphi^2_o(s_o, (-1, 0))$ and $\bar{\Pi}_o^{-1}(0) = \varphi^2_o(-\bar{s}_o, (-1, 0))$, where $s_o$ and $\bar{s}_o$ satisfy $\varphi^1_o( s_o, (-1, 0))=1$ and $\varphi^1_o(-\bar{s}_o, (-1, 0))=1$, respectively. Moreover $s_o$ and $\bar{s}_o$ are functions of $b_2$ and satisfy \begin{eqnarray} \Pi_o(0) & = & -2 e^{\gamma_o\tau_o}+ \left( -2\alpha_o+e^{\gamma_o\tau_o} \right) \left(b_2+1\right) +\dfrac{1}{4} e^{-\gamma_o\tau_o } \left( b_2 +1 \right) ^{2} \label{eq:9} \\ &+ & \mathcal{O}\left((b2+1)^3\right),\nonumber \\ \bar{\Pi}_o^{-1}(0) & = & 2 e^{-\gamma_o\bar{\tau}_o}- \left( 2\alpha_o+e^{-\gamma_o\bar{\tau}_o} \right) \left(b_2+1\right) -\dfrac{1}{4} e^{\gamma_o\bar{\tau}_o } \left( b_2 +1 \right) ^{2} \label{eq:8}\\ &+ & \mathcal{O}\left((b2+1)^3\right),\nonumber \end{eqnarray} where $\beta_os_o(-1)=\tau_o=\arctan \left(\dfrac{\beta_o}{\alpha_o}\right)$ and $\beta_o\bar{s}_o(-1)=\bar{\tau}_o=\arctan \left(-\dfrac{\beta_o}{\alpha_o}\right)$, i.e. for $b_2=-1$, $\varphi^1_o( \tau_o/\beta_o, (-1, 0))=1$ and $\varphi^1_o(-\bar{\tau}_o/\beta_o, (-1, 0))=1$, respectively. \end{lemma} \begin{proof} The first statement follows directly from the definitions of $\Pi_o(0)$ and $\bar{\Pi}_o^{-1}(0)$. We have that \begin{equation} \label{eq:10} \begin{aligned} \varphi_o^1 (s, (-1,0)) & = (b_2-1) e^{\alpha_o s}\cos (\beta_o s)-\gamma_o(b_2-1) e^{\alpha_o s} \sin(\beta_o s)-b_2, \\ \varphi_o^2 (s, (-1, 0)) & = (b2-1)\dfrac{e^{\alpha_o s}}{\beta_o}\sin (\beta_o s). \end{aligned} \end{equation} The equation $\varphi^1_o(s, (-1, 0))=1$ can be written as \begin{equation} \label{eq:11} (b_2-1)e^{\alpha_o s}\left[\cos (\beta_o s)-\gamma_o \sin (\beta_o s)\right]=b_2+1. \end{equation} Hence, for $b_2=-1$, this equation is equivalent to \[ \cos (\beta_o s)-\gamma_o \sin (\beta_o s)=0, \] whose solution $s_o(-1)$ is \[ s_o(-1)=\dfrac{1}{\beta_o}\arctan (\gamma_o^{-1})=\dfrac{\tau_o}{\beta_o}, \] i.e. $\tau_o=\arctan \left(\dfrac{\beta_o}{\alpha_o}\right)$. Note that, if $\alpha_o>0$, then $0<\tau_o<\dfrac{\pi}{2}$. Now, if $\alpha_o<0$, then $\dfrac{\pi}{2}<\tau_o<\pi$. Moreover it follows from equation \eqref{base system nf} that $\det A_o=1.$ This implies that $\alpha_o^2+\beta_o^2=1.$ So we have $\cos\tau_o=\alpha_o$ and $\sin\tau_o=\beta_o$. Computing the derivative with respect to $b_2$ in both sides of equation \eqref{eq:11}, treating $s$ as a function of $b_2$ and after substituting $b_2=-1$, we have \[ s_o'(-1)=\dfrac{e^{-\alpha_os_o(-1)}}{2\beta_o(\sin(\tau_o)+\gamma_o\cos(\tau_o))}=\frac{1}{2}e^{-\gamma_o\tau_o}. \] Analogously, we obtain \[ s_o''(-1)=\dfrac{1}{2}\left(1-\alpha_o e^{-\gamma_o\tau_o}\right)e^{-\gamma_o\tau_o}. \] Thus, the second order expansion of the function $s_o(b_2)$, solution of the equation $\varphi^1_o(s, (-1, 0))=1$, in a neighborhood of $b_2=-1$ is \[ s_o(b_2)=\dfrac{\tau_o}{\beta_o}+\frac{1}{2}e^{-\gamma_o\tau_o}(b_2+1)+\dfrac{1}{4}\left(1-\alpha_o e^{-\gamma_o\tau_o}\right)e^{-\gamma_o\tau_o}(b_2+1)^2+\mathcal{O} ((b_2+1)^3). \] Now, substituting the above expression in $\varphi_o^2 (s, (-1, 0)) $ and expanding it as a function of $b_2$ in a neighborhood of $b_2=-1$, we obtain the second order expansion of $\Pi_o(0)$, i.e. \[ \Pi_o(0) = -2 e^{\gamma_o\tau_o}+ \left( -2\alpha_o+e^{\gamma_o\tau_o} \right) \left(b_2+1\right) +\dfrac{1}{4} e^{-\gamma_o\tau_o } \left( b_2 +1 \right) ^{2} +\mathcal{O} \left((b2+1)^3\right). \] In a similar way, we can determine the second order expansion of the function $\bar{s}_o(b_2)$, solution of the equation $\varphi^1_o(-s, (-1, 0))=1$, in a neighborhood of $b_2=-1$. Just changing in \eqref{eq:10} $s$ by $-s$ and noting that in this case the equation $\varphi^1_o(-s, (-1, 0))=1$, for $b_2=-1$, is equivalent to \[ \cos (\beta_o s)+\gamma_o \sin (\beta_o s)=0, \] whose solution $\bar{s}_o(-1)$ is \[ \bar{s}_o(-1)=\dfrac{1}{\beta_o}\arctan (-\gamma_o^{-1})=\dfrac{\bar{\tau}_o}{\beta_o}, \] i.e. $\bar{\tau}_o=\arctan \left(-\dfrac{\beta_o}{\alpha_o}\right)$. Moreover, if $\alpha_o>0$, then $\dfrac{\pi}{2}<\tau_o<\pi$ and if $\alpha_o<0$, then $0<\tau_o<\dfrac{\pi}{2}$. In both cases, as $\alpha_o^2+\beta_o^2=1$, we have $\cos\tau_o=-\alpha_o$ and $\sin\tau_o=\beta_o$. Hence, we obtain \[ \bar{s}_o(b_2)=\dfrac{\bar{\tau}_o}{\beta_o}+\frac{1}{2}e^{\gamma_o\bar{\tau}_o}(b_2+1)+\dfrac{1}{4}\left(1+\alpha_o e^{\gamma_o\bar{\tau}_o}\right)e^{\gamma_o\bar{\tau}_o}(b_2+1)^2+\mathcal{O}((b_2+1)^3), \] and so \[ \bar{\Pi}_o^{-1}(0) = 2 e^{-\gamma_o\bar{\tau}_o}- \left( 2\alpha_o+e^{-\gamma_o\bar{\tau}_o} \right) \left(b_2+1\right) -\dfrac{1}{4} e^{\gamma_o\bar{\tau}_o } \left( b_2 +1 \right) ^{2} +\mathcal{O}\left((b2+1)^3\right). \] \end{proof} In what follows, we denote by $d_+^*$, $\beta_+^*$ and $\gamma^*_+$ the restrictions of $d_+$, $\beta_+$ and $\gamma_+$ to $b_2=-1$, respectively. \begin{lemma} Let $\varphi_+(s, q)=(\varphi^1_+(s, q), \varphi^2_+(s, q))$ be the flow of $X_+$ by the point $q=(1, \bar{\Pi}_o^{-1}(0))$, where $\bar{\Pi}_o^{-1}(0)$ is given by \eqref{eq:8}. Then $\Pi_+^{-1}\circ \bar{\Pi}_o^{-1}(0) =\varphi^2_+(-s_+, (1, $ $\bar{\Pi}_o^{-1}(0)))$, where $s_+$ satisfy $\varphi^1_+(- s_+, (1, \bar{\Pi}_o^{-1}(0)))=1$. Moreover $s_+$ is a function of $b_2$ and \begin{equation} \label{eq:13} \begin{aligned} \Pi_+^{-1}\circ \bar{\Pi}_o^{-1}(0) = & -2 e^{-(\gamma_o\bar{\tau}_o+\gamma^*_+\pi)} \\ &+\left(e^{-(\gamma_o\bar{\tau}_o+\gamma_+^*\pi)}\left[2e^{\gamma_o\bar{\tau}_o}\left(\alpha_o-\dfrac{\alpha_+}{d_+^*}\right)+\left(1-\dfrac{\pi\gamma_+^*}{(\beta_+^*)^2}\right)\right]\right. \\ & \left.-\dfrac{2\alpha_+}{d_+^*}\right)(b_2+1) \\ & +\frac{e^{-(\gamma_o\bar{\tau}_o+\gamma^*_+\pi)} }{4( d_+^*)^2 (\beta_+^*) ^6}\left[d_+^* (\beta_+^*) ^6 e^{2 \gamma_o\bar{\tau}_o} \left(d_+^*+e^{2\gamma_+^* \pi}-1\right) \right. \\ & \left.+4 \alpha_+ (\beta_+^*)^3 e^{\gamma_o\bar{\tau}_o} \left(\pi d_+^* (\alpha_o d_+^*-\alpha_+ )\right.+2 (\beta_+^*)^3 \left(e^{\gamma_+^*\pi \alpha}+1\right)\right)\\ &+\left. \pi (d_+^*)^2 \alpha_+ \left(2 (\beta_+^*)^3-\pi \alpha_+ +3 \beta_+^* \right)\right] (b_2+1)^2\\ & +\mathcal{O} ((b_2+1)^3). \end{aligned} \end{equation} \end{lemma} \begin{proof} The first statement follow directly from the definition of $\Pi_+^{-1}(0)$. We have that \[ \begin{aligned} \varphi_+^1 (-s_+, (1, \bar{\Pi}_o^{-1}(0))) = & \dfrac{(1+b_2)}{d_+} e^{-\alpha_+ s}\cos (\beta_+ s)\\ &+\dfrac{d_+ \bar{\Pi}_o^{-1}(0)+(1+b_2)\alpha_+}{\beta_+d_+} e^{-\alpha_+ s} \sin(\beta_+ s)\\ &+\dfrac{d_+-(1+b_2)}{d_+}, \\ \varphi_+^2 (-s_+, (1, \bar{\Pi}_o^{-1}(0))) = & \dfrac{d_+ \bar{\Pi}_o^{-1}(0)+c_{11}(1+b_2)}{d_+} e^{-\alpha_+ s}\cos (\beta_+ s) \\ &-\left[\dfrac{2(1+b_2)(d_+-c_{11}t_+)}{2\beta_+d_+}\right. \\ & \left.+\dfrac{ d_+(a_1-c_{11})\bar{\Pi}_o^{-1}(0)}{2\beta_+d_+}\right] e^{-\alpha_+ s} \sin(\beta_+ s)\\ &-\dfrac{c_{11}(1+b_2)}{d_+}. \end{aligned} \] Note that, for $b_2=-1$, $s_+(-1)=\dfrac{\pi}{\beta_+^*}$ is the solution of the equation $\varphi_+^1 (-s_+, (1, $ $\bar{\Pi}_o^{-1}(0)))=1$. Thus, if $s_+(b_2)=\dfrac{\pi}{\beta_+^*}+s_{+_1}(b_2+1)+s_{+_2}(b_2+1)^2+\cdots$ is the solution of this equation, substituting $\bar{\Pi}_o^{-1}(0)$ by its expansion \eqref{eq:8} and expanding $\varphi_+^1 (-s_+, (1, \bar{\Pi}_o^{-1}(0)))$ as a power series in $(b_2+1)$ in a neighborhood of $b_2=-1$, we obtain \[ \begin{aligned} s_+(b_2)= & \dfrac{\pi}{\beta_+^*}-\left( \dfrac{1+e^{-\gamma_+^*\pi}}{2d_+^*}\right)e^{\gamma_o\bar{\tau}_o+\gamma_+^*\pi}(b_2+1) \\ &+ e^{\gamma_o\bar{\tau}_o} \left(\left(\dfrac{e^{\gamma_+^*\pi}+1}{2(d_+^*)^2}\right) \left(\frac{1}{2} \alpha_+ e^{\gamma_o\bar{\tau}_o} \left(e^{\gamma_+^*\pi}+1\right)+1\right)\right. \\ & +\left.\dfrac{e^{\gamma_+^*\pi} }{4 d_+^*}\left(-2 \alpha_o e^{\gamma_o\bar{\tau}_o}+\frac{\pi \alpha_+ }{(\beta_+^*) ^3}-1\right)-\dfrac{2 \alpha_o e^{\gamma_o\bar{\tau}_o}+1}{4 d_+^*}\right)(b_2+1)^{2}+ \cdots. \end{aligned} \] Hence, substituting the above expression in $\varphi_+^2(-s_+(b_2),(1,\bar{\Pi}_o^{-1}(0)))$ and expanding it as a power series in $(b_2+1)$ at the neighborhood of $b_2=-1$, we obtain the second order expansion of $\Pi_+^{-1}\circ \bar{\Pi}_o^{-1}(0)$ in $(b_2+1)$ at the neighborhood of $b_2=-1$, i.e. we have the equality \eqref{eq:13} . \end{proof} \begin{lemma} Let $a_o^*=\pi_o(0)$ and $a_+^o=\pi_+^{-1}(b_o^*)$. Then the diference $a_o^*-a_o^+$ is equivalent to $(\Pi_o - \Pi_+^{-1}\circ \bar{\Pi}_o^{-1})(0)$ and \begin{equation} \label{eq:14} \begin{aligned} (\Pi_o - \Pi_+^{-1}\circ \bar{\Pi}_o^{-1})(0)= & 2e^{-(\gamma_o\bar{\tau}_o+\gamma_+^*\pi)}\left[1-e^{(\gamma_o+\gamma_+^*)\pi}\right] \\ & +e^{-(\gamma_o+\gamma_+^*)\pi}\left[ e^{\gamma_o\tau_o}\left(\dfrac{\pi t_+}{2(\beta_+^*)^3}+e^{(\gamma_o+\gamma_+^*)\pi}-1\right) \right. \\ & +\left.\left(\dfrac{t_+}{d_+^*}-t_o\right)\left(e^{(\gamma_o+\gamma_+^*)\pi}+e^{\gamma_o\pi}\right)\right](b_2+1)+\mathcal{O}((b_2+1)^2), \end{aligned} \end{equation} where $\Pi_o(0)$ and $ \Pi_+^{-1}\circ \bar{\Pi}_o^{-1}(0)$ are given by \eqref{eq:9} and \eqref{eq:13}, respectively. \end{lemma} \begin{proof} The first statement is a consequence of the definitions of $\Pi_o(0)$ and \linebreak$\left(\Pi_+^{-1}\circ \bar{\Pi}_o^{-1}\right)(0)$. Now, the equality \eqref{eq:14} follows directly from \eqref{eq:9} and \eqref{eq:13}, noting that \[ \tau_o+\bar{\tau}_o=\arctan \left(\dfrac{\beta_o}{\alpha_o}\right)+\arctan \left(-\dfrac{\beta_o}{\alpha_o}\right)=\pi. \] \end{proof} \begin{proposition}\label{prop.9.2} For $|b_2+1|$ small enough, we have: \begin{itemize} \item[(a)] if $\gamma_o+\gamma_+^*>0$, then $a_o^*-a_o^+>0$; \item[(b)] if $\gamma_o+\gamma_+^*<0$, then $a_o^*-a_o^+<0$; \item[({c})] if $\gamma_o+\gamma_+^*=0$, $\gamma_o>0$ and $b_2<-1$ (resp. $b_2>-1$), then $a_o^*-a_o^+<0$ (resp. $a_o^*-a_o^+>0$); \item[(d)] if $\gamma_o+\gamma_+^*=0$, $\gamma_o<0$ and $b_2<-1$ (resp. $b_2>-1$), then $a_o^*-a_o^+>0$ (resp. $a_o^*-a_o^+<0$). \end{itemize} \end{proposition} \begin{proof} Note that, when $(\Pi_o - \Pi_+^{-1}\circ \bar{\Pi}_o^{-1})(0)>0$ (resp. $(\Pi_o - \Pi_+^{-1}\circ \bar{\Pi}_o^{-1})(0)<0$), then $a_o^*-a_o^+<0$ (resp. $a_o^*-a_o^+>0$). Hence, by \eqref{eq:14}, if $\gamma_o+\gamma_+^*\neq 0$ the sign of $(\Pi_o - \Pi_+^{-1}\circ \bar{\Pi}_o^{-1})(0)$ is determined by sign of \[ 1-e^{(\gamma_o+\gamma_+^*)\pi}, \] and the statement (a) and (b) follow. Now, if $\gamma_o+\gamma_+^*= 0$, the sign of $(\Pi_o - \Pi_+^{-1}\circ \bar{\Pi}_o^{-1})(0)$ is determined by the sign of \[ \left[ e^{\gamma_o\tau_o}\left(\dfrac{\pi t_+}{2(\beta_+^*)^3}\right)+\left(\dfrac{t_+}{d_+^*}-t_o\right)\left(1+e^{\gamma_o\pi}\right)\right](b_2+1). \] Thus, as $d_+^*>0$, $\beta_o>0$, $\beta_+^*>0$, $t_o=2\gamma_o \beta_o$ and $t_+=2\gamma_+^* \beta_+^*$ with $t_o t_+<0$, the statement ({c}) and (d) follow. \end{proof} The proof of Theorem \ref{the_paper2:01} follows directly from the proof of the next proposition. \begin{proposition} \label{teo two limit cycle} Consider system {\rm (\ref{base system nf})}. Assume that $X_-$ has a center and $X_+$ and $X_o$ have foci with different stability. Then for $b_2<-1$, with $|b_2+1|$ small enough, if either $\gamma_o>0$ (equivalently $\gamma_+^*<0$) and $\gamma_o+\gamma_+^*>0$ or $\gamma_o<0$ (equivalently $\gamma_+^*>0$) and $\gamma_o+\gamma_+^*<0$, system {\rm (\ref{base system nf})} has at least two limit cycles. Moreover, one limit cycle visit only the two zones $R_o$ and $R_+$ and the other visit the three zones. \end{proposition} \begin{proof} Firstly we consider the case $\gamma_o>0$, $\gamma_+^*<0$ and $\gamma_o+\gamma_+^*>0$. So, by Proposition \ref{prop.9.2}, $a_o^*-a_o^+>0$, i.e. we have the case showed in Figure \ref{figura8ab} (a). Note that the orbit $\Gamma(t)$, of system {\rm (\ref{base system nf})} by the point associated to $a_o^+$, spirals toward the focus of $X_+$ when $t\rightarrow -\infty$. On the other hand, the focus of $X_+$ is an attractor. Then there is at least a limit cicle in two zones that pass by a point of $L_+^I$ between $a_+^*$ and $a_o^+$. Moreover this limit cycle is repeller. In the three zones of Figure \ref{figura8ab} (a), we consider the Poincar\'e map $\Pi:[0, \infty)\rightarrow [0, \infty)$ given by $\Pi=\bar\pi_o\circ\pi_+\circ\pi_o\circ\pi_-$. Now, by Proposition \ref{prop mau 1}~(b), $\pi_-$ is the identity map, so we can write $\Pi=\bar{\pi}_o\circ\pi_+\circ \pi_o$. Hence we have $$ \Pi'(c)= \bar{\pi}_o'(\pi_+(\pi_o({c})))\cdot\pi_+'(\pi_o({c}))\cdot\pi_o'({c}). $$ From Propositions \ref{prop.8}, \ref{prop.9}, \ref{prop.10} and \ref{prop.11}, we obtain \begin{equation} \label{eq:15} \displaystyle \lim_{c\rightarrow +\infty}\Pi'(c)=\dfrac{b_2+1}{b_2-1}e^{\gamma_+\pi}\dfrac{b_2-1}{b_2+1}=e^{\gamma_+\pi}. \end{equation} As $\gamma_+^*<0$, so $\gamma_+<0$ for $|b_2+1|$ small enough and $\Pi$ is decreasing in a neighborhood of infinity, i.e. the infinity is a repeller to system {\rm (\ref{base system nf})}. On the other hand, the orbit $\Gamma(t)$ spirals moving away from the focus. Therefore we have at least a limit cycle in the three zones. For the second case $\gamma_o<0$, $\gamma_+^*>0$ and $\gamma_o+\gamma_+^*<0$, by Proposition \ref{prop.9.2}, we have $a_o^*-a_o^+<0$, i.e. we have the case showed in Figure \ref{figura9ab} (b). Hence, as the focus of $X_+$ is repeller, similar the previous case it follows that there is at least a limit cicle in two zones that pass by a point of $L_+^O$ between $a_+^*$ and $b_o^*$, which is attractor. Now, for Figure \ref{figura9ab} (b), we consider the same Poincar\'e map $\Pi$ as in the case of Figure \ref{figura8ab} (a). However in this case the domain of $\Pi$ is $[c^{*}, \infty)$, where $\Pi(c^{*})=0$. Thus, as $\gamma_+^*>0$, it follows by \eqref{eq:15} that the infinity is an attractor to system {\rm (\ref{base system nf})} and as in the previous case we have at least a limit cycle in the three zones. Then in both cases we conclude the existence of at least two limit cycles, one visiting only two zones and the other visiting the three zones. \end{proof} \begin{remark} From above result, we conclude the existence of at least two limit cycles to system {\rm (\ref{base system nf})} assuming that $X_-$ has a center, $X_+$ and $X_o$ have foci with different stability and $b_2<-1$, with $|b_2+1|$ small enough. The same conclusion we have for the equivalent case, when $X_+$ has a center, $X_-$ and $X_o$ have foci with different stability and $b_2>1$, with $|b_2-1|$ small enough. However we have not been able to determine the exact number of limit cycles for this class of vector fields. In fact, for the cases determined in Figures \ref{figura8ab} (b) and \ref{figura9ab} (a) we also cannot say nothing about the existence or not of limit cycles for now. \end{remark} \section{Examples} \label{sec examples} In this section, we will illustrate, through two examples, the appearance of two limit cycles when the periodic annulus of Figures \ref{figura1a} is broken, obtaining the cases described in Figures \ref{figura8ab}(a) and \ref{figura9ab}(b). We assume that $X_+$ has a focus and, by the hypothesis (H1) and (H2), $X_-$ has a center and $X_o$ has a focus with different stability with respect to the focus of $X_o$. In this case, by the previous section, the phase portrait of vector field \eqref{base system nf} is equivalent to Figures \ref{figura1a} if and only if $b_2=-1$ and $\gamma_++\gamma_o=0$. The idea is to obtain two one-parameter families of vector fields, whose parameter is $b_2$, such that for $b_2=-1$ we have the periodic annulus of Figures \ref{figura1a}. On another words, we want to put the condition $\gamma_++\gamma_o=0$ as a function of $b_2$. By Lemma \ref{nf}, we have \[ \gamma_o=\dfrac{a_1}{\sqrt{4-a_1^2}}\;\; \mbox{and}\;\; \gamma_+=\dfrac{c_{11}+a_1}{\sqrt{4(1+b_2-f_2+a_1c_{11})-(c_{11}+a_1)^2}}. \] Now, for $b_2=-1$, $\gamma_o+\gamma_+=0$ if and only if $f_2=a_1c_{11}-\left(\dfrac{c_{11}+a_1}{a_1}\right)^2$. Hence, doing $f_2=a_1c_{11}-\left(\dfrac{c_{11}+a_1}{a_1}\right)^2-\varepsilon$, a simple calculation gives us \begin{equation} \label{eq:16} \gamma_+=\dfrac{c_{11}+a_1}{|c_{11}+a_1|} \dfrac{|a_1|}{\sqrt{4-a_1^2+\dfrac{4a_1^2}{(c_{11}+a_1)^2}(b_2+1+\varepsilon)}}. \end{equation} Then, as $\gamma_o\gamma_+<0$ or equivalently $a_1(c_{11}+a_1)<0$, we get \begin{equation}\label{eq:17} \gamma_+=- \dfrac{a_1}{\sqrt{4-a_1^2+\dfrac{4a_1^2}{(c_{11}+a_1)^2}(b_2+1+\varepsilon)}}. \end{equation} {For} $b_2+1+\varepsilon>0$, from \eqref{eq:17}, $\gamma_o+\gamma_+>0$ if $\gamma_o>0$ (i.e. $a_1>0$) and $\gamma_o+\gamma_+<0$ if $\gamma_o<0$ (i.e. $a_1<0$). Hence, from Proposition \ref{prop.9.2} (for $b_2<-1$ with $|b_2+1|$ small enough), we have in the first case the configuration of Figure \ref{figura8ab}(a) and in the second one the configuration of Figure \ref{figura9ab} (b). Moreover, by Proposition \ref{teo two limit cycle}, vector field \eqref{base system nf} has at least two limit cycles. We obtain the follows examples. \begin{example} \label{Ex:01} Consider system \eqref{base system nf} with $\gamma_o>0$, $\gamma_+<0$ (i.e. $a_1>0$ and $c_{11}+a_1<0$), $4-a_1^2>0$, $a_{11}=-a_1$, $1-a_1^2-b_2+d_2>0$ and $f_2=a_1c_{11}-\left(\dfrac{c_{11}+a_1}{a_1}\right)^2-\varepsilon$, then (for $(b_2+1)^2+\varepsilon^2$ small enough) we have that \begin{itemize} \item[(a)] if $b_2>-1$ and $b_2+1+\varepsilon<0$, by Proposition 4.2 of \cite {Mauricio} and Proposition \ref{prop.9.2}, the phase portrait of vector field \eqref{base system nf} is equivalent to Figure \ref{bifurca1}(a); \item[(b)] if $b_2=-1$ and $\varepsilon=0$, by Proposition 12 from \cite{ClauWeMau}, the phase portrait of vector field \eqref{base system nf} is equivalent to Figure \ref{bifurca1}(b); \item[(c)] if $b_2<-1$ and $b_2+1+\varepsilon>0$, by Proposition \ref{teo two limit cycle}, the phase portrait of vector field \eqref{base system nf} in a neighborhood of these limit cycles is equivalent to Figure \ref{bifurca1}({c}). \end{itemize} \begin{figure}[h!] \begin{center} \begin{overpic}[width=14cm,height=3cm]{bifurca1_2.eps} \put(1,23){\tiny $L_-$} \put(17,23){\tiny $L_+$} \put(38,23){\tiny $L_-$} \put(54,23){\tiny $L_+$} \put(72.5,23){\tiny $L_-$} \put(89,23){\tiny $L_+$} \put(-3,-6){\footnotesize (a) $b_2>-1$ and $b_2+1+\varepsilon<0$} \put(39,-6){\footnotesize(b) $b_2=-1$ and $\varepsilon=0$} \put(68,-6){\footnotesize({c}) $b_2<-1$ and $b_2+1+\varepsilon>0$} \end{overpic} \end{center} \vspace{1.5cm} \caption{Phase portrait of system \eqref{base system nf} with $a_1>0$ and $c_{11}+a_1<0$, $4-a_1^2>0$, $a_{11}=-a_1$, $1-a_1^2-b_2+d_2>0$ and $f_2=a_1c_{11}-\left(\dfrac{c_{11}+a_1}{a_1}\right)^2-\varepsilon$, for $(b_2+1)^2+\varepsilon^2$ small enough.} \label{bifurca1} \end{figure} We can use the software P5 to do numerical simulations of the phase portraits of system \eqref{base system nf} for specific parameter values. For instance, \begin{itemize} \item the Figure \ref{Ex5.1Fig7}(a) correspond to Figure \ref{bifurca1}(a) with $a_1=1$, $b_2=-0.9$, $c_{11}=-1.4$, $a_{11}=-1$, $d_2=4$ and $\varepsilon=-0.21$; \item the Figure \ref{Ex5.1Fig7}(b) correspond to Figure \ref{bifurca1}(b) with $a_1=1$, $b_2=-1$, $c_{11}=-1.4$, $a_{11}=-1$, $d_2=4$ and $\varepsilon=0$; \item the Figure \ref{Ex5.1Fig7}({c}) correspond to Figure \ref{bifurca1}({c}) with $a_1=1$, $b_2=-1.09$, $c_{11}=-1.4$, $a_{11}=-1$, $d_2=4$ and $\varepsilon=0.21$. \end{itemize} \begin{figure}[h!] \begin{center} \begin{overpic}[width=15cm,height=5cm]{Ex5_1Fig7.eps} \put(20.5,-6){(a)} \put(50.7,-6){(b)} \put(78,-6){({c})} \end{overpic} \end{center} \vspace{1.5cm} \caption{Numerical simulations of phase portraits of system \eqref{base system nf} using software P5 on the conditions of Example \ref{Ex:01}.} \label{Ex5.1Fig7} \end{figure} \end{example} \begin{example} \label{Ex:02} Consider system \eqref{base system nf} with $\gamma_o<0$, $\gamma_+>0$ (i.e. $a_1<0$ and $c_{11}+a_1>0$), $4-a_1^2>0$, $a_{11}=-a_1$, $1-a_1^2-b_2+d_2>0$ and $f_2=a_1c_{11}-\left(\dfrac{c_{11}+a_1}{a_1}\right)^2-\varepsilon$, then (for $(b_2+1)^2+\varepsilon^2$) we have that \begin{itemize} \item[(a)] if $b_2>-1$ and $b_2+1+\varepsilon<0$, by Proposition 4.2 of \cite {Mauricio} and Proposition \ref{prop.9.2}, the phase portrait of vector field \eqref{base system nf} is equivalent to Figure \ref{bifurca2}(a); \item[(b)] if $b_2=-1$ and $\varepsilon=0$, by Proposition 12 from \cite{ClauWeMau}, the phase portrait of vector field \eqref{base system nf} is equivalent to Figure \ref{bifurca2}(b); \item[(c)] if $b_2<-1$ and $b_2+1+\varepsilon>0$, by Proposition \ref{teo two limit cycle}, the phase portrait of vector field \eqref{base system nf} in a neighborhood of these limit cycles is equivalent to Figure \ref{bifurca2}({c}). \end{itemize} \begin{figure}[h!] \begin{center} \begin{overpic}[width=14cm,height=3cm]{bifurca2.eps} \put(-1,23){\tiny $L_-$} \put(15,23){\tiny $L_+$} \put(36.5,23){\tiny $L_-$} \put(52.5,23){\tiny $L_+$} \put(71.5,23){\tiny $L_-$} \put(88,23){\tiny $L_+$} \put(-5,-6){\footnotesize (a) $b_2>-1$ and $b_2+1+\varepsilon<0$} \put(36,-6){\footnotesize(b) $b_2=-1$ and $\varepsilon=0$} \put(68,-6){\footnotesize({c}) $b_2<-1$ and $b_2+1+\varepsilon>0$} \end{overpic} \end{center} \vspace{1.5cm} \caption{Phase portrait of system \eqref{base system nf} with $a_1<0$ and $c_{11}+a_1>0$, $4-a_1^2>0$, $a_{11}=-a_1$, $1-a_1^2-b_2+d_2>0$ and $f_2=a_1c_{11}-\left(\dfrac{c_{11}+a_1}{a_1}\right)^2+2(b_2+1)$, for $(b_2+1)^2+\varepsilon^2$ small enough.} \label{bifurca2} \end{figure} As in Example \ref{Ex:01}, we can use the software P5 (see \cite{P5}) to do numerical simulations of the phase portraits of system \eqref{base system nf} for specific parameter values. For instance, \begin{itemize} \item the Figure \ref{Ex5.2Fig8}(a) correspond to Figure \ref{bifurca2}(a) with $a_1=-0.8$, $b_2=-0.9$, $c_{11}=1.4$, $a_{11}=0.8$, $d_2=4$ and $\varepsilon=-0.35$; \item the Figure \ref{Ex5.2Fig8}(b) correspond to Figure \ref{bifurca2}(b) with $a_1=-0.8$, $b_2=-1$, $c_{11}=1.4$, $a_{11}=0.8$, $d_2=4$ and $\varepsilon=0$; \item the Figure \ref{Ex5.2Fig8}({c}) correspond to Figure \ref{bifurca2}({c}) with $a_1=-0.8$, $b_2=-1.09$, $c_{11}=1.4$, $a_{11}=0.8$, $d_2=4$ and $\varepsilon=0.43$. \end{itemize} \begin{figure}[h!] \begin{center} \begin{overpic}[width=15cm,height=5cm]{Ex5_2Fig8.eps} \put(20.5,-6){(a)} \put(50.7,-6){(b)} \put(78,-6){({c})} \end{overpic} \end{center} \vspace{1.5cm} \caption{Numerical simulations of phase portraits of system \eqref{base system nf} using software P5 on the conditions of Example \ref{Ex:02}.} \label{Ex5.2Fig8} \end{figure} \end{example} \section*{Acknowledgments} The first author is partially supported by Fapesp grant number 2013/15941-5 and grant FP7-PEOPLE-2012-IRSES 318999. The second and third authors are partially supported by a FAPESP grant 2013/34541--0. The second author is supported by a CAPES PROCAD grant 88881.068462/2014-01.

Wednesday, November 16, 2011 The National Kitchen & Bath Association (NKBA), working with MultiView, Inc., the leading provider of digital media for associations, has announced the launch of a new Kitchen & Bath Industry Directory, a valuable resource for kitchen and bath professionals. Full story Wednesday, November 09, 2011 The National Kitchen & Bath Association (NKBA), released the results of the Kitchen & Bath Market Report for the third quarter, reflecting a rise in showroom traffic and an increase in kitchen remodels from one year ago. Although the number of bathroom remodels declined from the second quarter, results show a significant increase over the bathroom sales of a year ago. Monday, October 03, 2011. Tuesday, September 20, 2011 The. Wednesday, August 03, 2011 The National Kitchen & Bath Association (NKBA) has announced the launch of a new non-design certification: the Certified Kitchen & Bath Professional (CKBP). Monday, August 01, 2011 According to the National Kitchen & Bath Association (NKBA), kitchen and bath dealers remain undeterred following a challenging quarter across various industry market conditions including showroom visits, sales volumes, remodeling budgets and revenue. The NKBA Kitchen & Bath Market Index (KBMI), which forecasts the confidence of kitchen and bath dealers on a scale of -60 to +60, is +21 for Q3 2011. While positive, the +21 KMBI is down slightly from +33 and +37 during the second and first quarter of 2011, respectively. Tuesday, June 14, 2011 For the sixth consecutive year, the National Kitchen & Bath Association has recognized NKBA-accredited schools that demonstrate the highest standards of kitchen and bath instruction. Wednesday, June 01, 2011 The National Kitchen & Bath Association (NKBA) recently recognized kitchen and bath design students with the 2011 NKBA Outstanding Student Awards, while also presenting eleven NKBA-accredited colleges with the 2010 Excellence in Education Awards. The NKBA presented these prestigious honors during the Educators’ Forum at the NKBA’s 2011 Kitchen & Bath Industry Show (KBIS) in Las Vegas. Tuesday, May 24, 2011 The National Kitchen & Bath Association (NKBA) is proud to announce the winner of the 2010 NKBA President’s Award. The winner, Cosentino North America, was recognized by NKBA Immediate Past President Mark L. Karas, CMKBD at the State of the Industry Address held in Las Vegas during the NKBA’s 2011 Kitchen & Bath Industry Show (KBIS). Monday, May 16, 2011 The National Kitchen & Bath Association (NKBA) reveals the latest standout design trends culled from the 2011 NKBA Design Competition.

TITLE: Discussion on even and odd perfect numbers. QUESTION [8 upvotes]: First of all thank you so much for answering my previous post. These are few interesting problems drawn from Prof. Gandhi lecture notes. kindly discuss: 1) If $n$ is even perfect number then $(8n +1)$ is always a perfect square. 2) Every odd perfect number has at least three different prime factors. This is by observation we can understand that. But, how to prove mathematically? 3) Every even perfect number (other than 6) can be expressible as sum of consecutive odd cubes. Thanks in advance. REPLY [6 votes]: As Andre mentioned, the even perfect numbers have the form $2^{p-1} (2^{p} -1 )$. Claim: The sum of the first $2^{(p-1)/2}$ odd cubes is equal to this perfect number. Proof: We know that the sum of the first $n$ odd cubes is: $\sum_{i=1}^{n} (2i-1)^3 = \sum_{i=1}^{2n} i^3 - \sum_{i=1}^{n} (2i)^3 \\= \left[ \frac {(2n)(2n+1)}{2} \right] ^2 - 8\left[ \frac {n(n+1)}{2} \right]^2 = \frac {n^2 [16n^2 + 16n + 4 - 8(n^2 + 2n+1)]}{4}\\ = n^2(2n^2 - 1) $ Now substitute $n = 2^{(p-1)/2}$.

TITLE: Billiards in a circular table QUESTION [0 upvotes]: This is a variation of Alhazen's Billiard Problem. Suppose we have a semicircular billiards table of radius r centered at the origin O, and a billiard ball placed somewhere on the 'x-axis' of the table. Let us call this point P, with coordinate $(0,p)$, with the stipulation that $0$ < P < r. Let the distance OP be $p$. On what point on the table should we aim at such that the billiard ball will bounce off the edge of the table once and into the other 'end' of the x-axis at $(0,-r)$? Is it also possible to generalize this for any point $(x,y)$ in the circle? Thanks. REPLY [1 votes]: Let $P=(p,0)$ be the starting point, $E=(-r,0)$ the end point, $Q$ the bouncing point and $\theta=\angle POQ$. Then we have $PQ^2=p^2+r^2-2pr\cos\theta$, $EQ^2=2r^2(1+\cos\theta)$ by the cosine rule, and $PO:OE=PQ:EQ$, because $QO$ is the bisector of $\angle PQE$. Combining these we get $$ p^2:r^2=(p^2+r^2-2pr\cos\theta):(2r^2(1+\cos\theta)) $$ whence $$ \cos\theta={r-p\over2p}. $$ Notice that a solution exists only if $p\ge r/3$.

Opened in 2018, the Home Inn (Dalian Pulandian Commercial Street, SBeilan Road Store) is a great accommodation choice in Dalian. The hotel is conveniently located just 2km from Pulandian Railway Station and 70km from Zhoushuizi International Airport. The nearby area boasts an abundance of attractions including Pulandian City Construction Archives Hall, Nan Mountain Park and Erlong Mountain National Forest Park. In their spare time, guests can explore the hotel's surroundings. This Dalian hotel provides parking on site. Our guests consider this hotel to have excellent service. For guests traveling on business, this hotel is consistently one of the most popular choices.

Fire at Jim Beam July 8, 2006 - Jed could see smoke from the Fields Ertel area, so we decided to drive past what we thought would be a fire near the Blue Ash Airport. We kept driving until we finally found that the smoke was coming from a fire way down at the Jim Beam plant on 75 South. According to the news, it was a fatal fire. The building was not actually part of the Jim Beam plant.

TITLE: Finding the inverse of an integral QUESTION [4 upvotes]: I'm looking for a computational approach here, since I don't think there is a closed-form solution. I have the following: $$ s(x) = \rho + \int_{\rho}^{x} \sqrt{ 1 + (\alpha \cos t - k)^2 } \, dt $$ $ k = \frac{\rho \sin \alpha - \alpha \sin(\rho \cos \alpha)}{\frac{\pi}{2} - \rho \cos \alpha} $ for $x<\frac{\pi}{2}$, otherwise $k=0$. The following are all constants: $0 < \rho < \frac{\pi}{2}$, $0 < \alpha < \frac{\pi}{2}$ (and thus $k$). I'd like to find $x$ such that $s(x) = r$ for some constant $0 \le r \le \pi$. The interesting thing is that $s(x)$ is an elliptic integral for $x \ge \frac{\pi}{2}$, because $k=0$. However, my lower limit $\rho$ means that I need to handle $k \ne 0$. My hunch is that I can use some kind of series expansion here. Any ideas? REPLY [1 votes]: The numerical approach I would use here is quite obvious: reducing the problem to an ODE. From the integral itself it's clear that: $$\frac{ds}{dx} = \sqrt{ 1 + (\alpha \cos x - k)^2 }$$ $$s(\rho)=\rho$$ Now we simply invert the function: $$\frac{dx}{ds} = \frac{1}{\sqrt{ 1 + (\alpha \cos x - k)^2 }}$$ $$x(\rho)=\rho$$ Now we use any suitable numerical scheme, for example, the simplest explicit Euler method. $$x_0=\rho$$ $$x_{n+1}=x_n+\frac{\Delta s}{\sqrt{ 1 + (\alpha \cos x_n - k)^2 }}$$ To reach $r$ we take: $$\Delta s=\frac{r-\rho}{N}$$ Then: $$x_{N} \approx x(r)$$

Two blog posts in two nights!!!! My budget is done and graduation practice starts tomorrow.... Stick a fork in me because the end is nigh. Like most middle school teachers, I love to end the year with something very interactive. For those of you that don't teach middle school, just close your eyes and imagine why you don't teach middle school. Now multiply it by 12. That is what we eighth grade teachers call June. Usually I end with the kiddos making plays or videos about the various groups that went west, but I decided to abbreviate my Manifest Destiny and add in a unit on Social Reformers. I didn't have time for plays so I went in my way back machine and pulled out an old favorite machine. They're called tableaux (defined as an interlude during a scene when all the performers on stage freeze in position and then resume action as before.) or living monuments. Students have to convey information through a single staged scene with no words. They are encouraged to use props, signs and we also use Google Presentation to create backgrounds for our monuments. The above are backgrounds one of my groups made for Frederick Douglass. They made the train he was on when he escaped slavery, a crowd he spoke to, and his newspaper. The project was perfect!!!! The kids told me they had a blast and it was a great way to keep them on track for the end of the year. Teachers Pay Teachers and you can buy the project there. It comes with rubrics and research graphics organizers. What do you do at the end of the year to keep your kids engaged? I LOVE this idea and am so excited to have found your great blog - I am your newest follower! fifthisfabulous.blogspot.com Thanks! I've done this with various topics and the kids always love it. It is sooooo much easier than staging plays:) Hi I'm Heather! Please email me when you get a chance, I have a question about your blog! LifesABanquet1(at)gmail.com Is this spam?

Ash Wednesday, March 1st – Start of Lent Millions of Christians are preparing to celebrate Lent –. Lent takes place every year in the 40 days leading up to Easter, and is treated as a period of reflection and a time for fasting from food and festivities. It symbolizes the days which lead up to Jesus’ crucifixion and subsequent resurrection, when Christ spent 40 days and nights alone in the Judaean Desert being tempted by Satan. When does Lent start?. What days are excluded from Lent? Lent.

‘Let her stay’Leave a Comment Campaigners gathered outside a Leeds court earlier this week to protest against the deportation of a Nigerian mother whose children face the threat of genital mutilation should they return to the African country. Afusat Saliu, 31, fled Nigeria in 2011 to protect her two children, four-year-old Bassy and Rashidat, two, from the brutal practice of female genital mutilation (FGM) and to escape a forced marriage to a man 40 years her senior. HELP: Roseline Akhalu has lived in the UK for 10 years and believes her friend, Afusat Saliu, is being mistreated by the UK Home Office After being turned down for asylum status in 2012, she was once again turned away last month despite new evidence showing her children’s very real threat of FGM. A group of protestors gathered outside the UKBA Waterside Court, Kirkstall Road, on 29th April, to show their support for the local mother. Kate Jennings helped organise the event, by group ‘Leeds No Borders’, and explained how her friend, Afusat, is in ‘real danger’ should she be forced to go back to Nigeria. “If she is deported back we fear that her children will have to face FGM,” she explained. “What’s totally crazy about this situation is that the government is about to pass a law that says FGM is illegal in this country and that we should do everything in our power to protect people who face it. PROTEST: Leeds No Borders hold monthly protests on Kirkstall Road in support of asylum seekers facing deportation “Yet, at the same time, they are here, trying to deport two children, whose risk is very evident that they would get cut should they go back. “Afusat is such an active member of the Leeds community. She volunteers in loads of places and we are going to do everything we can to help her stay in the city.” More than 110,000 people have now signed an online petition to keep Afusat’s family in the UK, whilst Leeds MP, George Mudie, has recently urged the Home Office minister, James Brokenshire, to review her case. Roseline Akhalu, also from Nigeria but living in Leeds, came to the UK 10 years ago and said she did not understand why Afusat was not given permission to live in the country. “We are so surprised to see that Afi and her children are still facing deportation,” she said. “We come here to survive, to escape persecution, so I do not know why they want to deport her. DANGER: Afusat Saliu (right) and her two children, Bassy and Rashidat, are fighting against their deportation to Nigeria due to the threat of FGM upon return “She is an FGM survivor herself and if she was to take her children back to Nigeria she would not be able to do anything to protect them. “This country has a right to protect them and that is why we are here.” The Leeds No Borders group were also protesting against the running of the country’s Yarl’s Woods Immigration Removal Centre which made headlines last month following the death of a detainee. Ms Akhalu, who has also been detained twice at the premises, added: “There is no freedom there. You are treated like a cow. Immigrants and asylum seekers are treated like illegals in that place. “The UN were recently banned from going in there so we are asking ‘what are they hiding?’ and are calling for an independent investigation to get it shut down.” If you would like to show your support to keep Afusat Saliu and her two children in the country please visit change.org.

Jonathan Bachman / AP Former New Orleans Mayor Ray Nagin arrives at the Hale Boggs Federal Building in New Orleans on Monday, Jan. 27, 2014. him. Nagin, 57, last-minute delay to give Nagin's attorney, Robert Jenkins, more time to prepare. Jenkins did not respond to a request for comment. Ray Nagin, former mayor of New Orleans, talks about his book, "Katrina's Secrets: Storms After the Storm," which tells how he struggled to handle the disaster that unfolded in his city. statements during the crisis that inflamed passions. Later, Nagin, who is black, was criticized for racial divisiveness for urging residents to rebuild a "chocolate New Orleans," referring to its majority African-American population.

Dawn Butler Labour MP for Brent Central Due to the evolving situation regarding the COVID-19 outbreak, otherwise known as the Coronavirus, we have been forced to take the regrettable decision of cancelling face to face advice surgeries. However, please be rest assured that all constituency and parliament operations will continue to operate but under different procedures. We will continue to monitor developments closely and will update constituents as soon as the situation changes. Opening new casework Myself and my office open on average between 250 and 350 every month, on a wide range of issues. If you believe that I may be able to assist you with a particular concern I would recommend the following steps are taken to ensure we are best placed to assist: 1. Verify that I am your Member of Parliament by visiting 2. Complete one of the following forms below and read my Privacy Policy 3. Send the completed form to dawnbutlermpoffice@parliament.ukwith any further supporting information that you believe will assist with your case. Email is the most efficient way for me to look into your case. If you have an existing case the Casework helpline will be available between the hours of 10:30 am -3 pm Monday to Friday. Ongoing casework We will continue to work on all ongoing casework and have measures in place, that are GDPR compliant, to ensure this can be completed, even if faced with further disruption to our Brent offices and or Parliament. COVID-19 advice For the latest up to date advice please visit the NHS website here: If you are suffering from a new continuous cough or a high temperature of 37.8 degrees or higher. The latest NHS guidance is to stay at home and self-isolate and do not leave your house for 14 days from when your symptoms first started. You should not go to a GP surgery, pharmacy or hospital. You do not need to contact 111 to tell them you’re staying at home. However, if you feel that your condition is worsening, you can’t cope with your symptoms at home, you have underlying health problems, or your condition doesn’t improve after 7 days then please call 111 to seek further advice.

SAM stands for Surface to Air Missile. SAMs are fast-travelling and highly accurate defenses, great for shooting down faster moving aircraft. While not seen as often as Flak, SAMs are essential for killing T3 air such as strategic bombers in order to stop their attacks. - Defence buildings - SAM defences defence launcher, belonging to the Seraphim faction: Unlike other factions' T3 static anti air, Seraphim use a high damage, high velocity projectile rather than tracking missiles. UnitsEdit SAMs are rarely built at sea, despite being aquatic. This is because SAMs are equipped on Aeon and UEF Cruisers, and a variant is aboard the Cybran Siren Class also. Certain aircraft carriers, such as the Aeon Keefer and UEF Experimental Aircraft Carrier Atlantis carry them as their AA defence . There are SAM-like launchers installed on individual units, such as the Cybran Experimental Spiderbot Monkeylord.

Narrative in the Life of Frederick Douglass "If there is no struggle, there is no progress." 1818-1895 Frederick Douglas-Isaac Myers Maritime Park Baltimore, MD First African American Shipyard "Going to live at Baltimore laid the foundation, and opened the gateway, to all my subsequent prosperity."--Ch. 5, A Narrative Douglass as a slave Douglass was born in Talbot County, Maryland in 1818. He was separated from his mother shortly after he was born and while the identity of his father is uncertain, it is thought that his father was a white slavemaster. Douglass was raised by his grandmother until age 7 when he was sent to Wye House Plantation. Wye House Plantation, home of Colonol Lloyd At the Wye House Plantation, Douglass became a companion for Lloyd's son, Daniel From the Wye House Plantation, Douglass was sent to Baltimore's Fells Point, to the home of Captain Auld, where he became a companion to Captain Auld's son. Douglass soon established a relationship with Captai Auld's wife who began to teach Douglass to read; however, because it was against the law to teach slaves to read, Captain Auld forbade his wife from continuing to teach Douglass. BUT this did not stop Douglass from reading; anything he could get his hands on, he would read. This is how he became an educated man and how he learned that slavery was not a common practice in other parts of the country. Douglass was soon sent to another master and it was there that Douglass began to teach other slaves how to read, holding classes every Sunday. When other slave owners in the area heard of this, they were outraged. Douglass was forced to stop holding the classes, and his old master, Captain Auld, took him back. Captain Auld, angry at Douglass's behavior, sent Douglass to work for a man named Mr. Covey. Mr. Covey was known as a harsh slave owner and had a repuation for being able to "break" slaves. It was during his time at Mr. Covey's that Douglass was forced to perform harsh physical labor and was whipped regularly. He joined an African Methodist Episcopal Church where he met free blacks and began to think about his own freedom. At this time, Douglass was also learning a trade in the shipyards of Baltimore In 1838, Douglass escaped slavery heading north to New York. He escaped with the financial support of a free black woman named Anna Murray, whom he later married when he settled in New Bedford, Massachusetts. "It is not that I loved Maryland less, I hated slavery Douglass No description byTweet Amy E. Countson 8 May 2014 Please log in to add your comment.

import pseudo_normed_group.LC import analysis.normed.group.hom_completion /-! # V-hat(M_c^n) One of the key players in the proof of the main theorem of this repo is the normed group V-hat(M-bar_r'(S)_{≤c}^n). This file constructs ## Key defintions - `CLCP V n`: the functor that sends a profinite set `S` to `V-hat(S^n)` - `CLFCP v r' c n`: the functor sending a profinitely-filtered `T⁻¹`-module `M` to `V-hat((M_c)^n)` -/ open_locale classical nnreal noncomputable theory local attribute [instance] type_pow open SemiNormedGroup opposite Profinite pseudo_normed_group category_theory breen_deligne open profinitely_filtered_pseudo_normed_group universe variable u variables (r : ℝ≥0) (V : SemiNormedGroup) (r' : ℝ≥0) variables (c c₁ c₂ c₃ c₄ : ℝ≥0) (l m n : ℕ) /-- `CLC V n` is the functor that sends a profinite set `S` to `V-hat(S^n)` -/ def CLC (V : SemiNormedGroup) : Profiniteᵒᵖ ⥤ SemiNormedGroup := LC V ⋙ Completion namespace CLC lemma map_norm_noninc {M₁ M₂} (f : M₁ ⟶ M₂) : ((CLC V).map f).norm_noninc := Completion.map_norm_noninc $ LC.map_norm_noninc _ _ def T [normed_with_aut r V] [fact (0 < r)] : CLC V ≅ CLC V := ((whiskering_right _ _ _).obj _).map_iso (LC.T r V) lemma norm_T_le [normed_with_aut r V] [fact (0 < r)] (A) : ∥(T r V).hom.app A∥ ≤ r := le_trans (normed_add_group_hom.norm_completion _).le $ LC.norm_T_le _ _ _ def T_inv [normed_with_aut r V] [fact (0 < r)] : CLC V ⟶ CLC V := whisker_right (LC.T_inv r V) Completion lemma T_inv_eq [normed_with_aut r V] [fact (0 < r)] : (T r V).inv = T_inv r V := rfl lemma norm_T_inv_le [normed_with_aut r V] [fact (0 < r)] (A) : ∥(T_inv r V).app A∥ ≤ r⁻¹ := le_trans (normed_add_group_hom.norm_completion _).le $ LC.norm_T_inv_le _ _ _ end CLC /-- `CLFCP v r' c n` is the functor sending a profinitely-filtered `T⁻¹`-module `M` to `V-hat((M_c)^n)` -/ def CLCFP (V : SemiNormedGroup) (r' : ℝ≥0) (c : ℝ≥0) (n : ℕ) : (ProFiltPseuNormGrpWithTinv r')ᵒᵖ ⥤ SemiNormedGroup := (FiltrationPow r' c n).op ⋙ CLC V theorem CLCFP_def (V : SemiNormedGroup) (r' : ℝ≥0) (c : ℝ≥0) (n : ℕ) : CLCFP V r' c n = LCFP V r' c n ⋙ Completion := rfl namespace CLCFP lemma map_norm_noninc {M₁ M₂} (f : M₁ ⟶ M₂) : ((CLCFP V r' c n).map f).norm_noninc := CLC.map_norm_noninc _ _ @[simps app] def res [fact (c₂ ≤ c₁)] : CLCFP V r' c₁ n ⟶ CLCFP V r' c₂ n := (whisker_right (nat_trans.op $ FiltrationPow.cast_le r' c₂ c₁ n) (CLC V) : _) lemma res_def [fact (c₂ ≤ c₁)] : res V r' c₁ c₂ n = whisker_right (nat_trans.op (FiltrationPow.cast_le r' c₂ c₁ n)) (CLC V) := rfl lemma res_def' [fact (c₂ ≤ c₁)] (M : ProFiltPseuNormGrpWithTinv r') : (res V r' c₁ c₂ n).app (op M) = (CLC V).map ((Filtration.cast_le ((ProFiltPseuNormGrpWithTinv.Pow r' n).obj M) c₂ c₁)).op := rfl lemma res_app' [fact (c₂ ≤ c₁)] (M : (ProFiltPseuNormGrpWithTinv r')ᵒᵖ) : (res V r' c₁ c₂ n).app M = (CLC V).map ((FiltrationPow.cast_le r' c₂ c₁ n).app (unop M)).op := rfl @[simp] lemma res_refl : res V r' c c n = 𝟙 _ := by { rw [res, FiltrationPow.cast_le_refl, nat_trans.op_id, whisker_right_id'], refl } lemma res_comp_res [fact (c₂ ≤ c₁)] [fact (c₃ ≤ c₂)] [fact (c₃ ≤ c₁)] : res V r' c₁ c₂ n ≫ res V r' c₂ c₃ n = res V r' c₁ c₃ n := by simp only [res, ← whisker_right_comp, FiltrationPow.cast_le_comp, ← nat_trans.op_comp] lemma res_norm_noninc [fact (c₂ ≤ c₁)] (M) : ((res V r' c₁ c₂ n).app M).norm_noninc := Completion.map_norm_noninc $ LCFP.res_norm_noninc _ _ _ _ _ _ section Tinv -- kmb commented out the next line --open profinitely_filtered_pseudo_normed_group_with_Tinv variables [fact (0 < r')] [fact (c₂ ≤ r' * c₁)] -- @[simps obj {fully_applied := ff}] def Tinv : CLCFP V r' c₁ n ⟶ CLCFP V r' c₂ n := (whisker_right (nat_trans.op $ FiltrationPow.Tinv r' c₂ c₁ n) (LocallyConstant.obj V ⋙ Completion) : _) . lemma Tinv_def : Tinv V r' c₁ c₂ n = (whisker_right (LCFP.Tinv V r' c₁ c₂ n) Completion : _) := rfl lemma Tinv_def' : Tinv V r' c₁ c₂ n = whisker_right (nat_trans.op $ FiltrationPow.Tinv r' c₂ c₁ n) (CLC V) := rfl lemma res_comp_Tinv [fact (c₂ ≤ c₁)] [fact (c₃ ≤ c₂)] [fact (c₃ ≤ r' * c₂)] : res V r' c₁ c₂ n ≫ Tinv V r' c₂ c₃ n = Tinv V r' c₁ c₂ n ≫ res V r' c₂ c₃ n := begin dsimp only [Tinv, res, CLC, LC], simp only [← whisker_right_comp, ← nat_trans.op_comp], refl end end Tinv section T_inv variables [normed_with_aut r V] [fact (0 < r)] @[simps {fully_applied := ff}] def T : CLCFP V r' c n ≅ CLCFP V r' c n := ((whiskering_left _ _ _).obj (FiltrationPow r' c n).op).map_iso (CLC.T r V) @[simps app_apply {fully_applied := ff}] def T_inv : CLCFP V r' c n ⟶ CLCFP V r' c n := whisker_left (FiltrationPow r' c n).op (CLC.T_inv r V) lemma T_inv_eq [normed_with_aut r V] [fact (0 < r)] : (T r V r' c n).inv = T_inv r V r' c n := rfl lemma T_inv_def : T_inv r V r' c n = (whisker_right (LCFP.T_inv r V r' c n) Completion : _) := rfl lemma T_inv_app [fact (0 < r)] (M : (ProFiltPseuNormGrpWithTinv r')ᵒᵖ) : (T_inv r V r' c n).app M = (CLC.T_inv r V).app ((FiltrationPow r' c n).op.obj M) := rfl lemma res_comp_T_inv [fact (c₂ ≤ c₁)] : res V r' c₁ c₂ n ≫ T_inv r V r' c₂ n = T_inv r V r' c₁ n ≫ res V r' c₁ c₂ n := begin ext M : 2, simp only [nat_trans.comp_app, res_app', T_inv_app], exact (CLC.T_inv r V).naturality _, end end T_inv end CLCFP namespace breen_deligne open CLCFP variables {l m n} namespace universal_map variables (ϕ ψ : universal_map m n) def eval_CLCFP [ϕ.suitable c₂ c₁] : CLCFP V r' c₁ n ⟶ CLCFP V r' c₂ m := (whisker_right (ϕ.eval_LCFP V r' c₁ c₂) Completion : _) lemma eval_CLCFP_of (f : basic_universal_map m n) [f.suitable c₂ c₁] : eval_CLCFP V r' c₁ c₂ (free_abelian_group.of f) = (whisker_right (nat_trans.op $ f.eval_FP r' c₂ c₁) (CLC V)) := by { rw [eval_CLCFP, eval_LCFP_of, basic_universal_map.eval_LCFP, whisker_right_twice], refl } @[simp] lemma eval_CLCFP_zero : (0 : universal_map m n).eval_CLCFP V r' c₁ c₂ = 0 := begin simp only [eval_CLCFP, eval_LCFP_zero], ext x : 2, exact Completion.map_zero _ _ end @[simp] lemma eval_CLCFP_add [ϕ.suitable c₂ c₁] [ψ.suitable c₂ c₁] : (ϕ + ψ : universal_map m n).eval_CLCFP V r' c₁ c₂ = ϕ.eval_CLCFP V r' c₁ c₂ + ψ.eval_CLCFP V r' c₁ c₂ := begin simp only [eval_CLCFP, eval_LCFP_add], ext x : 2, exact Completion.map_add end @[simp] lemma eval_CLCFP_sub [ϕ.suitable c₂ c₁] [ψ.suitable c₂ c₁] : (ϕ - ψ : universal_map m n).eval_CLCFP V r' c₁ c₂ = ϕ.eval_CLCFP V r' c₁ c₂ - ψ.eval_CLCFP V r' c₁ c₂ := begin simp only [eval_CLCFP, eval_LCFP_sub], ext x : 2, exact Completion.map_sub end open category_theory.limits lemma eval_CLCFP_comp (g : universal_map m n) (f : universal_map l m) [hg : g.suitable c₂ c₁] [hf : f.suitable c₃ c₂] : @eval_CLCFP V r' c₁ c₃ _ _ (comp g f) (suitable.comp c₂) = g.eval_CLCFP V r' c₁ c₂ ≫ f.eval_CLCFP V r' c₂ c₃ := by simp only [eval_CLCFP, ← whisker_right_comp, eval_LCFP_comp V r' c₁ c₂ c₃] lemma res_comp_eval_CLCFP [fact (c₂ ≤ c₁)] [ϕ.suitable c₄ c₂] [ϕ.suitable c₃ c₁] [fact (c₄ ≤ c₃)] : res V r' c₁ c₂ n ≫ ϕ.eval_CLCFP V r' c₂ c₄ = ϕ.eval_CLCFP V r' c₁ c₃ ≫ res V r' c₃ c₄ m := by { dsimp only [CLC, res], simp only [eval_CLCFP, ← whisker_right_comp, ← whisker_right_twice], congr' 1, apply res_comp_eval_LCFP } lemma Tinv_comp_eval_CLCFP [fact (0 < r')] [fact (c₂ ≤ r' * c₁)] [fact (c₄ ≤ r' * c₃)] [ϕ.suitable c₃ c₁] [ϕ.suitable c₄ c₂] : Tinv V r' c₁ c₂ n ≫ ϕ.eval_CLCFP V r' c₂ c₄ = ϕ.eval_CLCFP V r' c₁ c₃ ≫ Tinv V r' c₃ c₄ m := by simp only [eval_CLCFP, Tinv_def, ← whisker_right_comp]; congr' 1; apply Tinv_comp_eval_LCFP lemma T_inv_comp_eval_CLCFP [normed_with_aut r V] [fact (0 < r)] [ϕ.suitable c₂ c₁] : T_inv r V r' c₁ n ≫ ϕ.eval_CLCFP V r' c₁ c₂ = ϕ.eval_CLCFP V r' c₁ c₂ ≫ T_inv r V r' c₂ m := by simp only [eval_CLCFP, T_inv_def, ← whisker_right_comp, T_inv_comp_eval_LCFP] lemma norm_eval_CLCFP_le [normed_with_aut r V] [fact (0 < r)] [ϕ.suitable c₂ c₁] (N : ℕ) (h : ϕ.bound_by N) (M) : ∥(ϕ.eval_CLCFP V r' c₁ c₂).app M∥ ≤ N := le_trans (normed_add_group_hom.norm_completion _).le $ norm_eval_LCFP_le _ _ _ _ _ _ _ h _ end universal_map end breen_deligne

published: 12.05.2014, 07:29 updated: 12.05.2014 07:40 Prague - Russian President Vladimir Putin creates a situation that cannot be solved, in which he may become the only possible peacemaker able to propose and push through an agreement, Zbynek Petracek says in daily Lidove noviny (LN) today about the developments in Ukraine. Ruský prezident Vladimir Putin dohlíží na rozsáhlé cvičení ozbrojených sil se zapojením všech druhů vojsk, včetně sil jaderného odstrašení. Manévry prý nemají nic společného s ukrajinskou krizí, nicméně šéf Kremlu poznamenal, že vývoj v sousední zemi je výsledkem "nezodpovědné politiky" způsobující velké problémy. ČTK/AP Alexei Nikolsky Before the weekend referendums on sovereignty in eastern Ukraine, Putin formally called on the pro-Russian leaders to postpone the referenda in order to show that Moscow does not provoke the conflict. At the same time, he demanded that Kiev does not deal with the situation in the eastern part of the country by force, Petracek writes. He says this leads to an untenable situation that can be solved only by Putin. Putin used this strategy already last year in Syria, Petracek writes. Russia provided military support to the regime of Bashar Assad for a long time and Putin let U.S. President Barack Obama threaten Assad with a military strike and then he mediated an agreement and saved the Russian military base in Syria, Petracek writes. What Putin tested in Syria he can use in Ukraine, Petracek concludes. Czech Foreign Minister Lubomir Zaoralek lost his vitality and looked tired when he said pro-Russian separatists would take part in the talks on the future of southeastern Ukraine, Alexandr Mitrofanov writes in Pravo. Zaoralek said German Chancellor Angela Merkel supports this idea, Mitrofanov notes. The position of the European Union has changed and turned into Realpolitik, or diplomacy primarily based on power, Mitrofanov writes. Lubos Palata says in Mlada fronta Dnes (MfD) that there are no good solutions anymore in Ukraine, but only bad ones and even worse ones. The result of the referendum on sovereignty in the Donetsk and Luhansk regions is of no importance, but the fact that it was held is very significant for the separatists with guns who organised it, Palata writes. TV shots of people crowding outside the election rooms is a defeat of the Ukrainian government. A functional state cannot let such a referendum to be held. However, Ukraine is not a functional state, Palata says. Written by: ČTK

November 7, 2017 Image credit : Illustrative Image Kottayam: The Kerala Government has received nod from law authorities to proceed against former Chief Minister Oommen Chandy and other Congress leaders in connection with the sexual and corruption case in the solar scam. Former Supreme Court judge Arijit Pasayat gave the nod to the government which had sought the clearance after it had set up a investigation team to probe the charges. The clearance comes at a time when the government has convened a Assembly meeting to place the solar commission report before legislators. The solar case involves Sarita S Nair lobbying with political parties and the then Chief Minister office and other ministers for getting solar contracts to Team Solar floated by her and associate Biju Radhakrishnan in 2013. They allegedly collected advance amounts from people promising to make them partners or installing solar power system at different locations. Some officials from Chandy’s office made several phone calls to Saritha. The total amount involved in the scam was about Rs 10 crore. Five persons including Saritha Nair, Chandy’s personal aide Tenny Joppan, Biju Radhakrishnan and actor Shalu Menon were arrested during fag end of Congress-led United Democratic Front Government term. There are 30 cases pending against Sarita and Biju in different courts in connection with solar case. Friendship and tackling peer pressure 4 Natural Beauty Ingredients That Soothe The Skin 5 Jackets To Cosy Up In During Your Winter Vacation Sakkarai Pongal Recipe

\section{$\WF$-holonomicity}\label{sec:WF} \subsection{}\label{sec:WF1} In this section we elaborate on a notion of \blue{algebraic} $\WF$-holonomicity introduced by Aizenbud and Drinfeld in \cite{AizDr}. We rephrase a question by Aizenbud and Drinfeld for future research, based on Remark 3.2.2 and the surrounding text of \cite{AizDr}, and we make a first step in its direction in Theorem \ref{thm:AizDrthmA}. We first adapt the definition of \cite[Section 3.2, 3.2.1 and 3.2.2]{AizDr} of \blue{algebraic} $\WF$-holonomicity, to the situation of any non-archimedean local field $F$ (of arbirary characteristic). Say that a Zariski closed subset $C$ of $\AA^n$ has dimension $-\infty$ if it is empty, and has dimension $i$ for some $i\geq 0$ when one of the irreducible components of $C$ has dimension $i$ and any irreducible component of $C$ has dimension $\leq i$. \begin{defn}\label{def:WF-hol} Let $F$ be in $\Loc$ and let $X\subset F^m$ be a strict $C^1$ variety of dimension $n$. Let $u$ be a distribution on $X$, and let $\WF(u)$ be its $F^\times$-wave front set. Say that $u$ is \blue{algebraically} $\WF$-holonomic if \blue{$X=\cX(F)$ for some smooth subvariety $\cX$ of $\AA^m_F$ and there are finitely many smooth locally closed subvarieties $\cY_i$ of $\cX$ such that $\WF(u)$ is contained in $$ \bigcup_i CN_{\cY_i}^{\cX}(F), $$ where $CN_{\cY_i}^{\cX}$ is the co-normal bundle of $\cY_i\subset \cX$.} \blue{Say that $u$ is $F$-analytically $\WF$-holonomic if $X$ is an $F$-analytic submanifold of $F^m$ and there are finitely many $F$-analytic submanifolds $Y_i$ of $X$ such that $\WF(u)$ is contained in the union over $i$ of the co-normal bundles $CN_{Y_i}^X$ of $Y_i\subset X$. Say that $u$ is strictly $C^1$ $\WF$-holonomic if there are finitely many strict $C^1$ subvarieties $Y_i$ of $X$ such that $\WF(u)$ is contained in the union of the co-normal bundles $CN_{Y_i}^X$ of $Y_i\subset X$.} \end{defn} In \cite{AizDr}, the question is considered when the Fourier transform of a $\WF$-holonomic distribution is again a $\WF$-holonomic distribution, and for some distributions this property is shown. These distributions of \cite{AizDr} are all of $\cCexp$-class (up to working with charts to make the varieties affine), and it seems sensible to explore the mentioned question of \cite{AizDr} in the context of distributions of $\cCexp$-class, where more geometric tools are available than for abstract distributions, and which are stable under Fourier transforms by Theorem \ref{Fourier:p}. The following generalization of Theorem A of \cite{AizDr} to general distributions of $\cCexp$-class may be a first step towards this question. \begin{thm}\label{thm:AizDrthmA} Any $\cCexp$-class distribution is smooth on the complement of a proper Zariski closed subset. In more detail, let $Y$ and $W\subset Y\times \VF^m$ be definable sets for some $m\geq 0$ such that $W_{F,y}$ is a strict $C^1$ manifold of dimension $n$ for each $F$, each $y\in Y_F$ and some $n\leq m$. Let $E$ be in $\cCexp(W\times \ZZ)$. Then there are $G$ in $\cCexp(W)$ and a definable set $C\subset Y\times \VF^m$ such that for each $F$ in $\Loc'_{\gg 1}$ and for each $y\in Y_F$, the set $C_{F,y}$ is Zariski closed in $F^m$ of dimension at most $n-1$, the function $G_{F}(y,\cdot)$ is locally constant on $W_{F,y}\setminus C_{F,y}$, and, for $y\in {\rm Dis}(E,Y)_F$, the distribution $u_{F,y}$ associated to $E_{F,y}$ is smooth on $W_{F,y}\setminus C_{F,y}$ and moreover represented by $G_{F}(y,\cdot)$ on $W_{F,y}\setminus C_{F,y}$. Moreover, in the case that $Y$ is a definable subset of $\VG^N\times \RF_{N}^N$ for some $N\geq 0$, there is a single Zariski closed subset $C$ of $\AA^{m}_\ZZ$ such that one can take for the $C_{F,y}$ the set $C(F)$ (independently from $y$). \end{thm} \begin{proof} Let $G$ be in $\cCexp(W)$ such that $G_F(y,x)$ equals the limit for $r\to\infty$ of $$ q_F^{nr}\cdot E_{F}(y,x,r) $$ whenever this limit exists in $\CC$. Here, $F$ runs over $\Loc'_{\gg 1}$, $y$ over $Y_F$, and $x$ over $W_{F,y}$. Such $G$ exists by Theorem 3.1.3(2) of \cite{CGH5}. By Theorem 4.4.3 of \cite{CHallp}, the remark below its proof, by using charts (as at the end of the proof of Theorem \ref{Smo}), and by quantifier elimination result in the generalized Denef-Pas language from \cite{Rid}, there is a definable set $C_1 \subset Y \times \VF^m$ such that for every $F\in \Loc'_{\gg 1}$ and every $y\in Y_F$, the set $C_{1,F,y}$ is Zariski closed in $F^m$, has dimension at most $n-1$, and such that $G_{F}(y,\cdot)$ is locally constant on $W_{F,y}\setminus C_{1,F,y}$. Similarly, there is a definable set $C_2 \subset Y \times \VF^m$ such that for every $F\in \Loc'_{\gg 1}$ and every $y\in Y_F$, the set $C_{2,F,y}$ is Zariski closed in $F^m$, has dimension at most $n-1$, and such that the function sending $r\gg 1$ to $E_{F}(y,x,r)$ is locally independent of $x$ in $W_{F,y}\setminus C_{2,F,y}$. By additivity of distributions (and since $E_{F,y}$ is a $B$-function), this implies, for $y\in {\rm Dis}(E,Y)_F$, $x\in W_{F,y}\setminus C_{2,F,y}$ and for sufficiently large $r$, that $$ E_{F}(y,x,r+1) = q_F^{n}\cdot E_{F}(y,x,r), $$ and hence the above limit for $r\to\infty$ exists for all $x\in W_{F,y}\setminus C_{2,F,y}$. Now take $C_{F,y}$ to be the union of $C_{1,F,y}$ and $C_{2,F,y}$ to finish the proof, where the final simplified form for the $C_{f,y}$ follows again from the valued field quantifier elimination in the generalized Denef-Pas language from \cite{Rid}. \end{proof} \blue{ \subsection{Loci, their complements, and dimension} Note that $\cCexp$-loci (and also complements of $\cCexp$-loci) are more general than definable sets, and that their study is more subtle than that of definable sets (the latter situation is well understood by the quantifier elimination result for definable sets in the generalized Denef-Pas language from \cite{Rid}, see also Theorem 5.1.2 of \cite{CHallp}, generalizing \cite{Pas} and \cite{Denef2}). We continue this section with a study of dimensions of $\cCexp$-loci and their complements. \begin{def-prop}[Dimensions of loci]\label{prop:localstructure} Let $f$ be in $\cCexp(\VF^n)$. Fix $F$ in $\Loc'_{\gg 1}$. Write $A_1\subset F^n$ for the zero locus of $f_F$, and, $A_2$ for the complement of $A_1$ in $F^n$. Then there exist finitely many strict $C^1$-submanifolds $W_i$ of $F^n$ such that $A_1$ (resp.~$A_2$) equals $\bigcup_i W_i$. Moreover, for each $x\in W_i$ for any $i$, there exists an open neighborhood $U$ of $x$ in $F^n$ such that $W_i\cap U$ is $\gLPas(F)$-definable. Let $m$ be the maximum of the dimensions of the $W_i$, then we call $m$ the dimension of $A_1$ (resp.~of $A_2$). Finally, the Zariski closure of $\bigcup_i W_i$ in $F^n$ has dimension $m$ as well. \end{def-prop} \begin{proof} Given $f$ and $F$, there are finitely many $\gLPas(F)$-definable (that is, definable with parameters from $F$) pieces $D_i\subset F^n$ whose union equals $F^n$, such that each $D_i$ is a strict $C^1$ manifold, and such that the restriction of $f_F$ to $D_i$ is locally constant. (This follows from cell decomposition and the definition of functions of $\cCexp$-class, and it requires $F$ to be fixed.) Take $x\in A_1\cap D_i$ (resp.~in $A_2\cap D_i$). Let us put $W_i$ equal to $A_1\cap D_i$ (resp.~$A_2\cap D_i$). Since $f_F$ is locally constant on $D_i$, and since $D_i$ is a strict $C^1$ manifold, the $W_i$ are clearly strict $C^1$ submanifolds. Similarly, for any $x\in W_i$, one clearly has that there exists an open neighborhood $U$ of $x$ in $F^n$ such that $W_i\cap U$ is definable. Again by the local constancy on $D_i$, either $W_i$ is empty, or, the dimensions of $D_i$ and $W_i$ coincide. Since $D_i$ is $\gLPas(F)$-definable, it is contained in a Zariski closed of the same dimension by the quantifier elimination result for definable sets in the generalized Denef-Pas language from \cite{Rid}. \end{proof} } \begin{prop}\label{Zar} Let $g$ be a function in $\cCexp(X)$ for some definable set $X\subset Y\times \VF^\ell$, where $Y$ is a definable set, and let \blue{$k \le \ell$} be given. For each $F$ in $\Loc'_{\gg 1}$, write $A_{F}$ for the complement of the zero locus of $g_F$ in $X_F$. For each $y\in Y_F$, write $Z_{F,y}$ for the Zariski closure of $A_{F,y}$ in $F^\ell$. Then there exists a definable set $C\subset Y\times \VF^\ell$ such that for each $F$ in $\Loc'_{\gg 1}$ and for each $y\in Y_F$, the set $C_{F,y}$ is Zariski closed in $F^\ell$ of dimension at most $k$, and if $\dim Z_{F,y} \le k$, then $Z_{F,y} \subset C_{F,y}$. In particular, the set of $y$ for which we have $\dim Z_{F,y} \le k$ is a $\cCexp$-locus. \end{prop} \begin{proof} We may assume that $X = Y \times \VF^\ell$, by extending $g$ by $0$ on $(Y \times \VF^\ell) \setminus X$. First note that the ``in particular'' part indeed follows from the remainder of the proposition: From $Z_{F,y} \subset C_{F,y} \iff \dim Z_{F,y} \le k$, we obtain that $\dim Z_{F,y} \le k$ iff $g$ vanishes on the definable set $F^\ell \setminus C_{F,y}$. The set of $y$ for which this holds is a $\cCexp$-locus by \cite[Proposition 1.3.1]{CGH5}. Now consider the main part of the proposition. For $k = m$, there is nothing to prove. (Set $C := Y \times \VF^\ell$.) In the case $k = \ell - 1$, we use Theorem 4.4.3 of \cite{CHallp} and the remark below its proof to find a definable set $C \subset Y \times \VF^\ell$ such that for every $F\in \Loc'_{\gg 1}$ and every $y\in Y_F$, the set $C_{F,y}$ is Zariski closed in $F^\ell$, has dimension at most $\ell-1$, and contains the set of non-local constancy of $g_{F,y}$. Then clearly, if $Z_{F,y}$ has dimension at most $k$, it is contained in $C_{F,y}$. For $k < \ell - 1$, we use the case $k=\ell-1$ to find a definable set $C_{\ell-1}\subset Y \times \VF^\ell$ and one finishes the proof by induction on $\ell$, by using the restriction of $g$ to $C_{\ell-1}$ and charts on $C_{\ell-1}$ which are coordinate projections (as at the end of the proof of Theorem \ref{Smo}), and using the quantifier elimination result in the generalized Denef-Pas language from \cite{Rid} to get to definable sets which are Zariski closed as desired. \end{proof} \subsection{Holonomicity} The following result relates to Remark 3.2.2 of \cite{AizDr}, \blue{and is about witnessing algebraic holonomicity. Its first part states} that given a family of distributions $u_{F,y}$, the set of $F$ and $y$ such that $u_{F,y}$ is \blue{strict $C^1$} $\WF$-holonomic is a locus sets. Further, the $u_{F,y}$ that are $\WF$-holonomic are even ``uniformly algebraically $\WF$-holonomic'' in the sense that their wave front sets are contained in \blue{a finite union of the co-normal bundles of uniformly definable algebraic submanifolds. Finally, it says that, for $\cCexp$-class distributions, to be algebraically $\WF$-holonomic is essentially the same as being strict $C^1$ $\WF$-holonomic.} \begin{prop}\label{thm:WFh} Let $Y$ and $W\subset Y\times \VF^n$ be definable sets for some $n\geq 0$. Let $E$ be in $\cCexp(W\times \ZZ)$. For each $F$ in $\Loc'_{\gg 1}$ and each $y\in {\rm Dis}(E,Y)_F$, denote the distribution on $W_{F,y}$ associated to $E_{F,y}$ by $u_{F,y}$. Define the sets, for $F$ in $\Loc'_{\gg 1}$, $$ {\rm{Hol}}(E,Y)_F := \{y\in Y_F \mid y\in {\rm Dis}(E,Y)_F \mbox{ and } \mbox{$u_{F,y}$ is \blue{strict $C^1$} $\WF$-holonomic} \}. $$ Then ${\rm{Hol}}(E,Y)$ is a $\cCexp$-locus. Moreover, \blue{there exist finitely many definable subsets $X_i\subset W$ such that for each $F$ in $\Loc'_{\gg 1}$ and each $y\in {\rm Dis}(E,Y)_F$, each $X_{i,F,y}$ is the set of $F$-rational points on a smooth locally closed subvariety of $\AA_F^{n}$, $X_{0,F,y}$ has the same dimension as $W_{F,y}$, $X_{0,F,y}$ contains both $W_{F,y}$ and the $X_{i,F,y}$, and, $u_{F,y}$ is strict $C^1$ $\WF$-holonomic if and only if the $F^\times$-wave front set of $u_{F,y}$ is contained in $$ \bigcup_i CN_{X_{i,F,y}}^{X_{0,F,y}}. $$ Hence, for any $F$ in $\Loc'_{\gg 1}$ and any $y\in {\rm Dis}(E,Y)_F$, the distribution $u_{F,y}$ on $W_{F,y}$ is strict $C^1$ $\WF$-holonomic if and only if it can be extended to an algebraically $\WF$-holonomic distribution on $\cX(F)\supset W_{F,y}$ for some smooth algebraic subvariety $\cX$ of $\AA^n_F$ of the same dimension as $W_{F,y}$. } \end{prop} \begin{proof} \blue{ By Theorem \ref{Smo}, $\Smo(\Lambda,E,Y)$ is a $\cCexp$-locus, say, given as the locus of $g\in \cCexp (W\times \VF^n)$. We may suppose that the dimension of $W_{F,y}$ equals $k$ for each $F$ and each $y\in {\rm Dis}(E,Y)_F$. Let $C$ be as in Proposition \ref{Zar} for $g$ and $k$, and, let $C_0$ be as in Proposition \ref{Zar} for the characteristic function of $W$ and $k$. By Proposition \ref{prop:localstructure}, Chevalley's theorem and valued field quantifier elimination in the generalized Denef-Pas language, there are finitely many definable sets $X_{i}$ as postulated in the proposition. Namely, each $X_{i,F,y}$ equals the set of $F$-rational points on a locally closed smooth submanifold of $\AA_F^{n}$ of dimension at most $k$, and, for each $F$ and each $y\in {\rm Dis}(E,Y)_F$, the set $X_{0,F,y}$ contains both $W_{F,y}$ and the $X_{i,F,y}$, and, $u_{F,y}$ is strict $C^1$ $\WF$-holonomic if and only if $\WF(u_{F,y})$ is contained in $$ \bigcup_i CN_{X_{i,F,y}}^{X_{0,F,y}}. $$ By the formalism of \cite[Proposition 1.3.1]{CGH5}, this condition corresponds to a $\cCexp$-locus which finishes the proof. } \end{proof} We now get a transfer principle for $\WF$-holonomicity. \begin{cor}[Transfer Principle for $\WF$-holonomicity]\label{transfer} Let data and notation be as in Proposition~\ref{thm:WFh}. Then there is $M$ such that for all $F_1,F_2\in \Loc_M$ with isomorphic residue fields, one has that $$ Y_{F_1} = {\rm{Hol}}(E,Y)_{F_1,\psi} \mbox{ for all $\psi$ in $\cD_{F_1}$} $$ if and only if $$ Y_{F_2} = {\rm{Hol}}(E,Y)_{F_2,\psi} \mbox{ for all $\psi$ in $\cD_{F_2}$} $$ \end{cor} \begin{proof} By Proposition~\ref{thm:WFh}, to be $\WF$-holonomic is a $\cCexp$-locus condition. By Proposition 9.2.1 of \cite{CLexp}, one has transfer for any $\cCexp$-locus condition. \end{proof} \begin{remark} \blue{All results, statements and definitions of Sections \ref{sec:distCexpclass} and \ref{sec:WF} (modulo adaptations in the parts concerning algebraic varieties) hold when one consequently replaces $\gLPas$ by an enrichment obtained by adding some analytic structure as in \cite{CLip}, \cite{CLips} to $\gLPas$. Similarly, one can put additional structure on the residue rings $\Res_n$, and, one can add constants for a ring of integers $\cO$ of a number field in the sort $\VF$ and work uniformly in all finite field extensions of completions of the fraction field of $\cO$. Indeed, this corresponds to Remark A.3 of \cite{CGH5} and Section 4.7 of \cite{CHallp}.} \end{remark}

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TITLE: group of permutation - cycles of length $n$ - order of generated group QUESTION [3 upvotes]: We have all cycles of length $n$. For example when $n=4$ our set is following form: $$G = \{(1234);(1243);(1342);(1324);(1432);(1423)\} $$ Our task is to find order of the group generated by the set $G$. We know, that $|G|\ge (n-1)!$, but I have no clue how to begin. Try to help me, please. REPLY [2 votes]: Note that any subgroup of $S_n$ generated by a complete set of cycles of length $n$ will be a normal subgroup. For $n\gt 4$ there are just two possibilities - $A_n$ or $S_n$ (the trivial subgroup is impossible). If $n$ is odd, the cycles which generate the group are all even permutations, and can't generate an odd permutation, so the group is $A_n$. If $n$ is even, the cycles are odd permutations, so the group they generate contains an odd permutation, and must be $S_n$. For $n\leq 4$ it is necessary to use a different approach (e.g. direct computation in a small group). Note that essentially the same proof applies when the generators are a complete set of elements of any conjugacy class in $S_n$. However, when $n=4$ the elements conjugate to $(12)(34)$ generate the normal subgroup of order $4$ - so this approach cannot work in general for order $4$.

\begin{document} \begin{abstract} We prove a conjecture of Dipendra Prasad on the Ext-branching problem from $\GL_{n+1}(F)$ to $\GL_n(F)$, where $F$ is a $p$-adic field. \end{abstract} \maketitle \section{Introduction} Decomposing a smooth representation of $\GL_{n+1}(F)$, when restricted to $\GL_n(F)$, is a well known and well studied problem that was initiated in a paper of Prasad \cite{Pr1}. Today, this problem is a part of a large family of Gan-Gross-Prasad restriction problems \cite{GGP} that are at the center of much research in Representation Theory and Number Theory. In order to describe what is known, and what is new in this paper, let $G_n=\GL_n(F)$ and let $\Alg(G_n)$ be the category of smooth representations of $G_n$. For every $\pi\in \Alg(G_n)$, let $\Wh(\pi)$ be the space of Whittaker functionals on $\pi$. If $\pi$ is irreducible then $\Wh(\pi)$ is one or zero dimensional. We say that $\pi$ is generic or degenerate, respectively. Let $\pi_1$ be an irreducible representation of $G_{n+1}$. One of the most significant results in the subject is that the restriction of $\pi_1$ to $G_n$ is multiplicity free \cite{AGRS}, \cite{AG}, \cite{SZ}, that is, for every irreducible representation $\pi_2$ of $G_n$, \[ \dim \Hom_{G_n}(\pi_1, \pi_2) \leq 1\] and it is one if both representations are generic. On the other hand, Dipendra Prasad has proved in \cite{Pr} the following beautiful formula: \[ \EP(\pi_1, \pi_2) := \sum (-1)^i \dim\Ext^i_{G_n}(\pi_1,\pi_2) = \dim \Wh(\pi_1) \cdot \dim \Wh(\pi_2). \] In particular, the formula implies that $\EP(\pi_1, \pi_2)=1$ if both representations are generic. Since $\dim \Hom_{G_n}(\pi_1, \pi_2)=1$, Prasad has conjectured that $\Ext^i_{G_n}(\pi_1,\pi_2)$ vanish for $i>0$ if both representations are generic. \smallskip The first main result in this paper is a proof of this conjecture. The proof is based on the theory of Bernstein-Zelevinsky derivatives \cite{BZ1}, \cite{BZ} with the following, additional ingredient. The theory of derivatives describes how a smooth representation of $G_{n+1}$ restricts to the mirabolic subgroup $M_{n+1}$. However, instead of $M_{n+1}$ one can consider the transpose $M_{n+1}^{\top}$ of $M_{n+1}$, and develop a theory of derivatives with respect to $M_{n+1}^{\top}$. Thus we have two notions of derivatives: those with respect to $M_{n+1}$ are called right derivatives and those with respect to $M_{n+1}^{\top}$ are called left derivatives. Since $M_{n+1}^{\top}$ is not conjugated to $M_{n+1}$ in $G_{n+1}$, the information provided by left and right derivatives taken together is stronger, and is essential to our combinatorial arguments. Let us illustrate the argument when $\pi_1$ is the Steinberg representation of $\GL_2(F)$. Let $\nu(g)=|g|$ be a character of $\GL_1$. The theory of derivatives implies that the restriction of $\pi_1$ to $\GL_1(F)$ is given by the following Bernstein-Zelevinsky filtration \[ 0 \rightarrow C_c( F^{\times}) \rightarrow \pi_1 \rightarrow \mathbb C \rightarrow 0 \] where $C_c( F^{\times})$ is the space of locally constant, compactly supported functions on $F^{\times}$, and $\GL_1( F)$ acts on $\mathbb C$ by the character $\nu$ or $\nu^{-1}$, depending whether we use right or left derivatives, respectively. Thus, for a given character $\pi_2$ of $\GL_1(F)$, one can clearly arrange that the character on the quotient $\mathbb C$ in the above sequence is different from $\pi_2$. Now higher extension spaces vanish since $C_c( F^{\times})$ is projective. Even multiplicity one statement is clear since it holds for $C_c( F^{\times})$. The general case, restricting from $G_{n+1}$ to $G_n$, follows this strategy. The bottom piece of the Bernstein-Zelevinsky filtration of $\pi_1$ is the Gelfand-Graev representation of $G_n$, thus vanishing of higher extensions, and multiplicity one for generic representations, follow from projectivity \cite{CS}, and multiplicity one for the Gelfand-Graev representation of $G_n$, respectively. \smallskip Let $K_r$ be the $r$-th principal congruence subgroup in $G_n$. For every $\pi\in \Alg(G_n)$ generated by the subspace $\pi^{K_r}$ of $K_r$-fixed vectors, the left ${}^{(i)}\pi$ and the right $\pi^{(i)}$ derivative are related by the isomorphism $(\pi^{(i)})^{\vee} \cong {}^{(i)} (\pi^{\vee})$. This isomorphism is not true for all $\pi\in \Alg(G_n)$. We establish it as a consequence of a ``second adjointness isomorphism'' for Bernstein-Zelevinsky derivatives, proved in the appendix. \smallskip The second main result is projectivity of essentially square integrable representations of $G_{n+1}$, when restricted to $G_n$. If $\pi_1$ is projective as $G_n$-module then higher extension spaces vanish without assuming that $\pi_2$ is generic. On the other hand, if $\pi_2$ is degenerate, then $\EP(\pi_1, \pi_2) =0$ by the Prasad's formula. If $\pi_2$ is also a quotient of $\pi_1$ then $\Ext^i_{G_n}(\pi_1,\pi_2)\neq 0$ for some $i>0$. Thus a necessary condition for $\pi_1$ to be $G_n$-projective is not to have degenerate quotients. In this paper we show that this is also a sufficient condition. The proof relies heavily on the Hecke algebra methods from our earlier paper \cite{CS}. We also show that this condition is satisfied if $\pi_1$ is an essentially square integrable representation. Therefore essentially square integrable representations of $G_{n+1}$ are projective $G_n$-modules. Moreover, any two essentially square integrable representations of $G_{n+1}$ are isomorphic as $G_n$-modules. This result generalizes the classical result of Bernstein and Zelevinsky which says that any two cuspidal representations of $G_{n+1}$ are isomorphic when restricted to the mirabolic subgroup $M_{n+1}$ of $G_{n+1}$. We illustrate this result when $\pi_1$ is a representation of $\GL_2(F)$. If $\pi_1$ is cuspidal, then it is isomorphic to $C_c( F^{\times})$. On the other hand, if $\pi_1$ is the Steinberg representation, then it is isomorphic to $C_{c}( F)$. The isomorphism between these two $\GL_1(F)$-modules is obtained one Bernstein component at a time, that is, for the isotypic components under the action of $O^{\times}$ where $O$ is the ring of integers in $F$. It is clear that, given a character of $O^{\times}$, the corresponding components are naturally isomorphic except when the character is trivial. Note that $O^{\times}$-invariants are \[ C_c( F^{\times})^{O^{\times}} = \oplus_{n\in \mathbb Z} \mathbb C \varphi^{\times}_n \text{ and } C_c( F)^{O^{\times}} = \oplus_{n\in \mathbb Z} \mathbb C \varphi_n \] where $\varphi^{\times}_n$ and $\varphi_n$ are the characteristic functions of $\varpi^n O^{\times}$ and $\varpi^n O$, respectively, and $\varpi$ a uniformizing element. Now $\varphi^{\times}_n \leftrightarrow \varphi_n$ gives an isomorphism. In general, for any essentially square integrable representation $\pi_1$, we identify each Bernstein component of $\pi_1$ with an explicit projective Hecke algebra module, independent of $\pi_1$. In a forthcoming paper \cite{Ch}, we shall classify all irreducible representations which are projective when restricted from $G_{n+1}$ to $G_n$, and also study the indecomposibility of irreducible smooth representations under restriction. \smallskip This paper is devoted to study of the quotient restriction problem, but one can also consider the submodule restriction problem: $\Hom_{G_n}(\pi_2,\pi_1)$. The two problems are related by a cohomological duality \[ \Ext_{G_n}^i(\pi_2,\pi_1)^{\vee} \cong \Ext_{G_n}^{d(\pi_2) -i}(\pi_1, D(\pi_2)), \] where $d(\pi_2)$ is the cohomological dimension of $\pi_2$ and $D(\pi_2)$ is the Aubert involute of $\pi_2$, due to Nori and Prasad \cite{NP}. This duality gives an additional importance to the cohomological restriction problem studied in this paper. Since $d(\pi_2)>0$, due to the presence of one-dimensional center in $G_n$, it follows that $\Hom_{G_n}(\pi_2,\pi_1)=0$ for all irreducible $\pi_2$ if $\pi_1$ is projective, in particular, this is true if $\pi_1$ is an essentially square integrable representation, by the results of this paper. \section{Bernstein-Zelevinsky derivatives} In this section we study Bernstein-Zelevinsky derivatives, or simply derivatives, as functors from $\Alg(G_n)$ to $\Alg(G_{n-i})$. We state a ``second adjointness isomorphism'' for these functors, as well as an Ext version of the formula. Mirabolic group will appear in the next section. \subsection{Notation} Let $G_n=\GL_n(F)$, where $F$ is $p$-adic field. Let $\nu(g)=|\mathrm{det}(g)|$ be the character of $G_n$, where $|\cdot |$ is the absolute value on $F$. Let $B_n$ be the Borel subgroup of $G_n$ consisting of upper triangular matrices and let $U_n$ be the unipotent radical of $B_n$. Let \[ R_{n-i}= \left\{ \begin{pmatrix} g & x \\ 0 & u \end{pmatrix} : g \in G_{n-i}, u \in U_{i}, x \in \mathrm{Mat}_{n-i,i}(F) \right\} . \] We have an obvious Levi decomposition $R_{n-i}=G_{n-i}E_{n-i}$, where $E_{n-i}$ is the unipotent radical of $R_{n-i}$. Moreover, $E_{n-i}=N_{n-i}U_i$ where $N_{n-i}$ is the unipotent radical of the maximal parabolic subgroup $P_{n-i}$ consisting of block upper triangular matrices and Levi factor $G_{n-i} \times G_i$. Fix a non-zero additive character $\psi$ of $F$. Let $\psi_i$ be the character of $E_{n-i}$ defined by \[ \psi_i \left( \begin{smallmatrix} 1 & x \\ 0 & u \end{smallmatrix}\right) = \psi(u_{1,2}+ \ldots +u_{i-1,i}) \] where $u_{1,2}, \ldots , u_{i-1,i}$ are the entries of $u$ above the diagonal. Let $\delta_{R_i}$ be the modular character of $R_{n-i}$. The modular character is trivial on the unipotent radical $E_{n-i}$, and it is equal to $\nu^{i}$ on the Levi factor $G_{n-i}$. Let $\pi$ be a smooth representation of $G_n$ on a vector space $V$. The right $i$-th Bernstein-Zelevinsky derivative of $\pi$ is a smooth representation $\pi^{(i)}$ of $G_{n-i}$ on the vector space $V^{(i)}$ defined by \[ V^{(i)}=V / \langle \pi(e)v - \psi_i(e)v : e \in E_{n-i}, v \in V \rangle. \] The representation $\pi^{(i)}$ is the natural action of the Levi factor $G_{n-i}$ on $V^{(i)}$ twisted by $\delta_{R_{n-i}}^{-1/2}$, that is, Bernstein-Zelevinsky derivatives in this paper are normalized. \smallskip From the definition of derivatives, the factorization $R_{n-i}=G_{n-i}E_{n-i}$, and the Frobenius reciprocity, one can easily prove the ``first adjointness'' isomorphism for right Bernstein-Zelevinsky derivatives: For any smooth representation $\pi$ of $G_n$ and smooth representation $\sigma$ of $G_{n-i}$: \[ \Hom_{G_n} (\pi, \mathrm{Ind}_{R_{n-i}}^{G_n}(\sigma \otimes \psi_{i})) \cong \Hom_{G_{n-i}} (\pi^{(i)}, \sigma). \] We called the derivative right because there is also a left derivative, which is taken with respect to the transpose of the groups used to define right derivatives. More precisely, let $\theta_n(g) = (g^{-1})^{\top}$ be the outer automorphism of $G_n$ where $g^{\top}$ is the transpose of $g$ (over the usual diagonal here, but it may be over the opposite diagonal if more convenient). Then the left derivative of $\pi$ is defined by \[ {}^{(i)} \pi = \theta_{n-i}(\theta_n(\pi)^{(i)}) .\] In other words, the underlying vector space for ${}^{(i)} \pi$ is \[ {}^{(i)}V=V / \langle \pi(e)v - \psi^{\top}_i(e)v : e \in E^{\top}_{n-i}, v \in V \rangle, \] where $\psi_i^{\top}$ is the character of $E^{\top}_{n-i}$ defined by \[ \psi^{\top}_i \left( \begin{smallmatrix} 1 & 0 \\ x & u \end{smallmatrix}\right) = \bar\psi(u_{2,1}+ \ldots +u_{i,i-1}). \] \smallskip The following is the ``second adjointness'' isomorphism for left Bernstein-Zelevinsky derivatives, proved in the appendix. \begin{lemma} \label{lem second adjointness bz der} Let $K_r$ be the $r$-th principal congruence subgroup in $G_n$. For any representation $\pi$ of $G_n$ generated by $\pi^{K_r}$, the space of $K_r$-fixed vectors in $\pi$, and any smooth representation $\sigma$ of $G_{n-i}$, \[ \Hom_{G_n} (\mathrm{ind}_{R_{n-i}}^{G_n}(\sigma \otimes \bar\psi_{i}), \pi) \cong \Hom_{G_{n-i}} (\sigma, {}^{(i)}\pi). \] \end{lemma} \smallskip We now derive some consequences of the two adjointness isomorphisms. The first consequence is a relationship between right and left derivatives via the contragredient: \begin{lemma} \label{lem_key_isomorphism} Let $K_r$ be the $r$-th principal congruence subgroup in $G_n$. Let $\pi$ be a representation of $G_n$ generated by $\pi^{K_r}$. Then $(\pi^{(i)})^{\vee} \cong {}^{(i)} (\pi^{\vee})$. \end{lemma} \begin{proof} If we insert $\pi^{\vee}$ in Lemma \ref{lem second adjointness bz der} then \[ \Hom_{G_n} (\mathrm{ind}_{R_{n-i}}^{G_n}(\sigma \otimes \bar\psi_{i}), \pi^{\vee}) \cong \Hom_{G_{n-i}} (\sigma, {}^{(i)}(\pi^{\vee})). \] On the other hand, we have three isomorphisms \begin{align*} & \Hom_{G_n} (\mathrm{ind}_{R_{n-i}}^{G_n}(\sigma \otimes \bar \psi_{i}), \pi^{\vee}) \\ \cong & \Hom_{G_n} (\pi, \mathrm{Ind}_{R_{n-i}}^{G_n}((\sigma^{\vee} \otimes \psi_{i})) \\ \cong & \mathrm{Hom}_{G_{n-i}}(\pi^{(i)}, \sigma^{\vee}) \\ \cong & \mathrm{Hom}_{G_{n-i}}(\sigma, (\pi^{(i)})^{\vee}) \end{align*} where the first and the last are dualizing isomorphisms the second is the first adjointness isomorphism for Bernstein-Zelevinsky derivatives. Thus \[ \mathrm{Hom}_{G_{n-i}}(\sigma, (\pi^{(i)})^{\vee}) \cong \Hom_{G_{n-i}} (\sigma, {}^{(i)}(\pi^{\vee})) \] for every smooth representation $\sigma$ of $G_{n-i}$. The lemma follows from the Yoneda lemma. \end{proof} \smallskip \noindent {\bf Remark:} The statement of Lemma \ref{lem second adjointness bz der} is optimal in the sense that it cannot be extended to all smooth representations $\pi$. To that end, observe that Lemma \ref{lem_key_isomorphism}, the case $i=n$, says that we have an isomorphism of vector spaces $(\pi^{\vee})_{U_n,\psi_n} \cong (\pi_{U_n, \psi_n})^{\vee}$ for every $G_n$-module $\pi$ generated by $\pi^{K_r}$. Let $\pi$ be any smooth $G_n$-module. It can be written as a direct sum \[ \pi\cong \oplus_{r=1}^{\infty} \pi_r \] where $\pi_r$ is generated by $\pi_r^{K_r}$ and $\pi_r^{K_{r-1}}=0$. Then \[ \pi^{\vee}\cong \oplus_{r=1}^{\infty} \pi_r^{\vee} \] hence $\pi^{\vee}_{U_n,\psi_n}$ is a direct sum of $(\pi_r^{\vee})_{U_n,\psi_n} \cong ((\pi_r)_{U_n, \psi_n})^{\vee}$. But $(\pi_{U_n,\psi_n})^{\vee}$ is a product of $((\pi_r)_{U_n, \psi_n})^{\vee}$, hence much larger unless the sum over $r$ is finite. \smallskip The following is not needed in this work, however, we cannot resist not to state it. \begin{lemma} Let $\pi$ be an irreducible representation of $G_n$. If the irreducible subquotients of ${}^{(i)}\pi$ are multiplicity free, then ${}^{(i)}\pi$ is a direct sum of its irreducible subquotients. \end{lemma} \begin{proof} The key observation is that, in view of Lemma \ref{lem_key_isomorphism}, we have two ways to compute ${}^{(i)}\pi$: \[ {}^{(i)}\pi= \theta_{n-i}(\theta_n(\pi)^{(i)})= ((\pi^{\vee})^{(i)})^{\vee}. \] Since $\pi$ is irreducible, we have $\theta_n(\pi)\cong \pi^{\vee}$, and if we denote by $\sigma$ either of two isomorphic representations $\theta_n(\pi)^{(i)}$ and $(\pi^{\vee})^{(i)}$, we see that on one hand ${}^{(i)}\pi$ is obtained from $\sigma$ by applying a co-variant functor $\theta$, and on the other hand by applying the contra-variant functor taking the contragradient. Since these two functors coincide on irreducible representations, corollary follows. \end{proof} \begin{lemma} \label{L:sec_adj_ext} For any smooth representation $\pi$ of $G_n$ and smooth representation $\sigma$ of $G_{n-i}$, \[ \Ext^j_{G_n} (\mathrm{ind}_{R_{n-i}}^{G_n}(\sigma \otimes \bar\psi_{i}), \pi) \cong \Ext^j_{G_{n-i}} (\sigma, {}^{(i)}\pi). \] \end{lemma} \begin{proof} In order to compute the right hand side we need to use a projective resolution of $\sigma$. Using the induction in stages, \[ \mathrm{ind}_{R_{n-i}}^{G_n}(\sigma \otimes \bar\psi_{i}) \cong \mathrm{Ind}_{P_{n-i}}^{G_{n}}(\sigma \boxtimes \ind_{U_i}^{G_i}(\bar\psi_i)). \] The Gelfand-Graev representation $\ind_{U_i}^{G_i}(\bar \psi_i)$ is projective by \cite{CS}. Thus, if $\sigma$ is projective it follows that $\mathrm{ind}_{R_{n-i}}^{G_n}(\sigma \otimes \bar\psi_{i})$ is projective, since the parabolic induction takes projective modules into projective modules. So we have shown that taking a projective resolution of $\sigma$ also gives a projective resolution of $\mathrm{ind}_{R_{n-i}}^{G_n}(\sigma \otimes \bar\psi_{i})$. Hence lemma follows from Lemma \ref{lem second adjointness bz der}. \end{proof} \subsection{Zelevinsky segments} Here we follow \cite{Ze}. Let $\rho$ be a cuspidal representation of $G_r$. For any $a, b \in \mathbb{C}$ with $b-a \in \mathbb{Z}_{\geq0}$, we have a Zelevinsky segment let $\Delta=[ \nu^{a}\rho, \nu^{a+1}\rho, \ldots, \nu^{b}\rho ]$. The absolute length of $\Delta$ is defined to be $r(b-a+1)$, and the relative $b-a+1$. We can truncate $\Delta$ form each side to obtain two segments of absolute length $r(b-a)$: \[ {}^{-}\Delta= [\nu^{a+1}\rho, \ldots, \nu^{b}\rho ] \text{ and } \Delta^-= [ \nu^{a}\rho, \ldots, \nu^{b-1}\rho ]. \] Moreover, if we perform the truncation $k$-times, the resulting segments will be denoted by ${}^{(k)}\Delta$ and $\Delta^{(k)}$. The induced representation $\nu^{a}\rho \times \nu^{a+1}\rho \times \ldots \times \nu^{b}\rho$ contains a unique irreducible submodule denoted by $\langle\Delta\rangle$. \begin{proposition} \label{P:derivative_segment} Let $i>0$ be an integer. The $i$-th left and right derivatives of $\langle\Delta\rangle$ vanish unless $i=r$ when \[ {}^{(r)}\langle\Delta\rangle=\langle{}^{-}\Delta\rangle \text{ and } \langle\Delta\rangle^{(r)}= \langle \Delta^-\rangle. \] \end{proposition} \begin{corollary} \label{C:degenerate} Let $\pi$ be an irreducible subquotient of $\langle \Delta_1 \rangle \times \ldots \times \langle \Delta_k \rangle$. If a right derivative of $\pi$ is generic, then every $\Delta_j$ is of the relative length one or two, and if the relative length is two, then $\Delta_j^-$ contributes to the cuspidal support of the right derivative of $\pi$. Similarly, if a left derivative of $\pi$ is generic, then ${}^-\Delta_j$ contributes to the cuspidal support of the left derivative. \end{corollary} \begin{proof} Observe that $\langle\Delta\rangle$ is generic if and only if the relative length of $\Delta$ is one. A right derivative of $\langle \Delta_1 \rangle \times \ldots \times \langle \Delta_k \rangle$ has a filtration whose subquotients are $\langle \Delta'_1 \rangle \times \ldots \times \langle \Delta'_k \rangle$ where $\Delta_j'$ is $\Delta_j$ or $\Delta^-_j$. This representation is generic if and only if the relative length of every $\Delta_j$ is one or two, and if it is two then $\Delta_j'=\Delta^-_j$. \end{proof} We summarize some other results from \cite{Ze} that we shall need. The induced representation $\nu^{a}\rho \times \nu^{a+1}\rho \times \ldots \times \nu^{b}\rho$ also contains a unique irreducible quotient denoted by $\mathrm{St}(\Delta)$. This representation is an essentially square integrable representation i.e. its matrix coefficients are square integrable when restricted to the derived subgroup. Every essentially square integrable representation is isomorphic to $\St(\Delta)$ for some segment $\Delta$. \begin{proposition} Let $i>0$ be an integer. The $i$-th left and right derivatives of $\mathrm{St}(\Delta)$ vanish unless $i=jr$ when \[ {}^{(i)}\mathrm{St}(\Delta)=\St(\Delta^{(j)}) \text{ and } \St(\Delta)^{(i)}= \St({}^{(j)} \Delta). \] \end{proposition} Let $\mathfrak m=\{ \Delta_1, \ldots , \Delta_k\}$ be a multisegment i.e a multiset of segments. One say that $\mathfrak m$ is generic if no two segments are linked. Then $\St(\Delta_1) \times \ldots \times\St(\Delta_k)$ is an irreducible generic representation, and every such representation arises in this way. We shall denote this representation by $\St(\mathfrak m)$. \section{Bernstein-Zelevinsky filtration} In this section we begin our study of the restriction problem from $G_{n+1}$ to $G_n$. Using the second adjointness formula, for both left and right derivatives, we prove that degenerate representations of $G_n$ cannot be quotients of essentially square integrable representations of $G_{n+1}$. \subsection{Bernstein-Zelevinsky functors} Let $M_{n+1} \subseteq G_{n+1}$ be the mirabolic subgroup \[ M_{n+1}= \left\{ \begin{pmatrix} g & u \\ 0 & 1 \end{pmatrix} : g \in G_{n}, u \in \mathrm{Mat}_{n,1}(F) \right\} . \] be the mirabolic subgroup of $G_{n+1}$. We have an obvious Levi decomposition $M_{n+1}=G_n E_n$. Abusing notation, let $\psi$ be the character of $E_n$ defined by $\psi(u)=\psi(u_n)$ where $u_n$ is the bottom entry of the column vector $u$. Note that the stabilizer of $\psi$ in $G_n$ is $M_n$. We have a pair of functors \[ \Phi^- : \Alg(M_{n+1}) \rightarrow \Alg(M_n) \text { and } \Phi^+ : \Alg(M_{n}) \rightarrow \Alg(M_{n+1}) \] defined by $\Phi^-(\tau)= \tau_{E_n,\psi}$ and $\Phi^+(\tau)=\ind_{M_n E_n}^{M_{n+1}}( \tau \boxtimes \psi)$. We also have a pair of functors \[ \Psi^- : \Alg(M_{n+1}) \rightarrow \Alg(G_n) \text { and } \Psi^+ : \Alg(G_{n}) \rightarrow \Alg(M_{n+1}) \] where $\Psi^-(\tau) = \tau_{E_n}$ and $\Psi^+$ is simply the inflation. All functors are normalized as in \cite{BZ}. Any $\tau\in \Alg(M_{n+1})$ has an $M_{n+1}$-filtration \[ \tau_n \subset \ldots \subset \tau_0 =\tau \] where, $\tau_i= (\Phi^+)^{i}(\Phi^-)^{i}(\tau)$, and \[ \tau_i/\tau_{i+1}= (\Phi^+)^{i}\Psi^+ \Psi^-(\Phi^-)^{i}(\tau). \] Observe that $\Psi^-(\Phi^-)^{i}(\tau)= \tau^{(i+1)}$, is the $(i+1)$-th derivative, and the subquotients of the filtration, considered as $G_n$-modules, are \[ \tau_{i}/\tau_{i+1} \cong \mathrm{ind}_{R_{n-i}}^{G_{n}}(\nu^{1/2} \cdot \tau^{(i+1)} \boxtimes \psi_i), \] where we have used notation from the previous section. In particular, $\tau_n$ is a multiple of the Gelfand-Graev representation. We derive some consequences of this filtration that we shall need later. \begin{lemma} \label{L:finite_gen} Let $\tau \in \Alg(M_{n+1})$ such that its derivatives are all finitely generated. When $\tau$ is considered a $G_n$-module, its Bernstein components are finitely generated. \end{lemma} \begin{proof} Recall that $P_{n-i}\supseteq R_{n-i}$ is the maximal parabolic subgroup of $G_n$ with the Levi factor $G_i\times G_{n-i}$. Using induction in stages, the $i$-th subquotient in the Bernstein-Zelevinsky filtration of $\Pi$ can be written as \[ \mathrm{Ind}_{P_{n-i}}^{G_{n}}(\nu^{1/2} \cdot \tau^{(i+1)} \boxtimes \ind_{U_i}^{G_i}(\psi_i)). \] By the assumption $\tau^{(i+1)}$ is a finitely generated $G_{n-i}$-module and the Bernstein components of the Gelfand-Graev representation $\ind_{U_i}^{G_i}(\psi_i)$ are finitely generated \cite{BH}. Lemma follows since parabolic induction sends finitely generated modules to finitely generated modules by 3.11 in \cite{BD}. \end{proof} \begin{lemma} \label{L:key} Let $\pi_1 \in \Alg(G_{n+1})$ and $\pi_2$ an admissible representation of $G_{n}$. If $\pi_2$ is a quotient of $\pi_1$ then, for some $i,j\geq 0$. \[ \Hom_{G_{n-i}}(\nu^{1/2}\cdot \pi_1^{(i+1)}, {}^{(i)}\pi_2 ) \neq 0 \text{ and } \Hom_{G_{n-j}}(\nu^{-1/2}\cdot {}^{(j+1)}\pi_1, \pi_2^{(j)} ) \neq 0. \] \end{lemma} \begin{proof} In order to prove the first isomorphism, we restrict $\pi_1$ to $G_{n}$, by way of $M_{n+1}$, and use the second adjointness formula. For the second we restrict to $G_{n}$, by way of $M^{\top}_{n+1}$, i.e. we reverse the roles of left and right derivatives. \end{proof} \subsection{Essentially square integrable representations} \begin{theorem} \label{T:St_generic} Let $\Delta=[\nu^a\rho, \ldots , \nu^b \rho]$ be a segment of absolute length $n+1$, where $\rho$ is a cuspidal representation of $G_r$. Let $\pi$ be an irreducible $G_n$-module. If $\pi$ is a quotient of $\St(\Delta)$ then $\pi$ is generic. \end{theorem} \begin{proof} Let $l=b-a+1$, in particular, $n+1=lr.$ Assume that $\pi$ is degenerate. Let $\mathfrak m=\{ \Delta_1, \ldots, \Delta_k\}$ be a multi-segment, from the Zelevinsky classification, such that $\pi$ is the unique submodule of $\langle \Delta_1 \rangle \times \ldots \times \langle \Delta_k \rangle$. Since $\pi$ is degenerate, by 9.10 in \cite{Ze} one segment in $\mathfrak m$ has the relative length at least two. If $\pi$ is a quotient of $\St(\Delta)$ then, by Lemma \ref{L:key}, ${}^{(i)}\pi$ contains $\nu^{1/2}\cdot \St(\Delta)^{(i+1)}$ as a generic submodule for some $i$. Now we can apply Corollary \ref{C:degenerate} : the relative length of each segment in $\mathfrak m$ is one or two, and one of them is $[\nu^{c-1/2}\rho, \nu^{c+1/2}\rho]$ where $\nu^{c+1/2}\rho$ contributes to the cuspidal support of $\nu^{1/2}\cdot \St(\Delta)^{(i+1)}$. It follows that $\nu^{1/2}\cdot \St(\Delta)^{(i+1)}$ is a generalized Steinberg representation corresponding to a segment ending in $\nu^{b+1/2}$, and containing $\nu^{c+1/2}\rho$. Thus, for every $d=c, \ldots, b$, $\nu^{d+1/2}\rho$ contributes to the cuspidal support of ${}^{(i)}\pi$ as well as to the cuspidal support of $\pi$. Similarly, if we use the second identity in Lemma \ref{L:key}, then for every $d= a, \ldots, c$, $\nu^{d-1/2}\rho$ contributes to the cuspidal support of $\pi$. We see that $\mathfrak m$ contains segments of total relative length $\geq l$ and absolute length $(l+1)r=n+1+r >n$. This is a contradiction. \end{proof} \section{Vanishing of Ext's} The purpose of this section is to prove: \begin{theorem} Let $\pi_1$ be an irreducible generic representation of $G_{n+1}$ and $\pi_2$ an irreducible generic representation of $G_{n}$. Then \[ \Ext_{G_n}^i( \pi_1, \pi_2)=0 \text{ if } i >0 \text{ and } \cong \mathbb C \text{ if } i=0. \] \end{theorem} Let us explain the strategy of the proof. Fix $\pi_2$, and assume that $\pi_2$ is a subquotient of $\rho_1\times \ldots \times \rho_k$ where $\rho_i$ are cuspidal representations. Let $m(\pi_1)$ be the integer that counts the number of cuspidal representations $\rho$ in the support of $\pi_1$ such that $\rho$ is an unramified twist of a $\rho_i$, for some $1\leq i \leq k$. The proof is by induction on $m(\pi_1)$. The base case $m(\pi_1)=0$ is easy. It is deduced from the Bernstein-Zelevinsky filtration of $\pi_1$ where the bottom piece is the Gelfand-Graev representation of $G_n$. Assume now that $\pi_1=\St(\mathfrak m_1)$ and $\pi_2=\St(\mathfrak m_2)$ for a pair of generic multisegments $\mathfrak{m}_1$ and $\mathfrak{m}_2$ i.e. no two segments in $\mathfrak{m}_i$ are linked. Let $\Delta=[\nu^a\rho, \ldots , \nu^b\rho]$ be a segment in $\mathfrak m_1$ such that $\rho$ contributes to $m(\pi_2)$. Assume that $\Delta$ is also a shortest such segment. Write $\pi_1=\St(\Delta) \times \pi$ where $\pi=\St(\mathfrak m)$ and $\mathfrak m=\mathfrak m_1 \setminus \Delta$. Let $r$ be the integer such that $\rho \in \Alg(G_r)$. Let $\rho'\in \Alg(G_r)$ be another cuspidal representation such that no unramified twist of $\rho'$ appears in the cuspidal supports of $\pi_1$ and $\pi_2$. Now both $\rho'\times \St({}^{-}\Delta) \times \pi$ and $\rho'\times \St(\Delta^-) \times \pi \in \Alg(G_{n+1})$ are irreducible and satisfy the induction assumption. We shall use this information to prove the theorem for $\pi_1$. \subsection{Transfer} Let $l=s+r$. Recall that $P_s$ is the maximal parabolic of $G_l$ with the Levi $G_s\times G_{r}$. Starting with $\sigma \in \Alg(G_s)$ and $\tau\in \Alg(M_r)$ one can manufacture two representations of $M_l$. The first one is obtained by the (normalized) induction from $P_s\cap M_l$ and, abusing notation, denoted by $\sigma\times \tau$. The second is obtained by the normalized induction from $P_s ^{\top}\cap M_l$ but only after $\sigma$ is multiplied by $\nu^{-1/2}$, see \cite{BZ} page 457, where the definition uses a different subgroup, but conjugated in $M_l$. This representation is denoted by $\tau\times \sigma$. Our interest in these representations comes from the following, 4.13 Proposition in \cite{BZ}. \begin{proposition} \label{P:BZ} Let $\rho\in \Alg(G_r)$, $\sigma \in \Alg(G_s)$ and $\tau\in \Alg(M_r)$. Let $\rho\vert_M$ and $\sigma\vert_M$ denote restrictions to $M_r$ and $M_s$, respectively. \begin{enumerate} \item There exists an exact sequence in $\Alg(M_l)$ \[ 0\rightarrow (\rho\vert_{M}) \times \sigma \rightarrow \rho \times \sigma \rightarrow \rho \times (\sigma \vert_{M}) \rightarrow 0 \] \item If $\Omega$ is any of the four functors $\Phi^{\pm}$ and $\Psi^{\pm}$, then \[ \Omega(\sigma\times \tau)=\sigma\times \Omega(\tau). \] \item $\Psi^-(\tau\times \sigma)= \Psi^-(\tau) \times \sigma $, and there exists an exact sequence in $\Alg(M_{l-1})$ \[ 0\rightarrow \Phi^-(\tau)\times \sigma \rightarrow \Phi^-(\tau\times \sigma) \rightarrow \Psi^-(\tau) \times (\sigma \vert_{M}) \rightarrow 0 \] \end{enumerate} \end{proposition} \begin{proposition} \label{P:transfer} Let $\Delta=[\nu^a\rho, \ldots, \nu^b\rho]$ be a segment where $\rho\in \Alg(G_r)$. Let $\tau_r=(\Phi^+)^{r-1}(1) \in \Alg(M_r)$, the Gelfand-Graev module. Then $\St(\Delta)\vert_{M}$ is isomorphic to $\tau_r \times \St({}^-\Delta)$. \end{proposition} \begin{proof} Recall that $\rho\vert_{M}\cong \tau_r$ (this is true for every cuspidal representation). Note that $\St(\Delta)$ is a quotient of $\nu^a \rho \times \St({}^-\Delta)$. By Proposition \ref{P:BZ} (1), we have an exact sequence of mirabolic subgroup modules \[ 0\rightarrow \tau_r \times \St({}^-\Delta) \rightarrow \nu^a \rho \times \St({}^-\Delta) \rightarrow \nu^a\rho \times (\St({}^-\Delta) \vert_{M}) \rightarrow 0 \] By Proposition \ref{P:BZ} (2), any derivative of the quotient in the above sequence is equal to $\nu^a\rho \times \St({}^{(k)}\Delta)$ with $k>1$. Since $\nu^a\rho$ and ${}^{(k)}\Delta$ are not linked, the corresponding subquotients in the Bernstein-Zelevinsky filtration are irreducible as mirabolic subgroup modules. Observe that they are non-isomorphic to the subquotients of the Bernstein-Zelevinsky filtration of $\St(\Delta)$. Hence the projection from $\nu^a \rho \times \St({}^-\Delta)$ onto $\St(\Delta)$, restricted to $\tau_r \times \St({}^-\Delta)$ gives the desired isomorphism. \end{proof} Now we arrive to a key result: \begin{corollary} \label{C:transfer} Let $\rho, \rho'\in \Alg(G_r)$ be two irreducible cuspidal representations. Let $\Delta=[\nu^a\rho, \ldots, \nu^b\rho]$, and $\pi\in \Alg(G_s)$. Then we have an isomorphism of mirabolic modules \[ \St(\Delta)\vert_{M} \times \pi \cong \rho'\vert_{M} \times (\St({}^-\Delta)\times \pi). \] \end{corollary} \begin{proof} By Proposition \ref{P:transfer}, we can substitute $\St(\Delta)\vert_{M}= \tau_r \times \St({}^-\Delta)$. Furthermore, we have a natural isomorphism \[ (\tau_r \times \St({}^-\Delta)) \times \pi \cong \tau_r \times (\St({}^-\Delta)) \times \pi) \] given by the induction in stages in two different orders. We finish by observing that $\tau_r= \rho'\vert_{M}$. \end{proof} Now we continue with the proof of vanishing for $\pi_1=\St(\Delta)\times \pi$, notation as in the start of the section. By Proposition \ref{P:BZ} (1) there is an exact sequence in $\Alg(M_{n+1})$ \[ 0\rightarrow (\St(\Delta)\vert_{M}) \times \pi \rightarrow \St(\Delta)\times \pi \rightarrow \St(\Delta)\times (\pi \vert_{M}) \rightarrow 0. \] Likewise, there is an exact sequence in $\Alg(M_{n+1})$ \[ 0\rightarrow \rho'\vert_{M} \times (\St({}^-\Delta)\times \pi) \rightarrow \rho'\times (\St({}^-\Delta)\times \pi) \rightarrow \rho'\times( \St({}^-\Delta)\times \pi )\vert_{M} \rightarrow 0. \] Note that the submodules in the two sequences are isomorphic by Corollary \ref{C:transfer}. Furthermore, by the choice of $\rho'$, \[ \Ext_{G_n}^i( \rho'\times( \St({}^-\Delta)\times \pi )\vert_{M} , \pi_2)=0 \text{ if } i\geq 0. \] Now we can apply the induction assumption to $\rho'\times \St({}^-\Delta)\times \pi$ and conclude that \[ \Ext_{G_n}^i((\St(\Delta)\vert_{M}) \times \pi , \pi_2)=0 \text{ if } i >0 \text{ and } \cong \mathbb C \text{ if } i=0. \] Hence, in order to establish the conjecture for the pair $(\pi_1,\pi_2)$, it suffices to show that \[ \Ext_{G_n}^i( \St(\Delta)\times (\pi \vert_{M}), \pi_2)=0 \text{ if } i \geq 0, \] and to do this it suffices to show vanishing for each subquotient in the Bernstein-Zelevinsky filtration of $\St(\Delta)\times (\pi \vert_{M})$. By Proposition \ref{P:BZ} part (2), the derivatives of $\St(\Delta)\times (\pi \vert_{M})$ are computed on the second factor. Thus, combining with the second adjointness formula, it suffices to show that \begin{itemize} \item $\Ext_{G_n}^j( \nu^{1/2}\St(\Delta)\times \pi^{(i+1)}, {}^{(i)}\pi_2)=0$ for $i,j\geq 0$. \end{itemize} Alternatively, by reversing the roles of left and right derivatives, it suffices to show that \begin{itemize} \item $\Ext_{G_n}^j( \nu^{-1/2}\St(\Delta)\times {}^{(i+1)}\pi , \pi^{(i)}_2)=0$ for $i,j\geq 0$. \end{itemize} \smallskip Hence it suffices to show that the Bernstein center spectra of $\nu^{1/2}( \St(\Delta) \times \pi^{(i+1)})$ and of ${}^{(i)}\pi_2$ are different for all $i$, or, they are different for $\nu^{-1/2}( \St(\Delta) \times {}^{(i+1)} \pi)$ and ${}\pi_2^{(i)}$ for all $i$. The strategy is to show that, if both statements fail, then $\mathfrak m_2$ contains linked segments. \subsection{Combinatorics} Let $\mathfrak m=\{ \Delta_1, \ldots , \Delta_k\}$ be a multisegment. Then $\St(\mathfrak m)$ is generic but reducible if some segments are linked. However, if $\Delta_i$ and $\Delta_j$ are linked, then they can be replaced by $\Delta_i \cap \Delta_j$ and $\Delta_i\cup \Delta_j$. This process (called recombination henceforth) eventually leads to a generic segment such that the corresponding irreducible generic representation is the unique generic subquotient in $\St(\mathfrak m)$. Important observation is that the points in the spectrum of the Bernstein center are uniquely represented by generic multisegments! The following is a key lemma. \begin{lemma} Let $\mathfrak m$ be a generic multisegment and $\mathfrak m'$ a multisegment obtained by truncating $\mathfrak m$ from the right. Then the generic segment corresponding to $\mathfrak m'$ by recombination is also obtained from $\mathfrak m$ by truncating from the right. \end{lemma} \begin{proof} This is proved by induction on the number of steps in the recombination process. If that number is 0 there is nothing to prove. Otherwise there is a pair of linked segments $\Delta'$ and $\Delta''$ in $\mathfrak m'$ such that the first step in the recombination is replacing $\Delta'$ and $\Delta''$ by $\Delta' \cap \Delta''$ and $\Delta' \cup \Delta''$. It is trivial to see that the resulting multisegment is also obtained by right truncation from $\mathfrak m$. The proof follows by induction. \end{proof} \subsection{Finishing the proof} Let $l=b-a+1$ be the relative length of $\Delta$. We note that ${}^{(i)}\pi_2$ is glued from $\St(\mathfrak m'_2)$ where $\mathfrak m'_2$ runs over all multisegments obtained from $\mathfrak m_2$ by truncating from the right $i$-times (in the sense of absolute length). By the previous lemma the Bernstein center spectrum of ${}^{(i)}\pi_2$ is given by such generic multisegments. Likewise, $\St(\Delta)\times \pi^{(i+1)}$ is glued from $\St(\Delta) \times \St({}'\mathfrak m)$ where ${}'\mathfrak m$ runs over all multisegments obtained from $\mathfrak m$ by truncating from the left $i+1$-times, and to determine the Bernstein center spectrum we need to consider only generic ${}'\mathfrak m$. However, $\{\Delta\} \cup {}'\mathfrak m$ needs not be generic. There could be segments in ${}'\mathfrak m$ linked to $\Delta$. Since $\Delta$ is not linked to any segment in $\mathfrak m$ and ${}'\mathfrak m$ is obtained from $\mathfrak m$ by left truncation, it follows that linking occurs over the right end point of $\Delta$. Let $\Delta_0$ be the longest segment in ${}'\mathfrak m$ linked to $\Delta$. It is easy to see that $\Delta \cup \Delta_0$ is a segment in the generic multisegment corresponding to $\{\Delta\} \cup {}'\mathfrak m$ by the recombination process. Note that $\Delta \cup \Delta_0$ starts with $\nu^a\rho$ and is of relative length at least $l$. Thus the Bernstein spectra of $\nu^{1/2} (\St(\Delta)\times \pi^{(i+1)})$ and ${}^{(i)}\pi_2$ can have a point in common only if $\mathfrak m_2$ contains a segment starting with $\nu^{a+1/2}\rho$ and of relative length at least $l$. Similarly, the Bernstein spectra of $\nu^{-1/2}( \St(\Delta)\times {}^{(i+1)}\pi)$ and ${}\pi_2^{(i)}$ can have a point in common only if $\mathfrak m_2$ contains a segment ending with $\nu^{b-1/2}\rho$ and of length at least $l$. In other words we have constructed a pair of linked segments in $\mathfrak m_2$, a contradiction. This completes the proof of the Ext-vanishing theorem. \section{Hecke algebra methods} The main goal of this section is to prove projectivity of an essentially square integrable representation $\pi_1$ of $G_{n+1}$ when restricted to $G_n$. The proof uses Hecke algebras and identifies all Bernstein components of $\pi_1$ with the sign-projective module of the Hecke algebra corresponding to the Bushnell-Kutzko type \cite{BK1}, \cite{BK2}, \cite{BK3}. As a consequence, any two essentially square integrable representations of $G_{n+1}$ are isomorphic when restricted to $G_n$. \subsection{Hecke algebras} Let $\Delta=[ \nu^a\rho, \ldots, \nu^b\rho ]$ be a Zelevinsky segment. Let $m=b-a+1$. The Bernstein component of $\St(\Delta)$ is equivalent to the category of representations of a Hecke algebra $\mathcal H_m$ arising from a simple Bushnell-Kutzko type $\tau_{\Delta}$, that is, if $\pi$ is a smooth representation in the Bernstein component, then $\Hom(\tau_{\Delta}, \pi)$ is the corresponding $\mathcal H_m$-module. The algebra $\mathcal H_m$ is isomorphic to the Iwahori Hecke algebra of $\GL_m(E)$, for some field $E$. Thus, as an abstract algebra, $\mathcal H_m$ is generated by $\theta_1, \ldots, \theta_{m}$, and $T_w$ ($w \in S_m$) satisfying the following relations: \begin{enumerate} \item $\theta_k\theta_l=\theta_l\theta_k$ for any $k,l=1,\ldots, m$; \item $T_{s_k}\theta_k-\theta_{k+1}T_{s_k}=(q-1)\theta_{k}$, where $q$ is a prime power depending on $\tau_{\Delta}$ and $s_k$ is the transposition of numbers $k$ and $k+1$; \item $T_{s_k}\theta_l =\theta_lT_{s_k}$ if $l \neq k, k+1$; \item $(T_{s_k}-q)(T_{s_k}+1)=0$, where $s_k$ is as in (2), and $T_w$ satisfies a braid relation. \end{enumerate} Let $\mathcal A_m=\mathbb C[\theta_1^{\pm 1}, \ldots , \theta_m^{\pm 1}]$ and $\mathcal H_{S_m}$ be the span of $T_w$, $w\in S_m$. Then $\mathcal H_m \cong \mathcal A_m \otimes \mathcal H_{S_m}$. The finite dimensional algebra $\mathcal H_{S_m}$ has a one dimensional sign representation $\mathrm{sgn}(T_w)= (-1)^{\ell(w)}$, where $\ell$ is the length function on $S_m$. An irreducible representation $\pi$ in the component is Whittaker generic if and only if $\Hom(\tau_{\Delta},\pi)$ contains the sign type as $\mathcal H_{S_m}$-module \cite{CS}. \smallskip Let $\Delta_1, \ldots, \Delta_r$ be segments such that for $i\neq j$, the cuspidal representations $\rho_i$ and $\rho_j$ are not unramified twists of each other. The Bernstein component of $\St(\Delta_1)\times \cdots \times \St(\Delta_r)$ is equivalent to the category of representations of a Hecke algebra $\mathcal H$ arising from a semi-simple Bushnell-Kutzko type $\tau$. We have $\mathcal H\cong \mathcal H_{m_1} \otimes \cdots \otimes \mathcal H_{m_r}$ and $\mathcal H \cong \mathcal A \otimes \mathcal H_{S}$ where $\mathcal A\cong \mathcal A_{m_1} \otimes \cdots \otimes \mathcal A_{m_r}$ and $\mathcal H_{S}\cong \mathcal H_{S_{m_1}} \otimes \cdots \otimes \mathcal H_{S_{m_r}}$. The subalgebra $\mathcal A$ is isomorphic to the ring of Laurent polynomials in $m=m_1+ \ldots + m_r$ variables, while $\mathcal H_{S}$ is spanned by $T_w$, $w\in S=S_{m_1} \times \cdots \times S_{m_r}$. An irreducible representation $\pi$ in the component can be written as $\pi_1\times \cdots \times \pi_r$ where $\pi_i$ is in the component of $\St(\Delta_i)$, thus it clear that $\pi$ is Whittaker generic if and only if $\Hom(\tau, \pi)$ contains the sign type of $\mathcal H_S$. \subsection{Some projective modules} Let $\chi$ be a character of $\mathcal A$. The $\mathcal H$-module $\mathcal H \otimes_{\mathcal A} \chi$ is called the principal series representation of $\mathcal H$. A twisted Steinberg representation of $\mathcal H$ is any one-dimensional $\mathcal H$-module such that the restriction to $\mathcal H_S$ is the sign type. For example, if $\pi= \St(\Delta_1)\times \cdots \times \St(\Delta_r)$, then $\Hom(\tau, \pi) $ is a twisted Steinberg representation. The following is Theorem 2.1 in \cite{CS}. (It is stated there for $\mathcal H$ arising from the singleton partition $(m)$ but the proof is applicable to a general partition $(m_1, \ldots , m_r)$). \begin{theorem} \label{T:IMRN} Let $\Pi$ be an $\mathcal H$-module. Assume that \begin{enumerate} \item $\Pi$ is projective and finitely generated. \item $\dim \Hom_{\mathcal H}(\Pi, \pi) \leq 1$ for an irreducible principal series representation $\pi$. \item A twisted Steinberg representation is a quotient of $\Pi$. \end{enumerate} Then $\Pi \cong \mathcal H\otimes_{\mathcal H_S} \mathrm{sgn}$. \end{theorem} As in \cite{CS}, we have the following corollary. \begin{corollary} \label{C:GG} Let $\Gamma$ be the summand of the Gelfand-Graev representation corresponding to the Bernstein component of $\St(\Delta_1)\times \cdots \times \St(\Delta_r)$. Then we have an isomorphism $\Hom(\tau, \Gamma) \cong \mathcal H\otimes_{\mathcal H_S} \mathrm{sgn}$ of $\mathcal H$-modules. \end{corollary} \subsection{Projectivity for Hecke algebras} Let $\mathcal Z$ be the center of $\mathcal H$. Recall that $\mathcal Z=\mathcal A^S$, in particular, $\mathcal H$ is a finitely generated $\mathcal Z$-module. Let $\mathcal J$ be a maximal ideal in $\mathcal Z$. Let $\hat{\mathcal H}$ denote the $\mathcal J$-adic completion of $\mathcal H$ \cite{AMD}. For every $\mathcal H$-module $\Pi$, let $\hat\Pi$ denote the $\mathcal J$-adic completion of $\Pi$. If $\Pi$ is finitely generated, then $\hat \Pi\cong \hat{\mathcal H} \otimes_{\mathcal H} \Pi$. \begin{theorem} \label{T:local_projectivity} Let $\Pi$ be a finitely generated $\mathcal H$-module and $\mathcal J$ a maximal ideal in $\mathcal Z$. Let $\pi$ be the unique irreducible $\mathcal H$-module annihilated by $\mathcal J$ and containing the sign type. Assume that \begin{enumerate} \item $\dim \Hom_{\mathcal H}(\Pi, \pi) = 1$ \item $\Pi$ has no other irreducible quotients annihilated by $\mathcal J$. \item $\Pi$ contains a torsion free element for $\mathcal A$. \end{enumerate} Then $\hat\Pi \cong \hat{\mathcal H}\otimes_{\mathcal H_S} \mathrm{sgn}$. \end{theorem} \begin{proof} In order to simplify notation, write $\Sigma=\mathcal H\otimes_{\mathcal H_S} \mathrm{sgn}$. Since $\Pi$ is finitely generated, $\hat{\Pi}/\mathcal J\hat{\Pi} \cong \Pi/\mathcal J\Pi$ is a finite dimensional $\mathcal H$-module, annihilated by $\mathcal J$. By (2) it must be generated by the sign type subspace. Let $r$ be the dimension of the sign type in $\Pi/\mathcal J\Pi$. By Frobenius reciprocity, we have a surjection $f:\Sigma^{\oplus r} \rightarrow \Pi/\mathcal J\Pi$ which descends to a surjection $\bar f:(\Sigma/\mathcal J \Sigma)^{\oplus r} \rightarrow \Pi/\mathcal J\Pi$. Observe that $\bar f$ is bijective on the sign type, since the sign type in $\Sigma/\mathcal J\Sigma$ is one-dimensional. Since $\pi$ is the unique irreducible quotient of $\Sigma/\mathcal J\Sigma$ and $\bar f$ is bijective on the sign type, it follows that $\pi^r$ is a quotient of $\Pi/\mathcal J\Pi$. This forces $r=1$ by (1) and, by Nakayama lemma, we have a surjection $\hat f: \hat{\Sigma} \rightarrow \hat{\Pi}$. Since $\hat{\Sigma}\cong \hat{\mathcal A}$, as $\hat{\mathcal A}$-modules, (3) implies that the surjection is in fact an isomorphism. \end{proof} \begin{corollary} \label{C:local_vanishing} Let $\Pi$ be a finitely generated $\mathcal H$-module and $\mathcal J$ a maximal ideal in $\mathcal Z$. Assume that the conditions of Theorem \ref{T:local_projectivity} are satisfied. Then, for all $\mathcal H$-modules $\sigma$ annihilated by $\mathcal J$ and for all $i>0$, \[ \mathrm{Ext}^i_{\mathcal H} (\Pi, \sigma) =0. \] \end{corollary} \begin{proof} Since $\sigma$ is annihilated by $\mathcal J$, we have \[ \mathrm{Ext}^i_{\mathcal H} (\Pi, \sigma) \cong \mathrm{Ext}^i_{\hat{\mathcal H}} (\hat{\Pi}, \sigma). \] The latter spaces are trivial by projectivity of $\hat{\mathcal H}\otimes_{\mathcal H_S} \mathrm{sgn}$. \end{proof} \begin{corollary} \label{C:global_vanishing} Let $\Pi$ be a finitely generated $\mathcal H$-module. Assume that the conditions of Theorem \ref{T:local_projectivity} are satisfied for every maximal ideal in $\mathcal Z$. Then $\Pi \cong \mathcal H\otimes_{\mathcal H_S} \mathrm{sgn}$. \end{corollary} \begin{proof} Corollary \ref{C:local_vanishing} implies that $\mathrm{Ext}^i_{\mathcal H} (\Pi, \sigma) =0$ for all finite length modules. Since $\Pi$ is also finitely generated, it is projective by Theorem 8.1 in the appendix of \cite{CS}. Now we can apply Theorem \ref{T:IMRN}. \end{proof} \subsection{Projectivity for groups} Now we can apply the Hecke-module results to the restriction problem, one Bernstein component at the time. Let $\pi_1$ be an irreducible generic representation of $G_{n+1}$ and fix a Bushnell-Kutzko type $\tau$ for $G_n$. Let $\Pi= \Hom(\tau, \pi_1)$ be the corresponding $\mathcal H$-module. Note that the conditions (1) and (3) in Theorem \ref{T:local_projectivity} are satisfied for every maximal ideal $\mathcal J$. Indeed, the condition (1) because all irreducible generic $G_n$-representations are quotients of $\pi_1$ with multiplicity one and (3) because $\pi_1$, restricted to $G_n$, contains the Gelfand-Graev representation, a free $\mathcal A$-module. Theorem \ref{T:local_projectivity} implies the following local Ext vanishing result for groups. \begin{theorem} \label{T:vanish_local} Let $\pi_1$ be an irreducible generic representation of $G_{n+1}$. Let $\mathcal J$ be a maximal ideal of the Bernstein center of $G_n$. Assume that no degenerate irreducible representation of $G_n$ annihilated by $\mathcal J$ is a quotient of $\pi_1$. Then $\mathrm{Ext}^i_{G_n} (\pi_1, \pi_2)=0$, $i>0$, for all irreducible representation $\pi_2$ of $G_n$ annihilated by $\mathcal J$. \end{theorem} Finally since, by Theorem \ref{T:St_generic}, essentially square integrable representation have no degenerate quotients, Corollary \ref{C:global_vanishing} implies: \begin{theorem}\label{T:projective_general} Let $\pi_1$ be an essentially square integrable representation of $G_{n+1}$. Then $\pi_1$ considered a $G_n$-module, is projective. Moreover, if $\pi'_1$ is another essentially square integrable representation of $G_{n+1}$ then $\pi_1$ and $\pi'_1$ are isomorphic as $G_n$-modules. \end{theorem} \section{Appendix} In this appendix we prove Lemma \ref{lem second adjointness bz der}, that is, the second adjointness isomorphism for Bernstein-Zelevisky derivatives. The key ingredient is Rodier's approximation \cite{Ro} of the Whittaker character by characters of compact pro-$p$ groups. \subsection{Groups} Let $F$ be a $p$-adic field, $R$ its ring of integers and $P$ the maximal ideal generated by a prime $\varpi$. Let $\psi$ be the character of $F$ of conductor $R$. Let $G=\GL_n(F)$ and let $U$ be the group of unipotent upper triangular matrices in $G$. Let $\psi_U: U \rightarrow \mathbb C$ be a Whittaker character defined by \[ \psi_U(u)=\psi(u_{1,2} + \cdots +u_{n-1,n}) \] where $u_{i,j}$ denote the entries of the matrix $u$. \smallskip For every natural number $r$, let $L_r$ be the lattice in $M_n(F)$ consisting of all matrices whose entries are in $P^r$. Then \[ K_r = 1 + L_r \] is a principal congruence subgroup of $G$. Let $t=(t_i)\in G$ be a diagonal matrix such that $t_i/t_{i+1}=\varpi^2$ for $i=1,\ldots, n-1$. Let $H_r=t^{-r} K_r t^{r}$. Let $B^{\top}$ be the Borel subgroup of lower-triangular matrices. Then we have a parhoric decomposition \[ H_r= (H_r\cap B^{\top}) (H_r\cap U). \] The sequence of groups $H_r\cap B^{\top}$ is decreasing with trivial intersection, while the sequence of groups $H_r\cap U$ is increasing with union $U$. Let $\psi_r$ be a character of $H_r$ defined by \[ \psi_r(g) = \psi( g_{1,2} + \cdots +g_{n-1,n}). \] Observe that \[ \psi_r|_{H_r \cap U} = \psi_U|_{H_r\cap U}. \] \subsection{Representations} Let $\pi$ be a smooth $G$-module. For every non-negative integer $r$ we have a projection map $P_r : \pi \rightarrow \pi^{H_r, \psi_r}$ defined by \[ P_r(v) = \mathrm{vol}(H_r)^{-1} \int_{H_r} \bar\psi_r(u) \pi(g) v ~dg. \] For $r\leq s $ we have maps $i_r^s : \pi^{H_r, \psi_r} \rightarrow \pi^{H_s, \psi_s}$ defined by restricting $P_s$ to $\pi^{H_r, \psi_r}$. From the parahoric decomposition of $H_r$, it is easy to see that \[ i_r^s(v) = \mathrm{vol}(H_s\cap U)^{-1}\int_{H_s\cap U} \bar\psi_s(u) \pi(u) v ~du. \] This formula, in turn, implies that these maps form a direct system i.e. $i_s^t\circ i_r^s = i_r^t$, for $r \leq s \leq t$. We have natural maps $i_r : \pi^{H_r, \psi_r} \rightarrow \pi_{U,\psi_U}$. Observe that $i_s\circ i_r^s = i_r$. Hence we have a map \[ i_\pi : \lim_r \pi^{H_r, \psi_r} \rightarrow \pi_{U,\psi_U}. \] \begin{proposition} \label{P:compact_approximation} For every smooth $G$-module $\pi$ the map $i_{\pi}$ is an isomorphism of vector spaces. \end{proposition} \begin{proof} Surjectivity: Let $v\in \pi$. Since $H_r\cap B^{\top} \rightarrow \{1\}$ there exists $r$ such that $v$ is $H_r\cap B^{\top}$-invariant. Let \[ w=\mathrm{vol}(H_r\cap U)^{-1} \int_{H_r\cap U} \bar\psi_r(u) \pi(u) v ~du \in \pi^{H_r,\psi_r}. \] Then $v$ and $w$ have the same projection on $\pi_{U,\psi_U}$. Injectivity: Let $v\in \pi^{H_r,\psi_r}$ that projects to $0$ in $\pi_{U,\psi_U}$. Then there exists an open compact subgroup $U_c \subset U$ such that \[ \int_{U_c} \bar\psi_s(u) \pi(u) v ~du =0. \] Since $H_s\cap U\rightarrow U$ there exists $s\geq r$ such that $ H_s\cap U \supset U_c$. Then the above integral, with $U_c$ substituted by $H_s\cap U$, vanishes. In other words, $i_r^s(v)=0$ and hence $v=0$, viewed as an element of the direct limit. \end{proof} For $r\leq s$ we have maps $p_r^s : \pi^{H_s, \psi_s} \rightarrow \pi^{H_r, \psi_r}$, going in the opposite direction, defined by restricting $P_r$ to $\pi^{H_s \psi_s}$. From the parahoric decomposition of $H_r$, it is easy to see that \[ p_r^s(v) = \mathrm{vol}(H_r\cap B^{\top})^{-1}\int_{H_r\cap B^{\top}} \pi(g) v ~dg \] and this implies that these maps form an inverse system i.e. $p_r^s\circ p_s^t = p_r^t$, for $r \leq s \leq t$. \smallskip By Proposition 4 in \cite{Ro} (see also VI, page 169), there exists an integer $r_0$, independent of $\pi$, such that $p_r^s \circ i_r^s$ is a non-zero multiple of the identity on $\pi^{H_r,\psi_r}$, if $r_0 \leq r \leq s$. Thus $i_r^s$ is an injection and $p_s^r$ is a surjection. It follows, from Proposition \ref{P:compact_approximation}, that the maps $i_r : \pi^{H_r,\psi_r} \rightarrow \pi_{U,\psi_U}$ are injections, for all $r\geq r_0$. \smallskip We shall use surjectivity of the maps $p_r^s$ to construct a natural complement of $\pi^{H_r,\psi_r}$ in $\pi_{U,\psi_U}$. So fix $r\geq r_0$, and for every $s\geq r$, let $\tau_s$ be the kernel of $p_r^s$. Observe that $\tau_s$ is a complement of $\pi^{H_r,\psi_r}$ in $\pi^{H_s,\psi_s}$, where we have identified $\pi^{H_r,\psi_r}$ with its image in $\pi^{H_s,\psi_s}$. We claim that $\tau_s$, for $s\geq r$ form an injective subsystem. To that end, let $t\geq s$. We need to prove if $v\in \tau_s$ then $i_s^t(v) \in \tau_t$, that is, $p_r^t (i_s^t (v))=0$. Write $p_r^t = p_r^s \circ p_s^t $. Then \[ p_r^t (i_s^t (v))= p_r^s \circ p_s^t ( (i_s^t (v))= p_r^s ( p_s^t \circ i_s^t (v)) = 0 \] where for the last equality we used that $p_s^t \circ i_s^t (v)$ is a multiple of $v$. Hence \[ \pi^{H_r,\psi_r}_c := \lim_{s\geq r} \tau_s \] is a complement of $\pi^{H_r,\psi_r}$ in $\lim_{s\geq r}\pi^{H_r,\psi_r} \cong \pi_{U,\psi_U}$. \smallskip We apply the above considerations to $\pi=S(G)$, the space of locally constant, compactly supported functions on $G$, considered a $G$-module with respect to the action by left translations. In this case, the vector spaces $\pi^{H_r,\psi_r}$ and $\pi_{U,\psi_U}$ are naturally $G$-modules, coming from the right translation action of $G$ on $S(G)$ and the maps $i_r^s$, $p_r^s$ and $i_r$ are $G$-morphisms. Observe that $S(G)^{H_r,\psi_r}= \ind_{H_r}^G(\psi_r)$ and $S(G)_{U,\psi_U}\cong \ind_{U}^G(\psi)$, the Gelfand-Graev representation. Hence $\lim_r \ind_{H_r}^G(\psi_r) \cong \ind_{U}^G(\psi)$, as $G$-modules. Moreover, if $r\geq r_0$, then $\ind_{H_r}^G(\psi_r)$ is a direct summand of $\ind_{U}^G(\psi)$. We record this in the following: \begin{proposition} \label{P:summand} For every $r\geq r_0$, $\ind_{H_r}^G(\psi_r)$ is a direct $G$-invariant summand of $\ind_{U}^G(\psi)$: \[ \ind_{U}^G(\psi) \cong \ind_{H_r}^G(\psi_r) \oplus \ind_{H_r}^G(\psi_r)_c. \] \end{proposition} \begin{proposition} \label{P:vanishing} Fix $r\geq r_0$. For almost all $s \geq r$, $(\ind_{H_s}^G(\psi_s)_c)^{K_r}$ is trivial. \end{proposition} \begin{proof} The key is the following lemma: \begin{lemma} Let $r\geq r_0$. Let $\pi$ be an irreducible Whittaker generic $G$-module such that $\pi^{K_r}\neq 0$. There exists a positive integer $m$, independent of $\pi$, such that $\pi^{H_{mr}\psi_{mr}} \neq 0$. \end{lemma} \begin{proof} The first step in the proof is a reduction to supercuspidal representations. Let $P=MN$ be a standard parabolic subgroup of block upper-triangular matrices. Assume that $\pi$ is a Whittaker generic subquotient of ${\mathrm{ Ind}}_{P^{\top}}^G\sigma$, where $P^{\top}$ is the transpose of $P$. Let $K=\GL_n(R)$. Using $G=P^{\top} K$ and normality of $K_r$ in $K$, it is easy to see that $\pi^{K_r}\neq 0$ implies that $\sigma^{K^M_r}\neq 0$ where $K^M_r=K_r\cap M$. Now assume that $\sigma^{H_s^M, \psi_s^M}\neq 0$ where $H^M_s=H_s\cap M$ and $\psi_s^M$ is the restriction of $\psi_s$ to $H_s^M$. Let $v\in \sigma^{H_s^M, \psi_s^M}$, and define $f\in {\mathrm {Ind}}_{P^{\top}}^G\sigma$, supported on $P^{\top} (H_s\cap N)$, such that $f(1)=v$ and that it is right $(\psi_s)|_{H_s\cap N}$-invariant. Then $f\in ({\mathrm {Ind}}_{P^{\top}}^G\sigma)^{H_s,\psi_s}$. This type must belong to the Whittaker generic subquotient of the induced representation by injectivity of the map $i_s$. It remains to deal with supercuspidal $\pi$. Let $\ell$ be a Whittaker functional on $\pi$, and for every $v\in \pi$ we have a Whittaker function $f_v(g)= \ell(\pi(g)v)$. Let $T(r) \subset T$ be the subset of $t=(t_1, \ldots, t_n)$ such $1/q^{(2m-2)r} \leq | t_i/t_{i+1} | \leq q^{(2m-2)r}$, for all $i$. By Theorem 2.1 in \cite{La}, there exists $m$, independent of $\pi$, such that $f_v$ is supported on $UT(r) K$ for all $v\in \pi^{K_r}$. Since $f_v$ is non-zero, for a non-zero $v$, there exists $t\in T(r)$ and $k\in K$ such that $\ell(\pi(tk) v) \neq 0$. Since $K$ normalizes $K_r$, $\pi(k) v\in \pi^{K_r}$. It follows that $\pi(tk)v$ is fixed by $tK_r t^{-1}$. Observe that this group contains $H_{mr} \cap B^{\top}$, by the definition of $T(r)$, hence \[ w= \mathrm{vol}( H_{mr} \cap U)^{-1} \int_{H_{mr} \cap U} \bar \psi_U(u) \pi(u) \pi(tk) v \in \pi^{H_{mr},\psi_{mr}} \] and it is non-zero since $\ell(w)= \ell(\pi(tk)v) \neq 0$. The lemma is proved. \end{proof} Take $s \geq mr$, where $m$ is as in the lemma. Recall that, by \cite{BH}, Bernstein's components of $\ind_{U}^G \psi_U$ are finitely generated and hence admit irreducible quotients. Thus, if $(\ind_{H_s}^G(\psi_s)_c)^{K_r} \neq 0$ then $\ind_{H_s}^G(\psi_s)_c$ has an irreducible quotient $\pi$ such that $\pi^{K_r} \neq 0$. Then $\pi^{H_s,\psi_s} \neq 0$ by the lemma, and hence $\dim_G(\ind_{U}^G \psi_U, \pi) \geq 2$, by Proposition \ref{P:summand}, a contradiction. \end{proof} \begin{proposition} \label{C:rodier} For every $G$-module $\pi$ generated by $\pi^{K_r}$, and every vector space $\sigma$ \[ \Hom_G(\sigma\otimes \ind_U^G \psi_U, \pi) \cong \Hom(\sigma, \pi_{U,\psi_U}). \] \end{proposition} \begin{proof} By Propositions \ref{P:summand} and \ref{P:vanishing}, for almost all $s\geq r $, \[ \Hom_G (\sigma \otimes \ind_U^G \psi_U, \pi)\cong \Hom_G(\sigma\otimes \ind_{H_s}^G(\psi_s), \pi). \] Let $\mathbb C[G]$ denote the group algebra of $G$. Then we can write \[ \sigma\otimes \ind_{H_s}^G(\psi_s)\cong \ind_{H_s}^G(\sigma\otimes \psi_s) \cong \mathbb C[G]\otimes_{\mathbb C[H_s]} (\sigma \otimes \psi_s). \] Hence, by the Frobenius reciprocity, \[ \Hom_G(\sigma\otimes \ind_{H_s}^G(\psi_s), \pi) \cong \Hom(\sigma, \pi^{H_s, \psi_s}). \] Now observe that the starting space $\Hom_G (\sigma \otimes \ind_U^G \psi_U, \pi)$ does not depend on $s$. It follows that the spaces $\pi^{H_s, \psi_s}$ are isomorphic for almost all $s$. In particular, $\pi^{H_s, \psi_s}\cong \pi_{U,\psi_U}$ for such $s$. Hence \[ \Hom_G (\sigma \otimes \ind_U^G \psi_U, \pi) \cong \Hom(\sigma,(\pi)_{U,\psi_U}). \] \end{proof} \subsection{Second adjointness} Now we are ready to prove Lemma \ref{lem second adjointness bz der}. We resume using the notation from the main body of the paper, in particular, $G_n=\GL_n(F)$, $U_n$ is the group of upper-triangular unipotent matrices, and $P_{n-i}=M_{n-i} N_{n-i}$ the standard maximal parabolic subgroup of block upper-triangular matrices with the Levi $G_{n-i}\times G_i$. Let $\pi$ be a smooth representation of $G_n$ generated by vectors fixed by the $r$-th principal congruence subgroup in $G_n$, and $\sigma$ a smooth representation of $G_{n-i}$, as in the statement of the lemma. Using the induction in stages, \[ \mathrm{ind}_{R_{n-i}}^{G_n}(\sigma \otimes \bar\psi_{i}) \cong \mathrm{Ind}_{P_{n-i}}^{G_{n}}(\sigma \boxtimes \ind_{U_i}^{G_i}(\bar \psi_i)). \] By the second adjointness isomorphism for parabolic induction, due to Bernstein, \[ \Hom_{G_n}(\mathrm{Ind}_{P_{n-i}}^{G_{n}}(\sigma \boxtimes \ind_{U_i}^{G_i}(\bar \psi_i)),\pi )\cong \Hom_{G_{n-i}\times G_i} (\sigma \boxtimes \ind_{U_i}^{G_i}(\bar \psi_i), \pi_{N_{n-i}^{\top}}). \] It is easy to see that $\pi_{N_{n-i}^{\top}}$, as a $G_i$-module, is also generated by vectors fixed by the $r$-th principal congruence subgroup in $G_i$. Thus we can apply Proposition \ref{C:rodier} to $G_i$ to derive \[ \Hom_{G_{n-i}\times G_i} (\sigma \boxtimes \ind_{U_i}^{G_i}(\bar \psi_i), \pi_{N_{n-i}^{\top}}) \cong \Hom_{G_{n-i}} (\sigma, {}^{(i)}\pi). \] \section{Acknowledgment} A part of this collaboration was carried out at the Weizmann Institute of Science during a program on the representation theory of reductive groups in 2017. Both authors would like to thank the organizers, Avraham Aizenbud, Joseph Bernstein, Dmitry Gourevitch and Erez Lapid, for their kind invitation to participate in the program. The second author is partially supported by an NSF grant DMS-1901745.

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TITLE: Comprehensive and self-contained treatment of Algebraic Geometry using Functor of Points approach QUESTION [15 upvotes]: The book everyone seems to use to study Algebraic Geometry is Hartshorne's book. However, I hear a good number of people saying that this book totally misses the functorial point of view. Hence, could you please recommend a good source to learn AG using the Functor of Points approach? Thanks!! REPLY [8 votes]: Dear Brian, it seems that algebraic geometers who adopt your favoured approach are essentially specialists in algebraic groups. My favourite example would be Jantzen's Representations of Algebraic groups", Academic Press 1987, in which all of Chapter 1 (18 pages) is devoted to the functor approach you require. Let me emphasize that Jantzen doesn't limit himself to affine schemes nor to group schemes. He considers completely general schemes defined as local functors admitting an open covering (in the functor sense!) consisting of affine schemes . I am sure you'll love the ingenious but natural definitions of open subfunctor, closed subfunctor, base change ... introduced in this meaty chapter: good luck!

. 5 posts • Page 1 of 1 - Founding Member - Posts: 4138 - Joined: Fri Apr 10, 2009 7:59 am Re: Hello from the UK! Hi Zephyr, Welcome to Dharma Wheel. Regards, Welcome to Dharma Wheel. Regards, Re: Hello from the UK! Zephyr wrote: reddit Upvote I find that the Kagyu teachings strike a particular chord with me. Upvote I love art and music, and create both, usually using software. I also really enjoy video games. Upvote Brad Warner Downvote Just kidding. Welcome Zephyr! grasping the letter of the text and ignoring its intention! - Vasubandhu - David N. Snyder - Site Admin - Posts: 2816 - Joined: Sun Apr 05, 2009 4:23 pm - Location: Las Vegas, Nevada Re: Hello from the UK! Welcome to Dharma Wheel! Re: Hello from the UK! Your not the Zephyr, are you? Kevin Kevin 5 posts • Page 1 of 1 Return to “Introductions” Who is online Users browsing this forum: No registered users and 20 guests

TITLE: Does this definite integral exist? QUESTION [2 upvotes]: So I have the following definite integral: ${\int_{0}^{1}x(2x^2-1)^{-10}}dx$ I suppose I cannot just integrate over the interval (0,1) because of the discontinuity there. I used the substition method where $t=2x^2-1$ and then the new integral is: ${\int_{-1}^{1}t^{-10}}dt$ which of course still has the discontinuity at $t=0$ So what I did was I broke the integral into two pieces: ${\int_{-1}^{0}t^{-10}}dt$ and ${\int_{0}^{1}t^{-10}}dt$ and used limits to calculate each of them. Of course each of them diverges to $+\infty$ so the whole integral diverges. Is this correct? At first, I made the mistake of not realizing the discontinuity - I calculated the integral without breaking it up into the two integrals mentioned and I got the result $-1/18$. What is the correct answer here? Does the integral exist or not? I know there is something called the Cauchy´s principal value but it´s too advanced for me to understand. Thanks for any help. REPLY [0 votes]: Let $\epsilon<1$, then the function is symmetric on the interval $[-1,-\epsilon]$ and the interval $[\epsilon,1]$. Thus the integral can be rewritten as $\lim_{\epsilon\to0^+}2\int_\epsilon^1t^{-10}dt$, which diverges. The plot of $f(t)=t^{-10}$ looks as follows:

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Calling all boys and girls interested in cheering during the 2019-2020 school year!!! Clinics will be held Monday, March 11th-March 13th at the High School in the main gym from 3:15-5:15. During clinics all participants will learn the material needed for tryouts. Clinics are not mandatory but are highly recommended. Throughout the clinic times the coaches will spend time evaluating the participants progress and level of participation. All participants must have an active/current physical on file at the school in order attend clinics and tryout. All incoming 7th and 8th grade participants will have the ability to ride the activity bus from the middle school to the high school. When they arrive at the high school they will wait in the cafeteria until we escort them to the main gym for clinics. The incoming 7th and 8th graders will have tryouts on Thursday, March 14th beginning promptly at 3:30. All incoming 9th, 10th, 11th, and 12th graders, tryouts will be held on Friday, March 15th beginning promptly at 3:45 at the Middle School. By the end of clinics on Tuesday, we will have assigned all participants with a tryout time and partner. We recommend that you arrive approximately 30 minutes prior to your tryout time to check in and allow time to practice. Please notify the coaches as soon as possible if you may have a scheduling conflict so we can make special arrangements. On the day of their tryout participants should arrive with their hair and makeup already complete and be wearing red or black shorts and a plain white t-shirt. See below attachment of the judges score sheet to see how the cheerleaders will be scored during tryouts. Participants will display the material that they learned throughout clinics and perform a chant that they paired motions with. Below are the chant words for each grade (please remember, it’s the grade you will be going in to). 7th Grade Touchdown, Pirates Touchdown, Let’s Win 8th Grade Pirates Let’s yell it W-C-MS 9th Grade Red, Black, White Come on Pirates Let’s Fight 10th Grade Come on – Pirates Score – Six More 11th Grade Hey Fans Get Loud Yell Red & Black 12th Grade Red, Black, and White Back once again Let’s hear it Pirates G0 – Fight – Win Hey West Carrollton Stand up, Yell it loud Go Pirates, Say it proud Go Pirates XX Go Pirates XX Hey West Carrollton Stand up , Yell it loud WCHS, Say it proud WCHS XX WCHS XX Go Pirates WCHS Incoming cheerleaders will have the ability to preference a squad/season that they choose to cheer. Middle School cheerleaders will have the option to choose from football, basketball or both and High School cheerleaders will have the option to choose from football/soccer, basketball, competition or both/all. Competition cheerleaders do not have to cheer for a sideline squad. Results will be posted by 8pm on Saturday, March 16th. Coaches will immediately enter a “No Contact Period” as soon as results are posted. We will have a parent/player meeting on Monday, April 22nd at 5:30 at the High School (location TBD), at this time, there will be a $50 deposit for all High School cheerleaders that plan to attend summer camp. We look forward to the upcoming season and wish everyone the best of luck with tryouts!!! Cheer Permission Slip 2019-2020

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\begin{document} \title{On fusion categories} \begin{abstract}Using a variety of methods developed in the literature (in particular, the theory of weak Hopf algebras), we prove a number of general results about fusion categories in characteristic zero. We show that the global dimension of a fusion category is always positive, and that the S-matrix of any (not necessarily hermitian) modular category is unitary. We also show that the category of module functors between two module categories over a fusion category is semisimple, and that fusion categories and tensor functors between them are undeformable (generalized Ocneanu rigidity). In particular the number of such categories (functors) realizing a given fusion datum is finite. Finally, we develop the theory of Frobenius-Perron dimensions in an arbitrary fusion category. At the end of the paper we generalize some of these results to positive characteristic. \end{abstract} \author{Pavel Etingof} \address{Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139, USA} \email{etingof@math.mit.edu} \author{Dmitri Nikshych} \address{Department of Mathematics and Statistics, University of New Hampshire, Durham, NH 03824, USA} \email{nikshych@math.unh.edu} \author{Viktor Ostrik} \address{Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139, USA} \email{ostrik@math.mit.edu} \maketitle \section{Introduction} Throughout this paper (except for Section 9), $k$ denotes an algebraically closed field of characteristic zero. By a {\em fusion category} $\mC$ over $k$ we mean a $k$-linear semisimple rigid tensor (=monoidal) category with finitely many simple objects and finite dimensional spaces of morphisms, such that the endomorphism algebra of the neutral object is $k$ (see \cite{BaKi}). Fusion categories arise in several areas of mathematics and physics -- conformal field theory, operator algebras, representation theory of quantum groups, and others. This paper is devoted to the study of general properties of fusion categories. This has been an area of intensive research for a number of years, and many remarkable results have been obtained. However, many of these results were proved not in general but under various assumptions on the category. The goal of this paper is to remove such assumptions, and to give an account of the theory of fusion categories in full generality. The structure of the paper is as follows. In Section 2, we give our main results about squared norms, global dimensions, weak Hopf algebras, and Ocneanu rigidity. This section contains many results which are partially or fully due to other authors, and our own results are blended in at appropriate places. Sections 3-7 are mostly devoted to review of the technical tools and to proofs of the results of Section 2. Namely, in section 3, we prove one of the main theorems of this paper, saying that any fusion category has a nonzero global dimension (in fact, we show that for $k=\CC$, the dimension is positive). The proof relies in an essential way on the theorem that in any fusion category, the identity functor is isomorphic to ${****}: V \mapsto V^{****}$ (where $V \mapsto V^*$ is the duality in $\mC$), as a tensor functor, which is proved using the theory of weak Hopf algebras (more specifically, the formula for the fourth power of the antipode from \cite{N}). We also prove that the $SL_2({\mathbb Z})$-representation attached to any modular category over $\CC$ is unitary. In Section 4, we give a short review of the theory of weak Hopf algebras, which are both a tool and an object of study in this paper, and prove the isomorphism between the identity functor ${\rm Id}$ and ${****}$, which is crucial for the main theorem. In Section 5, we prove a formula for the trace of squared antipode of a semisimple connected regular weak Hopf algebra. Then we proceed to prove a number of results (partially due to M\"uger) about semisimplicity of the category of module functors between module categories over a fusion category, in particular of the dual and the Drinfeld center. We also generalize these results to multi-fusion categories (the notion obtained from that of a fusion category by relaxing the condition that $\End(\mathbf 1)=k$). In particular, we prove that a semisimple weak Hopf algebra over $k$ is cosemisimple. Finally, we prove a categorical version of the class equation of Kac and Zhu. In Section 6, we consider a pair of semisimple weak Hopf algebras $B\subset A$ and study the subcomplex of $B$-invariants in the co-Hochschild complex of $A$ (this complex is a weak analog of the complex considered in \cite{Sch}, \cite{EG1} for Hopf algebras). We prove that this complex is acyclic in positive degrees. In Section 7, we show that the complex considered in Section 6 coincides with the deformation complex of a monodial functor between two multi-fusion categories introduced independently by Yetter \cite{Y1,Y2} and Davydov \cite{Da}. Using this, we establish ``Ocneanu rigidity'' (absence of deformations) for arbitrary nondegenerate multi-fusion categories and tensor functors between them. The idea of the proof of this result is due to Ocneanu-Blanchard-Wassermann, and the fusion category case was worked out completely by Blanchard-Wassermann in \cite{Wa,BWa} under the assumption that the global dimension is non-zero. In Section 8 we discuss the notion of Frobenius-Perron dimensions of objects in a fusion category, show that they are additive and multiplicative, and have similar properties to categorical dimensions in a pivotal category. We prove a categorical analogue of the Nichols-Zoeller freeness theorem (saying that a finite dimensional Hopf algebra is a free module over a Hopf subalgebra), in particular prove the freeness theorem for semisimple quasi-Hopf algebras. We also show that the global dimension of a fusion category is divisible by its Frobenius-Perron dimension (in the ring of algebraic integers), and the ratio is $\le 1$. In particular, if the Frobenius-Perron dimension of a fusion category is an integer, then it coincides with the global dimension. This result may be regarded as a categorical version of the well known theorem of Larson and Radford, saying that the antipode of a semisimple Hopf algebra is involutive. Further, we show that the Frobenius-Perron dimension of a fusion category is divisible by the Frobenius-Perron dimension of its full subcategory, and use this result to show that any fusion category of Frobenius-Perron dimension $p$ (a prime) is the category of representations of a cyclic group with a 3-cocycle. We also classify fusion categories of dimension $p^2$. Finally, we show that the property of a fusion category to have integer Frobenius-Perron dimensions (which is equivalent to being the representation category of a quasi-Hopf algebra) is stable under basic operations with categories, in particular is ``weak Morita invariant'' (i.e. invariant under passing to the the opposite of the dual category). At the end of the section we define group-theoretical fusion categories, which is a subclass of fusion categories with integer Frobenius-Perron dimensions; they are constructed explicitly from finite groups. Many of the results of this paper have analogs in positive characteristic. These generalizations are given in Section 9. In particular, we show that a fusion category of nonzero global dimension over a field of positive characteristic can be lifted to characteristic zero. Throughout the paper, we are going to freely use the theory of rigid tensor categories. We refer the reader to the textbooks \cite{K},\cite{BaKi} for details. We also recommend the reader the expository paper \cite{CE}, where much of the content of this paper is discussed in detail. {\bf Acknowledgments.} The research of P.E. was partially supported by the NSF grant DMS-9988796, and was done in part for the Clay Mathematics Institute. The research of D.N. was supported by the NSF grant DMS-0200202. The research of V.O. was supported by the NSF grant DMS-0098830. We are grateful to A.~Davydov, A.~Kirillov Jr., M.~M\"uger, and A.~Wassermann for useful discussions, and to E.~Blanchard for giving us the formulation of the result of \cite{BWa} before publication. \section{Results on squared norms, global dimensions, weak Hopf algebras, and Ocneanu rigidity} Let $k$ be an algebraically closed field. By a {\em multi-fusion} category over $k$ we mean a rigid semisimple $k$-linear tensor category $\mC$ with finitely many simple objects and finite dimensional spaces of morphisms. If the unit object $\bold 1$ of $\mC$ is simple, then the category $\mC$ is said to be a {\em fusion} category. Otherwise (if $\bold 1$ is not simple), it is easy to see that we have $\mathbf 1=\oplus_{i\in J}\mathbf 1_i$, where $\mathbf 1_i$ are pairwise nonisomorphic simple objects. Let us list a few examples to keep in mind when thinking about fusion and multi-fusion categories. \vskip .05in {\bf Examples of fusion categories.} 1. The category ${\rm Vec}_{G}$ of finite dimensional vector spaces graded by a finite group $G$ (or, equivalently, finite dimensional modules over the function algebra ${\rm Fun}(G,k)$.) Simple objects in this category are evaluation modules $V_g$, $g\in G$, and the tensor product is given by $V_g\otimes V_h=V_{gh}$, with the associativity morphism being the identity. More generally, pick a 3-cocycle $\omega\in Z^3(G,k^\times)$. To this cocycle we can attach a twisted version ${\rm Vec}_{G,\omega}$ of ${\rm Vec}_{G}$: the simple objects and the tensor product functor are the same, but the associativity isomorphism is given by $\Phi_{V_g,V_h,V_k}=\omega(g,h,k)\textrm{id}$. The pentagon axiom then follows from the cocycle condition $$\omega(h,k,l)\omega(g,hk,l)\omega(g,h,k)=\omega(gh,k,l)\omega(g,h,kl).$$ Note that cohomologous cocycles define equivalent fusion categories. 2. The category of finite dimensional $k$-representations of a finite group $G$, whose order is relatively prime to the characteristic of $k$. 3. The category of integrable modules (from category $\mathcal O$) over the affine algebra $\widehat{sl}_2$ at level $l$ (see \cite{BaKi}). The tensor product in this category is the fusion product, defined at the level of objects by the Verlinde fusion rule $$V_i\otimes V_j=\sum_{\substack{k=\vert i-j\vert \\ k\equiv i+j~\textrm{mod}~2}}^{l-\vert i+j-l\vert}V_k $$ \vskip .05in {\bf Examples of multi-fusion categories.} 1. The category of finite dimensional bimodules over a finite dimensional semisimple $k$-algebra, with bimodule tensor product. It has simple objects $M_{ij}$ with ``matrix'' tensor product $M_{ij}\otimes M_{jk}=M_{ik}$; thus the identity object is $\bold 1=\oplus_i M_{ii}$. 2. The category of finite dimensional modules over the function algebra on a finite groupoid. \subsection{Squared norms and global dimensions} Let us introduce the notion of the global dimension of a fusion category. First of all, we have the following known result (see e.g., \cite{O}). \begin{proposition}\label{doubledual} In a fusion category, any simple object $V$ is isomorphic to its double dual $V^{**}$. \end{proposition} To prove this, it suffices to note that for any simple object $V$, the right dual $V^*$ is the unique simple object $X$ for which $V\otimes X$ contains the neutral object, while the left dual $^*V$ is the unique simple object $X$ for which $V\otimes X$ projects to the neutral object; so $^*V=V^*$ by semisimplicity of the category, and hence $V=V^{**}$ for any simple $V$. Next, recall (\cite{BaKi}, p.39) that if $V\in \mC$, and $g:V\to V^{**}$ is a morphism, then one can define its ``quantum trace'' $\Tr_V(g)\in k$ by the formula $$ \Tr_V(g)={\rm ev}_{V^*} \circ (g\otimes 1_{V^*})\circ {\rm coev}_V, $$ where $\mathbf 1$ is the unit object of $\mC$, ${\rm coev}_V: \mathbf 1\to V\otimes V^*$ and ${\rm ev}_V: V^*\otimes V\to \mathbf 1$ are the coevaluation and evaluation maps (see \cite{BaKi}). It is easy to show (and well known, see e.g. \cite{BaKi}) that for any simple object $V$ and a nonzero morphism $a: V\to V^{**}$ one has $\Tr_V(a)\ne 0$. Indeed, otherwise there is a sequence of nonzero maps $\mathbf 1\to V\otimes V^*\to \mathbf 1$ with zero composition, which would imply that the multiplicity of $\mathbf 1$ in $V\otimes V^*$ is at least 2 -- a contradiction. Now, following \cite{Mu1}, for every simple object $V$ of a fusion category $\mC$, define the {\em squared norm} $|V|^2\in k^\times$ of $V$ as follows. Fix an isomorphism $a: V\to V^{**}$ (which exists by Proposition \ref{doubledual}), and let $|V|^2=\Tr_V(a)\Tr_{V^*}((a^{-1})^*)$. It is clearly independent on the choice of $a$ (since $a$ is uniquely determined up to a scaling), and nonzero by the explanation in the previous paragraph. For example, for the neutral object $\mathbf 1$ we have $|\mathbf 1|^2=1$. \begin{definition}\cite{Mu1} The {\em global dimension} of a fusion category $\mC$ is the sum of squared norms of its simple objects. It is denoted by $\dim(\mC)$. \end{definition} One of our main results is \begin{theorem} \label{posi} If $k=\CC$ then $|V|^2>0$ for all simple objects $V\in \mC$; therefore $\dim(\mC)\ge 1$, and is $>1$ for any nontrivial $\mC$. In particular, for any fusion category $\mC$ one has $\dim(\mC)\ne 0$. \end{theorem} \begin{remark} Note that the second statement immediately follows from the first one, since $\mC$ is always defined over a finitely generated subfield $k'$ of $k$, which can be embedded into $\CC$. \end{remark} The proof of Theorem \ref{posi}, which relies on Theorem \ref{fourstars} below, is given in Section~3. Theorem \ref{fourstars} is proved in Section~4. \begin{remark} In the course of proof of Theorem \ref{posi} we show that for any simple object $V$, the number $|V|^2$ is an eigenvalue of an integer matrix. In particular, if $k=\CC$ then $|V|^2$ is an algebraic integer. Thus, for $k=\CC$ Theorem \ref{posi} actually implies that $|V|^2$ is a ``totally positive'' algebraic integer, i.e. all its conjugates are real and positive (since one can twist a fusion category by an automorphism of $\CC$ and get a new fusion category). Similarly, $\dim(\mC)$ is ``totally'' $\ge 1$, i.e. all its conjugates are $\ge 1$.\end{remark} One of the main tools in the proof of Theorem \ref{posi} is \begin{theorem} \label{fourstars} In any fusion category, the identity functor is isomorphic to the functor **** as a tensor functor. \end{theorem} For the category of representations of a Hopf algebra, this result follows from Radford's formula for $S^4$ \cite{R1}. In general, it follows from the analog of Radford's formula for weak Hopf algebras, which was proved by the second author in \cite{N} (see Section 4). \begin{definition} A pivotal structure on a fusion category $\mC$ is an isomorphism of tensor functors $i:\Id\to **$. A fusion category equipped with a pivotal structure is said to be a pivotal fusion category. \end{definition} We conjecture a stronger form of Theorem \ref{fourstars}: \begin{conjecture} Any fusion category admits a pivotal structure. \end{conjecture} For example, this is true for the representation category of a semisimple Hopf algebra over $k$, since in this case by the Larson-Radford theorem \cite{LR2}, the squared antipode is $1$ and hence $\Id=**$. Furthermore, in Section 8 we will show that the conjecture is true for the representation category of a semisimple quasi-Hopf algebra. \subsection{Results on pivotal, spherical, and modular categories} Now let $\mC$ be a pivotal fusion category. In such a category, one can define the dimension of an object $V$ by $\dim(V)=\Tr_V(i)$, and we have the following result which justifies the notation $|V|^2$. \begin{proposition}\label{pivo} In a pivotal fusion category one has $|V|^2=\dim(V)\dim(V^*)$ for any simple object $V$. Moreover, if $k=\CC$ then $\dim(V^*)=\overline{\dim(V)}$, so $|V|^2=|\dim(V)|^2$. \end{proposition} Proposition \ref{pivo} is proved in Section 3. This result can be further specialized to spherical (in particular, ribbon) categories, (see \cite{Mu1}). Namely, recall from \cite{Mu1} that a pivotal fusion category is spherical if and only if $\dim(V)=\dim(V^*)$ for all simple objects $V$. Thus we have the following corollary. \begin{corollary} \label{spheri} In a spherical category, $|V|^2=\dim(V)^2$. In particular, if $k=\CC$, then $\dim(V)$ is (totally) real. \end{corollary} Corollary \ref{spheri} readily follows from Proposition \ref{pivo}. \begin{remark} We note that in \cite{Mu1}, the number $|V|^2$ is called the squared dimension of $V$, and denoted $d^2(V)$. We do not use this terminology for the following reason. In a pivotal category over $\CC$, the dimensions of simple objects do not have to be real numbers (it suffices to consider the category of representations of a finite group $G$, where $i$ is given by a nontrivial central element of $G$). Thus, in general $\dim(V^*)\ne \dim(V)$, and $|V|^2\ne \dim(V)^2$. Therefore, the term ``squared dimension'', while being adequate in the framework of spherical categories of \cite{Mu1}, is no longer adequate in our more general setting.\end{remark} Now assume that $\mC$ is a modular category (i.e., a ribbon fusion category with a nondegenerate $S$-matrix, see \cite{BaKi}). In this case one can define matrices $S$ and $T$ of algebraic numbers, which yield a projective representation of the modular group $SL_2(\mathbb Z)$ (\cite{BaKi}). We note that the matrix $S$ is defined only up to a sign, since in the process of defining $S$ it is necessary to extract a square root of the global dimension of $\mC$. So to be more precise we should say that by a modular category we mean the underlying ribbon category together with a choice of this sign. \begin{proposition} \label{modu} If $k=\CC$ then the projective representation of $SL_2(\mathbb Z)$ associated to $\mC$ is unitary in the standard hermitian metric (i.e. the matrices $S$ and $T$ are unitary). \end{proposition} Proposition \ref{modu} is proved in Section 3. \begin{remark} As before, the proposition actually means that this representation is totally unitary, i.e. the algebraic conjugates of this representation are unitary as well. \end{remark} \begin{remark} It is interesting whether Proposition \ref{modu} generalizes to mapping class groups of higher genus Riemann surfaces.\end{remark} We note that the results of this subsection are known for hermitian categories \cite{BaKi}. Our results imply that these results are valid without this assumption. \subsection{Module categories, the dual category, the Drinfeld center} Let $\mM$ be an indecomposable left module category over a rigid tensor category $\mC$ \cite{O} (all module categories we consider are assumed semisimple). This means, $\mM$ is a module category which cannot be split in a direct sum of two nonzero module subcategories. In this case one can define the dual category $\mC^*_{\mM}$ to be the category of module functors from $\mM$ to itself: $\mC^*_{\mM}={\rm Fun}_\mC(\mM,\mM)$ \cite{O}. This is a rigid tensor category (the tensor product is the composition of functors, the right and left duals are the right and left adjoint functors). For example, let us consider $\mC$ itself as a module category over $\mC\otimes \mC^{op}$, via $(X,Y)\otimes Z=X\otimes Z\otimes Y$. Then the dual category is the Drinfeld center $Z(\mC)$ of $\mC$ (see \cite{O1, Mu2}; for the basic theory of the Drinfeld center see \cite{K}). The following result was proved by M\"uger \cite{Mu1,Mu2} under the assumption $\dim(\mC)\ne 0$ (which, as we have seen, is superfluous in zero characteristic) and minor additional assumptions. \begin{theorem}\label{dualcat} For any indecomposable module category $\mM$ over a fusion category $\mC$, the category $\mC^*_{\mM}$ is semisimple (so it is a fusion category), and $\dim(\mC^*_{\mM})=\dim(\mC)$. In particular, for any fusion category $\mC$ the category $Z(\mC)$ is a fusion category of global dimension $\dim(\mC)^2$. \end{theorem} In fact, we have the following more general results, which (as well as the results in the next subsection) are inspired by \cite{Mu1,Mu2}. \begin{theorem}\label{m1m2} For any module categories $\mM_1,\mM_2$ over a fusion category $\mC$, the category of module functors ${\rm Fun}_{\mC}(\mM_1,\mM_2)$ is semisimple. \end{theorem} Theorems \ref{dualcat} and \ref{m1m2} are proved in Section 5. \subsection{Multi-fusion categories} We say that a multi-fusion category $\mC$ is indecomposable if it is not a direct sum of two nonzero multi-fusion categories. Let $\mC$ be a multi-fusion category. Then for any simple object $X\in \mC$ there exist unique $i,j\in J$ such that $\mathbf 1_i\otimes X\ne 0$ and $X\otimes \mathbf 1_j\ne 0$; moreover, we have $\mathbf 1_i\otimes X=X\otimes \mathbf 1_j=X$. Thus, as an additive category, $\mC=\oplus_{m,n}\mC_{mn}$, where $\mC_{mn}$ is the full abelian subcategory of $\mC$ with simple objects having $i=m,j=n$. It is easy to check that $\mC_{ii}$ are fusion categories for all $i$, and $\mC_{ij}$ are $(\mC_{ii},\mC_{jj})$-bimodule categories, equipped with product functors $\mC_{ij}\times \mC_{jl}\to \mC_{il}$ satisfying some compatibility conditions. We will refer to $\mC_{ii}$ as {\it the component categories} of $\mC$. Since $\mC_{ii}$ are fusion categories, they have well defined global dimensions. In fact, the off-diagonal subcategories $\mC_{ij}$ can also be assigned global dimensions, even though they are not tensor categories. To do this, observe that for any simple object $V\in \mC_{ij}$, and any morphism $g:V\to V^{**}$ one can define $\Tr_V(g)\in \End(\mathbf 1_i)=k$. Therefore, we can define $|V|^2$ by the usual formula $|V|^2=\Tr_V(g)\Tr_{V^*}((g^{-1})^*)$, and set $\dim(\mC_{ij}):= \sum_{V\in {\rm Irr}\mC_{ij}}|V|^2$. \begin{proposition}\label{samedim} If $\mC$ is an indecomposable multi-fusion category, then all the categories $\mC_{ij}$ have the same global dimensions. \end{proposition} The setup of the previous section can be generalized to the multi-fusion case. Namely, we have the following generalization of Theorem \ref{m1m2} to the multi-fusion case: \begin{theorem}\label{multi-fus} If $\mC$ is a multi-fusion category, and $\mM_1,\mM_2$ are module categories over $\mC$, then the category ${\rm Fun}_{\mC}(\mM_1,\mM_2)$ is semisimple. In particular, the category $\mC^*_\mM$ is semisimple for any module category $\mM$. \end{theorem} Proposition \ref{samedim} and Theorem \ref{multi-fus} is proved in Section 5. \begin{remark} We note that it follows from the arguments of \cite{O,Mu1} that if $\mC$ is a multifusion category and $\mM$ is a faithful module category of $\mM$, then $\mM$ is a faithful module category over $\mC^*_\mM$, and $(\mC^*_\mM)^*_\mM=\mC$.\end{remark} \subsection{Results on weak Hopf algebras} A convenient way to visualize (multi-) fusion categories is using weak Hopf algebras (see Section 4 for the definitions). Namely, let $\mC$ be a multi-fusion category. Let $R$ be a finite dimensional semisimple $k$-algebra, and $F$ a fiber functor (i.e. an exact, faithful tensor functor) from $\mC$ to $R$-bimod. Let $A=\End_k(F)$ (i.e. the algebra of endomorphisms of the composition of $F$ with the forgetful functor to vector spaces). \begin{theorem} \label{Schlah} (\cite{Sz}) $A$ has a natural structure of a semisimple weak Hopf algebra with base $R$, and $\mC$ is equivalent, as a tensor category, to the category $\Rep(A)$ of finite dimensional representations of $A$. \end{theorem} \begin{remark} In order to lift the naturally existing map $A\to A\otimes_R A$ to a weak Hopf algebra coproduct $A\to A\otimes_k A$, one needs to use a separability idempotent in $R\otimes R$. We will always use the unique {\it symmetric} separability idempotent. In this case, the weak Hopf algebra $A$ satisfies an additional {\it regularity condition}, saying that the squared antipode is the identity on the base of $A$.\end{remark} One can show that for any multi-fusion category $\mC$, a fiber functor $F$ exists for a suitable $R$. Indeed, let $\mM$ be any semisimple faithful module category over $\mC$, i.e. such that any nonzero object of $\mC$ acts by nonzero (for example, $\mC$ itself). Let $R$ be a semisimple algebra whose blocks are labeled by simple objects of $\mM$: $R=\oplus_{M\in {\rm Irr}(\mM)}R_M$, and $R_M$ are simple. Now define a functor $F:\mC\to R-\text{bimod}$ as follows: for any object $X\in \mC$ set $$ F(X)=\oplus_{M,N\in {\rm Irr}(\mM)}\Hom_{\mM}(M,X\otimes N) \otimes B_{NM}, $$ where $B_{NM}$ is the simple $(R_N,R_M)$-bimodule. This functor has an obvious tensor structure (coming from composition of morphisms), and it is clearly exact and faithful. So it is a fiber functor. Therefore, we have \begin{corollary} \label{Schlah1} (\cite{H},\cite{Sz}) Any multi-fusion category is equivalent to the category of finite dimensional representations of a (non-unique) regular semisimple weak Hopf algebra. This weak Hopf algebra is connected if and only if $\mC$ is a fusion category. \end{corollary} \begin{remark} In particular, we can choose $R$ to be a commutative algebra (i.e., all blocks have size $1$). Thus, any multi-fusion category is equivalent to the category of representations of a weak Hopf algebra with a commutative base. This result was proved by Hayashi in \cite{H}, where weak Hopf algebras with commutative bases appear under the name ``face algebras''.\end{remark} \vskip .05in The language of weak Hopf algebras is convenient to visualize various categorical constructions. One of them is that of a dual category. Indeed, let $\mC$ be the category of representations of a weak Hopf algebra $A$. Let $R$ be the base of $A$. Then we have a natural fiber functor from $\mC$ to the category of $R$-bimodules -- the forgetful functor. This functor defines a natural structure of a module category over $\mC$ on the category $\mM=R-\text{bimod}$. In this case, the dual category $\mC^*_{\mM}$ is simply the representation category of the dual weak Hopf algebra $A^{*cop}$ with the opposite coproduct (see \cite{O}). Furthermore, as we just showed, this example is general, in the sense that any faithful module category over a multi-fusion category can be obtained in this way. The global dimension of the category of representations of a connected weak Hopf algebra is closely related to the trace of squared antipode in this weak Hopf algebra. This connection is expressed by the following theorem. Let a fusion category $\mC$ be the representation category of a regular semisimple weak Hopf algebra $A$. Let $S$ be the antipode of $A$. Let $q_i$ be the primitive idempotents of the center of the base $A_t$, and $n_i$ the block sizes of the matrix algebras $q_iA_t$. \begin{theorem}\label{traceofS2} One has $\Tr(S^2|_{q_jAS(q_i)})=\dim(\mC)n_i^2n_j^2$. In particular, over $\CC$ one always has $\Tr(S^2|_{q_jAS(q_i)})>0$. \end{theorem} \begin{remark}\label{tra} Theorem \ref{traceofS2} implies that $\Tr(S^2|_{A}) = \dim(\Rep(A)) \dim(A_t)^2$.\end{remark} We also have the following result, which generalizes the Larson-Radford theorem for Hopf algebras \cite{LR1}. \begin{theorem}\label{sscoss} A semisimple weak Hopf algebra over $k$ is also cosemisimple. \end{theorem} Theorem \ref{traceofS2} is proved in Section 5. Theorem \ref{sscoss} follows from Theorem \ref{multi-fus}, since the representation category of the dual weak Hopf algebra is the dual category. \subsection{Vanishing of the Yetter cohomology} The cohomology theory that controls deformations of tensor categories and tensor functors is the cohomology theory developed in \cite{CY,Y1,Y2,Da}, which we will call the Yetter cohomology theory. This theory associates to any tensor category $\mC$ and a tensor functor $F$ for $\mC$ to another tensor category $\mC'$, a sequence of cohomology spaces $H^i_F(\mC)$ (See Section 7). If $F$ is the identity functor, then these groups are simply denoted by $H^i(\mC)$. Namely, first order deformations of $\mC$ are elements of $H^3(\mC)$, while first order deformations of a tensor structure on a given functor $F$ between fixed tensor categories $\mC,\mC'$ are elements of $H^2_F(\mC)$. Obstructions to such deformations lie, respectively, in $H^4(\mC)$ and $H^3_F(\mC)$. For multi-fusion categories, the Yetter cohomology often vanishes. Namely, we have the following result. \begin{theorem}\label{van} Let $\mC$, $\mC'$ be multi-fusion categories, and $F: \mC\to \mC'$ be a (unital) tensor functor from $\mC$ to $\mC'$. Then $H^i_F(\mC)=0$ for $i>0$. \end{theorem} Theorem \ref{van} is proved in Section 7, using weak Hopf algebras. The idea of this proof is due to Ocneanu, Blanchard, and Wassermann. \subsection{Ocneanu rigidity} The statement that a fusion category over a field of characteristic zero cannot be nontivially deformed is known as Ocneanu rigidity, because its formulation and proof for unitary categories was suggested (but not published) by Ocneanu. Later Blanchard and Wassermann (\cite{BWa,Wa}) proposed an algebraic version of Ocneanu's argument, which proves the following result. \begin{theorem} \label{ocnrig} A fusion category does not admit nontrivial deformations. In particular, the number of such categories (up to equivalence) with a given Grothendieck ring (=fusion rules) is finite. \end{theorem} \begin{remark} This theorem is proved in \cite{BWa,Wa} under the assumption that $\dim(\mC)\ne 0$, but as we know from Theorem \ref{posi}, this assumption is superfluous in zero characteristic.\end{remark} In fact, one has the following somewhat more general theorem. \begin{theorem}\label{multocnrig} Theorem \ref{ocnrig} is valid for multi-fusion categories. \end{theorem} This theorem follows directly from Theorem \ref{van} for $i=3$ and $F=\Id$ (this is, essentially, the proof of Ocneanu-Blanchard-Wassermann). A sketch of proof is given in Section 7. Note that Theorem \ref{multocnrig} implies that any multi-fusion category is defined over an algebraic number field. {\bf Question.} Is any multi-fusion category defined over a cyclotomic field? Note that the answer is yes for fusion categories coming from quantum groups at roots of unity, for the categories of \cite{TY}, and for group-theoretical categories discussed in Section 8. Also, it follows from the results of subsection \ref{cy} that the global dimension of a fusion category belongs to a cyclotomic field. We also have \begin{theorem}\label{ocnrigfun} A (unital) tensor functor between multi-fusion categories does not have nontrivial deformations. In particular, the number of such functors (up to equivalence) for fixed source and target categories is finite. \end{theorem} \begin{remark} In particular, Theorem \ref{ocnrigfun} implies that a given fusion category has finitely many fiber functors. This is also shown in \cite{IK} in the special case of categories coming from subfactors (see Theorem 2.4 in \cite{IK}).\end{remark} \begin{remark} Theorem \ref{ocnrigfun} also implies Ocneanu's result that a given fusion category has finitely many braidings. Indeed, any braiding in particular defines an equivalence of tensor categories between $\mC$ and $\mC^{op}$.\end{remark} The proof of Theorem \ref{ocnrigfun} is analogous to the proof of Theorem \ref{multocnrig}. Namely, the first part of the result follows directly from Theorem \ref{van} for $i=2$. The second part is a consequence of a combinatorial fact that there are only finitely many homomorphisms between the Grothendieck rings of the categories under consideration; this fact is a consequence of Proposition 2.1 of \cite{O} since the second Grothendieck ring can be considered as a based module over the first one. \begin{remark} One says that an indecomposable multi-fusion category $\mC'$ is a quotient category of an indecomposable multi-fusion category $\mC$ if there exists a tensor functor $F:\mC \to \mC'$ such that any object of $\mC'$ is a subobject of $F(X),\; X\in \mC$ (See Section 5). It follows from the above that a given multi-fusion category $\mC$ has only finitely many quotient categories.\end{remark} \begin{corollary} \label{ocnrigmod} A module category $\mM$ over a multi-fusion category $\mC$ does not admit nontrivial deformations. In particular, the number of such module categories (up to equivalence) with a given number of simple objects is finite. \end{corollary} To prove Corollary \ref{ocnrigmod}, it suffices to choose a semisimple algebra $R$ with number of blocks equal to the number of simple objects in $\mM$, and apply Theorem \ref{ocnrigfun} to the functor $F:\mC \to R-\text{bimod}$ associated with $\mM$. \begin{remark} We note that this result was conjectured in \cite{O}. \end{remark} \section{Fusion categories over $\CC$} The goal of this subsection is to prove Theorem \ref{posi}, Proposition \ref{pivo}, Proposition \ref{modu}, and discuss their consequences. Let $\mC$ denote a fusion category over $\CC$. We will denote representatives of isomorphism classes of simple objects of $\mC$ by $X_i$, $i\in I$, with $0\in I$ labeling the neutral object $\mathbf{1}$. Denote by $*: I\to I$ the dualization map. (We note that at the level of labels, there is no distinction between right and left dualization, since by Proposition \ref{doubledual}, $V$ is isomorphic to $V^{**}$ for any simple object $V$). \subsection{Proof of Theorem \ref{posi}} By Theorem \ref{fourstars}, there exists an isomorphism of tensor functors $g: \Id\to ****$. Let us fix such an isomorphism. So for each simple $V\in \mC$ we have a morphism $g_V: V\to V^{****}$ such that $g_{V\otimes W}=g_V\otimes g_W$, and $g_{V^*}=(g_V^{-1})^*$. For all simple $V$, let us fix an isomorphism $a_V : V\to V^{**}$ such that $a_V^2=g_V$ ($a_V^2$ makes sense as ${\rm Hom}(V,V^{**})$ is canonically isomorphic to ${\rm Hom}(V^{**},V^{****})$, and so $a_V$ can also be regarded as an element of ${\rm Hom}(V^{**},V^{****})$). Then $a_{V^*}=\epsilon_V (a_V^{-1})^*$, where $\epsilon_V=\pm 1$, and $\epsilon_V\epsilon_{V^*}=1$. Define $d(V)=\Tr_V(a_V)$, in particular $d_i=d(X_i)$. These numbers are not at all canonical -- they depend on the choices we made. Nevertheless, it is easy to see from the definition that $|V|^2=\epsilon_V d(V)d(V^*)$. Now let us take any simple objects $V,W\in \mC$ and consider the tensor product $a_V\otimes a_W: V\otimes W\to (V\otimes W)^{**}$. We have $V\otimes W=\oplus_U {\rm Hom}(U,V\otimes W)\otimes U$, so we have $a_V\otimes a_W=\oplus b_{U}^{VW}\otimes a_U$, where $b_U^{VW}: {\rm Hom}(U,V\otimes W)\to {\rm Hom}(U^{**},V^{**}\otimes W^{**})$. The source and target of this map are actually canonically isomorphic, so we can treat $b_U^{VW}$ as an automorphism of ${\rm Hom}(U,V\otimes W)$. Also, since $a_V^2=g_V$ and $g_{V\otimes W}=g_V\otimes g_W$, we get $(b_U^{VW})^2=1$. Let $b_j^{ik}:=b_{X_j}^{X_iX_k}$. Let $T_j^{ik}=\trace(b_j^{ik})$. Since $(b_j^{ik})^2=1$, its eigenvalues are $\pm 1$, hence $T_j^{ik}$ are integers. By the definition of $b_j^{ik}$, we have $$ d_id_k=\sum_{j\in I}T_j^{ik}d_j $$ Let $T_i$ be the matrix such that $(T_i)_{kj}=T_j^{ik}$, and let $\mathbf d$ be the vector with components $d_i$. This vector is clearly nonzero (e.g., $d_0\ne 0$), and we have $$ T_i\mathbf d=d_i\mathbf d. $$ Thus, $\mathbf d$ is a common eigenvector of $T_i$ with eigenvalue $d_i$. Let $\epsilon_i=\epsilon_{X_i}$. Then we have: $\mathbf d$ is an eigenvector of the matrix $A_i=\epsilon_i T_iT_{i^*}$ with eigenvalue $|X_i|^2$. We claim that $T_{i^*}=\epsilon_i T_i^t$. This implies the theorem, as in this case $A_i=T_iT_i^t$, which is nonnegative definite (since the entries of $T_i$ are integer, hence real), so its eigenvalues, including $|X_i|^2$, have to be nonnegative. To prove that $T_{i^*}=\epsilon_i T_i^t$, we need to show that $T_{j}^{i^*k}=\epsilon_i T_k^{ij}$. But this, in turn, follows from the equality $b_j^{i^*k}=\epsilon_i (b_k^{ij})^*$ (which makes sense since we have canonical isomorphisms ${\rm Hom}(U,V\otimes W)\to {\rm Hom}(V^*\otimes U,W)\to {\rm Hom}(W,V^*\otimes U)^*$, the latter by semisimplicity). It remains to prove the last equality. Let $f: X_j\to X_i\otimes X_k$ be a morphism. By the definition, $b_j^{ik}f=(a_i\otimes a_k)fa_j^{-1}$. On the other hand, $f$ gives rise to an element $f': X_{i^*}\otimes X_j\to X_k$, given by $f'=({\rm ev}_i\otimes 1)(1\otimes f)$, and hence to a linear function on ${\rm Hom}(X_k,X_{i^*}\otimes X_j)$ given by $f''(h)=({\rm ev}_i\otimes 1)(1_{i^*}\otimes f)h\in {\rm End}(X_k)=\CC$ (for brevity we omit associativity maps and use the same notation for a morphism and its double dual). Now, we have \begin{eqnarray*} (b_j^{ik}f)''(h) &=& ({\rm ev}_i\otimes 1)(1_{i^*}\otimes (a_i\otimes a_k)fa_j^{-1})h \\ &=& \epsilon_i a_k({\rm ev}_i\otimes 1)(a_{i^*}^{-1}\otimes fa_j^{-1})h \\ &=& \epsilon_i ({\rm ev}_i\otimes 1)(1\otimes f)(a_{i^*}^{-1}\otimes a_j^{-1})ha_k \\ &=& \epsilon_i ({\rm ev}_i\otimes 1)(1\otimes f)(a_{i^*}\otimes a_j)ha_k^{-1}, \end{eqnarray*} where the last equality due to the facts that $g_{V\otimes W}=g_V\otimes g_W$, and $a_V=g_Va_V^{-1}$. Thus, $$ (b_j^{ik}f)''(h)= f''(b_k^{i^*j}h), $$ as desired. Theorem \ref{posi} is proved. \begin{remark}\label{pivotalization} Here is a modification of the above proof, which is simpler but requires an extension of the category $\mC$. The first step is to prove the result in the case of pivotal categories. This is a substantially simplified version of the above proof, since in the pivotal case $\epsilon_i$ and $b_j^{ik}$ are 1, so $T_j^{ik}$ are equal to $N_j^{ik}$, the multiplicity of $X_j$ in $X_i\otimes X_k$. The second step is reduction to the pivotal case. To do this, take the isomorphism $g: Id\to ****$ as tensor functors constructed in the proof of Theorem \ref{fourstars}. Define the fusion category $\tilde \mC$, whose simple objects are pairs $(X,f)$, where $X$ is a simple object of $\mC$, and $f:X\to X^{**}$ is an isomorphism, such that $f^{**}f=g$. It is easy to see that this fusion category has a canonical pivotal structure. Thus, $|X|^2>0$ in $\tilde\mC$. However, we have the forgetful functor $F:\tilde \mC\to \mC$, $F(X,f)=X$, which obviously preserves squared norms. Hence, $|V|^2>0$ for simple objects $V$ in $\mC$ (since for any $V$ there exists $f:V\to V^{**}$ such that $f^{**}f=g$), which completes the proof. \end{remark} \subsection{Proof of Proposition \ref{pivo}} The first statement is straightforward. To prove the second statement, let $N_i$ be the matrix whose entries are given by $(N_i)_{kj}={\rm dim}{\rm Hom}(X_j,X_i\otimes X_k)$. Then, like in the proof of Theorem \ref{posi}, $N_i\mathbf d=d_i\mathbf d$. Let us multiply this equation on the left by $\mathbf d^*$. (the hermitian adjoint). We have $\mathbf d^*N_i=(N_i^t\mathbf d)^*= (N_{i^*}\mathbf d)^*=\overline{d_{i^*}}\mathbf d^*$. Thus the multiplication will yield $$ \overline{d_{i^*}}|\mathbf d|^2=d_i|\mathbf d|^2. $$ Since $|\mathbf d|^2>0$, this yields the second statement. The proposition is proved. \subsection{Proof of Proposition \ref{modu} and properties of modular categories} It is well known (see \cite{BaKi}) that the matrix $T$ is diagonal with eigenvalues being roots of unity. So we only need to show that the S-matrix is unitary. The entries $s_{ij}$ of the S-matrix are known to satisfy $s_{ij}=s_{ji}$, and $\sum_j s_{i^*j}s_{jl}=\delta_{il}$ (\cite{BaKi}). So it suffices to show that $s_{i^*j}=\overline{s_{ij}}$. Recall that $s_{j0}=\pm \dim(V_j)/\sqrt{D}$, where $D$ is the global dimension of the category. So $s_{j0}$ is a real number. Further, by the Verlinde formula (\cite[3.1]{BaKi}), if $\mathbf v_m$ is the $m$-th column of $S$ then $$ N_i\mathbf v_m=\frac{s_{im}}{s_{0m}}\mathbf v_m, $$ for all $i$. Let $\mathbf v_m^*$ denote the row vector, Hermitian adjoint to $\mathbf v_m$. Then $$ \mathbf v_m^*N_i\mathbf v_m=\frac{s_{im}}{s_{0m}}|\mathbf v_m|^2. $$ One the other hand, since $N_i^T=N_{i^*}$, and $N_i$ is real, we have $$ \mathbf v_m^*N_i\mathbf v_m=(N_i^T\mathbf v_m)^*\mathbf v_m=(N_{i^*}\mathbf v_m)^*\mathbf v_m= (\frac{s_{i^*m}}{s_{0m}}\mathbf v_m)^*\mathbf v_m= \frac{\overline{s_{i^*m}}}{s_{0m}}|\mathbf v_m|^2. $$ Since $|\mathbf v_m|^2\ne 0$, the statement is proved. \begin{remark} An important property of modular categories is \begin{proposition}\label{moddivi} (\cite{EG4}, Lemma 1.2) The dimension of every simple object of a modular category divides $\sqrt{D}$, where $D$ is the global dimension of the category. \end{proposition} This implies that in a ribbon category, dimensions of objects divide the global dimension (this is seen by taking the Drinfeld center and using Theorem \ref{dualcat} and the well known fact that the Drinfeld center of a braided category contains the original category). This fact is useful in classification of modular categories (See section 8). \end{remark} In conclusion of this subsection let us give a simple application of Theorem \ref{posi} to modular categories. Let $c\in {\mathbb C}/8{\mathbb Z}$ be the Virasoro central charge of a modular category $\mC$, defined by the relation $(ST)^3=e^{2\pi ic/8}S^2$ (see \cite{BaKi}), and $D$ be the global dimension of $\mC$. Let $f: (-\infty,1]\to [1,\infty)$ be the inverse function to the monotonely decreasing function $g(x)=(3-x)\sqrt{x}/2$. \begin{proposition} One has $$ D\ge f({\rm sign}(s_{00})\cos(\pi c/4)) $$ \end{proposition} \begin{remark} The number $s_{00}$ equals $\pm 1/\sqrt{D}$.\end{remark} \begin{proof} We need to show that $$ {\rm sign}(s_{00})\cos(\pi c/4)\ge (3-D)\sqrt{D}/2 $$ To prove this, recall \cite{BaKi} that the number $e^{\pi ic/4}$ can be expressed via a Gaussian sum: $$ {\rm sign}(s_{00})e^{\pi i c/4}=\sum |X_i|^2\theta_i/\sqrt{D}, $$ where $\theta_i$ are the twists. This means that $$ {\rm sign}(s_{00})e^{\pi i c/4}-D^{-1/2}= \sum_{i\ne 0} |X_i|^2\theta_i/\sqrt{D}. $$ Taking squared absolute values of both sides, and using that $|\theta_i|=1$, and $|X_i|^2>0$ (by Theorem \ref{posi}), we get $$ |{\rm sign}(s_{00}) e^{\pi ic/4}-D^{-1/2}|^2\le (\sum_{i\ne 0}|X_i|^2)^2/D=\frac{(D-1)^2}{D}. $$ After simplifications, this yields the desired inequality. \end{proof} \begin{remark} Sometimes this inequality sharp. For example, for $c=\pm 2/5,\pm 22/5$, it yields an estimate $D\ge (5-\sqrt{5})/2$, which becomes an equality for the well known Yang-Lee category ($c=-22/5$, see \cite{DMS}) and three its Galois images. The same is true for $c=\pm 14/5,\pm 34/5$, in which case the estimate is $D\ge (5+\sqrt{5})/2$ (it is attained at four Galois images of the Yang-Lee category). Also, it goes without saying that the inequality is sharp for $c=0$, where it becomes an equality for the trivial category. \end{remark} These considerations inspire the following questions. {\bf Questions.} Does there exist a sequence of nontrivial fusion categories over $\CC$ for which the global dimensions tend to $1$? Does there exist such a sequence with a bounded number of simple objects? Is the set of global dimensions of fusion categories over $\CC$ a discrete subset of $\mathbb R$? We expect that the answer to the first question is ``no'', i.e. the point $1$ of the set of possible dimensions of fusion categories is an isolated point (``categorical property T''). \begin{section}{Weak Hopf algebras and their integrals} Below we collect the definition and basic properties of weak Hopf algebras. \subsection{Definition of a weak Hopf algebra} \begin{definition}[\cite{BNSz}] \label{finite weak Hopf algebra} A {\em weak Hopf algebra} is a vector space $A$ with the structures of an associative algebra $(A,\,m,\,1)$ with a multiplication $m:A\otimes_k A\to A$ and unit $1\in A$ and a coassociative coalgebra $(A,\,\Delta,\,\epsilon)$ with a comultiplication $\Delta:A\to A\otimes_k A$ and counit $\epsilon:A\to k$ such that: \begin{enumerate} \item[(i)] The comultiplication $\Delta$ is a (not necessarily unit-preserving) homomorphism of algebras: \begin{equation} \label{Delta m} \Delta(hg) = \Delta(h) \Delta(g), \qquad h,g\in A, \end{equation} \item[(ii)] The unit and counit satisfy the following identities: \begin{eqnarray} \label{Delta 1} (\Delta \otimes \id) \Delta(1) & =& (\Delta(1)\otimes 1)(1\otimes \Delta(1)) = (1\otimes \Delta(1))(\Delta(1)\otimes 1), \\ \label{eps m} \epsilon(fgh) &=& \epsilon(fg\1)\, \epsilon(g\2h) = \epsilon(fg\2)\, \epsilon(g\1h), \end{eqnarray} for all $f,g,h\in A$. \item[(iii)] There is a linear map $S: A \to A$, called an {\em antipode}, such that \begin{eqnarray} m(\id \otimes S)\Delta(h) &=&(\epsilon\otimes\id)(\Delta(1)(h\otimes 1)), \label{S epst} \\ m(S\otimes \id)\Delta(h) &=& (\id \otimes \epsilon)((1\otimes h)\Delta(1)), \label{S epss} \\ S(h) &=& S(h\1)h\2 S(h\3), \label{S id S} \end{eqnarray} for all $h\in A$. \end{enumerate} \end{definition} \begin{remark} We use Sweedler's notation for a comultiplication: $\Delta(c) = c\1 \otimes c\2$.\end{remark} Axioms (\ref{Delta 1}) and (\ref{eps m}) above are analogous to the usual bialgebra axioms of $\Delta$ being a unit preserving map and $\epsilon$ being an algebra homomorphism. Axioms (\ref{S epst}) and (\ref{S epss}) generalize the properties of the antipode in a Hopf algebra with respect to the counit. Also, it is possible to show that given (\ref{Delta m}) - (\ref{S epss}), axiom (\ref{S id S}) is equivalent to $S$ being both anti-algebra and anti-coalgebra map. The antipode of a finite dimensional weak Hopf algebra is bijective \cite[2.10]{BNSz}. \begin{remark} A weak Hopf algebra is a Hopf algebra if and only if the comultiplication is unit-preserving and if and only if $\epsilon$ is a homomorphism of algebras. \end{remark} A {\em morphism} between weak Hopf algebras $A_1$ and $A_2$ is a map $\phi : A_1 \to A_2$ which is both algebra and coalgebra homomorphism preserving $1$ and $\epsilon$ and which intertwines the antipodes of $A_1$ and $A_2$, i.e., $ \phi\circ S_1 = S_2\circ \phi$. The image of a morphism is clearly a weak Hopf algebra. When $\dim_k A < \infty$, there is a natural weak Hopf algebra structure on the dual vector space $A^*=\Hom_k(A,k)$ given by \begin{eqnarray} & & \phi\psi(h) = (\phi\otimes\psi)(\Delta(h)), \\ & & \Delta(\phi)(h\otimes g) = \phi(hg) \\ & & S(\phi)(h) = \phi(S(h)), \end{eqnarray} for all $\phi,\psi \in A^*,\, h,g\in A$. The unit of $A^*$ is $\epsilon$ and the counit is $\phi \mapsto \phi(1)$. The linear maps defined in (\ref{S epst}) and (\ref{S epss}) are called {\em target\,} and {\em source counital maps} and are denoted $\eps_t$ and $\eps_s$ respectively : \begin{equation} \eps_t(h) = \epsilon(1\1 h)1\2, \qquad \eps_s(h) = 1\1 \epsilon(h1\2), \end{equation} for all $h\in A$. The images of the counital maps \begin{equation} A_t = \eps_t(A), \qquad A_s = \eps_s(A) \end{equation} are separable subalgebras of $A$, called target and source {\em bases} or {\em counital subalgebras} of $A$. These subalgebras commute with each other; moreover \begin{eqnarray*} A_t &=& \{(\phi\otimes \id)\Delta(1) \mid \phi\in A^* \} = \{ h\in A \mid \Delta(h) = \Delta(1)(h\otimes 1) \}, \\ A_s &=& \{(\id \otimes \phi)\Delta(1) \mid \phi\in A^* \} = \{ h\in A \mid \Delta(h) = (1\otimes h)\Delta(1) \}, \end{eqnarray*} i.e., $A_t$ (respectively, $A_s$) is generated by the right (respectively, left) tensor factors of $\Delta(1)$ in the shortest possible presentation of $\Delta(1)$ in $A\otimes_k A$. The category $\Rep(A)$ of left $A$-modules is a rigid tensor category \cite{NTV}. The tensor product of two $A$-modules $V$ and $W$ is $V \otimes_{A_t} W$ with the $A$-module structure defined via $\Delta$. The unit object $\mathbf{1}$ of $\Rep(A)$ is the target counital algebra $A_t$ with the action $h\cdot z = \eps_t(hz)$ for all $h\in A,\, z\in A_t$. For any algebra $B$ we denote by $Z(B)$ the center of $B$. Then the unit object of $\Rep(A)$ is irreducible if and only if $Z(A)\cap A_t = k$. In this case we will say that $A$ is {\em connected}. We will say that that $A$ is {\em coconnected } if $A^*$ is connected, and that $A$ is {\em biconnected} if it is both connected and coconnected. If $p\neq 0$ is an idempotent in $A_t\cap A_s\cap Z(A)$, then $A$ is the direct sum of weak Hopf algebras $pA$ and $(1-p)A$. Consequently, we say that $A$ is {\em indecomposable} if $A_t\cap A_s\cap Z(A) = k1$. Every weak Hopf algebra $A$ contains a canonical {\em minimal\,} weak Hopf subalgebra $A_{{\rm min}}$ generated, as an algebra, by $A_t$ and $A_s$ \cite[Section 3]{N}. In other words, $A_{{\rm min}}$ is the minimal weak Hopf subalgebra of $A$ that contains $1$. Obviously, $A$ is an ordinary Hopf algebra if and only if $A_{{\rm min}} = k1$. Minimal weak Hopf algebras over $k$, i.e., those for which $A= A_{{\rm min}}$, were completely classified in \cite[Proposition 3.4]{N}. The restriction of $S^2$ on $A_{{\rm min}}$ is always an inner automorphism of $A_{{\rm min}}$, see \cite{N}. \begin{remark} Unless indicated otherwise, in what follows we will use only weak Hopf algebras satisfying the following natural property : \begin{equation} \label{regularity property} S^2|_{A_{{\rm min}}} = \id. \end{equation} This property has an easy categorical interpretation. Let $\mathbf{1} = A_t$ be the trivial $A$-module. Then \eqref{regularity property} is satisfied if and only if the canonical isomorphism $\mathbf{1} \to \mathbf{1}^{**}$ is the identity map.\end{remark} \begin{definition} We will call a weak Hopf algebra satisfying \eqref{regularity property} {\em regular}. \end{definition} \begin{remark} It was shown in \cite[6.1]{NV} that every weak Hopf algebra can be obtained as a twisting of some regular weak Hopf algebra with the same algebra structure. \end{remark} \subsection{Integrals} \label{subsection about integrals} The following notion of an integral in a weak Hopf algebra is a generalization of that of an integral in a usual Hopf algebra \cite{M}. \begin{definition}[\cite{BNSz}] \label{integral} A left {\em integral} in $A$ is an element $\ell\in A$ such that \begin{equation} h\ell =\eps_t(h)\ell, \qquad (rh = r\eps_s(h)) \qquad \mbox{ for all } h\in A. \end{equation} \end{definition} The space of left integrals in $A$ is a left ideal of $A$ of dimension equal to $\dim_k(A_t)$. Any left integral $\lambda$ in $A^*$ satisfies the following invariance property : \begin{equation} \label{left invariance} g\1 \lambda(hg\2) = S(h\1)\lambda(h\2g), \qquad g,h\in A. \end{equation} In what follows we use the Sweedler arrows $\actl$ and $\actr$ for the dual actions : \begin{equation} \label{Sweedler arrows} h\actl\phi (g) = \phi(gh) \quad \text { and } \quad \phi\actr h (g)= \phi(hg). \end{equation} for all $g,\,h\in A,\phi\in A^*$. Recall that a functional $\phi\in A^*$ is {\em non-degenerate} if $\phi\circ m$ is a non-degenerate bilinear form on $A$. Equivalently, $\phi$ is non-degenerate if the linear map $h\mapsto (h \actl \phi)$ is injective. An integral (left or right) in a weak Hopf algebra $A$ is called {\em non-degenerate} if it defines a non-degenerate functional on $A^*$. A left integral $\ell$ is called {\em normalized} if $\eps_t(\ell)=1$. It was shown by P.~Vecsernyes \cite{V} that a finite dimensional weak Hopf algebra always possesses a non-degenerate left integral. In particular, a finite dimensional weak Hopf algebra is a Frobenius algebra (this extends the well-known result of Larson and Sweedler for usual Hopf algebras). Maschke's theorem for weak Hopf algebras, proved in \cite[3.13]{BNSz}, states that a weak Hopf algebra $A$ is semisimple if and only if $A$ is separable, and if and only if there exists a normalized left integral in $A$. In particular, every semisimple weak Hopf algebra is finite dimensional. For a finite dimensional $A$ there is a useful notion of duality between non-degenerate left integrals in $A$ and $A^*$ \cite[3.18]{BNSz}. If $\ell$ is a non-degenerate left integral in $A$ then there exists a unique $\lambda\in A^*$ such that $\lambda\actl \ell = 1$. This $\lambda$ is a non-degenerate left integral in $A^*$. Moreover, $\ell\actl \lambda =\epsilon$. Such a pair of non-degenerate integrals $(\ell,\,\lambda)$ is called a pair of {\em dual} integrals. An invertible element $g\in A$ is called {\em group-like} if $\Delta(g) = (g\otimes g)\Delta(1) = \Delta(1)(g\otimes g)$ \cite[Definition 4.1]{N}. Group-like elements of $A$ form a group $G(A)$ under multiplication. This group has a normal subgroup $G_0(A) := G(A_{{\rm min}})$ of {\em trivial} group-like elements. If $A$ is finite dimensional, the quotient group $\widetilde{G}(A) = G(A)/ G_0(A)$ is finite. It was shown in \cite{N} that if $A$ is finite dimensional and $\ell\in A$ and $\lambda\in A^*$ is a dual pair of left integrals, then there exist group-like elements $\alpha\in G(A^*)$ and $a\in G(A)$, called {\em distinguished} group-like elements, whose classes in $\widetilde{G}(A^*)$ and $\widetilde{G}(A)$ do not depend on the choice of $\ell$ and $\lambda$, such that \begin{equation} S(\ell) = \alpha\actl \ell \qquad \mbox{ and } \qquad S(\lambda)=a \actl \lambda. \end{equation} (Note that $\alpha$ and $a$ themselves depend on the choice of $\ell$ and $\lambda$). The following result is crucial for this paper. It is an analogue of Radford's formula for usual Hopf algebras. \begin{theorem}\label{S4} \cite[Theorem 5.13]{N} One has \begin{equation} S^4(h) = a^{-1}(\alpha\actl h \actr \alpha^{-1})a. \end{equation} for all $h\in A$. \end{theorem} A weak Hopf algebra $A$ is said to be {\em unimodular} if it contains a non-degenerate $2$-sided integral; equivalently, $A$ is unimodular if the coset of $\alpha$ in $G(A^*)$ is trivial. A semisimple weak Hopf algebra is automatically unimodular \cite{BNSz}. \subsection{Proof of Theorem \ref{fourstars}}\label{pffourstars} Let $\mC$ be a fusion category over $k$. By Theorem \ref{Schlah1}, $\mC$ is the representation category of a semisimple connected weak Hopf algebra $A$. Since $A$ is semisimple, it is unimodular, and hence one can choose a left integral $\ell$ such that the corresponding element $\alpha=\epsilon$. Thus, by Theorem \ref{S4}, one has $S^4(h)=a^{-1}ha$, for some group-like element $a\in G(H)$, and hence for any simple object $V\in \mC=\Rep(A)$, the operator $a^{-1}|_V$ defines an isomorphism $V\to V^{****}$. Since $a$ is grouplike, this isomorphism respects the tensor product. The theorem is proved. \subsection{Canonical integrals} Using the squared antipode, one can construct integrals on a weak Hopf algebra $A$ called the {\em canonical integrals}. Namely, we have the following proposition, a particular case of which was proved in \cite{BNSz}. \begin{proposition} \label{chi} Let $A$ be a finite-dimensional weak Hopf algebra and let $\{q_i\}$ be primitive central idempotents in $A_t$. Then $\chi_i(h) := \Tr(L_h\circ S^2|_{AS(q_i)})$, where $L_h \in \End_k(A)$ is given by $L_h(x)=hx,\,h, x\in A$, defines a left integral in $A^*$ such that \begin{eqnarray} \label{chi1} \chi_i(q_j) &=& \Tr(S^2|_{q_jAS(q_i)}) \qquad \mbox{and } \\ \label{q-trace} \chi_i(gh) &=& \chi(h S^2(g)), \end{eqnarray} for all $g,h\in A$. \end{proposition} \begin{proof} The identities \eqref{chi1} and \eqref{q-trace} follow immediately from the definition of $\chi$, since $A$ is regular and $S^2\circ L_g = L_{S^2(g)}\circ S^2$ for all $g\in A$. Next, for all $\phi\in A^*$ we define $R_\phi \in \End_k(A)$ by $R_\phi(x) = x\actr \phi,\, x\in A^*$. Then $R_\phi(xS(q_i)) = R_\phi(x)S(q_i)$ so that $AS(q_i)$ is $R_\phi$-invariant. A direct computation shows that $L_{(h \actr \phi)} = R_{\phi\1}\circ L_h \circ R_{S^{-1}(\phi\2)}$ for all $h\in A,\, \phi\in A^*$, and hence, \begin{eqnarray*} (\phi \otimes \chi_i)(\Delta(h)) &=& \Tr(L_{(h \actr \phi)}\circ S^2|_{AS(q_i)}) \\ &=& \Tr( R_{\phi\1}\circ L_h \circ R_{S^{-1}(\phi\2)}\circ S^2|_{AS(q_i)}) \\ &=& \Tr( R_{\eps_t(\phi)}\circ L_h \circ S^2|_{AS(q_i)}), \end{eqnarray*} therefore, $ (\id \otimes \chi_i)\Delta(h) = (\eps_t \otimes \chi)\Delta(h)$ for all $h\in A$, i.e., $\chi_i$ is a left integral in $A^*$. \end{proof} \subsection{Sufficient conditions for semisimplicity} The next proposition is a refinement of \cite[Proposition 6.4]{N} and is analogous to the Larson-Radford theorem \cite{LR1} for usual Hopf algebras which says that if $A$ is a finite dimensional Hopf algebra with $\Tr(S^2|_A)\neq 0$ then $A$ is semisimple and cosemisimple. \begin{proposition} \label{semisimplicity condition} Let $A$ be a finite dimensional weak Hopf algebra. If for every primitive idempotent $p$ in $A_t \cap Z(A)$ there exist idempotents $q_p,\, q_p'$ in $pA_t$ such that \begin{equation} \label{Tr S2 neq 0} \Tr(S^2|_{q_pAS(q_p')}) \neq 0, \end{equation} then $A$ is semisimple. \end{proposition} \begin{proof} Let $\ell$ and $\lambda$ be a dual pair of non-degenerate left integrals in $A$ and $A^*$ respectively. Then $(\ell\2 \actl \lambda) \otimes q_p S^{-1}(\ell\1) S(q_p')$ is the dual bases tensor for $q_p A S(q_p')$, cf.\ \cite{N}. Therefore, \begin{eqnarray*} 0 &\neq& \Tr(S^2|_{q_pAS(q_p')}) \\ &=& \la \ell\2 \actl \lambda,\, q_p S(\ell\1)S(q_p') \ra \\ &=& \la \lambda,\, q_p \eps_s(q_p' \ell) \ra. \end{eqnarray*} Therefore, $y_p:=\eps_s(q_p'\ell) \neq 0$. Next, we compute \begin{eqnarray*} y_p \ell &=& S(\ell\1) S(q_p')\ell\2 \ell \\ &=& S(\ell\1) S(q_p') \eps_t(\ell\2) \ell \\ &=& S(\ell\1) \eps_t(\ell\2) S(q_p') \ell = S(\ell) y_p. \end{eqnarray*} Let $\xi_p$ be an element in $A^*_t$ (here and below we write $A^*_t:=(A^*)_t$, $A^*_s:=(A^*)_s$) such that $\xi_p\actl 1 =y_p$. Then from the previous computation we have \begin{equation} y_p \ell = (\alpha \actl \ell) y_p, \end{equation} where $\alpha$ is a distinguished group-like element of $A^*$. Arguing as in \cite[Proposition 6.4]{N} one gets \begin{equation} \xi_p = S(\xi_p) \alpha, \end{equation} whence $\alpha$ is trivial, i.e., $\alpha\in A^*_{\rm min}$. This means that one can choose $\ell$ in such a way that $\alpha =\epsilon$, i.e., $\ell = S(\ell)$. Hence, $y_p\in A_s \cap Z(A)$ is a non-zero multiple of $p$. We conclude that $y :=\eps_s(\ell) =\sum_p\, y_p$ is an invertible central element of $A$. Hence, $\ell' = y^{-1}\ell$ is a normalized right integral in $A$. By Maschke's theorem, $A$ is semisimple. \end{proof} Let $q$ qnd $q'$ be idempotents in $A_t$, and let $\rho = \epsilon \actr q$ and $\rho' = \epsilon \actr q'$. Then $\rho$ and $\rho'$ are idempotents in $A_t ^*$. \begin{lemma} \label{identify} The space $\rho A^* S(\rho')$ may be identified, in an $S^2$-invariant way, with the dual space of $qAS(q')$. \end{lemma} \begin{proof} It is easy to check that for all $h\in A$ and $\phi\in A^*$ one has $\la \phi,\, qhS(q') \ra = \la \rho \phi S(\rho'),\, h\ra$, whence the statement follows. \end{proof} The next three corollaries refine \cite[Corollaries 6.5 - 6.7]{N}. \begin{corollary} \label{dual semisimplicity condition} Let $A$ be a finite dimensional weak Hopf algebra. If for every primitive idempotent $\pi$ in $A^*_t \cap A^*_s$ there exist idempotents $\rho_\pi,\, \rho_\pi'$ in $\pi A^*_t$ such that \begin{equation} \label{Tr S2 neq 0 dual} \Tr(S^2|_{\rho_\pi A^* S(\rho_\pi')}) \neq 0, \end{equation} then $A$ is semisimple. \end{corollary} \begin{proof} This immediately follows from Proposition~\ref{semisimplicity condition} and Lemma~\ref{identify}, since the latter implies $\Tr(S^2|_{\rho_\pi A^* S(\rho_\pi')}) = \Tr(S^2|_{q_pAS(q_p')})$. \end{proof} \begin{corollary} \label{Tr S2 connected} Let $A$ be a connected finite dimensional weak Hopf algebra. If there exist idempotents $q,\, q'$ in $A_t$ such that $\Tr(S^2|_{qAS(q')})\neq 0$ then $A$ is semisimple. \end{corollary} \begin{corollary} \label{Tr S2 biconnected} Let $A$ be a biconnected finite dimensional weak Hopf algebra. If there exist idempotents $q,\, q'$ in $A_t$ such that $\Tr(S^2|_{qAS(q')})\neq 0$ then $A$ and $A^*$ are semisimple. \end{corollary} \end{section} \begin{section}{$\Tr(S^2)$ and the global dimension} \subsection{Proof of Theorem \ref{traceofS2}} Let $A$ be a connected semisimple weak Hopf algebra. Then $\mathcal{C} = \Rep(A)$ is a fusion category. In particular, $V^{**}$ is isomorphic to $V$ for any simple object $V$, which means that $S^2$ is an inner automorphism of $A$ (this is also proved in [BNSz, 3.22]). So let $g\in A$ be such that $S^2(x) = gxg^{-1}$ for all $x\in A$. Then $g|_V : V \to V^{**} : v \mapsto g\cdot v$ is an isomorphism of representations. The calculation of $\Tr_V(g)$ and $\Tr_V((g^{-1})^*)$ yields \begin{eqnarray} \label{Tr no 1} \Tr_V(g){\rm Id}_{A_t} &=& \Tr(S(g)1\1|_{V^*})1\2,\\ \label{Tr no 2} \Tr_{V^*}((g^{-1})^*){\rm Id}_{A_t}&=&\Tr(g^{-1}1\1|_{V})1\2. \end{eqnarray} \begin{remark} We note an important distinction between the quantum trace in an object $V$ of a morphism $V\to V^{**}$, which is denoted by $\Tr_V$, and the usual trace of a linear operator in the vector space $V$, which is denoted just by $\Tr$. \end{remark} Let $\{q_i\}$ be primitive idempotents in $Z(A_t)$ and for every $i$ let $n_i^2 = \dim_k(q_iA_t)$ (i.e., $n_i$ is the dimension of the irreducible representation of $A_t$ corresponding to $q_i$). Multiplying equations \eqref{Tr no 1} and \eqref{Tr no 2} on the right by $q_j$ and $q_i$ respectively, applying $\Tr|_{A_t}$, and using that for a regular weak Hopf algebra $\Tr(z|_{A_t}) = \epsilon(z)$ for all $z\in A_t$ \cite{N}, we get \begin{equation*} \Tr_V(g)n_j^2=\Tr(g|_{q_jV}),\quad \Tr_{V^*}((g^{-1})^*)n_i^2=\Tr(g^{-1}|_{S(q_i)V}), \end{equation*} and hence, \begin{equation} \Tr(g|_{q_jV})\Tr(g^{-1}|_{S(q_i)V})=|V|^2n_i^2n_j^2. \end{equation} Taking the sum over all isomorphism classes of simple $V$, we obtain : \begin{equation} \label{TrS2 vs global dimension refined} \Tr(S^2|_{q_jAS(q_i)}) = \dim(\Rep(A)) n_i^2 n_j^2. \end{equation} This proves the first statement of the theorem. The second one follows from Theorem \ref{posi}. Theorem \ref{traceofS2} is proved. \subsection{Proof of Theorem \ref{dualcat}} Let $\mC$ be a fusion category, and let $\mM$ be an indecomposable module category over $\mC$. Let us show that $\mC^*_{\mM}$ is semisimple. Choose a semisimple algebra $R$ with blocks labeled by simple objects of $\mM$. Let $F$ be the fiber functor $F:\mC\to R-\text{bimod}$ associated with $\mM,R$ as explained in Section 2. Let $A=\End_k(F)$ be the corresponding weak Hopf algebra. Consider the weak Hopf algebra $A^{*cop}$. The representation category of $A^{*cop}$ is $\mC^*_{\mM}$. Thus, $A^{*cop}$ is connected (as $\mM$ is indecomposable). So, $A$ is biconnected. Also, by Theorem \ref{traceofS2}, we have $\Tr(S^2|_{q_jAS(q_i)})=\dim(\mC)n_i^2n_j^2\ne 0$. Thus, by Corollary \ref{Tr S2 biconnected}, $A^{*cop}$, and hence $\mC^*_{\mM}$ is semisimple. The equality $\dim(\mC)=\dim(\mC^*_{\mM})$ follows from Theorem \ref{traceofS2} and Lemma \ref{identify} (noting that the block sizes corresponding to the idempotents $q$ and $\epsilon\actr q$ are the same). Theorem \ref{dualcat} is proved. \subsection{Proof of Theorem \ref{m1m2}} Let $\mM$ be a possibly decomposable module category over $\mC$. So $\mM=\oplus \mM_j$ is a direct sum of indecomposable subcategories. Now for each $i$ choose semisimple algebras $R_i$ with blocks labeled by simple objects of $\mM_i$, and let $R=\oplus R_i$. Let $F$ be the fiber functor $\mC\to R-\text{bimod}$, associated to $(\mM,R)$, and let $B=\End_k(F)$ be the corresponding weak Hopf algebra. Then by Theorem \ref{traceofS2} and Corollary~\ref{dual semisimplicity condition} applied to $A=B^*$, we find that $B^*$ is semisimple. In other words, the the category $\mC^*_{\mM}$ is semisimple. This implies Theorem \ref{m1m2}, since the category ${\rm Fun}_{\mC}(\mM_1,\mM_2)$ is contained in $\mC^*_{\mM_1\oplus \mM_2}$. Theorem \ref{m1m2} is proved. \subsection{A refinement of Theorem \ref{traceofS2}} Let $A$ be a connected, but not necessarily coconnected, semisimple weak Hopf algebra. Let $\pi_i$ be the primitive idempotents in $A^*_s\cap Z(A^*)$, and let $A_{ij}\subset A$ be defined by $A_{ij}^*=\pi_iA^*S(\pi_j)$. It is easy to see that $A_{ij}=p_jAp_i$, where $p_i$ are the primitive idempotents of $A_s\cap A_t$. In particular, $A_{ii}$ are semisimple algebras. In fact, for every $i$ the algebra $A_{ii}$ has a natural structure of a biconnected weak Hopf algebra, and its representation category $\Rep(A_{ii})$ is equivalent to $\Rep(A)$ via the usual Morita equivalence $V\to p_i V$, $V\in \Rep(A)$. Indeed, this map is a tensor functor since $\Delta(p_i) =(p_i\otimes p_i)\Delta(1)$. To prove that it is an equivalence it suffices to show that the central support $\hat p_i$ of $p_i$ equals $1$. From the formula for $\Delta(p_i)$ above we have $\Delta(\hat p_iA) \subset \hat p_iA\otimes \hat p_iA$. But $\hat{p_i}A$ contains the matrix block $q_0A$ corresponding to the trivial representation of $A$ (which is irreducible since $A$ is connected). Hence, $\Delta(q_0A)\subset \hat p_iA\otimes \hat p_iA$ and so $\hat p_i=1$. Let $q_{ir}$ be the primitive idempotents of $Z(A_{rr})$. Let $n_{ir}^2=\dim(q_{ir}(A_{rr})_t)$. We have the following refinement of Theorem \ref{traceofS2}. \begin{theorem}\label{refine} One has $\Tr(S^2|_{q_{jr}A_{rs}S(q_{is})})=\dim(\Rep(A))n_{jr}^2n_{is}^2$. \end{theorem} \begin{proof} The proof is completely analogous to the proof of Theorem \ref{traceofS2}. Namely, let $g$ be an element of $A$ such that $gxg^{-1}=S^2(x)$. Then $p_rgp_r$ realizes $S^2$ in $A_{rr}$. Hence, as was shown in the proof of Theorem \ref{traceofS2} for any $V\in \Rep(A)$, one has $$ \Tr(p_rgp_r|_{q_{ir}V})=\Tr_V(g)n_{ir}^2, $$ and $$ \Tr(p_rg^{-1}p_r|_{S(q_{ir})V})=\Tr_V((g^{-1})^*)n_{ir}^2. $$ Now, $A_{rs}=\oplus_{V}p_sV\otimes (p_rV)^*$, and $S^2|_{A_{rs}}=p_sgp_s\otimes (p_rg^{-1}p_r)^*$. Hence, taking the sum over all $V$, we arrive at the result. \end{proof} \subsection{Proof of Proposition \ref{samedim}} Let $\mC$ be an indecomposable multi-fusion category. Fix an index $r$ and let $\mM=\oplus_i \mC_{ir}$. It is clear that $\mM$ is an indecomposable $\mC$-module category, and the category $\mC^*_\mM:={\rm Fun}_\mC(\mM,\mM)$ equals $\mC_{rr}^{op}$. Choose a semisimple algebra $R$ with blocks labeled by simple objects of $\mM$. Let $R_j$ be the two sided ideal in $R$ whose blocks are labeled by simple objects of $\oplus_i \mC_{ij}\subset \mM$. Let $n_j^2=\dim(R_j)$. Let $F$ be the corresponding fiber functor, and $B=\End_k(F)$ the corresponding weak Hopf algebra, and let $A=B^*$. Then $A$ is connected and semisimple. Therefore, by Theorem \ref{refine}, $\Tr(S^2|_{q_{jr}A_{rs}S(q_{is})})=\dim(\Rep(A))n_{is}^2n_{jr}^2$. Thus, using Lemma \ref{identify}, we get \begin{eqnarray*} \dim(\mC_{rs})n_{is}^2n_{jr}^2 &=& \Tr(S^2|_{(\epsilon\actr q_{jr})A_{rs}^*S(\epsilon\actr q_{is})})\\ &=& \Tr(S^2|_{q_{jr}A_{rs}S( q_{is})})\\ &=& \dim(\Rep(A))n_{is}^2n_{jr}^2, \end{eqnarray*} so $\dim(\mC_{rs})=\dim(\Rep(A))$, as desired. \subsection{Proof of Theorem \ref{multi-fus}} Without loss of generality it can be assumed that the multi-fusion category $\mC$ is indecomposable. Let $\mM_1$, $\mM_2$ be module categories over $\mC$. Let $\mathbf 1_i$ be the simple constituents of the unit object of $\mC$. Then it is shown in the standard way that for each $i$, the restriction functor ${\rm Fun_{\mC}}(\mM_1,\mM_2)\to {\rm Fun_{\mC_i}}(\mathbf 1_i\mM_1, \mathbf 1_i\mM_2)$ is an equivalence of categories (one can construct a quasi-inverse functor by extending a functor from $\mathbf 1_i\mM_1$ to $\mathbf 1_j\mM_1$ by tensoring with objects of $\mC_{ij}$ and checking various compatibilities). Therefore, the result follows from Theorem \ref{m1m2}. \subsection{Surjective and injective functors} Let $\mC,\mD$ be multi-fusion categories, and $F:\mC\to \mD$ be a unital tensor functor. We will say that $F$ is injective if it is fully faithful (in this case $F$ identifies $\mC$ with a full tensor subcategory of $\mD$). We will say that $F$ is surjective if any simple object $X$ of $\mD$ is contained in an object of the form $F(T)$, $T\in \mC$. If $F$ is surjective and $\mC$,$\mD$ are indecomposable, then $\mD$ is said to be a quotient category of $\mC$. Now let $\mM$ be a faithful module category over $\mD$. Then it is also a faithful module category over $\mC$. Let $R$ be a semisimple algebra with blocks labeled by simple objects of $\mM$, and $G$ the corresponding fiber functor $\mD\to R$-bimod. Let $A=\End_k(G)$, $B=\End_k(G\circ F)$ be the corresponding weak Hopf algebras. Then the functor $F$ induces a morphism of weak Hopf algebras $\phi_F:A\to B$, and vice versa. It is easy to check that $F$ is surjective, respectively injective if and only if $\phi_F$ is injective, respectively surjective. Let us now consider the corresponding dual categories $\mC^*,\mD^*$ (we will drop the subscript $\mM$, since the only module category we will consider in this subsection will be $\mM$). We have a natural dual functor $F^*:\mD^*\to \mC^*$, since any $\mD$-linear functor $\Phi:\mM\to \mM$ has a natural $\mC$-linear structure, defined using $F$. It is clear that the morphism $\phi_{F^*}:B^*\to A^*$ is given by $\phi_{F^*}=\phi_F^*$. \begin{proposition}\label{star} Let $F:\mC\to \mD$ be a unital tensor functor between multi-fusion categories. If $F$ is surjective (injective) then $F^*$ is injective (surjective). \end{proposition} \begin{proof} The claim is equivalent to the statement that surjectivity of $\phi_F$ is equivalent to injectivity of $\phi_F^*$, and vice versa, which is obvious. \end{proof} \subsection{The induction functor and the class equation for fusion categories} \label{indfu} In finite group theory, an important role is played by the ``class equation'' $\sum_{s}\frac{1}{|G_s|}=1$, where the summation is over conjugacy classes of a finite group $G$, and $G_s$ are the corresponding centralizers. There is a similar formula for semisimple Hopf algebras (due to G.I.Kac and Y.Zhu, see \cite{Ka}, \cite{Zhu}), which is also very useful. Here we will prove a class equation for any fusion category, which generalizes the results of Kac and Zhu. Let $\mC$ be a fusion category. Let $I:\mC\to Z(\mC)$ be the induction functor, defined by the condition $\Hom_{Z(\mC)}(I(V),X)=\Hom_{\mC}(V,X)$, $V\in \mC$, $X\in Z(\mC)$. Let $[X:Y]$ denote the multiplicity of a simple object $Y$ in an object $X$. For $X\in Z(\mC)$, let $X|_\mC$ denote the corresponding object of $\mC$. Then $I(V)=\oplus_{X\in Irr(Z(\mC))}[X|_{\mC}:V]X$. \begin{proposition}\label{I(1)} One has $I(V)|_\mC=\oplus_{Y\in Irr(\mC)}Y\otimes V\otimes Y^*$. \end{proposition} \begin{example} If $G$ is a finite group and $\mC=\Rep(G)$, then $I(\mathbf 1)$ is the following representation of the quantum double $D(G)=\Bbb C[G]\ltimes \mathcal O(G)$: it is the regular representation of the function algebra $\mathcal O(G)$, on which $G$ acts by conjugation. In this case proposition \ref{I(1)} for trivial $V$ is the standard fact that for conjugation action, $\mathcal O(G)=\oplus_{Y\in Irrep(G)}Y\otimes Y^*$. \end{example} \begin{proof} Let $\mC=\Rep(A)$, where $A$ is a weak Hopf algebra. Then $Z(\mC)=\Rep(D(A))$, where $D(A)$ is the quantum double of $A$ (see \cite{NTV}). As a space, $D(A)=A^*\otimes_{A_{min}}A$. So we have $I(V)|_\mC=D(A)\otimes_A V=A^*\otimes_{A_{min}}V$, with the action of $A$ defined by $h(\phi\otimes v) = h\3 \actl \phi \actr S(h\1) \otimes h\2 v$ for all $h\in A,\, \phi\in A^*,\, v\in V$. This yields exactly $\oplus_{Y\in Irr(\mC)}Y\otimes V\otimes Y^*$, where the tensor product is taken in the representation category of $A$. The proposition is proved. \end{proof} \begin{proposition}\label{numobj} The number $\sum_{X\in Irr(Z(\mC))}[X|_{\mC}:\mathbf 1]^2$ is equal to the number of simple objects of $\mC$. \end{proposition} \begin{proof} This sum equals $\dim{\rm End}(I(\mathbf 1))$, which is the same as $[I(\mathbf 1)|_{\mC}:\mathbf 1]$. But by Proposition \ref{I(1)} we know that $I(\mathbf 1)|_{\mC}=\oplus_{Y\in Irr(\mC)}Y\otimes Y^*$, so the result follows. \end{proof} \begin{proposition} (The class equation)\label{cleq} Let $\mC$ be a spherical fusion category. Then one has $$ \sum_{X\in Irr(Z(\mC)):[X|_{\mC}:\mathbf 1]\ne 0}[X|_{\mC}:\mathbf 1]\frac{1}{m_X}=1, $$ where $m_X=\frac{\dim \mC}{\dim X}$ is an algebraic integer. \end{proposition} \begin{proof} The proposition is immediately obtained by computing the dimension of $I(\mathbf 1)$ in two ways. On the one hand, this dimension is equal to $\dim \mC$ by Proposition \ref{I(1)}. On the other hand, it equals $\sum [X|_\mC:\mathbf 1]\dim X$. The fact that $m_X$ is an algebraic integer follows from Proposition \ref{moddivi}. The proposition is proved. \end{proof} \begin{example} Let $\mC=\Rep(G)$. In this case $Z(\mC)=\Rep D(G)$, so simple objects are parametrized by pairs (conjugacy class, irreducible representation of the centralizer). The simple objects $X\in Z(\mC)$ for which $X|_\mC$ contains $\mathbf 1$ are exactly those for which the representation of the centralizer is trivial. Thus, the equation of Proposition \ref{cleq} is exactly the class equation for $G$. \end{example} \end{section} \subsection{Graded categories} Let $G$ be a finite group. \begin{definition} A $G$-grading on a tensor category $\mC$ is a decomposition $\mC=\oplus_{g\in G}\mC_g$, such that $\otimes: \mC_g\times \mC_h\to \mC_{gh}$, $\mathbf 1\in \mC_e$, and $*:\mC_g\to \mC_{g^{-1}}$. We will say that the grading is faithful if $\mC_g\ne 0$ for any $g\in G$. \end{definition} It is clear that if $\mC$ is a $G$-graded category, then $\mC_e$ is a tensor category, and $\mC_g$ are left and right module categories over $\mC_e$. Let $\mC$ be a fusion category, $G$ a finite abelian group. \begin{proposition}\label{gradi} Suppose that $Z(\mC)$ contains, as a full tensor subcategory, the category of modules $\tilde G$ over $\mathcal O(G)$, and that all simple objects of this subcategory map to the neutral object of $\mC$ under the functor $Z(\mC)\to \mC$. Then $\mC$ admits a faithful $G^\vee$-grading (where $G^\vee$ is the dual group to $G$). \end{proposition} \begin{proof} Let $g\in \tilde G$ be a simple object. Since $g$ is trivial as an object of $\mC$, it defines a tensor automorphism of the identity functor of $\mC$. Thus we have a homomorphism $G\to {\rm Aut}_{\otimes}(Id)$, and hence any simple object of $\mC$ defines an element of the dual group $G^\vee$. This is obviously a faithful grading. \end{proof} \subsection{Deformations of weak Hopf algebras} It is well known that a non-semisimple finite dimensional algebra can be perturbed into a semisimple one. However, for (connected) weak Hopf algebras the situation is different. Namely, we have the following result. \begin{proposition} Let $H_t$ be a continuous family of finite dimensional regular connected weak Hopf algebras over $\Bbb C$ depending of a real parameter $t\in (a,b)$ (i.e., $H_t$ is independent of $t$ as a vectors space, and the structure maps continuously depend on $t$). Then either $H_t$ is semisimple for all $t$ or $H_t$ is non-semisimple for all $t$. \end{proposition} \begin{proof} The set $U\subset (a,b)$ where $H_t$ is semisimple is open. Assume that this set is nonempty. The function $f(t)=\Tr(S^2|_{H_t})$ is a continuous function of $t$. By Corollary \ref{Tr S2 connected}, $f$ is zero on the complement $U^c$ of $U$. Also, by Theorem \ref{posi} and Remark \ref{tra}, $f$ is a nonzero algebraic integer for any $t\in U$. Since $f$ is continuous, this implies that $f$ is constant (and nonzero) on each connected component of $U$. Thus, $U^c$ is empty. We are done. \end{proof} \subsection{Sphericity of pivotalization} In this Section we allow the ground field to have a positive characteristic, different from $2$. Let $A$ be a quasitriangular weak Hopf algebra, i.e., such that $\Rep(A)$ is a braided monoidal category. Let $\R\in (A\otimes A)\Delta(1)$ be the $R$-matrix of $A$. It was shown in \cite{NTV} that the Drinfeld element $u = S(\R\II)\R\I$ is invertible, satisfies $S^2(h)=uhu^{-1}$ for all $h$ in $A$, and \begin{equation} \label{delta u} \Delta(u^{-1}) = \R_{21}\R (u^{-1}\otimes u^{-1}). \end{equation} Furthermore, the element $uS(u)^{-1}$ is group-like and implements $S^4$ via the adjoint action. Observe that for any group-like element $\eta\in G(A^*)$ the element $g_\eta = \R\I\la \eta,\, \R\II\ra$ is group-like in $A$ and the map $\eta\mapsto g_\eta$ is a group homomorphism from $G(A^*)$ to $G(A^{op})$. Recall that distingusihed elments of $A$ and $A^*$ were defined in Section~\ref{subsection about integrals}. In the next Lemma we extend a result of Radford \cite{R2} relating the distinguished elements and $uS(u)^{-1}$. \begin{lemma} \label{uS(u)-1 vs a} Let $a\in A$ and $\alpha\in A^*$ be a pair of distinguished group-like elements chosen as in Section~\ref{subsection about integrals}. Then $g_\alpha a = S(u)u^{-1}$. \end{lemma} \begin{proof} Let $\ell\in A$ be a non-degenerate integral from which $a$ and $\alpha$ are constructed. It follows from \cite[Section 5]{N} that \begin{eqnarray*} \ell\2 \otimes \ell\1 &=& \ell\1 \otimes S^2(\ell\2)a^{-1}\\ \ell\1 h \otimes \ell\2 &=& \ell\1 \otimes \ell\2 S(h\actr \alpha),\qquad h\in A. \end{eqnarray*} Using these identities and properties of integrals and $R$-matrices we compute \begin{eqnarray*} \ell\2 S(\R\II \actr\alpha) \R\I \otimes \ell\1 &=& \ell\2 \R\I \otimes \ell\1 \R\II \\ &=& \R\I \ell\1 \otimes \R\II \ell\2 \\ &=& \R\I S(\R\II) \ell\1 \otimes \ell\2 \\ &=& \R\I S(\R\II) S^2(\ell\2)a^{-1} \otimes \ell\1, \end{eqnarray*} whence $S(\R\II \actr\alpha) \R\I = \R\I S(\R\II) a^{-1} = S(u)a^{-1}$ by non-degeneracy of $\ell$. On the other hand, from the axioms of $R$-matrix we have $S(\R\II \actr\alpha) \R\I= S(\R\II) \R\I g_\alpha = u g_\alpha$, which completes the proof. \end{proof} Recall \cite[Section 7]{NTV} that a central $\nu\in A$ is a ribbon element if $\nu=S(\nu)$ and $\Delta(\nu) = \R_{21}\R (\nu\otimes \nu)$. A quasitriangular weak Hopf algebra $A$ has a ribbon element if and only if $\Rep(A)$ is a ribbon category. \begin{corollary} \label{ribonness} Suppose that $A$ is a unimodular quasitriangular weak Hopf algebra and that $\alpha$ is gauged to be $\epsilon$. If $g\in A$ is a group-like element such that $g^2 = a^{-1}$ and $S^2(h)=ghg^{-1}$ for all $h\in A$ then $\nu = u^{-1}g$ is a ribbon element of $A$. \end{corollary} \begin{proof} In this case we have $g_\alpha= 1$. Since both $u$ and $g$ implement $S^2$, the element $\nu$ is central. It follows from Equation~\eqref{delta u} that $\Delta(\nu) = \R_{21}\R (\nu\otimes \nu)$. Finally, $S(\nu) = S(u)^{-1}g^{-1} = u^{-1}g = \nu$ by Lemma~\ref{uS(u)-1 vs a}. \end{proof} Recall that in Remark \ref{pivotalization}, it was shown that to every fusion category $\mC$ one can associate a twice bigger fusion category $\tilde\mC$, which has a canonical pivotal structure. \begin{proposition}\label{sphpiv} If $\dim\mC\ne 0$ then the canonical pivotal stucture of the category $\tilde \mC$ is in fact a spherical structure. \end{proposition} \begin{proof} Let us realize $\tilde\mC$ as the representation category of a regular weak Hopf algebra $A$. In this case, the pivotal structure of $\tilde\mC$ is defined by a grouplike element $g\in A$, such that $g^2=a^{-1}$, where $a$ is the distinguished group-like element of $A$. Assume that the distinguished group-like element $\alpha\in A^*$ is gauged to be $\epsilon$. Consider the Drinfeld center $Z(\tilde \mC)$, i.e., the representation category of the Drinfeld double $D(A)$. One can check that $a$ is also the distinguished grouplike element for $D(A)$. Since $g$ is a grouplike element such that $gxg^{-1}=S^2(x)$ for $x\in D(A)$, and $g^2=a^{-1}$, it follows from Corollary~\ref{ribonness} that $D(A)$ is ribbon, hence, $Z(\tilde\mC)$ is ribbon. Thus, we have $\dim(V)=\dim(V^*)$ in $Z(\tilde\mC)$ \cite{K}. Hence, the same holds in $\tilde \mC$ (i.e., $\tilde \mC$ is spherical). Indeed, let $I: \tilde \mC\to Z(\tilde \mC)$ be the induction functor defined in subsection \ref{indfu}. Then by Proposition \ref{I(1)}, for any $V\in \tilde\mC$, $I(V)=\oplus_{X\in Irr(\tilde\mC)} X\otimes V\otimes X^*$ as an object of $\tilde\mC$. Hence, $\dim I(V)=\dim V\dim \tilde \mC$. On the other hand, $I(V)^*=I(V^*)$, which implies the statement. \end{proof} \begin{corollary}\label{posroot} In any pivotal fusion category $\mC$ over $\CC$, the dimension of a simple object is a positive number times a root of unity. \end{corollary} \begin{proof} Consider the category $\tilde\mC$. It has two pivotal structures $b_1,b_2: V\to V^{**}$. Namely, $b_1$ comes from $\mC$ and $b_2$ is the canonical pivotal structure. By Proposition \ref{sphpiv}, $b_2$ is spherical. On the other hand, $b_2^{-1}\circ b_1$ is a tensor endomorphism of the identity functor, which obviously acts by roots of unity on simple objects. This implies the statement. \end{proof} \begin{section}{The co-Hochschild complex of a cosemisimple unimodular weak Hopf algebra and its subcomplex of invariants} \subsection{The definition} Let $A$ be a weak Hopf algebra and let $B \subset A$ be a weak Hopf subalgebra. We will define a complex $C^\bullet(A, B)$ that generalizes the subcomplex of the co-Hochschild complex of a Hopf algebra consisting of invariants of the adjoint action, cf. \cite{S, Sch}. We set $C^0(A, B) = A_t \cap Z(A)$, the algebra of endomorphisms of the trivial $A$-module; and for all $n\geq 1$, we set \begin{equation} \label{Cn} C^n(A, B) := \{ f\in A^{\otimes n} \mid f = \Ad(1)f \mbox{ and } \Ad(h)f = \Ad(\eps_t(h))f,\, \forall h\in B \}, \end{equation} where $\Ad(h)$ denotes the adjoint action of $h$, i.e., $\Ad(h)f = \Delta^n(h\1) f \Delta^n(S(h\2))$ for all $h\in B$ and $f\in A^{\otimes n}$. Here $\Delta^n : A \to A^{\otimes n}$ is the iteration of the comultiplication (we set $\Delta^1 =\id$). Note that $A^{\otimes n}$ is a non-unital $B$-module with respect to $\Ad$. The conditions of \eqref{Cn} mean that $A$ lies in $A\otimes_{A_t}\cdots \otimes_{A_t} A$ (where $A$ is an $A_t$-bimodule via $z_1\cdot h\cdot z_2 = z_1 S(z_2) h$ for $h\in A,\, z_1,z_2, \in A_t$) viewed as a subspace of $A^{\otimes n}$ and that $f$ commutes with $\Delta ^n(h)$ for all $h\in A$. Let us define a linear map $\delta^n : C^n(A, B)\to A^{\otimes (n+1)}$ by $\delta^0(f) = f - S(f)$ and \begin{eqnarray} \label{delta} \delta^n(f) &=& \Ad(1) \{1\otimes f +\sum_{i=1}^n\, (-1)^i (\id_{i-1}\otimes \Delta \otimes \id_{n-i})(f) +(-1)^{n+1} f\otimes 1\} \end{eqnarray} for $n \geq 1$. \begin{lemma} For all $n\geq 0$ we have $\text{Im}(\delta^n) \subset C^{n+1}(A, B)$ and $\delta^{n+1}\circ \delta^n =0$, i.e., the collection $\{C^n(A, B),\,\delta^n\}_{n\geq 0}$ is a complex. \end{lemma} \begin{proof} That $\delta^n(f),\, f\in C^n(A, B)$ is $\Ad(1)$ stable is a part of definition \eqref{delta}. For all $h\in B$ we compute \begin{eqnarray*} \lefteqn{\Ad(h)\delta^n(f) =} \\ &=& \Ad(h)(1\otimes f) - \Ad(h)(\Delta\otimes \id_{n-1})f + \cdots \\ & & \cdots + (-1)^n\Ad(h) (\id_{n-1}\otimes \Delta)f + (-1)^{n+1}\Ad(h)(f\otimes 1) \\ &=& h\1 S(h\3) \otimes \Ad(h\2)f - (\Delta\otimes\id_{n-1})(\Ad(h)f) +\cdots\\ & & \cdots + (-1)^n (\id_{n-1}\otimes \Delta)\Ad(h)f + (-1)^{n+1} \Delta^n(h\1) f \Delta^n(S(h\3))\otimes \eps_t(h\2)\\ &=& 1\1\eps_t(h)\otimes \Ad(1\2)f - (\Delta\otimes\id_{n-1}) (\Ad(\eps_t(h))f) +\cdots\\ & & \cdots + (-1)^n (\id_{n-1}\otimes \Delta)\Ad(\eps_t(h))f + (-1)^{n+1} \Delta^n(1\1)\Ad(\eps_t(h))f \otimes 1\2 \\ &=& \Ad(\eps_t(h))\delta^n(f), \end{eqnarray*} where we used the identity $(\id\otimes\eps_t)\Delta(x) = (1\otimes x)\Delta(1)$ for all $x\in A$. Therefore, $\delta^n(f)\in {C}^{n+1}(A, B)$. Thus, $\delta^n(C^n(A, B)) \subset C^{n+1}(A, B)$. That $\delta^{n+1}\circ \delta^n =0$ can be checked in a direct standard way, by cancelling the similar terms with opposite signs among $(n+3)(n+2)$ tensors in $\delta^{n+1}\circ\delta^n(f)$. \end{proof} The $n$-th cohomology group of the pair $A,\, B$ is \begin{equation} \label{Hn} H^n (A,\,B) := H^n(C^\bullet(A, B))= \mbox{Ker}(\delta^n)/\mbox{Im}(\delta^{n-1}), \qquad \mbox{ for } n\geq 0. \end{equation} \subsection{The vanishing theorem for cohomology} Now let $A$ be a regular finite-dimensional semisimple (and hence, cosemisimple) weak Hopf algebra. \begin{lemma} \label{nondeg of chi} Suppose that $A$ is indecomposable. Let $q_i$ be a primitive idempotent in $Z(A_t)$ and let $\chi:=\chi_i$ be the corresponding canonical left integral in $A^*$ from Proposition~\ref{chi}. Then $u = 1\1 \chi(1\2)$ is an invertible scalar. \end{lemma} \begin{proof} Since $\chi$ is a left integral in $A^*$, we have \begin{equation} u = 1\1 \chi(1\2) = S(1\1)\chi(1\2) = S(u) \in A_s\cap A_t. \end{equation} Next, the following identities hold in every weak Hopf algebra $A$: \begin{equation*} \Delta(1)(h\otimes 1) = (\id \otimes \eps_t)\Delta(h), \qquad (h\otimes 1)\Delta(1) = (\id \otimes \eps_t^{{\text{op}}})\Delta(h),\qquad h\in A, \end{equation*} where $\eps_t^{\text{op}}$ is the target counital map in $A^{\text{op}}$. Using Proposition~\ref{chi} we compute for all $h\in A$ : \begin{eqnarray*} 1\1 h \chi(1\2) &=& h\1 \chi(\eps_t(h\2)) = h\1 \chi(h\2 S(h\3)) \\ &=& h\1 \chi (S^{-1}(h\3)h\2) = h\1\chi( \eps_t^{{\text{op}}}(h\2)) \\ &=& h 1\1 \chi(1\2), \end{eqnarray*} hence, $u\in A_s\cap A_t\cap Z(A) = k1_A$. For any primitive central idempotent $q_j$ in $A_t$ we have \begin{equation*} \chi(q_j) = \Tr(S^2|_{q_jAS(q_i)}) =\dim(\Rep(A)) n_i^2 n_j^2 \neq 0, \end{equation*} where $n_i^2 = \dim_k(q_iA_t)$ and $n_j^2 =\dim_k(q_jA_t)$. Therefore, $\chi|_{A_t}\neq 0$ and $u\neq 0$. \end{proof} \begin{theorem} \label{acyclic} Let ${C}$ be the complex associated to $A,\,B$ as above. Then $H^0(A, B) = A_t\cap A_s\cap Z(A)$ and $H^n(A, B) =0$ for all $n\geq 1$. \end{theorem} \begin{proof} It suffices to prove this theorem in the case when $A$ is indecomposable. In this case we have $H^0(A, B) = \text{Ker}(\delta^0) = A_t\cap A_s\cap Z(A) =k$. Let $f\in {C}^n(A, B),\, n\geq 1$ be a cycle, i.e., $\delta^n(f) =0$. We will show by a direct computation that \begin{equation} \tilde{f} =(\id_{n-1}\otimes \chi)(f) \end{equation} belongs to ${C}^{n-1}(A, B)$ and that $f$ is a scalar multiple of the differential $\delta^{n-1}(\tilde{f})$. We will write $f=f^{(1)}\otimes\cdots\otimes f^{(n)}$, where a sum of some set of tensors is understood. We have : \begin{eqnarray*} \lefteqn{ \Ad(1)\tilde{f} = }\\ &=& 1\1 f^{(1)} S(1_{(2n-2)})\otimes\cdots\otimes 1_{(n-1)} f^{(n-1)} S(1_{(n)})\chi(f^{(n)}) \\ &=& \Delta^{n-1}(1\1 1'\1)(f^{(1)}\otimes\cdots\otimes f^{(n-1)}) \Delta^{n-1}(S(1\2 1'\4)) \chi(1'\2 f^{(n)}S(1'\3)) \\ &=& \Delta^{n-1}(1'\1)(f^{(1)}\otimes\cdots\otimes f^{(n-1)}) \Delta^{n-1}(S(1'\4)) \chi(1'\2 f^{(n)}S(1'\3)) = \tilde{f}, \end{eqnarray*} where $1'$ stands for a second copy of $1$ and we use that in any regular weak Hopf algebra one has $(1\1 \otimes 1\otimes 1\otimes 1\2)\Delta^4(1) = \Delta^4(1)$. Next, we have for all $h\in B$ : \begin{eqnarray*} \lefteqn{ \Ad(h)\tilde{f} = }\\ &=& h\1 f^{(1)} S(h_{(2n-2)})\otimes\cdots\otimes h_{(n-1)} f^{(n-1)} S(h_{(n)})\chi(f^{(n)}) \\ &=& h\1 f^{(1)} S(h_{(2n-1)})\otimes\cdots\otimes h_{(n-1)}) f^{(n-1)} \eps_s(h_{(n)}) S(h_{(n+1)}) \chi(f^{(n)}) \\ &=& h\1 f^{(1)} S(h_{(2n-1)})\otimes\cdots\otimes h_{(n-1)} f^{(n-1)} S(h_{(n+1)})\chi(f^{(n)}S(\eps_s(h_{(n)}))) \\ &=& h\1 f^{(1)} S(h_{(2n)})\otimes\cdots\otimes h_{(n-1)} f^{(n-1)} S(h_{(n+2)})\chi(h_{(n)} f^{(n)}S(h_{(n+1)})) \\ &=& (\id_{n-1}\otimes \chi)(\Ad(h)f) = (\id_{n-1}\otimes \chi)(\Ad(\eps_t(h))f) = \Ad(\eps_t(h))\tilde{f}. \end{eqnarray*} Here we used that \begin{equation} \label{Delta1} f = (1\otimes\cdots\otimes 1\otimes \Delta(1))f \end{equation} for all $f\in {C}^n(A, B)$, the weak Hopf algebra identities \begin{equation*} x\1\eps_s(x\2)=x \quad \mbox{ and } \quad (y\otimes 1)\Delta(1) = (1\otimes S^{-1}(y))\Delta(1) \end{equation*} for all $x\in A$ and $y\in A_s$ (cf.\ \cite{NV}), and Proposition~\ref{chi}. Hence, $\tilde{f}\in {C}^{n-1}(A, B)$. Next, since $(\id_n\otimes \chi)(\delta^n(f))=0$, we get \begin{eqnarray*} 0 &=& (\id_n\otimes \chi)\Ad(1) \{ 1\otimes f^{(1)}\otimes\cdots\otimes f^{(n)} - \Delta(f^{(1)})\otimes\cdots\otimes f^{(n)} + \\ & & \cdots + (-1)^n f^{(1)}\otimes\cdots\otimes \Delta(f^{(n)}) + (-1)^{n+1} f^{(1)}\otimes\cdots\otimes f^{(n)} \otimes 1\} \\ &=& \Ad(1) \{ 1\otimes f^{(1)}\otimes\cdots\otimes f^{(n-1)} \chi(f^{(n)}) - \Delta(f^{(1)})\otimes\cdots\otimes f^{(n-1)} \chi(f^{(n)}) + \\ & & \cdots + (-1)^n f^{(1)}\otimes\cdots\otimes (\id\otimes\chi) \Delta(f^{(n)})\} + (-1)^{n+1} f \Delta^n(1\1) \chi(1\2). \end{eqnarray*} Let $u =1\1 \chi(1\2)$. Then \begin{eqnarray*} f \,\Delta^n(u) &=& (-1)^n \delta^{n-1}(\tilde{f}) \\ & & + \Ad(1) \{f^{(1)}\otimes\cdots\otimes (\id\otimes\chi) \Delta(f^{(n)}) - f^{(1)}\otimes\cdots\otimes f^{(n-1)} \otimes \chi(f^{(n)})1\}. \end{eqnarray*} Let us show that \begin{equation} \label{it remains} \Ad(1) \{ (\id_{n-1}\otimes \chi)(f) \otimes 1 - (\id_n \otimes \chi) (\id_{n-1}\otimes\Delta)(f) \} =0 \end{equation} for all $f\in C^n(A, B)$. Observe that both sides of \eqref{it remains} belong to $C^{n+1}(A, B)$. We compute : \begin{eqnarray*} \lefteqn{ \Ad(1)\{f^{(1)}\otimes\cdots\otimes f^{(n-1)} \otimes {f^{(n)}}\1 \chi({f^{(n)}}\2)\} = }\\ &=& f^{(1)}\otimes\cdots\otimes f^{(n-1)} \otimes S(1\1)\chi(1\2 f^{(n)}) \\ &=& f^{(1)}\otimes\cdots\otimes 1'\1 f^{(n-1)} \otimes 1'\2 S(1\1) \chi(1\2 f^{(n)}) \\ &=& f^{(1)}\otimes\cdots\otimes 1'\1 1\1 f^{(n-1)} \otimes 1'\2 \chi(1\2 f^{(n)}) \\ &=& f^{(1)}\otimes\cdots\otimes 1\1 f^{(n-1)} \otimes 1\2 \chi(f^{(n)}\\ &=& \Ad(1)\{f^{(1)}\otimes\cdots\otimes f^{(n-1)} \otimes \chi(f^{(n)})1 \}. \end{eqnarray*} Here $1'$ stands for another copy of $1$. We used the definition of a left integral, several identities in a weak Hopf algebra, and equation~\eqref{Delta1} applied to $f$ and $\tilde{f}$. Therefore, $f\Delta^n(u) = (-1)^n \delta^{n-1}(\tilde{f})$. Now $u$ is a non-zero scalar by Lemma~\ref{nondeg of chi}, hence $f = \delta^{n-1}((-1)^n u^{-1}\tilde{f})$, i.e., $\text{Ker}\, \delta^n \subset \text{Im}\, \delta^{n-1}$ and $H^n(A, B)=0$ for $n\geq 1$. \end{proof} \end{section} \begin{section} {The Yetter cohomology of a tensor category and Ocneanu rigidity} \subsection{The definition} The following cohomology of a tensor category with respect to a tensor functor was defined in \cite{Y1,Y2}, motivated by the previous work \cite{CY}, and independently in \cite{Da}. Let $\mC,\mC'$ be tensor categories over a field $k$, and $F:\mC\to \mC'$ be a (unital) tensor functor. For a nonnegative integer $n$, let $\mC^n$ denote the $n$-th Cartesian power of $\mC$. In particular, $\mC^0$ has one object (empty tuple $\emptyset$) and one morphism (identity). Define the functor $T_n: \mC^n\to \mC$ by $T_n(X_1,...,X_n):=X_1\otimes...\otimes X_n$. In particular, $T_0:\mC^0\to \mC$ is defined by $T_0(\emptyset)=\mathbf 1$, and $T_1=\Id$. Let $C^n_F(\mC)=\text{End}(T_n\circ F^{\otimes n})$ (so e.g., $C^0_F(\mC)=\End(\mathbf 1_{\mC'})$). We define a differential $d: C^n(\mC)\to C^{n+1}(\mC)$ by the formula $$ df=\Id\otimes f_{2,...,n+1}-f_{12,...,n+1}+f_{1,23,...,n+1}-...+ (-1)^nf_{1,...,nn+1}+(-1)^{n+1}f_{1,...,n}\otimes \Id, $$ where for instance $f_{12,3,...,n+1}$ is the endomorphism of the product of $n$ objects $F(X_1)\otimes F(X_2),F(X_3),...,F(X_{n+1})$, and we use the identification $F(X_1)\otimes F(X_2)\to F(X_1\otimes X_2)$ defined by the tensor structure on $F$. It is easy to show that $d^2=0$. Thus $(C^\bullet(\mC),d)$ is a complex. We will call the cohomology groups $H^n_F(\mC),n\ge 0$ of this complex the Yetter cohomology groups of $\mC$ with respect to $F$. In the important special case $\mC=\mC'$, $F=\Id$, we will denote this cohomology simply by $H^i(\mC)$ and call it the Yetter cohomology of $\mC$. As usual, low dimensional Yetter cohomology groups have an independent meaning \cite{Y1,Y2,Da}. The group $H^1_F(\mC)$ classifies derivations of $F$ as a tensor functor. The group $H^2_F(\mC)$ classifies first order deformations of the tensor structure on the functor $F$. The group $H^3(\mC)$ classifies first order defomations of the associativity constraint in $\mC$, i.e. of the structure of $\mC$ itself. As usual, obstructions to these deformations live in the cohomology groups one degree higher. To illustrate this definition, we give a few examples. \begin{example} If $\mC$ is $\Rep(H)$ for a Hopf algebra $H$ and $F$ the forgetful functor, then $H^i_F(\mC)$ is the usual Hochschild cohomology of $H$ as a coalgebra. \end{example} \begin{example} Let $G$ be an affine proalgebraic group over $k$, and $\mC=\Rep(G)$ be the category of algebraic representations of $G$. Then $H^i(\mC)=H^i(O(G))$, the Hochschild cohomology of $O(G)$ as a coalgebra, where $O(G)$ is the Hopf algebra of algebraic functions on $G$. In other words, $H^i(\mC)=H^i_{\rm alg}(G,k)$, the algebraic group cohomology of $G$. \end{example} \begin{example} Let $G$ be a reductive algebraic group over $k$ with Lie algebra $\g$. Let $\Rep(G)$ denote the category of algebraic representations of $G$. Then $H^i(\Rep(G))=(\Lambda^i\g)^G$ for all $i$. Indeed, $\Rep(G)=\Rep(O(G)^*)$. Therefore, $C^n(\Rep(G))=(O(G^n)^*)^G$, where $G$ acts diagonally by conjugation. Since $G$ is reductive, the cohomology of this complex is the $G$-invariants in the Hochschild cohomology of $O(G)$ with coefficients in the trivial representation (corresponding to $1\in G$). This cohomology is well known to be $\Lambda\g$, so the answer is $(\Lambda\g)^G$, as desired. Thus, if $\g$ is a simple Lie algebra, then there is no nontrivial derivations or tensor structure deformations of the identity functor, but there exists a unique (up to scaling) first order deformation of the associativity constraint, corresponding to a basis element in $(\Lambda^3\g)^G$. It is easy to guess that this deformation comes from an actual deformation, namely from the deformation of $O(G)$ to the quantum group $O_q(G)$. \end{example} \subsection{Comparison of two complexes} The following proposition shows that for categories and functors coming from weak Hopf algebras, the Yetter complex defined here is the same as the complex defined in the previous section. \begin{proposition} \label{twocomp} Let $A$ be a weak Hopf algebra, $B$ a weak Hopf subalgebra in $A$, $\mC=\Rep(A)$, $\mC'=\Rep(B)$, and $F:\mC\to \mC'$ be the restriction functor. Then the complex $C^\bullet_F(\mC)$ coincides with the complex $ C^\bullet(A,B)$ defined in Section 6. \end{proposition} \begin{proof} Let us show that $\text{End}(T_n\circ F^{\otimes n})= C^n(A,B)$. For this, define a map $\xi: C^n(A,B)\to \text{End}(T_n)$. Recall that for any representations $X_1,...,X_n$ of $A$, the representation $X_1\otimes_\mC ...\otimes_\mC X_n$ is defined to be the image of the projection $\Delta^n(1)$ acting on $X_1\otimes...\otimes X_n$ \cite{NTV}. Therefore, any element $a\in C^n(A,B)$ defines an operator on the usual tensor product $X_1\otimes...\otimes X_n$ which preserves the subspace $X_1\otimes_\mC ...\otimes_\mC X_n$ (since $a=\text{Ad}(1)a$). Let us denote by $\xi(a)$ its restriction to this subspace. Since $a$ is invariant under the adjoint action, we find that $\xi(a)$ is not only a linear map but also a morphism of representations. So we have defined the map $\xi$. Now we need to show that the map $\xi$ is an isomorphism. For this purpose, we construct the inverse map. Take $f\in \text{End}(T_n\circ F^{\otimes n})$. Extend it by zero to the space $\text{Ker} \Delta^n(1)$ in $X_1\otimes...\otimes X_n$. Then we get a linear operator $\tilde f$ on $X_1\otimes...\otimes X_n$, which is functorial in $X_1,...,X_n$. Thus, $\tilde f$ corresponds to a unique element $\hat f$ of $A^{\otimes n}$. It is clear that $\hat f=\text{Ad}(1)\hat f$, since both sides define the same operators in $X_1\otimes...\otimes X_n$. It is also clear that $\hat f$ is invariant under adjoint action, as $f$ is a morphism in the category. Finally, it is clear that $\xi(\hat f)=f$, and $\widehat{\xi(a)}=a$. Thus, $\xi$ is an isomorphism. Finally, it is easy to see that the map $\xi$ identifies the differentials of the two complexes. The proposition is proved. \end{proof} \subsection{Proof of Theorem \ref{van} and Theorem \ref{multocnrig}} Let $\mC,\mC'$ be multifusion categories, and $F:\mC\to \mC'$ a tensor functor between them. We can assume without loss of generality that $F$ is faithful and surjective. Let $R$ be a semisimple algebra, and let $G:\mC'\to R-\text{bimod}$ be a fiber functor on $\mC'$ (we know by now that it necessarily exists for a suitable $R$). Let $B=\End(G)$ be the corresponding weak Hopf algebra. Let $A=\End(G\circ F)$. Then $B$ is a weak Hopf subalgebra of $A$, $\mC'=\Rep(B)$, $\mC=\Rep(A)$, and $F$ is the restriction functor. Thus, by Proposition \ref{twocomp}, the complex $C^\bullet_F(\mC)$ is isomorphic to the complex $C^\bullet(A,B)$. Now, since $A$ is semisimple, it is also cosemisimple by Theorem \ref{sscoss}. Hence, by Theorem \ref{acyclic}, $C^\bullet_F(\mC)$ is acyclic in positive degrees. Theorem \ref{van} is proved. Now let us prove Theorem \ref{multocnrig}. Let $\mC$ be a multifusion category. By theorem \ref{van}, we have $H^i(\mC)=0$, $i>0$, in particular $H^3(\mC)=0$. Now consider the set of all admissible associativity constraints for a tensor category with the same Grothendieck ring as $\mC$. It is an affine algebraic variety. This variety is acted upon by the group of twists. By a standard argument, one finds that the quotient of the tangent space to this variety at the point $\mC$ by the tangent space to the orbit of $\mC$ is equal to $H^3(\mC)$. Since $H^3(\mC)=0$, the group action is locally transitive, and hence has finitely many orbits. We are done. In conclusion we give the formulation of Ocneanu rigidity for semisimple weak Hopf algebras. This is a generalization to the weak case of Stefan's rigidity theorem for semisimple Hopf algebras \cite{S}, which was conjectured in \cite[3.9]{N}. \begin{theorem}\label{rigweak} A regular semisimple finite dimensional weak Hopf algebra does not admit nontrivial deformations in the class of regular weak Hopf algebras. In particular, there are finitely many such weak Hopf algebras in each dimension. \end{theorem} \begin{proof} Let $A$ be a regular semisimple weak Hopf algebra, $\Rep(A)$ its representation category, $R$ its base, and let $F:\Rep(A)\to R-\text{bimod}$ be the forgetful functor. Suppose we have a deformation of $A$. Then by Theorem \ref{multocnrig}, this deformation produces a trivial deformation of $\Rep(A)$ as a tensor category. It is also clear that the base $R$ of $A$, being a semisimple algebra, is deformed trivially. This means that this deformation comes from a deformation of the forgetful functor $F$ as a tensor functor (indeed, a regular weak Hopf algebra is completely determined by the associated forgetful functor, since the symmetric separability idempotent is unique). But the deformation of $F$ must also be trivial, by Theorem \ref{ocnrigfun}. This implies the first statement of the theorem. The second statement follows from the first one as in the proof of Theorem \ref{multocnrig}. \end{proof} \end{section} \section{Frobenius-Perron dimensions} In this section we will give some applications of the Frobenius-Perron theorem to the theory of fusion categories. We work with categories over $\CC$. \subsection{Definition of the Frobenius-Perron dimensions} We start by recalling the Frobenius-Perron theorem (see \cite{Ga}). \begin{theorem} Let $A$ be a square matrix with nonnegative entries. \begin{enumerate} \item[(1)] $A$ has a nonnegative real eigenvalue. The largest nonnegative real eigenvalue $\lambda(A)$ of $A$ dominates absolute values of all other eigenvalues of $A$. \item[(2)] If $A$ has strictly positive entries then $\lambda(A)$ is a simple positive eigenvalue, and the corresponding eigenvector can be normalized to have strictly positive entries. \item[(3)] If $A$ has an eigenvector $\mathbf f$ with strictly positive entries, then the corresponding eigenvalue is $\lambda(A)$. \end{enumerate} \end{theorem} Now let $A$ be a unital based ring of finite rank in the sense of Lusztig (see, e.g., \cite{O} for a definition). Let $b_i$ be the basis of $A$ ($b_0=1$), and $[b_i]$ be the matrix of multiplication by $b_i\in A$ in the basis $b_j$. This matrix has nonnegative entries. Let $\lambda_i$ be the largest nonnegative eigenvalue of $[b_i]$. Since $b_i$ is clearly not nilpotent, this eigenvalue is actually positive. We will call $\lambda_i$ the {\it Frobenius-Perron dimension} of $b_i$. We note that Frobenius-Perron dimensions for commutative based rings were defined and used in the book \cite{FK}. \begin{theorem}\label{fp0} The assignment $b_i\to \lambda_i$ extends to a homomorphism of algebras $A\to \mathbb R$. \end{theorem} \begin{proof} For any $j$ define $z_j=\sum_i b_ib_jb_i^*$. We claim that $z_j$ is central. Indeed, let $b_ib_j=\sum_k N_{ij}^kb_k$. Then $N_{ri}^k=N_{k^*r}^{i^*}$, so $$ b_rz_j=b_r\sum_i b_ib_jb_i^*=\sum_{k,i}N_{ri}^kb_kb_jb_i^*= \sum_{k,i}N_{k^*r}^{i^*}b_kb_jb_{i^*}= \sum_k b_kb_jb_{k^*}b_r=z_jb_r. $$ Let $z=\sum_j z_j$. Since $z_j$ contain $b_j$ as a summand, the matrix $[z]$ of multiplication by $z= \sum_j z_j$ has strictly positive entries. Let $\mathbf f$ be the unique eigenvector of $[z]$ with positive entries and $f_0=1$ (its existence and uniqueness follows from the Frobenius-Perron theorem). Then $[b_j]\mathbf f$ must be a multiple of $\mathbf f$, since $\mathbf f$ has positive entries and $[z][b_j]=[b_j][z]$. So by the Frobenius-Perron theorem, $[b_j]\mathbf f = \lambda_j\mathbf f$, whence the statement follows. \end{proof} We note that $b_i\to \lambda_i$ is the unique homomorphism of $A$ to $\mathbb R$ such that images of the basis vectors are positive. This is demonstrated by the following simple and probably well known lemma. Let $A$ be a $\mathbb Z_+$-ring of finite rank with $\mathbb Z_+$-basis $b_i$ (see \cite{O} for definition). \begin{lemma}\label{uniq} If there exists a homomorphism $\phi:A\to \mathbb R$ such that $\phi(b_i)>0$ for all $i$, then $\phi$ is unique. \end{lemma} \begin{proof} Let $[b_i]$ be the matrix of multiplication by $b_i$ in the basis $b_j$, and $\mathbf f$ be the column vector with entries $\phi(b_j)$. Then $[b_i]\mathbf f=\phi(b_i)\mathbf f$. But $[b_i]$ is a matrix with nonnegative entries, and $\mathbf f$ has positive entries. Thus, by the Frobenius-Perron theorem, $\phi(b_i)$ is the largest positive eigenvalue of $[b_i]$, and hence is uniquely determined. Lemma is proved. \end{proof} \begin{remark} M.~M\"uger pointed out to us that for any $i$ one has $\lambda_i\ge 1$; moreover if $\lambda_i<2$ then $\lambda_i= 2\cos(\pi/n)$ for some integer $n\ge 3$ (this follows from the well known Kronecker theorem on the eigenvalues of positive integer matrices). \end{remark} One can also define the Frobenius-Perron dimensions of basis elements of a based module over a unital based ring. Indeed, let $M$ be an indecomposable based $A$-module with basis $m_j$ (see \cite{O} for definition). \begin{proposition}\label{frobmod0} There exists a unique up to scaling common eigenvector $\mathbf m$ of the matrices $[b_i]|_M$ of the operators $b_i$ in $M$, which has strictly positive entries. The corresponding eigenvalues are $\lambda_i$. \end{proposition} \begin{proof} Let $z$ be the element from the proof of Theorem \ref{fp0}. The matrix $[z]|_M$ has strictly positive entries. Let $\mathbf m$ be an eigenvector of $[z]|_M$ with positive entries. It exists and is unique up to a positive scalar, by the Frobenius-Perron theorem. Hence, it is an eigenvector of $[b_i]|_M$, and the eigenvalues are $\lambda_i$ by Lemma \ref{uniq}. \end{proof} Now let $\mC$ be a fusion category. Let $K(\mC)$ be the Grothendieck ring of $\mC$. For any object $X\in \mC$, define the Frobenius-Perron dimension of $X$, ${\rm FPdim}(X)$, to be the largest positive eigenvalue of the matrix $[X]$ of multiplication by $X$ in $K(\mC)$. Since $K(\mC)$ is a unital based ring of finite rank, Theorem \ref{fp0} implies the following result. \begin{theorem}\label{fp} The assignment $X\to {\rm FPdim}(X)$ extends to a homomorphism of algebras $K(\mC)\to \mathbb R$. \end{theorem} Similarly, one can define Frobenius-Perron dimensions of simple objects of a module category over a fusion category. Namely, let $\mM$ be an indecomposable module category over a fusion category $\mC$. For any $X\in \mC$ let $[X]_\mM$ be the matrix of action of $X$ on the Grothendieck group of $\mM$. Since the Grothendieck group $K(\mM)$ is an indecomposable based module over $K(\mC)$, Proposition \ref{frobmod0} implies \begin{proposition}\label{frobmod} There exists a unique up to scaling common eigenvector $\mathbf m$ of the matrices $[X]_\mM$, which has strictly positive entries. The corresponding eigenvalues are ${\rm FPdim}(X)$. \end{proposition} The entries $\mu_i$ of $\mathbf m$ are called the Frobenius-Perron dimensions of simple objects $M_i$ of $\mM$. Unlike Frobenius-Perron dimensions of objects of $\mC$, they are well defined only up to scaling. \subsection{Frobenius-Perron dimension of a fusion category} Let $\mC$ be a fusion category. Define the Frobenius-Perron dimension ${\rm FPdim}(\mC)$ of $\mC$ to be the sum of squares of Frobenius-Perron dimensions of simple objects of $\mC$. Also, define the regular representation $R_\mC$ to be the (virtual) object \linebreak $\oplus_{X\in Irr(\mC)} {\rm FPdim}(X)X$, where $Irr(\mC)$ is the set of isomorphism classes of simple objects of $\mC$. It is clear that ${\rm FPdim}(\mC)={\rm FPdim}(R_\mC)$. \begin{proposition} \label{freeness} Let $F:\mC\to \mD$ be a surjective tensor functor between fusion categories. Then $F(R_\mC)=\frac{{\rm FPdim}(\mC)}{{\rm FPdim}(\mD)}R_\mD$. \end{proposition} \begin{proof} Let $X$ be an object of $\mC$ such that $F(X)$ contains all simple objects of $\mD$. It exists since $F$ is surjective. The vectors of multiplicities of $F(R_\mC)$ and $R_\mD$ are eigenvectors of the matrix of multiplication by $F(X)$. The matrix and the vectors have strictly positive entries. By the Frobenius-Perron theorem, this implies that $F(R_\mC)$ and $R_\mD$ differ by a positive scalar. This scalar is readily found to be the claimed one by taking ${\rm FPdim}$ of $F(R_\mC)$ and $R_\mD$. \end{proof} \begin{corollary}\label{freeness1} Let $A\subset B$ be an inclusion of finite dimensional semisimple quasi-Hopf algebras. Then $B$ is free as a left $A$-module. \end{corollary} \begin{proof} In this case Frobenius-Perron dimensions coincide with the usual ones, so if $\mC=\Rep(B)$, $\mD=\Rep(A)$ then $R_\mC=B$, $R_\mD=A$, proving the claim. \end{proof} \begin{remark} For Hopf algebras, Corollary \ref{freeness1} is the well known freeness theorem of Nichols and Zoeller \cite{NZ} (it is true in the nonsemisimple case as well). In the quasi-Hopf case it was independently obtained by P. Schauenburg (without the semisimplicity assumption), \cite{Scha}. \end{remark} \begin{corollary}\label{divis} If $F:\mC\to \mD$ is surjective then ${\rm FPdim}(\mC)$ is divisible by ${\rm FPdim(\mD)}$ (i.e. the ratio is an algebraic integer). \end{corollary} \begin{proof} Taking the multiplicity of the neutral object of $\mD$ in Proposition \ref{freeness}, we find $\sum_{X\in Irr(\mC)}{\rm FPdim}(X)[F(X):\mathbf 1]= \frac{{\rm FPdim}(\mC)}{{\rm FPdim}(\mD)}$, which yields the statement. \end{proof} \begin{proposition}\label{squar} For any fusion category $\mC$, one has ${\rm FPdim}(Z(\mC))={\rm FPdim}(\mC)^2$. \end{proposition} \begin{remark} If $\mC$ is a representation category of a quasi-Hopf algebra $A$, then Proposition \ref{squar} is saying that the Frobenius-Perron dimension of $Z(\mC)$ is $\dim(A)^2$. This is a simple consequence of the result of Hausser and Nill \cite{HN}, who constructed a quasi-Hopf algebra $D(A)$ (the double of $A$), isomorphic to $A\otimes A^*$ as a vector space, such that $Z(\mC)=\Rep(D(A))$. \end{remark} \begin{proof} Recall from Section 5 that $V\to I(V)$ denotes the induction functor. By Proposition \ref{I(1)}, we have \begin{eqnarray*} {\rm FPdim}(I(V)) &=& \oplus_{X\in Irr(Z(\mC))}{\rm FPdim}(X)[I(V):X]\\ &=& \oplus_{X\in Irr(Z(\mC))}{\rm FPdim}(X)[X|_\mC:V] \\ &=& [R_{Z(\mC)}:V]=\frac{{\rm FPdim}Z(\mC)}{{\rm FPdim}\mC} [R_\mC:V]=\frac{{\rm FPdim}Z(\mC)}{{\rm FPdim}\mC}{\rm FPdim}(V). \end{eqnarray*} In particular, ${\rm FPdim}(I(\mathbf 1))=\frac{{\rm FPdim}Z(\mC)}{{\rm FPdim}\mC}$. On the other hand, by Proposition \ref{I(1)}, $I(\mathbf 1)=\oplus_{Y\in Irr(\mC)}Y\otimes Y^*$, which implies the desired statement (since the dimension of the right hand side is ${\rm FPdim}(\mC)$). The proposition is proved. \end{proof} \begin{corollary}\label{fpdual} Let $\mC$ be a fusion category and $\mC^*$ be its dual with respect to an indecomposable module category. Then ${\rm FPdim}(\mC)={\rm FPdim}(\mC^*)$. \end{corollary} \begin{proof} This follows from Proposition \ref{squar} and the fact that $Z(\mC)$ is equivalent to $Z(\mC^*)$ \cite{O1,Mu2}. \end{proof} \begin{proposition}\label{subdivi} Let $\mC$ be a full tensor subcategory of a fusion category $\mD$. Then ${\rm FPdim}(\mD)/{\rm FPdim}(\mC)$ is an algebraic integer. \end{proposition} \begin{proof} Consider the decomposition of $\mD$, as a left module category over $\mC$, in a direct sum of indecomposables. We have $\mD=\oplus_{i=1}^r \mM_r$. Let us consider the duals of $\mC$ and $\mD$ with respect to this module category. We have $\mD^*=\mD^{op}$, and $\mC^*$ is a multi-fusion category $\mC^*=\oplus_{i,j=1}^r \mC^*_{ij}$, where $\mC^*_{ij}:={\rm Fun}(\mM_j,\mM_i)$. Note that the Frobenius-Perron dimensions of $\mC_{ii}^*$ are equal to ${\rm FPdim}(\mC)$ by Corollary \ref{fpdual}. Let $E_{ij}$ denote nonzero virtual objects of $\mC^*_{ij}$, which satisfy the condition \linebreak $X\otimes E_{ij}\otimes Y= {\rm FPdim}(X){\rm FPdim}(Y)E_{ij}$, $X\in \mC^*_{ii},Y\in \mC^*_{jj}$. Since the category $\mC_{ij}^*$ is an indecomposable bimodule over $\mC_{ii}^*$ and $\mC_{jj}^*$, by the Frobenius-Perron theorem, the object $E_{ij}$ is unique up to scaling. We require that the coefficients in $E_{ij}$ be positive, which determines it uniquely up to a positive scalar. To fix the scaling completely, we will require that $E_{ii}$, which is a multiple of the regular representation $R_{ii}$ in $\mC^*_{ii}$, be actually equal to ${\rm FPdim}(\mC)^{-1}R_{ii}$. Further, we normalize $E_{1j}$ arbitrarily, then normalize $E_{j1}$ by the condition $E_{j1}\otimes E_{1j}=E_{jj}$, and finally define $E_{jk}$ as $E_{j1}\otimes E_{1k}$. It is easy to see that $E_{ij}$ satisfy the matrix unit relations $E_{ij}\otimes E_{kl}=\delta_{jk}E_{il}$. Now let $F:\mD^*\to \mC^*$ be the natural functor, $R$ be the regular representation of $\mD^*$, and consider $F(R)\in \mC^*$. We claim that $F(R)=\oplus_{i,j} b_{ij}E_{ij}$, where $b_{ij}>0$. Indeed, $F(R)^{\otimes 2}={\rm FPdim}(\mD)F(R)$, so for any $i$ the virtual object $L_i:=\oplus_j b_{ij}E_{ij}$ is an eigen-object of the functor of right multiplication by $F(R)$. Since the category $\oplus_j \mC_{ij}^*$ is an indecomoposable right module over $\mC^*$, by the Frobenius-Perron theorem such eigen-object with positive coefficients is unique up to scaling. Thus, for any $X\in \mC_{ii}^*$, $X\otimes L_i=c(X)L_i$, where $c(X)$ is a scalar. By the Frobenius-Perron theorem $c(X)={\rm FPdim}(X)$. Similarly, $P_j:=\oplus_i b_{ij}E_{ij}$ satisfies $P_j\otimes X={\rm FPdim}(X)P_j$, $X\in \mC^*_{jj}$. This implies the claim. Now let $B$ be the matrix $(b_{ij})$. Since $F(R)^{\otimes 2}={\rm FPdim}(\mD)F(R)$, and $E_{ij}$ behave like matrix units, we have $B^2={\rm FPdim}(\mD)B$. Thus eigenvalues of $B$ are ${\rm FPdim}(\mD)$ and zero, and $B$ is diagonalizable. Since $b_{ij}>0$, by the Frobenius-Perron theorem the multiplicity of the eigenvalue ${\rm FPdim}(\mD)$ is $1$, so the rank of $B$ is $1$. Hence $\Tr(B)={\rm FPdim}(\mD)$. On the other hand, let us compute $b_{ii}$. We have $$ \sum_{X\in Irr(\mD)}{\rm FPdim}(X)[F(X):\mathbf 1_{ii}]= b_{ii}/{\rm FPdim}(\mC) $$ (as $E_{ii}=R_{ii}/{\rm FPdim}(\mC)$). Thus, $b_{ii}/{\rm FPdim}(\mC)$ is an algebraic integer. Summing over all $i$, we find that $\Tr(B)/{\rm FPdim}(\mC)={\rm FPdim}(\mD)/{\rm FPdim}(\mC)$ is an algebraic integer, as desired. \end{proof} \begin{remark} Since $b_{ii}\ge 1$, we see that ${\rm FPdim}(\mD)/{\rm FPdim}(\mC)$ dominates the number of indecomposable blocks in $\mD$ as a left $\mC$-module category.\end{remark} \begin{remark} Let $\mM_1=\mC$. Then $\mC_{i1}^*=\mM_i$. It is easy to see that $F(R)_{i1}=\sum_{X\in Irr(\mM_i)} {\rm FPdim}(X)X$. By uniqueness of eigen-object of $F(R)$ in $\oplus_j\mC^*_{ji}$, we have $F(R)_{i1}\otimes F(R)_{1i}= {\rm FPdim}(\mC)F(R)_{ii}$. Thus, $$ [F(R)_{ii}:\mathbf 1_{ii}]=\dim{\rm End}(F(R)_{i1},F(R)_{i1})/{\rm FPdim}(\mC)={\rm FPdim}(\mM_i)/{\rm FPdim}(\mC). $$ This implies the refinement of Theorem \ref{subdivi} stating that ${\rm FPdim}(\mM_i)/{\rm FPdim}(\mC)$ is an algebraic integer. \end{remark} \begin{remark} We do not know if the global dimension of a fusion category is divisible by the global dimension of its full tensor subcategory. \end{remark} \begin{example} The following example shows that Proposition \ref{subdivi} is a property of tensor categories and not just of based rings. Let $G$ be a finite group and $n$ a nonnegative integer. Consider the based ring $A_n(G)$ with basis $G\cup \lbrace X\rbrace$, with multiplication given by the usual multiplication in $G$ and the relations $gX=Xg=X,g\in G$, and $X^2=nX+\sum_{g\in G}g$. Such rings for special $n,G$ can arise as Grothendieck rings of fusion categories (for example, the categories of \cite{TY}, or representation categories of affine groups ${\rm Aff}(\mathbb F_q)$). We claim that if $A_n(G)$ is categorifiable, then the number $(n+\sqrt{n^2+4|G|})^2/4|G|$ is an algebraic integer. In particular, for $n\ne 0$ one has $|G|\le 2n^2$. (For $n=1$ we find $|G|=1,2$, so $G=1$ or $G=\mathbb Z/2\mathbb Z$, which was proved by T.~Chmutova even for non-rigid categories). Indeed, let $\mC$ be a fusion category with Grothendieck ring $A_n(G)$. Then the Frobenius-Perron dimensions of elements of $G$ are $1$, while the Frobenius-Perron dimension of $X$ is the largest root of the equation $x^2-nx-m=0$, where $m=|G|$. Thus $x=\frac{1}{2}(n+\sqrt{n^2+4m})$. By Proposition \ref{subdivi}, $x^2/m$ should be an algebraic integer. \end{example} \begin{proposition}\label{dimgradi} If a fusion category $\mC$ is faithfully graded by a finite group $G$, then the Frobenius-Perron dimensions of $\mC_g$ are equal for all $g\in G$. In particular, $|G|$ divides ${\rm FPdim}(\mC)$. \end{proposition} \begin{proof} Let $R$ be the regular object of $\mC$. Then $R=\oplus R_g$, where $R_g$ is a virtual object of $\mC_g$. We have $R\otimes R_h={\rm FPdim}(R_h)R$. Taking the $gh$-th component of this equation, we find $R_g\otimes R_h={\rm FPdim}(R_h)R_{gh}$. Similarly, $R_g\otimes R={\rm FPdim}(R_g)R$, which yields $R_g\otimes R_h={\rm FPdim}(R_g)R_{gh}$. Since $R_{gh}\ne 0$, we get ${\rm FPdim}(R_g)={\rm FPdim}(R_h)$, i.e. ${\rm FPdim}(R_g)={\rm FPdim}(X)/|G|$. \end{proof} \subsection{Comparison of the global and Frobenius-Perron dimensions} \begin{proposition} \label{domi} Let $\mC$ be a fusion category. Then for any simple object $V$ of $\mC$ one has $|V|^2\le {\rm FPdim}(V)^2$, and hence $\dim(\mC)\le {\rm FPdim}(\mC)$. Hence, if $\dim(\mC)={\rm FPdim}(\mC)$ then $|V|^2={\rm FPdim}(V)^2$ for all simple objects $V$. \end{proposition} \begin{proof} Let $X_i$ be the $i$-th simple object of $\mC$, and $d_i=d(X_i)$ be its dimension introduced in the proof of Theorem \ref{posi}. Then $|d_i|^2=|V|^2$, and $d_i$ is an eigenvalue of the matrix $T_i$. It follows from Section 3 that the absolute values of the entries of $T_i$ are dominated by the corresponding entries of the matrix $N_i$ of multiplication by $X_i$. Let $f_i$ be the Frobenius-Perron dimension of $X_i$, and consider a norm on the Grothendieck group $Gr_{\CC}(\mC)$ given by $||\sum a_iX_i||=\sum f_i|a_i|$. Let us estimate the norm of the operator $T_i$. We have $$ ||T_i\mathbf a||=\sum_jf_j|\sum_k(T_i)_{jk}a_k|\le \sum_{j,k}f_j(N_i)_{jk}|a_k|= f_i\sum_kf_k|a_k|=f_i||\mathbf a|| $$ Thus, $||T_i||\le f_i$. This implies that $|V|^2\le f_i^2$, which proves the proposition. \end{proof} \begin{proposition}\label{divi} Let $\mC$ be a fusion category of global dimension $D$ and Frobenius-Perron dimension $\Delta$. Then $D/\Delta$ is an algebraic integer, which is $\le 1$. \end{proposition} \begin{proof} By Proposition \ref{domi}, we only need to prove that $D/\Delta$ is an algebraic integer. First of all, we can assume without loss of generality that $\mC$ is a spherical category. Indeed, if $\mC$ is not spherical, then, as explained in Proposition \ref{sphpiv}, we can construct a spherical fusion category $\tilde \mC$ which projects onto $\mC$. It is easy to see that this category has global dimension $2D$ and Frobenius-Perron dimension $2\Delta$, so the ratio of dimensions is the same. Now assume that $\mC$ is spherical. Consider the Drinfeld center $Z(\mC)$. This is a modular category of global dimension $D^2$, and by Proposition \ref{squar}, its Frobenius-Perron dimension is $\Delta^2$. Let $S=(s_{ij})$ be its S-matrix. It follows from Verlinde's formula (see \cite{BaKi}) that the commuting matrices $N_i$ of multiplication by simple objects $X_i$ of $Z(\mC)$ have common eigenvectors $\mathbf v_j$ with eigenvalues $s_{ij}/s_{0j}$. In particular, there exists a unique (by nondegeneracy of $S$) distinguished label $j=r$ such that $s_{ir}/s_{0r}=f_i$, the Frobenius-Perron dimension of $X_i$. Since $S$ is symmetric, and $S^2$ is the charge conjugation matrix, we have $\Delta^2=\sum_i f_i^2=\delta_{rr^*}/s_{0r}^2$. Thus $r=r^*$ and $\Delta^2=1/s_{0r}^2=D^2/d_r^2$, where $d_r$ is the categorical dimension of the $r$-th simple object. So $D^2/\Delta^2=d_r^2$ is an algebraic integer. \end{proof} \subsection{Pseudo-unitary categories} Let us say that a fusion category $\mC$ is {\em pseudo-unitary} if $\dim(\mC)={\rm FPdim}(\mC)$. This property is automatically satisfied for unitary categories, which occur in the theory of operator algebras, but it fails, for example, for the Yang-Lee category. \begin{proposition}\label{uniqpivo} Any pseudo-unitary fusion category $\mC$ admits a unique pivotal (in fact, spherical) structure, with respect to which the categorical dimensions of all simple objects are positive, and coincide with their Frobenius-Perron dimensions. \end{proposition} \begin{proof} Let $g: Id\to ****$ be an isomorphism of tensor functors, $a:Id\to **$ an isomorphism of additive functors, such that $a^2=g$, $d_i$ be the dimensions of simple objects $X_i$ associated to $a$ as in Section 3.1, and $\mathbf d$ the vector with components $d_i$. Then $T_i\mathbf d=d_i\mathbf d$, and $|d_i|=f_i$ is the Frobenius-Perron dimension of $X_i$. But $|(T_i)_{jk}|\le (N_i)_{jk}$. Thus, $$ f_if_j=|d_id_j|=|\sum (T_i)_{jk}d_k|\le \sum (N_i)_{jk}f_k=f_if_j $$ This means that the inequality in this chain is an equality. In particular $(T_i)_{jk}=\pm (N_i)_{jk}$, and the argument of $d_id_j$ equals the argument of $(T_i)_{jk}d_k$ whenever $(N_i)_{jk}>0$. This implies that whenever $X_k$ occurs in the tensor product $X_i\otimes X_j$, the ratio $d_i^2d_j^2/d_k^2$ is positive. Thus, the automorphism of the identity functor $\sigma$ defined by $\sigma|_{X_i}=d_i^2/|d_i|^2$ is a tensor automorphism. Let us twist $g$ by this automorphism. After this twisting, the new dimensions $d_i$ will be real. Thus, we can assume without loss of generality that $d_i$ were real from the beginning. It remains to twist the square root $a$ of $g$ by the automorphism of the identity functor $\tau$ given by $\tau|_{X_i}=d_i/|d_i|$. After this twisting, new $T_i$ is $N_i$ and new $d_k$ is $f_k$. This means that $\beta_i^{jk}=1$ and $a$ is a pivotal structure with positive dimensions. It is obvious that such a structure is unique. We are done. \end{proof} \subsection{Fusion categories with integer Frobenius-Perron dimension} \begin{proposition}\label{intFP} Let $\mC$ be a fusion category, such that $\Delta:={\rm FPdim}(\mC)$ is an integer. Then $\mC$ is pseudo-unitary. In particular, the category of representations of a semisimple finite dimensional quasi-Hopf algebra is pseudo-unitary. \end{proposition} \begin{proof} Let $D$ be the global dimension of $\mC$. Let $D_1=D,D_2,...,D_N$ be all the algebraic conjugates of $D$, and $g_1,...,g_N$ the elements of $Gal(\bar{\mathbb Q}/{\mathbb Q})$ such that $D_i=g_i(D)$. Applying Propostion \ref{divi} to the category $g_i(\mC)$, we find that $D_i/\Delta$ is an algebraic integer $\le 1$. Then $\prod_i (D_i/\Delta)$ is an algebraic integer $\le 1$. But this product is a rational number. So it equals $1$ and hence all factors equal $1$. Hence $D=\Delta$, as desired. \end{proof} \begin{remark} We note that Proposition \ref{intFP} may be regarded as a categorical analog of the well known theorem of Larson and Radford \cite{LR2}, saying that in a semisimple Hopf algebra, the square of the antipode is the identity. More precisely, using Proposition \ref{intFP}, one can obtain the following proof of the Larson-Radford theorem (which is close to the original proof). Let $A$ be a finite dimensional semisimple Hopf algebra with antipode $S$. Since $S$ has finite order, $S^2$ is semisimple, and its eigenvalues are roots of $1$. Now, for every simple $A$-module $V$, $S^2$ preserves ${\rm End}(V)\subset A$ (as $V^{**}$ is isomorphic to $V$), and $|V|^2=\Tr(S^2|_{{\rm End}(V)})$, while ${\rm FPdim}(V)^2= \dim(V)^2=\dim{\rm End}(V)$. Thus, by Proposition \ref{intFP}, $\Tr(S^2|_{{\rm End}(V)})=\dim{\rm End}(V)$. Since $S^2$ is semisimple and its eigenvalues are roots of unity, we have $S^2=1$. \end{remark} \begin{remark} In general, the Frobenius-Perron dimension of a fusion category may not only be different from its global dimension, but may in fact be not conjugate to the global dimension by the Galois group action. The same is true for the numbers ${\rm FPdim}(V)^2$ and $|V|^2$ for a simple object $V$. An example is a tensor product of the Yang-Lee category and its Galois conjugate. It has global dimension $5$, but Frobenius-Perron dimension $5(3+\sqrt{5})/2$. Also, this category has a simple object $V$ with $|V|^2=1$ but ${\rm FPdim}(V)^2=(7+3\sqrt{5})/2$. \end{remark} One may wonder if a fusion category $\mC$ with integer ${\rm FPdim}(\mC)$ must have integer Frobenius-Perron dimensions of simple objects. Unfortunately, the answer is ``no'' (see \cite{TY}, where dimensions of some simple objects are square roots of integers). However, the following weaker result is true. Let $\mC_{ad}$ be the full tensor subcategory of $\mC$ generated by objects occuring in $X\otimes X^*$, where $X\in \mC$ is simple (if $\mC=\Rep(G)$ for a group $G$ then $\mC_{ad}$ is the representation category of the adjoint group $G/Z(G)$ -- the quotient by the center). \begin{proposition} \label{roots} Suppose that ${\rm FPdim}(\mC)$ is an integer. The Frobenius-Perron dimensions of simple objects in $\mC_{ad}$ are integers. Furthermore, for any simple object $X\in \mC$, one has ${\rm FPdim}(X)=\sqrt{N}$, where $N$ is an integer. \end{proposition} \begin{proof} The second statement follows from the first one, since for any simple object $X\in \mC$ the object $X\otimes X^*$ lies in $\mC_{ad}$. To prove the first statement, consider the object $B=\oplus_i X_i\otimes X_i^*\in \mC_{ad}$, where $X_i$ are the simple objects of $\mC$. For some $n$ the matrix of multiplication by the object $B^{\otimes n}$ has strictly positive entries. On the other hand, the vector of dimensions of objects of $\mC_{ad}$ is an eigenvector of this matrix with integer eigenvalue ${\rm FPdim}(\mC)^n$. Thus, by the Frobenius-Perron theorem its entries are rational (as $d_0=1$), and hence integer (as they are algebraic integers). \end{proof} \subsection{Fusion categories of Frobenius-Perron dimension $p^n$} \begin{theorem}\label{filtrat} Let $\mC$ be a fusion category of Frobenius-Perron dimension $p^n$, where $n\ge 1$ is an integer, and $p$ is a prime. Then: (i) There exists a nontrivial automorphism $g\in {\rm Aut}_\otimes (Id_\mC)$ of order $p$, so that $\mC$ is $\Bbb Z/p\Bbb Z$ graded: $\mC=\oplus_{j=0}^{p-1}\mC_j$, where $\mC_j$ is spanned by simple objects $X$ such that $g|_X= e^{2\pi ij/p}$. The Frobenius-Perron dimension of the fusion category $\mC_0$ and module categories $\mC_j$ over it are equal to $p^{n-1}$. (ii) $\mC$ admits a filtration by fusion subcategories $\mC\supset \mC^{(1)}\supset...\supset \mC^{(n)}=<\mathbf 1>$, where $\dim \mC^{(i)}=p^{n-i}$. \end{theorem} \begin{remark} For representation categories of semisimple Hopf algebras, this theorem says that a semisimple Hopf algebra of order $p^n$ has a central group-like element of order $p$, so one can take a quotient and obtain a Hopf algebra of dimension $p^{n-1}$. This result is due to Masuoka \cite{Ma}. \end{remark} \begin{proof} It is clear that (ii) follows from (i), so it suffices to prove (i). Since the Frobenius-Perron dimension of $\mC$ is an integer, it is pseudo-unitary. In particular, it is spherical. By Proposition \ref{moddivi}, the dimensions of all objects $X\in Z(\mC)$ which occur in $I(\mathbf 1)$, are divisors of $p^n$. In addition, they are all integers, since the image of $X$ in $\mC$ belongs to $\mC_{ad}$. By the class equation (Proposition \ref{cleq}), there have to be at least $p$ such objects $X$, which are invertible. This means, the group of invertible objects of $Z(\mC)$ is nontrivial. By Proposition \ref{subdivi}, the order of this group must divide $p^{2n}$. In particular, this group contains a cyclic subgroup $G$ of order $p$. The group $G$ maps trivially to $\mC$, so by Proposition \ref{gradi} we get a faithful $G^\vee$-grading on $\mC$. The rest follows from Proposition \ref{dimgradi}. The theorem is proved. \end{proof} The following corollary of this result is a categorical generalization of (the semisimple case of) the well known result of Y.~Zhu \cite{Zhu} saying that a Hopf algebra of prime dimension is a group algebra. \begin{corollary}\label{zhu} Let $\mC$ be a fusion category of Frobenius-Perron dimension $p$, where $p$ is a prime number. Then $\mC$ is the category of representations of the group $\mathbb Z/p\mathbb Z$, in which associativity is defined by a cohomology class in $H^3(\mathbb Z/p\mathbb Z,\CC^*)=\mathbb Z/p\mathbb Z$. \end{corollary} \begin{proof} The corollary is immediate from Theorem \ref{filtrat}. Indeed, this theorem shows that $\mC$ is faithfully graded by $\Bbb Z/p\Bbb Z$, so by Proposition \ref{dimgradi}, all $\mC_g$ are 1-dimensional, i.e. $\mC_g$ contains a single simple object, which is invertible. Thus, $\mC$ has $p$ simple objects, which are invertible, and we are done. \end{proof} \begin{corollary} \label{quasih} A semisimple quasi-Hopf algebra of prime dimension $p$ is twist equivalent to the group algebra of a cyclic group of order $p$, with associator defined by a 3-cocycle. \end{corollary} The following proposition classifies fusion categories of dimension $p^2$. \begin{proposition} Let $\mC$ be a fusion category of Frobenius-Perron dimension $p^2$, where $p$ is a prime number. If $p$ is odd, then simple objects of $\mC$ are invertible, so $\mC$ is the category of representations of a group of order $p^2$ with associativity defined by a 3-cocycle. If $p=2$, then either $\mC$ is the representation category of a group of order $4$ with a 3-cocycle, or it has 3 objects of dimensions $1,1$ and $\sqrt{2}$, and thus belongs to the list of \cite{TY}. \end{proposition} \begin{proof} Assume that not all simple objects of $\mC$ are invertible. Our job is to show that $p=2$. Indeed, in this case the dimension is 4, and we have a subcategory of dimension 2 by Theorem \ref{filtrat}. Since dimensions of all simple objects are square roots of integers, there is room for only one new object of dimension $\sqrt{2}$. All such categories are listed in \cite{TY} (there are two of them, and they correspond in physics to ``Ising model''). Assume the contrary, i.e. that $p$ is odd. We have a grading of $\mC$ by $\Bbb Z/p\Bbb Z$, and for all $j\in \Bbb Z/p\Bbb Z$ the dimension of $\mC_j$ is $p$ by Proposition \ref{dimgradi}. Clearly, $\mC_0$ holds $p$ invertible simple objects, and no other simple objects. It is also clear that all invertible objects of $\mC$ are contained in $\mC_0$ (as by Proposition \ref{subdivi}, the order of the group of invertibles is a divisor of $p^2$). Further, any object $X$ in $\mC_j$, $j\ne 0$, must be invariant under tensoring (on either side) with invertibles from $\mC_0$, since otherwise the dimension of $\mC_j$ would be greater than $p$ (as the dimension of $X$ is greater than $1$). This means that $X\otimes X^*$ contains all the invertibles from $\mC_0$, and hence the dimension of $X$ is at least $\sqrt{p}$. Since the dimension of $\mC_j$ is $p$, we find that the dimension of $X$ is exactly $\sqrt{p}$, and $X$ is the only simple object of $\mC_j$. Let $j,k\in \Bbb Z/p\Bbb Z$ be nonzero and distinct (they exist since $p>2$). Let $X_j,X_k$ be the only simple objects of $\mC_j,\mC_k$. Then $X_j\otimes X_k^*$ is a $p$-dimensional object, which is an integer multiple of a $\sqrt{p}$-dimensional object, which spans $\mC_{j-k}$. Contradicton. The proposition is proved. \end{proof} \subsection{Categories with integer Frobenius-Perron dimensions of simple objects} \begin{theorem}\label{quasihopf} Let $\mC$ be a fusion category. The following two conditions are equivalent. \begin{enumerate} \item[(i)] for all $X\in \mC$, the number ${\rm FPdim}(X)$ is an integer. \item[(ii)] $\mC$ is a representation category of a finite dimensional quasi-Hopf algebra $A$. \end{enumerate} \end{theorem} \begin{remark} It is shown in \cite{EG2} that $A$ is determined uniquely up to twisting. \end{remark} \begin{proof} (i)$\to$ (ii). Let $F:\mC\to {\rm Vect}$ be the additive functor that maps simple objects $X_i$ of $\mC$ to vector spaces $F(X_i)$ of dimension ${\rm FPdim}(X_i)$. For any $i,j$, choose any isomorphism $J_{ij}: F(X_i)\otimes F(X_j)\to F(X_i\otimes X_j)$ (this is possible by Theorem \ref{fp}). Then by a standard reconstruction argument the algebra $\End_k(F)$ has a natural structure of a quasi-Hopf algebra (changing $J$ corresponds to twisting the quasi-Hopf algebra). We note that $A$ is not, in general, a Hopf algebra, since $J$ is not required to be a tensor structure on $F$. (ii)$\to$ (i). This is clear since for $\mC=\Rep(A)$, ${\rm FPdim}(X)={\rm Dim}(X)$, where ${\rm Dim}(X)$ denotes the dimension of the underlying vector space. \end{proof} It is obvious that the class of fusion categories with integer Frobenius-Perron dimensions is closed under the operations of tensor product, passing to the opposite category, and taking a fusion subcategory. The following theorem shows that it is also closed under taking duals (i.e., in the language of \cite{Mu1},\cite{O} the property of integer Frobenius-Perron dimensions is weak Morita invariant). \begin{theorem}\label{dualfrob} Let $\mC$ be a fusion category with integer Frobenius-Perron dimensions, $\mM$ be an indecomposable module category over $\mC$. Then the fusion category $\mC_{\mM}^*$ has integer Frobenius-Perron dimensions. \end{theorem} \begin{proof} We know from \cite{Mu1},\cite{O1} that $Z(\mC)$ is canonically equivalent to $Z(\mC_\mM^*)$. So let $V$ be an object of $Z(\mC)$ such that the matrix of multiplication by $V$ in both $\mC$ and $\mC^*_\mM$ has strictly positive entries (it exists, since both $\mC$ and $\mC^*_\mM$ are fusion categories, and so the natural functors $Z(\mC)\to \mC$, $Z(\mC^*)\to \mC^*_\mM$ are surjective). Let $[X], [X]_*$ be the matrices of multiplication by $X\in Z(\mC)$ in the Grothendieck groups $K(\mC),K(\mC^*_\mM)$, respectively. For a square matrix $A$ with nonnegative entries, let $t(A)$ be the largest real eigenvalue of $A$. By the assumptions of the theorem, $t([V])$ is an integer (it is the Frobenius-Perron dimension of $V$ as an object of $\mC$). Since the asssignments $X\to t([X])$, $X\to t([X]_*)$, $X\in Z(\mC)$, are both homomorphisms of algebras, Lemma \ref{uniq} implies that $t([V])=t([V]_*)$. Hence, $t([V]_*)$ is an integer. The vector of Frobenius-Perron dimensions of simple objects in $\mC_\mM^*$ is the unique, up to scaling, eigenvector of $[V]_*$ with eigenvalue $t([V]_*)$. Since the matrix $[V]_*$ has integer entries, the entries of this vector are rational numbers (as the zeroth entry is $1$). This means that they are integers (as they are eigenvalues of integer matrices and hence algebraic integers). The theorem is proved. \end{proof} \begin{corollary}\label{quotcat} The class of fusion categories with integer Frobenius-Perron dimensions is closed under the operation of taking a component of a quotient category. \end{corollary} \begin{proof} Let $\mC$ be a fusion category with integer Frobenius-Perron dimensions, $\mD$ an indecomposable multi-fusion category, and and $F:\mC\to \mD$ be a surjective tensor functor (so $\mD$ is a quotient of $\mC$). Let $\mM$ be an indecomposable module category over $\mD$. Then $F^*:\mD^*\to \mC^*$ is an injective tensor functor between the corresponding dual fusion categories ($\mC^*$ is fusion since $\mM$ is indecomposable over $\mC$, by surjectivity of $F$). By Theorem \ref{dualfrob}, $\mC^*$ has integer Frobenius-Perron dimensions. Hence, so does $\mD^*$ as its full subcategory. But any component category $\mD_{ii}$ of $\mD$ is dual to $\mD^*$ with respect to the $i$-th part of $\mM$ as a $\mD^*$-module. Thus, $\mD_{ii}$ has integer Frobenius-Perron dimensions by Theorem \ref{dualfrob}. \end{proof} \begin{proposition}\label{frobmod1} Let $\mC$ be a fusion category with integer Frobenius-Perron dimensions. Then for any indecomposable module category $\mM$ over $\mC$, the Frobenius-Perron dimensions of objects in $\mM$ are integers (under a suitable normalization). \end{proposition} \begin{proof} Since the matrix $[V]_\mM$ from Proposition \ref{frobmod} has integer entries, and its largest real eigenvalue ${\rm FPdim}(V)$ is an integer and has multiplicity $1$, the entries of $\mathbf m$ are rational. \end{proof} In \cite{O2}, it is asked whether there are finitely many fusion categories with a given number of simple objects, and proved that the answer is ``yes'' for representation categories of Hopf algebras. Using the above techniques, one can show that the answer is ``yes'' for fusion categories with integer Frobenius-Perron dimension (which includes Hopf and quasi-Hopf algebras). Namely, we have \begin{proposition} There are finitely many fusion categories of integer Frobenius-Perron dimension with a given number of simple objects. In particular, there are finitely many equivalence classes of semisimple quasi-Hopf algebras with a given number of matrix blocks. \end{proposition} \begin{proof} Let $\mC$ be such a category with $N$ simple objects. Since the Frobenius-Perron dimension of $\mC$ is an integer, it is pseudo-unitary, in particular has a canonical spherical structure. Let $d_i$ be the categorical (=Frobenius-Perron) dimensions of simple objects of $Z(\mC)$ which occur in $I(\mathbf 1)$, and $D$ be its global (=Frobenius-Perron) dimension. Then $d_i$ are integers (by Proposition \ref{roots}, since for $X_i$ occuring in $I(\mathbf 1)$, $X_i|_\mC$ lies in $\mC_{ad}$ by Proposition \ref{I(1)}). Thus, $D/d_i=m_i$ are (usual) integers by Proposition \ref{moddivi}, and $\sum_i[X_i|_\mC:1]\frac{1}{m_i}=1$. The number of summands $1/m_i$ here is $\sum [X_i|_{\mC}:\mathbf 1]$, which is at most $N$ by Proposition \ref{numobj}. By a classical argument of Landau (see \cite{O2} and references therein), the number of such collections $m_i$ is finite for any given $N$. On the other hand, $m_0=D$ (as $X_0$ is the neutral object). Thus, $D$ is bounded by some function of $N$. But for a given $D$, there are only finitely many fusion rings, and hence, by Ocneanu rigidity, finitely many fusion categories. \end{proof} \subsection{Group-theoretical fusion categories} An example of a fusion category with integer Frobenius-Perron dimensions is the category $\mC(G,H,\omega,\psi)$ associated to quadruples $(G,H,\omega,\psi)$, where $G$ is a finite group, $H$ is a subgroup, $\omega\in Z^3(G,\CC^*)$ is a 3-cocycle, $\psi\in C^2(H,\CC^*)$ a 2-cochain such that $d\psi=\omega|_H$ (see \cite{O1}). Namely, let ${\rm Vec}_{G,\omega}$ be the category of finite dimensional $G$-graded vector spaces with associativity defined by $\omega$. Let ${\rm Vec}_{G,\omega}(H)$ be the subcategory of ${\rm Vec}_{G,\omega}$ of objects graded by $H$. Consider the twisted group algebra $A=\CC_\psi[H]$. Since $d\psi=\omega|_H$, it is an associative algebra in this category. Then $\mC(G,H,\omega,\psi)$ is defined to be the category of $A$-bimodules in ${\rm Vec}_{G,\omega}$. \begin{remark} If $(\omega',\psi')$ is another pair of elements such that $\omega'=\omega+d\eta,\psi'=\psi+\eta|_H+d\chi$, $\eta\in C^2(G,\CC^*)$, $\chi\in C^1(H,\CC^*)$, then $\mC(G,H,\omega,\psi)$ is equivalent to $\mC(G,H,\omega',\psi')$. Thus, the categories $\mC(G,H,\omega,\psi)$ are parametrized by the set $S(G,H)$ of equivalence classes of pairs $(\omega,\psi)$ under the above equivalence, which is a fibration over ${\rm Ker}(H^3(G,\CC^*)\to H^3(H,\CC^*))$ with fiber being a torsor over ${\rm Coker}(H^2(G,\CC^*)\to H^2(H,\CC^*))$. \end{remark} \begin{definition} A fusion category is said to be group-theoretical if it is of the form $\mC(G,H,\omega,\psi)$. \end{definition} \begin{remark} We note that there may be more than one way to represent a given fusion category in the form $\mC(G,H,\omega,\psi)$. In particular, $G$ is not uniquely determined. However, $|G|$ is uniquely determined, since it equals the global dimension of $\mC(G,H,\omega,\psi)$. \end{remark} Group-theoretical fusion categories have the following simple characterization. Let us say that a fusion category $\mD$ is pointed if all its simple objects are invertible. Any pointed fusion category has the form ${\rm Vec}_{G,\omega}$. \begin{proposition}\label{1dim}\cite{O1}. A fusion category $\mC$ is group-theoretical if and only if it is dual to a pointed category with respect to some indecomposable module category. \end{proposition} \begin{corollary}\label{intdim} Group-theoretical fusion categories have integer Frobenius-Perron dimensions of simple objects. In particular, they are pseudo-unitary. \end{corollary} \begin{proof} This follows from Proposition \ref{1dim}, Theorem \ref{dualfrob}, and the fact that any invertible object has Frobenius-Perron dimension $1$. \end{proof} It is clear from the results of \cite{O1} that the class of group-theoretical categories is closed under tensor product, taking the opposite category, and taking the dual category with respect to an indecomposable module category. Moreover, the following result shows that this class is also closed under taking a subcategory or a component of a quotient category. \begin{proposition}\label{subquot} (i) A full fusion subcategory of a group-theoretical category is group-theoretical. (ii) A component in a quotient category of a group-theoretical category is group-theoretical. \end{proposition} \begin{proof} (i) Let $\mC\subset \mD$ be fusion categories ($\mC$ is a full subcategory), and let $\mD$ be group-theoretical. Let $\mM$ be an indecomposable module category over $\mD$ such that $\mD^*$ is ${\rm Vec}_{G,\omega}$. Then by Proposition \ref{star}, $\mC^*$ is a quotient of ${\rm Vec}_{G,\omega}$. Let $I$ label the component categories of $\mC^*$. It is clear that for each $g\in G$ and $i\in I$ there exists a unique $g(i)\in I$ such that the functor $g\otimes$ is an equivalence $\mC_{ir}\to \mC_{g(i)r}$ for each $r\in I$. So we have an action of $G$ on $I$. Let $j\in I$ and $H$ be the stabilizer of $j$ in $G$. For any $g\in H$, denote by $\bar g_{jj}$ the projection of $g$ to the fusion category $\mC^*_{jj}$. Then the assignment $g\to \bar g_{jj}$ is a surjective tensor functor ${\rm Vec}_{H,\omega}\to \mC^*_{jj}$. This functor must map invertible objects to invertible objects. Hence, all simple objects of $\mC^*_{jj}$ are invertible, so $\mC^*_{jj}$ is pointed, and hence $\mC$ is group-theoretical (as it is dual to $\mC^*_{jj}$ with respect to a module category). (ii) Let $\mD$ be group-theoretical, $F:\mD\to \mC$ be surjective ($\mC$ is indecomposable), and $\mC_{ii}$ a component of $\mC$. We need to show that $\mC_{ii}$ is group-theoretical. Let $\mM$ be an indecomposable module category over $\mC$. It suffices to show that $\mC^*$ is group-theoretical, as $\mC^*$ is dual to $\mC_{ii}$ with respect to the i-th part of $\mM$. But by Proposition \ref{star}, $\mC^*$ is embedded into $\mD^*$, so it suffices to know that $\mD^*$ is group-theoretical, which follows by duality from the fact that $\mD$ is group-theoretical. \end{proof} \subsection{A question} We think that the following question is interesting. \begin{question} \label{quest} Does there exist a finite dimensional semisimple Hopf algebra $H$ whose representation category is not group-theoretical? \end{question} \begin{remark} Hopf algebras with group theoretical category of representations can be completely classified in group theoretical terms (see \cite{O1}). Therefore, a negative answer to Question \ref{quest} would provide a full classification of semisimple Hopf algebras. \end{remark} \begin{remark} The answer to Question \ref{quest} is ``no'' for triangular Hopf algebras, as follows from \cite{EG3}. However, if the answer is ``yes'' in general, then it is ``yes'' already for quasitriangular Hopf algebras. Indeed, if $\Rep(D(H))$ is group-theoretical then so is $\Rep(H)\otimes \Rep(H^{*cop})$ (as a dual category to $\Rep D(H)$), from which it follows by Proposition \ref{subquot} that so is $\Rep(H)$. \end{remark} \begin{remark} It is interesting that there exists a quasi-Hopf algebra $H$ whose category of representations is not group-theoretical. Namely, recall that in \cite{TY}, there is a classification of fusion categories whose simple objects are elements of a group $G$ and an additional object $X=X^*$, with fusion rules $g\otimes X=X\otimes g=X, X\otimes X=\sum_{g\in G}g$ (and the usual multiplication rule in $G$). The result is that $G$ must be abelian, and for each $G$ the categories are parametrized by pairs $(q,r)$ where $q$ is a nondegenerate symmetric bilinear form on $G$ with coefficients in $\CC^*$, and $r$ is a choice of $\sqrt{|G|}$. Let us denote such category by $TY(G,q,r)$. It is clear that $TY(G,q,r)$ is a representation category of a quasi-Hopf algebra (i.e. has integer Frobenius-Perron dimensions) iff $|G|$ is a square. So let us take $G=(\mathbb Z/p)^2$, where $p>2$ is a prime, and let $q$ be given by $q((x,y),(x,y))=ax^2-by^2$, where $a,b\in \mathbb F_p$ and $a/b$ is a quadratic non-residue (i.e. the quadratic form $ax^2-by^2$ is elliptic). Then one can check by direct inspection that $TY(G,q,r)$ is not group-theoretical. (We note that $TY(G,q,r)$ is not a representation category of a Hopf algebra; on the other hand, if $q$ is a hyperbolic form, then $TY(G,q,+\sqrt{|G|})$ is a representation category of a Hopf algebra, but is group-theoretical). \end{remark} \subsection{Cyclotomicity of dimension functions}\label{cy} The forgetful functor $Z(\mC)\to \mC$ induces a homomorphism of rings $F:K(Z(\mC))\to K(\mC)$. It is clear that the image of this map is contained in the center $Z(K(\mC))$ of the ring $K(\mC)$. \begin{lemma} \label{centers} The map $K(Z(\mC))\otimes {\Bbb Q} \to Z(K(\mC))\otimes {\Bbb Q}$ is surjective. \end{lemma} \begin{proof} Recall that $I: \mC \to Z(\mC)$ denotes the induction functor, that is the left adjoint to the forgetful functor $Z(\mC)\to \mC$. We will denote the induced map $K(\mC)\to K(Z(\mC))$ by the same letter. Let $b_i$ be the basis of $K(\mC)$. We have by Proposition \ref{I(1)} $F(I(x))=\sum_ib_ixb_i^*$ for any $x\in K(\mC)$. In particular for $x\in Z(K(\mC))$ we have $F(I(x))=x\sum_ib_ib_i^*$. Finally note that the operator of multiplication by the element $\sum_ib_ib_i^*\in Z(K(\mC))$ is a self-adjoint positive definite operator $K(\mC)\to K(\mC)$ (with respect to the usual scalar product defined by $(b_i,b_j^*)=\delta_{ij}$) and hence invertible. The Lemma is proved. \end{proof} \begin{remark} In the special case $\mC =\Rep(H)$ for some Hopf algebra $H$ Lemma \ref{centers} was proved in \cite{KSZ} 6.3. \end{remark} \begin{theorem}\label{cyclotomic} Let $\mC$ be a fusion category over $\Bbb C$ and let $L$ be an irreducible representation of $K(\mC)$. There exists a root of unity $\xi$ such that for any object $X$ of $\mC$ one has $\Tr([X],L)\in {\mathbb Z}[\xi]$. \end{theorem} \begin{proof} First of all we can assume without loss of generality that the category $\mC$ is spherical. Indeed as explained in Proposition \ref{sphpiv} we can construct a spherical fusion category $\tilde \mC$ which projects onto $\mC$; moreover the map $K(\tilde \mC)\to K(\mC)$ is surjective and the simple objects of $\tilde \mC$ map to the simple objects of $\mC$. Now let $\mC$ be spherical and let $b_i$ be the basis of $K(\mC)$. The element $e_L=\sum_i\Tr(b_i,L)b_i^*$ is proportional to a primitive central idempotent in $K(\mC)$, see e.g. \cite{Lu} 19.2 (b). By Lemma \ref{centers} there exists a primitive central idempotent $\tilde e_L\in K(Z(\mC))$ such that $e_L$ is proportional to $F(\tilde e_L)$. The element $\tilde e_L$ can expressed in terms of $S-$matrix of the category $Z(\mC)$, see e.g. \cite{BaKi}. By Theorem \ref{app} there exists a root of unity $\xi$ such that the entries of $S-$matrix lie in ${\Bbb Q} (\xi)$ and thus $\tilde e_L\in K(Z(\mC))\otimes {\Bbb Q} (\xi)$. Hence $e_L$ is proportional to some element of $K(\mC)\otimes {\Bbb Q} (\xi)$. Now the coefficient of $b_0=1\in K(\mC)$ in $e_L$ equals to $\dim (L)\in {\Bbb Q} (\xi)$ and is nonzero. Hence $e_L\in K(\mC)\otimes {\Bbb Q} (\xi)$ and $\Tr(b_i,L)\in {\Bbb Q} (\xi)$. Obviously the number $\Tr (b_i,L)$ is an algebraic integer and the ring of integers in $\Bbb Q(\xi)$ is $\Bbb Z[\xi]$. The Theorem is proved. \end{proof} \begin{remark} Again in the case when $\mC =\Rep(H)$ for some Hopf algebra $H$ Theorem \ref{cyclotomic} is proved in \cite{KSZ} 6.3. \end{remark} \begin{corollary} Any irreducible representation of $K(\mC)$ is defined over some cyclotomic field. In particular for any homomorphism $\phi: K(\mC)\to \CC$ we have $\phi (b_i)\in {\Bbb Q} [\xi]$ for some root of unity $\xi$. \end{corollary} \begin{proof} Let ${\Bbb Q}^{ab}$ be the field of all cyclotomic numbers. We already proved that $K(\mC)\otimes {\Bbb Q}^{ab}$ decomposes into direct sum of simple algebras (such decomposition is the same as the decomposition of $1\in K(\mC)$ into the sum of primitive central idempotents). Now it is well known that the Brauer group of ${\Bbb Q}^{ab}$ is trivial. The result follows. \end{proof} \begin{corollary} Let $\mC$ be a fusion category over $\Bbb C$. There exists a root of unity $\xi$ such that for any object $X$ of $\mC$ one has ${\rm FPdim}(X)\in {\mathbb Z}[\xi]$. \end{corollary} \subsection{Module categories over products} Let $a,b$ be two algebraic integers. We say that $a,b$ are coprime if there exist algebraic integers $p,q$ such that $pa+qb=1$. \begin{proposition} Let $\mC,\mD$ be nondegenerate fusion categories with coprime Frobenius-Perron dimensions. Then any indecomposable module category over $\mC\otimes \mD$ is $\mM\otimes \mN$, where $\mM$ is an indecomposable module category over $\mC$, and $\mN$ over $\mD$. \end{proposition} \begin{proof} Let $\mB$ be an indecomposable module category over $\mC\otimes \mD$. Let us restrict it to $\mC$. We get $\mB|_{\mC}=\oplus_i\mM_i$, where $\mM_i$ are indecomposable. Thus the action of $\mD$ on $\mB$ can be regarded as a tensor functor $F: \mD\to \oplus_{i,j}{\rm Fun}_{\mC}(\mM_i,\mM_j)$. Let $\mD'$ be the image of this functor, i.e. the category generated by subquotients of $F(X)$. Then $\mD'$ is a multifusion category, which is indecomposable (as $\mB$ is indecomposable over $\mC\otimes \mD$). Since the functor $F: \mD\to \mD'$ is surjective, the Frobenius-Perron dimension $d$ of any component category of $\mD'$ divides ${\rm FPdim}(\mD)$. On the other hand, each component category of $\mD'$ is a tensor subcategory in $Fun_{\mC}(\mM_i,\mM_i)$ for some $i$. The latter is a dual category to $\mC$, so it has the same Frobenius-Perron dimension as $\mC$. Hence, $d$ must divide ${\rm FPdim}(\mC)$. Thus, $d=1$ by the coprimeness condition. Hence $\mD'=A-bimod$, where $A$ is a semisimple algebra, and furthermore, all $\mM_i$ are equivalent to a single module category $\mM$. The functor $F:\mD\to \mD'$ defines a module category $\mN$ over $\mD$, and it is clear that $\mB=\mM\otimes \mN$. We are done. \end{proof} {\bf Example.} Let $\mC_h$ be the fusion category associated to the quantum group $U_q(so_3)$ for $q=e^{\pi i/h}$. The Frobenius-Perron dimension of this category is $d(h)=\frac{h}{4\sin^2(\pi/h)}$ (this follows from [BaKi], (3.3.9)). The denominator of this fraction is an algebraic integer. Thus, our result implies that if $h_i$ are pairwise coprime then any indecomposable module category over $\otimes_i \mC_{h_i}$ is the product of those over $\mC_{h_i}$. \section{Fusion categories in positive characteristic} In this section we will consider fusion and multi-fusion categories over an algebraically closed field $k$ of positive characteristic $p$. We will show that many of the results of the previous sections continue to hold in this case after very minor modifications. \subsection{List of modifications} Fusion and multi-fusion categories in characteristic $p$ are defined in the same way as in charactersitic zero. Below we will describe the modifications which need to be made to validate the statements from the previous sections in this case. {\bf Section 2.} The statements of subsections 2.1, 2.2 which make sense generalize without changes, except the last statement of Theorem \ref{posi}, which claims that the global dimension of a fusion category is nonzero. This is in fact false in positive characteristic. For example, if $G$ is a finite group whose order is divisible by $p$, then the category of modules over the function algebra ${\rm Fun}(G,k)$ is a fusion category of global dimension $|G|=0$. This is the main difference between the zero and positive characteristic case. To deal with this difference, we introduce the following definition. \begin{definition} A fusion category over $k$ is said to be nondegenerate if its global dimension is nonzero. An indecomposable multi-fusion category is nondegenerate if at least one of its component categories is nondegenerate. A multi-fusion category is nondegenerate if all its indecomposable parts are nondegenerate. A regular semisimple weak Hopf algebra over $k$ is nondegenerate if its category of representations is a nondegenrate multi-fusion category. \end{definition} Theorems \ref{dualcat}, \ref{m1m2}, \ref{multi-fus} remain valid if the category $\mC$ is nondegenerate. Proposition \ref{samedim} remains true without any additional conditions. In Section 2.5 (and at other places throughout), the algebra $R$ has to be chosen in such a way that its block sizes are not divisible by $p$ (since only such algebras admit a symmetric separability idempotent). Theorem \ref{Schlah} is valid for such $R$. Corollary \ref{Schlah1} and Theorem \ref{traceofS2} also remain valid. Theorem \ref{sscoss} is valid for nondegenerate weak Hopf algebras. Theorems \ref{van}, \ref{ocnrig}, \ref{multocnrig}, \ref{ocnrigfun}, and Corollary \ref{ocnrigmod} remain valid for nondegenerate categories. More precisely, in Theorems \ref{van}, \ref{ocnrigfun} one only needs nondegeneracy of the category $\mC$. \begin{remark} We note that the nondegeneracy condition is essential. For example, if $A$ is the Hopf algebra ${\rm Fun}(G,k)$, where the order of $G$ is divisible by $p$, then Theorem \ref{dualcat} is false for $\Rep(A)$, and Theorem \ref{sscoss} is false for $A$. Similarly, Theorem \ref{van} is false for $\Rep(A)$ and the identity functor, since the corresponding cohomology is $H^i(G,k)$, which may be nonzero if the order of $G$ is divisible by $p$. However, we do not know a counterexample to Theorem \ref{ocnrig} in positive characteristic. \end{remark} {\bf Sections 3-7.} While Section 3 makes no sense in positive characteristic, the material of Section 4 generalizes without changes, as long as one considers regular weak Hopf algebras. The only change in Section 5 is that one should always choose nonzero block sizes for the algebra $R$ (except subsections 5.8,5.9 where the characteristic should be zero). Theorems \ref{acyclic} and \ref{rigweak} are valid for nondegenerate weak Hopf algebras. {\bf Section 8.} The results of Section 8.1 generalize without changes, since they have nothing to do with the characteristic of the ground field. So do the results of Section 8.2, except Proposition \ref{squar}, Corollary \ref{fpdual}, and Proposition \ref{subdivi}, which are valid for nondegenerate categories. Finally, let us point out that Theorem \ref{quasihopf} is valid in positive characteristic. {\bf Example.} A semisimple Hopf algebra $H$ is nondegenerate if and only if it is cosemisimple. Indeed, if $H$ is nondegenerate, then so is $\Rep(H)$, and hence $H^*$ is semisimple, since $\Rep(H^{*op})$ is a dual category to $\Rep(H)$. Conversely, if $H$ is cosemisimple, then $\Tr(S^2)\ne 0$ (\cite{LR1}), so the global dimension of $\Rep(H)$ is nonzero, and $H$ is nondegenerate. \subsection{Lifting theorems} In this subsection we will show that a nondegenerate multi-fusion category over a field of characteristic $p$, and a tensor functor between such categories, can be lifted to characteristic zero. It is an analog of the result of \cite{EG1}, where this was shown for semisimple cosemisimple Hopf algebras and morphisms between them. First of all, we need to define fusion categories over any commutative ring. To do this, recall that a fusion category over a field $k$ can be regarded as a collection of finite dimensional vector spaces $H_{ij}^k={\rm Hom}(X_i\otimes X_j,X_k)$ (multiplicity spaces), together with a collection of linear maps between tensor products of these spaces (associativity morphism, evaluation, coevaluation morphisms), satisfying some equations (axioms of a rigid tensor category). Now let $R$ be a commutative ring with unit, and define a fusion category over $R$ to be a collection of free finite rank $R$-modules $H_{ij}^k$ together with a collection of module homomorphisms satisfying the same equations. Tensor functors between fusion categories over $k$ can be defined in similar terms, as collections of linear maps satisifying algebraic equations; this allows one to define tensor functors over $R$ in an obvious way. Now let $W(k)$ be the ring of Witt vectors of $k$. Let $I$ be the maximal ideal in $W(k)$ generated by $p$. If $\mC$ is a fusion category over $k$, then a lifting of $\mC$ to $W(k)$ is a fusion category $\tilde \mC$ over $W(k)$, together with an equivalence $\tilde \mC/I\to \mC$, where $\mC/I$ is the reduction of $\mC$ modulo $p$. In a similar way one defines a lifting of a module category and, more generally, a lifting of a tensor functor. \begin{theorem}\label{lifting} A nondegenerate multi-fusion category $\mC$ over $k$ admits a unique lifting to $W(k)$. Moreover, a tensor functor $F$ between nondegenerate fusion categories $\mC,\mC'$ over $k$ admits a unique lifting to $W(k)$. In particular, a module category over a nondegenerate multi-fusion category $\mC$ has a unique lifting to $W(k)$. \end{theorem} \begin{proof} The theorem follows from the facts that $H^3(\mC)=0,H^4(\mC)=0$, $H^2_F(\mC)=0$, $H^3_F(\mC)=0$, proved in Section 7, in the same way as in \cite{EG1}. \end{proof} \begin{corollary}\label{brsym} A nondegenerate braided (symmetric) fusion category over $k$ is uniquely liftable to a braided (resp. symmetric) fusion category over $W(k)$. \end{corollary} \begin{proof} A braiding on $\mC$ is the same as a splitting $\mC\to Z(\mC)$ of the tensor functor $Z(\mC)\to \mC$. By Theorem \ref{lifting}, such splitting is uniquely liftable. Hence, a braiding on $\mC$ is uniquely liftable. Now let us prove the result in the symmetric case. A braiding gives rise to a categorical equivalence $B: \mC\to \mC^{op}$. A braiding is symmetric iff the composition of two categorical equivalences $B$ and $B^{21}$ is the identity. Thus, Corollary follows from Theorem \ref{lifting}. \end{proof} \begin{remark} Corollary \ref{brsym} is false for degenerate categories. For example, let $\tilde \mC$ be the fusion category corresponding to the quantum group $U_q(sl_2)$ at $q=e^{\pi i/p}$, where $p$ is an odd prime. This category is known to have semisimple reduction to $\bar{\mathbb F}_p$, which is symmetric. However, this reduction does not lift to a symmetic category, since it has non-integer Frobenius-Perron dimensions. This example also shows that there exist semisimple and cosemisimple triangular weak Hopf algebras over $k$ which do not come from group algebras. For usual Hopf algebras, this is impossible (see \cite{EG3}). \end{remark} \subsection{Faithfulness of lifting} In this section we will show that the lifting procedure of nondegenerate fusion (braided, symmetric) categories is faithful, i.e. if two such categories over $k$ have equivalent liftings, then they are equivalent. We will also show that if two tensor functors from a nondegenerate fusion category over $k$ to another fusion category have isomorphic liftings, then they are isomorphic. Let $R$ be a complete local ring, $\Bbb F$ the fraction field of $R$, $I\subset R$ the maximal ideal, and and $k=R/I$ the residue field. \begin{theorem}\label{faithlif} (i) Let $\mC_1,\mC_2$ be fusion categories defined over $R$, such that the fusion categories $\mC_1/I$, $\mC_2/I$ are nondegenerate, and the categories $\mC_1\otimes_R\Bbb F,\mC_1\otimes_R\Bbb F$. are tensor equivalent. Then $\mC_1,\mC_2$ (and hence $\mC_1/I,\mC_2/I$) are tensor equivalent. (ii) The same statement is valid for braided and symmetric fusion categories. \end{theorem} \begin{proof} (i) Pick a tensor equivalence $E: \mC_1\otimes_R \Bbb F\to \mC_1\otimes_R\Bbb F$. Now pick an equivalence of $\mC_1$, $\mC_2$ as {\it additive} categories, which is the same as $E$ at the level of isomorphism classes of objects. By doing this, we may assume without loss of generality that $\mC_1$ and $\mC_2$ are really one category $(\mC,\otimes)$ with two different associativity isomorphisms $\Phi_1,\Phi_2\in C^3(\mC)$ and evaluation maps ${\rm ev}_1,{\rm ev}_2$ (the coevaluation maps of simple objects of both categories may be identified with any fixed collection of such maps by renormalization). We know that there exists an invertible element $J$ of $C^2(\mC)\otimes_R \Bbb F$ such that $(\Phi_1,{\rm ev}_1)^J= (\Phi_2,{\rm ev}_2)$. (where ${}^J$ denotes the twisting by $J$). Our job is to show that such $J$ can be found in $C^2(\mC)$, without localization to $\Bbb F$. Consider the affine group scheme $T$ (defined over $R$) of all invertible elements of $C^2(\mC)$. Let $T_0$ be the subscheme of $T$ consisting of ``trivial'' twists, i.e. those of the form $\Delta(x)(x^{-1}\otimes x^{-1})$, where $x$ is an invertible element of $C^1(\mC)$ (i.e. the center of $\mC$). It is clear that $T_0$ is central in $T$. {\bf Remark.} Since the algebra $C^2(\mC)$ is a direct sum of matrix algebras (split over $R$), $T$ is a direct product of general linear groups. Let $Z$ be the affine scheme of all invertible associators $\Phi\in C^3(\mC)$ and collections of evaluation morphisms ${\rm ev}$ satisfying (together with the fixed coevaluation morphisms) the axioms for a rigid tensor category (i.e. the pentagon relation, the unit object axiom, and the duality axioms). The group scheme $T$ acts on $Z$ by twisting (and then renormalizing the coevaluations to be the fixed ones): $(\Phi,{\rm ev})\to (\Phi, {\rm ev})^J$. The restriction of this action to $T_0$ is trivial, so we get an action of $T/T_0$ on $Z$. Consider the regular mapping $\phi: T/T_0\to Z$, given by $J\to (\Phi_1,{\rm ev_1})^J$. \begin{lemma}\label{fin} The mapping $\phi$ is finite. \end{lemma} \begin{proof} By Nakayama's lemma, it is sufficient to check the finiteness of the reduction $\phi|_k$ of $\phi$ modulo $I$. First of all, we have $H^2(\mC_1/I)=0$, which implies that the stabilizer of $(\Phi_1,{\rm ev}_1)$ in $T/T_0$ is a finite group. Thus the map $\phi|_k: T/T_0\to \phi|_k(T/T_0)$ is a finite covering. It remains to show that $\phi|_k(T/T_0)$ is closed in $Z$. To do this, consider its closure $Z'$. Points of $Z'$ represent fusion categories over $k$, all of which are nondegenerate Indeed, the global dimension is a regular (in fact, locally constant) function on $Z$, and it is nonzero on $\mC_1/I$, so it must be nonzero for all points of $Z'$. But for a nondegenerate fusion category $\mD$ over $k$ we have $H^3(\mD)=0$, which implies that the action of $T/T_0$ on $Z'$ is locally transitive. This means that $Z'$ is in fact a single orbit of $T/T_0$, i.e. $\phi|_k(T/T_0)$ is closed. The lemma is proved. \end{proof} Thus we have a homomorphism $\phi^*: R[Z]\to R[T/T_0]$, which makes $R[T/T_0]$ into a finitely generated $R[Z]$-module. Now let $J\in (T/T_0)(\Bbb F)$ be such that $(\Phi_1,{\rm ev}_1)^J= (\Phi_2,{\rm ev}_2)$. Thus, $J$ can be regarded as a homomorphism $J: R[T/T_0]\to \Bbb F$. This homomorphism satisfies the condition $J(af)=\gamma(a)J(f)$, $f\in R[T/T_0]$, $a\in R[Z]$. where $\gamma: R[Z]\to R$ corresponds to the point $(\Phi_2,{\rm ev}_2)$. Now, by Lemma \ref{fin}, for any $f\in R[T/T_0]$ there exist $b_1,...,b_n\in R[Z]$ such that $f^n+b_1f^{n-1}+...+b_n=0$. But then $J(f)^n+\gamma(b_1)J(f)^{n-1}+...+\gamma(b_n)=0$. Now, since $\gamma(b_i)$ are integers, we see that $J(f)$ is an integer, so $J$ in fact belongs to $(T/T_0)(R)$. But it is easy to see that the map $T(R)\to (T/T_0)(R)$ is surjective. Hence, there exists $J\in T(R)$ such that $(\Phi_1,{\rm ev}_1)^J=(\Phi_2,{\rm ev}_2)$, as desired. The theorem is proved. (ii) The proof is the same, since, as we showed, the integrality of $J\in T/T_0$ follows already from the equation $(\Phi_1,{\rm ev}_1)^J=(\Phi_2,{\rm ev}_2)$. \end{proof} \begin{theorem}\label{faithfun} Let $\mC,\mC'$ be fusion categories defined over $R$, such that the fusion category $\mC/I$ is nondegenerate. Let $F_1,F_2: \mC\to \mC'$ be tensor functors, which become isomorphic after tensoring with $\Bbb F$. Then $F_1,F_2$ (and hence their reductions modulo $I$) are isomorphic. \end{theorem} \begin{proof} The proof is analogous to the proof of Theorem \ref{faithlif}. As in Theorem \ref{faithlif}, we may assume that $F_1=F_2=F$ as additive functors, with two different tensor structures $J_1,J_2$. Let $G$ be the group scheme of invertible elements of $C^1_F(\mC)$. Let $S$ be the scheme of tensor structures on $F$. The group scheme $G$ naturally acts on $S$ by ``gauge transformations'', $J\to J^g$. We know there exists $g\in G(\Bbb F)$ such that $J_1^g=J_2$, and need to show that this $g$ is in fact in $G(R)$. Using the fact that $H^1_F(\mC)=H^2_F(\mC)=0$, one finds that the action map $\chi:G\to S$ given by $g\to J_1^g$ is finite. Now the proof that $g\in G(R)$ runs analogously to the proof that $J\in (T/T_0)(R)$ in Theorem \ref{faithlif}. Theorem \ref{faithfun} is proved. \end{proof} Now let $K$ be the algebraic closure of the field of quotients of $W(k)$. \begin{corollary} \label{faithlif1} (i) If two nondegenerate fusion (braided, symmetric) categories over $k$ have equivalent liftings to $K$, then they are equivalent. (ii) If two tensor functors between nondegenerate fusion categories over $k$ have isomorphic liftings to $K$, then they are isomorphic. \end{corollary} \begin{proof} (i) follows easily from Theorem \ref{faithlif}, (ii) from Theorem \ref{faithfun}. \end{proof} To give the next corollary, recall that in \cite{EG1} it was proved that a semisimple cosemisimple Hopf algebra over $k$ has a unique lifting to $K$. In fact, this is a special case of the lifting theory developed in this paper, since a semisimple cosemisimple Hopf algebra is nothing but a pair $(\mC,F)$, where $\mC$ is a nondegenerate fusion category over $k$, and $F$ a fiber functor on $\mC$. \begin{corollary} \label{faithhopf} If two semisimple cosemisimple Hopf algebras over $k$ have isomorphic liftings to $K$ then they are isomorphic. \end{corollary} \begin{proof} We can regard these two Hopf algebras as pairs $(\mC_1,F_1)$ and $(\mC_2,F_2)$. By theorem \ref{faithlif1}(i), we can assume that $\mC_1=\mC_2$; then by theorem \ref{faithlif1}(ii), $F_1$ is isomorphic to $F_2$. The Corollary is proved. \end{proof} The same theorem applies to quasitriangular, triangular Hopf algebras. The proof is parallel. \subsection{Some applications of lifting} \begin{corollary}\label{fpnonzero} Let $\mC$ be a nondegenerate fusion category over $k$. Then its Frobenius-Perron dimension $\Delta$ is not divisible by $p$. \end{corollary} \begin{remark} We do not know if the converse statement is true. For Hopf algebras the converse statement is the well known conjecture that if $H$ is a semisimple Hopf algebra whose dimension is not divisible by $p$ then $H$ is cosemisimple. \end{remark} \begin{proof} Assume that $\Delta$ is divisible by $p$. Let $\tilde\mC$ be the lifting of $\mC$, and $\hat \mC=\tilde\mC\otimes_{W(k)}K$, where $K$ is the algebraic closure of the field of fractions of $W(k)$. Then by Theorem \ref{divi}, the global dimension $D$ of $\hat\mC$ is divisible by $\Delta$, hence by $p$. So the global dimension of $\mC$ is zero. Contradiction. \end{proof} \begin{corollary}\label{fpsubcat} Let $\mD$ be a nondegenerate fusion category over $k$, and $\mC$ be a full subcategory of $\mD$ of integer Frobenius-Perron dimension $N$. Then $\mC$ is nondegenerate. \end{corollary} \begin{remark} We do not know if this result is valid without assuming that $N$ is an integer. \end{remark} \begin{proof} Consider the lifting $\tilde \mD$ of $\mD$. This lifting necessarily contains a lifting $\tilde\mC$ of $\mC$. Consider the corresponding categories $\hat\mC$, $\hat\mD$ over the algebraically closed field $K$ of zero characteristic. Let $D$ be the global dimension of $\hat\mD$, $\Delta$ its Frobenius-Perron dimension. Then $D$ is divisible by $\Delta$ by Proposition \ref{divi}. Also, the Frobenius-Perron dimension of $\hat\mC$ is $N$ (an integer), hence it is equal to the global dimension of $\hat\mC$ by Proposition \ref{intFP}. But by Theorem \ref{subdivi}, $N$ divides $\Delta$, and hence $D$. But $D$ is not divisible by $p$, hence $N$ is not divisible by $p$. Thus, $\mC$ is nondegenerate, as desired. \end{proof} \section{Appendix: Galois properties of S-matrix} The remarkable result due to J.~de~Boere, J.~Goeree, A.~Coste and T.~Gannon states that the entries of the S-matrix of a semisimple modular category lie in a cyclotomic field, see \cite{dBG, CG}. This result is used in Section 8. For reader's convenience in this Appendix we reproduce a proof of this result. \begin{theorem}\label{app} Let $S=(s_{ij})_{i,j\in I}$ be the S-matrix of a modular fusion category $\mC$. There exists a root of unity $\xi$ such that $s_{ij}\in {\mathbb Q}(\xi)$. \end{theorem} \begin{proof} Let $\{ X_i\}_{i\in I}$ be the representatives of isomorphism classes of simple objects of $\mC$; let $0\in I$ be such that $X_0$ is the unit object of $\mC$ and the involution $i\mapsto i^*$ of $I$ be defined by $X_i^*\cong X_{i^*}$. By the definition of modularity, any homomorphism $f: K(\mC)\to {\mathbb C}$ is of the form $f([X_i])=s_{ij}/s_{0j}$ for some well defined $j\in I$. Hence for any automorphism $g$ of ${\mathbb C}$ one has $g(s_{ij}/s_{0j})= s_{ig(j)}/s_{0g(j)}$ for a well defined action of $g$ on $I$. Now we are going to use the following identities (see \cite{BaKi}): (i) $\sum_js_{ij}s_{jk}=\delta_{ik^*}$; (ii) $s_{ij}=s_{ji}$; (iii) $s_{0i^*}=s_{0i}\ne 0$. Thus, $\sum_js_{ij}s_{ji^*}=1$ and hence $(1/s_{0i})^2=\sum_j(s_{ji}/s_{0i}) (s_{ji^*}/s_{0i^*})$. Applying the automorphism $g$ to this equation we get \begin{equation} g(\frac{1}{s_{0i}^2})=g(\sum_j\frac{s_{ji}}{s_{0i}}\frac{s_{ji^*}}{s_{0i^*}})= \sum_j\frac{s_{jg(i)}}{s_{0g(i)}}\frac{s_{jg(i^*)}}{s_{0g(i^*)}}= \frac{\delta_{g(i)g(i^*)^*}}{s_{0g(i)}s_{0g(i^*)}}. \end{equation} It follows that $g(i^*)=g(i)^*$ and $g((s_{0i})^2)=(s_{0g(i)})^2$. Hence $$g((s_{ij})^2)=g((s_{ij}/s_{0j})^2\cdot (s_{0j})^2)=(s_{ig(j)})^2.$$ Thus $g(s_{ij})=\pm s_{ig(j)}$. Moreover the sign $\epsilon_g(i)=\pm 1$ such that $g(s_{0i})=\epsilon_g(i)s_{0g(i)}$ is well defined since $s_{0i}\ne 0$, and $g(s_{ij})=g((s_{ij}/s_{0j})s_{0j})=\epsilon_g(j)s_{ig(j)}=\epsilon_g(i) s_{g(i)j}$. In particular, the extension $L$ of ${\mathbb Q}$ generated by all entries $s_{ij}$ is finite and normal, that is Galois extension. Now let $h$ be another automorphism of ${\mathbb C}$. We have $$gh(s_{ij})=g(\epsilon_h(j)s_{ih(j)})=\epsilon_g(i)\epsilon_h(j)s_{g(i)h(j)}$$and $$hg(s_{ij})=h(\epsilon_g(i)s_{g(i)j})=\epsilon_h(j)\epsilon_g(i)s_{g(i)h(j)}= gh(s_{ij})$$ and the Galois group of $L$ over ${\mathbb Q}$ is abelian. Now the Kronecker-Weber theorem (see e.g. \cite{Ca}) implies the result. \end{proof} \bibliographystyle{ams-alpha}

Okay, no pictures today because it won't let me add them. Let's recap my day, shall we? Woke up, ate breakfast, went on the treadmill for 45min, made dinner to take to work, cleaned up, had a shower, ate lunch, went to work... worked... .... came home, found out TARA IS ENGAGED!!! and I have the honour of being in the wedding!!! Got off the phone with Tara and called my mom and Alex while I made tomorrow's lunch and dinner (yep, another late Tuesday of course tomorrow)... and here we are. SO, here's what I ate today: BF - fibre 1 bar, chocolate (2), blackberries (1), granola bites (2) - such a healthy breakfast. Lunch - Sante Fe Rice and beans smart ones (6) Snack - cadbury thins (2) Dinner - 1/2c chick peas (2), red pepper (0), green onion (0), cucumber (0), grape tomatoes (0), allegro cheese (1), 3 tbsp ff italian (0), romaine (0), apple (1), yogurt (1), hostes 100 calorie cakes (2) There we have it! Happy Tuesday everyone! 1 comment: What a busy day!! Good work with the treadmill today. We CAN do this :)

How To Get The Best Support Through The Holidays Before we begin the podcast today, it is Giving Tuesday. I hope you had a great Thanksgiving. We do have a goal of raising $5000 by the end of today. Please go to btr.org to help us reach this goal. Please mark your donation as recurring to support this podcast. Another exciting piece of news is that beginning December 1, our Betrayal Trauma Recovery Club is going to provide 6 sessions for women--one every weekday during various times and then on Tuesdays there will be an AM and PM session. We did a survey with current members of this club and this is what they asked for. We have women interested in joining in Australia and the UK so we need to provide different times. Go to our site btr.org to check on the new times and register right away by clicking on the schedule and join button. I'm so grateful to have one of our clients here today. Her name is Elsie. She found Betrayal Trauma Recovery and has been listening to the podcast and using our services. Welcome Elsie. Elsie: Thank you for having me. Marriage Before Discovering My Husband’s Pornography Use Anne: What was your marriage like before you found out about your husband's lying and emotional abuse and compulsive sexual behaviors? Elsie: We got along well. He was egotistical but he was fun-loving and generally polite. He did have some anger issues--at least that's what I thought it was. If I was to describe my marriage from the beginning in one word, it would be sexless. Because we were older, I think sex was not as much a priority for me as it may have been for someone younger but I did miss it. He didn't seem to, however. After a certain period of "starvation," I would mention it to him, he would defend himself, offer reasons for why it didn't happen, and then make a half-hearted advance which usually left me feeling a little like a beggar. There was no initiation towards me. There was usually compliance if I said something about it, but there was no initiation towards me. After a while, it hurt a lot. I gained a lot of weight and became depressed. I didn't know what was going on. I had no idea. Anne: How old are you? Elsie: 56. Anne: When did you get married? Elsie: 8 years ago but we were together for 10. I was around 45 when I met him and we married about 2 1/2 years later. Anne: Was this your first marriage? Elsie: No. We were not believers in God at the time. There was a conversion experience that came as this all happened. He suggested to me that we start going to church when I began to discover things. I went down this path and it was the best path I could have taken to help me deal with everything coming my way. I Thought His Porn Use Was Casual, Occasional And No Big Deal Anne: How did you find out about what was going on? Elsie: The first time I discovered something was quite by chance. I was moved to pick up his IPod and look at it--which I never did. I had total trust in him. But something moved me to look at his IPod and I discovered he was looking on Craigslist at the personals. I questioned him about it and he said it was free pornography. I remember telling him that this was close to home because it was a city about 30 minutes away. I asked him why he didn't look in Ontario, Canada. Anne: So at the time, did you not think pornography was a bad thing? Elsie: I looked at it as a casual-use and occasional thing for him and that it was no big deal. It made me uneasy; remember I was living in a virtually sexless marriage and he's looking at this...but I had no knowledge of anything that I know now. I wrote it off and eventually I was moved again. He had gone to take a shower and something moved me to pick up his phone. I went to the all-male section on his phone and found his ad--the ad he had placed. This was the start of the ball rolling. I discovered he had actually linked up with someone. This was the first real element of infidelity that I found. It was the first of a lot. I was very disturbed by it. Something prompted me to go to Craigslist and plug in the email address and see if I could access the account he had. I did and discovered he had been on it for 14 months. He was off shore--home a month and gone on the rig for a month--and he had posted 148 postings in all the various cities he had been in. It was shocking to say the least. I contacted him about it and he was immediately defensive but the defenses began to build from the time I made my first discovery in September. This was my first step forward in me becoming the enemy. Anne: I love that--there is no way you could get around his perception that you were the enemy. That's really good. My Husband Viewed Me As The Enemy Elsie: There's no question about it. I was. I was the one that had cracked the shell of secrecy. When I found his posting, I became a pretty determined bulldog in what I sunk my teeth into! I wasn't going to let go until I got some answers! I had been advised by Christian people to let it go, to forgive him...and like I have said to you, I believe a lot of this was God-led, for my safety and probably for the sole purpose of just disclosing it and getting it in the open...bringing to light what was happening. Anne:...because God loves you, right? Elsie: Yes, ma'am, he does and he showed that through this traumatic experience, over and over again. I Sought Out The Wrong Counselors, Untrained To Help With Sex Addiction Anne: You mentioned a few things that were not super helpful like Christian people who mentioned forgiveness or sweeping it under the rug. Can you talk more about the things you tried or where you turned for help? Elsie: I immediately sought out counseling. A counselor looked at me and said that what I was seeing was the tip of the ice berg. That resonated with me. He went on to some brash and harsh language and I immediately knew he was the wrong guy. So I sought out another counselor and found one. She basically took a bad situation and made it worse. she diagnosed him with PTSD, offered no counseling for me, on any level; every focus was towards him. I was not treating him properly, I needed to understand that he wasn't well because he had PTSD...essentially counseling wise, I was abandoned. I continued and over the course of time, in 2014, I sought out a local church, contacted the pastor, met with him and explained my circumstances, and he vowed to do all he could. Of course, that didn't help. I found that even he began to be bitter towards me because I was obviously "not forgiving" enough. This is a lot of what was called for--to be forgiving. I was never allowed to express anger. That would have been inappropriate. No one had any real answer at this point. I did eventually find a Christian counselor. We both went in individually and then together and it was some form of marriage counseling. She was ill-equipped, albeit a very good counselor, and she referred him over to someone else but the behaviors didn't stop and one relapse led to the termination of his job and we couldn't afford the counselors anymore. So it stopped and not long after that, I left for 3 months and I got a call from him one day saying he really wanted help. We got back together and met with a minister who began giving us spiritual counseling. I saw some change in my spouse. He seemed to be reaching towards God and it was the only change I saw. It lasted about 6 weeks. He saw a medical doctor who sought to help him with medication. This destroyed the peace that he had found--even though it was brief peace. The doctor treated him for low testosterone using a synthetic steroid or hormone. This was injectable toxicity when it comes to a sex addict! Anne: So the doctor begins to inject him with testosterone and things get worse. Elsie: He gave it to him to go home and do it himself. Anne: If I had to guess, and I'm not a therapist, this is a two-fold issue here: a doctor is telling him that this isn't a mental problem to work on and that all he needs to do is use a testosterone injection and it will solve his problem. So there is this mental shift and then there is the actual testosterone in his body. I've found that anything anyone suggests--no matter how small--to give them an excuse for their abusive behaviors removes the pressure from them to change and keeps them in the abuse cycle and the mental state of narcissism--or whatever the sexual mess of chaos they have going on in their brain. Any suggestion of it being something else if they are not in recovery can get them off track because they think they don't have to be accountable for their behaviors. Being Blamed For His Sex Addiction Elsie: And he was not. In my opinion, any PTSD he was suffering from stemmed directly from my reaction to the discoveries! He was traumatized at being "outed" because he didn't show any other signs of trauma; although serious childhood trauma that began years and years earlier, at a time in his life when it was well out of his control, play in. However, the general attitude was not one of recovery. It was "you're the problem. Stop badgering me. Just be happy where you are." And of course the constant promises that he wasn't doing it anymore of course were not true. Until I found BTR, no one out there validated my experiences in any of this. Any time we sought help elsewhere, all help focused on him. Any focus on me said, "You're making it worse." I felt blamed in some ways. I was already being blamed by him. Anne: Right. And then the help you sought out was also traumatizing. So how did you find BTR? How I Found Betrayal Trauma Recovery Elsie: This is very interesting. Let me preface it by saying that over the summer I took a course in Biblical human sexuality. During that time, I realized how very much I missed intimacy...gentle touch, kissing, flirtation, romance, and sex. And the class I followed this up with was a class on shepherding women in pain. Then I really began to recognize the abuse--this was just pure abuse. So, I've had these classes back to back, hours and hours of crying, still no one to validate anything, and I cried out, "God, where are you in all this pain." Tears were rolling down my face, snot bubbles, the whole shebang! I went into my room and got on my knees and begged God to bring me some relief. I know He was with me that night even though I didn't feel it at the time. I was too emotionally distraught. Eventually I fell asleep. I woke up the next morning feeling remarkably refreshed, went to my computer, went to Covenant Eyes for some reason, and someone posting on a forum on there commented about BTR for women. Related: We Recommend Covenant Eyes Internet Filtering & Accountability On Every Device I clicked on it and the rest is history! BTR was a God-send to me. This is to me, in my opinion, God's answer to "Where are you in all this pain?" He brought me to BTR which has really helped a lot in making sense of what I have been dealing with. There Is Validation For Us At BTR Anne: This is why I started BTR--to make sense of what I was dealing with! I prayed and prayed to know what to do...should I let my husband back into the house, start talking to him, or do I file for divorce...I didn't know. God's answer to me was, "Start a podcast." I was incredulous! Elsie: And I am so glad! I'm glad God did that. If he had not, he would not have prompted me to come to you. It is a God-send. BTR gives women an opportunity to be validated. If I was to summarize it all in one word, there is validation for us through BTR. Anne: I think it's what God wants and needs us to hear: that we are not the problem and that he loves us and that all the blame and gas lighting and everything we have experienced made us question our worth; I think God wants us to know we are enough and He loves us. How Betrayal Trauma Recovery Club Meets My Needs So, Elsie, you joined Betrayal Trauma Recovery Club. How did it meet your needs more than any of the options you had tried before? Elsie: I had never done any kind of group prior to beginning a BTR group. I don't have anything else to compare it to but I think the BTR groups stand quite well on their own merits. They give support, feedback, and are run by APSATS trained coaches. Coach Rae is excellent. If you want to listen, you can; there is no pressure. The ladies, as well as the coach who has had her own experience, understood my plight. They understood the exact predicament that I was in. They offered tremendous support. It helps to know you are not alone and that there are other people out there who have sorted through it--like the coach--and others who are sorting through it and sharing ideas. If you throw something out there, someone will give you a little feedback on it and maybe expound on it a little further. This is excellent for those of us who need connection with others, who need validation...BTR offers this. Anne: Women are trying to figure it out. They are working hard. They read books, are learning, and seeking out therapy. Unless someone who has been through it before and really understands it can help give them the words to say, it's hard to describe exactly what is happening. It's so liberating to hear someone else say the thing you were trying to figure out...and then you realize that is what you have been trying to say; you just didn't have the words to say before now. Reactive Abuse During Betrayal Trauma Elsie: Yes! You are right. Remember that at this point I had been studying for several years now and nowhere that I had studied was this ever really expounded upon. I had never heard the term "reactive abuse" until Coach Rae brought it to me. I was often accused by my spouse of being very damaging to him and lashing out after the constant onslaught of anger towards me or if I tried to communicate with him I was belittled or shut down; he would just get up and disengage or yell at me or break things...a wide variety of totally negative behaviors. Then in time I would lash back and all the fingers would be pointing at me because now I had bruised the narcissist and was now the "bad guy." I didn't know that it was reactive abuse until one of the BTR coaches, Coach Rae, defined it for me. She connected me with literature that defined it very clearly. I realized that it was wrong--that I needed to stop being reactive and be more proactive in my healing. Trying To Make Him Love Me Wouldn't Ever Result In Feeling Loved Anne: I think I did the same thing. All the ways that I really nit-picked him about cutting the tomatoes or his bonsai tree, or whatever it was...I think about those times and I would say to him then, "I just don't feel loved right now." He told me flat out: "I don't love you. I love the kids more than you." Or, "I can't love you because you're terrible." I just didn't feel secure enough. I think I was getting to this point where I was so irritable about little things. I've learned now that trying to make him love me wasn't ever going to result in me feeling loved. I needed to set boundaries around that. This is how I could feel secure. At the time, that was all I knew how to do. I can see now how unhealthy it was but back then. I felt like I was grasping for reassurance and security in the strangest ways. It's kind of embarrassing to think back on it! Elsie: Do not be embarrassed! We grasp for any number of things to try to make sense of what we are experiencing. I lost 80 pounds and underwent plastic surgery. So, talk about drastic! I was 232 pounds at 5'11". I went down to 153 pounds--my high school weight. My children were worried about me! I elected to have facial plastic surgery so I would look better. I worried about my aging...it was crazy...all the while believing that God wanted to heal our marriage. I still believe God wants to heal marriages but when it comes down to it, God heals the individuals as they draw more towards Him, and subsequently find the marriage healing. God does not place the institution of marriage above humans, in my opinion. Individual Healing Before Marriage Healing In Sex Addiction Anne: And there's no way the marriage can heal without the individual being healed because abuse and pornography use are not marriage issues. They aren't communication issues. They are abuse issues. Two people cannot resolve abuse. It has to be 100% the abuser taking accountability and making amends for his actions as well as seeking a change of heart. There just isn't any other way around it. There isn't any way to love and forgive and serve an abuser out of abusing you. It doesn’t work this way, unfortunately. I think we would like it to be this way because then we would have a little more control. I think all of us have tried the route of loving, serving, etc..., more. Finding Help To Make Sense Of The Betrayal Returning back to your experience with BTR, before we close today, is there anything you would like to share with our listeners about your experience with Betrayal Trauma Recovery? Elsie: Yes. I would like to say that this is not something that a woman can do alone. You have to have a community of women who understand what you are going through, what your experiences are, who, through your experiences and training can validate you and help you to make sense of this. My Christian counselor had encouraged me to get into a community--which I could not find, by the way. There were none in my local community. There wasn't anything around me. When I found BTR and went to the Facebook page first, I discovered numerous women who were having the same experience I was having...some having separated and moved on, some fully healed, some in the process of healing, some just starting out...But collectively there is a unified understanding that fit. It fit my circumstances. It gave me a place to go where I knew whatever I would say concerning these circumstances would be well received and understood because the women in the group were experiencing the same things. I was amazed at how similar our stories were even though the specifics were different; the underlying abuse and gas lighting were very similar. I found a sisterhood in the women in BTR. I really like Coach Rae. She is the one I speak to. I bring my head into the conversation but she gives me another head to help sound off ideas. Two heads are better than one! I really, really like her. The Facebook group is there all the time--24/7. You can express your thoughts and opinions in a post--even if it's 3 am and no one responds back right then. When you wake up in the morning, someone has stepped forward to say they are praying for you or they understand your pain or are validating it with a similar experience. It's made a difference to me. Free Betrayal Trauma Recovery Facebook Group Anne: I'm so glad to hear that. When I began BTR, I wanted to provide women with all the things we haven't had. I wanted to make it available all on-line so that women anywhere could find it. There are some women who are lucky enough to find an amazing therapist or some type of support group in their area. But there are more of us who have tried and tried and tried and been unable to find that community that we need. Elsie just talked about our free Facebook group. It is not moderated by our couches but it is moderated by women who are in recovery. They help to moderate the group to make sure it's safe. If you want to join this free group, go to btr.org and scroll down and enter your email. You will get a return email with instructions about how to join the free Facebook group. If you want to join Betrayal Trauma Recovery Club, go to btr.org/schedule and join. Our Club gives you access to 6 APSATS-led group sessions per week. That's one every weekday at different times with two on Tuesdays. Keep an eye out for this schedule as it will be posted very soon at btr.org. The more women who join Betrayal Trauma Recovery Club, the more sessions we will add. Elsie also talked about individual support calls with our coaches. Coach Rae is who Elsie sees; Coach Sarah, for example, is really good with helping women who are trying to help their children understand what is going on. Coach Lara is in a healthy and loving relationship with a recovered addict, as well as Coach Cat. Coach Rae is an expert at divorce. If you want to learn more, email Coach Rae at rae @btr.org. She is happy to answer questions about which coach may be best for you or we recommend scheduling with Coach Lara if you don't know where to start; she can give you a lot of different options which provides understanding about which coach or group to join. Recommended Reading: Your Sexually Addicted Spouse, by Steffens and Means Elsie, thank you so much for talking today about your experience. We appreciate you as a client and are so grateful you found us. We are also grateful to God for bringing all of us together. My hope is that through His grace and through His mercy as we become healthier, and that we can become a force for good in the world--which is what I believe He wants from us. I think this is why He is gathering us all together. Elsie: I agree. I encourage any woman out there who has the financial means to do so to contribute, to donate to BTR. This organization is truly a help and God-blessed. Anne: Thank you so much for that. I would love to provide our services for free. During this horrific time for women, they are having serious financial problems. We understand that. The training our coaches receive and the things we do at BTR cost money to provide this for women. If we could, we would provide free service! We are in similar situations to our listeners. I am a single mom now and this is how I make my living, as well as our coaches. Please know our hearts are with you. We have been in these very difficult situations. Even a $2 donation really helps us. Again, our goal for this Giving Tuesday is to raise $5000 that will help cover the costs of the website, podcast, the technical costs, and every service we provide at BTR. Go to BTR.org. Please make a generous contribution today and then a monthly donation to keep this podcast going and this information coming to you. If this podcast is helpful to you, please rate it on Itunes or whatever other podcasting service you use. Every single rating increases our ranking on search engines and helps women who are isolated to find us. I want to do one more shout out to our new schedule for the Betrayal Trauma Recovery Club starting December 1. It is only $100/month which is about the same or less than one therapy session! It allows you to have access to 6 APSATS-led group sessions per week. Have a great holiday season. Until next week, stay safe out there!